• Replacement of Cardinality

    From WM@21:1/5 to All on Fri Jul 26 16:31:19 2024
    It is strange that blatantly false results as the equinumerosity of prime numbers and algebraic numbers could capture mathematics and stay there for
    over a century. But by what meaningful mathematics can we replace Cantor's wrong bijection rules?

    Not all infinite sets can be compared by size, but we can establish some
    useful rules

    _The rule of subset_ proves that every proper subset has less elements
    than its superset. So there are more natural numbers than prime numbers,
    |ℕ| > |P|, and more complex numbers than real numbers. Even finitely
    many exceptions from the subset-relation are admitted for infinite
    subsets. Therefore there are more odd numbers than prime numbers.

    _The rule of construction_ yields the numbers of integers |Z| = 2|ℕ| + 1
    and of fractions |Q| = 2|ℕ|^2 + 1 (there are less rational numbers).
    Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers than rational numbers.

    _The rule of symmetry_ yields precisely the same number of reals in every interval (n, n+1] and with at most a small error same number of odd
    numbers and of even numbers in every finite interval and in the whole real line.

    Regards, WM

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  • From joes@21:1/5 to All on Fri Jul 26 21:57:44 2024
    XPost: sci.math

    Am Fri, 26 Jul 2024 16:31:19 +0000 schrieb WM:
    It is strange that blatantly false results as the equinumerosity of
    prime numbers and algebraic numbers could capture mathematics and stay
    there for over a century. But by what meaningful mathematics can we
    replace Cantor's wrong bijection rules?
    Juste because it doesn't match your intuition doesn't mean it's not
    useful.

    Not all infinite sets can be compared by size, but we can establish some useful rules
    that you would like instead.
    _The rule of subset_ proves that every proper subset has less elements
    than its superset. So there are more natural numbers than prime numbers, |ℕ| > |P|, and more complex numbers than real numbers. Even finitely
    many exceptions from the subset-relation are admitted for infinite
    subsets. Therefore there are more odd numbers than prime numbers.
    What exceptions do you mean?
    This immediately creates as many sizes as there are naturals, one for
    each of your endsegments.

    _The rule of construction_ yields the numbers of integers |Z| = 2|ℕ| + 1 and of fractions |Q| = 2|ℕ|^2 + 1 (there are less rational numbers).
    Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers than rational
    numbers.
    Only this leads to some contradictions depending on the construction.

    _The rule of symmetry_ yields precisely the same number of reals in
    every interval (n, n+1] and with at most a small error same number of
    odd numbers and of even numbers in every finite interval and in the
    whole real line.
    How small an error?

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

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  • From Richard Damon@21:1/5 to All on Fri Jul 26 22:23:07 2024
    On 7/26/24 12:31 PM, WM wrote:
    It is strange that blatantly false results as the equinumerosity of
    prime numbers and algebraic numbers could capture mathematics and stay
    there for over a century. But by what meaningful mathematics can we
    replace Cantor's wrong bijection rules?

    Not all infinite sets can be compared by size, but we can establish some useful rules

    _The rule of subset_ proves that every proper subset has less elements
    than its superset. So there are more natural numbers than prime numbers, |ℕ| > |P|, and more complex numbers than real numbers.  Even finitely
    many exceptions from the subset-relation are admitted for infinite
    subsets. Therefore there are more odd numbers than prime numbers.

    _The rule of construction_ yields the numbers of integers |Z| = 2|ℕ| + 1 and of fractions |Q| = 2|ℕ|^2 + 1 (there are less rational numbers).
    Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers than rational numbers. _The rule of symmetry_ yields precisely the same number of reals in
    every interval (n, n+1] and with at most a small error same number of
    odd numbers and of even numbers in every finite interval and in the
    whole real line.

    Regards, WM


    The problem is that there can be infinite sets constructed different
    ways that give different answers.

    By your logic, if you take a set and replace every element with a number
    that is twice that value, it would by the rule of construction say they
    must be the same size.

    But that resultant set is the evens, which can also be shown by your
    logic to have less elements than the Natural Numbers they were made from
    by doubling, so a set is smaller (or larger) than itself.

    This is the sort of error that comes when you try to use logic designed
    for "finite" sets with infinite sets.

    This shows that infinite sets just must have some different rules.

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  • From Mikko@21:1/5 to All on Sat Jul 27 11:21:43 2024
    On 2024-07-26 16:31:19 +0000, WM said:

    It is strange that blatantly false results as the equinumerosity of
    prime numbers and algebraic numbers could capture mathematics and stay
    there for over a century. But by what meaningful mathematics can we
    replace Cantor's wrong bijection rules?

    Not all infinite sets can be compared by size, but we can establish
    some useful rules

    _The rule of subset_ proves that every proper subset has less elements
    than its superset. So there are more natural numbers than prime
    numbers, |ℕ| > |P|, and more complex numbers than real numbers. Even finitely many exceptions from the subset-relation are admitted for
    infinite subsets. Therefore there are more odd numbers than prime
    numbers.

    _The rule of construction_ yields the numbers of integers |Z| = 2|ℕ| +
    1 and of fractions |Q| = 2|ℕ|^2 + 1 (there are less rational numbers). Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers than rational
    numbers.
    _The rule of symmetry_ yields precisely the same number of reals in
    every interval (n, n+1] and with at most a small error same number of
    odd numbers and of even numbers in every finite interval and in the
    whole real line.

    First you should study what Galilei said about infinities long before Cantor. Fix his errore, if any, and continue what he left unfinished.

    --
    Mikko

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  • From WM@21:1/5 to All on Sat Jul 27 11:11:52 2024
    Le 27/07/2024 à 10:21, Mikko a écrit :

    First you should study what Galilei said about infinities long before Cantor.

    I did.

    Fix his errore, if any, and continue what he left unfinished.

    He applied only potential infinity. There he is right. But I assume actual infinity.

    Regards, WM

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  • From WM@21:1/5 to All on Sat Jul 27 11:09:40 2024
    Le 26/07/2024 à 23:57, joes a écrit :
    Am Fri, 26 Jul 2024 16:31:19 +0000 schrieb WM:
    It is strange that blatantly false results as the equinumerosity of
    prime numbers and algebraic numbers could capture mathematics and stay
    there for over a century. But by what meaningful mathematics can we
    replace Cantor's wrong bijection rules?
    Juste because it doesn't match your intuition doesn't mean it's not
    useful.

    Cantor has been disproved in different ways. See for instance https://www.academia.edu/91188101/Proof_of_the_existence_of_dark_numbers_bilingual_version_

    Not all infinite sets can be compared by size, but we can establish some
    useful rules
    that you would like instead.

    There are more natural numbers than prime numbers. That is fact.

    _The rule of subset_ proves that every proper subset has less elements
    than its superset. So there are more natural numbers than prime numbers,
    |ℕ| > |P|, and more complex numbers than real numbers. Even finitely
    many exceptions from the subset-relation are admitted for infinite
    subsets. Therefore there are more odd numbers than prime numbers.
    What exceptions do you mean?

    The exception prime number 2 is not an odd number.

    This immediately creates as many sizes as there are naturals, one for
    each of your endsegments.

    _The rule of symmetry_ yields precisely the same number of reals in
    every interval (n, n+1] and with at most a small error same number of
    odd numbers and of even numbers in every finite interval and in the
    whole real line.
    How small an error?

    Only 1 or 2 depending on the chosen interval. In the interval (0, 3] there
    are two odd natnumbers but only one even natnumber.

    Regards, WM

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  • From WM@21:1/5 to All on Sat Jul 27 11:13:57 2024
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a number
    that is twice that value, it would by the rule of construction say they
    must be the same size.

    That is true in potential infinity. But I assume actual infinity.

    But that resultant set is the evens, which can also be shown by your
    logic to have less elements than the Natural Numbers they were made from
    by doubling,

    So it is.

    Regards, WM

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  • From Richard Damon@21:1/5 to All on Sat Jul 27 07:27:30 2024
    On 7/27/24 7:13 AM, WM wrote:
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a
    number that is twice that value, it would by the rule of construction
    say they must be the same size.

    That is true in potential infinity. But I assume actual infinity.


    So, what part is not true? Are you stating that replacing every element
    with another unique distinct element something that make the set change
    size?


    But that resultant set is the evens, which can also be shown by your
    logic to have less elements than the Natural Numbers they were made
    from by doubling,

    So it is.

    WHich shows the logic to be wrong.


    Regards, WM



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  • From WM@21:1/5 to All on Sat Jul 27 12:18:20 2024
    Le 27/07/2024 à 13:27, Richard Damon a écrit :
    On 7/27/24 7:13 AM, WM wrote:
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a
    number that is twice that value, it would by the rule of construction
    say they must be the same size.

    That is true in potential infinity. But I assume actual infinity.


    So, what part is not true?

    In potential infinity there is no ω.

    Are you stating that replacing every element
    with another unique distinct element something that make the set change
    size?

    In actual infinity the number of elements of any infinite set is fixed. Doubling all elements of the set ℕ U ω = {1, 2, 3, ..., ω} yields the
    set
    {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.

    Regards, WM

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  • From WM@21:1/5 to All on Sat Jul 27 12:16:24 2024
    Le 27/07/2024 à 13:27, Richard Damon a écrit :
    On 7/27/24 7:13 AM, WM wrote:
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a
    number that is twice that value, it would by the rule of construction
    say they must be the same size.

    That is true in potential infinity. But I assume actual infinity.


    So, what part is not true?

    In potential infinity there is no ω.

    Are you stating that replacing every element
    with another unique distinct element something that make the set change
    size?

    In actual infinity the number of elements of any infinite set is fixed. Doubling all elements of the set ℕ U ω = {2, 4, 6, ..., ω} yields the
    set
    {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.

    Regards, WM

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  • From joes@21:1/5 to All on Sat Jul 27 12:48:59 2024
    XPost: sci.math

    Am Sat, 27 Jul 2024 12:16:24 +0000 schrieb WM:
    Le 27/07/2024 à 13:27, Richard Damon a écrit :
    On 7/27/24 7:13 AM, WM wrote:
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a
    number that is twice that value, it would by the rule of construction
    say they must be the same size.

    That is true in potential infinity. But I assume actual infinity.

    So, what part is not true?
    In potential infinity there is no ω.
    Neither is there in actual infinity.

    Are you stating that replacing every element with another unique
    distinct element something that make the set change size?
    Yes they are.
    In actual infinity the number of elements of any infinite set is fixed. Doubling all elements of the set ℕ U ω = {2, 4, 6, ..., ω} yields the
    set {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.
    I wonder how you get the second infinity. What is the preimage of all
    the omegas?

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

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  • From Richard Damon@21:1/5 to All on Sat Jul 27 08:55:55 2024
    On 7/27/24 8:16 AM, WM wrote:
    Le 27/07/2024 à 13:27, Richard Damon a écrit :
    On 7/27/24 7:13 AM, WM wrote:
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a
    number that is twice that value, it would by the rule of
    construction say they must be the same size.

    That is true in potential infinity. But I assume actual infinity.


    So, what part is not true?

    In potential infinity there is no ω.

    Are you stating that replacing every element with another unique
    distinct element something that make the set change size?

    In actual infinity the number of elements of any infinite set is fixed. Doubling all elements of the set ℕ U ω = {2, 4, 6, ..., ω} yields the set {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.


    Why?

    Note, ω is NOT a member of the Natural Numbers, it is just the "least
    upper bound" that isn't in the set.

    That seems to be your problem in understanding.

    There is no Natural Number that is ω/2 so that doubling it get you to ω,
    as every Natural Number when doubled gets you another Natural Number.

    Your "logic" just seems to be that ω is just some very big, an perhaps unexpressed, value of a Natural Number, because you "logic" can't
    actually handle infinite values.


    The fact that you can't understand this, doesn't make it not true, just
    that you own logic is too limited.

    Regards, WM


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  • From Jim Burns@21:1/5 to All on Sat Jul 27 13:34:48 2024
    XPost: sci.math

    sci.logic and sci.math

    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_ proves that
    every proper subset has
    less elements than its superset.

    ⎛ Each non.{}.set A of ordinals holds min.A

    ⎜ Ordinal j = {i:i<j} set of ordinals before j

    ⎜ Finite ordinal j has fewer elements than j∪{j}

    ⎝ ℕⁿᵒᵗᐧᵂᴹ is the set of ALL finite ordinals.

    ℕⁿᵒᵗᐧᵂᴹ breaks the rule of subset.

    If ℕ has fewer elements than ℕ∪{ℕ}
    then
    |ℕ| ∈ ℕ
    |ℕ∪{ℕ}| ∈ ℕ
    ℕ ≠⊂ {i:i<ℕ∪{ℕ}} ⊂ ℕ
    ℕ ≠⊂ ℕ
    and
    ℕ has fewer elements than ℕ

    Because ℕ does not have fewer elements than ℕ
    ℕ does not have fewer elements than ℕ∪{ℕ}
    and the rule of subsets is broken.

    _The rule of subset_ proves that

    To make a claim
    is not sufficient
    to make a proof.

    To make a finite sequence of claims
    such that no claim is first.false
    is sufficient
    to make a proof.

    The most that is true here is that
    the rule of subset _claims_ without proof that

    every proper subset has
    less elements than its superset.

    ⎛ In English, grammatically speaking,
    ⎜ it is never correct to say "less <plural.noun>"

    ⎜ English has mass nouns (Stoffnamen)
    ⎜ "less rock" ...
    ⎜ and count nouns (zählbare Substantive)
    ⎜ "one rock", "fewer rocks" ...
    ⎜ Only count nouns have a plural.
    ⎜ Only mass nouns are modified by "less".
    ⎝ "Less rocks" and "lescs elements" are never correct.

    ⎛ ...as you (WM) know, since you've been told.
    ⎜ Why you say what you say is often more mysterious
    ⎜ than what you say.

    ⎜ Perhaps you see a mass noun as more appropriate
    ⎜ to your darkᵂᴹ numbers -- to number purée?
    ⎜ That's not the way it would be said.

    ⎜⎛ The smaller bucket holds fewer rocks and less rock.
    ⎜⎝ Every proper subset has less element than its superset.

    ⎜ Perhaps you intentionally insert small errors
    ⎜ in order to draw attention away from large errors:
    ⎜ dropping verbal chaff.
    https://en.wikipedia.org/wiki/Chaff_(countermeasure)

    ⎝ Perhaps looking ignorant is your kink.

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  • From WM@21:1/5 to All on Sun Jul 28 11:43:55 2024
    XPost: sci.math

    Le 27/07/2024 à 14:48, joes a écrit :
    Am Sat, 27 Jul 2024 12:16:24 +0000 schrieb WM:
    Le 27/07/2024 à 13:27, Richard Damon a écrit :
    On 7/27/24 7:13 AM, WM wrote:
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a
    number that is twice that value, it would by the rule of construction >>>>> say they must be the same size.

    That is true in potential infinity. But I assume actual infinity.

    So, what part is not true?
    In potential infinity there is no ω.
    Neither is there in actual infinity.

    That is a solitary opinion.

    Are you stating that replacing every element with another unique
    distinct element something that make the set change size?
    Yes they are.
    In actual infinity the number of elements of any infinite set is fixed.
    Doubling all elements of the set ℕ U ω = {2, 4, 6, ..., ω} yields the

    Here I made a mistake: ℕ U ω = {1, 2, 3, ..., ω}

    set {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.
    I wonder how you get the second infinity. What is the preimage of all
    the omegas?

    See my correction.

    Regards, WM

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  • From WM@21:1/5 to I know. Therefore I on Sun Jul 28 11:55:17 2024
    Le 27/07/2024 à 14:55, Richard Damon a écrit :
    On 7/27/24 8:16 AM, WM wrote:
    Le 27/07/2024 à 13:27, Richard Damon a écrit :
    On 7/27/24 7:13 AM, WM wrote:
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a
    number that is twice that value, it would by the rule of
    construction say they must be the same size.

    That is true in potential infinity. But I assume actual infinity.


    So, what part is not true?

    In potential infinity there is no ω.

    Are you stating that replacing every element with another unique
    distinct element something that make the set change size?

    In actual infinity the number of elements of any infinite set is fixed.
    Doubling all elements of the set ℕ U ω = {2, 4, 6, ..., ω}

    Mistake! ℕ U ω = {1, 2, 3, ..., ω}

    yields the set
    {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.


    Why?

    See the correction.

    Note, ω is NOT a member of the Natural Numbers, it is just the "least
    upper bound" that isn't in the set.

    I know. Therefore I wrote ℕ U ω, or better ℕ U {ω}.

    There is no Natural Number that is ω/2 so that doubling it get you to ω,
    as every Natural Number when doubled gets you another Natural Number.

    There is no definable natural number ω/2. But if there are all elements,
    then there is no gap before ω but ω-1.

    Your "logic" just seems to be that ω is just some very big, an perhaps unexpressed, value of a Natural Number,

    No, it is the first transfinite number like 0 is the first non-positive
    number.

    The fact that you can't understand this is deplorable but does not make my theory wrong.
    Using the unit fractions itelligent readers understand that there must be
    a first one after zero. Others must believe in the magical appearance of infinitely many unit fractions.

    Regards, WM

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  • From WM@21:1/5 to All on Sun Jul 28 12:17:51 2024
    XPost: sci.math

    Le 27/07/2024 à 19:34, Jim Burns a écrit :

    If ℕ has fewer elements than ℕ∪{ℕ}
    then
    |ℕ| ∈ ℕ

    |ℕ| = ω-1 ∈ ℕ

    ℕ has fewer elements than ℕ

    ℕ has ω-1 elements.

    Because ℕ does not have fewer elements than ℕ
    ℕ does not have fewer elements than ℕ∪{ℕ}
    and the rule of subsets is broken.

    ℕ = {1, 2, 3, ..., ω-1} = {1, 2, 3, ..., |ℕ|}

    _The rule of subset_ proves that

    To make a claim
    is not sufficient
    to make a proof.

    To make a finite sequence of claims
    such that no claim is first.false
    is sufficient
    to make a proof.

    First false is your claim that |ℕ| is larger than all elements of ℕ.
    ℕ counts its elements.

    The most that is true here is that
    the rule of subset _claims_ without proof that

    every proper subset has
    less elements than its superset.

    The proof is easy. Since the superset has at least one more element than
    its proper subset, it has more elements than its proper subset.


    ⎛ In English, grammatically speaking,
    ⎜ it is never correct to say "less <plural.noun>"

    ⎜ English has mass nouns (Stoffnamen)
    ⎜ "less rock" ...
    ⎜ and count nouns (zählbare Substantive)
    ⎜ "one rock", "fewer rocks" ...
    ⎜ Only count nouns have a plural.
    ⎜ Only mass nouns are modified by "less".
    ⎝ "Less rocks" and "lescs elements" are never correct.

    Thank you, I will try to remember it.

    Regards, WM

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  • From Richard Damon@21:1/5 to All on Sun Jul 28 13:07:59 2024
    On 7/28/24 7:55 AM, WM wrote:
    Le 27/07/2024 à 14:55, Richard Damon a écrit :
    On 7/27/24 8:16 AM, WM wrote:
    Le 27/07/2024 à 13:27, Richard Damon a écrit :
    On 7/27/24 7:13 AM, WM wrote:
    Le 27/07/2024 à 04:23, Richard Damon a écrit :

    By your logic, if you take a set and replace every element with a
    number that is twice that value, it would by the rule of
    construction say they must be the same size.

    That is true in potential infinity. But I assume actual infinity.


    So, what part is not true?

    In potential infinity there is no ω.

    Are you stating that replacing every element with another unique
    distinct element something that make the set change size?

    In actual infinity the number of elements of any infinite set is fixed.
    Doubling all elements of the set ℕ U ω = {2, 4, 6, ..., ω}

    Mistake! ℕ U ω = {1, 2, 3, ..., ω}

    And who was using that set?


    yields the set
    {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.


    Why?

    See the correction.

    But what number became ω when doubled?

    Every natural number when doubled is a Natural Number.

    so the result should be { 2, 4, 6, ... 2*ω}.


    Note, ω is NOT a member of the Natural Numbers, it is just the "least
    upper bound" that isn't in the set.

    I know. Therefore I wrote ℕ U ω, or better ℕ U {ω}.

    But why? we were talking about the infinite set of the Naturals.


    There is no Natural Number that is ω/2 so that doubling it get you to
    ω, as every Natural Number when doubled gets you another Natural Number.

    There is no definable natural number ω/2. But if there are all elements, then there is no gap before ω but ω-1.

    Which isn't a Natural Number, as if it was, then that set of Natural
    Numbers would have a maximun member and be finite.

    Note, ω-1 doesn't exist in the base transfinite numbers, just as -1
    doesn't exist in the Natural Numbers, you can't go below the first element.

    But, just as we can expand the Natural Numbers to the Integers, and get negative numbers, we also might be able to define an extention to the transfinite numbers that can have a ω-1 element.


    Your "logic" just seems to be that ω is just some very big, an perhaps
    unexpressed, value of a Natural Number,

    No, it is the first transfinite number like 0 is the first non-positive number.

    And thus you can't have ω-1, just like you can't have -1 in the Natural Numbers.


    The fact that you can't understand this is deplorable but does not make
    my theory wrong.

    The fact that your theory is inconsistant makes it wrong.

    Using the unit fractions itelligent readers understand that there must
    be a first one after zero. Others must believe in the magical appearance
    of infinitely many unit fractions.


    Nope, since that implies there is a highest Natural Number, which breaks
    their definition,

    It just shows you mind can't handle proper logic of unbounded sets, but
    is stuck with bounded logic that just breaks when used with unbounded sets.

    Regards, WM




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  • From Jim Burns@21:1/5 to All on Sun Jul 28 14:17:45 2024
    XPost: sci.math

    On 7/28/2024 8:17 AM, WM wrote:
    Le 27/07/2024 à 19:34, Jim Burns a écrit :
    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_ proves that
    every proper subset has less elements than its superset.

    If ℕ has fewer elements than ℕ∪{ℕ}
    then
    |ℕ| ∈ ℕ

    |ℕ| = ω-1 ∈ ℕ

    ⎛ Each non.{}.set A of ordinals holds min.A

    ⎜ Ordinal j = {i:i<j} set of ordinals before j

    ⎜ Finite ordinal j has fewer elements than j∪{j}

    ⎝ ℕⁿᵒᵗᐧᵂᴹ is the set of ALL finite ordinals.

    No finite.ordinal is last.finite,
    no visibleᵂᴹ finite.ordinal,
    no darkᵂᴹ finite.ordinal.
    In particular, no finite.ordinal is ω-1

    Also, no before.first infinite.ordinal is
    before the first infinite.ordinal ω
    In particular, no infinite.ordinal is ω-1

    ----
    Consider ordinals i j k such that
    i∪{i} = j and j∪{j} = k

    Obviously, their order is i < j < k

    Either they're all finite
    |i| < |j| < |k|
    or they're all infinite
    |i| = |j| = |k|

    No finite.to.infinite step exists.
    no visibleᵂᴹ finite.to.infinite step,
    no darkᵂᴹ finite.to.infinite step.

    Defining declares the meaning of one's words.
    'Defining into existence' that which doesn't exist
    makes nonsense of whatever meaning one's words have.

    ⎛ if
    ⎜ g: j∪{j}→i∪{i}: 1.to.1
    ⎜ then
    ⎜ f(x) := (g(x)=i ? g(j) : g(x))
    ⎜ (Perl ternary conditional operator)
    ⎜ f: j→i: 1.to.1

    ⎜ if
    ⎜ f: j→i: 1.to.1
    ⎜ then
    ⎜ g(x) := (x=j ? i : f(x))
    ⎝ g: j∪{j}→i∪{i}: 1.to.1

    Therefore,
    i has fewer than j iff j has fewer than k

    ℕ has fewer elements than ℕ

    ℕ has ω-1 elements.

    ℕⁿᵒᵗᐧᵂᴹ holds all finite ordinals.

    Finite doesn't need to be small.
    ℕⁿᵒᵗᐧᵂᴹ holds ordinals which
    are big compared to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ,
    but those big ordinals have an immediate predecessor,
    and each non.0.ordinal before them has
    an immediate predecessor.
    That makes them finite, but not necessarily small.

    Because ℕ does not have fewer elements than ℕ
    ℕ does not have fewer elements than ℕ∪{ℕ}
    and the rule of subsets is broken.

    ℕ = {1, 2, 3, ..., ω-1} = {1, 2, 3, ..., |ℕ|}

    ∀j ∈ ℕⁿᵒᵗᐧᵂᴹ:
    ∃k ∈ ℕⁿᵒᵗᐧᵂᴹ\{0}:
    k = j+1 ∧ ¬∃kₓ≠k: kₓ=j+1

    '+1': ℕⁿᵒᵗᐧᵂᴹ→ℕⁿᵒᵗᐧᵂᴹ\{0}: 1.to.1
    and the rule of subset is broken.

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  • From Moebius@21:1/5 to All on Mon Jul 29 02:04:26 2024
    XPost: sci.math

    Am 27.07.2024 um 19:34 schrieb Jim Burns:
    sci.logic and sci.math

    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_ proves that [bla bla bla]

    Where did Mückenheim get this "rule" from? Any source?

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  • From Mikko@21:1/5 to All on Mon Jul 29 12:01:43 2024
    On 2024-07-27 11:11:52 +0000, WM said:

    Le 27/07/2024 à 10:21, Mikko a écrit :

    First you should study what Galilei said about infinities long before Cantor.

    I did.

    Fix his errore, if any, and continue what he left unfinished.

    He applied only potential infinity. There he is right. But I assume
    actual infinity.

    In "Two new scineces" Galilei discusses both and concludes that one
    is not possible without the other.

    He also notes that what we have learned from finite quatities does
    not apply to infinity.

    --
    Mikko

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  • From Jim Burns@21:1/5 to Moebius on Mon Jul 29 05:28:06 2024
    XPost: sci.math

    On 7/28/2024 8:04 PM, Moebius wrote:
    Am 27.07.2024 um 19:34 schrieb Jim Burns:
    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_ proves that [bla bla bla]

    Where did Mückenheim get this "rule" from?
    Any source?

    Mückenheim is wishcasting.
    The rule is that no sets are infiniteⁿᵒᵗᐧᵂᴹ.
    The counter.argument is the roster of
    the usual suspects: ℕⁿᵒᵗᐧᵂᴹ ℤⁿᵒᵗᐧᵂᴹ ℚⁿᵒᵗᐧᵂᴹ ℝⁿᵒᵗᐧᵂᴹ

    It's underwhelming.
    And yet,
    his rule is better exposition than I can recall,
    out of years of much.the.same,
    to himself of his own thinking.

    Perhaps one old dog can teach himself one new trick?

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  • From WM@21:1/5 to All on Mon Jul 29 13:11:59 2024
    Le 28/07/2024 à 19:07, Richard Damon a écrit :
    On 7/28/24 7:55 AM, WM wrote:


    Mistake! ℕ U ω = {1, 2, 3, ..., ω}

    And who was using that set?

    I.


    yields the set
    {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.


    Why?

    See the correction.

    But what number became ω when doubled?

    ω*2

    Every natural number when doubled is a Natural Number.

    No.

    Note, ω-1 doesn't exist in the base transfinite numbers, just as -1
    doesn't exist in the Natural Numbers, you can't go below the first element.

    If all natural numbers exist, then ω-1 exists.

    But, just as we can expand the Natural Numbers to the Integers, and get negative numbers, we also might be able to define an extention to the transfinite numbers that can have a ω-1 element.

    ω-1 is not transfinite but cisfinite.

    Using the unit fractions itelligent readers understand that there must
    be a first one after zero. Others must believe in the magical appearance
    of infinitely many unit fractions.

    Nope, since that implies there is a highest Natural Number, which breaks their definition,

    That is unavoidable. You believe in the magical appearance of infinitely
    many unit fractions. That breaks logic and mathematics.

    Regards, WM

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  • From Jim Burns@21:1/5 to Ross Finlayson on Mon Jul 29 08:32:40 2024
    XPost: sci.math

    On 7/28/2024 7:42 PM, Ross Finlayson wrote:
    On 07/28/2024 04:32 PM, Ross Finlayson wrote:
    On 07/28/2024 04:25 PM, Ross Finlayson wrote:
    On 07/28/2024 11:17 AM, Jim Burns wrote:

    [...]
    [...]
    [...]

    about ubiquitous ordinals

    What are ubiquitous ordinal?

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  • From Python@21:1/5 to All on Mon Jul 29 15:15:21 2024
    Le 29/07/2024 à 15:11, Wolfgang Mückenheim a écrit :
    Le 28/07/2024 à 19:07, Richard Damon a écrit :
    ...
    Every natural number when doubled is a Natural Number.

    No.


    ω-1 is not transfinite but cisfinite.

    This is worse and worse. How can any country allow such a
    kook to teach?!!!

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  • From WM@21:1/5 to All on Mon Jul 29 13:15:36 2024
    Le 29/07/2024 à 11:01, Mikko a écrit :

    He also notes that what we have learned from finite quatities does
    not apply to infinity.

    Unit fractions are isolated. They have distances. The function "unit
    fractions between 0 and x" can nowhere grow by more than 1.

    Regards, WM

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  • From WM@21:1/5 to All on Mon Jul 29 13:25:39 2024
    XPost: sci.math

    Le 29/07/2024 à 02:04, Moebius a écrit :
    Am 27.07.2024 um 19:34 schrieb Jim Burns:
    sci.logic and sci.math

    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_

    Where did Mückenheim get this "rule" from? Any source?

    Logic! If A contains all elements of B, but B does not contain all
    elements of A, then A has more elements than B.

    Regards, WM

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  • From WM@21:1/5 to All on Mon Jul 29 13:23:02 2024
    XPost: sci.math

    Le 28/07/2024 à 20:17, Jim Burns a écrit :
    On 7/28/2024 8:17 AM, WM wrote:
    Le 27/07/2024 à 19:34, Jim Burns a écrit :
    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_ proves that
    every proper subset has less elements than its superset.

    If ℕ has fewer elements than ℕ∪{ℕ}
    then
    |ℕ| ∈ ℕ

    |ℕ| = ω-1 ∈ ℕ

    No finite.ordinal is last.finite,
    no visibleᵂᴹ finite.ordinal,
    no darkᵂᴹ finite.ordinal.
    In particular, no finite.ordinal is ω-1

    NUF(x) cannot grow by more than 1 at any x. Therefore there is a first
    step. First unit fraction implies last natural number.

    Also, no before.first infinite.ordinal is
    before the first infinite.ordinal ω
    In particular, no infinite.ordinal is ω-1

    Don't claim. Explain how you imagine the positions of the unit fractions
    such that NUF does not have a first step.

    Regards, WM

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  • From Tom Bola@21:1/5 to All on Mon Jul 29 15:40:38 2024
    XPost: sci.math

    Am 29.07.2024 15:25:39 the clown WM drivels bullshit - as always:
    Le 29/07/2024 à 02:04, Moebius a écrit :
    Am 27.07.2024 um 19:34 schrieb Jim Burns:
    sci.logic and sci.math

    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_

    Where did Mückenheim get this "rule" from? Any source?

    Logic! If A contains all elements of B, but B does not contain all
    elements of A, then A has more elements than B.

    Cardinality within Math is not just a matter of "more", goofy...

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  • From Tom Bola@21:1/5 to All on Mon Jul 29 16:05:40 2024
    XPost: sci.math

    Am 29.07.2024 15:43:19 the clown WM drivels again his bullshit:

    Le 29/07/2024 à 15:40, Tom Bola a écrit :

    Logic! If A contains all elements of B, but B does not contain all
    elements of A, then A has more elements than B.

    Cardinality within Math is not just a matter of "more",

    No, it is simply nonsense.

    Not for 99,9999999 percent of the educated mankind, but just for
    your few parts of that sort of fully idiotic very low IQ clowns...

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  • From WM@21:1/5 to All on Mon Jul 29 13:43:19 2024
    XPost: sci.math

    Le 29/07/2024 à 15:40, Tom Bola a écrit :

    Logic! If A contains all elements of B, but B does not contain all
    elements of A, then A has more elements than B.

    Cardinality within Math is not just a matter of "more",

    No, it is simply nonsense.

    Regards, WM

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  • From Tom Bola@21:1/5 to All on Mon Jul 29 17:59:55 2024
    XPost: sci.math

    Am 29.07.2024 17:52:18 FromTheRafters schrieb:
    WM formulated the question :
    Le 29/07/2024 à 02:04, Moebius a écrit :
    Am 27.07.2024 um 19:34 schrieb Jim Burns:
    sci.logic and sci.math

    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_

    Where did Mückenheim get this "rule" from? Any source?

    Logic! If A contains all elements of B, but B does not contain all elements >> of A, then A has more elements than B.

    Have you reviewed 'Cardinal Arithmetic' lately? I know it has been
    pointed out to you several times. The aforementioned does not work for infinite sets. Addition like this simply doesn't affect the 'size' of
    the set if the set is infinite.

    A billion times "discussed" under the buzzword "Dedekind-Infinity"...

    The point is, that, for 50++ years, WM does not WANT (accept) our Math.

    -
    https://de.wikipedia.org/wiki/Unendliche_Menge#Dedekind-Unendlichkeit

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  • From Moebius@21:1/5 to All on Mon Jul 29 19:16:02 2024
    XPost: sci.math

    Am 28.07.2024 um 20:17 schrieb Jim Burns:
    On 7/28/2024 8:17 AM, WM wrote:

    ω-1 ∈ ℕ

    Holy shit!

    What does "ω-1" even mean?

    A "reasonable" interpretation might be:

    ω-1 is the natural number k such that k+1 = ω.

    The only problem with this interpretation is that there is no natural
    number k such that k+1 = ω.

    So (in a mathematical context) we can't define:

    ω-1 =df the natural number k such that k+1 = ω.

    'Defining into existence' that which doesn't exist
    makes nonsense of whatever meaning one's words have.

    Right.

    ℕ = {1, 2, 3, ..., ω-1}

    *sigh*

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  • From Jim Burns@21:1/5 to All on Mon Jul 29 14:33:31 2024
    XPost: sci.math

    On 7/29/2024 9:23 AM, WM wrote:
    Le 28/07/2024 à 20:17, Jim Burns a écrit :
    On 7/28/2024 8:17 AM, WM wrote:
    Le 27/07/2024 à 19:34, Jim Burns a écrit :
    On 7/26/2024 12:31 PM, WM wrote:

    _The rule of subset_ proves that
    every proper subset has less elements than its superset.

    If ℕ has fewer elements than ℕ∪{ℕ}
    then
    |ℕ| ∈ ℕ

    |ℕ| = ω-1 ∈ ℕ

    No finite.ordinal is last.finite,
    no visibleᵂᴹ finite.ordinal,
    no darkᵂᴹ finite.ordinal.
    In particular, no finite.ordinal is ω-1

    NUF(x) cannot grow by more than 1 at any x.

    NUF(x) = |⅟ℕ∩(0,x]|
    NUF(x) cannot grow by more than 1 at any x > 0
    ¬(0 > 0)

    ⅟ℕ∩(0,x] has
    each non.{}.subset maximummed
    each unit.fraction down.stepped
    each non.maximum up.stepped
    and therefore
    ℵ₀.many unit.fractions

    x > 0 ⇒ NUF(x) = ℵ₀

    Therefore there is a first step.

    0 < ⅟⌈1+⅟x⌉ < ⅟⌈⅟x⌉ ≤ x
    ⅟⌈⅟x⌉ ∈ ⅟ℕ∩(0,x]
    ⅟⌈1+⅟x⌉ ∈ ⅟ℕ∩(0,x]
    NUF(x) = |⅟ℕ∩(0,x]| ≠ 1

    First unit fraction implies last natural number.

    Ordinal λ = the set of ordinals < λ
    λ = [0,λ)ᴼʳᵈ

    ( k is a finite.ordinal

    ( each non.{}.subset of [0,k)ᴼʳᵈ is 2.ended

    _Finite doesn't need to be small_

    ( k is a finite.ordinal

    ⎛ [0,k)ᴼʳᵈ is 2.ended and
    ⎝ ∀j ∈ (0,k)ᴼʳᵈ: [0,j)ᴼʳᵈ is 2.ended

    if
    ( k is a finite.ordinal
    then
    ( k+1 is a finite ordinal

    ...because
    if
    ⎛ [0,k)ᴼʳᵈ is 2.ended and
    ⎝ ∀j ∈ (0,k)ᴼʳᵈ: [0,j)ᴼʳᵈ is 2.ended
    then
    ⎛ [0,k+1)ᴼʳᵈ is 2.ended and
    ⎝ ∀j ∈ (0,k+1)ᴼʳᵈ: [0,j)ᴼʳᵈ is 2.ended

    ω is defined to be the first transfinite.ordinal.
    ω is not a reallyreallyreally big finite.ordinal.

    ω = [0,ω)ᴼʳᵈ

    if ω-1 exists
    then
    ⎛ ω-1 = max.[0,ω)ᴼʳᵈ and
    ⎜⎛ [0,ω)ᴼʳᵈ is 2.ended and
    ⎜⎝ ∀j ∈ (0,ω)ᴼʳᵈ: [0,j)ᴼʳᵈ is 2.ended
    ⎜ and
    ⎝ ω is finite.

    However,
    ω is not finite, not even a reallyreallyreally big one.
    ω is transfinite (the first).
    [0,ω)ᴼʳᵈ isn't 2.ended.
    ω-1 doesn't exist.

    Also, no before.first infinite.ordinal is
    before the first infinite.ordinal ω
    In particular, no infinite.ordinal is ω-1

    Don't claim.

    Before the first star, no stars existed.
    Before the first mammal, no mammal existed.
    Before the first infinite.ordinal, no infinite ordinal exists.
    What ω means is the first infinite ordinal.

    Explain how you imagine
    the positions of the unit fractions such that
    NUF does not have a first step.

    ℚ the rationals are fractal.

    For each rational p/q in [0,⅟n]
    there is rational n⋅p/q in [0,1]

    Zoom from [0,1] in to [0,⅟n]
    [0,⅟n] looks just like [0,1]

    Zoom from [0,⅟n] in to [0,⅟n²]
    [0,⅟n²] looks just like [0,⅟n] and
    [0,⅟n] looks just like [0,1]

    No zoom.level ⅟nₓ exists such that
    [0,⅟nₓ] does NOT look like [0,1]

    No zoom.level ⅟nₓ exists such that
    some unit.fraction.free gap opens.

    [0,⅟nₓ] looks like [0,1] and
    [0,1] isn't unit.fraction.free.

    No zoom.level ⅟nₓ exists such that
    [0,⅟nₓ] holds fewer unit.fractions than [0,1]

    No x > 0 exists such that
    NUF(x) ≠ NUF(1) = ℵ₀

    ----
    A lagniappe.

    ℚ the rationals are fractal.

    ℝ the reals are fractal AND each split is situated in ℝ:
    For each ℝ.split F′ ᵃˡˡ<ᵃˡˡ H′
    some x′ ∈ ℝ is last.in.F′ or first.in.H′

    With each zoom.level ⅟n
    x′ is at the split F′ ᵃˡˡ<ᵃˡˡ H′

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  • From Jim Burns@21:1/5 to Ross Finlayson on Mon Jul 29 17:12:10 2024
    XPost: sci.math

    On 7/29/2024 3:44 PM, Ross Finlayson wrote:
    On 07/29/2024 05:32 AM, Jim Burns wrote:
    On 7/28/2024 7:42 PM, Ross Finlayson wrote:

    about ubiquitous ordinals

    What are ubiquitous ordinal?

    Well, you know that ORD, is, the order type of ordinals,
    and so it's an ordinal, of all the ordinals.

    If a ubiquitous ordinal is an ordinal,
    then I recommend referring to as an ordinal.

    The "ubiquitous ordinals", sort of recalls Kronecker's
    "G-d made the integers, the rest is the work of Man",
    that the Integer Continuum, is the model and ground
    model, of any sort of language of finite words,
    like set theory.

    If a ubiquitous ordinal is
    a finite ordinal ==
    a natural number ==
    a non.negative integer,
    then
    (I bet you see where I'm headed here)
    I recommend that you refer to it as
    a finite ordinal or
    a natural number or
    a non.negative integer.

    It's like the universe of set theory,

    Do you and I mean the same by "universe of set theory"?

    I am most familiar with theories of
    well.founded sets without urelements.

    In the von Neumann hierarchy of hereditary well.founded sets
    V[0] = {}
    V[β+1] = 𝒫(V[β])
    V[γ] = ⋃[β<γ] V[β]

    V[ω] is the universe of hereditarily finite sets.

    For the first inaccessible ordinal κ
    V[κ] is a model of ZF+Choice.

    For first inaccessible ordinal κ
    [0,κ) holds an uncountable ordinal and
    is closed under cardinal arithmetic.

    then as that there's _always_ an arithmetization, or
    as with regards to ordering and numbering
    as a bit weaker property than collecting and counting,
    so that "ubiquitous ordinals" is
    what you get from a discrete world.

    Is a ubiquitous ordinal a finite ordinal?
    I would appreciate a "yes" or a "no" in your response.

    Then there's that
    according to the set-theoretic Powerset theorem of Cantor,
    that when the putative function is successor,
    in ubiquitous ordinals
    where order type is powerset is successor,
    then there's no missing element.

    So, "ubiquitous ordinals" is exactly what it says.

    I find it concerning that you (Ross Finlayson) think that
    "what it says" answers "What does it say?" in any useful way.

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  • From Richard Damon@21:1/5 to All on Mon Jul 29 21:20:34 2024
    On 7/29/24 9:15 AM, WM wrote:
    Le 29/07/2024 à 11:01, Mikko a écrit :

    He also notes that what we have learned from finite quatities does
    not apply to infinity.

    Unit fractions are isolated. They have distances. The function "unit fractions between 0 and x" can nowhere grow by more than 1.

    Regards, WM



    Unless it just jumps to infinity because it isn't actually correctly
    defined.

    Since it uses transfinite numbers (which your mathematics can't handle)
    then it can change in the transfinite gap between 0 and the finite numbers.

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  • From Richard Damon@21:1/5 to All on Mon Jul 29 21:18:57 2024
    On 7/29/24 9:11 AM, WM wrote:
    Le 28/07/2024 à 19:07, Richard Damon a écrit :
    On 7/28/24 7:55 AM, WM wrote:


    Mistake! ℕ U ω = {1, 2, 3, ..., ω}

    And who was using that set?

    I.


    yields the set
    {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.


    Why?

    See the correction.

    But what number became ω when doubled?

    ω*2

    No, that is w double, what number in the first set became w?


    Every natural number when doubled is a Natural Number.

    No.

    Why not? WHich ones don't?


    Note, ω-1 doesn't exist in the base transfinite numbers, just as -1
    doesn't exist in the Natural Numbers, you can't go below the first
    element.

    If all natural numbers exist, then ω-1 exists.

    Why?


    But, just as we can expand the Natural Numbers to the Integers, and
    get negative numbers, we also might be able to define an extention to
    the transfinite numbers that can have a ω-1 element.

    ω-1 is not transfinite but cisfinite.

    Nope, it is not finite, but beyound the finite.


    Using the unit fractions itelligent readers understand that there
    must be a first one after zero. Others must believe in the magical
    appearance of infinitely many unit fractions.

    Nope, since that implies there is a highest Natural Number, which
    breaks their definition,

    That is unavoidable. You believe in the magical appearance of infinitely
    many unit fractions. That breaks logic and mathematics.

    Nope, the fact that you admit that you logic breaks the definition of
    the Natural Numbers says you admit your logic can't be used with them.

    It may be that you logic just can't handle a mathematics are big and
    powerful as the full definition of the Natural Numbers, but only
    arbirary but finite subsets of them.

    You just need to understand your limitations.


    Regards, WM




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  • From Mikko@21:1/5 to All on Tue Jul 30 10:26:47 2024
    On 2024-07-29 13:15:36 +0000, WM said:

    Le 29/07/2024 à 11:01, Mikko a écrit :

    He also notes that what we have learned from finite quatities does
    not apply to infinity.

    Unit fractions are isolated. They have distances. The function "unit fractions between 0 and x" can nowhere grow by more than 1.

    The number of unit fractions between 0 and x is infinite and the number
    of unit fractions between 0 and y is infinite. Whether one infinity is
    bigger or smaller than another infinity is not answerable merely on the
    basis of our experinece about finite quantities but only after a carful analysis if at all.

    --
    Mikko

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Tue Jul 30 14:41:15 2024
    Le 30/07/2024 à 09:26, Mikko a écrit :
    On 2024-07-29 13:15:36 +0000, WM said:

    Le 29/07/2024 à 11:01, Mikko a écrit :

    He also notes that what we have learned from finite quatities does
    not apply to infinity.

    Unit fractions are isolated. They have distances. The function "unit
    fractions between 0 and x" can nowhere grow by more than 1.

    The number of unit fractions between 0 and x is infinite

    Not for every x. Growth by more than 1 is contradicting mathematics.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

    and the number
    of unit fractions between 0 and y is infinite. Whether one infinity is
    bigger or smaller than another infinity is not answerable merely on the
    basis of our experinece about finite quantities but only after a carful analysis if at all.

    If (0, x) is a subset of (0, y), the relative quantities are clear.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Tue Jul 30 17:39:50 2024
    Le 30/07/2024 à 03:20, Richard Damon a écrit :
    On 7/29/24 9:15 AM, WM wrote:
    Le 29/07/2024 à 11:01, Mikko a écrit :

    He also notes that what we have learned from finite quatities does
    not apply to infinity.

    Unit fractions are isolated. They have distances. The function "unit
    fractions between 0 and x" can nowhere grow by more than 1.

    Unless it just jumps to infinity because it isn't actually correctly
    defined.

    It is correctly defined. Try to define the function Number of
    UnitFractions between 0 and x as you like it but in accordance with
    mathematics ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Tue Jul 30 17:37:09 2024
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?

    ω/2

    No, that is w double, what number in the first set became w?


    Every natural number when doubled is a Natural Number.

    No.

    Why not? WHich ones don't?

    ω/2 and larger.


    Note, ω-1 doesn't exist in the base transfinite numbers, just as -1
    doesn't exist in the Natural Numbers, you can't go below the first
    element.

    If all natural numbers exist, then ω-1 exists.

    Why?

    Because otherwise there was a gap below ω.

    That is unavoidable. You believe in the magical appearance of infinitely
    many unit fractions. That breaks logic and mathematics.

    Nope,

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the u niversal quantifier.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Tue Jul 30 17:30:17 2024
    XPost: sci.math

    Le 29/07/2024 à 20:33, Jim Burns a écrit :
    On 7/29/2024 9:23 AM, WM wrote:

    NUF(x) cannot grow by more than 1 at any x.

    NUF(x) = |⅟ℕ∩(0,x]|
    NUF(x) cannot grow by more than 1 at any x > 0

    Correct. At x < 0 or x = 0 NUF(x) = 0 and remains so.

    ¬(0 > 0)

    Correct.

    ⅟ℕ∩(0,x] has
    each non.{}.subset maximummed
    each unit.fraction down.stepped
    each non.maximum up.stepped
    and therefore
    ℵ₀.many unit.fractions

    x > 0 ⇒ NUF(x) = ℵ₀

    That implies a growth between [0, 1] and (0, 1].
    Between [0, 1] and (0, 1] NUF(x) cannot grow because no unit fraction fits
    in between.
    Strange how void of logical thinking matheologians are.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Tue Jul 30 14:37:43 2024
    XPost: sci.math

    On 7/30/2024 1:30 PM, WM wrote:
    Le 29/07/2024 à 20:33, Jim Burns a écrit :
    On 7/29/2024 9:23 AM, WM wrote:

    NUF(x) cannot grow by more than 1 at any x.

    NUF(x) = |⅟ℕ∩(0,x]|
    NUF(x) cannot grow by more than 1 at any x > 0

    Correct. At x < 0 or x = 0 NUF(x) = 0 and remains so.

    ¬(0 > 0)

    Correct.

    ⅟ℕ∩(0,x] has
    each non.{}.subset maximummed
    each unit.fraction down.stepped
    each non.maximum up.stepped
    and therefore
    ℵ₀.many unit.fractions

    x > 0  ⇒  NUF(x) = ℵ₀

    That implies a growth between [0, 1] and (0, 1].

    No.
    x > 0 ⇒ NUF(x) = ℵ₀
    does not imply a growth between [0,1] and (0,1]

    (Check your work for a quantifier shift.)

    NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
    does not imply any unit fraction outside (0,x]

    The unit.fractions in (0,x] are
    each non.{}.subset maximummed
    each unit.fraction down.stepped
    each non.maximum up.stepped.

    Things with that order.type are ℵ₀.many.

    Finite doesn't need to be small.
    Infinite is beyond all big.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Wed Jul 31 00:49:18 2024
    XPost: sci.math

    Am 30.07.2024 um 20:37 schrieb Jim Burns:
    On 7/30/2024 1:30 PM, WM wrote:

    [...] between [0, 1] and (0, 1].

    Could you please explain to me the meaning of the phrase "between [0, 1]
    and (0, 1]"?

    What IS "between [0, 1] and (0, 1]". (Which real numbers are "between
    [0, 1] and (0, 1]"? 0? So why not just talk about 0 in this case?)

    Thank you.

    (Actually, it sounds like meaningless mumbo-jumbo to me. Something a
    psychotic asshole full of shit might utter.)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Tue Jul 30 21:28:48 2024
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?

    ω/2

    And where is that in {1, 2, 3, ... w} ?


    No, that is w double, what number in the first set became w?


    Every natural number when doubled is a Natural Number.

    No.

    Why not? WHich ones don't?

    ω/2 and larger.

    Which is what number?

    The input set was the Natural Numbers and w, so you are just proving you
    are lying,



    Note, ω-1 doesn't exist in the base transfinite numbers, just as -1
    doesn't exist in the Natural Numbers, you can't go below the first
    element.

    If all natural numbers exist, then ω-1 exists.

    Why?

    Because otherwise there was a gap below ω.

    But you combined two different sets, so why can't there be a gap?

    That is just the nature of unbounded sets.


    That is unavoidable. You believe in the magical appearance of
    infinitely many unit fractions. That breaks logic and mathematics.

    Nope,

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the u niversal quantifier.

    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit fraction 1/n, there exists another unit fraction smaller than itself.
    The only way to have a smallest unit fraction is to have a largest
    natural number, at which point they aren't even "potentially infinite"
    as you have established a finite limit to them.

    Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.


    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Tue Jul 30 21:31:56 2024
    On 7/30/24 1:39 PM, WM wrote:
    Le 30/07/2024 à 03:20, Richard Damon a écrit :
    On 7/29/24 9:15 AM, WM wrote:
    Le 29/07/2024 à 11:01, Mikko a écrit :

    He also notes that what we have learned from finite quatities does
    not apply to infinity.

    Unit fractions are isolated. They have distances. The function "unit
    fractions between 0 and x" can nowhere grow by more than 1.

    Unless it just jumps to infinity because it isn't actually correctly
    defined.

    It is correctly defined. Try to define the function Number of
    UnitFractions between 0 and x as you like it but in accordance with mathematics ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

    Regards, WM

    And you equation proves that there is no smallest unit fraction as your mathematics also says that you statement is equivalent to

    ∀n ∈ ℕ: 1/n > 1/(n+1) so there does not exist a smallest unit fraction, since part of the definition of the Natural numbers is that
    ∀n ∈ ℕ: there exist n+1 ∈ ℕ

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to All on Wed Jul 31 11:51:40 2024
    On 2024-07-30 14:41:15 +0000, WM said:

    Le 30/07/2024 à 09:26, Mikko a écrit :
    On 2024-07-29 13:15:36 +0000, WM said:

    Le 29/07/2024 à 11:01, Mikko a écrit :

    He also notes that what we have learned from finite quatities does
    not apply to infinity.

    Unit fractions are isolated. They have distances. The function "unit
    fractions between 0 and x" can nowhere grow by more than 1.

    The number of unit fractions between 0 and x is infinite

    Not for every x.

    The context where you mentioned unit fractions was a discussion about
    Galileis opinions. At that time the concept of quantity did not include
    absurd quantities and usually not zero, either. Therefore the expression
    "unit fractions between 0 and x" should be interpreted to assume that
    x > 0. Of course if x is zero there is nothing between 0 and x.

    Growth by more than 1 is contradicting mathematics.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

    Irrelevant. But do you mean that 0 is not in ℕ?

    and the number of unit fractions between 0 and y is infinite. Whether one
    infinity is bigger or smaller than another infinity is not answerable merely >> on the basis of our experinece about finite quantities but only after a
    carful analysis if at all.

    If (0, x) is a subset of (0, y), the relative quantities are clear.

    As Galilei pointed out, it is not. That one is a subset of the other
    makes one look bigger but the existence of a bijection between the two
    makes them look equal. Bot appearences cannot be true, so the clarity
    is a hallusination.

    However, although Galilei was happy with all infinities being equalt
    there is a way out, and in fact more than one way: at least cardinals, ordinals, and measures, perhaps something else.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Jul 31 13:59:36 2024
    XPost: sci.math

    Le 30/07/2024 à 20:37, Jim Burns a écrit :

    NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
    does not imply any unit fraction outside (0,x]

    But it implies that you can use only x having ℵ₀ smaller positive
    points, in fact even ℵ₀*2ℵ₀.

    ∀ x > 0: NUF(x) = ℵ₀ would be wrong.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Jul 31 14:05:38 2024
    XPost: sci.math

    Le 31/07/2024 à 00:49, Moebius a écrit :
    Am 30.07.2024 um 20:37 schrieb Jim Burns:
    On 7/30/2024 1:30 PM, WM wrote:

    [...] between [0, 1] and (0, 1].

    Could you please explain to me the meaning of the phrase "between [0, 1]
    and (0, 1]"?

    What IS "between [0, 1] and (0, 1]". (Which real numbers are "between
    [0, 1] and (0, 1]"? 0?

    There is nothing. But ℵ₀ unit fractions require ℵ₀*2ℵ₀ smaller positive points.
    Therefore ∀x > 0: NUF(x) = ℵ₀ is wrong.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Jul 31 14:20:00 2024
    Le 31/07/2024 à 10:51, Mikko a écrit :
    On 2024-07-30 14:41:15 +0000, WM said:

    Growth by more than 1 is contradicting mathematics.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

    Irrelevant.

    Mathematics is not irrelevant? ? ?

    If (0, x) is a subset of (0, y), the relative quantities are clear.

    As Galilei pointed out, it is not. That one is a subset of the other
    makes one look bigger but the existence of a bijection between the two
    makes them look equal. Bot appearences cannot be true, so the clarity
    is a hallusination.

    The bijection is not true.

