It is strange that blatantly false results as the equinumerosity ofJuste because it doesn't match your intuition doesn't mean it's not
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 rulesthat you would like instead.
_The rule of subset_ proves that every proper subset has less elementsWhat exceptions do you mean?
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).Only this leads to some contradictions depending on the construction.
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 inHow small an error?
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.
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
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.
Am Fri, 26 Jul 2024 16:31:19 +0000 schrieb WM:
It is strange that blatantly false results as the equinumerosity ofJuste because it doesn't match your intuition doesn't mean it's not
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?
useful.
Not all infinite sets can be compared by size, but we can establish somethat you would like instead.
useful rules
_The rule of subset_ proves that every proper subset has less elementsWhat exceptions do you mean?
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.
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 inHow small an error?
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.
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,
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
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?
Le 27/07/2024 à 13:27, Richard Damon a écrit :Neither is there in actual infinity.
On 7/27/24 7:13 AM, WM wrote:In potential infinity there is no ω.
Le 27/07/2024 à 04:23, Richard Damon a écrit :So, what part is not true?
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.
Yes they are.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 theI wonder how you get the second infinity. What is the preimage of all
set {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.
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
_The rule of subset_ proves that
every proper subset has
less elements than its superset.
_The rule of subset_ proves that
every proper subset has
less elements than its superset.
Am Sat, 27 Jul 2024 12:16:24 +0000 schrieb WM:
Le 27/07/2024 à 13:27, Richard Damon a écrit :Neither is there in actual infinity.
On 7/27/24 7:13 AM, WM wrote:In potential infinity there is no ω.
Le 27/07/2024 à 04:23, Richard Damon a écrit :So, what part is not true?
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.
Yes they are.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}.I wonder how you get the second infinity. What is the preimage of all
the omegas?
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.
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,
If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ
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.
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
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
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|}
sci.logic and sci.math
On 7/26/2024 12:31 PM, WM wrote:
_The rule of subset_ proves that [bla bla bla]
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.
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?
On 7/28/24 7:55 AM, WM wrote:
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.
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.
Using the unit fractions itelligent readers understand that there mustNope, since that implies there is a highest Natural Number, which breaks their definition,
be a first one after zero. Others must believe in the magical appearance
of infinitely many unit fractions.
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
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.
He also notes that what we have learned from finite quatities does
not apply to infinity.
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?
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
Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1
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.
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.
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",
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.
On 7/28/2024 8:17 AM, WM wrote:
ω-1 ∈ ℕ
'Defining into existence' that which doesn't exist
makes nonsense of whatever meaning one's words have.
ℕ = {1, 2, 3, ..., ω-1}
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.
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.
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.
It's like the universe of set theory,
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.
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.
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
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 thereNope, since that implies there is a highest Natural Number, which
must be a first one after zero. Others must believe in the magical
appearance of infinitely many unit fractions.
breaks their definition,
That is unavoidable. You believe in the magical appearance of infinitely
many unit fractions. That breaks logic and mathematics.
Regards, WM
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.
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.
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.
On 7/29/24 9:11 AM, WM wrote:
But what number became ω when doubled?
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?
That is unavoidable. You believe in the magical appearance of infinitely
many unit fractions. That breaks logic and mathematics.
Nope,
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
¬(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
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].
On 7/30/2024 1:30 PM, WM wrote:
[...] between [0, 1] and (0, 1].
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
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
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.
NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
does not imply any unit fraction outside (0,x]
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?
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.
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.
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} ?
The input set was the Natural Numbers and w,
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?
∀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.
Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.
Le 31/07/2024 à 03:28, Richard Damon a écrit :That is a bit imprecise. Even though you keep on talking about
On 7/30/24 1:37 PM, WM wrote:In the midst, far beyond all definable numbers, far beyond ω/10^10.
Le 30/07/2024 à 03:18, Richard Damon a écrit :And where is that in {1, 2, 3, ... w} ?
On 7/29/24 9:11 AM, WM wrote:ω/2
But what number became ω when doubled?
Completeness of N? No number n reaches omega.The input set was the Natural Numbers and w,ω/10^10 and ω/10 are dark natural numbers.
I assume completness.But you combined two different sets, so why can't there be a gap?Because otherwise there was a gap below ω.If all natural numbers exist, then ω-1 exists.Why?
That is not a contradiction.No. My formula says ∀n ∈ ℕ.∀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.
Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.Not for all dark numbers.