    However, although Galilei was happy with all infinities being equalt

    Whether he was happy is irrelevant. When you claim as many natural numbers
    as algebraic numbers, you are a crank of mathology.

    Regards, WM


    there is a way out, and in fact more than one way: at least cardinals, ordinals, and measures, perhaps something else.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to No. My formula on Wed Jul 31 14:27:06 2024
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?

    ω/2

    And where is that in {1, 2, 3, ... w} ?

    In the midst, far beyond all definable numbers, far beyond ω/10^10.

    The input set was the Natural Numbers and w,

    ω/10^10 and ω/10 are dark natural numbers.

    If all natural numbers exist, then ω-1 exists.

    Why?

    Because otherwise there was a gap below ω.

    But you combined two different sets, so why can't there be a gap?

    I assume completness.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit fraction 1/n, there exists another unit fraction smaller than itself.

    No. My formula says ∀n ∈ ℕ.

    Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.

    Not for all dark numbers.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From joes@21:1/5 to All on Wed Jul 31 16:20:14 2024
    XPost: sci.math

    Am Wed, 31 Jul 2024 14:27:06 +0000 schrieb WM:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?
    ω/2
    And where is that in {1, 2, 3, ... w} ?
    In the midst, far beyond all definable numbers, far beyond ω/10^10.
    That is a bit imprecise. Even though you keep on talking about
    consecutive infinities, you can't compare natural and "dark" numbers.

    The input set was the Natural Numbers and w,
    ω/10^10 and ω/10 are dark natural numbers.

    If all natural numbers exist, then ω-1 exists.
    Why?
    Because otherwise there was a gap below ω.
    But you combined two different sets, so why can't there be a gap?
    I assume completness.
    Completeness of N? No number n reaches omega.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.
    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit
    fraction 1/n, there exists another unit fraction smaller than itself.
    No. My formula says ∀n ∈ ℕ.
    That is not a contradiction.

    Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.
    Not for all dark numbers.

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Wed Jul 31 13:33:25 2024
    XPost: sci.math

    On 7/31/2024 9:59 AM, WM wrote:
    Le 30/07/2024 à 20:37, Jim Burns a écrit :

    NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
    does not imply any unit fraction outside (0,x]

    But it implies that you can use only
    x having ℵ₀ smaller positive points,
    in fact even ℵ₀*2ℵ₀.

    ∀ x > 0: NUF(x) = ℵ₀ would be wrong.

    Each non.{}.subset of ⅟ℕ∩(0,1] is maximummed.
    ∀S ⊆ ⅟ℕ∩(0,1]: S ≠ {} ⇒ ∃u = max.S

    Each unit.fraction in ⅟ℕ∩(0,1] is down.stepped.
    ∀u ∈ ⅟ℕ∩(0,1] ∃v = ⅟(1+⅟u) = max.⅟ℕ∩(0,u)

    Each unit.fraction in ⅟ℕ∩(0,1] is non.max.up.stepped.
    ∀u ∈ ⅟ℕ∩(0,1]: u ≠ max.⅟ℕ∩(0,1] ⇒
    ∃v = ⅟(-1+⅟u) = min.⅟ℕ∩(u,1]

    ⎛ Whatever else is
    ⎜ maximummed and down.stepped and non.max.up.stepped
    ⎜ has as many as
    ⎜ maximummed and down.stepped and non.max.up.stepped
    ⎝ ⅟ℕ∩(0,1]

    Each non.{}.subset of ⅟ℕ∩(0,x] is
    a non.{}.subset of ⅟ℕ∩(0,1]
    and is maximummed.
    ∀S ⊆ ⅟ℕ∩(0,x]: S ≠ {} ⇒ ∃u = max.S

    Each unit.fraction in ⅟ℕ∩(0,x] is
    a unit.fraction in ⅟ℕ∩(0,1]
    and is down.stepped.
    ∀u ∈ ⅟ℕ∩(0,x] ∃v = ⅟(1+⅟u) = max.⅟ℕ∩(0,u)

    Each unit.fraction in ⅟ℕ∩(0,x] is
    a unit.fraction in ⅟ℕ∩(0,1]
    and is non.max.up.stepped.
    ∀u ∈ ⅟ℕ∩(0,x]: u ≠ max.⅟ℕ∩(0,x] ⇒
    ∃v = ⅟(-1+⅟u) = min.⅟ℕ∩(u,x]

    ⅟ℕ∩(0,x] is
    maximummed and down.stepped and non.max.up.stepped
    ⅟ℕ∩(0,x] has as many as ⅟ℕ∩(0,1]
    |⅟ℕ∩(0,x]| = |⅟ℕ∩(0,1]| = ℵ₀


    ∀ᴿx > 0: NUF(x) = ℵ₀
    because of
    maximumming and down.stepping and non.max.up.stepping.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Wed Jul 31 20:09:53 2024
    On 7/31/24 10:27 AM, WM wrote:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?

    ω/2

    And where is that in {1, 2, 3, ... w} ?

    In the midst, far beyond all definable numbers, far beyond ω/10^10.

    In other words, outside the Natural Nubmer, all of which are defined and definable.


    The input set was the Natural Numbers and w,

    ω/10^10 and ω/10 are dark natural numbers.

    They may be "dark" but they are not Natural Numbers.

    Natural numbers, by their definition, are reachable by a finite number
    of successor operations from 0.

    If all natural numbers exist, then ω-1 exists.

    Why?

    Because otherwise there was a gap below ω.

    But you combined two different sets, so why can't there be a gap?

    I assume completness.

    I guess you definition of "completeness" is incorrect.

    If I take the set of all cats, and the set of all doges, can there not
    be a gap between them?


    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every
    unit fraction 1/n, there exists another unit fraction smaller than
    itself.

    No. My formula says ∀n ∈ ℕ.

    Right, for ALL n in ℕ, there exist another number in ℕ that is n+1, and that one has an n+2, which has an n+3 and so on continuing without bound.

    If you want there to be a n that doesn't have an n+1 in the set, then
    you are not talking about the actual Natural Numbers, because you logic
    system just can't handle them.

    That just proves you are using too primative of tools for the task at
    hand, which has been obvious for years.


    Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.

    Not for all dark numbers.

    Maybe not for dark numbers, but it does for all Natural Numbers, as that
    is part of their DEFINITION.

    Of course, if you logic is based on lying about definitions, it isn't
    good for must.


    Regards, WM



    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 1 12:02:52 2024
    XPost: sci.math

    Le 31/07/2024 à 18:20, joes a écrit :
    Am Wed, 31 Jul 2024 14:27:06 +0000 schrieb WM:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?
    ω/2
    And where is that in {1, 2, 3, ... w} ?
    In the midst, far beyond all definable numbers, far beyond ω/10^10.
    That is a bit imprecise. Even though you keep on talking about
    consecutive infinities, you can't compare natural and "dark" numbers.

    Dark natnumbers are larger than defined natnumbers. Even dar natnumbers
    can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.

    ω/10^10 and ω/10 are dark natural numbers.

    If all natural numbers exist, then ω-1 exists.
    Why?
    Because otherwise there was a gap below ω.
    But you combined two different sets, so why can't there be a gap?
    I assume completness.
    Completeness of N? No number n reaches omega.

    What is immediately before ω? Is it a blasphemy to ask such questions?

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.
    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit
    fraction 1/n, there exists another unit fraction smaller than itself.
    No. My formula says ∀n ∈ ℕ.
    That is not a contradiction.

    It is not a contradiction to my formula if some n has no n+1.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 1 12:13:39 2024
    XPost: sci.math

    Le 31/07/2024 à 19:33, Jim Burns a écrit :
    On 7/31/2024 9:59 AM, WM wrote:
    Le 30/07/2024 à 20:37, Jim Burns a écrit :

    NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
    does not imply any unit fraction outside (0,x]

    But it implies that you can use only
    x having ℵ₀ smaller positive points,
    in fact even ℵ₀*2ℵ₀.

    ∀ x > 0: NUF(x) = ℵ₀ would be wrong.

    Each non.{}.subset of ⅟ℕ∩(0,1] is maximummed.
    ∀S ⊆ ⅟ℕ∩(0,1]: S ≠ {} ⇒ ∃u = max.S

    In the dark domain the maximum cannot be discerned.

    Each unit.fraction in ⅟ℕ∩(0,1] is down.stepped.
    ∀u ∈ ⅟ℕ∩(0,1] ∃v = ⅟(1+⅟u) = max.⅟ℕ∩(0,u)

    Assuming all this would lead to potential infinity.
    I assume actual infinity however. There are dark numbers. There is a
    greates natnumber. There is no chance to lose Bob.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 1 12:27:27 2024
    Le 01/08/2024 à 02:09, Richard Damon a écrit :
    On 7/31/24 10:27 AM, WM wrote:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?

    ω/2

    And where is that in {1, 2, 3, ... w} ?

    In the midst, far beyond all definable numbers, far beyond ω/10^10.

    In other words, outside the Natural Nubmer, all of which are defined and definable.

    That is simply nonsense. Do you know what an accumalation point is? Every
    eps interval around 0 contains unit fractions which cannot be separated
    from 0 by any eps. Therefore your claim is wrong.


    The input set was the Natural Numbers and w,

    ω/10^10 and ω/10 are dark natural numbers.

    They may be "dark" but they are not Natural Numbers.

    They are natural numbers.

    Natural numbers, by their definition, are reachable by a finite number
    of successor operations from 0.

    That is the opinion of Peano and his disciples. It holds only for potetial infinity, i.e., definable numbers.

    I assume completness.

    I guess you definition of "completeness" is incorrect.

    If I take the set of all cats, and the set of all doges, can there not
    be a gap between them?

    What is the reason for the gap before omega? How large is it? Are these questions a blasphemy?


    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every >>> unit fraction 1/n, there exists another unit fraction smaller than
    itself.

    No. My formula says ∀n ∈ ℕ.

    Right, for ALL n in ℕ, there exist another number in ℕ that is n+1,

    That does ny formula not say. It says for all n which have successors,
    there is distance between 1/n and 1/(n+1).

    Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.

    Not for all dark numbers.

    Maybe not for dark numbers, but it does for all Natural Numbers, as that
    is part of their DEFINITION.

    It is the definition of definable numbers. Study the accumulation point.
    Define (separate by an eps from 0) all unit fractions. Fail.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From joes@21:1/5 to All on Thu Aug 1 16:04:04 2024
    XPost: sci.math

    Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:
    Le 01/08/2024 à 02:09, Richard Damon a écrit :
    On 7/31/24 10:27 AM, WM wrote:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?
    ω/2
    And where is that in {1, 2, 3, ... w} ?
    In the midst, far beyond all definable numbers, far beyond ω/10^10.
    In other words, outside the Natural Nubmer, all of which are defined
    and definable.
    That is simply nonsense. Do you know what an accumalation point is?
    Every eps interval around 0 contains unit fractions which cannot be
    separated from 0 by any eps. Therefore your claim is wrong.
    No. There is ALWAYS an epsilon.

    The input set was the Natural Numbers and w,
    ω/10^10 and ω/10 are dark natural numbers.
    They may be "dark" but they are not Natural Numbers.
    They are natural numbers.
    How are they defined?

    Natural numbers, by their definition, are reachable by a finite number
    of successor operations from 0.
    That is the opinion of Peano and his disciples. It holds only for
    potetial infinity, i.e., definable numbers.

    I assume completness.
    I guess you definition of "completeness" is incorrect.
    If I take the set of all cats, and the set of all doges, can there not
    be a gap between them?
    What is the reason for the gap before omega? How large is it? Are these questions a blasphemy?
    A "gap" implies some sort of space that is not filled. There is no such
    space (it would be filled with infinitely many natural numbers).
    We just condense the whole of N into one concept and call that omega,
    or add it on the next level of infinity.
    Your questions are only a display of your unwillingness to understand
    infinity, even though you would like to imagine yourself as some sort
    of martyr.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.
    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every >>>> unit fraction 1/n, there exists another unit fraction smaller than
    itself.
    No. My formula says ∀n ∈ ℕ.
    Right, for ALL n in ℕ, there exist another number in ℕ that is n+1,
    That does ny formula not say. It says for all n which have successors,
    there is distance between 1/n and 1/(n+1).
    If k did not have a successor, what would k+1 be?

    Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists. >>> Not for all dark numbers.
    Maybe not for dark numbers, but it does for all Natural Numbers, as
    that is part of their DEFINITION.
    It is the definition of definable numbers. Study the accumulation point. Define (separate by an eps from 0) all unit fractions. Fail.

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Thu Aug 1 12:43:56 2024
    XPost: sci.math

    On 8/1/2024 8:02 AM, WM wrote:
    Le 31/07/2024 à 18:20, joes a écrit :
    Am Wed, 31 Jul 2024 14:27:06 +0000 schrieb WM:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
    Note the universal quantifier.

    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1),
    so that for every unit fraction 1/n, there exists
    another unit fraction smaller than itself.

    No. My formula says ∀n ∈ ℕ.

    That is not a contradiction.

    It is not a contradiction to my formula
    if some n has no n+1.

    No, it literally contradicts your formula
    for some n e N to not.have n+1

    Your formula uses 'n+1'
    Along with saying other things,
    using 'n+1' asserts that 'n+1' exists, ∀n ∈ ℕ

    That assertion is so often uncontroversial
    that it's easy to overlook.
    Nonetheless, it is there.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Thu Aug 1 14:12:37 2024
    XPost: sci.math

    On 8/1/2024 8:13 AM, WM wrote:
    Le 31/07/2024 à 19:33, Jim Burns a écrit :
    On 7/31/2024 9:59 AM, WM wrote:
    Le 30/07/2024 à 20:37, Jim Burns a écrit :

    NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
    does not imply any unit fraction outside (0,x]

    But it implies that you can use only
    x having ℵ₀ smaller positive points,
    in fact even ℵ₀*2ℵ₀.

    ∀ x > 0: NUF(x) = ℵ₀ would be wrong.

    Each non.{}.subset of ⅟ℕ∩(0,1] is maximummed.
      ∀S ⊆ ⅟ℕ∩(0,1]: S ≠ {}  ⇒  ∃u = max.S

    In the dark domain the maximum cannot be discerned.

    ⅟ℕᶠⁱⁿ∩(0,1] is the set of finite.unit.fractions.

    Each non.{}.subset of ⅟ℕᶠⁱⁿ∩(0,1] is maximummed.
    ∀S ⊆ ⅟ℕᶠⁱⁿ∩(0,1]: S ≠ {} ⇒ ∃u = max.S

    Each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,1] is down.stepped.
    ∀u ∈ ⅟ℕᶠⁱⁿ∩(0,1] ∃v = ⅟(1+⅟u) = max.⅟ℕᶠⁱⁿ∩(0,u)

    Each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,1] is non.max.up.stepped.
    ∀v ∈ ⅟ℕᶠⁱⁿ∩(0,1]: v ≠ max.⅟ℕᶠⁱⁿ∩(0,1] ⇒
    ∃u = ⅟(-1+⅟v) = min.⅟ℕᶠⁱⁿ∩(v,1]

    Each non.{}.subset of ⅟ℕᶠⁱⁿ∩(0,1] is maximummed.
    Each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,1] is down.stepped.
    Each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,1] is non.max.up.stepped.

    Therefore,
    the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,1] are ℵ₀.many.

    ∀ᴿx > 0: ⅟ℕᶠⁱⁿ∩(0,x] ≠ {}

    ⎛ Assume otherwise.
    ⎜ Assume ⅟ℕᶠⁱⁿ∩(0,x] = {}

    ⎜ β ≥ x > 0 situates the split between
    ⎜ lower.bounds of ⅟ℕᶠⁱⁿ∩(0,1] and
    ⎜ not.lower.bounds of ⅟ℕᶠⁱⁿ∩(0,1]
    ⎜ 2⋅β > β > ½⋅β > 0

    ⎜ ½⋅β < β is a lower.bound of ⅟ℕᶠⁱⁿ∩(0,1]

    ⎜ 2⋅β > β is a not.lower.bound of ⅟ℕᶠⁱⁿ∩(0,1]
    ⎜ finite.unit.fraction ⅟k < 2⋅β exists
    ⎜ finite.unit.fraction ¼⋅⅟k < ¼⋅2⋅β = ½⋅β exists
    ⎜ ½⋅β is a not.lower.bound of ⅟ℕᶠⁱⁿ∩(0,1]
    ⎝ Contradiction.

    Therefore,
    ∀ᴿx > 0: ⅟ℕᶠⁱⁿ∩(0,x] ≠ {}

    (0,x] inherits from its superset (0,1] properties by which
    each non.{}.subset of ⅟ℕᶠⁱⁿ∩(0,x] is maximummed, and
    each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,x] is down.stepped, and
    each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,x] is non.max.up.stepped.

    Therefore,
    the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.

    ∀ᴿx > 0: NUFᶠⁱⁿ(x) = ℵ₀

    NUFᶠⁱⁿ(x) ≥ NUF(x)

    ∀ᴿx > 0: NUF(x) ≥ ℵ₀

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 1 21:35:30 2024
    XPost: sci.math

    Am 01.08.2024 um 18:43 schrieb Jim Burns:
    On 8/1/2024 8:02 AM, WM wrote:
    Le 31/07/2024 à 18:20, joes a écrit :
    Am Wed, 31 Jul 2024 14:27:06 +0000 schrieb WM:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
    Note the universal quantifier.

    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1),
    so that for every unit fraction 1/n, there exists
    another unit fraction smaller than itself.

    No. My formula says ∀n ∈ ℕ.

    That is not a contradiction.

    It is not a contradiction to my formula
    if some n has no n+1.

    No, it literally contradicts your formula
    for some n e N to not.have n+1

    Your formula uses 'n+1'
    Along with saying other things,
    using 'n+1' asserts*) that 'n+1' exists [for all] n ∈ ℕ

    That assertion is so often uncontroversial
    that it's easy to overlook.
    Nonetheless, it is there.

    Yes, I already told him that in dsm, but to no avail.

    ________________________________

    *) or presupposes

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  • From Moebius@21:1/5 to All on Thu Aug 1 23:21:58 2024
    XPost: sci.math

    Am 01.08.2024 um 20:12 schrieb Jim Burns:

    In the dark domain

    I'd recommend

    https://www.youtube.com/watch?v=1-QEpRgs9sM

    as a cure in this case.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Thu Aug 1 19:45:29 2024
    XPost: sci.math

    On 8/1/24 8:02 AM, WM wrote:
    Le 31/07/2024 à 18:20, joes a écrit :
    Am Wed, 31 Jul 2024 14:27:06 +0000 schrieb WM:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?
    ω/2
    And where is that in {1, 2, 3, ... w} ?
    In the midst, far beyond all definable numbers, far beyond ω/10^10.
    That is a bit imprecise. Even though you keep on talking about
    consecutive infinities, you can't compare natural and "dark" numbers.

    Dark natnumbers are larger than defined natnumbers. Even dar natnumbers
    can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.

    ω/10^10 and ω/10 are dark natural numbers.

    If all natural numbers exist, then ω-1 exists.
    Why?
    Because otherwise there was a gap below ω.
    But you combined two different sets, so why can't there be a gap?
    I assume completness.
    Completeness of N? No number n reaches omega.

    What is immediately before ω? Is it a blasphemy to ask such questions?

    It has no predicessor, just like in the Natural Numbers 0 has nothing
    before it.

    You can expand your number system to define number there, which seems to
    be what you "dark numbers" are, numbers bigger than all the finite
    Natural Numbers, but smaller than w.


    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.
    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit
    fraction 1/n, there exists another unit fraction smaller than itself.
    No. My formula says ∀n ∈ ℕ.
    That is not a contradiction.

    It is not a contradiction to my formula if some n has no n+1.

    It is a violation of the DEFINITION of the Natural Numbers.

    Now, if you admit that you system doesn't actaully HAVE the actaual
    Natura Numbers, but some other set built by some other methd, go ahead
    and try to actually define that set and see what you can do with it.

    Just don't claim that anyone else needs to use your inferior number system.


    Regards, WM


    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Thu Aug 1 19:53:02 2024
    On 8/1/24 8:27 AM, WM wrote:
    Le 01/08/2024 à 02:09, Richard Damon a écrit :
    On 7/31/24 10:27 AM, WM wrote:
    Le 31/07/2024 à 03:28, Richard Damon a écrit :
    On 7/30/24 1:37 PM, WM wrote:
    Le 30/07/2024 à 03:18, Richard Damon a écrit :
    On 7/29/24 9:11 AM, WM wrote:

    But what number became ω when doubled?

    ω/2

    And where is that in {1, 2, 3, ... w} ?

    In the midst, far beyond all definable numbers, far beyond ω/10^10.

    In other words, outside the Natural Nubmer, all of which are defined
    and definable.

    That is simply nonsense. Do you know what an accumalation point is?
    Every eps interval around 0 contains unit fractions which cannot be
    separated from 0 by any eps. Therefore your claim is wrong.

    And thus there is no "smallest" unit fraction, as for any eps, there are
    unit fractions smaller, and thus NUF(x) does have a value in the finites
    where it has the value of 1.



    The input set was the Natural Numbers and w,

    ω/10^10 and ω/10 are dark natural numbers.

    They may be "dark" but they are not Natural Numbers.

    They are natural numbers.

    Natural numbers, by their definition, are reachable by a finite number
    of successor operations from 0.

    That is the opinion of Peano and his disciples. It holds only for
    potetial infinity, i.e., definable numbers.

    No, it holds for ALL his numbers.


    I assume completness.

    I guess you definition of "completeness" is incorrect.

    If I take the set of all cats, and the set of all doges, can there not
    be a gap between them?

    What is the reason for the gap before omega? How large is it? Are these questions a blasphemy?

    Because it is between two different sorts of number.

    There is a gap between 1 and 2, but that doesn't bother you.



    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

    Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every >>>> unit fraction 1/n, there exists another unit fraction smaller than
    itself.

    No. My formula says ∀n ∈ ℕ.

    Right, for ALL n in ℕ, there exist another number in ℕ that is n+1,

    That does ny formula not say. It says for all n which have successors,
    there is  distance between 1/n and 1/(n+1).

    Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists. >>>
    Not for all dark numbers.

    Maybe not for dark numbers, but it does for all Natural Numbers, as
    that is part of their DEFINITION.

    It is the definition of definable numbers. Study the accumulation point. Define (separate by an eps from 0) all unit fractions. Fail.

    So, which Unit fraction doesn't have an eps that seperates it from 0?

    You just get your order of conditions reversed.

    For all 1/n, there is a eps that is smaller than it (like 1/(n+1) )

    And for every eps, there is a unit fraction smaller than it

    1 / (ceil (1/eps) + 1)

    So we have an unlimited number of Unit fractions, and no smallest one.


    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to All on Fri Aug 2 09:48:45 2024
    On 2024-07-31 14:20:00 +0000, WM said:

    Le 31/07/2024 à 10:51, Mikko a écrit :
    On 2024-07-30 14:41:15 +0000, WM said:

    Growth by more than 1 is contradicting mathematics.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

    Irrelevant.

    Mathematics is not irrelevant? ? ?

    Most of mathematics is irrelevant to most purposes.

    If (0, x) is a subset of (0, y), the relative quantities are clear.

    As Galilei pointed out, it is not. That one is a subset of the other
    makes one look bigger but the existence of a bijection between the two
    makes them look equal. Bot appearences cannot be true, so the clarity
    is a hallusination.

    The bijection is not true.

    Of course not, just like the Moon is not true. It just is there.

    --
    Mikko

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Fri Aug 2 11:02:24 2024
    Le 02/08/2024 à 08:48, Mikko a écrit :
    On 2024-07-31 14:20:00 +0000, WM said:

    The bijection is not true.

    Of course not, just like the Moon is not true. It just is there.

    The bijection concerns only the potentially infinite initial segments.
    Every natural number that can be verified in the bijection has ∀n ∈ ℕ_verif: |ℕ \ {1, 2, 3, ..., n}| = ℵo successors, ℵo of which
    cannot be verified.
    Proof: Ther are more algebraic numbers than prime numbers. Every not
    completely confused brain recognizes this.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 2 11:40:18 2024
    XPost: sci.math

    Le 01/08/2024 à 18:43, Jim Burns a écrit :
    On 8/1/2024 8:02 AM, WM wrote:

    It is not a contradiction to my formula
    if some n has no n+1.

    No, it literally contradicts your formula
    for some n e N to not.have n+1

    My formula is explicitly valid only for natural numbers.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 2 11:38:33 2024
    XPost: sci.math

    Le 01/08/2024 à 18:04, joes a écrit :
    Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:

    separated from 0 by any eps. Therefore your claim is wrong.
    No. There is ALWAYS an epsilon.

    Failing to separate almost all unit fractions.
    Don't claim the contrary. Define (separate by an eps from 0) all unit fractions. Fail.

    What is the reason for the gap before omega? How large is it? Are these
    questions a blasphemy?
    A "gap" implies some sort of space that is not filled. There is no such
    space (it would be filled with infinitely many natural numbers).

    Hence there is only the sequence of natnumbers.

    We just condense the whole of N into one concept and call that omega,

    That is nonsense. ω is the first number following upon all natural
    numbers.

    or add it on the next level of infinity.

    Yes.

    Your questions are only a display of your unwillingness to understand infinity,

    They prove that I understand the real infinity while are (con)fusing
    potential and actual infinity.

    If k did not have a successor, what would k+1 be?

    ω

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 2 14:50:22 2024
    XPost: sci.math

    Le 01/08/2024 à 18:04, joes a écrit :

    A "gap" implies some sort of space that is not filled. There is no such
    space (it would be filled with infinitely many natural numbers).
    We just condense the whole of N into one concept and call that omega,
    or add it on the next level of infinity.

    The limit is not a next level but extends the ordinals just like 0 extends their reciprocals.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 2 14:58:59 2024
    XPost: sci.math

    Le 02/08/2024 à 01:45, Richard Damon a écrit :
    On 8/1/24 8:02 AM, WM wrote:

    What is immediately before ω? Is it a blasphemy to ask such questions?

    It has no predicessor, just like in the Natural Numbers 0 has nothing
    before it.

    0 has a continuum above it, no gap! Likewise there must be no gap below
    ω.

    You can expand your number system to define number there, which seems to
    be what you "dark numbers" are, numbers bigger than all the finite
    Natural Numbers, but smaller than w.

    Thank you.


    It is not a contradiction to my formula if some n has no n+1.

    It is a violation of the DEFINITION of the Natural Numbers.

    Otherwise there is a contradiction of mathematics: separated unit
    fractions.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 2 15:09:18 2024
    Le 02/08/2024 à 01:53, Richard Damon a écrit :
    On 8/1/24 8:27 AM, WM wrote:

    And thus there is no "smallest" unit fraction, as for any eps, there are
    unit fractions smaller,

    Your eps cannot be chosen small enough.

    That is the opinion of Peano and his disciples. It holds only for
    potetial infinity, i.e., definable numbers.

    No, it holds for ALL his numbers.

    Not for ℵo, i.e., for most it is wrong:
    ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

    What is the reason for the gap before omega? How large is it? Are these
    questions a blasphemy?

    Because it is between two different sorts of number.

    There is no gap above zero but e real continuum.

    There is a gap between 1 and 2, but that doesn't bother you.

    All gaps of size 1 do not bother me..

    It is the definition of definable numbers. Study the accumulation point.
    Define (separate by an eps from 0) all unit fractions. Fail.

    So, which Unit fraction doesn't have an eps that seperates it from 0?

    There are infinitely many by the definition of accumulation point. You
    cannot find them. Therefore they are dark.

    You just get your order of conditions reversed.

    I get it the only corect way. Every eps that you can chose belongs to a
    set of chosen eps. This set has a minimum - at every time. It is finite. Quantifiers therefore can be reversed.

    For all 1/n, there is a eps that is smaller than it (like 1/(n+1) )

    For all 1/n that you can define.

    And for every eps, there is a unit fraction smaller than it

    There are infinitely many, namely almost all.

    So we have an unlimited number of Unit fractions, and no smallest one.

    But you have a limited number of eps.

    Regards, WM

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  • From WM@21:1/5 to All on Fri Aug 2 15:36:07 2024
    XPost: sci.math

    Le 01/08/2024 à 20:12, Jim Burns a écrit :

    ∀ᴿx > 0: NUF(x) ≥ ℵ₀

    This nonsense will not become true by repeating it. ℵ₀ unit fractions
    need ℵ₀*2ℵ₀ points above zero. For those x > 0 your claim is
    wrong.

    Regards, WM

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  • From Richard Damon@21:1/5 to All on Fri Aug 2 11:17:57 2024
    XPost: sci.math

    On 8/2/24 10:58 AM, WM wrote:
    Le 02/08/2024 à 01:45, Richard Damon a écrit :
    On 8/1/24 8:02 AM, WM wrote:

    What is immediately before ω? Is it a blasphemy to ask such questions?

    It has no predicessor, just like in the Natural Numbers 0 has nothing
    before it.

    0 has a continuum above it, no gap! Likewise there must be no gap below ω.

    You can expand your number system to define number there, which seems
    to be what you "dark numbers" are, numbers bigger than all the finite
    Natural Numbers, but smaller than w.

    Thank you.


    It is not a contradiction to my formula if some n has no n+1.

    It is a violation of the DEFINITION of the Natural Numbers.

    Otherwise there is a contradiction of mathematics: separated unit
    fractions.

    Regards, WM


    No, all Unit fractions are separated in distance by a finite amount, it
    is just that you can't specify any finite distance that separates all
    unit fractions.

    The reversal of the order of qualifications causes the problem.

    This is the nature of unbounded numbers, something your logic doesn't
    seem to be able to handle.

    Every number 1/n is separated from the next smaller unit fraction,
    1/(n+1) by a distance of 1/(n*(n+1)) which is a value that is greater
    than zero, so we always have a finite difference between all unit
    fractions, but that distance gets arbitrarily small, so we can't choose
    a single finite eps that all unit fractions are seperated by, even
    though all unit fractions are seperated by a finite distance.

    This just shows that this spacing, APPROACHES 0, as a limit, as n
    increases. But approaching a limit of 0 is not the same as being 0.

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  • From WM@21:1/5 to All on Fri Aug 2 15:57:56 2024
    XPost: sci.math

    Le 02/08/2024 à 17:17, Richard Damon a écrit :
    On 8/2/24 10:58 AM, WM wrote:

    It is not a contradiction to my formula if some n has no n+1.

    It is a violation of the DEFINITION of the Natural Numbers.

    Otherwise there is a contradiction of mathematics: separated unit
    fractions.

    No, all Unit fractions are separated in distance by a finite amount, it
    is just that you can't specify any finite distance that separates all
    unit fractions.

    I don't try. But I know that one unit fraction must be the first.

    The reversal of the order of qualifications causes the problem.

    Distances on the real line imply an order, even it it cannot be discerned.

    Every number 1/n is separated from the next smaller unit fraction,
    1/(n+1) by a distance of 1/(n*(n+1)) which is a value that is greater
    than zero, so we always have a finite difference between all unit
    fractions, but that distance gets arbitrarily small,

    Can they get smaller than 2^ℵ₀ points?

    If not ℵ₀ unit fractions need ℵ₀*2^ℵ₀ points above zero. For
    those x > 0 the claim
    ∀x > 0: NUF(x) ≥ ℵ₀
    is wrong.

    Regards, WM

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  • From joes@21:1/5 to All on Fri Aug 2 15:42:59 2024
    Am Fri, 02 Aug 2024 15:09:18 +0000 schrieb WM:
    Le 02/08/2024 à 01:53, Richard Damon a écrit :
    On 8/1/24 8:27 AM, WM wrote:

    And thus there is no "smallest" unit fraction, as for any eps, there
    are unit fractions smaller,
    Your eps cannot be chosen small enough.
    All unit fractions are larger than zero, so an epsilon can be chosen

    That is the opinion of Peano and his disciples. It holds only for
    potetial infinity, i.e., definable numbers.
    No, it holds for ALL his numbers.
    Not for ℵo, i.e., for most it is wrong:
    ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

    What is the reason for the gap before omega? How large is it? Are
    these questions a blasphemy?
    Because it is between two different sorts of number.
    There is no gap above zero but e real continuum.

    There is a gap between 1 and 2, but that doesn't bother you.
    All gaps of size 1 do not bother me..

    It is the definition of definable numbers. Study the accumulation
    point.
    Define (separate by an eps from 0) all unit fractions. Fail.
    So, which Unit fraction doesn't have an eps that seperates it from 0?
    There are infinitely many by the definition of accumulation point. You
    cannot find them. Therefore they are dark.
    That is not the definition. The "infinitely many" are not the same ones
    for every epsilon. You can't seem to imagine different infinities.

    You just get your order of conditions reversed.
    I get it the only corect way. Every eps that you can chose belongs to a
    set of chosen eps. This set has a minimum - at every time. It is finite. Quantifiers therefore can be reversed.
    The set of reals is infinite and does not have a minimum.

    For all 1/n, there is a eps that is smaller than it (like 1/(n+1) )
    For all 1/n that you can define.

    And for every eps, there is a unit fraction smaller than it
    There are infinitely many, namely almost all.

    So we have an unlimited number of Unit fractions, and no smallest one.
    But you have a limited number of eps.
    WTF?

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

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  • From WM@21:1/5 to All on Fri Aug 2 16:05:44 2024
    XPost: sci.math

    Le 02/08/2024 à 17:59, Richard Damon a écrit :
    On 8/2/24 7:38 AM, WM wrote:
    Le 01/08/2024 à 18:04, joes a écrit :
    Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:

    separated from 0 by any eps. Therefore your claim is wrong.
    No. There is ALWAYS an epsilon.

    Failing to separate almost all unit fractions.
    Don't claim the contrary. Define (separate by an eps from 0) all unit
    fractions. Fail.

    Improperly revesing the conditionals.

    Not at all! Recognizing that eps must be chosen. You cannot choose a eps
    that separates more than few unit fractions. That is why most cranks claim
    ∀x > 0: NUF(x) ≥ ℵ₀. It holds or all x that can be chosen. How
    should there be always an epsilon smaller than every x > 0 which fails? ?
    ?

    Regards, WM

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  • From Richard Damon@21:1/5 to All on Fri Aug 2 11:59:22 2024
    XPost: sci.math

    On 8/2/24 7:38 AM, WM wrote:
    Le 01/08/2024 à 18:04, joes a écrit :
    Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:

    separated from 0 by any eps. Therefore your claim is wrong.
    No. There is ALWAYS an epsilon.

    Failing to separate almost all unit fractions.
    Don't claim the contrary. Define (separate by an eps from 0) all unit fractions. Fail.


    Improperly revesing the conditionals.

    What is the reason for the gap before omega? How large is it? Are these
    questions a blasphemy?
    A "gap" implies some sort of space that is not filled. There is no such
    space (it would be filled with infinitely many natural numbers).

    Hence there is only the sequence of natnumbers.

    Nope, just proof of the ignorance of WM.


    We just condense the whole of N into one concept and call that omega,

    That is nonsense. ω is the first number following upon all natural numbers.

    No, it is the first TRANSFINITE number beyond all the natural numbers,
    one which has no predicesors.


    or add it on the next level of infinity.

    Yes.

    Your questions are only a display of your unwillingness to understand
    infinity,

    They  prove that I understand the real infinity while are (con)fusing potential and actual infinity.

    Nope, just that you use broken logic,


    If k did not have a successor, what would k+1 be?

    ω

    Regards, WM





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  • From Richard Damon@21:1/5 to All on Fri Aug 2 12:12:08 2024
    On 8/2/24 11:09 AM, WM wrote:
    Le 02/08/2024 à 01:53, Richard Damon a écrit :
    On 8/1/24 8:27 AM, WM wrote:

    And thus there is no "smallest" unit fraction, as for any eps, there
    are unit fractions smaller,

    Your eps cannot be chosen small enough.

    But you have the wrong definition of eps.

    Between any two unit fractions, there is a finite non-zero eps that as
    smaller than their distance.

    That does not say that one eps works for all unit fractions, that is
    just your invalid reversal of the conditions, that shows that your logic
    is broken.


    That is the opinion of Peano and his disciples. It holds only for
    potetial infinity, i.e., definable numbers.

    No, it holds for ALL his numbers.

    Not for ℵo, i.e., for most it is wrong:
    ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

    ℵo is not a "Natural Number" or a number in Peano.

    The fact that for any Natural Number, there are ℵo Natural Numbers above
    it is not a problem, unless you are using broken logic that think that
    ℵo obeys ALL the laws of finite mathematics.


    What is the reason for the gap before omega? How large is it? Are
    these questions a blasphemy?

    Because it is between two different sorts of number.

    There is no gap above zero but e real continuum.

    Because the real continuum is a single type of number.


    There is a gap between 1 and 2, but that doesn't bother you.

    All gaps of size 1 do not bother me..

    Good, Then when talking about omega, gaps between numbers isn't a
    problem either, we get the sets of 0*Omega + n, as the Natural Numbers,
    then 1*Omega + n as the first set of transfinite numbers, nd those
    having a gap of Omega shouldn't be a problem.

    It is the definition of definable numbers. Study the accumulation
    point. Define (separate by an eps from 0) all unit fractions. Fail.

    So, which Unit fraction doesn't have an eps that seperates it from 0?

    There are infinitely many by the definition of accumulation point. You
    cannot find them. Therefore they are dark.

    Nope, we can find any one of them we want.

    Again, you are changing the conditional incorrectly because you logic
    can't handle unbounded values.


    You just get your order of conditions reversed.

    I get it the only corect way. Every eps that you can chose belongs to a
    set of chosen eps. This set has a minimum - at every time. It is finite. Quantifiers therefore can be reversed.

    And that is your problem, the set of eps doesn't HAVE a minimum, because
    it is unbounded.

    You logic just can't handle unbounded sets, and thus can't actually have
    any infinity,


    For all 1/n, there is a eps that is smaller than it (like 1/(n+1) )

    For all 1/n that you can define.

    For *ALL* 1/n, PERIOD, since I can define any of them.


    And for every eps, there is a unit fraction smaller than it

    There are infinitely many, namely almost all.

    No, FOR ALL.

    Name the one that isn't


    So we have an unlimited number of Unit fractions, and no smallest one.

    But you have a limited number of eps.


    Nope, why do you say I have a limited number of eps?

    That is just you proving your stupidity, and that you are stuck in a
    finite and bounded logic system trying to handle something that is
    actually unbounded.

    Regards, WM



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  • From WM@21:1/5 to All on Fri Aug 2 16:24:44 2024
    Le 02/08/2024 à 17:42, joes a écrit :
    Am Fri, 02 Aug 2024 15:09:18 +0000 schrieb WM:
    Le 02/08/2024 à 01:53, Richard Damon a écrit :
    On 8/1/24 8:27 AM, WM wrote:

    And thus there is no "smallest" unit fraction, as for any eps, there
    are unit fractions smaller,
    Your eps cannot be chosen small enough.
    All unit fractions are larger than zero, so an epsilon can be chosen

    Set theory says:
    ∀eps > 0: NUF(eps) ≥ ℵ₀.

    Who is right? You or set theory?

    By the way, set theory is wrong for all x occupied by unit fractions, at
    least ℵ₀*2^ℵ₀ points x > 0.

    There is a gap between 1 and 2, but that doesn't bother you.
    All gaps of size 1 do not bother me..

    It is the definition of definable numbers. Study the accumulation
    point.
    Define (separate by an eps from 0) all unit fractions. Fail.
    So, which Unit fraction doesn't have an eps that seperates it from 0?
    There are infinitely many by the definition of accumulation point. You
    cannot find them. Therefore they are dark.
    That is not the definition. The "infinitely many" are not the same ones
    for every epsilon.

    Strawman! Nobody claimed that. The claim is simply infimitely many. Not infinitely many the same.

    You just get your order of conditions reversed.
    I get it the only corect way. Every eps that you can chose belongs to a
    set of chosen eps. This set has a minimum - at every time. It is finite.
    Quantifiers therefore can be reversed.
    The set of reals is infinite and does not have a minimum.

    The set of chosen reals has a minimum at every time you choose.

    So we have an unlimited number of Unit fractions, and no smallest one.
    But you have a limited number of eps.
    WTF?

    You can choose as many as you like. You will never have chosen infinitely
    many eps. Therefore your claim that for every 1/n there is an eps
    separating it, is wrong.

    Of course for every 1/n there is a smaller real number. But you cannot
    choose it. Most are dark.

    Regards, WM

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  • From Richard Damon@21:1/5 to All on Fri Aug 2 12:24:50 2024
    XPost: sci.math

    On 8/2/24 12:05 PM, WM wrote:
    Le 02/08/2024 à 17:59, Richard Damon a écrit :
    On 8/2/24 7:38 AM, WM wrote:
    Le 01/08/2024 à 18:04, joes a écrit :
    Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:

    separated from 0 by any eps. Therefore your claim is wrong.
    No. There is ALWAYS an epsilon.

    Failing to separate almost all unit fractions.
    Don't claim the contrary. Define (separate by an eps from 0) all unit
    fractions. Fail.

    Improperly revesing the conditionals.

    Not at all! Recognizing that eps must be chosen. You cannot choose a eps
    that separates more than few unit fractions. That is why most cranks
    claim ∀x > 0:  NUF(x) ≥ ℵ₀. It holds or all x that can be chosen. How
    should there be always an epsilon smaller than every x > 0 which fails? ? ?

    Regards, WM

    But you have the condition backwards.

    The claim is that there is a finite difference that seperates any two
    specific unit fractions,

    Not that there is a single finite difference that seperates any two
    arbirtry unit fractions.

    Not understanding the order of the arguement just blow your logic up.

    For your example, for any x, there is an eps = 1/(ceil(1/x)+1) that
    provides a unit fraction smaller than x, and thus NUF(x) can not be 1
    for any finite x.

    The fact that we get a different eps for every x is not a problem, it is
    just a property of the unbounded nature of the numbers we are using.

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  • From Jim Burns@21:1/5 to All on Fri Aug 2 12:35:30 2024
    XPost: sci.math

    On 8/2/2024 7:40 AM, WM wrote:
    Le 01/08/2024 à 18:43, Jim Burns a écrit :
    On 8/1/2024 8:02 AM, WM wrote:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
    Note the universal quantifier.

    It is not a contradiction to my formula
    if some n has no n+1.

    No, it literally contradicts your formula
    for some n e N to not.have n+1

    My formula is explicitly valid only for natural numbers.

    Your formula means
    ∀n ∈ ℕ: ∃k ∈ ℕ: 1/n - 1/k > 0 ∧
    k=n+1 ∧ ¬∃k₂≠k: k₂=n+1

    Operations have unique values.
    '+' is an operation.
    ∀n ∈ ℕ: ∃k ∈ ℕ: k=n+1

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  • From Richard Damon@21:1/5 to All on Fri Aug 2 12:19:14 2024
    On 8/2/24 7:02 AM, WM wrote:
    Le 02/08/2024 à 08:48, Mikko a écrit :
    On 2024-07-31 14:20:00 +0000, WM said:

    The bijection is not true.

    Of course not, just like the Moon is not true. It just is there.

    The bijection concerns only the potentially infinite initial segments.
    Every natural number that can be verified in the bijection has ∀n ∈ ℕ_verif: |ℕ \ {1, 2, 3, ..., n}| = ℵo successors, ℵo of which cannot be
    verified.
    Proof: Ther are more algebraic numbers than prime numbers. Every not completely confused brain recognizes this.

    Regards, WM



    No, YOU THINK there are more algebraic numbers than prime numbers
    because you don't understand that there are exactly ℵo of both of them.

    All countably infinite sets are the same size, and any logic that tries
    to order the sizes of such sets gets into contradictions, and blows
    itself up.

    This of course, has happened to your logic system, and apparently taken
    out your brain.

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  • From WM@21:1/5 to All on Fri Aug 2 16:32:50 2024
    XPost: sci.math

    Le 02/08/2024 à 18:24, Richard Damon a écrit :
    On 8/2/24 12:05 PM, WM wrote:

    The claim is that there is a finite difference that seperates any two specific unit fractions,

    Of course.

    Not that there is a single finite difference that seperates any two
    arbirtry unit fractions.

    Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points.

    Not understanding the order of the arguement

    Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points.
    No order necessary.

    For your example, for any x, there is an eps = 1/(ceil(1/x)+1) that
    provides a unit fraction smaller than x,

    No, eps has to be chosen. x has to be chosen. My claim is that most are
    dark and cannot be chosen.
    Note: For every 1/n there exists a smaller real number. But they cannot be chosen because most 1/n already cannot be chosen. Proof: For every chosen
    eps you fail to separate infinitely many unit fractions.

    and thus NUF(x) can not be 1
    for any finite x.

    But for x belonging to the first ℵ₀*2ℵ₀ points we have less than
    ℵ₀ unit fractions.

    Regards, WM

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  • From WM@21:1/5 to All on Fri Aug 2 16:38:13 2024
    Le 02/08/2024 à 18:19, Richard Damon a écrit :

    No, YOU THINK there are more algebraic numbers than prime numbers
    because you don't understand that there are exactly ℵo of both of them.

    I know that ℵo is nonsense, because all prime numbers are algebraic but
    not all algebraic numbers are prime. This does not change in the infinite.

    All countably infinite sets are the same size,

    That proves that ℵo is nonsense.

    Regareds, WM

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  • From WM@21:1/5 to All on Fri Aug 2 16:42:31 2024
    Le 02/08/2024 à 18:12, Richard Damon a écrit :
    On 8/2/24 11:09 AM, WM wrote:

    There are infinitely many by the definition of accumulation point. You
    cannot find them. Therefore they are dark.

    Nope, we can find any one of them we want.

    But you cannot want any one.

    Again, you are changing the conditional incorrectly because you logic
    can't handle unbounded values.

    What you want is bounded. The set of all eps you ever choose is finite.
    You will never separate infinitely many unit fractions.

    Regards, WM

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  • From Jim Burns@21:1/5 to All on Fri Aug 2 13:06:17 2024
    XPost: sci.math

    On 8/2/2024 11:36 AM, WM wrote:
    Le 01/08/2024 à 20:12, Jim Burns a écrit :

    ∀ᴿx > 0:  NUF(x) ≥ ℵ₀

    This nonsense will not become true by repeating it.

    It does not become false by deleting it.

    (0,x] inherits from its superset (0,1] properties by which,
    for ⅟ℕᶠⁱⁿ∩(0,x] finite.unit.fractions in (0,x]
    each non.{}.subset is maximummed, and
    each finite.unit.fraction is down.stepped, and
    each finite.unit.fraction in is non.max.up.stepped.

    Therefore,
    the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.

    ∀ᴿx > 0: NUFᶠⁱⁿ(x) = ℵ₀

    NUF(x) ≥ NUFᶠⁱⁿ(x)

    ∀ᴿx > 0: NUF(x) ≥ ℵ₀

    ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

    ℵ₀ is the cardinality of all final ordinals.
    Final ordinal α is smaller than α∪{α}

    For each final ordinal α: α∪{α} is also a final ordinal.

    The cardinality ℵ₀ of final ordinals can't be a final ordinal.
    Otherwise,
    there would be at least ℵ₀∪{ℵ₀} final ordinals,
    which would be more final ordinals than final ordinals.

    Therefore,
    ℵ₀ is NOT smaller than ℵ₀∪{ℵ₀}

    For those x > 0 your claim is wrong.

    Finite doesn't need to be small.
    Infinite is beyond big.

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  • From Moebius@21:1/5 to All on Fri Aug 2 19:08:07 2024
    XPost: sci.math

    Am 02.08.2024 um 18:35 schrieb Jim Burns:
    On 8/2/2024 7:40 AM, WM wrote:
    Le 01/08/2024 à 18:43, Jim Burns a écrit :
    On 8/1/2024 8:02 AM, WM wrote:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
    Note the universal quantifier.

    It is not a contradiction to my formula
    if some n has no n+1.

    No, it literally contradicts your formula
    for some n e N to not.have n+1

    My formula is explicitly valid only for natural numbers.

    This is one of Mückenheim's brainfarts.

    I guess what he means is:

    "1/n - 1/(n+1) > 0" is (especially) valid for (all) natural numbers n ,

    or something like that.

    To state that a closed formula / sentence / statement "is valid ... for
    natural numbers" is just mumbo-jumbo.

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  • From Richard Damon@21:1/5 to All on Fri Aug 2 13:08:42 2024
    XPost: sci.math

    On 8/2/24 12:32 PM, WM wrote:
    Le 02/08/2024 à 18:24, Richard Damon a écrit :
    On 8/2/24 12:05 PM, WM wrote:

    The claim is that there is a finite difference that seperates any two
    specific unit fractions,

    Of course.

    Not that there is a single finite difference that seperates any two
    arbirtry unit fractions.

    Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points.

    So? that number is just ℵo (you don't understand the mathematics of transfinite values)


    Not understanding the order of the arguement

    Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points. No order necessary.

    So? That just shows that between any two unit fractions there IS a
    distance, and thus they are seperated, as required.


    For your example, for any x, there is an eps = 1/(ceil(1/x)+1) that
    provides a unit fraction smaller than x,

    No, eps has to be chosen. x has to be chosen. My claim is that most are
    dark and cannot be chosen.

    It has to be chosen for a given pair of unit fractions.

    Note: For every 1/n there exists a smaller real number. But they cannot
    be chosen because most 1/n already cannot be chosen. Proof: For every
    chosen eps you fail to separate infinitely many unit fractions.

    WHY CAN'T IT BW CHOSEN? That seems to be the flaw in your logic, you
    think there exist numbers that are defined mathematically, but can't be
    chosen.

    If there WAS just a single eps that seperated ALL unit fractions, then
    the could be no more than 1/eps unit fractions, and thus that would be
    larger than the largest Natural Number.

    But EVERY Natural Number has a successor, so there is no largest Natural Number.

    This just shows that your logic is limited and doesn't handle the sets
    you are trying to us it one, and it breaks.

    and thus NUF(x) can not be 1 for any finite x.

    But for x belonging to the first ℵ₀*2ℵ₀ points we have less than ℵ₀ unit
    fractions.

    Nope, because ℵo-1 = ℵo, and thus there are ℵo points below (and above) every point. All values of ℵo are not ordered within themselves.

    If you math can't handle that, it can't handle the infinities. Since it
    doesn't handle the unbounded numbers, that it doesn't handle infinities
    isn't surprizing.



    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Fri Aug 2 13:12:39 2024
    On 8/2/24 12:42 PM, WM wrote:
    Le 02/08/2024 à 18:12, Richard Damon a écrit :
    On 8/2/24 11:09 AM, WM wrote:

    There are infinitely many by the definition of accumulation point.
    You cannot find them. Therefore they are dark.

    Nope, we can find any one of them we want.

    But you cannot want any one

    Why not?


    Again, you are changing the conditional incorrectly because you logic
    can't handle unbounded values.

    What you want is bounded. The set of all eps you ever choose is finite.
    You will  never separate infinitely many unit fractions.

    Regards, WM

    Nope, Maybe YOU are the one that is limited.

    Numbers are not limited by us, but we might be limited by numbers.

    The Numbers are not limited by the numbers we happen to choose, since we
    are not limited to choosing numbers that have already been choosen.

    This is just the flaw in your logic.

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Fri Aug 2 13:19:23 2024
    On 8/2/24 12:38 PM, WM wrote:
    Le 02/08/2024 à 18:19, Richard Damon a écrit :

    No, YOU THINK there are more algebraic numbers than prime numbers
    because you don't understand that there are exactly ℵo of both of them.

    I know that ℵo is nonsense, because all prime numbers are algebraic but
    not all algebraic numbers are prime. This does not change in the infinite.

    Nope, infinite sets do not obey the same set of rules that finite sets
    do. Failure to understand that is YOUR problem, not the problem of those
    sets.


    All countably infinite sets are the same size,

    That proves that ℵo is nonsense.

    Regareds, WM



    No, it proves that your logic can't handle it.

    The fact you don't understand something doesn't make it wrong.

    It just shows that your understanding is limited.

    If you can show a contradiction in the system, USING THE RULES OF THE
    SYSTEM, then you might have something, but you can't try to impose rules
    you think "must be true" on the system.

    Under your rules of "bounded logic" you can perhaps look into "potential infinity" but not fully understand it, but that logic totally breaks if
    you try to look at "actual infinity" as it creates concepts just totally foreign to it.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Fri Aug 2 19:31:24 2024
    XPost: sci.math

    Am 02.08.2024 um 19:06 schrieb Jim Burns:
    On 8/2/2024 11:36 AM, WM wrote:

    ℵ₀ unit fractions need ℵ₀ [corrected --Moebius] points above zero.

    Right. And where is the problem?

    For those x > 0 your claim is wrong.

    Nope. For each and every of these points [here referred to with the
    variable "x"]: NUF(x) = ℵ₀ .

    --- SoupGate-Win32 v1.05
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  • From Mikko@21:1/5 to All on Sat Aug 3 13:00:08 2024
    On 2024-08-02 11:02:24 +0000, WM said:

    Le 02/08/2024 à 08:48, Mikko a écrit :
    On 2024-07-31 14:20:00 +0000, WM said:

    The bijection is not true.

    Of course not, just like the Moon is not true. It just is there.

    The bijection concerns only the potentially infinite initial segments.
    Every natural number that can be verified in the bijection has ∀n ∈ ℕ_verif: |ℕ \ {1, 2, 3, ..., n}| = ℵo successors, ℵo of which cannot be verified.
    Proof: Ther are more algebraic numbers than prime numbers. Every not completely confused brain recognizes this.

    Which is irrelevant to the meaning of "true" and its inapplicablility
    to a bijection.

    --
    Mikko

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Sat Aug 3 13:33:37 2024
    Le 03/08/2024 à 12:00, Mikko a écrit :
    On 2024-08-02 11:02:24 +0000, WM said:

    Le 02/08/2024 à 08:48, Mikko a écrit :
    On 2024-07-31 14:20:00 +0000, WM said:

    The bijection is not true.

    Of course not, just like the Moon is not true. It just is there.

    The bijection concerns only the potentially infinite initial segments.
    Every natural number that can be verified in the bijection has ∀n ∈
    ℕ_verif: |ℕ \ {1, 2, 3, ..., n}| = ℵo successors, ℵo of which cannot >> be verified.
    Proof: There are more algebraic numbers than prime numbers. Every not
    completely confused brain recognizes this.

    Which is irrelevant to the meaning of "true" and its inapplicablility
    to a bijection.

    If arguments are irrelevant for you, there is no reason to continue.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Sat Aug 3 14:31:32 2024
    Le 02/08/2024 à 19:12, Richard Damon a écrit :
    On 8/2/24 12:42 PM, WM wrote:

    There are infinitely many by the definition of accumulation point.
    You cannot find them. Therefore they are dark.

    Nope, we can find any one of them we want.