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.
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
Am Wed, 31 Jul 2024 14:27:06 +0000 schrieb WM:
Le 31/07/2024 à 03:28, Richard Damon a écrit :That is a bit imprecise. Even though you keep on talking about
On 7/30/24 1:37 PM, WM wrote:In the midst, far beyond all definable numbers, far beyond ω/10^10.
Le 30/07/2024 à 03:18, Richard Damon a écrit :And where is that in {1, 2, 3, ... w} ?
On 7/29/24 9:11 AM, WM wrote:ω/2
But what number became ω when doubled?
consecutive infinities, you can't compare natural and "dark" numbers.
ω/10^10 and ω/10 are dark natural numbers.Completeness of N? No number n reaches omega.
I assume completness.But you combined two different sets, so why can't there be a gap?Because otherwise there was a gap below ω.If all natural numbers exist, then ω-1 exists.Why?
That is not a contradiction.No. My formula says ∀n ∈ ℕ.∀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.
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)
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.
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,
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.
Le 01/08/2024 à 02:09, Richard Damon a écrit :No. There is ALWAYS an epsilon.
On 7/31/24 10:27 AM, WM wrote:That is simply nonsense. Do you know what an accumalation point is?
Le 31/07/2024 à 03:28, Richard Damon a écrit :In other words, outside the Natural Nubmer, all of which are defined
On 7/30/24 1:37 PM, WM wrote:In the midst, far beyond all definable numbers, far beyond ω/10^10.
Le 30/07/2024 à 03:18, Richard Damon a écrit :And where is that in {1, 2, 3, ... w} ?
On 7/29/24 9:11 AM, WM wrote:ω/2
But what number became ω when doubled?
and definable.
Every eps interval around 0 contains unit fractions which cannot be
separated from 0 by any eps. Therefore your claim is wrong.
How are they defined?They are natural numbers.They may be "dark" but they are not Natural Numbers.The input set was the Natural Numbers and w,ω/10^10 and ω/10 are dark natural numbers.
A "gap" implies some sort of space that is not filled. There is no suchNatural numbers, by their definition, are reachable by a finite numberThat is the opinion of Peano and his disciples. It holds only for
of successor operations from 0.
potetial infinity, i.e., definable numbers.
What is the reason for the gap before omega? How large is it? Are these questions a blasphemy?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?
If k did not have a successor, what would k+1 be?That does ny formula not say. It says for all n which have successors,Right, for ALL n in ℕ, there exist another number in ℕ that is n+1,No. My formula says ∀n ∈ ℕ.∀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.
there is distance between 1/n and 1/(n+1).
It is the definition of definable numbers. Study the accumulation point. Define (separate by an eps from 0) all unit fractions. Fail.Maybe not for dark numbers, but it does for all Natural Numbers, asRemember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists. >>> Not for all dark numbers.
that is part of their DEFINITION.
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.
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.
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.
In the dark domain
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 :That is a bit imprecise. Even though you keep on talking about
On 7/30/24 1:37 PM, WM wrote:In the midst, far beyond all definable numbers, far beyond ω/10^10.
Le 30/07/2024 à 03:18, Richard Damon a écrit :And where is that in {1, 2, 3, ... w} ?
On 7/29/24 9:11 AM, WM wrote:ω/2
But what number became ω when doubled?
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.Completeness of N? No number n reaches omega.
I assume completness.But you combined two different sets, so why can't there be a gap?Because otherwise there was a gap below ω.If all natural numbers exist, then ω-1 exists.Why?
What is immediately before ω? Is it a blasphemy to ask such questions?
That is not a contradiction.No. My formula says ∀n ∈ ℕ.∀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.
It is not a contradiction to my formula if some n has no n+1.
Regards, 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.
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).
Not for all dark numbers.
Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists. >>>
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
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.
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.
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
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.
What is the reason for the gap before omega? How large is it? Are theseA "gap" implies some sort of space that is not filled. There is no such
questions a blasphemy?
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,
If k did not have a successor, what would k+1 be?
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.
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.
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.
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.
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,
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.
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.
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
So we have an unlimited number of Unit fractions, and no smallest one.
∀ᴿx > 0: NUF(x) ≥ ℵ₀
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
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.
The reversal of the order of qualifications causes the problem.