    But you cannot want any one

    Why not?

    Because most are dark.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sat Aug 3 14:23:52 2024
    XPost: sci.math

    Le 02/08/2024 à 19:06, Jim Burns a écrit :

    (0,x] inherits from its superset (0,1] properties by which,
    for ⅟ℕᶠⁱⁿ∩(0,x] finite.unit.fractions in (0,x]
    each non.{}.subset is maximummed, and
    each finite.unit.fraction is down.stepped, and
    each finite.unit.fraction in is non.max.up.stepped.

    Therefore,
    the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.

    ∀ᴿx > 0: NUFᶠⁱⁿ(x) = ℵ₀

    I recognized lately that you use the wrong definition of NUF.
    Here is the correct definition:
    There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
    Note that the order is ∃ u ∀ y.
    NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already
    is wrong since there is no unit fraction smaller than all unit fractions.

    ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Sat Aug 3 14:25:03 2024
    XPost: sci.math

    Le 02/08/2024 à 19:31, Moebius a écrit :
    For each and every of these points [here referred to with the
    variable "x"]: NUF(x) = ℵ₀ .

    I recognized lately that you use the wrong definition of NUF.
    Here is the correct definition:
    There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
    Note that the order is ∃ u ∀ y.
    NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already
    is wrong since there is no unit fraction smaller than all unit fractions.

    ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sat Aug 3 14:30:18 2024
    XPost: sci.math

    Le 02/08/2024 à 19:08, Richard Damon a écrit :
    On 8/2/24 12:32 PM, WM wrote:

    Note: For every 1/n there exists a smaller real number. But they cannot
    be chosen because most 1/n already cannot be chosen. Proof: For every
    chosen eps you fail to separate infinitely many unit fractions.

    WHY CAN'T IT BW CHOSEN?

    Because it is dark.

    If there WAS just a single eps that seperated ALL unit fractions, then
    the could be no more than 1/eps unit fractions,

    But there is not even an eps that separates half of all unit fractions.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sat Aug 3 14:35:46 2024
    Le 02/08/2024 à 19:19, Richard Damon a écrit :
    On 8/2/24 12:38 PM, WM wrote:
    Le 02/08/2024 à 18:19, Richard Damon a écrit :

    No, YOU THINK there are more algebraic numbers than prime numbers
    because you don't understand that there are exactly ℵo of both of them. >>
    I know that ℵo is nonsense, because all prime numbers are algebraic but
    not all algebraic numbers are prime. This does not change in the infinite.

    Nope, infinite sets do not obey the same set of rules that finite sets
    do.

    An excuse of cranks.


    All countably infinite sets are the same size,

    That proves that ℵo is nonsense.

    No, it proves that your logic can't handle it.

    In fact, logic can't agree with nonsense.

    If you can show a contradiction in the system, USING THE RULES OF THE
    SYSTEM,

    Every system of mathematics agrees that there are more natural numbers
    than prime numbers.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Sat Aug 3 12:00:40 2024
    On 8/3/24 10:31 AM, WM wrote:
    Le 02/08/2024 à 19:12, Richard Damon a écrit :
    On 8/2/24 12:42 PM, WM wrote:

    There are infinitely many by the definition of accumulation point.
    You cannot find them. Therefore they are dark.

    Nope, we can find any one of them we want.

    But you cannot want any one

    Why not?

    Because most are dark.

    Nope, NONE of the Natural Numbers are Dark to me.

    Maybe you shut your eyes to them and can't find them, but most of us can.

    Your "darkness" is just your own ignorance shining brightly as you try
    ti give your mistakes cover.


    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Sat Aug 3 11:56:39 2024
    XPost: sci.math

    On 8/3/24 10:30 AM, WM wrote:
    Le 02/08/2024 à 19:08, Richard Damon a écrit :
    On 8/2/24 12:32 PM, WM wrote:

    Note: For every 1/n there exists a smaller real number. But they
    cannot be chosen because most 1/n already cannot be chosen. Proof:
    For every chosen eps you fail to separate infinitely many unit
    fractions.

    WHY CAN'T IT BE CHOSEN?

    Because it is dark.

    Why is it dark, we know a possible value for it, and thus can choose it.

    You just incorrectly assume that some numbers just can't be reached
    because some 'magic" made them dark.

    No, your dark numbers are non-finite numbers that you invent to try to
    explain why your logic doesn't work with unbounded numbers.


    If there WAS just a single eps that seperated ALL unit fractions, then
    the could be no more than 1/eps unit fractions,

    But there is not even an eps that separates half of all unit fractions.

    Because such a question is meaningles, as there isn't a finite number
    that is half of the count of unit fractions.

    This is the problem with your logic, it doens't understand that the size
    of the unbounded set isn't a finite number that you can do normal math with.


    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Sat Aug 3 12:02:26 2024
    On 8/3/24 10:35 AM, WM wrote:
    Le 02/08/2024 à 19:19, Richard Damon a écrit :
    On 8/2/24 12:38 PM, WM wrote:
    Le 02/08/2024 à 18:19, Richard Damon a écrit :

    No, YOU THINK there are more algebraic numbers than prime numbers
    because you don't understand that there are exactly ℵo of both of them. >>>
    I know that ℵo is nonsense, because all prime numbers are algebraic
    but not all algebraic numbers are prime. This does not change in the
    infinite.

    Nope, infinite sets do not obey the same set of rules that finite sets
    do.

    An excuse of cranks.

    Nope, but yours is.



    All countably infinite sets are the same size,

    That proves that ℵo is nonsense.

    No, it proves that your logic can't handle it.

    In fact, logic can't agree with nonsense.


    Right, logic rejects your darkness.


    If you can show a contradiction in the system, USING THE RULES OF THE
    SYSTEM,

    Every system of mathematics agrees that there are more natural numbers
    than prime numbers.

    Nope. Only ones that are broken.


    Regards, WM

    Sorry, you are just too stupid to argue with.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Sat Aug 3 15:54:12 2024
    XPost: sci.math

    On 8/3/2024 10:23 AM, WM wrote:
    Le 02/08/2024 à 19:06, Jim Burns a écrit :

    (0,x] inherits from its superset (0,1] properties by which,
    for ⅟ℕᶠⁱⁿ∩(0,x] finite.unit.fractions in (0,x]
    each non.{}.subset is maximummed,  and
    each finite.unit.fraction is down.stepped,  and
    each finite.unit.fraction in is non.max.up.stepped.

    Therefore,
    the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.

    ∀ᴿx > 0:  NUFᶠⁱⁿ(x) = ℵ₀

    I recognized lately that you use
    the wrong definition of NUF.

    Here is the correct definition:
    There exist NUF(x) unit fractions u, such that
    for all y >= x: u < y.

    ∀ᴿy ≥ x: y > u ⟺ x > u
    ⎛ Assume otherwise.
    ⎜ Assume y ≥ x ∧ ¬(y > u) ∧ x > u

    ⎜ However, '>' is transitive.
    ⎜ y ≥ x ∧ x > u ⇒ y > u
    ⎝ Contradiction.

    Here is an equivalent definition:
    There exist NUF(x) unit fractions u, such that
    u < x

    Note that the order is ∃ u ∀ y.

    The order is ∀x ∃u ∀y
    ∃u ∀x ∀y is an unreliable quantifier shift.

    NUF(x) = ℵ₀ for all x > 0 is wrong.
    NUF(x) = 1 for all x > 0 already is wrong since
    there is no unit fraction smaller than all unit fractions.

    NUF(x) > 1 for all x > 0 is correct since
    each unit.fraction is larger than at least two unit.fractions.
    and
    a positive lower.bound of finite unit.fractions
    implies
    finite unit.fractions below a lower.bound,
    a contradiction.

    ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

    The finite unit.fractions in (0,x] are
    maximummed and down.stepped and non.max.up.stepped.
    The finite unit.fractions in (0,x] are
    ℵ₀.many.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 4 02:09:41 2024
    XPost: sci.math

    Am 03.08.2024 um 23:51 schrieb Chris M. Thomasson:
    On 8/3/2024 7:25 AM, WM wrote:

    I recognized lately that you use the wrong definition of NUF.
    Mückenheim, Du hirnloser Spinner: Deine Definition von NUF(x) war
    bisher: "die Anzahl (Kardinalzahl) aller Stammbrüche, die kleiner-gleich
    x sind".

    In Zeichen: NUF(x) := card({s e SB : s <= x}) (x e IR)

    mit SB := {1/n : n e IN}.

    Demenz ist schon ne üble Sache.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 4 02:14:40 2024
    XPost: sci.math

    Am 03.08.2024 um 21:54 schrieb Jim Burns:
    On 8/3/2024 10:23 AM, WM wrote:

    NUF(x) = ℵ₀ for all x > 0 is wrong.

    Nonsense.

    Actually, Ax > 0: NUF(x) = ℵ₀.

    NUF(x) = 1 for all x > 0 [...] is wrong

    Hey, that's INDEED true!

    On the other hand:

    NUF(x) > 1 for all x > 0 is correct

    Right.

    WM is deluded.

    "When you're deluded, you don't know you're deluded." (Wikipedia)

    :-)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 4 02:15:10 2024
    XPost: sci.math

    Am 03.08.2024 um 21:54 schrieb Jim Burns:
    On 8/3/2024 10:23 AM, WM wrote:

    NUF(x) = ℵ₀ for all x > 0 is wrong.

    Nonsense.

    Actually, Ax > 0: NUF(x) = ℵ₀.

    NUF(x) = 1 for all x > 0 [...] is wrong

    Hey, that's INDEED true!

    On the other hand:

    NUF(x) > 1 for all x > 0 is correct

    Right.

    WM is deluded.

    "When you're deluded, you don't know you're deluded." (Internet)

    :-)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sun Aug 4 15:29:07 2024
    XPost: sci.math

    Le 03/08/2024 à 21:54, Jim Burns a écrit :
    On 8/3/2024 10:23 AM, WM wrote:

    I recognized lately that you use
    the wrong definition of NUF.

    Here is the correct definition:
    There exist NUF(x) unit fractions u, such that
    for all y >= x: u < y.

    Here is an equivalent definition:
    There exist NUF(x) unit fractions u, such that
    u < x

    Note that the order is ∃ u ∀ y.

    The order is ∀x ∃u ∀y
    ∃u ∀x ∀y is an unreliable quantifier shift.

    It is not a shift but it is the definition of NUF. It excludes that ∃u ∀x>0: u < x,

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sun Aug 4 15:23:09 2024
    XPost: sci.math

    Le 03/08/2024 à 17:56, Richard Damon a écrit :
    On 8/3/24 10:30 AM, WM wrote:

    But there is not even an eps that separates half of all unit fractions.

    Because such a question is meaningles, as there isn't a finite number
    that is half of the count of unit fractions.

    If there are all, then there is half of all.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sun Aug 4 15:32:49 2024
    XPost: sci.math

    Le 04/08/2024 à 02:09, Moebius a écrit :


    I recognized lately that you use the wrong definition of NUF.
    Deine Definition von NUF(x) war
    bisher: "die Anzahl (Kardinalzahl) aller Stammbrüche, die kleiner-gleich
    x sind".

    Sie war und ist folgende: There exist NUF(x) unit fractions u, such that
    for all y >= x: u < y.
    Note that the order is ∃ u ∀ y.
    NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already
    is wrong since there is no unit fraction smaller than all unit fractions.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sun Aug 4 15:35:22 2024
    XPost: sci.math

    Le 04/08/2024 à 02:15, Moebius a écrit :
    Am 03.08.2024 um 21:54 schrieb Jim Burns:
    On 8/3/2024 10:23 AM, WM wrote:

    NUF(x) = ℵ₀ for all x > 0 is wrong.

    Nonsense.

    Actually, Ax > 0: NUF(x) = ℵ₀.

    You mean that there are ℵ₀ unit fractions smaller than all positive x? Impossible. ℵ₀ unit fractions occupy ℵ₀*2^ℵ₀ points. Not even
    one unit fraction cann be smaller than all positive x.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Sun Aug 4 11:53:23 2024
    XPost: sci.math

    On 8/4/2024 11:35 AM, WM wrote:
    Le 04/08/2024 à 02:15, Moebius a écrit :
    Am 03.08.2024 um 21:54 schrieb Jim Burns:
    On 8/3/2024 10:23 AM, WM wrote:

    NUF(x) = ℵ₀ for all x > 0 is wrong.

    Nonsense.

    Actually, Ax > 0: NUF(x) = ℵ₀.

    You mean that there are
    ℵ₀ unit fractions smaller than all positive x?

    No, he doesn't mean that.
    You (WM) have mad an unreliable quantifier shift.

    Impossible.

    Misunderstood.

    ℵ₀ unit fractions occupy ℵ₀*2^ℵ₀ points.
    Not even one unit fraction cann be
    smaller than all positive x.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 4 17:57:37 2024
    XPost: sci.math

    Am 04.08.2024 um 17:53 schrieb Jim Burns:

    You (WM) have mad [sic!] an unreliable quantifier shift.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Moebius on Sun Aug 4 12:55:40 2024
    XPost: sci.math

    On 8/4/2024 11:57 AM, Moebius wrote:
    Am 04.08.2024 um 17:53 schrieb Jim Burns:

    You (WM) have mad [sic!] an unreliable quantifier shift.

    Typo.
    Better:
    You (WM) are mad from an unreliable quantifier shift.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Sun Aug 4 12:39:06 2024
    XPost: sci.math

    On 8/4/2024 11:29 AM, WM wrote:
    Le 03/08/2024 à 21:54, Jim Burns a écrit :
    On 8/3/2024 10:23 AM, WM wrote:

    I recognized lately that you use
    the wrong definition of NUF.

    Here is the correct definition:
    There exist NUF(x) unit fractions u, such that
    for all y >= x: u < y.

    Here is an equivalent definition:
    There exist NUF(x) unit fractions u, such that
    u < x

    Note that the order is ∃ u ∀ y.

    The order is ∀x ∃u ∀y

    The order of the claim which you (WM) address
    in an attempt to "prove" dark numbers is
    ∀ᴿx > 0:
    ∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
    ∀ᴿy ≥ x:
    y >ᵉᵃᶜʰ U

    That claim and the following claim are
    either both true or both false.
    ∀ᴿx > 0:
    ∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
    x >ᵉᵃᶜʰ U

    ∃u ∀x ∀y is an unreliable quantifier shift.

    It is not a shift but it is the definition of NUF.

    Your recently corrected definition of NUF is
    NUF(x) =
    |{u ∈ ⅟ℕ: ∀ᴿy ≥ x: y > u}|

    That definition is equivalent to
    NUF(x) =
    |{u ∈ ⅟ℕ: x > u}|

    Note that,
    for x > 0, {u ∈ ⅟ℕ: x > u}
    is maximummed and down.stepped and non.max.up.stepped.
    For x > 0: |{u ∈ ⅟ℕ: x > u}| = ℵ₀

    The claim you (WM) use
    ∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
    ∀ᴿx > 0:
    ∀ᴿy ≥ x:
    y >ᵉᵃᶜʰ U

    is an unreliable quantifier shift from
    the claim we make
    ∀ᴿx > 0:
    ∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
    ∀ᴿy ≥ x:
    y >ᵉᵃᶜʰ U

    It excludes that ∃u ∀x>0: u < x,

    Only you (WM) think that ∃u ∀x>0: u < x
    follows from ∀x>0 ∃u: u < x,

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 4 19:28:47 2024
    XPost: sci.math

    Am 04.08.2024 um 18:39 schrieb Jim Burns:
    On 8/4/2024 11:29 AM, WM wrote:

    It excludes that ∃u ∈ ⅟ℕ: ∀x > 0: u < x,

    Only you (WM) think [?!] that ∃u ∈ ⅟ℕ: ∀x > 0: u < x
    follows from ∀x > 0: ∃u ∈ ⅟ℕ: u < x.

    ...sort of.

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  • From WM@21:1/5 to All on Sun Aug 4 18:13:22 2024
    XPost: sci.math

    Le 04/08/2024 à 18:39, Jim Burns a écrit :
    On 8/4/2024 11:29 AM, WM wrote:
    Le 03/08/2024 à 21:54, Jim Burns a écrit :
    On 8/3/2024 10:23 AM, WM wrote:

    I recognized lately that you use
    the wrong definition of NUF.

    Here is the correct definition:
    There exist NUF(x) unit fractions u, such that
    for all y >= x: u < y.

    Here is an equivalent definition:
    There exist NUF(x) unit fractions u, such that
    u < x

    Note that the order is ∃ u ∀ y.

    The order is ∀x ∃u ∀y

    When all x are involved, the universal quantifier is usually not written.

    The order of the claim which you (WM) address
    in an attempt to "prove" dark numbers is
    ∀ᴿx > 0:
    ∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
    ∀ᴿy ≥ x:
    y >ᵉᵃᶜʰ U

    That claim and the following claim are
    either both true or both false.
    ∀ᴿx > 0:
    ∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
    x >ᵉᵃᶜʰ U

    More of interest are these two claims which are not both true or both
    false:
    For every x there is u < x.
    There is u < x for every x.
    The latter is close to my function:
    There are NUF(x) u < x.

    Your recently corrected definition of NUF is
    NUF(x) =
    |{u ∈ ⅟ℕ: ∀ᴿy ≥ x: y > u}|

    That definition is equivalent to
    NUF(x) =
    |{u ∈ ⅟ℕ: x > u}|

    Note that,
    for x > 0, {u ∈ ⅟ℕ: x > u}
    is maximummed and down.stepped and non.max.up.stepped.
    For x > 0: |{u ∈ ⅟ℕ: x > u}| = ℵ₀

    The claim you (WM) use
    ∃U ⊆ ⅟ℕ ∧ |U| = ℵ₀:
    ∀ᴿx > 0:
    ∀ᴿy ≥ x:
    y >ᵉᵃᶜʰ U

    is an unreliable quantifier shift from
    the claim we make

    What claim you make is not of interest to me. I express that no u can be smaller than all x but that some u can be smaller than many x.
    You express that for all x, there is a smaller u. Both are very different.

    Only you (WM) think that ∃u ∀x>0: u < x
    follows from ∀x>0 ∃u: u < x,

    Not at all! Please spare these insults! Your claim concerns only definable
    x. For ℵo*2^ℵo undefinable points x it is wrong. My claim concerns all
    x.

    Regards, WM

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  • From WM@21:1/5 to All on Sun Aug 4 17:55:06 2024
    XPost: sci.math

    Le 04/08/2024 à 17:53, Jim Burns a écrit :
    On 8/4/2024 11:35 AM, WM wrote:
    Le 04/08/2024 à 02:15, Moebius a écrit :
    Am 03.08.2024 um 21:54 schrieb Jim Burns:
    On 8/3/2024 10:23 AM, WM wrote:

    NUF(x) = ℵ₀ for all x > 0 is wrong.

    Nonsense.

    Actually, Ax > 0: NUF(x) = ℵ₀.

    You mean that there are
    ℵ₀ unit fractions smaller than all positive x?

    No, he doesn't mean that.

    NUF(x) = ℵ₀ means exactly that.

    You (WM) have mad an unreliable quantifier shift.

    I have defined my function this way.

    ℵ₀ unit fractions occupy ℵ₀*2^ℵ₀ points.
    Not even one unit fraction can be
    smaller than all positive x.

    Can you (JB) understand that?
    If so, can you define a function expressing just that?
    Or do you feel it a sacrilege to explicitly express that?

    Regards, WM

    Regards, WM

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  • From Richard Damon@21:1/5 to All on Sun Aug 4 15:18:48 2024
    XPost: sci.math

    On 8/4/24 11:23 AM, WM wrote:
    Le 03/08/2024 à 17:56, Richard Damon a écrit :
    On 8/3/24 10:30 AM, WM wrote:

    But there is not even an eps that separates half of all unit fractions.

    Because such a question is meaningles, as there isn't a finite number
    that is half of the count of unit fractions.

    If there are all, then there is half of all.


    Not with infinities.

    Sorry, your logic just breaks with unbounded sets.

    That is the problem of using the wrong tools, they tend to just blow up
    on you.



    Regards, WM



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  • From Moebius@21:1/5 to All on Sun Aug 4 22:48:49 2024
    XPost: sci.math

    Am 04.08.2024 um 22:16 schrieb Jim Burns:
    On 8/4/2024 2:13 PM, WM wrote:

    [...] two claims which are
    not both true or both false:

    For every x there is u < x.
    There is u < x for every x.

    Mückenheim, Du dummer Spinner: BEIDE Aussagen werden selbstverständlich
    wie folgt formalisiert:

    ∀x ∃u u < x .

    Vermutlich ist hier aber eigentlich

    ∀x > 0: ∃u ∈ ⅟ℕ: u < x (*)

    gemeint.

    The latter is close to [...]:
    There are NUF(x) [u ∈ ⅟ℕ:] u < x.

    Letzteres formalisiert man wie folgt:

    ∃^NUF(x) u ∈ ⅟ℕ: u < x .

    Hier fehlt noch der Allquantor für x, um eine WAHRE AUSSAGE zu erhalten:

    ∀x > 0: ∃^NUF(x) u ∈ ⅟ℕ: u < x .

    Mit ∀x > 0: NUF(x) = ℵ₀ ergibt sich dann daraus:

    ∀x > 0: ∃^ℵ₀ u ∈ ⅟ℕ: u < x .

    Ja, diese Formel ist in der Tat "close to" (*). Nur besagt sie natürlich
    noch etwas mehr als (*).

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  • From Richard Damon@21:1/5 to All on Sun Aug 4 16:23:50 2024
    XPost: sci.math

    On 8/4/24 11:23 AM, WM wrote:
    Le 03/08/2024 à 17:56, Richard Damon a écrit :
    On 8/3/24 10:30 AM, WM wrote:

    But there is not even an eps that separates half of all unit fractions.

    Because such a question is meaningles, as there isn't a finite number
    that is half of the count of unit fractions.

    If there are all, then there is half of all.

    Regards, WM



    But half of aleph_0 is just aleph_0, so doesn't actually get smaller.

    That is how the logic of infinite sets works.

    The "even" numbers are in once sense 1/2 the size of the Natural Numbers
    by dropping all the odds, but are also the same size by doubling each
    Natural Number.

    Yes, that seems counter-intuitive, but that is just the math of infinite numbers.

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  • From Jim Burns@21:1/5 to All on Sun Aug 4 16:16:05 2024
    XPost: sci.math

    On 8/4/2024 2:13 PM, WM wrote:
    Le 04/08/2024 à 18:39, Jim Burns a écrit :
    On 8/4/2024 11:29 AM, WM wrote:
    Le 03/08/2024 à 21:54, Jim Burns a écrit :
    On 8/3/2024 10:23 AM, WM wrote:

    I recognized lately that you use
    the wrong definition of NUF.

    Here is the correct definition:
    There exist NUF(x) unit fractions u, such that
    for all y >= x: u < y.

    Here is an equivalent definition:
    There exist NUF(x) unit fractions u, such that
    u < x

    Note that the order is ∃ u ∀ y.

    The order is ∀x ∃u ∀y

    When all x are involved,
    the universal quantifier is usually not written.

    When a universal quantifier is not written,
    it is implicit, and
    it can only stand implicitly outside the formula.
    ∃k≠j:j<k is implicitly ∀j(∃k≠j:j<k)

    When a universal quantifier is implicit,
    a quantifier shift is impossible to write.
    ∃k≠j:j<k always means ∀j∃k≠j:j<k
    ∃k≠j:j<k never means ∃k∀j≠k:j<k

    More of interest are these two claims which are
    not both true or both false:
    For every x there is u < x.
    There is u < x for every x.
    The latter is close to my function:
    There are NUF(x) u < x.

    NUFᵉᵃᶜʰ := |{u∈⅟ℕᵈᵉᶠ: u <ᵉᵃᶜʰ ℝ⁺}|

    NUFᵉᵃᶜʰ = 0

    NUFᵈᵉᶠ(x) := |{u∈⅟ℕᵈᵉᶠ:u<x}|

    ∀ᴿx>0: NUFᵈᵉᶠ(x) = ℵ₀

    From ∀x∃U to ∃U∀x is unreliable,
    is not.always.correct, is sometimes.incorrect.

    Your claim concerns only definable x.

    There is no positive lower bound of ⅟ℕᵈᵉᶠ
    because
    a _positive_ lower.bound of ⅟ℕᵈᵉᶠ implies
    there are lower.bounds which don't bound ⅟ℕᵈᵉᶠ

    There is no darkᵂᴹ x: 0 < x <ᵉᵃᶜʰ ⅟ℕᵈᵉᶠ
    because
    darkᵂᴹ x would be a positive lower bound of ⅟ℕᵈᵉᶠ

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  • From Moebius@21:1/5 to All on Mon Aug 5 01:12:21 2024
    XPost: sci.math

    Am 04.08.2024 um 22:16 schrieb Jim Burns:
    On 8/4/2024 2:13 PM, WM wrote:

    When all x are involved,
    the universal quantifier is usually not written.

    When a universal quantifier is not written,
    it is implicit, and
    it can only stand implicitly outside the formula.

    Right. Still there's a distinct problem with "implicit quantification"
    IN THIS CASE. (Hence WM's "argument" is nonsense anyway.)

    Here we need "∀x > 0". Clearly an "implicit quantifier" does not know
    that he's restricted to "x > 0".

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  • From Jim Burns@21:1/5 to Moebius on Sun Aug 4 19:41:53 2024
    XPost: sci.math

    On 8/4/2024 7:12 PM, Moebius wrote:
    Am 04.08.2024 um 22:16 schrieb Jim Burns:
    On 8/4/2024 2:13 PM, WM wrote:

    When all x are involved,
    the universal quantifier is usually not written.

    When a universal quantifier is not written,
    it is implicit, and
    it can only stand implicitly outside the formula.

    Right.
    Still there's a distinct problem with "implicit quantification"
    IN THIS CASE.
    (Hence WM's "argument" is nonsense anyway.)

    Yes.

    Here we need "∀x > 0".
    Clearly an "implicit quantifier" does not know
    that he's restricted to "x > 0".

    We can re.write
    ∀ᴿx>0: NUFᵈᵉᶠ(x) = ℵ₀
    as
    ∀x: x∈ℝ ∧ x>0 ⇒ NUFᵈᵉᶠ(x) = ℵ₀
    and, implicitly quantified, as
    x∈ℝ ∧ x>0 ⇒ NUFᵈᵉᶠ(x) = ℵ₀

    The quantification is still there, even if
    it isn't written down.
    But Some People like it like that. <sigh>

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  • From Moebius@21:1/5 to All on Mon Aug 5 02:03:03 2024
    XPost: sci.math

    Am 05.08.2024 um 01:41 schrieb Jim Burns:
    On 8/4/2024 7:12 PM, Moebius wrote:
    Am 04.08.2024 um 22:16 schrieb Jim Burns:
    On 8/4/2024 2:13 PM, WM wrote:

    When all x are involved,
    the universal quantifier is usually not written.

    When a universal quantifier is not written,
    it is implicit, and
    it can only stand implicitly outside the formula.

    Right.
    Still there's a distinct problem with "implicit quantification"
    IN THIS CASE.
    (Hence WM's "argument" is nonsense anyway.)

    Yes.

    Here we need "∀x > 0".
    Clearly an "implicit quantifier" does not know that he's restricted to
    "x > 0".

    We can re.write
    ∀ᴿx>0: NUFᵈᵉᶠ(x) = ℵ₀

    Let's (for simplicity) assume that our domain of discourse is IR:

    ∀x>0: NUFᵈᵉᶠ(x) = ℵ₀

    Then we can re.write

    ∀x>0: NUFᵈᵉᶠ(x) = ℵ₀

    as
    ∀x: x > 0 ⇒ NUFᵈᵉᶠ(x) = ℵ₀
    and, implicitly quantified, as
    x > 0 ⇒ NUFᵈᵉᶠ(x) = ℵ₀

    Sure, but WM's "implicitely quantified" formula did not consist of an implication with antecedence "x > 0".

    To make a long story short: "implicit quantification" does not work for
    (with?) "restricted quantifiers".


    Again: I may claim that

    NUFᵈᵉᶠ(x) = ℵ₀

    is "implicitely (all)quantified".

    But

    ∀x: NUFᵈᵉᶠ(x) = ℵ₀

    would be wrong.

    Nuff said.

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  • From Moebius@21:1/5 to All on Mon Aug 5 02:24:13 2024
    XPost: sci.math

    Am 05.08.2024 um 01:07 schrieb Chris M. Thomasson:
    On 8/4/2024 8:35 AM, WM wrote:
    Le 04/08/2024 à 02:15, Moebius a écrit :
    Am 03.08.2024 um 21:54 schrieb Jim Burns:
    On 8/3/2024 10:23 AM, WM wrote:

    NUF(x) = ℵ₀ for all x > 0 is wrong.

    Nonsense.

    Actually, Ax > 0: NUF(x) = ℵ₀.

    You mean that there are ℵ₀ unit fractions smaller than all positive x?

    Obviously not.

    What I mean is that for all positive x there are ℵ₀ unit fractions
    smaller than x.

    Impossible. [...] Not even one unit fraction can be smaller than all positive x.

    No one (except WM) claimed that there's a unit fraction which is smaller
    than all positive x.

    Huh?

    WM is constantly mixing up

    ∀x > 0: ∃^ℵ₀ u ∈ ⅟ℕ: u < x (true)
    with
    ∃^ℵ₀ u ∈ ⅟ℕ: ∀x > 0: u < x (false) .

    (Here ⅟ℕ = {1/n : n e IN} is the set of all unit fractions.)

    Say x = 1/2, there are infinite smaller unit fractions, say, 1/4,
    1/5, 1/6, ect... However there is only one larger one, 1/1. See? No
    smallest one for 1/0 is not a unit fraction! There is a largest one, 1/1...

    They tend to zero, but there is no smallest one...

    Yeah.

    Proof: If s is a unit fraction then 1/(1/s + 1) is a unit fraction which
    is smaller than s (for each and every s).

    See?

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  • From Moebius@21:1/5 to All on Mon Aug 5 10:12:34 2024
    XPost: sci.math

    Am 05.08.2024 um 08:44 schrieb Chris M. Thomasson:
    On 8/4/2024 5:24 PM, Moebius wrote:
    Am 05.08.2024 um 01:07 schrieb Chris M. Thomasson:

    [...] there is no smallest [unit fraction]...

    Yeah.

    Proof: If s is a unit fraction then 1/(1/s + 1) is a unit fraction
    which is smaller than s (for each and every s).

    Yup.

    always_smaller(i) = (1/(i + 1))

    So, starting at 1/1:

    always_smaller(1) = 1/2
    always_smaller(2) = 1/3
    always_smaller(3) = 1/4
    always_smaller(4) = 1/5
    ...

    Let's try

    next_unit_fraction(u) = (1/(1/u + 1))

    So, starting with 1/1:

    next_unit_fraction(1/1) = 1/2
    next_unit_fraction(1/2) = 1/3
    next_unit_fraction(1/3) = 1/4
    next_unit_fraction(1/4) = 1/5
    :

    :-P

    Though in Mückenland we will have

    next_unit_fraction(1/n) = NaN

    for some natural number n, I guess.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Btw. we might implement this function (in Python) the followimg way:

    def next_unit_fraction(n, m):
    return (1, m//n + 1)

    Testcode:

    (n, m) = (1, 1); print((n, m))
    (n, m) = next_unit_fraction(n, m); print((n, m))
    (n, m) = next_unit_fraction(n, m); print((n, m))
    (n, m) = next_unit_fraction(n, m); print((n, m))
    (n, m) = next_unit_fraction(n, m); print((n, m))

    Output:

    (1, 1)
    (1, 2)
    (1, 3)
    (1, 4)
    (1, 5)

    Works! :-P

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  • From Moebius@21:1/5 to All on Mon Aug 5 10:03:26 2024
    XPost: sci.math

    Am 05.08.2024 um 08:44 schrieb Chris M. Thomasson:
    On 8/4/2024 5:24 PM, Moebius wrote:
    Am 05.08.2024 um 01:07 schrieb Chris M. Thomasson:

    [...] there is no smallest [unit fration]...

    Yeah.

    Proof: If s is a unit fraction then 1/(1/s + 1) is a unit fraction
    which is smaller than s (for each and every s).

    Yup.

    always_smaller(i) = (1/(i + 1))

    So, starting at 1/1:

    always_smaller(1) = 1/2
    always_smaller(2) = 1/3
    always_smaller(3) = 1/4
    always_smaller(4) = 1/5
    ...

    Let's try

    next_unit_fraction(u) = (1/(1/u + 1))

    So, starting with 1/1:

    next_unit_fraction(1/1) = 1/2
    next_unit_fraction(1/2) = 1/3
    next_unit_fraction(1/3) = 1/4
    next_unit_fraction(1/4) = 1/5
    ;

    :-P

    Though in Mückenland we will have

    next_unit_fraction(1/n) = NaN

    for some natural number n, I guess.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Btw. we might implement this function (in Python) the followimg way:

    def next_unit_fraction(n, m):
    return (1, m//n + 1)

    Testcode:

    (n, m) = (1, 1); print((n, m))
    (n, m) = next_unit_fraction(n, m); print((n, m))
    (n, m) = next_unit_fraction(n, m); print((n, m))
    (n, m) = next_unit_fraction(n, m); print((n, m))
    (n, m) = next_unit_fraction(n, m); print((n, m))

    Output:

    (1, 1)
    (1, 2)
    (1, 3)
    (1, 4)
    (1, 5)

    Works! :-P

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  • From WM@21:1/5 to All on Mon Aug 5 19:04:14 2024
    XPost: sci.math

    Le 04/08/2024 à 22:16, Jim Burns a écrit :
    On 8/4/2024 2:13 PM, WM wrote:

    More of interest are these two claims which are
    not both true or both false:
    For every x there is u < x.
    There is u < x for every x.
    The latter is close to my function:
    There are NUF(x) u < x.

    From ∀x∃U to ∃U∀x is unreliable,

    There is no from to. NUF(x) is so defined.
    For every number of unit fractions NUF(x) gives the smallest interval (0,
    x).

    Regards, WM

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  • From WM@21:1/5 to All on Mon Aug 5 19:12:17 2024
    XPost: sci.math

    Le 04/08/2024 à 22:48, Moebius a écrit :
    Am 04.08.2024 um 22:16 schrieb Jim Burns:
    On 8/4/2024 2:13 PM, WM wrote:

    [...] two claims which are
    not both true or both false:

    For every x there is u < x.
    There is u < x for every x.

    BEIDE Aussagen werden selbstverständlich
    wie folgt formalisiert:

    ∀x ∃u u < x .

    Vermutlich ist hier aber eigentlich



    ∀x > 0: ∃u ∈ ⅟ℕ: u < x (*)

    gemeint.

    The latter is close to [...]:
    There are NUF(x) [u ∈ ⅟ℕ:] u < x.

    Letzteres formalisiert man wie folgt:

    ∃^NUF(x) u ∈ ⅟ℕ: u < x .

    Hier fehlt noch der Allquantor für x, um eine WAHRE AUSSAGE zu erhalten:

    ∀x > 0: ∃^NUF(x) u ∈ ⅟ℕ: u < x .

    Mit ∀x > 0: NUF(x) = ℵ₀ ergibt sich dann daraus:

    ∀x > 0: ∃^ℵ₀ u ∈ ⅟ℕ: u < x .

    Ja, diese Formel ist in der Tat "close to" (*). Nur besagt sie natürlich noch etwas mehr als (*).

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  • From WM@21:1/5 to All on Mon Aug 5 18:58:40 2024
    XPost: sci.math

    Le 04/08/2024 à 21:18, Richard Damon a écrit :
    On 8/4/24 11:23 AM, WM wrote:
    Le 03/08/2024 à 17:56, Richard Damon a écrit :
    On 8/3/24 10:30 AM, WM wrote:

    But there is not even an eps that separates half of all unit fractions. >>>
    Because such a question is meaningles, as there isn't a finite number
    that is half of the count of unit fractions.

    If there are all, then there is half of all.

    Not with infinities.

    Always.

    Regards, WM

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  • From WM@21:1/5 to All on Mon Aug 5 19:21:49 2024
    XPost: sci.math

    Le 05/08/2024 à 02:24, Moebius a écrit :

    ∃^ℵ₀ u ∈ ⅟ℕ: ∀x > 0: u < x (false) .

    ∃^NUF(x) u ∈ ⅟ℕ: ∀x > 0: u < x true.

    Regards, WM

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  • From WM@21:1/5 to All on Mon Aug 5 19:17:42 2024
    XPost: sci.math

    Le 04/08/2024 à 22:48, Moebius a écrit :
    Am 04.08.2024 um 22:16 schrieb Jim Burns:
    On 8/4/2024 2:13 PM, WM wrote:

    [...] two claims which are
    not both true or both false:

    For every x there is u < x.
    There is u < x for every x.

    BEIDE Aussagen werden selbstverständlich
    wie folgt formalisiert:

    ∀x ∃u u < x .

    No.

    ∀x ∈ ℝ, ∀u ∈ 1/ℕ, ∃^NUF(x) u, ∀y >= x: u < y.

    NUF(x) = n is given. The interval (0, x) containing n unit fractions is
    asked for.

    Regards, WM

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  • From Moebius@21:1/5 to All on Mon Aug 5 22:47:31 2024
    XPost: sci.math

    Am 05.08.2024 um 10:12 schrieb Moebius:


    Btw. we might implement this function (in Python) the followimg way:

    def next_unit_fraction(n, m):

    Better: next_smaller_unit_fraction

        return (1, m//n + 1)

    :-P

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  • From Moebius@21:1/5 to All on Mon Aug 5 23:25:43 2024
    XPost: sci.math

    Am 05.08.2024 um 22:47 schrieb Moebius:
    Am 05.08.2024 um 10:12 schrieb Moebius:

    Btw. we might implement this function (in Python) the followimg way:

    next_smaller_unit_fraction(n, m):
         return (1, m//n + 1)

    It's strange, if (1, bla) is a "representation" of WM's "smallest unit fraction" 1/bla, next_smaller_unit_fraction(1, bla) returns (the
    representation of) a smaller unit fraction (if we assume that our
    computational ressources are unlimited). :-)

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  • From Jim Burns@21:1/5 to All on Mon Aug 5 18:35:16 2024
    XPost: sci.math

    On 8/5/2024 3:21 PM, WM wrote:
    Le 05/08/2024 à 02:24, Moebius a écrit :

    ∃^ℵ₀ u ∈ ⅟ℕ: ∀x > 0: u < x   (false) .

    ∃^NUF(x) u ∈ ⅟ℕ: ∀x > 0: u < x true.

    ¬∃u ∈ ⅟ℕ: ∀x > 0: u < x

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  • From Jim Burns@21:1/5 to All on Mon Aug 5 18:19:28 2024
    XPost: sci.math

    On 8/5/2024 3:04 PM, WM wrote:
    Le 04/08/2024 à 22:16, Jim Burns a écrit :
    On 8/4/2024 2:13 PM, WM wrote:

    More of interest are these two claims which are
    not both true or both false:
    For every x there is u < x.
    There is u < x for every x.
    The latter is close to my function:
    There are NUF(x) u < x.

     From ∀x∃U to ∃U∀x is unreliable,

    There is no from to.
    NUF(x) is so defined.

    NUFᵉᵃᶜʰ := |{u∈⅟ℕᵈᵉᶠ: u <ᵉᵃᶜʰ ℝ⁺}| NUFᵈᵉᶠ(x) := |{u∈⅟ℕᵈᵉᶠ: u < x}|

    ∀v ∈ ⅟ℕᵈᵉᶠ:
    ⎛ ¬(v < u/2 ∈ ℝ⁺)
    ⎜ ¬(v <ᵉᵃᶜʰ ℝ⁺)
    ⎝ ¬(v ∈ {u∈⅟ℕᵈᵉᶠ: u <ᵉᵃᶜʰ ℝ⁺})

    NUFᵉᵃᶜʰ = |{}| = 0

    ⅟ℕᵈᵉᶠ∩(0,1] is maximummed.
    ⅟ℕᵈᵉᶠ∩(0,x) is maximummed.

    ⅟ℕᵈᵉᶠ∩(0,1] is down.stepped.
    ⅟ℕᵈᵉᶠ∩(0,x) is down.stepped.

    ⅟ℕᵈᵉᶠ∩(0,1] is non.max.up.stepped.
    ⅟ℕᵈᵉᶠ∩(0,x) is non.max.up.stepped.

    ⅟ℕᵈᵉᶠ∩(0,1] is ℵ₀.many.
    ⅟ℕᵈᵉᶠ∩(0,x) is ℵ₀.many.

    NUF(1) = ℵ₀
    NUF(x) = ℵ₀

    For every number of unit fractions
    NUF(x) gives the smallest interval (0, x).

    For each real x > 0
    for each finite cardinal k > 0
    there is a finite.unit.fraction uₖ such that
    |⅟ℕᵈᵉᶠ∩[uₖ,x)| = k
    uₖ = ⅟⌊k+⅟x⌋

    For each x > 0
    for each finite cardinal k > 0
    there is a finite.unit.fraction uₖ such that
    |⅟ℕᵈᵉᶠ∩[uₖ,x)| = k
    ⅟(1+⅟uₖ) ∈ ⅟ℕᵈᵉᶠ∩(0,uₖ)
    k < |⅟ℕᵈᵉᶠ∩(0,x)|

    For each x > 0
    for each finite number k
    k ≠ |⅟ℕᵈᵉᶠ∩(0,x)|

    For each x > 0
    ⅟ℕᵈᵉᶠ∩(0,x) is not finite.

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  • From Moebius@21:1/5 to All on Tue Aug 6 01:22:39 2024
    XPost: sci.math

    Am 06.08.2024 um 00:35 schrieb Jim Burns:
    On 8/5/2024 3:21 PM, WM wrote:
    Le 05/08/2024 à 02:24, Moebius a écrit :

    ∃^ℵ₀ u ∈ ⅟ℕ: ∀x > 0: u < x   (false) .

    ∃^NUF(x) u ∈ ⅟ℕ: ∀x > 0: u < x true.

    Nonsense. Mückenheim, Du bist selbst zum Scheißen zu blöde.

    Actually,

    ¬∃u ∈ ⅟ℕ: ∀x > 0: u < x

    Something's very wrong with Mückenheim.

    Even this idiot should understand that there can't be an u in ⅟ℕ c IR+
    such that for all x e IR, x > 0: x < u. After all, this would imply the existence of an u0 in ⅟ℕ such that u0 < u0.

    Btw. like your simplified notation. Improves readability, imho. :-P

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  • From Moebius@21:1/5 to All on Tue Aug 6 03:55:05 2024
    XPost: sci.math

    Am 06.08.2024 um 00:19 schrieb Jim Burns:
    On 8/5/2024 3:04 PM, WM wrote:

    For every number of unit fractions
    NUF(x) gives the smallest interval (0, x).

    Huh?!

    IIRC: NUF(x) := |{u ∈ ⅟ℕ : u < x}| (x e IR)

    Hence NUF(x) does NOT "give the smallest interval (0, x)", but the
    (cardinal) number of (the set of) unit fractions which are smaller than
    x (with x e IR).

    Seems that Mückenheim is panicking.

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  • From Richard Damon@21:1/5 to All on Mon Aug 5 21:35:32 2024
    XPost: sci.math

    On 8/5/24 2:58 PM, WM wrote:
    Le 04/08/2024 à 21:18, Richard Damon a écrit :
    On 8/4/24 11:23 AM, WM wrote:
    Le 03/08/2024 à 17:56, Richard Damon a écrit :
    On 8/3/24 10:30 AM, WM wrote:

    But there is not even an eps that separates half of all unit
    fractions.

    Because such a question is meaningles, as there isn't a finite
    number that is half of the count of unit fractions.

    If there are all, then there is half of all.

    Not with infinities.

    Always.

    Regards, WM




    Then you believe that x can be less than x, as the number of even
    natural numbers is both 1/2 the number of natural numbers and the same
    size as it.

    Your problem is your brain has exploded with the inconsistency of your
    flawed logic.

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  • From WM@21:1/5 to All on Tue Aug 6 08:26:14 2024
    XPost: sci.math

    Le 06/08/2024 à 00:19, Jim Burns a écrit :
    On 8/5/2024 3:04 PM, WM wrote:

    NUF(1) = ℵ₀
    NUF(x) = ℵ₀

    NUF(x) gives the number of unit fractions smaller than x.

    Following unreadable symbols.

    For each x > 0
    ⅟ℕᵈᵉᶠ∩(0,x) is not finite.

    For NUF(x) = 3
    ⅟ℕᵈᵉᶠ∩(0,x) is finite, namely 3.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 6 08:38:18 2024
    XPost: sci.math

    Le 06/08/2024 à 03:35, Richard Damon a écrit :
    On 8/5/24 2:58 PM, WM wrote:

    If there are all, then there is half of all.

    Not with infinities.

    Always.

    Then you believe that x can be less than x, as the number of even
    natural numbers is both 1/2 the number of natural numbers and the same
    size as it.

    No. This "size" is nonsense.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 6 08:35:35 2024
    XPost: sci.math

    Le 06/08/2024 à 00:35, Jim Burns a écrit :
    On 8/5/2024 3:21 PM, WM wrote:
    Le 05/08/2024 à 02:24, Moebius a écrit :

    ∃^ℵ₀ u ∈ ⅟ℕ: ∀x > 0: u < x   (false) .

    ∃^NUF(x) u ∈ ⅟ℕ: ∀x > 0: u < x true.

    ¬∃u ∈ ⅟ℕ: ∀x > 0: u < x

    Right. But with NUF(x) = 1 ==> INVNUF(1) = x we get

    ∃u ∈ ⅟ℕ, u < x, ∀y > x = INVNUF(1).

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 6 08:48:46 2024
    XPost: sci.math

    Le 06/08/2024 à 03:55, Moebius a écrit :
    Am 06.08.2024 um 00:19 schrieb Jim Burns:
    On 8/5/2024 3:04 PM, WM wrote:

    For every number of unit fractions
    NUF(x) gives the smallest interval (0, x).

    Huh?!

    IIRC: NUF(x) := |{u ∈ ⅟ℕ : u < x}| (x e IR)

    Hence NUF(x) does NOT "give the smallest interval (0, x)", but the
    (cardinal) number of (the set of) unit fractions which are smaller than
    x (with x e IR).

    But with NUF(x) = k ==> INVNUF(k) = x we get

    ∃^k u ∈ ⅟ℕ, u < x, ∀y > x = INVNUF(k).

    Regards, WM

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  • From Jim Burns@21:1/5 to All on Tue Aug 6 06:32:18 2024
    XPost: sci.math

    On 8/6/2024 4:26 AM, WM wrote:
    Le 06/08/2024 à 00:19, Jim Burns a écrit :

    NUF(1) = ℵ₀
    NUF(x) = ℵ₀

    NUF(x) gives
    the number of unit fractions smaller than x.

    Following unreadable symbols.

    For  each x > 0
    ⅟ℕᵈᵉᶠ∩(0,x) is not finite.

    For NUF(x) = 3
    ⅟ℕᵈᵉᶠ∩(0,x) is finite, namely 3.

    For NUF(x) = 3.5
    ⅟ℕᵈᵉᶠ∩(0,x) is fractional, namely 3.5, however,
    no such x with NUF(x) = 3.5 exists.

    Also,
    no such x with NUF(x) = 3 exists.

    | Assume otherwise.
    | Assume NUF(x₃) = 3
    |
    | u₁ < u₂ < u₃ are all of
    | the finite unit fractions in (0,x₃)
    |
    | However,
    | ⅟(1+⅟u₁) < u₁ is also
    | a finite unit fraction in (0,x₃)
    | 0 < ⅟(1+⅟u₁) < u₁ < u₂ < u₃ < x₃
    |
    | NUF(x₃) > 3
    | Contradiction.

    Therefore,
    no such x with NUF(x) = 3 exists.

    Counter.argument:
    What about x₃′ = (u₂+u₃)/2 ?
    0 < ⅟(1+⅟u₁) < u₁ < u₂ < x₃′

    Counter.counter.argument:
    0 < ⅟(2+⅟u₁) < ⅟(1+⅟u₁) < u₁ < u₂ < x₃′
    NUF(x₃′) > 3

    Which unit.fractions changes.
    How many unit.fractions stays infinite.

    Infinite is not simply big. It's different.

    Finite can be big, too, even reallyreally big,
    but it's a reallyreally big finite which
    shares "common sense" properties with 3 ...

    ... "common sense" properties which,
    upon a more careful look,
    may be held in common with reallyreally big finites
    but are not held in common with, for example,
    how many unit.fractions there are.

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  • From Jim Burns@21:1/5 to All on Tue Aug 6 08:38:04 2024
    XPost: sci.math

    On 8/6/2024 4:35 AM, WM wrote:
    Le 06/08/2024 à 00:35, Jim Burns a écrit :
    On 8/5/2024 3:21 PM, WM wrote:
    Le 05/08/2024 à 02:24, Moebius a écrit :

    ∃^ℵ₀ u ∈ ⅟ℕ: ∀x > 0: u < x   (false) .

    ∃^NUF(x) u ∈ ⅟ℕ: ∀x > 0: u < x true.

    ¬∃u ∈ ⅟ℕ: ∀x > 0: u < x

    Right. But with
    NUF(x) = 1 ==> INVNUF(1) = x
    we get

    ∃u ∈ ⅟ℕ, u < x, ∀y > x = INVNUF(1).

    material conditional ⇒ means
    p q p⇒q ¬p∨q
    T T T T
    f T T T
    T f f f
    f f T T

    NUF(x) = 1 ⇒ INVNUF(1) = x
    means
    NUF(x) ≠ 1 ∨ INVNUF(1) = x

    Because
    NUF(x) ≠ 1
    is true everywhere
    NUF(x) = 1 ⇒ INVNUF(1) = x
    is true everywhere
    However,
    its truth doesn't imply INVNUF(1) exists.

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  • From Moebius@21:1/5 to All on Tue Aug 6 15:33:11 2024
    XPost: sci.math

    Am 06.08.2024 um 04:24 schrieb Chris M. Thomasson:

    There are infinite[ely many] even[ numbers] and there are infinite[ly many] odd[ numbers].

    On the other hand, some even numbers are odd. :-)

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  • From WM@21:1/5 to All on Tue Aug 6 13:55:35 2024
    XPost: sci.math

    Le 06/08/2024 à 14:38, Jim Burns a écrit :
    On 8/6/2024 4:35 AM, WM wrote:

    NUF(x) ≠ 1
    is true everywhere
    NUF(x) = 1 ⇒ INVNUF(1) = x
    is true everywhere
    However,
    its truth doesn't imply INVNUF(1) exists.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 implies its existence.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 6 13:52:42 2024
    XPost: sci.math

    Le 06/08/2024 à 12:32, Jim Burns a écrit :
    On 8/6/2024 4:26 AM, WM wrote:
    Le 06/08/2024 à 00:19, Jim Burns a écrit :

    NUF(x) gives
    the number of unit fractions smaller than x.

    For NUF(x) = 3
    ⅟ℕᵈᵉᶠ∩(0,x) is finite, namely 3.

    For NUF(x) = 3.5
    ⅟ℕᵈᵉᶠ∩(0,x) is fractional, namely 3.5, however,
    no such x with NUF(x) = 3.5 exists.

    That is not of interest. (We could however subdivide the distance between
    u_3 and u_4.)

    Also,
    no such x with NUF(x) = 3 exists.

    At least we have found now a way to express finitely many unit fractions without the accusation of quantifier shift and without the insane result
    that for all x > 0 NUF(x) = ℵo. That would be wrong even when no gaps
    between the unit fractions existed.

    | Assume otherwise.
    | Assume NUF(x₃) = 3
    |
    | u₁ < u₂ < u₃ are all of
    | the finite unit fractions in (0,x₃)
    |
    | However,
    | ⅟(1+⅟u₁) < u₁ is also
    | a finite unit fraction in (0,x₃)
    | 0 < ⅟(1+⅟u₁) < u₁ < u₂ < u₃ < x₃
    |
    | NUF(x₃) > 3
    | Contradiction.

    Therefore,
    no such x with NUF(x) = 3 exists.

    All that is in vain if you accept mathematics, in particular
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

    Regards, WM

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  • From Jim Burns@21:1/5 to Moebius on Tue Aug 6 12:05:12 2024
    XPost: sci.math

    On 8/6/2024 9:33 AM, Moebius wrote:
    Am 06.08.2024 um 04:24 schrieb Chris M. Thomasson:

    There are infinite[ely many] even[ numbers] and
    there are infinite[ly many] odd[ numbers].

    On the other hand, some even numbers are odd. :-)

    0 is the oddest even number.
    0+n = n
    0⋅n = 0
    Weird!

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  • From Moebius@21:1/5 to All on Tue Aug 6 18:35:36 2024
    XPost: sci.math

    Am 06.08.2024 um 18:05 schrieb Jim Burns:
    On 8/6/2024 9:33 AM, Moebius wrote:
    Am 06.08.2024 um 04:24 schrieb Chris M. Thomasson:

    There are infinite[ly many] even[ numbers] and
    there are infinite[ly many] odd[ numbers].

    On the other hand, some even numbers are odd. :-)

    0 is the oddest even number.
    0+n = n
    0⋅n = 0
    Weird!

    Right! This explains why for Mückenheim 0 is not a natural number - it's unnatural!

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  • From Jim Burns@21:1/5 to All on Tue Aug 6 12:33:55 2024
    XPost: sci.math

    On 8/6/2024 9:55 AM, WM wrote:
    Le 06/08/2024 à 14:38, Jim Burns a écrit :

    NUF(x) ≠ 1
    is true everywhere
    NUF(x) = 1  ⇒  INVNUF(1) = x
    is true everywhere
    However,
    its truth doesn't imply INVNUF(1) exists.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    implies its existence.

    No.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    implies its NONexistence.

    | Assume x₁ exists: NUF(x₁) = 1
    | ⅟n is the one unit.fraction in (0,x₁)
    | 0 < ⅟n < x₁
    |
    | However,
    | ∀n ∈ ℕ: ⅟n - ⅟(n+1) > 0
    | 0 < ⅟(n+1) < ⅟n < x₁
    | NUF(x₁) > 1
    | Contradiction.

    Therefore,
    there is no x₁: NUF(x₁) = 1
    ∀ᴿx: NUF(x) ≠ 1

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  • From Jim Burns@21:1/5 to All on Tue Aug 6 13:44:22 2024
    XPost: sci.math

    On 8/6/2024 9:52 AM, WM wrote:
    Le 06/08/2024 à 12:32, Jim Burns a écrit :
    On 8/6/2024 4:26 AM, WM wrote:

    NUF(x) gives
    the number of unit fractions smaller than x.

    For NUF(x) = 3
    ⅟ℕᵈᵉᶠ∩(0,x) is finite, namely 3.

    For NUF(x) = 3.5
    ⅟ℕᵈᵉᶠ∩(0,x) is fractional, namely 3.5, however,
    no such x with NUF(x) = 3.5  exists.

    That is not of interest.

    How sad for you.

    You missed an example of something getting named
    which does not then pop into existence.
    You could have benefited from a little interest.

    (We could however subdivide the distance
    between u_3 and u_4.)

    Suppose we subdivided the distance between u_3 and u_4.
    Would there be an x with NUF(x) = 3.5 ?

    | Assume otherwise.
    | Assume NUF(x₃) = 3
    |
    | u₁ < u₂ < u₃  are all of
    | the finite unit fractions in (0,x₃)
    |
    | However,
    | ⅟(1+⅟u₁) < u₁ is also
    | a finite unit fraction in (0,x₃)
    | 0 < ⅟(1+⅟u₁) < u₁ < u₂ < u₃ < x₃
    |
    | NUF(x₃) > 3
    | Contradiction.

    Therefore,
    no such x with NUF(x) = 3  exists.

    All that is in vain if you accept mathematics,
    in particular ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

    ...which is equivalent to
    ∀n ∈ ℕ: 0 < 1/(n+1) < 1/n

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  • From Moebius@21:1/5 to All on Tue Aug 6 23:02:54 2024
    XPost: sci.math

    Am 06.08.2024 um 22:21 schrieb Chris M. Thomasson:
    On 8/6/2024 6:33 AM, Moebius wrote:
    Am 06.08.2024 um 04:24 schrieb Chris M. Thomasson:

    There are infinite[ly many] even[ numbers] and
    there are infinite[ly many] odd[ numbers].

    On the other hand, some even numbers are odd. :-)

    ;^D 666?

    Ahh zero. I think its even... ;^)

    1, 2, 3, 4, ...

    odd, even, odd, even, ...

    So, the pattern:

    -2, -1, 0, 1, 2

    even, odd, (even), odd, even

    Yes. An integer z is even, iff there is am integer k such that z = 2k.

    Clearly for z = 0 there is an integer k (namely 0) such that z = 2k.

    In other words, an integer z is even if it can be deviede by 2 "without
    a remainder =/= 0".

    Clearly 0 can be devided by 2 "without a remainder =/= 0": 0 / 2 = 0.

    So 0 is an odd even number? :-)

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  • From Moebius@21:1/5 to All on Tue Aug 6 23:07:57 2024
    XPost: sci.math

    Am 06.08.2024 um 23:02 schrieb Moebius:

    So 0 is an odd even number? :-)

    Jim Burns: "0 is the oddest even number."

    :-)

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  • From Moebius@21:1/5 to All on Wed Aug 7 02:14:07 2024
    XPost: sci.math

    Am 07.08.2024 um 01:46 schrieb Chris M. Thomasson:
    On 8/6/2024 2:02 PM, Moebius wrote:
    Am 06.08.2024 um 22:21 schrieb Chris M. Thomasson:
    On 8/6/2024 6:33 AM, Moebius wrote:
    Am 06.08.2024 um 04:24 schrieb Chris M. Thomasson:

    There are infinite[ly many] even[ numbers] and
    there are infinite[ly many] odd[ numbers].

    On the other hand, some even numbers are odd. :-)

    ;^D 666?

    Ahh zero. I think its even... ;^)

    1, 2, 3, 4, ...

    odd, even, odd, even, ...

    So, the pattern:

    -2, -1, 0, 1, 2

    even, odd, (even), odd, even

    Yes. An integer z is even, iff there is am integer k such that z = 2k.

    Clearly for z = 0 there is an integer k (namely 0) such that z = 2k.

    In other words, an integer z is even if it can be deviede by 2
    "without a remainder =/= 0".

    Clearly 0 can be devided by 2 "without a remainder =/= 0": 0 / 2 = 0.

    So 0 is an odd even number? :-)

    Hint: This was a joke. :-)

    "odd" here is meant in a colloquial sense, while "even" is meant in a
    technical (math.) sense. :-)

    I say even?

    Right. :-)

    Though still odd. lol.

    Hint: "0 is the oddest even number." (Jim Burns)

    Well, both (odd and even) at the same time?

    Nope. Just even (math).

    This makes me think of signed zero...

    +0
    -0

    and just, 0?

    Right. (Try it in some computer language of your coice: If 0 == +0 ...
    amnd If 0 == -0 ...)

    But when used in the context of "lim" there is a certain difference.
    (Still "in a pure technical sense" +0 = -0 = 0."

    0 is just zero?

    Right: 0 = +0 = -0.

    But the NOTATIONS "+0" and "-0" are used in the context of the "lim"
    notation, to signify if 0 is "approached" vom "right", or from "left".

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  • From Moebius@21:1/5 to All on Wed Aug 7 02:16:33 2024
    XPost: sci.math

    Am 07.08.2024 um 02:14 schrieb Moebius:
    Am 07.08.2024 um 01:46 schrieb Chris M. Thomasson:
    On 8/6/2024 2:02 PM, Moebius wrote:
    Am 06.08.2024 um 22:21 schrieb Chris M. Thomasson:
    On 8/6/2024 6:33 AM, Moebius wrote:

    On the other hand, some even numbers are odd. :-)

    ;^D 666?

    Exactly! :-)

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  • From Richard Damon@21:1/5 to All on Tue Aug 6 22:36:10 2024
    XPost: sci.math

    On 8/6/24 4:35 AM, WM wrote:
    Le 06/08/2024 à 00:35, Jim Burns a écrit :
    On 8/5/2024 3:21 PM, WM wrote:
    Le 05/08/2024 à 02:24, Moebius a écrit :

    ∃^ℵ₀ u ∈ ⅟ℕ: ∀x > 0: u < x   (false) .

    ∃^NUF(x) u ∈ ⅟ℕ: ∀x > 0: u < x true.

    ¬∃u ∈ ⅟ℕ: ∀x > 0: u < x

    Right. But with NUF(x) = 1 ==> INVNUF(1) = x we get

    ∃u ∈ ⅟ℕ, u < x, ∀y > x = INVNUF(1).

    Regards, WM



    But INVNUF(1) can't exist, as it will be bigger than

    1/ ( ceil(1/INVNUF(1)) +1 ), and 1/ ( ceil(1/INVNUF(1)) +2 )
    which are two different unit fractions.

    And thus proving that it can't exist.

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  • From Richard Damon@21:1/5 to All on Tue Aug 6 23:41:12 2024
    XPost: sci.math

    On 8/6/24 4:38 AM, WM wrote:
    Le 06/08/2024 à 03:35, Richard Damon a écrit :
    On 8/5/24 2:58 PM, WM wrote:

    If there are all, then there is half of all.

    Not with infinities.

    Always.

    Then you believe that x can be less than x, as the number of even
    natural numbers is both 1/2 the number of natural numbers and the same
    size as it.

    No. This "size" is nonsense.
    Regards, WM



    As is a size that is half of infinity, Your logic just can't handle it.

    Sorry, you are just working with a blown up brain that has been
    destroyed by your logic's inconsistancy.

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  • From WM@21:1/5 to All on Wed Aug 7 12:31:18 2024
    XPost: sci.math

    Le 06/08/2024 à 18:33, Jim Burns a écrit :
    On 8/6/2024 9:55 AM, WM wrote:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    implies its existence.

    No.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    implies its NONexistence.

    Do you agree that all unit fractions with no exception have gaps on the
    real line?

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 7 12:38:40 2024
    XPost: sci.math

    Le 06/08/2024 à 19:44, Jim Burns a écrit :
    On 8/6/2024 9:52 AM, WM wrote:

    Suppose we subdivided the distance between u_3 and u_4.
    Would there be an x with NUF(x) = 3.5 ?

    No, the unit fractions are quantized.

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 7 13:05:42 2024
    XPost: sci.math

    Le 07/08/2024 à 04:36, Richard Damon a écrit :
    On 8/6/24 4:35 AM, WM wrote:

    Right. But with NUF(x) = 1 ==> INVNUF(1) = x we get

    ∃u ∈ ⅟ℕ, u < x, ∀y > x = INVNUF(1).

    But INVNUF(1) can't exist, as it will be bigger than

    1/ ( ceil(1/INVNUF(1)) +1 ), and 1/ ( ceil(1/INVNUF(1)) +2 )
    which are two different unit fractions.

    Peano is not valid for all dark numbers.

    Regards, WM

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  • From Jim Burns@21:1/5 to All on Wed Aug 7 13:07:09 2024
    XPost: sci.math

    On 8/7/2024 8:38 AM, WM wrote:
    Le 06/08/2024 à 19:44, Jim Burns a écrit :
    On 8/6/2024 9:52 AM, WM wrote:
    Le 06/08/2024 à 12:32, Jim Burns a écrit :
    On 8/6/2024 4:26 AM, WM wrote:
    Le 06/08/2024 à 00:19, Jim Burns a écrit :

    NUF(x) gives
    the number of unit fractions smaller than x.

    For NUF(x) = 3
    ⅟ℕᵈᵉᶠ∩(0,x) is finite, namely 3.

    For NUF(x) = 3.5
    ⅟ℕᵈᵉᶠ∩(0,x) is fractional, namely 3.5, however,
    no such x with NUF(x) = 3.5 exists.

    (We could however subdivide
    the distance between u_3 and u_4.)

    Suppose we subdivided
    the distance between u_3 and u_4.
    Would there be an x with NUF(x) = 3.5 ?

    No, the unit fractions are quantized.

    For NUF(x) = 3
    ⅟ℕᵈᵉᶠ∩(0,x) is finite, namely 3.

    For NUF(x) = 3.5
    |⅟ℕᵈᵉᶠ∩(0,x)| is fractional, namely 3.5
    By definition, INVNUF(3.5) = x

    We agree that
    saying "INVNUF(3.5)" doesn't prove
    INVNUF(3.5) exists

    It is equally true that
    saying "INVNUF(3)" doesn't prove
    INVNUF(3) exists

    But we have
    proof INVNUF(3) does not exist.

    ⎛ Assume INVNUF(3) exists.
    ⎜ NUF(INVNUF(3)) = 3

    ⎜ INVNUF(3) >
    ⎜ ⅟ ⌊⅟INVNUF(3) +1⌋ >
    ⎜ ⅟ ⌊⅟INVNUF(3) +2⌋ >
    ⎜ ⅟ ⌊⅟INVNUF(3) +3⌋ >
    ⎜ ⅟ ⌊⅟INVNUF(3) +4⌋ > 0

    ⎜ NUF(INVNUF(3)) > 3
    ⎝ Contradiction.

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  • From WM@21:1/5 to All on Wed Aug 7 17:47:14 2024
    XPost: sci.math

    Le 07/08/2024 à 19:07, Jim Burns a écrit :
    On 8/7/2024 8:38 AM, WM wrote:

    We agree that
    saying "INVNUF(3.5)" doesn't prove
    INVNUF(3.5) exists

    It is equally true that
    saying "INVNUF(3)" doesn't prove
    INVNUF(3) exists

    Correct.

    But we have
    proof INVNUF(3) does not exist.

    We have proof that INVNUF(3) exists.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
    Why do you not consider this argument?

    Regards, WM

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  • From Jim Burns@21:1/5 to All on Wed Aug 7 14:03:55 2024
    XPost: sci.math

    On 8/7/2024 8:31 AM, WM wrote:
    Le 06/08/2024 à 18:33, Jim Burns a écrit :
    On 8/6/2024 9:55 AM, WM wrote:
    Le 06/08/2024 à 14:38, Jim Burns a écrit :

    NUF(x) ≠ 1
    is true everywhere
    NUF(x) = 1 ⇒ INVNUF(1) = x
    is true everywhere
    However,
    its truth doesn't imply INVNUF(1) exists.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    implies its existence.

    No.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    implies its NONexistence.

    Do you agree that
    all unit fractions with no exception
    have gaps on the real line?

    Each unit fraction ⅟n has,
    for n ≠ 1, a gap between ⅟n and ⅟(n-1)
    a gap between ⅟n and ⅟(n+1) and
    a gap between ⅟(n+1) and ⅟(n+2)
    thus
    NUF(⅟n) > 1
    ⅟n ≠ INNUF(1)

    Each unit fraction ⅟n ≠ INVNUF(1)


    Finite doesn't need to be small.
    Finite can be big compared to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ
    and, if it shares finiteness.properties with 1
    and with Avogadroᴬᵛᵒᵍᵃᵈʳᵒ, it is finite.

    Infinite is beyond all finites, even big.finites.
    Infinite does not have finiteness.properties.

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  • From Moebius@21:1/5 to All on Wed Aug 7 20:15:32 2024
    XPost: sci.math

    Am 07.08.2024 um 20:03 schrieb Jim Burns:
    On 8/7/2024 8:31 AM, WM wrote:

    Do you agree that
    all unit fractions with no exception
    have gaps on the real line?

    For WM this fact "implies" that there is a first/smallest unit fraction. :-)

    "∀n ∈ ℕ: 1/n - 1/(n+1) > 0 implies its existence." (WM)

    This idiot is not able to comprehend that (1) (1/n - 1/(n+1))_(n e IN)
    is a null sequence and (2) SUM_(n=1..oo) 1/n - 1/(n+1) = 1.

    <facepalm>

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  • From Moebius@21:1/5 to All on Wed Aug 7 20:20:31 2024
    XPost: sci.math

    Am 07.08.2024 um 20:16 schrieb Chris M. Thomasson:
    On 8/7/2024 6:05 AM, WM wrote:

    Peano is not valid for all dark numbers.

    Since "Peano" is "valid" for ALL natural numbers (i.e. concerns the SET
    of ALL natural numbers) this implies that SOME dark numbers aren't
    natural numbers.

    Good to know. :-)

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  • From Jim Burns@21:1/5 to All on Wed Aug 7 14:49:22 2024
    XPost: sci.math

    On 8/7/2024 9:05 AM, WM wrote:
    Le 07/08/2024 à 04:36, Richard Damon a écrit :
    On 8/6/24 4:35 AM, WM wrote:

    Right. But with NUF(x) = 1 ==> INVNUF(1) = x we get
    ∃u ∈ ⅟ℕ, u < x, ∀y > x = INVNUF(1).

    But INVNUF(1) can't exist, as it will be bigger than
    1/ ( ceil(1/INVNUF(1)) +1 ), and 1/ ( ceil(1/INVNUF(1)) +2 )
    which are two different unit fractions.

    Peano is not valid for all dark numbers.

    For each real x > 0
    there are ℵ₀.many visibleᵂᴹ unit.fractions
    between x and 0

    There is no real x > 0 such that
    there are fewer than ℵ₀.many visibleᵂᴹ unit fractions
    between x and 0

    Darkᵂᴹ numbers do not make _fewer_ visibleᵂᴹ numbers,
    do they?

    If darkᵂᴹ numbers are what's between [0,1] and (0,1]
    then darkᵂᴹ numbers do not exist,
    because
    a positive lower.bound of unit.fractions
    is contradictory.

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  • From Jim Burns@21:1/5 to All on Wed Aug 7 14:29:54 2024
    XPost: sci.math

    On 8/7/2024 1:47 PM, WM wrote:
    Le 07/08/2024 à 19:07, Jim Burns a écrit :

    We agree that
    saying "INVNUF(3.5)" doesn't prove
    INVNUF(3.5) exists

    It is equally true that
    saying "INVNUF(3)" doesn't prove
    INVNUF(3) exists

    Correct.

    But we have
    proof INVNUF(3) does not exist.

    We have proof that INVNUF(3) exists.
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
    Why do you not consider this argument?

    ∀n ∈ ℕ: ⅟n - ⅟(n+1) >
    ⅟(n+1) - ⅟(n+2) >
    ⅟(n+2) - ⅟(n+3) >
    ⅟(n+3) - ⅟(n+4) > 0
    ⅟n ≠ INVNUF(3)

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    It seems likely that you're using
    an unreliable quantifier shift
    (which doesn't become reliable by staying implicit),
    but
    you (WM) find silence with regard to
    the rest of your argument
    more advantageous, apparently.

    Your silence suggest that
    you also find your argument unpersuasive.
    Perhaps there is hope for you.

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  • From WM@21:1/5 to All on Wed Aug 7 18:51:25 2024
    XPost: sci.math

    Le 07/08/2024 à 20:03, Jim Burns a écrit :
    On 8/7/2024 8:31 AM, WM wrote:

    Do you agree that
    all unit fractions with no exception
    have gaps on the real line?

    Each unit fraction ⅟n has,
    for n ≠ 1, a gap between ⅟n and ⅟(n-1)
    a gap between ⅟n and ⅟(n+1) and
    a gap between ⅟(n+1) and ⅟(n+2)
    thus

    Never more than one unit fraction can be added simultaneously to NUF(x).

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 7 18:55:32 2024
    XPost: sci.math

    Le 07/08/2024 à 20:15, Moebius a écrit :
    Am 07.08.2024 um 20:03 schrieb Jim Burns:
    On 8/7/2024 8:31 AM, WM wrote:

    Do you agree that
    all unit fractions with no exception
    have gaps on the real line?

    For WM this fact "implies" that there is a first/smallest unit fraction.

    Of course. Never two can be added simultaneously to NUF(x)

    "∀n ∈ ℕ: 1/n - 1/(n+1) > 0 implies its existence." (WM)

    (1) (1/n - 1/(n+1))_(n e IN)
    is a null sequence

    Nevertheless no unit fractions are without gap to their neighbours.

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 7 19:01:19 2024
    XPost: sci.math

    Le 07/08/2024 à 20:29, Jim Burns a écrit :
    On 8/7/2024 1:47 PM, WM wrote:

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    That is the decisive part. Never two or more unit fractions are added to
    NUF.

    It seems likely that you're using
    an unreliable quantifier shift

    In case you have no arguments claim quantifier shift. Please elaborate.

    (which doesn't become reliable by staying implicit),
    but
    you (WM) find silence with regard to
    the rest of your argument
    more advantageous, apparently.

    There is no rest.

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 7 19:05:05 2024
    XPost: sci.math

    Le 07/08/2024 à 20:49, Jim Burns a écrit :
    On 8/7/2024 9:05 AM, WM wrote:
    Le 07/08/2024 à 04:36, Richard Damon a écrit :
    On 8/6/24 4:35 AM, WM wrote:

    Right. But with NUF(x) = 1 ==> INVNUF(1) = x we get
    ∃u ∈ ⅟ℕ, u < x, ∀y > x = INVNUF(1).

    But INVNUF(1) can't exist, as it will be bigger than
    1/ ( ceil(1/INVNUF(1)) +1 ), and 1/ ( ceil(1/INVNUF(1)) +2 )
    which are two different unit fractions.

    Peano is not valid for all dark numbers.

    For each real x > 0
    there are ℵ₀.many visibleᵂᴹ unit.fractions
    between x and 0

    For each visible real.

    There is no real x > 0 such that
    there are fewer than ℵ₀.many visibleᵂᴹ unit fractions
    between x and 0

    They exist but are dark.

    Darkᵂᴹ numbers do not make _fewer_ visibleᵂᴹ numbers,
    do they?

    No.

    If darkᵂᴹ numbers are what's between [0,1] and (0,1]

    No that is not the case.

    Regards, WM

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  • From Jim Burns@21:1/5 to All on Wed Aug 7 16:51:36 2024
    XPost: sci.math

    On 8/7/2024 3:05 PM, WM wrote:
    Le 07/08/2024 à 20:49, Jim Burns a écrit :
    On 8/7/2024 9:05 AM, WM wrote:
    Le 07/08/2024 à 04:36, Richard Damon a écrit :
    On 8/6/24 4:35 AM, WM wrote:

    Right. But with NUF(x) = 1 ==> INVNUF(1) = x we get
    ∃u ∈ ⅟ℕ, u < x, ∀y > x = INVNUF(1).

    But INVNUF(1) can't exist, as it will be bigger than
    1/ ( ceil(1/INVNUF(1)) +1 ), and 1/ ( ceil(1/INVNUF(1)) +2 )
    which are two different unit fractions.

    Peano is not valid for all dark numbers.

    For each real x > 0
    there are ℵ₀.many visibleᵂᴹ unit.fractions
    between x and 0

    For each visible real.

    There is no real x > 0 such that
    there are fewer than ℵ₀.many visibleᵂᴹ unit fractions
    between x and 0

    They exist but are dark.

    If darkᵂᴹ numbers exist,
    then there are
    visibleᵂᴹ unit.fractions between 0 and the darkᵂᴹ.

    There is no positive lower.bound of
    visibleᵂᴹ unit.fractions.

    Any hypothetically.darkᵂᴹ positive number not so
    is a positive lower.bound of visibleᵂᴹ unit.fractions.
    It, as with all positive lower.bounds, doesn't exist.

    There is no positive lower.bound of
    visibleᵂᴹ unit.fractions.

    ⎛ Assume otherwise.
    ⎜ Assume β is a positive greatest.lower.bound

    ⎜ ½⋅β is a lower.bound

    ⎜ 2⋅β is not.a.lower.bound
    ⎜ ⅟k < 2⋅β a visibleᵂᴹ unit.fraction exists
    ⎜ ¼⋅⅟k < ¼⋅2⋅β a visibleᵂᴹ unit.fraction exists
    ⎜ ½⋅β = ¼⋅2⋅β is not.a.lower.bound
    ⎝ Contradiction.

    Darkᵂᴹ numbers do not make _fewer_ visibleᵂᴹ numbers,
    do they?

    No.

    ∀ᴿx>0: NUFᵛⁱˢ(x) = ℵ₀

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  • From Jim Burns@21:1/5 to Arithmetic on Wed Aug 7 17:29:04 2024
    XPost: sci.math

    On 8/7/2024 3:01 PM, WM wrote:
    Le 07/08/2024 à 20:29, Jim Burns a écrit :

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    That is the decisive part.
    Never two or more unit fractions are added to NUF.

    Arithmetic says:
    ⅟n >
    ⅟(n+1) >
    ⅟(n+2) >
    ⅟(n+3) >
    ⅟(n+4) >
    ...

    It seems likely that you're using
    an unreliable quantifier shift

    In case you have no arguments
    claim quantifier shift.
    Please elaborate.

    You do not disclose why you think that
    the equation which proves you are wrong
    proves that you are right.

    "Quantifier shift" is my best guess at
    your undisclosed thinking:
    ⎛ ∀ᴿx > 0: ∃S ⊆ ⅟ℕ: S ᵉᵃᶜʰ< x ∧ |S| = ℵ₀
    ⎜ 🖙 SHIFT 🖘
    ⎝ ∃S ⊆ ⅟ℕ: ∀ᴿx > 0: S ᵉᵃᶜʰ< x ∧ |S| = ℵ₀

    And hypothetical shifted.S holds your darkᵂᴹ numbers.

    Tell me I haven't read your mind!
    I've never claimed to be psychic.
    But then tell me what you ARE thinking, instead.

    (which doesn't become reliable by staying implicit),
    but
    you (WM) find silence with regard to
    the rest of your argument
    more advantageous, apparently.

    There is no rest.

    Then there is no argument.

    Think about it, before you admit that.
    I'd like to address your _best_ argument.
    Can you come up with even bad reasons
    for shifted.S to exist?

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  • From Moebius@21:1/5 to All on Wed Aug 7 23:55:46 2024
    XPost: sci.math

    Am 07.08.2024 um 23:29 schrieb Jim Burns:
    On 8/7/2024 3:01 PM, WM wrote:
    Le 07/08/2024 à 20:29, Jim Burns a écrit :

    you (WM) find silence with regard to
    the rest of your argument
    more advantageous, apparently.

    There is no rest.

    Hint: WM's "arguments" usually have the form: <statement> is true.

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  • From Moebius@21:1/5 to All on Thu Aug 8 00:17:09 2024
    XPost: sci.math

    Am 07.08.2024 um 23:29 schrieb Jim Burns:
    On 8/7/2024 3:01 PM, WM wrote:
    Le 07/08/2024 à 20:29, Jim Burns a écrit :

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    That is the decisive part.

    Actually, his "thinking process" is simple:

    "Since there is a gap (space) between adjacent unit fractions and all
    unit fractions are in the interval (0, 1], there must be FINITELY MANY
    of them (i.e. a first/smallest one)." (Don't ask! At least this
    "argument" seems to be CLEAR FOR HIM).

    And of course, in this case:

    Never two or more unit fractions are added to NUF [starting with NUF(0) = 0].

    Right. And if x1 e IR is "the position" of the first/smallest unit
    fraction, then NUF(x1) = 1.

    you (WM) find silence with regard to
    the rest of your argument
    more advantageous, apparently.

    There is no rest.

    Right!

    I mean, it's an OBVIOUS FACT (in Mückenheim's world).

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  • From Jim Burns@21:1/5 to All on Thu Aug 8 00:38:34 2024
    XPost: sci.math

    On 8/7/2024 2:51 PM, WM wrote:
    Le 07/08/2024 à 20:03, Jim Burns a écrit :
    On 8/7/2024 8:31 AM, WM wrote:

    Do you agree that
    all unit fractions with no exception
    have gaps on the real line?

    Each unit fraction ⅟n has,
    for n ≠ 1, a gap between ⅟n and ⅟(n-1)
    a gap between ⅟n and ⅟(n+1)  and
    a gap between ⅟(n+1) and ⅟(n+2)
    thus

    Never more than one unit fraction
    can be added simultaneously to NUF(x).

    ⅟n and ⅟(n+1) and ⅟(n+2) aren't
    simultaneous [colocated]

    NUF(⅟n) > 1
    ⅟n ≠ INNUF(1)

    Each unit fraction ⅟n ≠ INVNUF(1)


    Finite doesn't need to be small.
    Finite can be big compared to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ
    and, if it shares finiteness.properties with 1
    and with Avogadroᴬᵛᵒᵍᵃᵈʳᵒ, it is finite.

    Infinite is beyond all finites, even big.finites.
    Infinite does not have finiteness.properties.

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 8 08:05:12 2024
    XPost: sci.math

    Le 08/08/2024 à 00:17, Moebius a écrit :

    Actually, his "thinking process" is simple:

    "Since there is a gap (space) between adjacent unit fractions and all
    unit fractions are in the interval (0, 1], there must be FINITELY MANY
    of them (i.e. a first/smallest one)."

    No, that is nonsense. There are not finitely many unit fractions. The
    argument is this:
    NUF(0) = 0. Never more than one unit fraction can be added simultaneously
    to NUF(x).
    Exists x with x = INVNUF(1).

    Regards, WM

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  • From WM@21:1/5 to All on Thu Aug 8 08:00:42 2024
    XPost: sci.math

    Le 08/08/2024 à 06:38, Jim Burns a écrit :
    On 8/7/2024 2:51 PM, WM wrote:

    Never more than one unit fraction
    can be added simultaneously to NUF(x).

    ⅟n and ⅟(n+1) and ⅟(n+2) aren't
    simultaneous [colocated]

    NUF(0) = 0. Never more than one unit fraction can be added simultaneously
    to NUF(x).

    Infinite is beyond all finites, even big.finites.
    Infinite does not have finiteness.properties.

    But logic remains valid.
    NUF(0) = 0 and NUF(x) cannot increase by more than one at any x.
    There exists x = INVNUF(1).

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 8 08:20:56 2024
    XPost: sci.math

    Le 07/08/2024 à 23:29, Jim Burns a écrit :
    On 8/7/2024 3:01 PM, WM wrote:
    Le 07/08/2024 à 20:29, Jim Burns a écrit :

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    That is the decisive part.
    Never two or more unit fractions are added to NUF.

    Arithmetic says:
    ⅟n >
    ⅟(n+1) >
    ⅟(n+2) >
    ⅟(n+3) >
    ⅟(n+4) >
    ...

    NUF(0) = 0. Never more than one unit fraction can be added simultaneously
    to NUF(x).
    Exists x with x = INVNUF(1).

    There are two contradicting arguments. One of them must be wrong. Or
    actual infinity is wrong.

    You do not disclose why you think that
    the equation which proves you are wrong
    proves that you are right.

    There is no rest.

    Then there is no argument.

    Think about it, before you admit that.
    I'd like to address your _best_ argument.
    Can you come up with even bad reasons
    for shifted.S to exist?

    Sorry, I cannot see what you are missing.
    NUF(0) = 0. Never more than one unit fraction can be added simultaneously
    to NUF(x).
    Exists x with x = INVNUF(1).

    Regards, WM

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  • From Jim Burns@21:1/5 to All on Thu Aug 8 13:13:56 2024
    XPost: sci.math

    On 8/7/2024 3:01 PM, WM wrote:
    Le 07/08/2024 à 20:29, Jim Burns a écrit :
    On 8/7/2024 1:47 PM, WM wrote:

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    That is the decisive part.
    Never two or more unit fractions are added to NUF.

    you (WM) find silence with regard to
    the rest of your argument
    more advantageous, apparently.

    There is no rest.

    I am finding it difficult to imagine you (WM)
    as a student of physics.
    You were a student of physics, right?

    ⎛ <teacher>
    ⎜ How long does it take the polar bear
    ⎜ to slide down the frictionless ice.slope?

    ⎜ <Wölfchen>
    ⎜ 37 seconds.

    ⎜ <teacher>
    ⎜ Please show your work.

    ⎜ <Wölfchen>
    ⎝ 37 seconds, you idiot.
    ?

    No, I just can't see it.

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Thu Aug 8 12:58:27 2024
    XPost: sci.math

    On 8/8/2024 4:20 AM, WM wrote:
    Le 07/08/2024 à 23:29, Jim Burns a écrit :
    On 8/7/2024 3:01 PM, WM wrote:
    Le 07/08/2024 à 20:29, Jim Burns a écrit :

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    That is the decisive part.
    Never two or more unit fractions are added to NUF.

    Arithmetic says:
    ⅟n  >
    ⅟(n+1)  >
    ⅟(n+2)  >
    ⅟(n+3)  >
    ⅟(n+4)  >
    ...

    NUF(0) = 0.
    Never more than one unit fraction can be
    added simultaneously to NUF(x).
    Exists x with x = INVNUF(1).

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1

    There are two contradicting arguments.
    One of them must be wrong.
    Or actual infinity is wrong.

    ⎛ A finite order is trichotomous and
    ⎜ each non.{} subset with min and max

    ⎜ An infinite order is trichotomous and
    ⎜ has one or more non.{} subsets without min or max.

    ⎝ No set has both a finite and an infinite order.

    A subset of a set with a finite order
    has only non.{} subsets with min and max
    has a finite order.

    A superset of a set with an infinite order
    has a non.{} subset without min or max
    has an infinite order.

    An actuallyᵂᴹ infinite set
    ( a finite.ordered set with an infinite.ordered subset
    does not exist.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 9 01:59:35 2024
    XPost: sci.math

    Am 08.08.2024 um 22:36 schrieb Chris M. Thomasson:
    On 8/8/2024 3:30 AM, FromTheRafters wrote:
    on 8/8/2024, WM supposed :
    Le 08/08/2024 à 00:17, Moebius a écrit :

    Actually, his "thinking process" is simple:

    "Since there is a gap (space) between adjacent unit fractions and
    all unit fractions are in the interval (0, 1], there must be
    FINITELY MANY of them (i.e. a first/smallest one)."

    No, that is nonsense. There are not finitely many unit fractions.

    Then stop assuming that there is a first [i.e. smallest] element.

    The first unit fraction is 1/1, there is no last one...

    Nope. We are using the usual order < defined on IR to determine if there
    is a first (smallest) / last (largest) unit fraction.

    So there is no first/smallest unit fraction and there is a last/largest
    unit farction (namely 1/1), in respect to <.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Fri Aug 9 02:01:23 2024
    XPost: sci.math

    Am 08.08.2024 um 22:44 schrieb Chris M. Thomasson:

    you [WM] have a HYPER finite mind that simply cannot deal with
    infinity in any way, shape or from in the first place...

    Exactly.

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  • From Moebius@21:1/5 to All on Fri Aug 9 02:04:17 2024
    XPost: sci.math

    Am 09.08.2024 um 00:14 schrieb Chris M. Thomasson:

    The main point is that there is no smallest unit fraction

    Right. Except in Mückenheim's world.

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  • From Moebius@21:1/5 to All on Fri Aug 9 02:07:35 2024
    XPost: sci.math

    Am 09.08.2024 um 00:21 schrieb Chris M. Thomasson:

    Sorry for any confusion. ;^o

    *We* aren't confused, imho. Keep cool and try to learn from your betters (knowledge wise). :-P

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Fri Aug 9 02:09:50 2024
    XPost: sci.math

    Am 09.08.2024 um 01:59 schrieb Moebius:
    Am 08.08.2024 um 22:36 schrieb Chris M. Thomasson:
    On 8/8/2024 3:30 AM, FromTheRafters wrote:
    on 8/8/2024, WM supposed :
    Le 08/08/2024 à 00:17, Moebius a écrit :

    Actually, his "thinking process" is simple:

    "Since there is a gap (space) between adjacent unit fractions and
    all unit fractions are in the interval (0, 1], there *must* be
    FINITELY MANY of them (i.e. a first/smallest one)."

    No, that is nonsense. There are not finitely many unit fractions.

    Then stop assuming that there is a first [i.e. smallest] element.

    The first unit fraction is 1/1, there is no last one...

    Nope. We are using the usual order < defined on IR to determine if there
    is a first (smallest) / last (largest) unit fraction.

    So there is no first/smallest unit fraction and there is a last/largest
    unit farction (namely 1/1), in respect to <.

    Hint: WM is just a complete idiot.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Fri Aug 9 02:14:20 2024
    XPost: sci.math

    Am 08.08.2024 um 19:13 schrieb Jim Burns:
    On 8/7/2024 3:01 PM, WM wrote:
    Le 07/08/2024 à 20:29, Jim Burns a écrit :
    On 8/7/2024 1:47 PM, WM wrote:

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    That is the decisive part.
    Never two or more unit fractions are added to NUF.

    you (WM) find silence with regard to
    the rest of your argument
    more advantageous, apparently.

    There is no rest.

    I am finding it difficult to imagine you (WM)
    as a student of physics.
    You were a student of physics, right?

    Actually, he was a professor of physics (sort of)! :-)

    ⎛ <teacher>
    ⎜ How long does it take the polar bear
    ⎜ to slide down the frictionless ice.slope?

    ⎜ <Wölfchen>
    ⎜ 37 seconds.

    ⎜ <teacher>
    ⎜ Please show your work.

    ⎜ <Wölfchen>
    ⎝ 37 seconds, you idiot.
    ?

    No, I just can't see it.

    You silly idiot, can't you comprehend the significance of

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    ?! (!!!)

    Hint: WM never really *studied* mathematics ... you see.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Fri Aug 9 03:10:48 2024
    XPost: sci.math

    Am 09.08.2024 um 02:32 schrieb Jim Burns:

    A physics argument and a mathematics argument
    are different, but
    there are things they both are NOT.

    I've to agree since I've studied both (at least partwise).

    I think I would feel some of the same confusion
    if WM was a practicing lawyer.
    Lawyers must argue, too.
    Surely, a lawyer wouldn't think that
    "Boom! Here's the conclusion"
    is an _argument_ ?

    *lol* There once was a philosopher (Meinong) who claimed:

    "Es gibt Dinge von denen gilt, dass es sie gar nicht gibt."

    [There are things that are considered not to exist at all.]

    Seams that WM is a Meinongian. :-)

    Lit.: Bertrand Russell, On Denoting (1905)
    http://bactra.org/Russell/denoting/

    and

    Willard Van Orman Quine, On What There Is (1948) https://rintintin.colorado.edu/~vancecd/phil375/Quine.pdf

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Fri Aug 9 02:27:59 2024
    XPost: sci.math

    Am 09.08.2024 um 02:09 schrieb Moebius:
    Am 09.08.2024 um 01:59 schrieb Moebius:
    Am 08.08.2024 um 22:36 schrieb Chris M. Thomasson:
    On 8/8/2024 3:30 AM, FromTheRafters wrote:
    on 8/8/2024, WM supposed :
    Le 08/08/2024 à 00:17, Moebius a écrit :

    Actually, his "thinking process" is simple:

    "Since there is a gap (space) between adjacent unit fractions and
    all unit fractions are in the interval (0, 1], there *must* be
    FINITELY MANY of them (i.e. a first/smallest one)."

    No, that is nonsense. There are not finitely many unit fractions.

    Then stop assuming that there is a first [i.e. smallest] element.

    The first unit fraction is 1/1, there is no last one...

    Nope. We are using the usual order < defined on IR to determine if
    there is a first (smallest) / last (largest) unit fraction.

    So there is no first/smallest unit fraction and there is a last/
    largest unit farction (namely 1/1), in respect to <.

    Hint: WM is just a complete idiot.

    Hmmm... writing this John Gabriel comes to mind ...

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to Moebius on Thu Aug 8 20:32:33 2024
    XPost: sci.math

    On 8/8/2024 8:14 PM, Moebius wrote:
    Am 08.08.2024 um 19:13 schrieb Jim Burns:
    On 8/7/2024 3:01 PM, WM wrote:
    Le 07/08/2024 à 20:29, Jim Burns a écrit :

    The only part of your argument which you've shared is
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    That is the decisive part.
    Never two or more unit fractions are added to NUF.

    you (WM) find silence with regard to
    the rest of your argument
    more advantageous, apparently.

    There is no rest.

    I am finding it difficult to imagine you (WM)
    as a student of physics.
    You were a student of physics, right?

    Actually, he was a professor of physics (sort of)! :-)

    I've heard that.
    That is the source of my confusion.

    ⎛ <teacher>
    ⎜ How long does it take the polar bear
    ⎜ to slide down the frictionless ice.slope?

    ⎜ <Wölfchen>
    ⎜ 37 seconds.

    ⎜ <teacher>
    ⎜ Please show your work.

    ⎜ <Wölfchen>
    ⎝ 37 seconds, you idiot.
    ?

    No, I just can't see it.

    You silly idiot,
    can't you comprehend the significance of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    ?! (!!!)

    Hint:
    WM never really *studied* mathematics ...
    you see.

    A physics argument and a mathematics argument
    are different, but
    there are things they both are NOT.
    I think I would feel some of the same confusion
    if WM was a practicing lawyer.
    Lawyers must argue, too.
    Surely, a lawyer wouldn't think that
    "Boom! Here's the conclusion"
    is an _argument_ ?

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Fri Aug 9 03:12:02 2024
    XPost: sci.math

    Am 09.08.2024 um 02:32 schrieb Jim Burns:

    A physics argument and a mathematics argument
    are different, but
    there are things they both are NOT.

    I've to agree since I've studied both (at least partwise).

    I think I would feel some of the same confusion
    if WM was a practicing lawyer.
    Lawyers must argue, too.
    Surely, a lawyer wouldn't think that
    "Boom! Here's the conclusion"
    is an _argument_ ?

    *lol* There once was a philosopher (Meinong) who claimed:

    "Es gibt Dinge von denen gilt, dass es sie gar nicht gibt."

    [There are things that are considered not to exist at all.]

    Seems that WM is a Meinongian. :-)

    Lit.: Bertrand Russell, On Denoting (1905)
    http://bactra.org/Russell/denoting/

    and

    Willard Van Orman Quine, On What There Is (1948) https://rintintin.colorado.edu/~vancecd/phil375/Quine.pdf

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to FromTheRafters on Thu Aug 8 23:34:32 2024
    XPost: sci.math

    On 8/8/2024 5:56 PM, FromTheRafters wrote:
    Chris M. Thomasson has brought this to us :
    On 8/8/2024 3:30 AM, FromTheRafters wrote:
    on 8/8/2024, WM supposed :
    Le 08/08/2024 à 00:17, Moebius a écrit :

    Actually, his "thinking process" is simple:
    "Since there is a gap (space) between adjacent unit fractions and
    all unit fractions are in the interval (0, 1], there must be
    FINITELY MANY of them (i.e. a first/smallest one)."

    No, that is nonsense. There are not finitely many unit fractions.

    Then stop assuming that there is a first and last element.

    The first unit fraction is 1/1, there is no last one...

    ... or there is a last one but no first one.

    ... or
    (here is the "infinite" difference:)
    each linear order has SOME SUBSET which is
    without a first or without a last.

    My opinion is that WM's idea of 'finite' omits
    the 'subset' part of that,
    and he thinks that
    a set ordered with two ends is
    more complete than
    the same set with one or zero ends.

    So, we say 'finite' and 'infinite' meaning one thing
    and WM says 'finite' and 'infinite' meaning another.
    And then we're off to the races.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Sat Aug 10 00:36:20 2024
    XPost: sci.math

    Am 09.08.2024 um 05:34 schrieb Jim Burns:
    On 8/8/2024 5:56 PM, FromTheRafters wrote:
    Chris M. Thomasson has brought this to us :
    On 8/8/2024 3:30 AM, FromTheRafters wrote:
    on 8/8/2024, WM supposed :
    Le 08/08/2024 à 00:17, Moebius a écrit :

    Actually, his "thinking process" is simple:
    "Since there is a gap (space) between adjacent unit fractions and
    all unit fractions are in the interval (0, 1], there must be
    FINITELY MANY of them (i.e. a first/smallest one)."

    No, that is nonsense. There are not finitely many unit fractions.

    Then stop assuming that there is a first [...] element.

    The first unit fraction is 1/1, there is no last one...

    ... or there is a last one but no first one.

    Which is the case in respect to the usual < defined on ÌR.

    1/1 is the <-largest/last unit fraction and there is no <-smallst/first
    unit fraction. Rather simple facts. Expect in Mückenhausen of course,

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Sat Aug 10 02:42:33 2024
    XPost: sci.math

    Am 10.08.2024 um 02:29 schrieb Chris M. Thomasson:
    On 8/8/2024 4:59 PM, Moebius wrote:
    Am 08.08.2024 um 22:36 schrieb Chris M. Thomasson:
    On 8/8/2024 3:30 AM, FromTheRafters wrote:
    on 8/8/2024, WM supposed :
    Le 08/08/2024 à 00:17, Moebius a écrit :

    Actually, his "thinking process" is simple:

    "Since there is a gap (space) between adjacent unit fractions and
    all unit fractions are in the interval (0, 1], there must be
    FINITELY MANY of them (i.e. a first/smallest one)."

    No, that is nonsense. There are not finitely many unit fractions.

    Then stop assuming that there is a first [i.e. smallest] element.

    The first unit fraction is 1/1, there is no last one...

    Nope. We are using the usual order < defined on IR to determine if
    there is a first (smallest) / last (largest) unit fraction.

    So there is no first/smallest unit fraction and there is a last/
    largest unit farction (namely 1/1), in respect to <.

    Yup. Instead if using first, I should say largest... Any better?

    It depends on the <-relation we are referring to. First and smallest as
    well as last and largest just mean the same.

    0 ----------------------- 1
    first/ last/
    smallest largst
    element in [0, 1] element in [0, 1]

    (when referring to the usual <-relation defined on IR).

    hat you mean (I guess) is the fact that 1/1 is ***the first term*** in
    the sequence (1/1, 1/2, 1/3, ...).

    Still we have: ... < 1/3 < 1/2 < 1/1.

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  • From WM@21:1/5 to All on Sat Aug 10 15:54:38 2024
    XPost: sci.math

    Le 09/08/2024 à 02:32, Jim Burns a écrit :

    Surely, a lawyer wouldn't think that
    "Boom! Here's the conclusion"
    is an _argument_ ?

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    Therefore there can only be a single first unit fraction.
    What else do you need?

    Regards, WM

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  • From WM@21:1/5 to All on Sat Aug 10 15:26:47 2024
    XPost: sci.math

    Le 08/08/2024 à 12:30, FromTheRafters a écrit :
    on 8/8/2024, WM supposed :

    No, that is nonsense. There are not finitely many unit fractions.

    Then stop assuming that there is a first and last element.

    Finitely many means that you can count from first to last. You cannot
    count through the unit fractions.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Sat Aug 10 15:59:05 2024
    XPost: sci.math

    Le 09/08/2024 à 05:34, Jim Burns a écrit :

    and he thinks that
    a set ordered with two ends is
    more complete than
    the same set with one or zero ends.

    The set of unit fractions has two ends, namely at 1 and before 0.

    Regards, WM

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  • From Jim Burns@21:1/5 to All on Sat Aug 10 13:39:46 2024
    XPost: sci.math

    On 8/10/2024 11:59 AM, WM wrote:
    Le 09/08/2024 à 05:34, Jim Burns a écrit :

    and he thinks that
    a set ordered with two ends is
    more complete than
    the same set with one or zero ends.

    The set of unit fractions has two ends,
    namely at 1 and before 0.

    No.

    1 is the largest unit.fraction.
    1 is an end.

    ∀n ∈ ℕ: ⅟n - ⅟(n+1) > 0
    ∀n ∈ ℕ: ⅟n is not a second end.
    No unit fraction is a second end.

    The set of unit fractions has only one end.

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Sat Aug 10 13:28:11 2024
    XPost: sci.math

    On 8/10/2024 11:54 AM, WM wrote:
    Le 09/08/2024 à 02:32, Jim Burns a écrit :

    Surely, a lawyer wouldn't think that
    "Boom! Here's the conclusion"
    is an _argument_ ?

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    <place for argument>

    Therefore there can only be a single first
    unit fraction.

    No one has said there are two first unit.fractions.
    What forbids zero first unit.fractions?

    What causes an exception: nₓ ∈ ℕ without ⅟(nₓ+1) ?

    What else do you need?

    That which you've not yet done,
    show the opposite of this argument:
    ⎛ ∀n ∈ ℕ: ⅟n - ⅟(n+1) > 0

    ⎜ ∀n ∈ ℕ: ∃k ∈ ℕ: ⅟k = ⅟(n+1) ∧ ⅟n > ⅟k

    ⎜ ∀n ∈ ℕ: ∃k ∈ ℕ: ⅟n > ⅟k

    ⎜ ⅟n is the first unit fraction ⇔
    ⎜ ¬∃k ∈ ℕ: ⅟n > ⅟k

    ⎜ ∀n ∈ ℕ: ¬(⅟n is the first unit fraction)

    ⎜ ¬∃n ∈ ℕ: ⅟n is the first unit fraction

    ⎝ There is no first unit fraction.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sat Aug 10 21:59:11 2024
    XPost: sci.math

    Am 09.08.2024 um 05:34 schrieb Jim Burns:

    and [WM] thinks that
    a set ordered with two ends is
    more complete than [...]

    Well, the set {-1/1, -1/2, -1/3, ... ..., 1/3, 1/2, 1/1}

    Clearly has two ends: -1/1 which is the smallest element in this set and
    1/1 which is the largest element in this set. It must be very /Mückenheim-complete/.

    Ok, the following set would do too: {0, 1/1, 1/2, 1/3, ...}: 0 is the
    smallest element in this set and 1/1 is the largest element in this set.
    Two ends.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Sun Aug 11 00:06:27 2024
    XPost: sci.math

    Am 10.08.2024 um 19:39 schrieb Jim Burns:

    The set of unit fractions has only one end.

    Agree! :-)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 11 00:50:56 2024
    XPost: sci.math

    Am 11.08.2024 um 00:40 schrieb Chris M. Thomasson:
    On 8/10/2024 8:59 AM, WM wrote:

    The set of unit fractions has two ends, namely at 1 and before 0.

    They have no end

    Nope. They have an end at 1/1.

    Hint: 1/1 is the largest/last unit fractions, Hence the (set of the)
    unit fractions have (has) an end, namely 1/1.

    Hope this helps. :-P

    _______________________________

    ... 1/3 < 1/2 < 1/1 << this is the end.

    See: https://www.youtube.com/watch?v=VScSEXRwUqQ

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 11 00:47:53 2024
    XPost: sci.math

    Am 11.08.2024 um 00:40 schrieb Chris M. Thomasson:
    On 8/10/2024 8:59 AM, WM wrote:

    The set of unit fractions has two ends, namely at 1 and before 0.

    They have no end

    Nope. They have an end at 1/1.

    Hint: 1/1 is the largest/last unit fractions, Hence the (set of the)
    unit fractions has an end, namely 1/1.

    Hope this helps. :-P

    _______________________________

    ... 1/3 < 1/2 < 1/1 << this is the end.

    See: https://www.youtube.com/watch?v=VScSEXRwUqQ

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 11 01:00:18 2024
    XPost: sci.math

    Am 11.08.2024 um 00:52 schrieb Chris M. Thomasson:
    On 8/10/2024 3:47 PM, Moebius wrote:
    Am 11.08.2024 um 00:40 schrieb Chris M. Thomasson:
    On 8/10/2024 8:59 AM, WM wrote:

    The set of unit fractions has two ends, namely at 1 and before 0.

    They have no end

    Nope. They have an end at 1/1.

    Well, I needed to give more context. 1/1, in my mind makes me think of
    the first, [...]

    Again, 1/1 is the first term in the sequence (1/1, 1/2, 1/3, ...), but
    it's the largest/last unit fraction in respect to < (as defined on IR,
    and hence on the unit fractions).

    ... < 1/3 < 1/2 < 1/1.

    In the other hand the first term of (1/1, 1/2, 1/3, ...) is 1/1, the
    second is 1/2, etc.

    Actually, the most important lesson is taught here:

    See: https://www.youtube.com/watch?v=VScSEXRwUqQ

    Please study it closely!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 11 01:03:07 2024
    XPost: sci.math

    Am 11.08.2024 um 01:00 schrieb Moebius:

    See: https://www.youtube.com/watch?v=VScSEXRwUqQ

    Please study it closely!

    A comment: "I consider this one of the best songs ever made"

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sun Aug 11 12:24:46 2024
    XPost: sci.math

    Le 10/08/2024 à 19:16, FromTheRafters a écrit :
    WM expressed precisely :
    Le 09/08/2024 à 05:34, Jim Burns a écrit :

    and he thinks that
    a set ordered with two ends is
    more complete than
    the same set with one or zero ends.

    The set of unit fractions has two ends, namely at 1 and before 0.

    Wrong,

    Name unit fractions larger than 1 or smaller than 0.
    Note: Before domains without unit fractions the set has ended.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sun Aug 11 12:29:13 2024
    XPost: sci.math

    Le 10/08/2024 à 19:28, Jim Burns a écrit :
    On 8/10/2024 11:54 AM, WM wrote:
    Le 09/08/2024 à 02:32, Jim Burns a écrit :

    Surely, a lawyer wouldn't think that
    "Boom! Here's the conclusion"
    is an _argument_ ?

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    If an interval contains unit fractions, then it contains a first one.

    Therefore there can only be a single first
    unit fraction.

    No one has said there are two first unit.fractions.
    What forbids zero first unit.fractions?

    The existence of unit frations enforces one or more first unit fractions.

    What causes an exception: nₓ ∈ ℕ without ⅟(nₓ+1) ?

    The end of the positivee axis.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sun Aug 11 14:09:41 2024
    XPost: sci.math

    Le 10/08/2024 à 19:39, Jim Burns a écrit :
    On 8/10/2024 11:59 AM, WM wrote:
    Le 09/08/2024 à 05:34, Jim Burns a écrit :

    and he thinks that
    a set ordered with two ends is
    more complete than
    the same set with one or zero ends.

    The set of unit fractions has two ends,
    namely at 1 and before 0.

    No.

    Note: And end is there, or before the point, after which no elements follow.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Sun Aug 11 13:56:16 2024
    XPost: sci.math

    On 8/11/2024 10:09 AM, WM wrote:
    Le 10/08/2024 à 19:39, Jim Burns a écrit :
    On 8/10/2024 11:59 AM, WM wrote:
    Le 09/08/2024 à 05:34, Jim Burns a écrit :

    and he thinks that
    a set ordered with two ends is
    more complete than
    the same set with one or zero ends.

    The set of unit fractions has two ends,
    namely at 1 and before 0.

    No.

    Note:
    And end is there, or before the point,
    after which no elements follow.

    ⎛ Something not.in a set cannot be
    ⎜ an end of the set,
    ⎜ although it might be a bound of the set.

    ⎜ A greatest.lower.bound which is in the set
    ⎜ is its lower.end.

    ⎜ A greatest.lower.bound which isn't in the set
    ⎜ isn't its lower.end, and
    ⎜ nothing else is its lower.end,
    ⎜ because
    ⎜ if it were its lower.end,
    ⎜ it would be its greatest.lower.bound,
    ⎜ which it isn't.

    ⎜⎛ If the greatest.lower.bound isn't an element,
    ⎝⎝ no lower.end exists.

    ⎛ The greatest.lower.bound of the unit.fractions
    ⎜ isn't a unit.fraction.

    ⎝ _No lower.end of the unit.fractions exists_

    ⎛ The greatest.lower.bound of the unit.fractions
    ⎜ isn't a unit.fraction.

    ⎜ Each unit.fraction is positive.
    ⎜ 0 is a lower.bound of the unit.fractions.

    ⎜( Each x > 0 is not a lower.bound of unit.fractions.

    ⎜ 0 is the greatest.lower.bound of unit.fractions.

    ⎝ 0 isn't a unit.fraction.

    ⎛ Each x > 0 is not a lower.bound of unit.fractions.

    ⎜⎛ Assume otherwise.
    ⎜⎜ Assume x > 0 is a lower.bound of unit.fractions.
    ⎜⎜
    ⎜⎜ Greatest.lower.bound β ≥ x > 0
    ⎜⎜
    ⎜⎜ Lower.bound ½⋅β < β
    ⎜⎜
    ⎜⎜ Not.lower.bound 2⋅β > β
    ⎜⎜ Unit.fraction ⅟k < 2⋅β
    ⎜⎜ Unit.fraction ¼⋅⅟k < ¼⋅2⋅β = ½⋅β
    ⎜⎜ Not.lower.bound ½⋅β
    ⎜⎝ Contradiction.

    ⎜ Therefore,
    ⎝ each x > 0 is not a lower.bound of unit.fractions.

    ⎛ Each x > 0 is not a lower.bound of unit.fractions.

    ⎜ 0 is the greatest.lower.bound of unit.fractions,
    ⎜ and 0 isn't a unit fraction.

    ⎝ _No lower.end of the unit.fractions exists_

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Mon Aug 12 13:47:18 2024
    XPost: sci.math

    Le 11/08/2024 à 19:56, Jim Burns a écrit :
    On 8/11/2024 10:09 AM, WM wrote:

    And end is there, or before the point,
    after which no elements follow.

    ⎛ Something not.in a set cannot be
    ⎜ an end of the set,
    ⎜ although it might be a bound of the set.

    Before the bound there is the end, perhaps dark.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Mon Aug 12 13:50:09 2024
    XPost: sci.math

    Le 11/08/2024 à 19:56, Jim Burns a écrit :
    On 8/11/2024 8:29 AM, WM wrote:

    What causes an exception: nₓ ∈ ℕ without ⅟(nₓ+1) ?

    The end of the positivee axis.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    ∀n ∈ ℕ: 1/n > 1/(n+1) > 0

    If 1/(n+1) exists.

    Each positive unit fraction is not
    the first positive unit fraction.

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Mon Aug 12 13:23:21 2024
    XPost: sci.math

    On 8/12/24 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :
    On 8/11/2024 8:29 AM, WM wrote:

    What causes an exception: nₓ ∈ ℕ without ⅟(nₓ+1) ?

    The end of the positivee axis.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    ∀n ∈ ℕ: 1/n > 1/(n+1) > 0

    If 1/(n+1) exists.

    Which, by the definition of the Natural Numbers, it does.

    Of course, if your logic doesn't actually allow the Natural Numbers to
    exist as a set, then you shouldn't be using it.


    Each positive unit fraction is not
    the first positive unit fraction.

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Which, by the definition of the Natural Numbers, doesn't exist.

    Of course, if your logic doesn't actually allow the Natural Numbers to
    exist as a set, then you shouldn't be using it.

    Since it is clear that you logic can't handle the Natuarl Numbers, you
    are just al Number, you use of them just shows you don't mind using
    broken logic.


    Regards, WM



    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Mon Aug 12 13:44:10 2024
    XPost: sci.math

    On 8/12/2024 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :
    On 8/11/2024 8:29 AM, WM wrote:
    Le 10/08/2024 à 19:28, Jim Burns a écrit :

    What causes an exception:
    nₓ ∈ ℕ without ⅟(nₓ+1) ?

    The end of the positivee axis.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    ∀n ∈ ℕ: 1/n > 1/(n+1) > 0

    If 1/(n+1) exists.

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    asserts 1/(n+1) exists, '∀n ∈ ℕ'
    along with asserting other things.

    Each positive unit fraction is not
    the first positive unit fraction.

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    There is no ⅟nₓ before the end of the positive axis
    without ⅟(nₓ+1) before the end of the positive axis.

    ⎛ Assume otherwise.
    ⎜ Assume ⅟nₓ > 0 ≥ ⅟(nₓ+1)

    ⎜ nₓ⋅⅟nₓ = 1
    ⎜ ⅟nₓ > 0 ∧ 1 > 0 ⇒ nₓ > 0
    ⎜ nₓ > 0 ⇒ nₓ+1 > 0
    ⎜ (nₓ+1)⋅⅟(nₓ+1) = 1
    ⎜ nₓ+1 > 0 ∧ 1 > 0 ⇒ ⅟(nₓ+1) > 0
    ⎝ Contradiction.

    There is no exception to not.being the lower end.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Mon Aug 12 14:34:33 2024
    XPost: sci.math

    On 8/12/2024 9:47 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :
    On 8/11/2024 10:09 AM, WM wrote:

    And end is there, or before the point,
    after which no elements follow.

    ⎛ Something not.in a set cannot be
    ⎜ an end of the set,
    ⎜ although it might be a bound of the set.

    Before the bound there is the end,
    perhaps dark.

    No.

    0 is
    greatest.lower.bound β of visibleᵂᴹ unit.fractions
    otherwise,
    ½⋅β is both lower.bound and not.lower.bound

    0 not a visibleᵂᴹ unit.fraction.

    No point ε > 0 is the end of
    visibleᵂᴹ.or.darkᵂᴹ unit.fractions.

    ⎛ Assume otherwise.
    ⎜ Assume ε > 0 is the end of
    ⎜ visibleᵂᴹ.or.darkᵂᴹ unit.fractions.

    ⎜ ε > 0 is a lower.bound of
    ⎜ visibleᵂᴹ.or.darkᵂᴹ unit.fractions.

    ⎜ ε > 0 is a lower.bound of
    ⎜ the visibleᵂᴹ unit.fractions.

    ⎜ However
    ⎜ 0 is
    ⎜ greatest.lower.bound of visibleᵂᴹ unit.fractions
    ⎝ Contradiction.

    Therefore,
    no point ε > 0 is the end of
    visibleᵂᴹ.or.darkᵂᴹ unit.fractions.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Mon Aug 12 20:24:04 2024
    XPost: sci.math

    Am 12.08.2024 um 19:44 schrieb Jim Burns:
    On 8/12/2024 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 (*) [WM]

    ∀n ∈ ℕ: 1/n > 1/(n+1) > 0 (**) [JB]

    If 1/(n+1) exists. [WM]

    Mückenheim, wenn Du meinst, dass das eine Voraussetzung für (**) ist,
    dann ist es doch auch eine für (*). Warum hast Du das VORHER nicht dazu gesagt? Du redest wirklich nur saudummen Scheißdreck daher, Mückenheim.

    In der Mathematik ist es so:

    Es gibt das allgemein akzeptierte Peano-Axiom:

    An e IN: s(n) e IN.

    Daraus folgt zusammen mit der Definition

    n+1 =df s(n) (n e IN)

    sofort:

    An e IN: s+1 e IN.

    Und daraus natürlich:

    An e IN: Em e IN: m = n+1 .

    "Mückensprech: (for each and every n in IN) n+1 exists and is an element in IN."

    Mit der (üblichen) Definition

    n < m =df Ek e IN: n+k = m

    ergibt sich zusammen mit dem ebenfalls allgemein akzeptierte Peano-Axiom:

    1 e IN

    auch

    An e IN: n+1 e IN & n < n+1 .

    Daraus folgt dann im Kontext der rationalen Zahlen (Q):

    An e IN: 1/(n+1) < 1/n .

    Man kann natürlich ebenso gut beweisen:

    An e IN: Eq,q' e Q: q = 1/n & q' = 1/(n+1) & q' < q ,

    oder auch nur:

    An e IN: Eq e Q: q = 1/(n+1) & q < 1/n .

    "Mückensprech: (for each and every n in IN) 1/(n+1) exists and is
    an element in Q."

    Was genau verstehst Du daran nicht?

    Also ja:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    asserts 1/(n+1) exists, '∀n ∈ ℕ'
    along with asserting other things.

    Right.

    Wie schon ZIG mal erwähnt, Mückenheim: Wenn 1/(n+1) nicht für jedes n e
    IN definiert wäre (also existieren würde), dann dürftest Du es in einer einschlägigen Behauptung wie

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    gar nicht verwenden, Depp.

    "Ebensogut" könntest Du behaupten:

    ∀x ∈ IR: |1/x| >= 0 . (***)

    Nur ist 1/x für x = 0 nicht definiert, daher ist der Ausdruck (***) als Behauptung nicht zulässig (da 0 e IR ist).

    Um aber zum eigentlichen Thema zurück zu kommen:

    Each positive unit fraction is not
    the first positive unit fraction.

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Nö. Natürlich nicht. Denn es gilt: An e IN: 0 < 1/(n+1) < 1/n.

    Dass also die "positive axis" bei 0 "aufhört", ist KEIN Problem.

    There is no ⅟nₓ before the end of the positive axis
    without ⅟(nₓ+1) before the end of the positive axis.

    ⎛ Assume otherwise.
    ⎜ Assume ⅟nₓ > 0 ≥ ⅟(nₓ+1)

    ⎜ nₓ⋅⅟nₓ = 1
    ⎜ ⅟nₓ > 0  ∧  1 > 0  ⇒  nₓ > 0
    ⎜ nₓ > 0  ⇒  nₓ+1 > 0
    ⎜ (nₓ+1)⋅⅟(nₓ+1) = 1
    ⎜ nₓ+1 > 0  ∧  1 > 0  ⇒  ⅟(nₓ+1) > 0
    ⎝ Contradiction.

    There is no exception to not.being the lower end.

    Jep. Außer in Mückenheims Wahnwelt. :-)

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Mon Aug 12 20:47:04 2024
    XPost: sci.math

    Am 12.08.2024 um 20:24 schrieb Moebius:
    Am 12.08.2024 um 19:44 schrieb Jim Burns:
    On 8/12/2024 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :

    Each positive unit fraction is not
    the first positive unit fraction.
    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Nö. Natürlich nicht. Denn es gilt: An e IN: 0 < 1/(n+1) < 1/n.

    Dass also die "positive axis" bei 0 "aufhört", ist KEIN Problem.

    Ist jetzt seit kurzem die Menge {1/n : n e IN} wieder endlich,
    Mückenheim? Oder wollen Sie behaupten, dass es ein n e IN gibt, so dass
    1/n NICHT größer als 0 ist? <faceplam>

    Jedenfalls erlaubt *ihre* Behauptung

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ,

    sofort auf

    ∀n ∈ ℕ: 1/(n+1) < 1/n

    zu schließen.

    Und - wegen Ak e IN: 0 < 1/k - damit auch auf

    ∀n ∈ ℕ: 0 < 1/(n+1) < 1/n ,

    wie von JB behauptet. (Wie schon erwähnt, gilt in der Mathematik ja An e
    IN: n+1 e IN.)

    Es gibt dazu eine schöne Entsprechung, die die natürlichen Zahlen (statt
    den Stammbrüchen) betrifft:

    ∀n ∈ ℕ: omega > n+1 > n .

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Mon Aug 12 20:55:30 2024
    XPost: sci.math

    Am 12.08.2024 um 20:47 schrieb Moebius:

    Jedenfalls erlaubt *ihre* Behauptung

            ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 ,

    sofort auf

            ∀n ∈ ℕ: 1/(n+1) < 1/n

    zu schließen.

    Und - wegen Ak e IN: 0 < 1/k - damit auch auf

            ∀n ∈ ℕ: 0 < 1/(n+1) < 1/n ,

    wie von JB behauptet. (Wie schon erwähnt, gilt in der Mathematik ja An e
    IN: n+1 e IN.)

    Es gibt dazu eine schöne Entsprechung, die die natürlichen Zahlen (statt den Stammbrüchen) betrifft:

             ∀n ∈ ℕ: omega > n+1 > n .

    Ein Unterschied zwischen den beiden Folgen (n)_(n e IN) und (1/n)_(n e
    IN) besteht natürlich darin, dass die Abstände zwischen zwei aufeinander folgenden Folgengliedern im einem Fall konstant sind, im anderen aber
    eine Nullfolge bilden. D. h. die natürlichen Zahlen "bevölkern" das
    Intervall [0, oo), während für die Stammbrüche das Intervall (0, 1] ausreicht. Iw. scheinen Sie mit diesem (an sich trivialen) Sachverhalt überfordert zu sein, Mückenheim.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Mon Aug 12 23:38:19 2024
    XPost: sci.math

    Am 12.08.2024 um 22:54 schrieb Chris M. Thomasson:
    On 8/12/2024 6:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :
    On 8/11/2024 8:29 AM, WM wrote:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    Hence Jim concluded (essentially):

    ∀n ∈ ℕ: 1/n > 1/(n+1)

    If 1/(n+1) exists. [WM]

    Uhhhh..... WHAT!?

    If the earth is a planet. If the sun exists. If 0 =/= 1, etc.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Mon Aug 12 23:43:45 2024
    XPost: sci.math

    Am 12.08.2024 um 23:38 schrieb Moebius:
    Am 12.08.2024 um 22:54 schrieb Chris M. Thomasson:
    On 8/12/2024 6:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :
    On 8/11/2024 8:29 AM, WM wrote:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    Hence Jim concluded (essentially):

    ∀n ∈ ℕ: 1/n > 1/(n+1)

    If 1/(n+1) exists. [WM]

    Uhhhh..... WHAT!?

    If the earth is a planet. If the sun exists. If 0 =/= 1, etc.

    On the other hand:

    "It is not guaranteed that n+1 exists for every n." (WM, sci.math)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Tue Aug 13 14:21:57 2024
    XPost: sci.math

    Le 12/08/2024 à 19:44, Jim Burns a écrit :
    On 8/12/2024 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    There is no ⅟nₓ before the end of the positive axis
    without ⅟(nₓ+1) before the end of the positive axis.

    You cannot see it. It is dark. The function NUF(x) is a step-function. It
    can increase from 0 at x = 0 to greater values, either in a step of size 1
    or in a step of size more than 1. But increase by more than 1 is excluded
    by the gaps between unit fractions. (Note the universal quantifier there, according to which never – in no limit and in no accumulation point –
    two unit fractions occupy the same point x.) Therefore the step size can
    only be 1, resulting in a real coordinate x with NUF(x) = 1.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Tue Aug 13 14:27:43 2024
    XPost: sci.math

    Le 12/08/2024 à 20:34, Jim Burns a écrit :
    On 8/12/2024 9:47 AM, WM wrote:

    Before the bound there is the end,
    perhaps dark.

    0 is
    greatest.lower.bound β of visibleᵂᴹ unit.fractions

    Yes. But the smallest unit fractions comes before.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Tue Aug 13 14:25:12 2024
    XPost: sci.math

    Le 12/08/2024 à 20:24, Moebius a écrit :

    In der Mathematik ist es so:

    Es gibt das allgemein akzeptierte Peano-Axiom:

    An e IN: s(n) e IN.

    Und alle paar tausend Jahre kommt eine neue Idee und wirft den
    verkrusteten Trödel um.

    Gruß, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Tue Aug 13 14:17:31 2024
    XPost: sci.math

    Le 12/08/2024 à 19:23, Richard Damon a écrit :
    On 8/12/24 9:50 AM, WM wrote:

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Which, by the definition of the Natural Numbers, doesn't exist.

    The end of the positive axis exists.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Tue Aug 13 18:35:53 2024
    XPost: sci.math

    Am 13.08.2024 um 18:26 schrieb Jim Burns:
    On 8/13/2024 10:27 AM, WM wrote:

    the smallest unit fractions

    only exist in Mückenland.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Tue Aug 13 12:26:45 2024
    XPost: sci.math

    On 8/13/2024 10:27 AM, WM wrote:
    Le 12/08/2024 à 20:34, Jim Burns a écrit :
    On 8/12/2024 9:47 AM, WM wrote:

    Before the bound there is the end,
    perhaps dark.

    0 is
    greatest.lower.bound β of visibleᵂᴹ unit.fractions

    Yes.

    Thank you.

    But the smallest unit fractions comes before.

    No visibleᵂᴹ.or.darkᵂᴹ point ε > 0 is the end of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    Otherwise,
    ⎛ ε > 0 is a lower.bound of
    ⎜ both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    ⎜ ε > 0 is a lower.bound of
    ⎜ visibleᵂᴹ unit.fractions.

    ⎜ 0 is the greatest.lower.bound of
    ⎜ visibleᵂᴹ unit.fractions.

    ⎜ ε is a lower.bound greater than the greatest.
    ⎝ Contradiction.


    You (WM) don't seem to want it, but
    we could stretch the definition of darkᵂᴹ so that
    ε > 0 is the end of darkᵂᴹ but not visibleᵂᴹ
    ( darkᵂᴹ in the cracks between some visibleᵂᴹ?
    However, such an ε > 0 would still leave
    the unit.fractions one.ended.

    ----
    0 is the greatest.lower.bound of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    0 isn't a unit.fraction.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Tue Aug 13 13:02:40 2024
    XPost: sci.math

    On 8/13/2024 10:21 AM, WM wrote:
    Le 12/08/2024 à 19:44, Jim Burns a écrit :
    On 8/12/2024 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    There is no ⅟nₓ before the end of the positive axis
    without ⅟(nₓ+1) before the end of the positive axis.

    You cannot see it. It is dark.

    The function NUF(x) is a step-function.
    It can increase from 0 at x = 0 to greater values,

    0 isn't a unit.fraction.

    either in a step of size 1
    or in a step of size more than 1.
    But increase by more than 1 is excluded by
    the gaps between unit fractions.

    0 isn't a unit.fraction.

    (Note the universal quantifier there,

    quantified over unit.fractions

    according to which never –
    in no limit and in no accumulation point –
    two unit fractions occupy the same point x.)

    Each unit fraction has GLB β > 0
    which other unit.fractions are at least as far as.

    0 is not a unit.fraction.
    0 does not have GLB β > 0
    which unit fractions are at least as far as.
    Otherwise,
    ½⋅β is lower.bound and not.lower.bound.

    Therefore
    the step size can only be 1,

    ...at a unit.fraction.
    0 isn't a unit.fraction.

    resulting in a real coordinate x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Tue Aug 13 19:32:17 2024
    XPost: sci.math

    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:
    Le 12/08/2024 à 19:44, Jim Burns a écrit :

    There is no ⅟nₓ before the end of the positive axis
    without ⅟(nₓ+1) before the end of the positive axis.

    You cannot see it. It is dark.

    No number can be seen, dark or not.

    resulting in a real coordinate x with NUF(x) = 1.

    Assume NUF(x0) = 1 with x0 e IR. This means that there is exactly one
    unit fraction u such that u < x0. Let's call this unit fraction u0. Then
    (by definition) there is a (actually exactly one) natural number n such
    that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by
    definition) 1/(n0 + 1) is a unit fraction which is smaller than u0 and
    hence smaller than x0. Hence NUF(x0) > 1. Contradiction!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Moebius on Tue Aug 13 13:42:49 2024
    XPost: sci.math

    On 8/13/2024 12:35 PM, Moebius wrote:
    Am 13.08.2024 um 18:26 schrieb Jim Burns:
    On 8/13/2024 10:27 AM, WM wrote:

    the smallest unit fractions

    only exist in Mückenland.

    The existence of the smallest unit fractions
    is contradictory in the land of
    rationals with
    countable.to numerators and denominators
    with each split situated ==
    a last point in the foresplit or
    a first point in the hindsplit.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Tue Aug 13 19:51:19 2024
    XPost: sci.math

    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    either in a step of size 1
    or in a step of size more than 1.

    Let's "assume" that this is true (sort of) for the sake of the argument.

    But increase by more than 1 is excluded by
    the gaps between unit fractions.

    I can't see any argument for this claim.

    Actually, for each and every real number x > 0 there are infinitely many
    unit fractions smaller than x: 1/ceil(1/x + 1)), 1/ceil(1/x + 2)),
    1/ceil(1/x + 3)), ...

    Hence the difference between NUF(0) (i.e. 0) and NUF(x) is "infinite"
    for each and every x e IR, x > 0.

    In fact, Ax > 0: NUF(x) = aleph_0, while NUF(0) = 0.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Tue Aug 13 19:47:35 2024
    XPost: sci.math

    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    either in a step of size 1
    or in a step of size more than 1.

    Let's "assume" that this is true (sort of) for the sake of the argument.

    But increase by more than 1 is excluded by
    the gaps between unit fractions.

    I can't see any argument for this claim.

    Actually, for each and every real number x > 0 there are infinitely many
    unit fractions smaller than x: 1/cail(1/x + 1)), 1/cail(1/x + 2)),
    1/cail(1/x + 3)), ...

    Hence the difference between NUF(0) (i.e. 0) and NUF(x) is "infinite"
    for each and every x e IR, x > 0.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Jim Burns on Tue Aug 13 14:39:08 2024
    XPost: sci.math

    On 8/13/2024 2:34 PM, Jim Burns wrote:
    On 8/13/2024 10:17 AM, WM wrote:
    Le 12/08/2024 à 19:23, Richard Damon a écrit :
    On 8/12/24 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Which,
    by the definition of the Natural Numbers,
    doesn't exist.

    The end of the positive axis exists.

    No point in the positive axis is
    largest in the positive axis or
    smallest in the positive axis.
    A point not.largest and not.smallest
    is not an end.

    No point in the positive axis is
    an upper.end or a lower.end.

    If an end of the positive exists,
    it is in the positive axis.
    A bound not.in is not an end.

    No point not.in the positive axis is
    an upper.end or a lower.end.

    No point is
    an upper.end or a lower.end.

    The end of the positive axis
    does not exist.

    ----
    In the land of rationals only with
    countable.to numerators and denominators
    and with each split situated
    ⎛ a last point in the foresplit or
    ⎝ a first point in the hindsplit,
    no point in the positive axis is
    largest in the positive axis or
    smallest in the positive axis.

    for each positive rational p/q
    p > 0, q > 0
    (p+1)/q > p/q > p/(q+1) > 0

    for each positive rational p/q
    p/q is not the upper.end or lower.end
    of the positive rationals.

    for each point x situating a split F,H
    of the positive axis,
    there is a rational < x in F
    and a rational > x in H
    and x is not the upper.end or lower.end
    of the positive rationals.

    s/the positive rationals/the positive axis

    No other points are in the positive axis.
    No points not.in are ends.



    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Tue Aug 13 14:34:50 2024
    XPost: sci.math

    On 8/13/2024 10:17 AM, WM wrote:
    Le 12/08/2024 à 19:23, Richard Damon a écrit :
    On 8/12/24 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Which,
    by the definition of the Natural Numbers,
    doesn't exist.

    The end of the positive axis exists.

    No point in the positive axis is
    largest in the positive axis or
    smallest in the positive axis.
    A point not.largest and not.smallest
    is not an end.

    No point in the positive axis is
    an upper.end or a lower.end.

    If an end of the positive exists,
    it is in the positive axis.
    A bound not.in is not an end.

    No point not.in the positive axis is
    an upper.end or a lower.end.

    No point is
    an upper.end or a lower.end.

    The end of the positive axis
    does not exist.

    ----
    In the land of rationals only with
    countable.to numerators and denominators
    and with each split situated
    ⎛ a last point in the foresplit or
    ⎝ a first point in the hindsplit,
    no point in the positive axis is
    largest in the positive axis or
    smallest in the positive axis.

    for each positive rational p/q
    p > 0, q > 0
    (p+1)/q > p/q > p/(q+1) > 0

    for each positive rational p/q
    p/q is not the upper.end or lower.end
    of the positive rationals.

    for each point x situating a split F,H
    of the positive axis,
    there is a rational < x in F
    and a rational > x in H
    and x is not the upper.end or lower.end
    of the positive rationals.

    No other points are in the positive axis.
    No points not.in are ends.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Tue Aug 13 22:29:07 2024
    XPost: sci.math

    Am 13.08.2024 um 21:25 schrieb Chris M. Thomasson:

    It depends on context. Are we going from 1/1 all the way down, or from
    0 all the way up to 1/1.

    Again, 1/1 is the first term in the sequence (1/1, 1/2, 1/3, ...), but
    it's the largest/last unit fraction in respect to < (as defined on IR,
    and hence on the unit fractions).

    ... < 1/3 < 1/2 < 1/1.

    0 is not a unit fraction, which means there is no smallest one... Fair enough?

    Right. There is no smallest unit fraction (in respect to < as defined on
    the IR).

    Proof: If u is a unit fraction, 1/(1/u + 1) is a smaller one.

    Using symbols: Au e {1/n : n e IN}: 1/(1/u + 1) e {1/n : n e IN} &
    1/(1/u + 1) < u.

    Or with the definition:

    1/IN := {1/n : n e IN}
    just:
    Au e 1/IN: 1/(1/u + 1) e 1/IN & 1/(1/u + 1) < u.

    This implies:

    Au e 1/IN: Eu' e 1/IN: u' < u.

    "For each and every unit fraction, there is a smaller one."

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Tue Aug 13 22:16:27 2024
    XPost: sci.math

    Am 13.08.2024 um 21:21 schrieb Chris M. Thomasson:
    On 8/13/2024 7:27 AM, WM wrote:

    ... the smallest unit fractions ...

    This irks me a bit. The smallest unit fractions? How do you define them?

    Consider a totally ordered set M, say ordered by <=.

    Then if there is an element m in M such that for all elements x in M: m
    <= x, we call that element _the_ /smallest/ element in m. (Reason: There
    can only be ONE such element in M, if there is such an Element in M at all.)

    By "the smallest unit fractions" WM is referring to the smalles unit
    fraction u_WM and some of the unit fractions which are slightly larger:

    u_WM < u_WM' < u_WM'' < ...

    Though it is not quite clear how many of these unit fractions are
    considered "smallest unit fractions" by WM. The first 2, 3? Who knows?!

    Mückenheim's World is a strange place.

    Especialy, since there is not even ONE smallest unit fraction in
    mathematics.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Chris M. Thomasson on Tue Aug 13 17:54:22 2024
    XPost: sci.math

    On 8/13/2024 3:21 PM, Chris M. Thomasson wrote:
    On 8/13/2024 7:27 AM, WM wrote:
    Le 12/08/2024 à 20:34, Jim Burns a écrit :
    On 8/12/2024 9:47 AM, WM wrote:

    Before the bound there is the end,
    perhaps dark.

    0 is
    greatest.lower.bound β of visibleᵂᴹ unit.fractions

    Yes. But the smallest unit fractions comes before.

    This irks me a bit.
    The smallest unit fractions?
    How do you define them?
    What if
    your version of "small" is _larger_ than
    somebody else's version of "small"?
    Is hyper small smaller than super small?

    "It depends".

    Note that super.duper small is
    between hyper.small and super.small.

    OMG.small and WTF.small are left as
    an exercise for the reader.

    I think that
    delta.epsilonics finesses the question
    by first picking what 'small' is, ε > 0
    then defining
    what continuity and its ilk need to be
    in relation to _the picked small_ ε > 0

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Wed Aug 14 00:09:31 2024
    XPost: sci.math

    Am 13.08.2024 um 23:54 schrieb Jim Burns:
    On 8/13/2024 3:21 PM, Chris M. Thomasson wrote:
    On 8/13/2024 7:27 AM, WM wrote:
    Le 12/08/2024 à 20:34, Jim Burns a écrit :
    On 8/12/2024 9:47 AM, WM wrote:

    Before the bound there is the end,
    perhaps dark.

    0 is
    greatest.lower.bound β of visibleᵂᴹ unit.fractions

    Yes. But the smallest unit fractions comes before.

    This irks me a bit.
    The smallest unit fractions?
    How do you define them?
    What if
    your version of "small" is _larger_ than
    somebody else's version of "small"?
    Is hyper small smaller than super small?

    "It depends".

    Note that super.duper small is
    between hyper.small and super.small.

    OMG.small and WTF.small are left as
    an exercise for the reader.

    You've forgotten about "extra small" (XS).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Aug 14 12:09:19 2024
    XPost: sci.math

    Le 13/08/2024 à 18:26, Jim Burns a écrit :
    On 8/13/2024 10:27 AM, WM wrote:
    Le 12/08/2024 à 20:34, Jim Burns a écrit :
    On 8/12/2024 9:47 AM, WM wrote:

    Before the bound there is the end,
    perhaps dark.

    0 is
    greatest.lower.bound β of visibleᵂᴹ unit.fractions

    Yes.

    Thank you.

    But the smallest unit fractions comes before.

    No visibleᵂᴹ.or.darkᵂᴹ point ε > 0 is the end of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    Otherwise,
    ⎛ ε > 0 is a lower.bound of
    ⎜ both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    ⎜ ε > 0 is a lower.bound of
    ⎜ visibleᵂᴹ unit.fractions.

    ⎜ 0 is the greatest.lower.bound of
    ⎜ visibleᵂᴹ unit.fractions.

    ⎜ ε is a lower.bound greater than the greatest.
    ⎝ Contradiction.


    You (WM) don't seem to want it, but
    we could stretch the definition of darkᵂᴹ so that
    ε > 0 is the end of darkᵂᴹ but not visibleᵂᴹ
    ( darkᵂᴹ in the cracks between some visibleᵂᴹ?
    However, such an ε > 0 would still leave
    the unit.fractions one.ended.

    ----
    0 is the greatest.lower.bound of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    0 isn't a unit.fraction.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Aug 14 12:14:21 2024
    XPost: sci.math

    Le 13/08/2024 à 18:26, Jim Burns a écrit :
    On 8/13/2024 10:27 AM, WM wrote:

    No visibleᵂᴹ.or.darkᵂᴹ point ε > 0 is the end of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    Every ε > 0 is visible, because it is chosen.

    But ℵo unit fractions occupy an interval y > 0 to settle there. Not
    every point of y has ℵo smaller unit fractions. Every chosen ε > 0 is
    larger than this interval y.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Aug 14 12:17:56 2024
    XPost: sci.math

    Le 13/08/2024 à 18:35, Moebius a écrit :
    Am 13.08.2024 um 18:26 schrieb Jim Burns:
    On 8/13/2024 10:27 AM, WM wrote:

    the smallest unit fractions

    only exist

    It is dark but it exists.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Aug 14 12:26:27 2024
    XPost: sci.math

    Le 13/08/2024 à 19:32, Moebius a écrit :
    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:
    Le 12/08/2024 à 19:44, Jim Burns a écrit :

    There is no ⅟nₓ before the end of the positive axis
    without ⅟(nₓ+1) before the end of the positive axis.

    You cannot see it. It is dark.

    No number can be seen, dark or not.

    Numbers can be seen.

    resulting in a real coordinate x with NUF(x) = 1.

    Assume NUF(x0) = 1 with x0 e IR. This means that there is exactly one
    unit fraction u such that u < x0. Let's call this unit fraction u0. Then
    (by definition) there is a (actually exactly one) natural number n such
    that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 + 1) is a unit fraction which is smaller than u0 and
    hence smaller than x0. Hence NUF(x0) > 1. Contradiction!

    Therefore 1/(n0 + 1) does not exist.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Aug 14 12:23:42 2024
    XPost: sci.math

    Le 13/08/2024 à 19:02, Jim Burns a écrit :
    On 8/13/2024 10:21 AM, WM wrote:

    the step size can only be 1,

    ...at a unit.fraction.
    0 isn't a unit.fraction.

    Therefore there is no step at 0.

    resulting in a real coordinate x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    No.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Aug 14 12:28:09 2024
    XPost: sci.math

    Le 13/08/2024 à 19:42, Jim Burns a écrit :
    On 8/13/2024 12:35 PM, Moebius wrote:

    The existence of the smallest unit fractions
    is contradictory in the land of
    rationals with
    countable.to numerators and denominators
    with each split situated ==
    a last point in the foresplit or
    a first point in the hindsplit.

    The existence of a smallest unit fraction is the only alternative to the existence of more than one at a real point.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Wed Aug 14 12:32:41 2024
    XPost: sci.math

    Le 13/08/2024 à 19:51, Moebius a écrit :
    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    either in a step of size 1
    or in a step of size more than 1.

    Let's "assume" that this is true (sort of) for the sake of the argument.

    But increase by more than 1 is excluded by
    the gaps between unit fractions.

    I can't see any argument for this claim.

    Deplorable.

    Actually, for each and every real number x > 0 there are infinitely many
    unit fractions smaller than x:

    That is wrong because ℵo unit fractions occupy a nonvanishig real
    interval, call it y. Then not every point of y has infinitely many unit fractions smaller than it self.

    In fact, Ax > 0: NUF(x) = aleph_0, while NUF(0) = 0.

    A very stupid statement. It claims unit fractions smaller than every
    positive x. "Every positive x" is tantamount to the interval (0, oo).

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 14 12:48:14 2024
    XPost: sci.math

    Le 13/08/2024 à 21:10, "Chris M. Thomasson" a écrit :
    On 8/13/2024 7:17 AM, WM wrote:
    Le 12/08/2024 à 19:23, Richard Damon a écrit :
    On 8/12/24 9:50 AM, WM wrote:

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Which, by the definition of the Natural Numbers, doesn't exist.

    The end of the positive axis exists.

    Nope.

    It exists and is an element of the open interval (0, oo).

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Wed Aug 14 12:34:31 2024
    XPost: sci.math

    Le 13/08/2024 à 20:34, Jim Burns a écrit :
    On 8/13/2024 10:17 AM, WM wrote:
    Le 12/08/2024 à 19:23, Richard Damon a écrit :
    On 8/12/24 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Which,
    by the definition of the Natural Numbers,
    doesn't exist.

    The end of the positive axis exists.

    No point in the positive axis is
    largest in the positive axis or
    smallest in the positive axis.

    What is smaller than every positive x is smaller than the interval (0,
    oo).

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Wed Aug 14 12:43:47 2024
    XPost: sci.math

    On 8/14/2024 8:23 AM, WM wrote:
    Le 13/08/2024 à 19:02, Jim Burns a écrit :
    On 8/13/2024 10:21 AM, WM wrote:

    The function NUF(x) is a step-function.
    It can increase from 0 at x = 0 to greater values,
    either in a step of size 1
    or in a step of size more than 1.
    But increase by more than 1 is excluded by
    the gaps between unit fractions.
    (Note the universal quantifier there,
    according to which never –
    in no limit and in no accumulation point –
    two unit fractions occupy the same point x.)
    Therefore
    the step size can only be 1,

    ...at a unit.fraction.
    0 isn't a unit.fraction.

    Therefore there is no step at 0.

    There is a step at 0.

    Each unit.fractionᵈᵉᶠ has GLB β > 0
    which other unit.fractionsᵈᵉᶠ are
    at least as far as.

    However,
    0 is not a unit.fractionᵈᵉᶠ.
    0 does not have GLB β > 0
    which unit.fractionsᵈᵉᶠ are
    at least as far as.
    ⎛ Otherwise,
    ⎝ ½⋅β is lower.bound and not.lower.bound.

    ∀ᴿx > 0: NUFᵈᵉᶠ(x) > 0

    ∀ᴿx > 0: NUF(x) ≥ NUF(x)ᵈᵉᶠ > 0

    There is a step at 0.


    ⎛ Otherwise,
    ⎝ ½⋅β is lower.bound and not.lower.bound.

    ⎜ Assume otherwise.
    ⎜ Assume β > 0 for
    ⎜ β greatest.lower.bound of unit.factionsᵈᵉᶠ

    ⎜ ½⋅β < β lower.bound

    ⎜ 2⋅β > β not.lower.bound
    ⎜ 2⋅β > ⅟k smaller unit.fractionᵈᵉᶠ
    ⎜ ¼⋅2⋅β > ¼⋅⅟k smaller unit.fractionᵈᵉᶠ
    ⎜ ¼⋅2⋅β = ½⋅β
    ⎜ ½⋅β > ¼⋅⅟k not.lower.bound

    ⎝ ½⋅β is lower.bound and not.lower.bound.

    resulting in a real coordinate x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
    NUF(INVNUF(1)) > 1

    No.

    Is "No" an argument?

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Wed Aug 14 13:35:17 2024
    XPost: sci.math

    On 8/14/2024 8:14 AM, WM wrote:
    Le 13/08/2024 à 18:26, Jim Burns a écrit :
    On 8/13/2024 10:27 AM, WM wrote:
    Le 12/08/2024 à 20:34, Jim Burns a écrit :
    On 8/12/2024 9:47 AM, WM wrote:

    Before the bound there is the end,
    perhaps dark.

    0 is
    greatest.lower.bound β of visibleᵂᴹ unit.fractions

    Yes.

    But the smallest unit fractions comes before.

    No visibleᵂᴹ.or.darkᵂᴹ point ε > 0  is the end of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    Every ε > 0 is visible, because it is chosen.

    No visibleᵂᴹ point ε > 0 is the end of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    No darkᵂᴹ point ε > 0 is the end of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    There is no lower.end of
    both visibleᵂᴹ and darkᵂᴹ unit.fractions.

    But ℵo unit fractions occupy an interval y > 0
    to settle there.

    No.
    0.many unit.fractions settle.
    ⎛ Related:
    ⎝ The intersection of infinite.end.segments is empty.

    For each unit fraction ⅟k
    there is an initial sub.segment of (0,1]
    which ⅟k is not.in.

    There are ℵ₀ unit.fractions in the ⅟k.free segment,
    but not ⅟k

    Not every point of y has ℵo smaller unit fractions.

    No.
    ∀ᴿy > 0: ∀k ∈ ℕ₁: 0 < ⅟⌊k+⅟y⌋ < y

    Every chosen ε > 0 is larger than this interval y.

    No.
    ¬∃ᴿy > 0: ¬∀k ∈ ℕ₁: 0 < ⅟⌊k+⅟y⌋ < y

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Wed Aug 14 14:04:22 2024
    XPost: sci.math

    On 8/14/2024 8:28 AM, WM wrote:
    Le 13/08/2024 à 19:42, Jim Burns a écrit :

    The existence of the smallest unit fractions
    is contradictory in the land of
    rationals with
    countable.to numerators and denominators
    with each split situated ==
    a last point in the foresplit or
    a first point in the hindsplit.

    The existence of a smallest unit fraction is
    the only alternative to the existence of
    more than one at a real point.

    The NONexistence of a smallest unit fraction is why,
    for each unit fraction,
    there are infinitely.many smaller unit fractions.
    And with no two at one point.

    In a finite ordered set,
    each non.{} subset has two ends.

    For each unit fraction,
    its smaller unit fractions have an upper end,
    which is that unit fraction,
    but a lower end doesn't exist,
    the smallest unit fraction doesn't exist.

    For each unit fraction,
    its smaller unit fractions are one.ended,
    they _aren't_ finitely.many.

    ∀ᴿx > 0: NUF(x) = ℵ₀
    because each x > 0 is not above a second end.

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Wed Aug 14 16:04:43 2024
    XPost: sci.math

    On 8/14/2024 8:34 AM, WM wrote:
    Le 13/08/2024 à 20:34, Jim Burns a écrit :
    On 8/13/2024 10:17 AM, WM wrote:
    Le 12/08/2024 à 19:23, Richard Damon a écrit :
    On 8/12/24 9:50 AM, WM wrote:
    Le 11/08/2024 à 19:56, Jim Burns a écrit :

    What causes an exception: nₓ ∈ ℕ:
    ⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?

    The end of the positive axis.

    Which,
    by the definition of the Natural Numbers,
    doesn't exist.

    The end of the positive axis exists.

    No point in the positive axis is
    largest in the positive axis or
    smallest in the positive axis.

    What is smaller than every positive x
    is smaller than the interval (0, oo).

    That which is not.in the interval (0,∞)
    is not an end of (0,∞).

    ----
    No point in the positive axis is
    largest in the positive axis or
    smallest in the positive axis.

    No countable.to number is largest countable.to.

    No rat. > 0 with countable.to num. and den.
    is largest rat. > 0 with countable.to num. and den.
    or smallest rat. > 0 with countable.to num. and den.
    ∀p/q ∈ ℚ⁺: ℚ⁺ ∋ p/(q+1) < p/q < (p+1)/q ∈ ℚ⁺

    No (0,∞).split F,H has a situating.point x
    (x last.in.F or x first.in.H) such that
    x is first in F∪H = (0,∞) or
    x is last in F∪H = (0,∞)

    No rational > 0 with countable.to num. and den.
    and no situating point of any (0,∞).split
    is an end of (0,∞)
    There are no other points in (0,∞)

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Thu Aug 15 01:04:13 2024
    XPost: sci.math

    Am 14.08.2024 um 18:43 schrieb Jim Burns:
    On 8/14/2024 8:23 AM, WM wrote:

    No.

    Is "No" an argument?

    Mr. Mückenheim does not have better arguments.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Thu Aug 15 01:28:24 2024
    XPost: sci.math

    Am 15.08.2024 um 01:26 schrieb Moebius:
    Am 14.08.2024 um 21:04 schrieb Chris M. Thomasson:
    On 8/13/2024 3:09 PM, Moebius wrote:
    Am 13.08.2024 um 23:54 schrieb Jim Burns:
    On 8/13/2024 3:21 PM, Chris M. Thomasson wrote:
    On 8/13/2024 7:27 AM, WM wrote:

    [...] the smallest unit fractions comes before [0].

    This irks me a bit.
    The smallest unit fractions?
    How do you define them?
    What if
    your version of "small" is _larger_ than
    somebody else's version of "small"?
    Is hyper small smaller than super small?

    "It depends".

    Note that super.duper small is
    between hyper.small and super.small.

    OMG.small and WTF.small are left as
    an exercise for the reader.

    You've forgotten about "extra small" (XS).

    oh my... Humm... What about the good ol' Super Extra Extra Small?
    (SEES)? ;^D

    Did you mean the good ol' Super Extra [small] (SEX)?

    SEX.small.

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Thu Aug 15 01:26:14 2024
    XPost: sci.math

    Am 14.08.2024 um 21:04 schrieb Chris M. Thomasson:
    On 8/13/2024 3:09 PM, Moebius wrote:
    Am 13.08.2024 um 23:54 schrieb Jim Burns:
    On 8/13/2024 3:21 PM, Chris M. Thomasson wrote:
    On 8/13/2024 7:27 AM, WM wrote:

    [...] the smallest unit fractions comes before [0].

    This irks me a bit.
    The smallest unit fractions?
    How do you define them?
    What if
    your version of "small" is _larger_ than
    somebody else's version of "small"?
    Is hyper small smaller than super small?

    "It depends".

    Note that super.duper small is
    between hyper.small and super.small.

    OMG.small and WTF.small are left as
    an exercise for the reader.

    You've forgotten about "extra small" (XS).

    oh my... Humm... What about the good ol' Super Extra Extra Small?
    (SEES)? ;^D

    Did you mean the good ol' Super Extra [small] (SEX)?

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Thu Aug 15 14:43:07 2024
    XPost: sci.math

    Am 13.08.2024 um 19:02 schrieb Jim Burns:

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    How is INVNUF defined? (Since you are using it too, you schould know its definition, right?)

    WM is now referring to INVNUF in dsm: He claims that INVNUF(1) and
    INVNUF(2) are two different unit fractions.

    Can you second that claim?

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 15 13:52:11 2024
    XPost: sci.math

    Le 14/08/2024 à 18:43, Jim Burns a écrit :
    On 8/14/2024 8:23 AM, WM wrote:


    Therefore there is no step at 0.

    There is a step at 0.

    Nonsense.

    However,
    0 is not a unit.fractionᵈᵉᶠ.

    Therefore there is no step. NUF(x) steps only at unit fractions x.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 15 13:55:13 2024
    XPost: sci.math

    Le 14/08/2024 à 20:04, Jim Burns a écrit :
    On 8/14/2024 8:28 AM, WM wrote:
    Le 13/08/2024 à 19:42, Jim Burns a écrit :

    The existence of the smallest unit fractions
    is contradictory in the land of
    rationals with
    countable.to numerators and denominators
    with each split situated ==
    a last point in the foresplit or
    a first point in the hindsplit.

    The existence of a smallest unit fraction is
    the only alternative to the existence of
    more than one at a real point.

    The NONexistence of a smallest unit fraction is why,
    for each unit fraction,
    there are infinitely.many smaller unit fractions.
    And with no two at one point.

    That is a self-contradiction.

    The first point with unit fractions is x = INVNUF(1).

    Regrads, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 15 13:57:34 2024
    XPost: sci.math

    Le 14/08/2024 à 21:01, "Chris M. Thomasson" a écrit :
    On 8/14/2024 5:28 AM, WM wrote:

    The existence of a smallest unit fraction is the only alternative to the
    existence of more than one at a real point.

    There is no smallest unit fraction,

    There is a smallest one because all are separated:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 15 13:59:49 2024
    XPost: sci.math

    Le 14/08/2024 à 22:04, Jim Burns a écrit :
    On 8/14/2024 8:34 AM, WM wrote:

    What is smaller than every positive x
    is smaller than the interval (0, oo).

    That which is not.in the interval (0,∞)
    is not an end of (0,∞).

    The smallest point of (0, oo) belongs to (0, oo).
    It cannot be seen. It is dark.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to Moebius on Thu Aug 15 11:53:11 2024
    XPost: sci.math

    On 8/15/2024 8:43 AM, Moebius wrote:
    Am 13.08.2024 um 19:02 schrieb Jim Burns:

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    How is INVNUF defined?
    (Since you are using it too,
    you should know its definition, right?)

    When I first saw WM use 'INVNUF'
    he said something that meant
    INVNUF(n) = _first_ x with NUF(x) = n

    When I use INVNUF(n) I only assume that
    INVNUF(n) = x with NUF(x) = n
    That's why I use ⌊x⌋ = floor(x)

    'Any x' still falls within WM's definition and,
    in our little whack.a.mole game,
    whacks more moles.

    Elsethread: <WM>
    The first point with unit fractions is x = INVNUF(1).
    </WM>
    Date: Thu, 15 Aug 24 13:55:13 +0000

    WM is now referring to INVNUF in dsm:
    He claims that INVNUF(1) and INVNUF(2)
    are two different unit fractions.

    Can you second that claim?

    Do you think I might second it,
    when I've just proved INVNUF(1) not.exists?

    You asked about a definition.
    A definition isn't a claim something exists.
    A definition is a claim how a word is used.

    ( p/q = √2 in lowest terms
    is not
    a claim that √2 is rational.
    It is part of proving √2 is irrational.
    It, with the proof, is a nonexistence claim for p/q

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Thu Aug 15 12:16:06 2024
    XPost: sci.math

    On 8/15/2024 9:52 AM, WM wrote:
    Le 14/08/2024 à 18:43, Jim Burns a écrit :
    On 8/14/2024 8:23 AM, WM wrote:

    Therefore there is no step at 0.

    There is a step at 0.

    Nonsense.

    ⎛ Assume otherwise.
    ⎜ Assume NUF(x) = 0 and x > 0

    ⎜ β ≥ x > 0 for
    ⎜ β greatest.lower.bound of unit.fractionsᵈᵉᶠ
    ⎜ 2⋅β > β ⇒ not.lower.bound 2⋅β
    ⎜ ½⋅β < β ⇒ lower.bound ½⋅β

    ⎜ 2⋅β > ⅟k ⇐ not.lower.bound 2⋅β
    ⎜ ¼⋅2⋅β > ¼⋅⅟k
    ⎜ ¼⋅2⋅β = ½⋅β
    ⎜ ½⋅β > ¼⋅⅟k ⇒ not.lower.bound ½⋅β

    ⎜ lower.bound ½⋅β and not.lower.bound ½⋅β
    ⎝ Contradiction.

    However,
    0 is not a unit.fractionᵈᵉᶠ.

    Therefore there is no step.
    NUF(x) steps only at unit fractions x.

    If NUF(x) doesn't step at 0
    then lower.bound ½⋅β and not.lower.bound ½⋅β

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Thu Aug 15 18:38:30 2024
    XPost: sci.math

    Am 15.08.2024 um 17:53 schrieb Jim Burns:
    On 8/15/2024 8:43 AM, Moebius wrote:
    Am 13.08.2024 um 19:02 schrieb Jim Burns:


    When I first saw WM use 'INVNUF'
    he said something that meant
    INVNUF(n) = _first_ x with NUF(x) = n

    This "definition" is lacking the domain of n.

    Hence it's technically not a proper definition i.e. not a definition at
    all).

    In fact, if n e IN = {1, 2, 3, ...} then there IS NO _first_ x e IR with
    NUF(x) = n (after all in this case there is no x e IR with NUF(x) = n AT
    ALL).

    With the knowledge that img(NUF) = {0, aleph_0} we may consider if
    there's a _first_ x e IR with NUF(x) = aleph_0. But no, there isn't such
    an x e IR.

    Finally, we may consider if there's a _first_ x e IR with NUF(x) = 0.
    And yes, there is such an x e IR, namely 0.

    So the domain of that "function" is just {0}.

    Hence a proper definition of it would have to be stated the following
    (or a similar or equivalent) way:

    INVNUF(n) = _first_ x with NUF(x) = n (n e {0}).

    So the only value for which INVNUF is defined is 0 (i.e. dom(INVNUF) = {0}).

    That's why you CAN'T use "INVNUF(1)" in any of your proofs. Since in
    this case you would use an undefined "term". And no, you can't just
    "assume that INVNUF(1) exists" - that's just mumbo-jumbo, but not math.

    I've just proved INVNUF(1) not.exists?

    Mumbo-Jumbo.*)

    Hint: You can't prove that there is no x e IR such that x = 1/0.

    Beacuse you can't even USE "1/0" in the statement you want to prove
    _just because_ the symbol "is undefined".

    ________________________________________________

    *) What you may try to prove is: 1 !e dom(INVNUF). But you can't do that
    by using "INVNUF(1)" in your proof.

    How about: Assume that 1 e dom(INVNUF), then 1 e {0} (since {0} is the
    domain of INVNUF). But then 1 = 0. Contradiction!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 18:43:39 2024
    XPost: sci.math

    Am 15.08.2024 um 17:53 schrieb Jim Burns:
    On 8/15/2024 8:43 AM, Moebius wrote:
    Am 13.08.2024 um 19:02 schrieb Jim Burns:


    When I first saw WM use 'INVNUF'
    he said something that meant
    INVNUF(n) = _first_ x with NUF(x) = n

    This "definition" is lacking the "domain" of n (i.e. the values it can
    take).

    Hence it's technically not a proper definition i.e. not a definition at
    all).

    In fact, if n e IN = {1, 2, 3, ...} then there IS NO _first_ x e IR with
    NUF(x) = n (after all in this case there is no x e IR with NUF(x) = n AT
    ALL).

    With the knowledge that img(NUF) = {0, aleph_0} we may consider if
    there's a _first_ x e IR with NUF(x) = aleph_0. But no, there isn't such
    an x e IR.

    Finally, we may consider if there's a _first_ x e IR with NUF(x) = 0.
    And yes, there is such an x e IR, namely 0.

    So the domain of that "function" is just {0}.

    Hence a proper definition of it would have to be stated the following
    (or a similar or equivalent) way:

    INVNUF(n) = _first_ x with NUF(x) = n (n e {0}).

    So the only value for which INVNUF is defined is 0 (i.e. dom(INVNUF) = {0}).

    That's why you CAN'T use "INVNUF(1)" in any of your proofs. Since in
    this case you would use an undefined "term". And no, you can't just
    "assume that INVNUF(1) exists" - that's just mumbo-jumbo, but not math.

    I've just proved INVNUF(1) not.exists?

    Mumbo-Jumbo.*)

    Hint: You can't prove that there is no x e IR such that x = 1/0.

    Beacuse you can't even USE "1/0" in the statement you want to prove
    _just because_ the symbol "is undefined".

    ________________________________________________

    *) What you may try to prove is: 1 !e dom(INVNUF). But you can't do that
    by using "INVNUF(1)" in your proof.

    How about: Assume that 1 e dom(INVNUF), then 1 e {0} (since {0} is the
    domain of INVNUF). But then 1 = 0. Contradiction!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 18:44:03 2024
    XPost: sci.math

    Am 15.08.2024 um 17:53 schrieb Jim Burns:
    On 8/15/2024 8:43 AM, Moebius wrote:
    Am 13.08.2024 um 19:02 schrieb Jim Burns:


    When I first saw WM use 'INVNUF'
    he said something that meant
    INVNUF(n) = _first_ x with NUF(x) = n

    This "definition" is lacking the "domain" of n (i.e. the values it may
    take).

    Hence it's technically not a proper definition i.e. not a definition at
    all).

    In fact, if n e IN = {1, 2, 3, ...} then there IS NO _first_ x e IR with
    NUF(x) = n (after all in this case there is no x e IR with NUF(x) = n AT
    ALL).

    With the knowledge that img(NUF) = {0, aleph_0} we may consider if
    there's a _first_ x e IR with NUF(x) = aleph_0. But no, there isn't such
    an x e IR.

    Finally, we may consider if there's a _first_ x e IR with NUF(x) = 0.
    And yes, there is such an x e IR, namely 0.

    So the domain of that "function" is just {0}.

    Hence a proper definition of it would have to be stated the following
    (or a similar or equivalent) way:

    INVNUF(n) = _first_ x with NUF(x) = n (n e {0}).

    So the only value for which INVNUF is defined is 0 (i.e. dom(INVNUF) = {0}).

    That's why you CAN'T use "INVNUF(1)" in any of your proofs. Since in
    this case you would use an undefined "term". And no, you can't just
    "assume that INVNUF(1) exists" - that's just mumbo-jumbo, but not math.

    I've just proved INVNUF(1) not.exists?

    Mumbo-Jumbo.*)

    Hint: You can't prove that there is no x e IR such that x = 1/0.

    Beacuse you can't even USE "1/0" in the statement you want to prove
    _just because_ the symbol "is undefined".

    ________________________________________________

    *) What you may try to prove is: 1 !e dom(INVNUF). But you can't do that
    by using "INVNUF(1)" in your proof.

    How about: Assume that 1 e dom(INVNUF), then 1 e {0} (since {0} is the
    domain of INVNUF). But then 1 = 0. Contradiction!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 18:57:19 2024
    XPost: sci.math

    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    [There is] a real [number] x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    Well, no, this just isn't a proof, sorry about that, Jim.

    (Hint: The term "INVNUF(1)" is not defined. Hence you may not even use
    it in a proof by contradiction. Actually, you didn't state an assumption
    in your "proof".)

    Proof by contradiction:

    Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such
    that NUF(x0) = 1. This means that there is exactly one unit fraction u
    such that u < x0. Let's call this unit fraction u0. Then (by definition)
    there is a (actually exactly one) natural number n such that u0 = 1/n.
    Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 +
    1) is a unit fraction which is smaller than u0 and hence smaller than
    x0. Hence NUF(x0) > 1. Contradiction!

    (Of course, it's clear that I'm using the same "proof idea" as you used
    in your attempt of a proof.)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Python@21:1/5 to All on Thu Aug 15 19:06:15 2024
    XPost: sci.math

    Le 15/08/2024 à 19:01, Moebius a écrit :
    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    [There is] a real [number] x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    Well, no, this just isn't a proof, sorry about that, Jim.

    (Hint: The term "INVNUF(1)" is not defined. Hence you may not even use
    it in a proof by contradiction. Actually, you didn't state an assumption
    in your "proof".)

    Proof by contradiction:

    Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such
    that NUF(x0) = 1. This means that there is exactly one unit fraction u
    such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n.
    Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 +
    1) is an unit fraction which is smaller than u0 and hence smaller than
    x0. Hence NUF(x0) > 1. Contradiction!

    (Of course, it's clear that I'm using the same "proof idea" as you used
    in your attempt of a proof.)

    Of course you are right. 100% right. Any marginally decent high school
    student could sketch up the very same proof you've posted.

    Our problem, as a community, is not proving Wolfgang Mückenheim wrong.

    Our problem, as a community, is that this crook is actually teaching
    in an academic institution.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 19:01:42 2024
    XPost: sci.math

    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    [There is] a real [number] x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    Well, no, this just isn't a proof, sorry about that, Jim.

    (Hint: The term "INVNUF(1)" is not defined. Hence you may not even use
    it in a proof by contradiction. Actually, you didn't state an assumption
    in your "proof".)

    Proof by contradiction:

    Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such
    that NUF(x0) = 1. This means that there is exactly one unit fraction u
    such that u < x0. Let's call this unit fraction u0. Then (by definition)
    there is a (actually exactly one) natural number n such that u0 = 1/n.
    Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 +
    1) is an unit fraction which is smaller than u0 and hence smaller than
    x0. Hence NUF(x0) > 1. Contradiction!

    (Of course, it's clear that I'm using the same "proof idea" as you used
    in your attempt of a proof.)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 19:34:10 2024
    XPost: sci.math

    Am 15.08.2024 um 19:01 schrieb Moebius:
    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    [There is] a real [number] x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    Well, no, this just isn't a proof, sorry about that, Jim.

    (Hint: The term "INVNUF(1)" is not defined. Hence you may not even use
    it in a proof by contradiction. Actually, you didn't state an assumption
    in your "proof".)

    Proof by contradiction:

    Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such
    that NUF(x0) = 1. This means that there is exactly one unit fraction u
    such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n.
    Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 +
    1) is an unit fraction which is smaller than u0 and hence smaller than
    x0. Hence NUF(x0) > 1. Contradiction!

    (Of course, it's clear that I'm using the same "proof idea" as you used
    in your attempt of a proof.)

    Ok, I'll give it a try.

    Assume that there is an x e IR such that NUF(x) = 1.

    May we (by using this assumption) define INVNUF in such a way that 1 is
    in its domain?

    I mean can we define

    INVNUF(n) = _first_ x with NUF(x) = n for n e {1, ...}

    now? (This would allow to use the term INVNUF(1) in our proof.)

    For this we would have to show/prove that there is a _first_ x with
    NUF(x) = 1. And for this we would have to show that (a) there is an x e
    IR such that NUF(x) = 1 (which is our assumption, so nothing to do here)
    and (b) that there is no smaller real number x' such that NUF(x') = 1.
    But I DOUBT that we will be able to prove/show that (even with our
    assumption).

    So I consider this a dead end.

    ---------------------------------------------------------------------

    Oh, wait! From our assumtion we can derive a contradiction (see my proof above)! Hence we can even prove (b) now (in the context of this proof by contradiction)! For that we assume ~(b) and immediately get ~~(b) and
    hence (b) from the assumption ~(b) and our original assumption, by contradiction. (The assumption ~(b) is now "discharged").

    Now we can define INVNUF such that 1 is in its domain. And hence now we
    may use your line of thought:

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    And hence we finally get: there is no x e IR such that NUF(x) = 1. qed

    Well...

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 19:35:12 2024
    XPost: sci.math

    Am 15.08.2024 um 19:06 schrieb Python:

    Our problem, as a community, is that this crook is actually teaching
    in an academic institution.

    Yes, it's a shame!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 19:31:01 2024
    XPost: sci.math

    Am 15.08.2024 um 19:01 schrieb Moebius:
    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    [There is] a real [number] x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    Well, no, this just isn't a proof, sorry about that, Jim.

    (Hint: The term "INVNUF(1)" is not defined. Hence you may not even use
    it in a proof by contradiction. Actually, you didn't state an assumption
    in your "proof".)

    Proof by contradiction:

    Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such
    that NUF(x0) = 1. This means that there is exactly one unit fraction u
    such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n.
    Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 +
    1) is an unit fraction which is smaller than u0 and hence smaller than
    x0. Hence NUF(x0) > 1. Contradiction!

    (Of course, it's clear that I'm using the same "proof idea" as you used
    in your attempt of a proof.)

    Ok, I'll give it a try.

    Assume that there is an x e IR such that NUF(x) = 1.

    May we (by using this assumption) define INVNUF in such a way that 1 is
    in its domain?

    I mean can we define

    INVNUF(n) = _first_ x with NUF(x) = n for n e {1, ...}

    now? (This would allow to use the term INVNUF(1) in our proof.)

    For this we would have to show/prove that there is a _first_ x with
    NUF(x) = 1. And for this we would have to show that (a) there is an x e
    IR such that NUF(x) = 1 (which is our assumption, so nothing to do here)
    and (b) that there is no smaller real number x' such that NUF(x') = 1.
    But I DOUBT that we will be able to prove/show that (even with our
    assumption).

    So I consider this a dead end.

    ---------------------------------------------------------------------

    Oh, wait! From our assumtion we can derive a contradiction (see my proof above)! Hence we can even prove (b) now (in the context of this proof by contradiction)! For that we assume ~(b) and immediately get ~~(b) and
    hence (b) from the assumption ~(b) and our original assumption, by contradiction. (The assumption ~(b) is no "discharged").

    Now we can define INVNUF such that 1 is in its domain. And hence now we
    may use your line of thought:

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    And hence we finally get: there is no x e IR such that NUF(x) = 1. qed

    Well...

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 19:41:22 2024
    XPost: sci.math

    Am 15.08.2024 um 19:34 schrieb Moebius:
    Am 15.08.2024 um 19:01 schrieb Moebius:
    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    [There is] a real [number] x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    Well, no, this just isn't a proof, sorry about that, Jim.

    (Hint: The term "INVNUF(1)" is not defined. Hence you may not even use
    it in a proof by contradiction. Actually, you didn't state an
    assumption in your "proof".)

    Proof by contradiction:

    Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such
    that NUF(x0) = 1. This means that there is exactly one unit fraction u
    such that u < x0. Let's call this unit fraction u0. Then (by
    definition) there is a (actually exactly one) natural number n such
    that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by
    definition) 1/(n0 + 1) is an unit fraction which is smaller than u0
    and hence smaller than x0. Hence NUF(x0) > 1. Contradiction!

    (Of course, it's clear that I'm using the same "proof idea" as you
    used in your attempt of a proof.)

    Ok, I'll give it a try.

    Assume that there is an x e IR such that NUF(x) = 1.

    May we (by using this assumption) define INVNUF in such a way that 1 is
    in its domain?

    I mean can we define

           INVNUF(n) = _first_ x with NUF(x) = n     for n e {1, ...}

    now? (This would allow to use the term INVNUF(1) in our proof.)

    For this we would have to show/prove that there is a _first_ x with
    NUF(x) = 1. And for this we would have to show that (a) there is an x e
    IR such that NUF(x) = 1 (which is our assumption, so nothing to do here)
    and (b) that there is no smaller real number x' such that NUF(x') = 1.
    But I DOUBT that we will be able to prove/show that (even with our assumption).

    So I consider this a dead end.

    ---------------------------------------------------------------------

    Oh, wait! From our assumtion we can derive a contradiction (see my proof above)! Hence we can even prove (b) now (in the context of this proof by contradiction)! For that we assume ~(b) and immediately get ~~(b) and
    hence (b) from the assumption ~(b) and our original assumption, by contradiction. (The assumption ~(b) is now "discharged").

    Now we can define INVNUF such that 1 is in its domain. And hence now we
    may use your line of thought:

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    And hence we finally get: there is no x e IR such that NUF(x) = 1.  qed

    Well...

    Im mean, in this case we would need (the mayor part of) my original proof

    [...] Let x0 e IR such that NUF(x0) = 1. This means that there is
    exactly one unit fraction u such that u < x0. Let's call this unit
    fraction u0. Then (by definition) there is a (actually exactly one)
    natural number n such that u0 = 1/n. Let n0 e IN such that u0 = 1/n0.
    But then (again by definition) 1/(n0 + 1) is an unit fraction which is
    smaller than u0 and hence smaller than x0. Hence NUF(x0) > 1.
    Contradiction!

    as a "subproof". You see... (unnecessarily complicated).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Moebius on Thu Aug 15 14:36:48 2024
    XPost: sci.math

    On 8/15/2024 12:44 PM, Moebius wrote:
    Am 15.08.2024 um 17:53 schrieb Jim Burns:

    I've just proved INVNUF(1) not.exists?

    Mumbo-Jumbo.*)

    Proven.to.be mumbo.jumbo.
    Which is a whole different thing from
    floating.around.the.universe mumbo.jumbo.

    You can't prove that
    there is no x e IR such that x = 1/0.

    Beacuse you can't even USE "1/0" in
    the statement you want to prove
    _just because_ the symbol "is undefined".

    This is why I drone on and on about definitions:
    If we give a definition of
    what a non.existing thing allegedly is,
    in a language which describes existing things,
    we can prove by contradiction that
    it isn't one of those existing things.

    Translate ¬∃ᴿx = 1/0
    to ¬∃ᴿx: 0⋅x = 1
    Prove that.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 21:37:41 2024
    XPost: sci.math

    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...] BEFORE stating a/the proper definition.

    '⅟𝔊' can be used to derive contradictions,

    No, it can't.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Thu Aug 15 15:28:39 2024
    XPost: sci.math

    On 8/15/2024 9:59 AM, WM wrote:
    Le 14/08/2024 à 22:04, Jim Burns a écrit :
    On 8/14/2024 8:34 AM, WM wrote:

    What is smaller than every positive x
    is smaller than the interval (0, oo).

    That which is not.in the interval (0,∞)
    is not an end of (0,∞).

    The smallest point of (0, oo) belongs to (0, oo).
    It cannot be seen. It is dark.

    A definition isn't a claim something exists.
    A definition is a claim how a word is used.

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.
    '⅟𝔊' can be used to derive contradictions,
    which prove that ⅟𝔊 does not exist.

    for each n ∈ ℕᵈᵉᶠ:
    n+1 disproves by counter.example that n is largest

    max.ℕᵈᵉᶠ doesn't exist,
    even though we can define max.ℕᵈᵉᶠ

    In mathematics,
    we _propose_ the existence of abstract objects:
    numbers, sets, etc.
    This is different from defining them into existence.

    You are not free to reject my definitions.
    That is what I know I mean by what I say.
    But that doesn't make what I mean true.

    You are free to reject my proposals,
    but the usual practice is to frame a proposal
    in maximally difficult.to.reject terms.

    That's what I do when I propose Boolos's ST
    ⎛ ∃{}
    ⎜ ∀x∀y∃z=x∪{y}
    ⎝ extensionality

    You can reject those existential claims.
    But, if you do, the discussion ends and
    you look more than a little unreasonable.
    Not logic, but forceful in its own way.

    If you accept those existential claims,
    I can follow them with definitions --
    which you _aren't_ free to reject --
    of what.I.mean.by the natural numbers,
    (spoiler: the usual naturals, '+', and '×').
    Iron.clad support of our usual arithmetic,
    but _not_ by definition, instead by
    maximally.acceptable.proposal, Boolos's ST.

    The smallest point of (0, oo) belongs to (0, oo).
    It cannot be seen. It is dark.

    The rationals in (0,∞) aren't ends of (0,∞).

    For each p/q ∈ ℚ⁺
    there is disproof by counter.example p/(q+1)
    to the claim p/q is smallest in (0,∞), and
    there is disproof by counter.example (p+1)/q
    to the claim p/q is largest in (0,∞).
    Rational p/q is not largest or smallest in (0,∞)

    The split.situaters in (0,∞) aren't ends of (0,∞).

    For each x ∈ ℝ⁺\ℚ⁺
    x situates the non.{} split Fₓ,Hₓ of ℚ⁺
    Fₓ ᵉᵃᶜʰ< x <ᵉᵃᶜʰ Hₓ
    non.{} Fₓ ∋ r₋
    non.{} Hₓ ∋ s₊
    r₋ < x < s₊
    Irrational x is not largest or smallest in (0,∞)

    By definition of the real numbers,
    only rationals and split.situaters are in (0,∞)

    No ends of (0,∞) exist in (0,∞)
    No ends of (0,∞) exist not.in (0,∞)
    No ends of (0,∞) exist.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 23:34:34 2024
    XPost: sci.math

    Am 15.08.2024 um 20:36 schrieb Jim Burns:

    Translate ¬∃ᴿx(x = 1/0) to

    There is nothing to translate. "¬∃ᴿx = 1/0" is just a meaningless expression, because "1/0" is a non-denoting (undefined) term/name.

    ¬∃ᴿx: 0⋅x = 1

    Now this is a meaningful statement.

    Prove that.

    Indeed! :-)

    For this we micht assume

    ∃ᴿx: 0⋅x = 1

    and try to derive a contradiction from this assumption.

    Proof by contradiction (RRA).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 23:36:09 2024
    XPost: sci.math

    Am 15.08.2024 um 20:36 schrieb Jim Burns:

    Translate ¬∃ᴿx(x = 1/0) to

    There is nothing to translate. "¬∃ᴿx = 1/0" is just a meaningless expression, because "1/0" is a undefined (non-denoting) term/name.

    ¬∃ᴿx: 0⋅x = 1

    Now this is a meaningful statement.

    Prove that.

    Indeed! :-)

    For this we might assume

    ∃ᴿx: 0⋅x = 1

    and try to derive a contradiction from this assumption.

    Proof by contradiction (RRA).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 23:25:23 2024
    XPost: sci.math

    Am 15.08.2024 um 21:37 schrieb Moebius:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...] BEFORE stating a/the proper definition.

    When defining a term we have to observe certain "definition rules".

    For example, if you want to define/introduce the constant c as

    c is the xxx

    you have 1. to prove that there IS an xxx (->existence) and 2. that
    there is not more than one xxx (->uniqueness).

    Otherwise one risks to introduce a contradiction by the definition.
    (This NOT something we would like to do in math.)

    '⅟𝔊' can be used to derive contradictions,

    Well, ... not really. See above.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    For example, we can't define/introduce "c" the following way:

    c = the real number x such that x * 0 = 1.

    (or c = x iff x e IR and x * 0 = 1).

    You see, after defining c that way, we could derive

    c e IR & c * 0 = 1.

    From this we could derive the "theorem":

    Ex e IR: x * 0 = 1.

    Which clearly is not the case.

    Please try to understand that a definition is not an "assumption". Hence
    if we get a contradiction (after using a "definition" which is not
    proper) we can't just "discarge" that definition (as if it were an
    assumption).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 23:35:45 2024
    XPost: sci.math

    Am 15.08.2024 um 20:36 schrieb Jim Burns:

    Translate ¬∃ᴿx(x = 1/0) to

    There is nothing to translate. "¬∃ᴿx = 1/0" is just a meaningless expression, because "1/0" is a undefined non-denoting) term/name.

    ¬∃ᴿx: 0⋅x = 1

    Now this is a meaningful statement.

    Prove that.

    Indeed! :-)

    For this we might assume

    ∃ᴿx: 0⋅x = 1

    and try to derive a contradiction from this assumption.

    Proof by contradiction (RRA).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Thu Aug 15 23:43:44 2024
    XPost: sci.math

    Am 15.08.2024 um 23:36 schrieb Moebius:
    Am 15.08.2024 um 20:36 schrieb Jim Burns:

    Translate ¬∃ᴿx(x = 1/0) to

    There is nothing to translate. "¬∃ᴿx = 1/0" is just a meaningless expression, because "1/0" is a undefined (non-denoting) term/name.

    ¬∃ᴿx: 0⋅x = 1

    Now this is a meaningful statement.

    Prove that.

    Indeed! :-)

    For this we might assume

         ∃ᴿx: 0⋅x = 1

    and try to derive a contradiction from this assumption.

    Proof by contradiction (RRA).

    Hint: And BECAUSE we can prove:

    ¬∃x(x e IR & 0⋅x = 1)

    we CAN'T define "1/0" the following way:

    1/0 := the x e IR such that 0⋅x = 1 .

    Nuff said.

    (It should be clear that the function x |-> 1/x is not defined for x =
    0, hence even we have defined x |-> 1/x (for x e IR, x =/= 0), we may
    not write "1/0", based on THIS definition.)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Moebius on Thu Aug 15 18:51:01 2024
    XPost: sci.math

    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]
    BEFORE stating a/the proper definition.

    Proving that mimsy borogoves exist
    before defining mimsy borogoves
    seems like a pretty steep hill to climb.
    That's a hill which I doubt we need to climb.

    '⅟𝔊' can be used to derive contradictions,

    No, it can't.

    Half or more of my proofs to WM say
    "Assume otherwise... However... Contradiction."

    But maybe I'm crazy.

    There's an old, old proof that √2 is irrational.
    If that's wrong, everyone but you is crazy.

    √2 is irrational.

    ⎛ Assume otherwise.
    ⎜ Assume p₃,q₃ ∈ ℕ₁: p₃⋅p₃ = 2⋅q₃⋅q₃

    ⎜ p₃ ∈ {p ∈ ℕ₁: ∃q ∈ ℕ₁: p⋅p = 2⋅q⋅q}
    ⎜ p₂ = min.{p ∈ ℕ₁: ∃q ∈ ℕ₁: p⋅p = 2⋅q⋅q}
    ⎜ ∃q₂ ∈ ℕ₁: p₂⋅p₂ = 2⋅q₂⋅q₂
    ⎜ ¬∃p₁ < p₂: ∃q₁ ∈ ℕ₁: p₁⋅p₁ = 2⋅q₁⋅q₁

    ⎜ However,
    ⎜ p₂⋅p₂ = 2⋅q₂⋅q₂
    ⎜ 2 is prime.
    ⎜ 2|p₂ or 2|p₂
    ⎜ p₂ = 2⋅p₁
    ⎜ 2⋅p₁⋅2⋅p₁ = 2⋅q₂⋅q₂
    ⎜ 2⋅p₁⋅p₁ = q₂⋅q₂
    ⎜ 2|q₂ or 2|q₂
    ⎜ q₂ = 2⋅q₁
    ⎜ 2⋅p₁⋅p₁ = 2⋅q₁⋅2⋅q₁
    ⎜ p₁⋅p₁ = 2⋅q₁⋅q₁ and p₁ < p₂
    ⎝ Contradiction.

    The proof uses properties such as
    well.order and unique.prime.factorization
    which p₂ and q₂ would have if they existed
    to show that they don't exist.

    Doing that isn't a problem.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 00:59:01 2024
    XPost: sci.math

    Am 16.08.2024 um 00:51 schrieb Jim Burns:
    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]
    BEFORE stating a/the proper definition.

    Proving that [a unique] mimsy borogove exist[s]
    before defining [the] mimsy borogove
    seems like a pretty steep hill to climb.

    No mater what you think, in math we HAVE TO climb this hill, and we do.

    That's a hill which I doubt we need to climb.

    Fuck you, asshole. Want't to be the the next crank in sci.math? (After
    all the other cranks have left?)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 00:55:54 2024
    XPost: sci.math

    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    A definition isn't a claim something exists.
    Yeah, if you define a "property", say,

    x is a /Mückenheim number/ iff x e IR & x * 0 = 1.

    Using symols:

    WM(x) :<-> x e IR & x * 0 = 1 .

    THEN we can prove:

    ~Ex(WM(x)). (*)

    But this "approach" does not work for constants (or "functions").

    We may express (*) (using normal language) the following way:

    There are no Mückenheim numbers.

    But this "approach" does not work for say, the "Mückenheim number" WM:

    WM = the smallest unit fraction ,

    SINCE THERE IS NO /smallest unit fraction/.

    Again, we may define the property /smallest unit fraction/:

    WM(x) :<-> x e SB & Ay e SB: x <= y.
    "x is a smallest unit fraction."

    Then we can prove:

    ~Ex(WM(x)).
    "There is no smallest unit fraction."

    HENCE we CAN'T define:

    WM = ix(x e SB & Ay e SB: x <= y)
    "WM is the smallest unit fraction."

    The rules concerning "proper definitions" do not allow for such a
    definition.

    Hope this helps.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:04:42 2024
    XPost: sci.math

    Am 16.08.2024 um 00:59 schrieb Chris M. Thomasson:
    On 8/15/2024 10:06 AM, Python wrote:
    Le 15/08/2024 à 19:01, Moebius a écrit :
    Am 13.08.2024 um 19:02 schrieb Jim Burns:
    On 8/13/2024 10:21 AM, WM wrote:

    [There is] a real [number] x with NUF(x) = 1.

    INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋

    NUF(INVNUF(1)) > 1
    Contradiction.

    Well, no, this just isn't a proof, sorry about that, Jim.

    (Hint: The term "INVNUF(1)" is not defined. Hence you may not even
    use it in a proof by contradiction. Actually, you didn't state an
    assumption in your "proof".)

    Proof by contradiction:

    Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such
    that NUF(x0) = 1. This means that there is exactly one unit fraction
    u such that u < x0. Let's call this unit fraction u0. Then (by
    definition) there is a (actually exactly one) natural number n such
    that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by
    definition) 1/(n0 + 1) is an unit fraction which is smaller than u0
    and hence smaller than x0. Hence NUF(x0) > 1. Contradiction!

    (Of course, it's clear that I'm using the same "proof idea" as you
    used in your attempt of a proof.)

    Of course you are right. 100% right. Any marginally decent high school
    student could sketch up the very same proof you've posted.

    [...]

    DING!!!! Yup! wow. ;^o

    Of course I know (believe for good reasons) that I am 100% right.
    Otherwise I wouldn't post such diatribes. :-P

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:06:29 2024
    XPost: sci.math

    Am 16.08.2024 um 01:05 schrieb Chris M. Thomasson:
    On 8/15/2024 6:57 AM, WM wrote:
    Le 14/08/2024 à 21:01, "Chris M. Thomasson" a écrit :

    There is no smallest unit fraction,

    There is a smallest one because all are separated:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

    My god.... There is no smallest unit fraction! They get closer and
    closer to zero, but no unit fraction ever equals zero and there are infinitely many of them... Got it? WOW!!! You should be ashamed of
    yourself for teaching the crap to students. just, holy shit, wow!


    100% ACK.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:11:54 2024
    XPost: sci.math

    Am 16.08.2024 um 00:51 schrieb Jim Burns:
    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]
    BEFORE stating a/the proper definition.

    Proving that [a unique] mimsy borogove exist[s]
    before defining [the] mimsy borogove
    seems like a pretty steep hill to climb.

    No mater what you think, in math we HAVE TO climb this hill, and we do.

    That's a hill which I doubt we need to climb.

    Fuck you, asshole. Want to be the the next crank in sci.math? (After all
    the other cranks have left?)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:32:50 2024
    XPost: sci.math

    Am 16.08.2024 um 01:11 schrieb Moebius:

    Want to be the the next crank in sci.math? (After all
    the other cranks have left?)

    Together with Mückenheim the last of your kind?

    See: The Man Trap (->"the last of its kind"). https://per-ineptia-ad-astra.tumblr.com/post/180046750716/episode-11-the-man-trap

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:16:31 2024
    XPost: sci.math

    Am 16.08.2024 um 00:51 schrieb Jim Burns:

    Half or more of my proofs to WM say
    "Assume otherwise... However... Contradiction."

    Yeah, to be precise a proof by contradiction assumes a STATEMENT/CLAIM.

    But maybe I'm crazy.

    No, you aren't, I guess.

    [nonsense deleted]

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:24:47 2024
    XPost: sci.math

    Am 16.08.2024 um 01:16 schrieb Moebius:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:

    Half or more of my proofs to WM say
    "Assume otherwise... However... Contradiction."

    Yeah, to be precise: a proof by contradiction assumes a STATEMENT/CLAIM and etc.

    But maybe I'm crazy.

    No, you aren't, I guess (or hope).

    [nonsense deleted]

    Hint: If "√2" is not defined, the "statement" (rather expression)

    √2 is irrational

    is not a meanigful statement (i.e. mathematical claim). (Got it?)

    What you (most certainly) mean is:

    There is no rational number x such that x^2 = 2.

    On the ohter hand, in the context of the real numbers we usually have
    defined "√2" (namely as the real number x > 0 such that x^2 = 2). Hence
    in this context we may claim

    √2 is irrational [i.e. not a rational number]

    AND actually prove this claim.

    Hope this helps.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:36:14 2024
    XPost: sci.math

    Am 16.08.2024 um 01:32 schrieb Moebius:
    Am 16.08.2024 um 01:11 schrieb Moebius:

    Want to be the the next crank in sci.math? (After all the other cranks
    have left?)

    Together with Mückenheim the last of your kind?

    See: https://64.media.tumblr.com/39e4df2a56de2a5136fb1736015d4f36/7cacdb637823a3b1-9d/s1280x1920/6649f33327cae138d87792b29bb04cf16154a2d7.png

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:53:27 2024
    XPost: sci.math

    Am 16.08.2024 um 01:39 schrieb Moebius:
    Am 16.08.2024 um 01:35 schrieb Chris M. Thomasson:
    On 8/15/2024 3:59 PM, Moebius wrote:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:
    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]
    BEFORE stating a/the proper definition.

    Proving that [a unique] mimsy borogove exist[s]
    before defining [the] mimsy borogove
    seems like a pretty steep hill to climb.

    No mater what you think, in math we HAVE TO climb this hill, and we do.

    That's a hill which I doubt we need to climb.

    Fuck you, asshole. Want't to be the the next crank in sci.math?
    (After all the other cranks have left?)

    Wrt sqrt(2), afaict anytime we create a unit square, we have the
    sqrt(2) in all of its irrational infinite glory as a diagonal.

    Right. In the context of geometry [...]

    Actually, we see/know that there is a "diagonal" of a unit square, but
    it's length and and the length 1 are "incommensurable".

    See: https://en.wikipedia.org/wiki/Commensurability_(mathematics)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:39:18 2024
    XPost: sci.math

    Am 16.08.2024 um 01:35 schrieb Chris M. Thomasson:
    On 8/15/2024 3:59 PM, Moebius wrote:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:
    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]
    BEFORE stating a/the proper definition.

    Proving that [a unique] mimsy borogove exist[s]
    before defining [the] mimsy borogove
    seems like a pretty steep hill to climb.

    No mater what you think, in math we HAVE TO climb this hill, and we do.

    That's a hill which I doubt we need to climb.

    Fuck you, asshole. Want't to be the the next crank in sci.math? (After
    all the other cranks have left?)

    Wrt sqrt(2), afaict anytime we create a unit square, we have the sqrt(2)
    in all of its irrational infinite glory as a diagonal.

    Right. In the context of geometry (with real "distances"/lengths of line segments).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:52:49 2024
    XPost: sci.math

    Am 16.08.2024 um 01:39 schrieb Moebius:
    Am 16.08.2024 um 01:35 schrieb Chris M. Thomasson:
    On 8/15/2024 3:59 PM, Moebius wrote:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:
    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]
    BEFORE stating a/the proper definition.

    Proving that [a unique] mimsy borogove exist[s]
    before defining [the] mimsy borogove
    seems like a pretty steep hill to climb.

    No mater what you think, in math we HAVE TO climb this hill, and we do.

    That's a hill which I doubt we need to climb.

    Fuck you, asshole. Want't to be the the next crank in sci.math?
    (After all the other cranks have left?)

    Wrt sqrt(2), afaict anytime we create a unit square, we have the
    sqrt(2) in all of its irrational infinite glory as a diagonal.

    Right. In the context of geometry [...]

    Actually, we see/know that there is a "diagonal" of a unit square, but
    it's length and and the length 1 is "incommensurable".

    See: https://en.wikipedia.org/wiki/Commensurability_(mathematics)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 01:57:21 2024
    XPost: sci.math

    Am 16.08.2024 um 01:11 schrieb Moebius:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:
    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]

    and a proof of proof of uniqueness

    BEFORE stating a/the proper definition.

    Proving that [a unique] mimsy borogove exist[s]
    before defining [the] mimsy borogove
    seems like a pretty steep hill to climb.

    Hint: That's exactly, what we have to do in math.

    Seriously. :-)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 02:03:26 2024
    XPost: sci.math

    Am 16.08.2024 um 01:39 schrieb Moebius:
    Am 16.08.2024 um 01:35 schrieb Chris M. Thomasson:
    On 8/15/2024 3:59 PM, Moebius wrote:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:
    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]
    BEFORE stating a/the proper definition.

    Proving that [a unique] mimsy borogove exist[s]
    before defining [the] mimsy borogove
    seems like a pretty steep hill to climb.

    No mater what you think, in math we HAVE TO climb this hill, and we do.

    That's a hill which I doubt we need to climb.

    Fuck you, asshole. Want't to be the the next crank in sci.math?
    (After all the other cranks have left?)

    Wrt sqrt(2), afaict anytime we create a unit square, we have the
    sqrt(2) in all of its irrational infinite glory as a diagonal.

    Right. In the context of geometry [...]

    Actually, we see/know that there is a "diagonal" of a unit square, but
    its length and and the length 1 are "incommensurable".

    See: https://en.wikipedia.org/wiki/Commensurability_(mathematics)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Moebius on Fri Aug 16 01:05:40 2024
    XPost: sci.math

    On 8/15/2024 7:16 PM, Moebius wrote:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:

    Half or more of my proofs to WM say
    "Assume otherwise... However... Contradiction."

    Yeah, to be precise a proof by contradiction
    assumes a STATEMENT/CLAIM.

    Many times, a false existence claim.

    But maybe I'm crazy.

    No, you aren't, I guess.

    [nonsense deleted]

    Well.known, ancient proof undeleted (below).
    If that is nonsense, everything is nonsense.

    That proof uses properties such as
    well.order and unique.prime.factorization
    which p₂ and q₂ would have if they existed
    in order to show that they don't exist.

    Yes,
    reasoning from those properties of p₂ and q₂
    generates nonsense, contradiction.
    However,
    the nonsense is the point,
    the reason that we conclude that p₂ and q₂ don't exist.


    Another example.
    ZFC+"An inaccessible ordinal exists" proves
    ZFC

    In order to make that proof,
    we need the _definition_ of 'inaccessible cardinal'. https://en.wikipedia.org/wiki/Inaccessible_cardinal

    You will not prove that an inaccessible cardinal _exists_
    not from ZFC+"An inaccessible ordinal exists" --
    or, if you do, Gödel shows that proves _inconsistency_

    ----
    √2 is irrational.

    ⎛ Assume otherwise.
    ⎜ Assume p₃,q₃ ∈ ℕ₁: p₃⋅p₃ = 2⋅q₃⋅q₃

    ⎜ p₃ ∈ {p ∈ ℕ₁: ∃q ∈ ℕ₁: p⋅p = 2⋅q⋅q}
    ⎜ p₂ = min.{p ∈ ℕ₁: ∃q ∈ ℕ₁: p⋅p = 2⋅q⋅q}
    ⎜ ∃q₂ ∈ ℕ₁: p₂⋅p₂ = 2⋅q₂⋅q₂
    ⎜ ¬∃p₁ < p₂: ∃q₁ ∈ ℕ₁: p₁⋅p₁ = 2⋅q₁⋅q₁

    ⎜ However,
    ⎜ p₂⋅p₂ = 2⋅q₂⋅q₂
    ⎜ 2 is prime.
    ⎜ 2|p₂ or 2|p₂
    ⎜ p₂ = 2⋅p₁
    ⎜ 2⋅p₁⋅2⋅p₁ = 2⋅q₂⋅q₂
    ⎜ 2⋅p₁⋅p₁ = q₂⋅q₂
    ⎜ 2|q₂ or 2|q₂
    ⎜ q₂ = 2⋅q₁
    ⎜ 2⋅p₁⋅p₁ = 2⋅q₁⋅2⋅q₁
    ⎜ p₁⋅p₁ = 2⋅q₁⋅q₁ and p₁ < p₂
    ⎝ Contradiction.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 07:57:27 2024
    XPost: sci.math

    Am 16.08.2024 um 07:50 schrieb Moebius:

    Of course, _if_ we already have introduced the real numbers (i.e. IR)
    we may define

          √2 = the real number x such that x^2 = 2 ,    (*)

    _after_ we have shown that that there is exactly one x e IR such that
    x^2 = 2.

    From (*) we get immediately: (√2)^2 = 2

    Now we may assume that there are natural numbers n,m such that √2 = n/m.

    (...)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 07:50:12 2024
    XPost: sci.math

    Am 16.08.2024 um 07:05 schrieb Jim Burns:
    On 8/15/2024 7:16 PM, Moebius wrote:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:

    Half or more of my proofs to WM say
    "Assume otherwise... However... Contradiction."

    Yeah, to be precise a proof by contradiction
    assumes a STATEMENT/CLAIM.

    Many times, a false existence claim.

    Right.

    √2 is irrational.

    This statement is just nonsense, _if_ "√2" is not already defined.*)

    ⎛ Assume otherwise.

    Nope. You clearly don't assume

    √2 is rational ,

    but:

    ⎜ Assume p₃,q₃ ∈ ℕ₁:  p₃⋅p₃ = 2⋅q₃⋅q₃
    :
    ⎝ Contradiction.

    ________________________________

    *) Of course, _if_ we already have introduced the real numbers (i.e. IR)
    we may define

    √2 = the real number x such that x*x = 2 , (*)

    _after_ we have shown that that there is exactly one x e IR such that
    x*x = 2.

    From (*) we get immediately: √2*√2 = 2.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Fri Aug 16 09:22:55 2024
    XPost: sci.math

    Am 16.08.2024 um 08:48 schrieb Chris M. Thomasson:
    On 8/15/2024 10:57 PM, Moebius wrote:
    Am 16.08.2024 um 07:50 schrieb Moebius:

    Of course, _if_ we already have introduced the real numbers (i.e. IR)
    we may define

           √2 = the real number x such that x^2 = 2 ,    (*)

    _after_ we have shown that there is exactly one x e IR such that
    x^2 = 2.

    From (*) we get immediately: (√2)^2 = 2

    Now we may assume that there are natural numbers n,m such that √2 = n/m. >>
    (...)

    The infinite convergents of continuous fractions can describe sqrt(2) up
    to any desired precision?

    Sure.

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to Moebius on Fri Aug 16 07:45:34 2024
    XPost: sci.math

    On 8/16/2024 1:50 AM, Moebius wrote:
    Am 16.08.2024 um 07:05 schrieb Jim Burns:
    On 8/15/2024 7:16 PM, Moebius wrote:
    Am 16.08.2024 um 00:51 schrieb Jim Burns:
    On 8/15/2024 3:37 PM, Moebius wrote:
    Am 15.08.2024 um 21:28 schrieb Jim Burns:

    Defining ⅟𝔊 to be the smallest unit fraction
    isn't a claim that ⅟𝔊 exists.

    You need an existence proof [...]
    BEFORE stating a/the proper definition.

    Half or more of my proofs to WM say
    "Assume otherwise... However... Contradiction."

    Yeah, to be precise a proof by contradiction
    assumes a STATEMENT/CLAIM.

    Many times, a false existence claim.

    Right.

    About things without an existence proof.

    √2 is irrational.

    This statement is just nonsense,
    _if_ "√2" is not already defined.*)

    Delete less.

    ⎛ Assume otherwise.

    Nope. You clearly don't assume

    √2 is rational ,

    but:

    ⎜ Assume p₃,q₃ ∈ ℕ₁:  p₃⋅p₃ = 2⋅q₃⋅q₃
    :
    ⎝ Contradiction.

    I assume a statement equivalent to
    ⎜ Rational √2 exists.

    ________________________________

    *) Of course,
    _if_ we already have introduced the real numbers
    (i.e. IR)
    we may define

    √2 = the real number x such that x*x = 2 ,    (*)

    _after_ we have shown that that
    there is exactly one x e IR such that x*x = 2.

    Upthread is not a proof that real √2 exists.

    Upthread is a proof that rational √2 not.exists,
    which never mentions Dedekind completeness,
    a never.mentioning which shows that,
    whether or not real √2 exists,
    rational √2 not.exists.

    No, I am not proving that, here.
    It makes my proof weaker and less focused, here.

    From (*) we get immediately: √2*√2 = 2.

    You're complaining I didn't show you √2*√2 = 2.

    Consider it left as an exercise for the reader.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Fri Aug 16 12:08:33 2024
    XPost: sci.math

    On 8/15/24 9:57 AM, WM wrote:
    Le 14/08/2024 à 21:01, "Chris M. Thomasson" a écrit :
    On 8/14/2024 5:28 AM, WM wrote:

    The existence of a smallest unit fraction is the only alternative to
    the existence of more than one at a real point.

    There is no smallest unit fraction,

    There is a smallest one because all are separated:

    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
    Regards, WM

    Actually, that statement says that there is no smallest, as for every
    1/n there exist a smaller 1/(n+1).

    If n+1 didn't exist, then your statement wouldn't be true for all n.

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Fri Aug 16 12:07:10 2024
    XPost: sci.math

    On 8/15/24 9:55 AM, WM wrote:
    Le 14/08/2024 à 20:04, Jim Burns a écrit :
    On 8/14/2024 8:28 AM, WM wrote:
    Le 13/08/2024 à 19:42, Jim Burns a écrit :

    The existence of the smallest unit fractions
    is contradictory in the land of
    rationals with
    countable.to numerators and denominators
    with each split situated ==
    a last point in the foresplit or
    a first point in the hindsplit.

    The existence of a smallest unit fraction is
    the only alternative to the existence of
    more than one at a real point.

    The NONexistence of a smallest unit fraction is why,
    for each unit fraction,
    there are infinitely.many smaller unit fractions.
    And with no two at one point.

    That is a self-contradiction.

    The first point with unit fractions is x = INVNUF(1).

    Regrads, WM

    Which doesn't have a value (at least not in the finite rationals or reals).

    That is your problem, NUF isn't properly defined, and thus INVNUF can't
    be properly defined.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 16 16:19:17 2024
    XPost: sci.math

    Le 15/08/2024 à 18:16, Jim Burns a écrit :
    On 8/15/2024 9:52 AM, WM wrote:

    ⎜ Assume NUF(x) = 0 and x > 0

    We assume that NUF(0) = 0 and many unit fractions are within (0, 1].
    We can reduce the interval to (0, x) c [0, 1]. Let x converge to 0.
    Then the number of unit fractions diminishes. Finally there is none
    remaining. But never, for no interval (0, x), more than one unit fraction
    is lost. Therefore there is only one last unit fraction.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 16 16:42:30 2024
    XPost: sci.math

    Le 15/08/2024 à 19:06, Python a écrit :

    Our problem, as a community, is not proving Wolfgang Mückenheim wrong.

    Your problem is that you can't think straight. If NUF grows from 0 to
    more, then there is a first one or more. The latter is excluded by ∀n
    ∈ ℕ: 1/n - 1/(n+1) > 0 .

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 16 16:45:29 2024
    XPost: sci.math

    Le 15/08/2024 à 19:01, Moebius a écrit :

    Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such
    that NUF(x0) = 1. This means that there is exactly one unit fraction u
    such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n.
    Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 +
    1) is an unit fraction which is smaller than u0 and hence smaller than
    x0. Hence NUF(x0) > 1. Contradiction!

    We can reduce the interval (0, x) c [0, 1] such that x converges to 0.
    Then the number of unit fractions diminishes. Finally there is none
    remaining. But never, for no interval (0, x), more than one unit fraction
    is lost. Therefore there is only one last unit fraction.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Fri Aug 16 12:57:09 2024
    XPost: sci.math

    On 8/16/24 12:42 PM, WM wrote:
    Le 15/08/2024 à 19:06, Python a écrit :

    Our problem, as a community, is not proving Wolfgang Mückenheim wrong.

    Your problem is that you can't think straight. If NUF grows from 0 to
    more, then there is a first one or more. The latter is excluded by ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .

    Regards, WM



    But that only holds if NUF actually works. If there isn't a first unit fraction, NUF(x) can't actually be defined, so you are just working a
    circular argument, which in fact proves that your NUF(x) can't be
    correctly defined.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Fri Aug 16 13:39:54 2024
    XPost: sci.math

    On 8/16/2024 12:19 PM, WM wrote:
    Le 15/08/2024 à 18:16, Jim Burns a écrit :

    ⎜ Assume NUF(x) = 0 and x > 0

    ⎜ lower.bound ½⋅β and not.lower.bound ½⋅β
    ⎝ Contradiction.

    We assume that NUF(0) = 0 and
    many unit fractions are within (0, 1].
    We can reduce the interval to (0, x) c [0, 1].
    Let x converge to 0.
    Then the number of unit fractions diminishes.

    For some sets,
    sets for which non.{} subsets always have both ends,
    removing one element diminishes them

    For other sets,
    sets for which one non.{} subset has 0 or 1 end,
    removing one element does not diminish them.

    The set ℕᵈᵉᶠ def of diminishable ordinals is
    one of the other sets.
    It is not
    a set for which non.{} subsets always have both ends.
    ℕᵈᵉᶠ has a lower.end, but

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    ℕᵈᵉᶠ is one of the other sets.
    Removing one element does not diminish ℕᵈᵉᶠ

    Finally there is none remaining.
    But never, for no interval (0, x),
    more than one unit fraction is lost.
    Therefore there is only one last unit fraction.

    ... and ½⋅β both is and is not a lower.bound to
    the visible unit.fractions,
    which apparently you (WM) accept.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Sat Aug 17 13:28:30 2024
    XPost: sci.math

    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases, but at no point x it increases by
    more than 1 because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
    Therefore there is a smallest unit fractions and vice versa a greatest natnumber.
    What can't you understand?

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Sat Aug 17 10:29:14 2024
    XPost: sci.math

    On 8/17/24 9:28 AM, WM wrote:
    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases, but at no point x it increases
    by more than 1 because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Therefore there is a smallest unit fractions and vice versa a greatest natnumber.
    What can't you understand?

    Regards, WM

    But there is no point (>0) where it has a finite value, so it just isn't
    a defined function for x>0.

    Sorry, you just don't understand how math works, because it seems you
    only know counting on your fingers.

    The can NOT be a greatest Natural Number because of how they are defined.

    If you natnumber are something different, then you need to try to define
    them and work out their properties,

    Then tell people what they can do with your fake "natnumbers" that is
    better than the real Natural Numbers.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Python@21:1/5 to All on Sat Aug 17 17:02:07 2024
    XPost: sci.math

    Le 17/08/2024 à 16:29, Richard Damon a écrit :
    On 8/17/24 9:28 AM, WM wrote:
    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases, but at no point x it
    increases by more than 1 because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Therefore there is a smallest unit
    fractions and vice versa a greatest natnumber.
    What can't you understand?

    Regards, WM

    But there is no point (>0) where it has a finite value, so it just isn't
    a defined function for x>0.

    Sorry, you just don't understand how math works, because it seems you
    only know counting on your fingers.

    There is a Mückenheimish way to "prove" that you do not need fingers
    to count up to 4 on your fingers.

    Check the index finger : not necessary. One can count using the other
    ones.
    Check the middle finger : not necessary. One can count using the other
    ones.
    Check the ring finger : not necessary. One can count using the other
    ones.
    Check the little finger : not necessary. One can count using the other
    ones.
    Check the thumb : not necessary. One can count using the other ones.

    Therefore *all* fingers are unnecessary: one can count up to four even
    if he has no fingers at all.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Sat Aug 17 17:39:30 2024
    XPost: sci.math

    On 8/17/2024 9:28 AM, WM wrote:
    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases,
    but at no point x it increases by more than 1
    because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
    Therefore there is a smallest unit fractions
    and vice versa a greatest natnumber.
    What can't you understand?

    How can ½⋅β
    ⎛ half the allegedly.positive greatest.lower.bound β of
    ⎝ visibleᵂᴹ.unit.fractions
    be both lower.bound ( ½⋅β < β)
    and not.lower.bound ( 2⋅β > ⅟k and ½⋅β > ¼⋅⅟k)
    of the visibleᵂᴹ.unit.fractions?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Aug 18 03:51:42 2024
    XPost: sci.math

    Am 17.08.2024 um 23:39 schrieb Jim Burns:
    On 8/17/2024 9:28 AM, WM wrote:

    SBZ(x) starts with 0 at 0 and increases,

    It does not "increase" (in the usual sense of the word) but _jump_ "at 0".

    but [bla bla bla].

    Therefore there is a smallest unit fractions and vice versa a greatest
    natnumber.

    Yeah!

    A typical Mückenheim argument: <nonsense>. Therefore <falsehood>.

    What can't you understand?

    Dass sich noch irgendjemand ERNSTHAFT mit dem saudummen Scheißdreck auseinandersetzt, den Du hier von Dir gibst.

    Mückenheim, Du bist für jede Art von Mathematik zu dumm und zu blöde und dazu vermutlich auch noch geisteskrank. Hau endlich ab, Du Spinner!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Mon Aug 19 11:32:44 2024
    XPost: sci.math

    Le 17/08/2024 à 16:29, Richard Damon a écrit :
    On 8/17/24 9:28 AM, WM wrote:
    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases, but at no point x it increases
    by more than 1 because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Therefore there is a smallest unit fractions
    and vice versa a greatest natnumber.
    What can't you understand?

    But there is no point (>0) where it has a finite value,

    You can't see it and you are unable to derive it from mathematics. But blindness is not an argument.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Mon Aug 19 11:40:37 2024
    XPost: sci.math

    Le 17/08/2024 à 23:39, Jim Burns a écrit :
    On 8/17/2024 9:28 AM, WM wrote:
    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases,
    but at no point x it increases by more than 1
    because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
    Therefore there is a smallest unit fractions
    and vice versa a greatest natnumber.
    What can't you understand?

    How can ½⋅β
    ⎛ half the allegedly.positive greatest.lower.bound β of
    ⎝ visibleᵂᴹ.unit.fractions

    Visible unit fractions have the lower bound 0.
    Dark unit fractions have a smallest element. No smaller unit fractions is existing because no larger natnumber is existing.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Mon Aug 19 12:58:10 2024
    XPost: sci.math

    On 8/19/2024 7:40 AM, WM wrote:
    Le 17/08/2024 à 23:39, Jim Burns a écrit :
    On 8/17/2024 9:28 AM, WM wrote:
    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases,
    but at no point x it increases by more than 1
    because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
    Therefore there is a smallest unit fractions
    and vice versa a greatest natnumber.
    What can't you understand?

    How can ½⋅β
    ⎛ half the allegedly.positive greatest.lower.bound β of
    ⎝ visibleᵂᴹ.unit.fractions
    be both lower.bound ( ½⋅β < β)
    and not.lower.bound ( 2⋅β > ⅟k and ½⋅β > ¼⋅⅟k)
    of the visibleᵂᴹ.unit.fractions?

    Visible unit fractions have the lower bound 0.

    Visibleᵂᴹ unit fractions have the greatest.lower.bound 0.
    Otherwise, ½⋅β > 0 is contradictory.
    ⎛ not.bound 2⋅β > ⅟k
    ⎜ not.bound ½⋅β > ¼⋅⅟k
    ⎝ bound ½⋅β < β

    Dark unit fractions have a smallest element.

    Darkᵂᴹ unit.fractions are positive. Correct?
    I don't want to alter your darkᵂᴹ

    No smaller unit fractions is existing

    Is that smallest darkᵂᴹ unit.fraction not a unit.fraction?
    Is it not positive?

    Each positive point, thus,
    that smallest darkᵂᴹ unit.fraction
    is not a lower.bound, thus,
    has smaller unit.fractions, visibleᵂᴹ ones.

    Are visibleᵂᴹ unit.fractions not existing?

    No smaller unit fractions is existing
    because no larger natnumber is existing.

    For visibleᵂᴹ natural k
    k∪{k} = k+1 disproves no larger.than.k existing.

    Darkᵂᴹ natural 𝔊 has darkᵂᴹ unit.fraction ⅟𝔊
    Darkᵂᴹ ⅟𝔊 is not.lower.bound of visiblesᵂᴹ ⅟k
    Exists visibleᵂᴹ ⅟k < ⅟𝔊, thus k > 𝔊
    For darkᵂᴹ natural 𝔊
    k > 𝔊 disproves no larger.than.𝔊 existing.

    Each natural, visibleᵂᴹ or darkᵂᴹ, is disproved
    from no larger.than.it existing.

    Each unit.fraction, visibleᵂᴹ or darkᵂᴹ, is disproved
    from no smaller.than.it existing.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Mon Aug 19 19:17:34 2024
    XPost: sci.math

    Am 19.08.2024 um 18:58 schrieb Jim Burns:

    You got it totally wrong!

    The dark unit fractions are smaller than the (all) visible ones.

    Now: The visible unit fraction don't have a smallest one (of course),
    but the dark unit fraction do (at least in mückenmath)!

    WM: "Dark unit fractions have a smallest element."

    See?!

    WM: "Visible unit fractions have the [or rather *a*] lower bound 0."

    Right, but the dark unit fractions ARE BETWEEN 0 and all the visible
    unit fractions.

    See: https://en.wikipedia.org/wiki/Infinitesimal
    and: https://en.wikipedia.org/wiki/Hyperreal_number

    WM: "No smaller [dark] unit fractions [than the smallest one] is
    existing, because no larger [dark] natnumber {than the largest one] is existing."

    Right!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Moebius on Mon Aug 19 14:12:38 2024
    XPost: sci.math

    On 8/19/2024 1:17 PM, Moebius wrote:
    Am 19.08.2024 um 18:58 schrieb Jim Burns:

    [...]

    You got it totally wrong!

    The dark unit fractions are smaller than
    the (all) visible ones.

    No positive point is a lower.bound of
    all the visibleᵂᴹ unit.fractions. (lemma)

    Each darkᵂᴹ unit.fraction is a positive point.

    No darkᵂᴹ unit.fraction is a lower.bound of
    all the visibleᵂᴹ unit.fractions.

    No darkᵂᴹ unit.fraction is smaller than
    all the visibleᵂᴹ unit.fractions.

    ----
    Lemma.
    No positive point is a lower.bound (lb) of
    all the visibleᵂᴹ unit.fractions.

    ⎛ Assume 0 < lb.⅟ℕᵈᵉᶠ ≤ glb.⅟ℕᵈᵉᶠ = β
    ⎜ not.bound 2⋅β > ⅟k ∈ ⅟ℕᵈᵉᶠ
    ⎜ not.bound ½⋅β > ¼⋅⅟k ∈ ⅟ℕᵈᵉᶠ
    ⎜ bound ½⋅β < β
    ⎝ Contradiction.

    ¬(lb.⅟ℕᵈᵉᶠ > 0)

    Now:
    The visible unit fraction don't have
    a smallest one (of course),

    More than that.
    The visibleᵂᴹ unit fractions don't have
    a positive lower bound.
    WM has not quite conceded that.
    The last I've seen, he omits 'greatest'.
    If he ever does, it is game over.

    With or without his concession,
    there is no positive lower bound of
    visibleᵂᴹ unit fractions, and
    there is no darkᵂᴹ unit.fraction.

    but the dark unit fraction do
    (at least in mückenmath)!

    WM: "Dark unit fractions have a smallest element."

    WM says a lot of things.
    If dark unit.fractions are positive lower bounds of
    visible unit fractions,
    then they don't exist,
    and {} doesn't have a smallest element.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Mon Aug 19 21:43:24 2024
    XPost: sci.math

    Am 19.08.2024 um 20:12 schrieb Jim Burns:

    WM: "Dark unit fractions have a smallest element."

    WM says a lot of things.

    Agree. :-)

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  • From Richard Damon@21:1/5 to All on Mon Aug 19 20:40:41 2024
    XPost: sci.math

    On 8/19/24 7:32 AM, WM wrote:
    Le 17/08/2024 à 16:29, Richard Damon a écrit :
    On 8/17/24 9:28 AM, WM wrote:
    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases, but at no point x it
    increases by more than 1 because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Therefore there is a smallest unit
    fractions and vice versa a greatest natnumber.
    What can't you understand?

    But there is no point (>0) where it has a finite value,

    You can't see it and you are unable to derive it from mathematics. But blindness is not an argument.

    Regards, WM




    But, you can't just claim sommething is there that isn't.

    The Number system is DERIVED from rules. Those rules DEFINE which
    numbers are, (and thus what are not).

    You can not derive a first number > 0 in any of the Number System that
    we have been talking about, Unit Fractions, Rationals or Reals, so you
    can't claim it to exist.

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  • From Mikko@21:1/5 to All on Tue Aug 20 13:36:21 2024
    On 2024-08-19 11:32:44 +0000, WM said:

    Le 17/08/2024 à 16:29, Richard Damon a écrit :
    On 8/17/24 9:28 AM, WM wrote:
    Le 16/08/2024 à 19:39, Jim Burns a écrit :

    no element of ℕᵈᵉᶠ is its upper.end,
    because
    for each diminishable k
    diminishable k+1 disproves by counter.example
    that k is the upper.end of ℕᵈᵉᶠ

    SBZ(x) starts with 0 at 0 and increases, but at no point x it increases
    by more than 1 because of
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Therefore there is a smallest unit fractions
    and vice versa a greatest natnumber.
    What can't you understand?

    But there is no point (>0) where it has a finite value,

    You can't see it and you are unable to derive it from mathematics. But blindness is not an argument.

    Blindness is indeed much weeker than a proof but you may use it
    as an argument as long as you can't find anything better.

    --
    Mikko

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  • From WM@21:1/5 to All on Tue Aug 20 13:00:46 2024
    XPost: sci.math

    Le 19/08/2024 à 19:17, Moebius a écrit :

    WM: "Dark unit fractions have a smallest element."

    See?!

    The function NUF(x) grows by 1 at every unit fraction. It starts from 0.

    WM: "Visible unit fractions have the [or rather *a*] lower bound 0."

    It is the greatest lower bound because every greater number can be
    undercut by a visible unit fraction.

    Right, but the dark unit fractions ARE BETWEEN 0 and all the visible
    unit fractions.

    So it is.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 20 12:53:27 2024
    XPost: sci.math

    Le 19/08/2024 à 18:58, Jim Burns a écrit :
    On 8/19/2024 7:40 AM, WM wrote:

    Dark unit fractions have a smallest element.

    Darkᵂᴹ unit.fractions are positive. Correct?

    Yes, all unit fractions are positive.

    No smaller unit fractions is existing

    Is that smallest darkᵂᴹ unit.fraction not a unit.fraction?
    Is it not positive?

    It is both positive and a unit fraction.

    Each positive point, thus,
    that smallest darkᵂᴹ unit.fraction
    is not a lower.bound, thus,
    has smaller unit.fractions, visibleᵂᴹ ones.

    No. The function NUF(x) grows by 1 at every unit fraction. It starts from
    0.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 20 13:07:25 2024
    XPost: sci.math

    Le 19/08/2024 à 20:12, Jim Burns a écrit :
    On 8/19/2024 1:17 PM, Moebius wrote:
    Am 19.08.2024 um 18:58 schrieb Jim Burns:

    [...]

    You got it totally wrong!

    The dark unit fractions are smaller than
    the (all) visible ones.

    No positive point is a lower.bound of
    all the visibleᵂᴹ unit.fractions. (lemma)

    Each darkᵂᴹ unit.fraction is a positive point.

    No darkᵂᴹ unit.fraction is a lower.bound of
    all the visibleᵂᴹ unit.fractions.

    No darkᵂᴹ unit.fraction is smaller than
    all the visibleᵂᴹ unit.fractions.

    ----
    Lemma.
    No positive point is a lower.bound (lb) of
    all the visibleᵂᴹ unit.fractions.

    ⎛ Assume 0 < lb.⅟ℕᵈᵉᶠ ≤ glb.⅟ℕᵈᵉᶠ = β
    ⎜ not.bound 2⋅β > ⅟k ∈ ⅟ℕᵈᵉᶠ
    ⎜ not.bound ½⋅β > ¼⋅⅟k ∈ ⅟ℕᵈᵉᶠ
    ⎜ bound ½⋅β < β
    ⎝ Contradiction.

    ¬(lb.⅟ℕᵈᵉᶠ > 0)

    Now:
    The visible unit fraction don't have
    a smallest one (of course),

    More than that.
    The visibleᵂᴹ unit fractions don't have
    a positive lower bound.
    WM has not quite conceded that.
    The last I've seen, he omits 'greatest'.
    If he ever does, it is game over.

    0 is the greatest lower bound because every greater number can be undercut
    by a visible unit fraction.

    With or without his concession,
    there is no positive lower bound of
    visibleᵂᴹ unit fractions, and
    there is no darkᵂᴹ unit.fraction.

    Dark unit fractions cannot be defined. The GLB that can be defined is 0.

    WM: "Dark unit fractions have a smallest element."

    WM says a lot of things.
    If dark unit.fractions are positive lower bounds of
    visible unit fractions,

    Impossible. Lower bounds must be definable.

    then they don't exist,

    The function NUF(x) grows by 1 at every unit fraction. It starts from 0.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 20 13:19:06 2024
    XPost: sci.math

    Le 20/08/2024 à 02:40, Richard Damon a écrit :

    But, you can't just claim sommething is there that isn't.

    This is: The function NUF(x) grows by 1 at every unit fraction. It starts
    from 0.

    The Number system is DERIVED from rules.

    The definable numbers.

    You can not derive a first number > 0 in any of the Number System that
    we have been talking about, Unit Fractions, Rationals or Reals, so you
    can't claim it to exist.

    This is fact: The function NUF(x) grows by 1 at every unit fraction. It
    starts from 0.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 20 13:21:44 2024
    XPost: sci.math

    Le 20/08/2024 à 12:31, FromTheRafters a écrit :
    on 8/19/2024, Richard Damon supposed :

    You can not derive a first number > 0 in any of the Number System that we
    have been talking about, Unit Fractions, Rationals or Reals, so you can't
    claim it to exist.

    Not in their natural ordering.

    Dark numbers have no discernible order. It is impossible to find the
    smallest unit fraction or the next one or the next one. It is only
    possible to prove that NUF(x) grows by 1 at every unit fraction. It starts
    from 0.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 20 13:25:36 2024
    XPost: sci.math

    Le 20/08/2024 à 15:21, Python a écrit :
    Le 20/08/2024 à 15:19, WM a écrit :

    This is fact: The function NUF(x) grows by 1 at every unit fraction. It
    starts from 0.

    But 0 is not a unit fraction. sigh.

    Therefore NUF(0) = 0. NUF(u) = 1 only at the first unit fraction u.

    Are you about to teach this at Technische Hochschule Augsburg

    Regards, WM

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  • From Python@21:1/5 to All on Tue Aug 20 15:21:26 2024
    XPost: sci.math

    Le 20/08/2024 à 15:19, WM a écrit :
    Le 20/08/2024 à 02:40, Richard Damon a écrit :

    But, you can't just claim sommething is there that isn't.

    This is: The function NUF(x) grows by 1 at every unit fraction. It
    starts from 0.

    The Number system is DERIVED from rules.

    The definable numbers.

    You can not derive a first number > 0 in any of the Number System that
    we have been talking about, Unit Fractions, Rationals or Reals, so you
    can't claim it to exist.

    This is fact: The function NUF(x) grows by 1 at every unit fraction. It starts from 0.

    But 0 is not a unit fraction. sigh.

    Are you about to "teach" this nonsense at Hochschule Augsburg again next
    year, crank Wolfgang Mückenheim or did they have fired you as you
    deserve?

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  • From Python@21:1/5 to All on Tue Aug 20 15:28:28 2024
    XPost: sci.math

    Le 20/08/2024 à 15:25, WM a écrit :
    Le 20/08/2024 à 15:21, Python a écrit :
    Le 20/08/2024 à 15:19, WM a écrit :

    This is fact: The function NUF(x) grows by 1 at every unit fraction.
    It starts from 0.

    But 0 is not a unit fraction. sigh.

    Therefore NUF(0) = 0. NUF(u) = 1 only at the first unit fraction u.

    There in NO such first unit fraction. EOS.

    Are you about to teach this at Technische Hochschule Augsburg

    Are you about to "teach" this nonsense at Hochschule Augsburg again next
    year, crank Wolfgang Mückenheim or did they have fired you as you
    deserve?

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  • From Python@21:1/5 to All on Tue Aug 20 15:41:39 2024
    XPost: sci.math

    Le 20/08/2024 à 15:39, WM a écrit :
    Le 20/08/2024 à 15:28, Python a écrit :
    Le 20/08/2024 à 15:25, WM a écrit :
    Le 20/08/2024 à 15:21, Python a écrit :
    Le 20/08/2024 à 15:19, WM a écrit :

    This is fact: The function NUF(x) grows by 1 at every unit
    fraction. It starts from 0.

    But 0 is not a unit fraction. sigh.

    Therefore NUF(0) = 0. NUF(u) = 1 only at the first unit fraction u.

    There in NO such first unit fraction.

    There are unit fractions.

    Definitely.

    NUF(x) counts them all. But it nowhere counts
    more than 1.

    False.

    Try to think.

    Don't talk about something you do not experience.

    Are you about to teach this at Technische Hochschule Augsburg

    If you can explain how more than one first unit fraction can bolster
    NUF(x), I will adopt your explanation. But only if it does not violate
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

    Are you about to "teach" this nonsense at Hochschule Augsburg again next
    year, crank Wolfgang Mückenheim or did they have fired you as you
    deserve?

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  • From WM@21:1/5 to All on Tue Aug 20 13:39:25 2024
    XPost: sci.math

    Le 20/08/2024 à 15:28, Python a écrit :
    Le 20/08/2024 à 15:25, WM a écrit :
    Le 20/08/2024 à 15:21, Python a écrit :
    Le 20/08/2024 à 15:19, WM a écrit :

    This is fact: The function NUF(x) grows by 1 at every unit fraction.
    It starts from 0.

    But 0 is not a unit fraction. sigh.

    Therefore NUF(0) = 0. NUF(u) = 1 only at the first unit fraction u.

    There in NO such first unit fraction.

    There are unit fractions. NUF(x) counts them all. But it nowhere counts
    more than 1. Try to think.

    Are you about to teach this at Technische Hochschule Augsburg

    If you can explain how more than one first unit fraction can bolster
    NUF(x), I will adopt your explanation. But only if it does not violate
    ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

    Regards, WM

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  • From WM@21:1/5 to All on Tue Aug 20 13:58:20 2024
    XPost: sci.math

    Le 20/08/2024 à 15:41, Python a écrit :
    Le 20/08/2024 à 15:39, WM a écrit :

    There are unit fractions.

    Definitely.

    NUF(x) counts them all. But it nowhere counts
    more than 1.

    False.

    Where?

    Regards, WM

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  • From Python@21:1/5 to All on Tue Aug 20 16:04:47 2024
    XPost: sci.math

    Le 20/08/2024 à 15:58, WM a écrit :
    Le 20/08/2024 à 15:41, Python a écrit :
    Le 20/08/2024 à 15:39, WM a écrit :

    There are unit fractions.

    Definitely.

    NUF(x) counts them all. But it nowhere counts more than 1.

    False.

    Where?

    At any x > 0. This has been shown to you numerous times.



    Are you about to "teach" this nonsense at Hochschule Augsburg again next
    year, crank Wolfgang Mückenheim or did they have fired you as you
    deserve?

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  • From Jim Burns@21:1/5 to All on Tue Aug 20 16:05:59 2024
    XPost: sci.math

    On 8/20/2024 9:07 AM, WM wrote:
    Le 19/08/2024 à 20:12, Jim Burns a écrit :

    Lemma.
    No positive point is a lower.bound (lb) of
    all the visibleᵂᴹ unit.fractions.
    ¬(lb.⅟ℕᵈᵉᶠ > 0)

    0 is the greatest lower bound
    because
    every greater number can be undercut by
    a visible unit fraction.

    Dark unit fractions cannot be defined.
    The GLB that can be defined is 0.

    And 0 is glb.⅟ℕᵈᵉᶠ not.in ⅟ℕᵈᵉᶠ

    ⎛ For set S,
    ⎜ min.S is defined to be glb.S in S

    ⎜ Because two greatest.lb.S cannot exist
    ⎜ if glb.S is not.in S
    ⎜ then no glb.S in S exists
    ⎝ and no min.S exists.

    No glb.⅟ℕᵈᵉᶠ in ⅟ℕᵈᵉᶠ exists.
    No min.⅟ℕᵈᵉᶠ exists.

    It starts from 0.

    If ⅟ℕᵈᵉᶠ starts,
    ⅟ℕᵈᵉᶠ starts from min.⅟ℕᵈᵉᶠ

    No min.⅟ℕᵈᵉᶠ exists.
    ⅟ℕᵈᵉᶠ does not start.

    Because 0 is glb.⅟ℕᵈᵉᶠ not.in ⅟ℕᵈᵉᶠ

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  • From Moebius@21:1/5 to All on Wed Aug 21 02:16:56 2024
    XPost: sci.math

    Am 20.08.2024 um 22:05 schrieb Jim Burns:
    On 8/20/2024 9:07 AM, WM wrote:

    0 is the greatest lower bound
    because
    every greater number can be undercut by
    a [...] unit fraction.

    Right.

    Dark unit fractions cannot be defined.

    Right. Since there are no such unit fractions. Simple as that.

    The GLB that can be defined is 0.

    Errr... Well, yes. And now?

    And 0 is glb.⅟ℕᵈᵉᶠ not.in ⅟ℕᵈᵉᶠ

    Indeed! After all 0 is not a unit fraction.

    It starts from 0.

    It? WM's asshole?

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  • From Richard Damon@21:1/5 to All on Tue Aug 20 22:02:18 2024
    XPost: sci.math

    On 8/20/24 9:19 AM, WM wrote:
    Le 20/08/2024 à 02:40, Richard Damon a écrit :

    But, you can't just claim sommething is there that isn't.

    This is: The function NUF(x) grows by 1 at every unit fraction. It
    starts from 0.

    The Number system is DERIVED from rules.

    The definable numbers.

    You can not derive a first number > 0 in any of the Number System that
    we have been talking about, Unit Fractions, Rationals or Reals, so you
    can't claim it to exist.

    This is fact: The function NUF(x) grows by 1 at every unit fraction. It starts from 0.

    Regards, WM



    No, that is your madness that imagines things that are not.

    The contradictions it creates has blown your logic, and your brain to smithereens.

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  • From WM@21:1/5 to All on Wed Aug 21 10:43:17 2024
    XPost: sci.math

    Le 20/08/2024 à 22:05, Jim Burns a écrit :
    On 8/20/2024 9:07 AM, WM wrote:

    No min.⅟ℕᵈᵉᶠ exists.

    The reason is potential infinity. But dark unit fractions are assumed to
    be actually infinite.

    It starts from 0.

    If ⅟ℕᵈᵉᶠ starts,
    ⅟ℕᵈᵉᶠ starts from min.⅟ℕᵈᵉᶠ

    No min.⅟ℕᵈᵉᶠ exists.
    ⅟ℕᵈᵉᶠ does not start.

    Because 0 is glb.⅟ℕᵈᵉᶠ not.in ⅟ℕᵈᵉᶠ

    All that is correct for definable unit fractions.
    But NUF(x) starts from 0 and cannot avoid to take the values 1, 2, 3, ...
    .

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 21 10:47:16 2024
    XPost: sci.math

    Le 21/08/2024 à 02:16, Moebius a écrit :
    Am 20.08.2024 um 22:05 schrieb Jim Burns:

    Dark unit fractions cannot be defined.

    Right. Since there are no such unit fractions. Simple as that.

    If all are existing, then there are dark ones.

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 21 10:44:59 2024
    XPost: sci.math

    Le 20/08/2024 à 23:25, FromTheRafters a écrit :
    WM explained :
    Le 20/08/2024 à 12:31, FromTheRafters a écrit :
    on 8/19/2024, Richard Damon supposed :

    You can not derive a first number > 0 in any of the Number System that we >>>> have been talking about, Unit Fractions, Rationals or Reals, so you can't >>>> claim it to exist.

    Not in their natural ordering.

    Dark numbers have no discernible order. It is impossible to find the smallest
    unit fraction or the next one or the next one. It is only possible to prove >> that NUF(x) grows by 1 at every unit fraction. It starts from 0.

    Normally, the unit fractions are listed in the sequence one over one,
    one over two, one over three etcetera. There is a first but no last.
    Now you have started from the wrong 'end'

    No, I have started from the other end. It exists at x > 0 because NUF(0) =
    0.

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 21 10:39:03 2024
    XPost: sci.math

    Le 20/08/2024 à 16:04, Python a écrit :
    Le 20/08/2024 à 15:58, WM a écrit :
    Le 20/08/2024 à 15:41, Python a écrit :
    Le 20/08/2024 à 15:39, WM a écrit :

    There are unit fractions.

    Definitely.

    NUF(x) counts them all. But it nowhere counts more than 1.

    False.

    Where?

    At any x > 0. This has been shown to you numerous times.

    You are stupid or a liar. All unit fractions are separated. The increase
    of NUF(x) from 0 at x = 0 cannot avoid 1, 2, 3, ... .

    Are you about to "teach" this nonsense at Hochschule Augsburg again next year,

    I will teach mathematics this year: All unit fractions are separated.
    Proof: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .

    Regards, WM

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  • From Richard Damon@21:1/5 to All on Wed Aug 21 07:32:51 2024
    XPost: sci.math

    On 8/21/24 6:44 AM, WM wrote:
    Le 20/08/2024 à 23:25, FromTheRafters a écrit :
    WM explained :
    Le 20/08/2024 à 12:31, FromTheRafters a écrit :
    on 8/19/2024, Richard Damon supposed :

    You can not derive a first number > 0 in any of the Number System
    that we have been talking about, Unit Fractions, Rationals or
    Reals, so you can't claim it to exist.

    Not in their natural ordering.

    Dark numbers have no discernible order. It is impossible to find the
    smallest unit fraction or the next one or the next one. It is only
    possible to prove that NUF(x) grows by 1 at every unit fraction. It
    starts from 0.

    Normally, the unit fractions are listed in the sequence one over one,
    one over two, one over three etcetera. There is a first but no last.
    Now you have started from the wrong 'end'

    No, I have started from the other end. It exists at x > 0 because NUF(0)
    = 0.

    Regards, WM



    But the other end doesn't "begin" with a first Natural Number Unit
    fraction, if it has a beginning that will be a trans-finite number.

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  • From Richard Damon@21:1/5 to All on Wed Aug 21 07:31:26 2024
    XPost: sci.math

    On 8/21/24 6:43 AM, WM wrote:
    Le 20/08/2024 à 22:05, Jim Burns a écrit :
    On 8/20/2024 9:07 AM, WM wrote:

    No min.⅟ℕᵈᵉᶠ exists.

    The reason is potential infinity. But dark unit fractions are assumed to
    be actually infinite.

    It starts from 0.

    If ⅟ℕᵈᵉᶠ starts,
    ⅟ℕᵈᵉᶠ starts from min.⅟ℕᵈᵉᶠ

    No min.⅟ℕᵈᵉᶠ exists.
    ⅟ℕᵈᵉᶠ does not start.

    Because 0 is glb.⅟ℕᵈᵉᶠ not.in ⅟ℕᵈᵉᶠ

    All that is correct for definable unit fractions.
    But NUF(x) starts from 0 and cannot avoid to take the values 1, 2, 3, ... .

    Regards, WM



    And thus must be doing so at things that are not the reciprical of
    Natural Numbers, since there is no such thing as the first unit fraction
    of Natural numbers, since there is no highest Natural Number.

    Thus your NUF must be counting trans-finite unit fractions, which you
    call dark so you don't see that they are not real unit fractions.

    Maybe the first dark unit fraction is at 1/Omega or 1/(Omega-1)

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  • From Richard Damon@21:1/5 to All on Wed Aug 21 07:26:09 2024
    XPost: sci.math

    On 8/21/24 6:47 AM, WM wrote:
    Le 21/08/2024 à 02:16, Moebius a écrit :
    Am 20.08.2024 um 22:05 schrieb Jim Burns:

    Dark unit fractions cannot be defined.

    Right. Since there are no such unit fractions. Simple as that.

    If all are existing, then there are dark ones.

    Regards, WM

    But the dark ones are not recipricals of Natural Numbers, because those
    aren't dark.

    These "Dark Numbers" seem to live in the trans-finite gaps between 0 and
    finite x > 0, and have the aleph_0 values that NUF(x) consider to be
    unit fractions, but are not reciprocal of Natural Numbers, but of some trans-finite values greater than the finite natural numbers but below Omega.

    Of course, since you mathematics can't actually handle the Natural
    Numbers, it can't handle the mathematic of these dark numbers.

    You only think your dark numbers are part of the Natual Numbers, because
    your math can handle that full of the Natual Numbers.

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Wed Aug 21 12:29:35 2024
    XPost: sci.math

    Le 21/08/2024 à 13:26, Richard Damon a écrit :
    On 8/21/24 6:47 AM, WM wrote:

    If all are existing, then there are dark ones.

    But the dark ones are not recipricals of Natural Numbers, because those aren't dark.

    There are as many dark natnumbers as dark unit fractions, namely ℵo,
    i.e. much more than visible.
    But since omega is not such an accepted fixed number as zero, the end of
    the dark natnumbers is not as obvious as the end of the unit fractions.

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 21 12:24:14 2024
    XPost: sci.math

    Le 21/08/2024 à 12:58, FromTheRafters a écrit :
    WM wrote :
    Le 20/08/2024 à 23:25, FromTheRafters a écrit :
    WM explained :
    Le 20/08/2024 à 12:31, FromTheRafters a écrit :
    on 8/19/2024, Richard Damon supposed :

    You can not derive a first number > 0 in any of the Number System that >>>>>> we have been talking about, Unit Fractions, Rationals or Reals, so you >>>>>> can't claim it to exist.

    Not in their natural ordering.

    Dark numbers have no discernible order. It is impossible to find the
    smallest unit fraction or the next one or the next one. It is only
    possible to prove that NUF(x) grows by 1 at every unit fraction. It starts >>>> from 0.

    Normally, the unit fractions are listed in the sequence one over one, one >>> over two, one over three etcetera. There is a first but no last. Now you >>> have started from the wrong 'end'

    No, I have started from the other end.

    Correct, that is the wrong end.

    It is an end which proves dark numbers.

    It exists at x > 0 because NUF(0) = 0.

    Since no unit fraction is below or at zero, the end is before. What else?

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 21 12:32:30 2024
    XPost: sci.math

    Le 21/08/2024 à 13:32, Richard Damon a écrit :
    On 8/21/24 6:44 AM, WM wrote:
    Le 20/08/2024 à 23:25, FromTheRafters a écrit :
    WM explained :
    Le 20/08/2024 à 12:31, FromTheRafters a écrit :
    on 8/19/2024, Richard Damon supposed :

    You can not derive a first number > 0 in any of the Number System
    that we have been talking about, Unit Fractions, Rationals or
    Reals, so you can't claim it to exist.

    Not in their natural ordering.

    Dark numbers have no discernible order. It is impossible to find the
    smallest unit fraction or the next one or the next one. It is only
    possible to prove that NUF(x) grows by 1 at every unit fraction. It
    starts from 0.

    Normally, the unit fractions are listed in the sequence one over one,
    one over two, one over three etcetera. There is a first but no last.
    Now you have started from the wrong 'end'

    No, I have started from the other end. It exists at x > 0 because NUF(0)
    = 0.

    But the other end doesn't "begin" with a first Natural Number Unit
    fraction, if it has a beginning that will be a trans-finite number.

    No, it is a finite number. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds for all
    and only reciprocals of natural numbers.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Wed Aug 21 14:20:44 2024
    XPost: sci.math

    On 8/21/2024 6:43 AM, WM wrote:
    Le 20/08/2024 à 22:05, Jim Burns a écrit :

    No min.⅟ℕᵈᵉᶠ exists.

    The reason is potential infinity.

    Is potentialᵂᴹ.infinity mathematicsᵂᴹ?

    Judging from your edicts on this topic,
    the answer looks to be both 'yes' and 'no'.


    We are agreed, though, that
    no min.⅟ℕᵈᵉᶠ exists.

    It follows that ⅟ℕᵈᵉᶠ has properties that
    the unit.fractionsᵈᵉᶠ > x > 0
    don't have.
    Each non.empty subset S ⊆ ⅟ℕᵈᵉᶠ(x,1]
    holds two ends, max.S min.S

    Contrary to that,
    ⅟ℕᵈᵉᶠ as a whole has
    a non.empty subset which does not hold two ends,
    itself, ⅟ℕᵈᵉᶠ

    But dark unit fractions are assumed to be
    actually infinite.

    As we move along, I find out more about
    what darkᵂᴹ unit.fractions and darkᵂᴹ numbers are NOT.

    We agree that darkᵂᴹ unit.fractions are NOT
    positive lower.bounds of the visibleᵂᴹ unit.fractions,
    NOT zero, NOT negative, NOT larger than
    any visibleᵂᴹ unit.fraction.

    NOT lower.bound and NOT not.lower.bound.
    I don't see anything left for the darkᵂᴹ to NOT not.be.

    [...] actually infinite.

    Consider finiteⁿᵒᵗᐧᵂᴹ sets as
    sets orderable with each non.{}.subset 2.ended.

    Each subset of a finiteⁿᵒᵗᐧᵂᴹ set is finiteⁿᵒᵗᐧᵂᴹ.

    Each superset of an infiniteⁿᵒᵗᐧᵂᴹ set is infiniteⁿᵒᵗᐧᵂᴹ.

    In order to know those, we only need to know
    what is a subset, what is a superset.

    ⎛ Each non.{}.subset of
    ⎜ a non.{}.subset of a finiteⁿᵒᵗᐧᵂᴹ set
    ⎜ is
    ⎜ a non.{}.subset of a finiteⁿᵒᵗᐧᵂᴹ set
    ⎜ and thus is
    ⎜ 2.ended.

    ⎜ Each non.{}.subset of a finiteⁿᵒᵗᐧᵂᴹ set
    ⎜ is
    ⎝ finiteⁿᵒᵗᐧᵂᴹ.

    ⎛ Each superset of
    ⎜ a set orderable with a non.2.ended subset
    ⎜ is
    ⎜ a set orderable with a non.2.ended subset

    ⎜ Each superset of an infiniteⁿᵒᵗᐧᵂᴹ set
    ⎜ is
    ⎝ infiniteⁿᵒᵗᐧᵂᴹ.

    It seems to be that
    potentiallyᵂᴹ.infinite sets and
    infiniteⁿᵒᵗᐧᵂᴹ sets are the same.

    Only potentiallyᵂᴹ.infinite sets have
    potentiallyᵂᴹ.infinite subsets.

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Wed Aug 21 20:10:54 2024
    XPost: sci.math

    On 8/21/24 8:32 AM, WM wrote:
    Le 21/08/2024 à 13:32, Richard Damon a écrit :
    On 8/21/24 6:44 AM, WM wrote:
    Le 20/08/2024 à 23:25, FromTheRafters a écrit :
    WM explained :
    Le 20/08/2024 à 12:31, FromTheRafters a écrit :
    on 8/19/2024, Richard Damon supposed :

    You can not derive a first number > 0 in any of the Number System >>>>>>> that we have been talking about, Unit Fractions, Rationals or
    Reals, so you can't claim it to exist.

    Not in their natural ordering.

    Dark numbers have no discernible order. It is impossible to find
    the smallest unit fraction or the next one or the next one. It is
    only possible to prove that NUF(x) grows by 1 at every unit
    fraction. It starts from 0.

    Normally, the unit fractions are listed in the sequence one over
    one, one over two, one over three etcetera. There is a first but no
    last. Now you have started from the wrong 'end'

    No, I have started from the other end. It exists at x > 0 because
    NUF(0) = 0.

    But the other end doesn't "begin" with a first Natural Number Unit
    fraction, if it has a beginning that will be a trans-finite number.

    No, it is a finite number. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds for all and only reciprocals of natural numbers.

    Regards, WM



    Can't be, because if it WAS 1/n, then 1/(n+1) would be before it, and
    thus your claim is wrong. If 1/(n+1) wasn't smaller than 1/n, then we
    just have that 1/n - 1/(n+1) wouldn't be > 0, so it can't be.

    BY DEFINITION, there is no "Highest" Natural Number, if n exists, so
    does n+1, and your formula says you accept that n+1 exists, or you
    couldn't use it.

    If you don't have that property, you don't have the Natural Numbers.

    PERIOD.

    DEFINITION.

    If you claim your mathematics say it can't be, then your mathematics
    were just proven to not be abble to handle the unbounded set of the
    Natural Numbers.

    Sorry, that is just the facts.

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to Ross Finlayson on Wed Aug 21 21:49:26 2024
    XPost: sci.math

    On 8/21/24 9:23 PM, Ross Finlayson wrote:
    On 08/21/2024 05:10 PM, Richard Damon wrote:
    On 8/21/24 8:32 AM, WM wrote:
    Le 21/08/2024 à 13:32, Richard Damon a écrit :
    On 8/21/24 6:44 AM, WM wrote:
    Le 20/08/2024 à 23:25, FromTheRafters a écrit :
    WM explained :
    Le 20/08/2024 à 12:31, FromTheRafters a écrit :
    on 8/19/2024, Richard Damon supposed :

    You can not derive a first number > 0 in any of the Number
    System that we have been talking about, Unit Fractions,
    Rationals or Reals, so you can't claim it to exist.

    Not in their natural ordering.

    Dark numbers have no discernible order. It is impossible to find >>>>>>> the smallest unit fraction or the next one or the next one. It is >>>>>>> only possible to prove that NUF(x) grows by 1 at every unit
    fraction. It starts from 0.

    Normally, the unit fractions are listed in the sequence one over
    one, one over two, one over three etcetera. There is a first but no >>>>>> last. Now you have started from the wrong 'end'

    No, I have started from the other end. It exists at x > 0 because
    NUF(0) = 0.

    But the other end doesn't "begin" with a first Natural Number Unit
    fraction, if it has a beginning that will be a trans-finite number.

    No, it is a finite number. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds for all and
    only reciprocals of natural numbers.

    Regards, WM



    Can't be, because if it WAS 1/n, then 1/(n+1) would be before it, and
    thus your claim is wrong. If 1/(n+1) wasn't smaller than 1/n, then we
    just have that 1/n - 1/(n+1) wouldn't be > 0, so it can't be.

    BY DEFINITION, there is no "Highest" Natural Number, if n exists, so
    does n+1, and your formula says you accept that n+1 exists, or you
    couldn't use it.

    If you don't have that property, you don't have the Natural Numbers.

    PERIOD.

    DEFINITION.

    If you claim your mathematics say it can't be, then your mathematics
    were just proven to not be abble to handle the unbounded set of the
    Natural Numbers.

    Sorry, that is just the facts.

    Perhaps you'd like to learn about Conway's "Surreal Numbers",
    which make one for omega and further fill out a "non-Archimedean"
    field, that otherwise it's the same usual definition since Archimedes,
    in terms of field reals, the Archimedean field.

    I know about the surreal numbers. But that would just further confuse
    the idiot. Just pointing out that his "dark numbers' can't be a part of
    the Natural Numbers breaks his logic.

    Of course, since is logic can't even handle the simple unbounded Natural Numbers, he has no hope in understanding the Surreal numbers.



    So, the usual idea of a "non-Archimedean field" is Conway's "sur-reals".


    Then, there's also a logical argument that if there are infinitely-many integers then that there are concomitantly infinitely-grand integers,
    that it belies the definition and makes it so that inductive inference
    by itself doesn't suffice, that "Eudoxus doesn't suffice", that
    "if there are infinite integers there are infinite integers"
    and so on, that logic automatically provides.

    So, you can find ways to make the points that there is a
    fixed-point, to the integers, that the integers _are_ compact,
    that there is an infinite member of otherwise the finite set,
    and these kinds of things, while at the same time the usual
    formalism's only use is that inductive inference never ends,
    abruptly.



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  • From WM@21:1/5 to All on Thu Aug 22 10:54:46 2024
    XPost: sci.math

    Le 21/08/2024 à 18:20, FromTheRafters a écrit :
    WM presented the following explanation :

    Since no unit fraction is below or at zero, the end is before. What else?

    No end, Duh!!

    In a linear order of elements which all have distances from each other,
    there is necessarily a last one (if nothing follows) because the only alternative would be more than one. Matheologial conjuring trick is
    outside of mathematics.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 22 11:17:25 2024
    XPost: sci.math

    Le 21/08/2024 à 20:20, Jim Burns a écrit :
    On 8/21/2024 6:43 AM, WM wrote:
    Le 20/08/2024 à 22:05, Jim Burns a écrit :

    No min.⅟ℕᵈᵉᶠ exists.

    The reason is potential infinity.

    Is potentialᵂᴹ.infinity mathematicsᵂᴹ?

    All of classical mathematics is embedded into potential infinity.

    Judging from your edicts on this topic,
    the answer looks to be both 'yes' and 'no'.

    I am interested in actual infinity. We cannot prove its existence but only apply logic like this: If a complete chain exists which does not continue
    into the negative, then it ends before. Either at one or at more members.
    The latter is excluded by ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .


    We are agreed, though, that
    no min.⅟ℕᵈᵉᶠ exists.

    Of course. After each one there will appear more. Choose the smallest one
    you can, recognize that there are many more following.

    It follows that ⅟ℕᵈᵉᶠ has properties that
    the unit.fractionsᵈᵉᶠ > x > 0
    don't have.

    There is no set ℕᵈᵉᶠ because of lacking completeness. Sets of set theory are complete.

    Regards, WM

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  • From Python@21:1/5 to All on Thu Aug 22 13:28:28 2024
    XPost: sci.math

    Le 22/08/2024 à 13:17, Crank Wolfgang Mückenheim a écrit :
    ...
    There is no set ℕᵈᵉᶠ because of lacking completeness. Sets of set theory
    are complete.

    We know, in mückenmath sets are so complete that you can remove one
    of their members without changing them. Like in e \in S and S \ {e} = S
    as you claimed numerous times.

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Thu Aug 22 12:27:45 2024
    XPost: sci.math

    Le 22/08/2024 à 13:28, Python a écrit :
    Le 22/08/2024 à 13:17, Crank Wolfgang Mückenheim a écrit :
    ...
    There is no set ℕᵈᵉᶠ because of lacking completeness. Sets of set theory

    are complete.

    We know, in mückenmath sets are so complete that you can remove one
    of their members without changing them. Like in e \in S and S \ {e} = S
    as you claimed numerous times.

    That is true for collections in potential infinity. It has nothing to do
    with the present topic.

    Regards, WM

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  • From Python@21:1/5 to All on Thu Aug 22 14:48:41 2024
    XPost: sci.math

    Le 22/08/2024 à 14:27, crank Wolfgang Mückenheim a écrit :
    Le 22/08/2024 à 13:28, Python a écrit :
    Le 22/08/2024 à 13:17, Crank Wolfgang Mückenheim a écrit :
    ...
    There is no set ℕᵈᵉᶠ because of lacking completeness. Sets of set theory
    are complete.

    We know, in mückenmath sets are so complete that you can remove one
    of their members without changing them. Like in e \in S and S \ {e} = S
    as you claimed numerous times.

    That is true

    You should advise you non-fellows mathematicians at University of
    Augsburg. I mean "University", not the bad joke that is Hochschule
    Augsburg.

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  • From Mikko@21:1/5 to All on Thu Aug 22 15:27:37 2024
    On 2024-08-21 12:24:14 +0000, WM said:

    Le 21/08/2024 à 12:58, FromTheRafters a écrit :
    WM wrote :
    Le 20/08/2024 à 23:25, FromTheRafters a écrit :
    WM explained :
    Le 20/08/2024 à 12:31, FromTheRafters a écrit :
    on 8/19/2024, Richard Damon supposed :

    You can not derive a first number > 0 in any of the Number System that >>>>>>> we have been talking about, Unit Fractions, Rationals or Reals, so you >>>>>>> can't claim it to exist.

    Not in their natural ordering.

    Dark numbers have no discernible order. It is impossible to find the >>>>> smallest unit fraction or the next one or the next one. It is only
    possible to prove that NUF(x) grows by 1 at every unit fraction. It
    starts from 0.

    Normally, the unit fractions are listed in the sequence one over one,
    one over two, one over three etcetera. There is a first but no last.
    Now you have started from the wrong 'end'

    No, I have started from the other end.

    Correct, that is the wrong end.

    It is an end which proves dark numbers.

    It exists at x > 0 because NUF(0) = 0.

    Since no unit fraction is below or at zero, the end is before. What else?

    Seems that you are making hidden assumptions about the end of endless.

    --
    Mikko

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  • From WM@21:1/5 to All on Thu Aug 22 12:19:58 2024
    XPost: sci.math

    Le 22/08/2024 à 02:10, Richard Damon a écrit :
    On 8/21/24 8:32 AM, WM wrote:

    No, it is a finite number. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds for all and >> only reciprocals of natural numbers.

    Can't be, because if it WAS 1/n, then 1/(n+1) would be before it,

    That is tadopted from definable numbers. It is not true for all dark
    numbers.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Thu Aug 22 14:38:23 2024
    XPost: sci.math

    On 8/22/2024 7:17 AM, WM wrote:
    Le 21/08/2024 à 20:20, Jim Burns a écrit :
    On 8/21/2024 6:43 AM, WM wrote:
    Le 20/08/2024 à 22:05, Jim Burns a écrit :

    No min.⅟ℕᵈᵉᶠ exists.

    The reason is potential infinity.

    Is potentialᵂᴹ.infinity mathematicsᵂᴹ?

    All of classical mathematics is
    embedded into potential infinity.

    What you didn't say is that
    potentialᵂᴹ.infinity is embedded into mathematicsᵂᴹ.

    Should I draw the implicature from your response that
    it isn't?
    Is mathematicsᵂᴹ within potentialᵂᴹ.infinity, but potentialᵂᴹ.infinity not within mathematicsᵂᴹ?

    You (WM) have a personal vocabulary,
    mathematicsᵂᴹ and potentialᵂᴹ and actualᵂᴹ.
    We use the same words you do,
    mathematicsⁿᵒᵗᐧᵂᴹ and potentialⁿᵒᵗᐧᵂᴹ and actualⁿᵒᵗᐧᵂᴹ,
    but not in the way you do.

    Work has been done during
    the last few thousand years
    to extend mathematicsⁿᵒᵗᐧᵂᴹ over infinityⁿᵒᵗᐧᵂᴹ.
    The distinction potentialⁿᵒᵗᐧᵂᴹ::actualⁿᵒᵗᐧᵂᴹ
    has turned out to be less than useful

    Judging from your edicts on this topic,
    the answer looks to be both 'yes' and 'no'.

    I am interested in actual infinity.

    It seems that your potentialᵂᴹ.infinity is
    our infinityⁿᵒᵗᐧᵂᴹ.

    It seems that your actualᵂᴹ.infinity isn't
    potentialᵂᴹ.infinity but it has a subset of
    potentialᵂᴹ.infinity.

    There is no superset of
    a potentiallyᵂᴹ.infinite (infiniteⁿᵒᵗᐧᵂᴹ) set which isn't
    a potentiallyᵂᴹ.infinite (infiniteⁿᵒᵗᐧᵂᴹ) superset.

    There is no actuallyᵂᴹ.infinite set.

    Augmenting with darkᵂᴹ.numbers cannot change
    the potentiallyᵂᴹ.infinite into the actuallyᵂᴹ.infinite
    because
    any non.2.ended subset which a potentiallyᵂᴹ.infinite has
    is also a non.2.ended subset of any superset.

    It follows that ⅟ℕᵈᵉᶠ has properties that
    the unit.fractionsᵈᵉᶠ > x > 0
    don't have.

    There is no set ℕᵈᵉᶠ because of lacking completeness.

    ⎛ Each non.empty S ⊆ ℕᵈᵉᶠ holds min.S in S.
    ⎜ Each k ∈ ℕᵈᵉᶠ has k+1 and k-1 next in ℕᵈᵉᶠ, except for ⎝ min.ℕᵈᵉᶠ = 0 which has 0+1 but not 0-1 in ℕᵈᵉᶠ

    Whatever class ℕᵈᵉᶠ belongs in, the above is true of ℕᵈᵉᶠ.
    It's what we mean by ℕᵈᵉᶠ.

    A finite sequence of not.first.false claims
    cannot have a first false claim and thus
    cannot have a false claim.
    It's part of what we mean by finite.

    We can assemble finite sequences of not.first.false claims
    which include the above claims about ℕᵈᵉᶠ.
    Each of its claims about ℕᵈᵉᶠ is true.

    If that's incompleteᵂᴹ,
    it'll do quite well until completeᵂᴹ comes along.

    Sets of set theory are complete.

    A setⁿᵒᵗᐧᵂᴹ does not change.
    A setⁿᵒᵗᐧᵂᴹ which has been changed is no longer
    the setⁿᵒᵗᐧᵂᴹ it was.
    The setⁿᵒᵗᐧᵂᴹ it was has not changed.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Thu Aug 22 21:23:33 2024
    XPost: sci.math

    On 8/22/24 6:54 AM, WM wrote:
    Le 21/08/2024 à 18:20, FromTheRafters a écrit :
    WM presented the following explanation :

    Since no unit fraction is below or at zero, the end is before. What
    else?

    No end, Duh!!

    In a linear order of elements which all have distances from each other,
    there is necessarily a last one (if nothing follows) because the only alternative would be more than one. Matheologial conjuring trick is
    outside of mathematics.

    Regards, WM

    But that is just a false statement that tries to assume the conclusion.

    Yes, if there is an element with nothing following, it will be the last one.

    But, if every element has elements following, there will not be a last
    one. Of course, that can only happen with an infinite set, which it
    seems your logic is incapable of actually handling.

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Fri Aug 23 14:49:34 2024
    XPost: sci.math

    Le 23/08/2024 à 03:23, Richard Damon a écrit :
    On 8/22/24 6:54 AM, WM wrote:

    In a linear order of elements which all have distances from each other,
    there is necessarily a last one (if nothing follows) because the only
    alternative would be more than one. Matheologial conjuring trick is
    outside of mathematics.

    But that is just a false statement that tries to assume the conclusion.

    NUF(x) increases from 0 to more. But it cannot increase by more than 1
    without shzowing a constant level.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Python@21:1/5 to All on Fri Aug 23 16:32:48 2024
    XPost: sci.math

    Le 23/08/2024 à 16:28, WM a écrit :
    Le 22/08/2024 à 16:22, FromTheRafters a écrit :
    WM used his keyboard to write :

    In a linear order of elements which all have distances from each
    other, there is necessarily a last one (if nothing follows) because
    the only alternative would be more than one.

    You are thinking finitely again.

    Logic is finite.

    For you, crank Wolfgang Mückenheim, logic is finished. It never
    actually started.

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Fri Aug 23 14:28:13 2024
    XPost: sci.math

    Le 22/08/2024 à 16:22, FromTheRafters a écrit :
    WM used his keyboard to write :

    In a linear order of elements which all have distances from each other, there
    is necessarily a last one (if nothing follows) because the only alternative >> would be more than one.

    You are thinking finitely again.

    Logic is finite.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Fri Aug 23 14:13:00 2024
    XPost: sci.math

    On 8/23/2024 10:28 AM, WM wrote:
    Le 22/08/2024 à 16:22, FromTheRafters a écrit :
    WM used his keyboard to write :

    In a linear order of elements which
    all have distances from each other,
    there is necessarily a last one
    (if nothing follows)
    because
    the only alternative would be more than one.

    You are thinking finitely again.

    Logic is finite.

    All finiteⁿᵒᵗᐧᵂᴹ ordered sets _and their subsets_
    are 2.ended, except for {}.

    An understanding of 'finiteⁿᵒᵗᐧᵂᴹ'
    is essential to extending mathematicsⁿᵒᵗᐧᵂᴹ
    from finite objects
    to finite.length claims about infinite objects.

    Claim Q₂ is not.first.false in
    claim.sequence ⟨Q₁ Q₁⇒Q₂ Q₂⟩
    ⎛ Either Q₂ is not false,
    ⎜ or a claim before Q₂ (Q₁ or Q₁⇒Q₂) is false.
    ⎜ That there is no other possibility
    ⎝ can be demonstrated finitely.

    Consider
    the finite claim sequence ⟨Q₁ Q₂ ... Qₖ⟩ and
    its sub.sequence ⟨F₁ F₂ ... Fⱼ⟩ of false claims
    in ⟨Q₁ Q₂ ... Qₖ⟩

    If we know ⟨F₁ F₂ ... Fⱼ⟩ is empty,
    we know each claim in ⟨Q₁ Q₂ ... Qₖ⟩ is true.
    ⎛ Even if Qₖ is something like
    ⎜( Each natural ≥2 has a unique prime factorization
    ⎜ which no finite being can check individually,
    ⎝ we know that Qₖ is true.

    If ⟨F₁ F₂ ... Fⱼ⟩ is non.empty,
    we know ⟨F₁ F₂ ... Fⱼ⟩ is finiteⁿᵒᵗᐧᵂᴹ
    and it has a first.end F₁
    and F₁ is first.false in ⟨Q₁ Q₂ ... Qₖ⟩
    ⎛ We might not know which claim is first.false.
    ⎝ It's enough to know that a first.false exists.

    If we know ⟨Q₁ Q₂ ... Qₖ⟩ is each not.first.false
    we know its first.false F₁ doesn't exist
    we know ⟨F₁ F₂ ... Fⱼ⟩ is empty
    we know ⟨Q₁ Q₂ ... Qₖ⟩ is each true.

    Even though Qₖ is about infinitely.many in ℕᵈᵉᶠ,
    because
    finite each.not.first.false ⟨Q₁ Q₂ ... Qₖ⟩ exists
    we know that Qₖ is true of infinitely.many in ℕᵈᵉᶠ

    Logic is finite.

    Logic is a finite telescope with which
    we can view (but not enter) the infinite realm.

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Sat Aug 24 14:17:58 2024
    XPost: sci.math

    Le 23/08/2024 à 20:13, Jim Burns a écrit :

    All finiteⁿᵒᵗᐧᵂᴹ ordered sets _and their subsets_
    are 2.ended, except for {}.

    Every set of unit fractions extended over an interval has two ends in that interval.

    Regards, WM

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  • From Richard Damon@21:1/5 to All on Sat Aug 24 13:06:45 2024
    XPost: sci.math

    On 8/23/24 10:28 AM, WM wrote:
    Le 22/08/2024 à 16:22, FromTheRafters a écrit :
    WM used his keyboard to write :

    In a linear order of elements which all have distances from each
    other, there is necessarily a last one (if nothing follows) because
    the only alternative would be more than one.

    You are thinking finitely again.

    Logic is finite.

    Regards, WM



    Logic uses finite steps, but can process the infinite, unless you logic
    is limited by rules that limit it to processing finite things.

    This is the whole idea of induction, that can prove in a finite number
    of steps what would otherwise require an infinite number of steps.

    I guess by your finite logic, Achilles can't pass the tortoise, since
    everytime he gets to where the tortoise was, the tortoise has moved
    forward, so no matter how many times your resolve that cycle, he never
    gets to the tortoise.

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Sun Aug 25 19:47:46 2024
    XPost: sci.math

    Le 24/08/2024 à 19:06, Richard Damon a écrit :

    I guess by your finite logic, Achilles can't pass the tortoise,

    11.1 Achilles and the tortoise [From my 5th book appearing 2025]

    Achilles and the tortoise run a race. The tortoise gets a start and the
    race begins (state 0). When Achilles reaches this point, the tortoise has advanced further already (state 1). When Achilles reaches that point, the tortoise has advanced again (state 2). And so on (states 3, 4, 5, ...).
    Since Achilles runs much faster than the tortoise, he will overtake (state ), but only after infinitely many finitely indexed states of the
    described kind. Their number must be completed. Otherwise Achilles will
    not overtake. But there must not be a last visible finitely indexed state.
    (The last 1000 states Achilles remembers have indices much smaller than
    .) This can only be realized by means of dark states.
    According to set theory, all states can be put in bijection with all
    natural numbers. This is impossible as completeness and well-order require
    a last mark. The three notions "all" and "infinite" and "well-ordered" do
    not match. This dilemma can only be solved by refraining from well-order
    of the set Y of dark-numbered states.

    Regards, WM

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  • From WM@21:1/5 to All on Sun Aug 25 19:45:22 2024
    XPost: sci.math

    Le 24/08/2024 à 19:06, Richard Damon a écrit :
    On 8/23/24 10:49 AM, WM wrote:

    NUF(x) increases from 0 to more. But it  cannot increase by more than 1
    without shzowing a constant level.

    And, since there is no finite value of x that makes NUF(x) equal to 1,
    since for every finite x, their exists an x/2

    For all visisble x but not for all dark x.

    Just because you have verbally defined a function doesn't mean it exists
    and has proper values.

    This function exists because nothing contradicts its existence.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Sun Aug 25 17:28:24 2024
    XPost: sci.math

    On 8/25/24 3:47 PM, WM wrote:
    Le 24/08/2024 à 19:06, Richard Damon a écrit :

    I guess by your finite logic, Achilles can't pass the tortoise,

        11.1 Achilles and the tortoise [From my 5th book appearing 2025]

    Achilles and the tortoise run a race. The tortoise gets a start and the
    race begins (state 0). When Achilles reaches this point, the tortoise
    has advanced further already (state 1). When Achilles reaches that
    point, the tortoise has advanced again (state 2). And so on (states 3,
    4, 5, ...). Since Achilles runs much faster than the tortoise, he will overtake (state ), but only after infinitely many finitely indexed
    states of the described kind. Their number must be completed. Otherwise Achilles will not overtake. But there must not be a last visible
    finitely indexed state. (The last 1000 states Achilles remembers have
    indices much smaller than .) This can only be realized by means of dark states.
    According to set theory, all states can be put in bijection with all
    natural numbers. This is impossible as completeness and well-order
    require a last mark. The three notions "all" and "infinite" and "well-ordered" do not match. This dilemma can only be solved by
    refraining from well-order of the set Y of dark-numbered states.

    Regards, WM

    So, your logic is contradictory.

    You say the number IS completed to allow it to reach the final state,
    but it also must not be complete.

    Why do you say that it is "1000" states that are dark?

    Either your sum of "defined" steps is finite, and thus stops before
    getting to the Tortoise, and then we can do the operation your 1000 more
    times, and find Achilles still short of the Tortoise, or your "defined"
    steps are infinite without an "max value" that destroys your theory.

    Thus, it seems your logic just "punts" when it needs to deal with
    infinity and just admits that it can't handle it.

    Your final answer is basically just admitting that your logic can't
    supply the needed properties of the Natural Numbers.

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Sun Aug 25 17:31:48 2024
    XPost: sci.math

    On 8/25/24 3:45 PM, WM wrote:
    Le 24/08/2024 à 19:06, Richard Damon a écrit :
    On 8/23/24 10:49 AM, WM wrote:

    NUF(x) increases from 0 to more. But it  cannot increase by more than
    1 without shzowing a constant level.

    And, since there is no finite value of x that makes NUF(x) equal to 1,
    since for every finite x, their exists an x/2

    For all visisble x but not for all dark x.

    But all finite values ARE visible, as for ANY finite x, there exists an x/2.

    Your "Dark" numbers are just not finite values.



    Just because you have verbally defined a function doesn't mean it
    exists and has proper values.

    This function exists because nothing contradicts its existence.

    Regards, WM

    Except that there is not value for it to be 1 at.

    You can't use the assumption of the existance of the function to prove
    that it must exist, especially when that creates the contradictions.

    All you are proving is that you logic can't have the Natural Numbers as
    part of there domain, and thus ANYTHING you try to prove with your logic
    must be in a set smaller than that, and to claim otherwise is just a LIE.

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Mon Aug 26 20:06:29 2024
    XPost: sci.math

    On 8/25/2024 3:45 PM, WM wrote:
    Le 24/08/2024 à 19:06, Richard Damon a écrit :
    On 8/23/24 10:49 AM, WM wrote:

    NUF(x) increases from 0 to more.
    But it  cannot increase by more than 1
    without shzowing a constant level.

    And,
    since there is no finite value of x that
    makes NUF(x) equal to 1,
    since for every finite x, their exists an x/2

    For all visible x but not for all dark x.

    Just because you have verbally defined a function
    doesn't mean it exists and has proper values.

    This function exists because
    nothing contradicts its existence.

    Except for the contradicting consequences of
    its existence.

    ∃ᴿx>0: ¬∃¹ᐟᴺu<x ⟹
    ∃ᴿβ≥x>0:
    (∀ᴿγ>β: ∃¹ᐟᴺu<γ>β ∧ ∀ᴿα<β: ¬∃¹ᐟᴺu<α<β) ∧ (∃¹ᐟᴺu<½⋅β: 4⋅u<2⋅β>β ∧ ¬∃¹ᐟᴺu<½⋅β<β) [contradiction]

    Therefore,
    ¬∃ᴿx>0: ¬∃¹ᐟᴺu<x

    ¬∃ᴿx>0: NUF(x) = 1

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Tue Aug 27 19:11:44 2024
    XPost: sci.math

    Le 27/08/2024 à 02:06, Jim Burns a écrit :
    On 8/25/2024 3:45 PM, WM wrote:

    This function exists because
    nothing contradicts its existence.

    Except for the contradicting consequences of
    its existence.

    The function exists if actual infinity exists.
    The function does not exist if only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist and infinity is not actual and sets are not complete.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Tue Aug 27 19:29:43 2024
    XPost: sci.math

    Le 25/08/2024 à 23:28, Richard Damon a écrit :

    Why do you say that it is "1000" states that are dark?

    A joke. In fact ℵo states are dark.

    Your final answer is basically just admitting that your logic can't
    supply the needed properties of the Natural Numbers.

    No logic can treat the complete set of natural numbers without dark
    numbers.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Moebius@21:1/5 to All on Tue Aug 27 21:31:00 2024
    XPost: sci.math

    Am 27.08.2024 um 21:22 schrieb Chris M. Thomasson:
    On 8/27/2024 12:11 PM, WM wrote:
    Le 27/08/2024 à 02:06, Jim Burns a écrit :
    On 8/25/2024 3:45 PM, WM wrote:

    This function exists because
    nothing contradicts its existence.

    Right. But the values of the function NUF are:

    0 for all real arguments <= 0 and aleph_0 for all real arguments > 0.

    The function exists if actual infinity exists.

    That's for sure.

    The function does not exist if only potential infinity exists.

    Right. card(...) can't be aleph_0 in this case. :-)

    Hence we are glad that in the context of axiomatic set theory the AoI
    ensures for infinite sets. (Cantor ->transfinite set theory).

    --- SoupGate-Win32 v1.05
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  • From WM@21:1/5 to All on Tue Aug 27 19:57:39 2024
    XPost: sci.math

    Le 27/08/2024 à 21:31, Moebius a écrit :
    On 8/25/2024 3:45 PM, WM wrote:

    This function exists because
    nothing contradicts its existence.

    Right. But the values of the function NUF are:

    0 for all real arguments <= 0 and aleph_0 for all real arguments > 0.

    Correction: For all ends of real intervals containing ℵo unit fractions.
    The interval (0, x) cannot contain ℵo unit fractions unless x is
    sufficiently large.

    Regards, WM

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  • From Moebius@21:1/5 to All on Tue Aug 27 22:03:56 2024
    XPost: sci.math

    Am 27.08.2024 um 21:34 schrieb Chris M. Thomasson:
    On 8/27/2024 12:29 PM, WM wrote:
    Le 25/08/2024 à 23:28, Richard Damon a écrit :

    Why do you say that it is "1000" states that are dark?

    A joke.

    Try not to project yourself on others. You are the joke, right?

    Bad joke, though.

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Tue Aug 27 22:13:04 2024
    XPost: sci.math

    On 8/27/24 3:11 PM, WM wrote:
    Le 27/08/2024 à 02:06, Jim Burns a écrit :
    On 8/25/2024 3:45 PM, WM wrote:

    This function exists because
    nothing contradicts its existence.

    Except for the contradicting consequences of
    its existence.

    The function exists if actual infinity exists.
    The function does not exist if only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist and infinity is not actual and sets are not complete.

    Regards, WM



    Well NUF(x) does not exist, but that doesn't say that infiity is not
    actual, or that sets are not complete, just that your logic can't handle it.

    Your IGNORANCE doesn't prove that something can't exist for people who
    can actually think.

    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Tue Aug 27 22:08:47 2024
    XPost: sci.math

    On 8/27/24 3:29 PM, WM wrote:
    Le 25/08/2024 à 23:28, Richard Damon a écrit :

    Why do you say that it is "1000" states that are dark?

    A joke. In fact ℵo states are dark.

    Which since you logic doesn't allow you to complete, means he can't
    actually pass it.


    Your final answer is basically just admitting that your logic can't
    supply the needed properties of the Natural Numbers.

    No logic can treat the complete set of natural numbers without dark
    numbers.

    Of course it can.

    Just not logic as primative as the one you want to use.

    MILLIONS (if nt Billions) of people do it daily.


    Regards, WM



    --- SoupGate-Win32 v1.05
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  • From Richard Damon@21:1/5 to All on Tue Aug 27 22:16:16 2024
    XPost: sci.math

    On 8/27/24 3:57 PM, WM wrote:
    Le 27/08/2024 à 21:31, Moebius a écrit :
    On 8/25/2024 3:45 PM, WM wrote:

    This function exists because
    nothing contradicts its existence.

    Right. But the values of the function NUF are:

    0 for all real arguments <= 0 and aleph_0 for all real arguments > 0.

    Correction: For all ends of real intervals containing ℵo unit fractions. The interval (0, x) cannot contain ℵo unit fractions unless x is sufficiently large.

    Regards, WM

    But sufficiently large is just being any finite positive number.

    Then it contains the Aleph_0 unit fractions starting at 1/ceil(1/x)

    Since it is a fact that for ANY Natural Number n, there are aleph_0
    other Natural Numbers after it, since aleph_0 is not exhausted by trying
    to reduce it by finite amounts.

    --- SoupGate-Win32 v1.05
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  • From Jim Burns@21:1/5 to All on Wed Aug 28 02:13:19 2024
    XPost: sci.math

    On 8/27/2024 3:11 PM, WM wrote:
    Le 27/08/2024 à 02:06, Jim Burns a écrit :
    On 8/25/2024 3:45 PM, WM wrote:

    This function exists because
    nothing contradicts its existence.

    Except for the contradicting
    consequences of its existence.

    The function exists if
    actual infinity exists.
    The function does not exist if
    only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist
    and infinity is not actual
    and sets are not complete.

    In a finiteⁿᵒᵗᐧᵂᴹ order ⟨B,<⟩
    each non.empty S ⊆ B is 2.ended.

    In a finiteᵂᴹ order ⟨B,<⟩
    no one can say
    what a setᵂᴹ is,
    what an orderᵂᴹ is,
    what finiteᵂᴹ is.
    Perhaps, in 30 more years,
    these question will have answers.

    An infiniteⁿᵒᵗᐧᵂᴹ order ⟨B,◁⟩ is
    trichotomous and not finiteⁿᵒᵗᐧᵂᴹ.

    No set B has both
    finiteⁿᵒᵗᐧᵂᴹ ⟨B,<⟩ and infiniteⁿᵒᵗᐧᵂᴹ ⟨B,◁⟩

    A potentially.infiniteᵂᴹ set is
    an infiniteⁿᵒᵗᐧᵂᴹ set.
    Perhaps. When it's convenient.

    An actually.infiniteᵂᴹ set is
    a not.potentially.infiniteᵂᴹ set with
    a potentially.infiniteᵂᴹ subset.
    Perhaps. When it's convenient.

    However,
    no actually.infiniteᵂᴹ set is
    a not.infiniteⁿᵒᵗᐧᵂᴹ set with
    an infiniteⁿᵒᵗᐧᵂᴹ subset.
    Definitely. Always.
    Because
    no setⁿᵒᵗᐧᵂᴹ of any kind is
    a not.infiniteⁿᵒᵗᐧᵂᴹ set with
    an infiniteⁿᵒᵗᐧᵂᴹ subset.
    Definitely. Always.

    The quest continues for
    any kind of meaning that would fit your words.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist

    What exists?

    I propose a very conservative answer:
    that we accept at least
    the empty set existsᴲ,
    adjuncts of existingᴲ sets existᴲ,
    definable pluralities of existingᴲ sets existᴲᴲ,
    definable meta.pluralities of
    existingᴲᴲ pluralities existᴲᴲᴲ.

    ⎛ ∃{}
    ⎜ ∀x∀y∃z=x∪{y}
    ⎜ ∃∃{z:P(z)}: ∀x:(x∈{z:P(z)} ⇔ P(x))
    ⎜ ∃∃∃{zz:P(zz)}: ∀∀xx:(xx∈{zz:P(zz)} ⇔ P(xx))
    ⎜ set.extensionalityᴲ
    ⎜ plurality.extensionalityᴲᴲ
    ⎝ meta.plurality.extensionalityᴲᴲᴲ

    From that very conservative proposal,
    making the usual definitions,
    there follows definitely always
    -- because not.first.falsely --
    our usual claims, claims which contradict
    your usual claims, those allegedly.uncontradicted claims.

    ----
    https://en.wikipedia.org/wiki/General_set_theory
    | Axioms
    | ST is GST with
    | the axiom schema of specification replaced by
    | the axiom of empty set.

    https://en.wikipedia.org/wiki/Plural_quantification

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  • From WM@21:1/5 to All on Wed Aug 28 12:57:31 2024
    XPost: sci.math

    Le 28/08/2024 à 04:13, Richard Damon a écrit :


    Well NUF(x) does not exist, but that doesn't say that infinity is not
    actual,

    So the unit fractions are actually existing but their number isn't?
    Strange!

    Regards, WM

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  • From WM@21:1/5 to All on Wed Aug 28 12:55:34 2024
    XPost: sci.math

    Le 28/08/2024 à 04:08, Richard Damon a écrit :
    On 8/27/24 3:29 PM, WM wrote:
    Le 25/08/2024 à 23:28, Richard Damon a écrit :

    Why do you say that it is "1000" states that are dark?

    A joke. In fact ℵo states are dark.

    Which since you logic doesn't allow you to complete, means he can't
    actually pass it.

    He does pass. Enlighten the state where he passes. Fail.

    Regards, WM

    --- SoupGate-Win32 v1.05
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  • From Python@21:1/5 to All on Wed Aug 28 15:30:31 2024
    XPost: sci.math

    Le 28/08/2024 à 15:25, WM a écrit :
    Le 28/08/2024 à 08:13, Jim Burns a écrit :
    On 8/27/2024 3:11 PM, WM wrote:

    The function exists if
    actual infinity exists.
    The function does not exist if
    only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist
    and infinity is not actual
    and sets are not complete.

    A potentially.infiniteᵂᴹ set is
    an infiniteⁿᵒᵗᐧᵂᴹ set.

    A collection.

    An actually.infiniteᵂᴹ set is
    a not.potentially.infiniteᵂᴹ set with
    a potentially.infiniteᵂᴹ subset.

    Subcollection.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist

    What exists?

    I propose a very conservative answer:
    that we accept at least
    the empty set existsᴲ,

    Does it?

    Bernard Bolzano, the inventor of the notion set (Menge) in mathematics
    would not have named a nothing an empty set. In German the word "Menge"
    has the meaning of many or great quantity. Often we find in German texts
    the expression "große (great or large) Menge", rarely the expression
    "kleine (small) Menge". Therefore Bolzano apologizes for using this word
    in case of sets having only two elements: "Allow me to call also a
    collection containing only two parts a set." [B. Bolzano: "Einleitung
    zur Grössenlehre", J. Berg (ed.), Friedrich Frommann Verlag, Stuttgart (1975) p. 152]

    Also Richard Dedekind discarded the empty set. But he accepted the
    singleton, i.e., the non-empty set of less than two elements: "For the uniformity of the wording it is useful to permit also the special case
    that a system S consists of a single (of one and only one) element a,
    i.e., that the thing a is element of S but every thing different from a
    is not an element of S. The empty system, however, which does not
    contain any element, shall be excluded completely for certain reasons, although it may be convenient for other investigations to fabricate
    such." [R. Dedekind: "Was sind und was sollen die Zahlen?" Vieweg, Braunschweig (1887), 2nd ed. (1893) p. 2]

    Bertrand Russell considered an empty class as not existing: "An existent class is a class having at least one member." [B. Russell: "On some difficulties in the theory of transfinite numbers and order types",
    Proc. London Math. Soc. (2) 4 (1906) p. 47]

    Gottlob Frege shared his opinion: "If, according to our previous use of
    the word, a class consists of things, is a collection, a collective
    union of them, then it must disappear when these things disappear. If we
    burn down all the trees of a forest, then we burn down the forest. Thus
    an empty class cannot exist." [G. Frege: "Kleine Schriften", I. Agelelli (ed.), 2nd ed., Olms, Hildesheim (1990) p. 195]

    Georg Cantor mentioned the empty set with some reservations and only
    once in all his work: "Further it is useful to have a symbol expressing
    the absence of points. We choose for that sake the letter O; P  O means that the set P does not contain any single point. So it is, strictly speaking, not existing as such." [Cantor, p. 146]
    And even Ernst Zermelo who made the "Axiom II. There is an (improper)
    set, the 'null-set' 0 which does not contain any elements" [E. Zermelo: "Untersuchungen über die Grundlagen der Mengenlehre I", Mathematische Annalen 65 (1908) p. 263], this same author himself said in private correspondence: "It is not a genuine set and was introduced by me only
    for formal reasons." [E. Zermelo, letter to A. Fraenkel (1 Mar 1921)] "I increasingly doubt the justifiability of the 'null set'. Perhaps one can dispense with it by restricting the axiom of separation in a suitable
    way. Indeed, it serves only the purpose of formal simplification." [E. Zermelo, letter to A. Fraenkel (9 May 1921)]     So it is all the more courageous that Zermelo based his number system completely on the empty
    set: { } = 0, {{ }} = 1, {{{ }}} = 2, and so on. He knew that there is
    only one empty set. But many ways to create the empty set can be
    devised, like the empty set of numbers, the empty set of bananas, the uncountably many empty sets of all real singletons, and the empty set of
    all these empty sets. Is it the emptiest set? Anyhow, "zero things"
    means "no things". So we can safely say (pun intended): Nothing is named
    the empty set.

    "pun intended"

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Aug 28 13:25:57 2024
    XPost: sci.math

    Le 28/08/2024 à 08:13, Jim Burns a écrit :
    On 8/27/2024 3:11 PM, WM wrote:

    The function exists if
    actual infinity exists.
    The function does not exist if
    only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist
    and infinity is not actual
    and sets are not complete.

    A potentially.infiniteᵂᴹ set is
    an infiniteⁿᵒᵗᐧᵂᴹ set.

    A collection.

    An actually.infiniteᵂᴹ set is
    a not.potentially.infiniteᵂᴹ set with
    a potentially.infiniteᵂᴹ subset.

    Subcollection.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist

    What exists?

    I propose a very conservative answer:
    that we accept at least
    the empty set existsᴲ,

    Does it?

    Bernard Bolzano, the inventor of the notion set (Menge) in mathematics
    would not have named a nothing an empty set. In German the word "Menge"
    has the meaning of many or great quantity. Often we find in German texts
    the expression "große (great or large) Menge", rarely the expression
    "kleine (small) Menge". Therefore Bolzano apologizes for using this word
    in case of sets having only two elements: "Allow me to call also a
    collection containing only two parts a set." [B. Bolzano: "Einleitung zur Grössenlehre", J. Berg (ed.), Friedrich Frommann Verlag, Stuttgart (1975)
    p. 152]

    Also Richard Dedekind discarded the empty set. But he accepted the
    singleton, i.e., the non-empty set of less than two elements: "For the uniformity of the wording it is useful to permit also the special case
    that a system S consists of a single (of one and only one) element a,
    i.e., that the thing a is element of S but every thing different from a is
    not an element of S. The empty system, however, which does not contain any element, shall be excluded completely for certain reasons, although it may
    be convenient for other investigations to fabricate such." [R. Dedekind:
    "Was sind und was sollen die Zahlen?" Vieweg, Braunschweig (1887), 2nd ed. (1893) p. 2]

    Bertrand Russell considered an empty class as not existing: "An existent
    class is a class having at least one member." [B. Russell: "On some difficulties in the theory of transfinite numbers and order types", Proc. London Math. Soc. (2) 4 (1906) p. 47]

    Gottlob Frege shared his opinion: "If, according to our previous use of
    the word, a class consists of things, is a collection, a collective union
    of them, then it must disappear when these things disappear. If we burn
    down all the trees of a forest, then we burn down the forest. Thus an
    empty class cannot exist." [G. Frege: "Kleine Schriften", I. Agelelli
    (ed.), 2nd ed., Olms, Hildesheim (1990) p. 195]

    Georg Cantor mentioned the empty set with some reservations and only once
    in all his work: "Further it is useful to have a symbol expressing the
    absence of points. We choose for that sake the letter O; P  O means
    that the set P does not contain any single point. So it is, strictly
    speaking, not existing as such." [Cantor, p. 146]

    And even Ernst Zermelo who made the "Axiom II. There is an (improper) set,
    the 'null-set' 0 which does not contain any elements" [E. Zermelo: "Untersuchungen über die Grundlagen der Mengenlehre I", Mathematische
    Annalen 65 (1908) p. 263], this same author himself said in private correspondence: "It is not a genuine set and was introduced by me only for formal reasons." [E. Zermelo, letter to A. Fraenkel (1 Mar 1921)] "I increasingly doubt the justifiability of the 'null set'. Perhaps one can dispense with it by restricting the axiom of separation in a suitable way. Indeed, it serves only the purpose of formal simplification." [E. Zermelo, letter to A. Fraenkel (9 May 1921)]
    So it is all the more courageous that Zermelo based his number system completely on the empty set: { } = 0, {{ }} = 1, {{{ }}} = 2, and so on.
    He knew that there is only one empty set. But many ways to create the
    empty set can be devised, like the empty set of numbers, the empty set of bananas, the uncountably many empty sets of all real singletons, and the
    empty set of all these empty sets. Is it the emptiest set? Anyhow, "zero things" means "no things". So we can safely say (pun intended): Nothing is named the empty set.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Wed Aug 28 17:12:18 2024
    XPost: sci.math

    Le 28/08/2024 à 18:55, Jim Burns a écrit :
    On 8/28/2024 9:25 AM, WM wrote:
    Le 28/08/2024 à 08:13, Jim Burns a écrit :
    On 8/27/2024 3:11 PM, WM wrote:

    The function exists if
    actual infinity exists.
    The function does not exist if
    only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist
    and infinity is not actual
    and sets are not complete.

    A potentially.infiniteᵂᴹ set is
    an infiniteⁿᵒᵗᐧᵂᴹ set.

    A collection.

    A flying.rainbow.sparkle.pony.

    An actually.infiniteᵂᴹ set is
    a not.potentially.infiniteᵂᴹ set with
    a potentially.infiniteᵂᴹ subset.

    Subcollection.

    Sub.flying.rainbow.sparkle.pony.

    Merely changing a term doesn't change
    what is referred to.

    Potentially infinite sets are called collections in set theory.

    I propose a very conservative answer:
    that we accept at least
    the empty set existsᴲ,

    Does it?

    Georg Cantor [...]
    "Further it is useful to have
    a symbol expressing the absence of points. [...]

    Exactly. "It is useful".

    However,
    ordaining a symbol as "means this thing"
    does not assert that this thing exists.

    {} means "the absence of points".
    Is there an absence of points?

    Is this absence a set?

    𝔊 means "the last natural number".
    Is there a last natural number?

    What is immediately before ω? Nothing? The empty set?

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Wed Aug 28 13:15:03 2024
    XPost: sci.math

    On 8/27/2024 3:29 PM, WM wrote:
    Le 25/08/2024 à 23:28, Richard Damon a écrit :

    Your final answer is basically just
    admitting that your logic can't supply
    the needed properties of the Natural Numbers.

    No logic can treat
    the complete set of natural numbers
    without dark numbers.

    ∀S ⊆ ℕ: S ≠ {} ⇔ ∃k ∈ S: k = min.S

    ∀k ∈ ℕ: k ≠ 0 ⇔ ∃j ∈ ℕ: j+1 = k

    ∀j ∈ ℕ: ∃k ∈ ℕ: j+1 = k

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Wed Aug 28 12:55:24 2024
    XPost: sci.math

    On 8/28/2024 9:25 AM, WM wrote:
    Le 28/08/2024 à 08:13, Jim Burns a écrit :
    On 8/27/2024 3:11 PM, WM wrote:

    The function exists if
    actual infinity exists.
    The function does not exist if
    only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist
    and infinity is not actual
    and sets are not complete.

    A potentially.infiniteᵂᴹ set is
    an infiniteⁿᵒᵗᐧᵂᴹ set.

    A collection.

    A flying.rainbow.sparkle.pony.

    An actually.infiniteᵂᴹ set is
    a not.potentially.infiniteᵂᴹ set with
    a potentially.infiniteᵂᴹ subset.

    Subcollection.

    Sub.flying.rainbow.sparkle.pony.

    Merely changing a term doesn't change
    what is referred to.

    Also,
    'set', 'collection' and 'flying.rainbow.sparkle.pony'
    don't even enter the formal language.
    Being absent, their change is no.change.

    ∀S ⊆ ℕ: S ≠ {} ⇒ ∃k ∈ S: k = min.S

    What are those?
    Sets?
    Collections?
    Flying.rain.bow.sparkle.ponies?
    Something else?
    All of the above, at once?

    1. There is no way to answer.
    2. There is no need to answer.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist

    What exists?

    I propose a very conservative answer:
    that we accept at least
    the empty set existsᴲ,

    Does it?

    Georg Cantor [...]
    "Further it is useful to have
    a symbol expressing the absence of points. [...]

    Exactly. "It is useful".

    However,
    ordaining a symbol as "means this thing"
    does not assert that this thing exists.

    {} means "the absence of points".
    Is there an absence of points?

    𝔊 means "the last natural number".
    Is there a last natural number?

    Questions about the language, answered.
    Questions about what the language is about,
    not answered, not yet.

    I propose a very conservative answer:
    that we accept at least
    the empty set existsᴲ,

    Does it?

    Is there an absence of points?
    Set aside useful '{}':
    Do you (WM) reject the axiom ∃x∀y:y∉x ?

    It isn't logic which encourages you to accept the Axiom of Empty.
    Not all consequences are logical consequences.
    You have it within your power to make
    what you are talking about useless, pointless,
    completely uninteresting.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Wed Aug 28 20:18:55 2024
    XPost: sci.math

    Am 28.08.2024 um 15:30 schrieb Python:
    Le 28/08/2024 à 15:25, WM a écrit :

    Gottlob Frege shared his opinion:

    *Holy shit*!!! No, he didn't "share (t)his opinion".

    This was a CRITICAL remark concerning one of Schröder's books and the
    theory expressed in this book

    "If, according to our previous use of the word, a class consists of
    things,
    is a collection, a collective union of them, then it must disappear
    when
    these things disappear. If we burn down all the trees of a forest,
    then
    we burn down the forest. Thus an empty class cannot exist."
    [G. Frege: "Kleine Schriften", I. Agelelli (ed.), 2nd ed., Olms,
    Hildesheim (1990) p. 195]

    OF COURSE, Frege's system had an empty "set" (Begriffsumfang). you (WM)
    fucking idiot full of shit!

    Anyhow, "zero things" means "no things". So we can
    safely say <bla bla bla>

    Yeah, we can safely say:

    |{x : Fx}| = 0 <-> ~ExFx

    Mückenheim, Du bist wirklich dümmer als die Polizei erlaubt!

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Wed Aug 28 14:41:10 2024
    XPost: sci.math

    On 8/28/2024 1:12 PM, WM wrote:
    Le 28/08/2024 à 18:55, Jim Burns a écrit :
    On 8/28/2024 9:25 AM, WM wrote:
    Le 28/08/2024 à 08:13, Jim Burns a écrit :
    On 8/27/2024 3:11 PM, WM wrote:

    The function exists if
    actual infinity exists.
    The function does not exist if
    only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist
    and infinity is not actual
    and sets are not complete.

    A potentially.infiniteᵂᴹ set is
    an infiniteⁿᵒᵗᐧᵂᴹ set.

    A collection.

    A flying.rainbow.sparkle.pony.

    An actually.infiniteᵂᴹ set is
    a not.potentially.infiniteᵂᴹ set with
    a potentially.infiniteᵂᴹ subset.

    Subcollection.

    Sub.flying.rainbow.sparkle.pony.

    Merely changing a term doesn't change
    what is referred to.

    Potentially infinite sets are called collections
    in set theory.

    And I call them flying.rainbow.sparkle.ponies.
    It doesn't matter what they're called.

    Except that it matters that you are claiming that
    something is potentially.infiniteᵂᴹ.

    What do you mean by potentially.infiniteᵂᴹ, exactly?
    I think you (WM) mean

    finiteⁿᵒᵗᐧᵂᴹ:
    each subset {} or 2.ended

    potentially.infiniteᵂᴹ:
    total order not finite
    one subset not( {} or 2.ended )

    actually.infiniteᵂᴹ:
    not potentially.infiniteᵂᴹ and also
    one subset potentially.infiniteᵂᴹ

    I suspect that finiteⁿᵒᵗᐧᵂᴹ isn't
    what you want finiteⁿᵒᵗᐧᵂᴹ to be.

    Definition.
    For each B ⊆ A,finiteⁿᵒᵗᐧᵂᴹ: B is {} or 2.ended.

    For each B ⊆ A,finiteⁿᵒᵗᐧᵂᴹ:
    for each C ⊆ B: C ⊆ A,finiteⁿᵒᵗᐧᵂᴹ
    for each C ⊆ B: C is {} or 2.ended
    B,finiteⁿᵒᵗᐧᵂᴹ

    For each B ⊆ A,finiteⁿᵒᵗᐧᵂᴹ: B,finiteⁿᵒᵗᐧᵂᴹ

    Each subset of a finiteⁿᵒᵗᐧᵂᴹ is finiteⁿᵒᵗᐧᵂᴹ. Likewise,
    each superset of an infiniteⁿᵒᵗᐧᵂᴹ is infiniteⁿᵒᵗᐧᵂᴹ.

    Potential.infinityᵂᴹ can't completeᵂᴹ.

    I propose a very conservative answer:
    that we accept at least
    the empty set existsᴲ,

    Does it?

    Georg Cantor [...]
    "Further it is useful to have
    a symbol expressing the absence of points. [...]

    Exactly. "It is useful".

    However,
    ordaining a symbol as "means this thing"
    does not assert that this thing exists.

    {} means "the absence of points".
    Is there an absence of points?

    Is this absence a set?

    Is there an object in our domain of discourse
    which "this absence" refers to?

    Maybe. It depends on the discourse.
    Is there a broader consideration preventing it?
    If so, what is the broader consideration?

    𝔊 means "the last natural number".
    Is there a last natural number?

    What is immediately before ω?
    Nothing?
    The empty set?

    {} = 0 (von Neumann)

    The set {ω-1} = {} of ordinals immediately before ω
    is not an ordinal immediately before ω

    I think that a good.sized chunk of mathematics
    becomes unintelligible if we darkenᵂᴹ the distinction
    between set and element.

    I can't tell if that darkeningᵂᴹ is your goal or
    merely a side.effect of the WM.cannot.be.wrong axiom.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jeff Barnett@21:1/5 to All on Wed Aug 28 13:04:48 2024
    XPost: sci.math

    T24gOC8yOC8yMDI0IDEyOjEzIEFNLCBKaW0gQnVybnMgd3JvdGU6DQo+IE9uIDgvMjcvMjAy NCAzOjExIFBNLCBXTSB3cm90ZToNCj4+IExlIDI3LzA4LzIwMjQgw6AgMDI6MDYsIEppbSBC dXJucyBhIMOpY3JpdCA6DQo+Pj4gT24gOC8yNS8yMDI0IDM6NDUgUE0sIFdNIHdyb3RlOg0K PiANCj4+Pj4gVGhpcyBmdW5jdGlvbiBleGlzdHMgYmVjYXVzZQ0KPj4+PiBub3RoaW5nIGNv bnRyYWRpY3RzIGl0cyBleGlzdGVuY2UuDQo+Pj4NCj4+PiBFeGNlcHQgZm9yIHRoZSBjb250 cmFkaWN0aW5nDQo+Pj4gY29uc2VxdWVuY2VzIG9mIGl0cyBleGlzdGVuY2UuDQo+Pg0KPj4g VGhlIGZ1bmN0aW9uIGV4aXN0cyBpZg0KPj4gYWN0dWFsIGluZmluaXR5IGV4aXN0cy4NCj4+ IFRoZSBmdW5jdGlvbiBkb2VzIG5vdCBleGlzdCBpZg0KPj4gb25seSBwb3RlbnRpYWwgaW5m aW5pdHkgZXhpc3RzLg0KPj4NCj4+PiDCrOKIg+G0v3g+MDogTlVGKHgpID0gMQ0KPj4NCj4+ IFRoZW4gTlVGKHgpIGRvZXMgbm90IGV4aXN0DQo+PiBhbmQgaW5maW5pdHkgaXMgbm90IGFj dHVhbA0KPj4gYW5kIHNldHMgYXJlIG5vdCBjb21wbGV0ZS4NCj4gDQo+IEluIGEgZmluaXRl 4oG/4bWS4bWX4ZCn4bWC4bS5IG9yZGVyIOKfqEIsPOKfqQ0KPiBlYWNoIG5vbi5lbXB0eSBT IOKKhiBCIGlzIDIuZW5kZWQuDQo+IA0KPiBJbiBhIGZpbml0ZeG1guG0uSBvcmRlciDin6hC LDzin6kNCj4gbm8gb25lIGNhbiBzYXkNCj4gd2hhdCBhIHNldOG1guG0uSBpcywNCj4gd2hh dCBhbiBvcmRlcuG1guG0uSBpcywNCj4gd2hhdCBmaW5pdGXhtYLhtLkgaXMuDQo+IFBlcmhh cHMsIGluIDMwIG1vcmUgeWVhcnMsDQo+IHRoZXNlIHF1ZXN0aW9uIHdpbGwgaGF2ZSBhbnN3 ZXJzLg0KPiANCj4gQW4gaW5maW5pdGXigb/htZLhtZfhkKfhtYLhtLkgb3JkZXIg4p+oQizi l4Hin6kgaXMNCj4gdHJpY2hvdG9tb3VzIGFuZCBub3QgZmluaXRl4oG/4bWS4bWX4ZCn4bWC 4bS5Lg0KDQpUcmljaG90b21vdXM/IERpZCB5b3UgYWN0dWFsbHkgbWVhbiB0byBzYXkgImRl bnNlIG9yZGVyIiwgImRlbnNlIGluIA0KaXRzZWxmIiwgb3Igc29tZXRoaW5nIGVsc2UgYWtp bj8NCg0KPFNOSVBQSU5HPg0KLS0gDQpKZWZmIEJhcm5ldHQNCg0K

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Jeff Barnett on Wed Aug 28 16:10:32 2024
    XPost: sci.math

    On 8/28/2024 3:04 PM, Jeff Barnett wrote:
    On 8/28/2024 12:13 AM, Jim Burns wrote:
    On 8/27/2024 3:11 PM, WM wrote:
    Le 27/08/2024 à 02:06, Jim Burns a écrit :
    On 8/25/2024 3:45 PM, WM wrote:

    This function exists because
    nothing contradicts its existence.

    Except for the contradicting
    consequences of its existence.

    The function exists if
    actual infinity exists.
    The function does not exist if
    only potential infinity exists.

    ¬∃ᴿx>0: NUF(x) = 1

    Then NUF(x) does not exist
    and infinity is not actual
    and sets are not complete.

    In a finiteⁿᵒᵗᐧᵂᴹ order ⟨B,<⟩
    each non.empty S ⊆ B is 2.ended.

    In a finiteᵂᴹ order ⟨B,<⟩
    no one can say
    what a setᵂᴹ is,
    what an orderᵂᴹ is,
    what finiteᵂᴹ is.
    Perhaps, in 30 more years,
    these question will have answers.

    An infiniteⁿᵒᵗᐧᵂᴹ order ⟨B,◁⟩ is
    trichotomous and not finiteⁿᵒᵗᐧᵂᴹ.

    Trichotomous?
    Did you actually mean to say "dense order",
    "dense in itself", or something else akin?

    I meant "trichotomous".

    I could equally.well have said "total ⟨B,◁⟩"
    but "trichotomous" seems less misunderstood by WM.

    I didn't bother to say "trichotomous and finite ⟨B,◁⟩"
    because trichotomy and transitivity follow from
    each non.empty S ⊆ B is 2.ended.

    I've proved these elsethread:

    ⎛ if
    ⎜ ⟨B,◁⟩: each S ⊆ B: {} or 2.ended and
    ⎜ ⟨B,<⟩: NOT each S ⊆ B: {} or 2.ended
    ⎜ then
    ⎜ sets S, {x} exist such that
    ⎜ ⟨S,<⟩: each T ⊆ S: {} or 2.ended
    ⎝ ⟨S∪{x},<⟩: NOT each T ⊆ S∪{x}: {} or 2.ended

    ⎛ no sets S, {x}, trichotomous '<' exist such that
    ⎜ ⟨S,<⟩: each T ⊆ S: {} or 2.ended
    ⎝ ⟨S∪{x},<⟩: NOT each T ⊆ S∪{x}: {} or 2.ended

    "Trichotomous" avoids x not being comparable to
    the contents of S, and {x} being not.2.ended.

    That is how I prove:
    because
    ℕ has ONE order with ONE non.2.ended non.{} subset
    ℕ has NO order WITHOUT one non.2.ended non.{} subset.

    That's why any all.2.ended.non.{}.subsets ORDER works
    as a way to distinguish a finite SET from infinite.

    I think that some of those in this thread
    are skeptical of that result.
    Thank you for the excuse to repeat it.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Wed Aug 28 22:35:25 2024
    XPost: sci.math

    Am 28.08.2024 um 21:06 schrieb Chris M. Thomasson:
    On 8/28/2024 5:57 AM, WM wrote:
    Le 28/08/2024 à 04:13, Richard Damon a écrit :


    Well NUF(x) does not exist,

    RD seems to be a crank too.

    Hint: NUF(x) := card({s e SB : s < x}) (x e IR)

    Hence NUF(x) = 0 for all x e IR, x <= 0, and NUF(x) = aleph_0 for all x
    e IR, x > 0.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Wed Aug 28 19:48:58 2024
    XPost: sci.math

    On 8/28/24 8:57 AM, WM wrote:
    Le 28/08/2024 à 04:13, Richard Damon a écrit :


    Well NUF(x) does not exist, but that doesn't say that infinity is not
    actual,

    So the unit fractions are actually existing but their number isn't?
    Strange!

    Regards, WM




    There number exists, it is aleph_0.

    You just can't count them from the "end" that doesn't have an end.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jeff Barnett@21:1/5 to All on Wed Aug 28 20:36:25 2024
    XPost: sci.math

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    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Thu Aug 29 13:27:49 2024
    XPost: sci.math

    Le 29/08/2024 à 01:48, Richard Damon a écrit :
    On 8/28/24 8:57 AM, WM wrote:
    Le 28/08/2024 à 04:13, Richard Damon a écrit :


    Well NUF(x) does not exist, but that doesn't say that infinity is not
    actual,

    So the unit fractions are actually existing but their number isn't?
    Strange!

    There number exists, it is aleph_0.

    You just can't count them from the "end" that doesn't have an end.

    Why not?

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Thu Aug 29 13:26:00 2024
    XPost: sci.math

    Le 28/08/2024 à 22:35, Moebius a écrit :

    Hint: NUF(x) := card({s e SB : s < x}) (x e IR)

    Hence NUF(x) = 0 for all x e IR, x <= 0, and NUF(x) = aleph_0


    for all x > the minimum distance between many unit fractions which is not
    0.

    Regards, QM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Thu Aug 29 13:38:36 2024
    XPost: sci.math

    Le 28/08/2024 à 19:15, Jim Burns a écrit :
    On 8/27/2024 3:29 PM, WM wrote:
    Le 25/08/2024 à 23:28, Richard Damon a écrit :

    Your final answer is basically just
    admitting that your logic can't supply
    the needed properties of the Natural Numbers.

    No logic can treat
    the complete set of natural numbers
    without dark numbers.

    ∀S ⊆ ℕ: S ≠ {} ⇔ ∃k ∈ S: k = min.S

    ∀k ∈ ℕ: k ≠ 0 ⇔ ∃j ∈ ℕ: j+1 = k

    ∀j ∈ ℕ: ∃k ∈ ℕ: j+1 = k

    That treats the potentially infinite collection.

    REgards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Thu Aug 29 13:56:06 2024
    XPost: sci.math

    On 8/29/2024 9:38 AM, WM wrote:
    Le 28/08/2024 à 19:15, Jim Burns a écrit :
    On 8/27/2024 3:29 PM, WM wrote:
    Le 25/08/2024 à 23:28, Richard Damon a écrit :

    Your final answer is basically just
    admitting that your logic can't supply
    the needed properties of the Natural Numbers.

    No logic can treat
    the complete set of natural numbers
    without dark numbers.

    ∀S ⊆ ℕ: S ≠ {} ⇔ ∃k ∈ S: k = min.S

    ∀k ∈ ℕ: k ≠ 0 ⇔ ∃j ∈ ℕ: j+1 = k

    ∀j ∈ ℕ: ∃k ∈ ℕ: j+1 = k

    That treats the potentially infinite collection.

    Call it ℕᴾᴵꟲ.
    The name doesn't matter.

    There is no natural number not.in ℕᴾᴵꟲ.

    ⎛ Assume otherwise.
    ⎜ Assume that darkᴹᵂ 𝔊 ≥ᵉᵃᶜʰ ℕᴾᴵꟲ exists.

    ⎜ Consider the corresponding unit fractions.
    ⎜ ⅟𝔊 is between 0 and each visibleᴹᵂ in ⅟ℕᴾᴵꟲ
    ⎜ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ ⅟𝔊 > 0

    ⎜ However,
    ⎜ 0 = greatest.lower.bound.⅟ℕᴾᴵꟲ
    ⎜ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ ⅟𝔊 > 0

    ⎜ 𝔊 ≱ᵉˣⁱˢᵗˢ ℕᴾᴵꟲ
    ⎝ Contradiction.

    Therefore,
    there is no natural number not.in ℕᴾᴵꟲ.

    ℕᴾᴵꟲ is
    the complete flying.rainbow.sparkle.pony
    of natural numbers.

    ----
    0 = greatest.lower.bound.⅟ℕᴾᴵꟲ
    0 > α ⇒ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ α
    γ > 0 ⇒ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ γ

    ⎛ Assume otherwise.
    ⎜ Assume greatest.lower.bound.⅟ℕᴾᴵꟲ = β > 0

    ⎜ β > ½⋅β
    ⎜ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ ½⋅β

    ⎜ 2⋅β > β
    ⎜ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ 2⋅β
    ⎜ exists visibleᵂᴹ ⅟k ≱ 2⋅β
    ⎜ exists visibleᵂᴹ ¼⋅⅟k ≱ ½⋅β
    ⎜ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ ½⋅β
    ⎝ Contradiction.

    Therefore,
    there is no higher lower.bound than 0

    Also,
    0 is a lower.bound.
    ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ 0

    0 = greatest.lower.bound.⅟ℕᴾᴵꟲ

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Jeff Barnett on Thu Aug 29 16:09:32 2024
    XPost: sci.math

    On 8/28/2024 10:36 PM, Jeff Barnett wrote:
    On 8/28/2024 2:10 PM, Jim Burns wrote:

    [...]

    "Trichotomous" generates the right mental images,
    I must admit.
    But I'm not sure how one could show that
    a relation R on S, where R is Trichotomous,
    implies that
    I can't 1-1 map a strict subset of R to R.

    I agree that one can't prove that.
    I am doing something else here.


    My apology for having stripped so much context.
    I get complaints about the length of my posts.

    And, as much as stripping context is not to my taste,
    if my chief correspondent won't read my brilliant text,
    Something Must Be Done.

    The Something which Must Be Done is that
    my side of this multi.year correspondence has become
    a search for perfect zingers by which to transmit
    principles of predicate logic, finiteness, infiniteness,
    sets, natural numbers, real numbers, and other topics,
    as appropriate.

    The search for zingers has kept me tolerably amused,
    so far, enough to keep me plugging away, at least.
    The downside is that a random passerby, such as yourself,
    might (understandably) find whatever.we.have.here opaque.
    Again, my apology.

    Context:

    Wolfgang Mückenheim has been rejecting
    Dedekind.infinite sets, sets R such that
    I can 1-1 map a strict subset of R to R.
    WM calls them "potentially infinite", by which
    WM means that these sets change, which means that
    WM is not talking about _those sets_
    which do not change.

    A typical WM.argument sets up some sequence and
    asserts by mathematics, by logic
    (in reality, by "common sense", by "obviousness")
    that the sequence has two ends.
    Since the sequence has only one _visible_ end,
    WM considers that proof of a second end which is _dark_

    It is an argument which focuses on sequences.

    There is another definition of finiteness/infiniteness
    which doesn't immediately call for Dedekind.infinite sets
    and which does focus on sequences.
    ⎛ A finite set has an order which is well.ordered
    ⎜ in both directions.
    ⎝ -- Paul Stäckel, 1862...1919

    I have been re.deriving familiar (to us) results
    for Stäckel.finite and Stäckel.infinite sets, instead of
    for Dedekind.finite and Dedekind.infinite sets.

    'Stäckel.finite' is messier.
    I frame it in terms of Stäckel.finite orders and
    Stäckel.infinite orders, but what do we say about
    sets which have both?

    My answer to that is what you, Random Passerby,
    have before you, here.

    Lemma.
    ⎛ A set cannot have both a Stäckel.finite order and
    ⎝ a Stäckel.infinite order.

    That's what I prove, which allows me to me to offer
    a clearer (or at least a different) view of
    finiteness and infiniteness.

    But I'm not sure how one could show that
    a relation R on S, where R is Trichotomous,
    implies that
    I can't 1-1 map a strict subset of R to R.

    And I don't show that.

    I show that,
    if one order of B is Stäckel.finite
    and a second order is trichotomous,
    then the second order is also Stäckel.finite.

    Thus,
    because ℕ has a standard order with one end,
    there is no Stäckel.finite order of ℕ.
    There is no superset of ℕ with a Stäckel.finite order,
    either.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Thu Aug 29 21:09:28 2024
    XPost: sci.math

    On 8/29/24 9:27 AM, WM wrote:
    Le 29/08/2024 à 01:48, Richard Damon a écrit :
    On 8/28/24 8:57 AM, WM wrote:
    Le 28/08/2024 à 04:13, Richard Damon a écrit :


    Well NUF(x) does not exist, but that doesn't say that infinity is
    not actual,

    So the unit fractions are actually existing but their number isn't?
    Strange!

    There number exists, it is aleph_0.

    You just can't count them from the "end" that doesn't have an end.

    Why not?

    Regards, WM



    Because you can't start at an end that isn't there.

    Try to tell me the actual number you are going to start at.

    Go ahead, try.

    You can say your logic says there must be one, but that shows your logic
    is wrong, as there isn't a smallest unit fraction or largest Natural
    Number.

    Logic that depends on the existance of something that doesn't exist is
    just broken.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jeff Barnett@21:1/5 to All on Thu Aug 29 22:20:55 2024
    XPost: sci.math

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    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 30 12:55:25 2024
    XPost: sci.math

    Le 29/08/2024 à 20:51, "Chris M. Thomasson" a écrit :
    On 8/29/2024 6:27 AM, WM wrote:
    Le 29/08/2024 à 01:48, Richard Damon a écrit :

    You just can't count them from the "end" that doesn't have an end.

    Why not?

    Because it does not have an end.

    0 lies below the end. Hence there is an end, even if you cannot see it.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 30 12:53:58 2024
    XPost: sci.math

    Le 29/08/2024 à 19:56, Jim Burns a écrit :
    On 8/29/2024 9:38 AM, WM wrote:

    That treats the potentially infinite collection.

    Call it ℕᴾᴵꟲ.
    The name doesn't matter.

    There is no natural number not.in ℕᴾᴵꟲ.

    Maybe if Bob can disappear. But logic prevents that.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 30 13:06:07 2024
    XPost: sci.math

    Le 30/08/2024 à 03:09, Richard Damon a écrit :
    On 8/29/24 9:26 AM, WM wrote:
    Le 28/08/2024 à 22:35, Moebius a écrit :

    Hint: NUF(x) := card({s e SB : s < x})   (x e IR)

    Hence NUF(x) = 0 for all x e IR, x <= 0, and NUF(x) = aleph_0


    for all x > the minimum distance between many unit fractions which is
    not 0.

    No, for all x > 0,

    Easier to answer: Are there two unit fractions lesorequal than all unit fractions?

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Fri Aug 30 09:01:41 2024
    XPost: sci.math

    On 8/30/24 8:55 AM, WM wrote:
    Le 29/08/2024 à 20:51, "Chris M. Thomasson" a écrit :
    On 8/29/2024 6:27 AM, WM wrote:
    Le 29/08/2024 à 01:48, Richard Damon a écrit :

    You just can't count them from the "end" that doesn't have an end.

    Why not?

    Because it does not have an end.

    0 lies below the end. Hence there is an end, even if you cannot see it.

    Regards, WM



    No, there does not need to be an "end" for an infinite sequence in that sequence.

    0 is not "below" the end, but IS the end for the unit fractions, as we
    can get as close to it as we want with a unit fraction.

    A sequence whose end is not in the sequence doesn't have an end in the sequence.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From WM@21:1/5 to All on Fri Aug 30 13:08:14 2024
    XPost: sci.math

    Le 30/08/2024 à 03:09, Richard Damon a écrit :
    On 8/29/24 9:27 AM, WM wrote:
    Le 29/08/2024 à 01:48, Richard Damon a écrit :
    On 8/28/24 8:57 AM, WM wrote:
    Le 28/08/2024 à 04:13, Richard Damon a écrit :


    Well NUF(x) does not exist, but that doesn't say that infinity is
    not actual,

    So the unit fractions are actually existing but their number isn't?
    Strange!

    There number exists, it is aleph_0.

    You just can't count them from the "end" that doesn't have an end.

    Why not?

    Because you can't start at an end that isn't there.

    0 is below the end. Hence there is an end.

    Try to tell me the actual number you are going to start at.

    I start with NUF(0) = 0.

    Logic that depends on the existance of something that doesn't exist is
    just broken.

    Mathematics is the science which allows to prove things which cannot be
    seen.

    Regards, WM

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Fri Aug 30 09:28:55 2024
    XPost: sci.math

    On 8/30/24 9:06 AM, WM wrote:
    Le 30/08/2024 à 03:09, Richard Damon a écrit :
    On 8/29/24 9:26 AM, WM wrote:
    Le 28/08/2024 à 22:35, Moebius a écrit :

    Hint: NUF(x) := card({s e SB : s < x})   (x e IR)

    Hence NUF(x) = 0 for all x e IR, x <= 0, and NUF(x) = aleph_0


    for all x > the minimum distance between many unit fractions which is
    not 0.

    No, for all x > 0,

    Easier to answer: Are there two unit fractions lesorequal than all unit fractions?

    Regards, WM



    No, but there are two unit fractions less than any given unit fraction,
    which is all that is needed to show that NUF(x) is broken, and can't
    have the value 1 at any finite unit fraction.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to All on Fri Aug 30 09:32:25 2024
    XPost: sci.math

    On 8/30/24 9:08 AM, WM wrote:
    Le 30/08/2024 à 03:09, Richard Damon a écrit :
    On 8/29/24 9:27 AM, WM wrote:
    Le 29/08/2024 à 01:48, Richard Damon a écrit :
    On 8/28/24 8:57 AM, WM wrote:
    Le 28/08/2024 à 04:13, Richard Damon a écrit :


    Well NUF(x) does not exist, but that doesn't say that infinity is
    not actual,

    So the unit fractions are actually existing but their number isn't?
    Strange!

    There number exists, it is aleph_0.

    You just can't count them from the "end" that doesn't have an end.

    Why not?

    Because you can't start at an end that isn't there.

    0 is below the end. Hence there is an end.

    No, 0 IS the end that isn't a part of the set, since the set has now end
    on that side.


    Try to tell me the actual number you are going to start at.

    I start with NUF(0) = 0.

    But then you can't move forward, as there can be no finite x such that
    NUF(x) is 1, since NUF(x/2) would need to be NUF(0) <= NUF(x/2) <=
    NUF(x), but there will be a unit fraction below x/2, and a unit fraction between x and x/2 (unless x >= 1) so NUF(x) can't be 1.


    Logic that depends on the existance of something that doesn't exist is
    just broken.

    Mathematics is the science which allows to prove things which cannot be
    seen.

    Regards, WM



    Right, but can not be seen is not the same as doesn't exist.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Fri Aug 30 12:09:46 2024
    XPost: sci.math

    On 8/30/2024 9:08 AM, WM wrote:
    Le 30/08/2024 à 03:09, Richard Damon a écrit :

    Logic that depends on
    the existance of something that doesn't exist
    is just broken.

    Mathematics is the science which allows to prove
    things which cannot be seen.

    Mathematics is the science which allows to prove
    things which cannot be seen and which do not exist.

    Mathematics is the science in which
    a set in which each element is not its end is
    a set for which its end does not exist.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to All on Fri Aug 30 11:44:39 2024
    XPost: sci.math

    On 8/30/2024 8:53 AM, WM wrote:
    Le 29/08/2024 à 19:56, Jim Burns a écrit :
    On 8/29/2024 9:38 AM, WM wrote:

    That treats the potentially infinite collection.

    Call it ℕᴾᴵꟲ.
    The name doesn't matter.

    There is no natural number not.in ℕᴾᴵꟲ.

    ⎛ Assume otherwise.
    ⎜ Assume darkᴹᵂ 𝔊 ≥ᵉᵃᶜʰ ℕᴾᴵꟲ

    ⎜ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ ⅟𝔊 > 0

    ⎜ β = greatest.lower.bound.⅟ℕᴾᴵꟲ
    ⎜ β > α ⇒ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ α
    ⎜ γ > β ⇒ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ γ
    ⎜ β ≥ ⅟𝔊 > 0

    ⎜ 2⋅β > β > ½⋅β > 0
    ⎜ β > ½⋅β ⇒ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ ½⋅β
    ⎜ 2⋅β > β ⇒ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ 2⋅β

    ⎜ However,
    ⎜ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ 2⋅β
    ⎜ visibleᵂᴹ ⅟k ≱ 2⋅β
    ⎜ visibleᵂᴹ ¼⋅⅟k ≱ ½⋅β
    ⎜ ⅟ℕᴾᴵꟲ ᵉˣⁱˢᵗˢ≱ ½⋅β

    ⎜ ...which contradicts
    ⎝ ⅟ℕᴾᴵꟲ ᵉᵃᶜʰ≥ ½⋅β

    Therefore,
    darkᴹᵂ 𝔊 ≱ᵉˣⁱˢᵗˢ ℕᴾᴵꟲ

    Maybe if Bob can disappear.

    ⎛ ∀S ⊆ ℕ: S ≠ {} ⇔ ∃k ∈ S: k = min.S

    ⎜ ∀k ∈ ℕ: k ≠ 0 ⇔ ∃j ∈ ℕ: j+1 = k

    ⎝ ∀j ∈ ℕ: ∃k ∈ ℕ: j+1 = k

    Move Bob into room 0.
    In order, swap guests in rooms n and n+1.

    If Bob is in room 𝔊, not all swaps are swapped.
    If all swaps are swapped, Bob is not in room 𝔊.
    Or in any room.
    'Bye, Bob.

    But logic prevents that.

    The next time you go further than saying
    "There is logic" to saying
    "Here is the logic..."
    will be the first time.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jeff Barnett@21:1/5 to All on Fri Aug 30 13:45:09 2024
    XPost: sci.math

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    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jeff Barnett@21:1/5 to All on Sat Aug 31 12:44:06 2024
    XPost: sci.math

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    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jim Burns@21:1/5 to Jeff Barnett on Sat Aug 31 14:20:55 2024
    XPost: sci.math

    On 8/30/2024 12:20 AM, Jeff Barnett wrote:
    On 8/29/2024 2:09 PM, Jim Burns wrote:

    Wolfgang Mückenheim has been rejecting
    Dedekind.infinite sets, sets R such that
    I can 1-1 map a strict subset of R to R.
    WM calls them "potentially infinite", by which
    WM means that these sets change, which means that
    WM is not talking about _those sets_
    which do not change.

    A typical WM.argument sets up some sequence and
    asserts by mathematics, by logic
    (in reality, by "common sense", by "obviousness")
    that the sequence has two ends.
    Since the sequence has only one _visible_ end,
    WM considers that proof of a second end which is _dark_

    I've put my nose into a few of
    the same newsgroups that you have and
    I am aware of the aroma.
    I no longer even think about interacting with
    WM, PO, or a few others cut from the same sad cloth.
    Of course I occasionally post something *about* them.
    There are several possibilities for
    the WM of these threads:
    1) He's an idiot;
    2) He's extremely lonesome and
    these interactions pass for "being engaged";
    3) He's not who he claims to be
    (an instructor at a (junior?) college);
    4) he's simply an outright troll;
    5) He's religious;
    6) He's delusional.

    7) He's a victim of the Dunning.Kruger effect.

    Of course any subset of these is possible.

    In any event
    no interactions with him will change his behavior --
    that is one of his most interesting similarities to PO.
    It's really too bad that some newsgroups
    that were both interesting and educational
    have fallen so far.
    I remember the great hopes we all had
    back in the late 1960s and early 1970s
    for a different, better informed, and
    just plain better world
    because of the supportable vision
    the ARPA research net seemed to offer.

    Please, do not surrender your vision of
    a just plain better world.
    These things take time.

    There's a phrase I heard somewhere that
    explained so much to me that
    it has become something of a catch.phrase of mine.
    ⎛ Technology is everything that doesn't work yet.
    ⎝ -- W. Danny Hillis

    https://www.wired.com/2010/06/cognitive-surplus/
    | Cognitive Surplus,
    | the new book by internet guru Clay Shirky,
    | begins with a brilliant analogy.
    | He starts with a description of London in the 1720s,
    | when the city was in the midst of a gin binge.
    | A flood of new arrivals from the countryside meant
    | the metropolis was crowded, filthy, and violent.
    | As a result,
    | people sought out the anesthesia of alcohol
    | as they tried to collectively forget
    | the early days of the Industrial Revolution.

    Overflowing cities, distilled spirits, at the time,
    were part of everything that doesn't work yet.
    And, to some degree, they're still part --
    but we've learned a lot in the last few centuries
    about how to deal with all that.

    I have an abiding faith that answers exist to
    the idiot/troll/Dunning.Kruger problem.
    And there are enough people who dearly want to
    find those answers, that I feel confident in
    my prediction that answers will be found.

    What we all forgot was
    the old wisdom that stated
    "Character / ethics / personnel worth / soul / etc.
    is what you are when no one can see you.
    Unfortunately
    the internet made the magic shield that
    exposed all of human nastiness while
    masking the faces of the perpetrators.

    It wasn't that long ago that everyone understood
    and agreed with
    ⎛ Never pick a fight with
    ⎜ a man who buys ink by the barrel.
    ⎝ -- possibly Mark Twain

    These days, essentially everyone buys ink by
    the metaphorical barrel.

    It is a situation which has its downsides,
    but, even so, I prefer it to the previous arrangement.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Sep 1 01:19:36 2024
    XPost: sci.math

    Am 30.08.2024 um 21:18 schrieb Chris M. Thomasson:
    On 8/30/2024 5:55 AM, WM wrote:
    Le 29/08/2024 à 20:51, "Chris M. Thomasson" a écrit :
    On 8/29/2024 6:27 AM, WM wrote:
    Le 29/08/2024 à 01:48, Richard Damon a écrit :

    You just can't count them from the "end" that doesn't have an end.

    Why not?

    Because it does not have an end.

    0 lies below the end. Hence there is an end, even if you cannot see it.

    1/0 is NOT a unit fraction damn it! wow.

    Wrong end.

    WM is talking about a SMALLEST unit fraction s with: 0 < s < ... 1/1.

    Of couse, there is no smallest unit fractions. If s is a unit fraction
    then 1/(1/s + 1) is a smaller one.

    1/0 would be the LARGEST unit fraction, if it were a unit fraction. :-)

    Actually, the largst unit fraction is 1/1 = 1.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Sep 1 01:26:29 2024
    XPost: sci.math

    Am 30.08.2024 um 21:22 schrieb Chris M. Thomasson:
    On 8/30/2024 12:18 PM, Chris M. Thomasson wrote:

    0 lies below the end. Hence there is an end, even if you cannot see it.

    1/0 is NOT a unit fraction damn it! wow.

    We can all see 0.

    Can we?

    But its not a unit fraction. If it was, then 1/0 would
    be defined as a unit fraction, but

    You are mixing up things here.

    Right 0 is not a unit fraction.

    On the other hand, if s is a unit fraction, then 1/s is a natural number
    (and vice versa).

    "1/0" doesn't denote a real number. (Usually it's not even a defined term.)

    In general we have for natural numbers n,m: n < m -> 1/n > 1/m.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Sep 1 05:21:12 2024
    XPost: sci.math

    Am 01.09.2024 um 04:54 schrieb Chris M. Thomasson:
    On 8/31/2024 4:19 PM, Moebius wrote:
    Am 30.08.2024 um 21:18 schrieb Chris M. Thomasson:
    On 8/30/2024 5:55 AM, WM wrote:
    Le 29/08/2024 à 20:51, "Chris M. Thomasson" a écrit :
    On 8/29/2024 6:27 AM, WM wrote:
    Le 29/08/2024 à 01:48, Richard Damon a écrit :

    You just can't count them from the "end" that doesn't have an end. >>>>>>
    Why not?

    Because it does not have an end.

    0 lies below the end. Hence there is an end, even if you cannot see it. >>>
    1/0 is NOT a unit fraction damn it! wow.

    Wrong end.

    Yeah. I still don't know how WM is going to count the opposite way wrt:

    1/1, 1/2, 1/3, 1/4, ...

    That would be:

    ..., 1/4, 1/3, 1/2, 1/1

    He can't do it.

    Yeah, but a function can do it.

    NUF(x) := the cardinal number of unit fractions that are smaller than x
    (where x is a real number)

    Then we get, say, NUF(0) = 0 and, say, NUF(1/1) = aleph_0, NUF(1/2) =
    aleph_0, NUF(1/3) = aleph_0, etc.

    In general, for all x e IR, x > 0: NUF(1/3) = aleph_0.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Sep 1 05:25:57 2024
    XPost: sci.math

    Am 01.09.2024 um 04:56 schrieb Chris M. Thomasson:
    On 8/31/2024 4:26 PM, Moebius wrote:
    Am 30.08.2024 um 21:22 schrieb Chris M. Thomasson:

    We can all see 0.

    Can we?

    We can see the symbol used to [denote, refer to] zero? ;^)

    Indeed (if we are not blind)! :-)

    In WM's world there are visible and dark (nonvisible) as well as grey
    NUMBERS. :-P Well...

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Sep 1 05:27:29 2024
    XPost: sci.math

    Am 01.09.2024 um 04:54 schrieb Chris M. Thomasson:

    I still don't know how WM is going to count the opposite way wrt:

    1/1, 1/2, 1/3, 1/4, ...

    That would be:

    ..., 1/4, 1/3, 1/2, 1/1

    He can't do it.

    Yeah, but a function can do it.

    NUF(x) := the cardinal number of unit fractions that are smaller than x
    (where x is a real number)

    Then we get, say, NUF(0) = 0 and, say, NUF(1/1) = aleph_0, NUF(1/2) =
    aleph_0, NUF(1/3) = aleph_0, etc.

    In general, for all x e IR, x > 0: NUF(x) = aleph_0.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Moebius@21:1/5 to All on Sun Sep 1 21:18:55 2024
    XPost: sci.math

    Am 01.09.2024 um 21:09 schrieb Chris M. Thomasson:
    On 8/31/2024 8:27 PM, Moebius wrote:
    Am 01.09.2024 um 04:54 schrieb Chris M. Thomasson:

    I still don't know how WM is going to count the opposite way wrt:

    1/1, 1/2, 1/3, 1/4, ...

    That would be:

    ..., 1/4, 1/3, 1/2, 1/1

    He can't do it.

    Yeah, but a function can do it.

    NUF(x) := the cardinal number of unit fractions that are smaller than
    x   (where x is a real number)

    Then we get, say, NUF(0) = 0 and, say, NUF(1/1) = aleph_0, NUF(1/2) =
    aleph_0, NUF(1/3) = aleph_0, etc.

    In general, for all x e IR, x > 0: NUF(x) = aleph_0.

    How does that fit with WM who thinks there is a smallest unit fraction
    to start counting from?

    It doesn't. After all, it's mathematcs.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)