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,
Le 02/08/2024 à 01:53, Richard Damon a écrit :All unit fractions are larger than zero, so an epsilon can be chosen
On 8/1/24 8:27 AM, WM wrote:
And thus there is no "smallest" unit fraction, as for any eps, thereYour eps cannot be chosen small enough.
are unit fractions smaller,
That is not the definition. The "infinitely many" are not the same onesNot for ℵo, i.e., for most it is wrong:That is the opinion of Peano and his disciples. It holds only forNo, it holds for ALL his numbers.
potetial infinity, i.e., definable numbers.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
There is no gap above zero but e real continuum.What is the reason for the gap before omega? How large is it? AreBecause it is between two different sorts of number.
these questions a blasphemy?
There is a gap between 1 and 2, but that doesn't bother you.All gaps of size 1 do not bother me..
There are infinitely many by the definition of accumulation point. YouIt is the definition of definable numbers. Study the accumulationSo, which Unit fraction doesn't have an eps that seperates it from 0?
point.
Define (separate by an eps from 0) all unit fractions. Fail.
cannot find them. Therefore they are dark.
The set of reals is infinite and does not have a minimum.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.
WTF?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 itThere 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.
On 8/2/24 7:38 AM, WM wrote:
Le 01/08/2024 à 18:04, joes a écrit :Improperly revesing the conditionals.
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.
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 theseA "gap" implies some sort of space that is not filled. There is no such
questions a blasphemy?
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
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
Am Fri, 02 Aug 2024 15:09:18 +0000 schrieb WM:
Le 02/08/2024 à 01:53, Richard Damon a écrit :All unit fractions are larger than zero, so an epsilon can be chosen
On 8/1/24 8:27 AM, WM wrote:Your eps cannot be chosen small enough.
And thus there is no "smallest" unit fraction, as for any eps, there
are unit fractions smaller,
That is not the definition. The "infinitely many" are not the same onesThere is a gap between 1 and 2, but that doesn't bother you.All gaps of size 1 do not bother me..
There are infinitely many by the definition of accumulation point. YouIt is the definition of definable numbers. Study the accumulationSo, which Unit fraction doesn't have an eps that seperates it from 0?
point.
Define (separate by an eps from 0) all unit fractions. Fail.
cannot find them. Therefore they are dark.
for every epsilon.
The set of reals is infinite and does not have a minimum.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.
WTF?So we have an unlimited number of Unit fractions, and no smallest one.But you have a limited number of eps.
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 :Improperly revesing the conditionals.
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.
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
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.
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
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,
Not that there is a single finite difference that seperates any two
arbirtry unit fractions.
Not understanding the order of the arguement
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.
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,
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.
Again, you are changing the conditional incorrectly because you logic
can't handle unbounded values.
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.
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.
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
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
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
On 8/2/2024 11:36 AM, WM wrote:
ℵ₀ unit fractions need ℵ₀ [corrected --Moebius] points above zero.
For those x > 0 your claim is wrong.
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.
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.
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?
(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
For each and every of these points [here referred to with the
variable "x"]: NUF(x) = ℵ₀ .
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?
If there WAS just a single eps that seperated ALL unit fractions, then
the could be no more than 1/eps unit fractions,
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 numbersI know that ℵo is nonsense, because all prime numbers are algebraic but
because you don't understand that there are exactly ℵo of both of them. >>
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.
All countably infinite sets are the same size,
That proves that ℵo is nonsense.
No, it proves that your logic can't handle it.
If you can show a contradiction in the system, USING THE RULES OF THE
SYSTEM,
Le 02/08/2024 à 19:12, Richard Damon a écrit :
On 8/2/24 12:42 PM, WM wrote:
Because most are dark.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?
Regards, WM
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.
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
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 numbersI know that ℵo is nonsense, because all prime numbers are algebraic
because you don't understand that there are exactly ℵo of both of them. >>>
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
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.
On 8/3/2024 7:25 AM, WM wrote:Mückenheim, Du hirnloser Spinner: Deine Definition von NUF(x) war
I recognized lately that you use the wrong definition of NUF.
On 8/3/2024 10:23 AM, WM wrote:
NUF(x) = ℵ₀ for all x > 0 is wrong.
NUF(x) = 1 for all x > 0 [...] is wrong
NUF(x) > 1 for all x > 0 is correct
On 8/3/2024 10:23 AM, WM wrote:
NUF(x) = ℵ₀ for all x > 0 is wrong.
NUF(x) = 1 for all x > 0 [...] is wrong
NUF(x) > 1 for all x > 0 is correct
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.
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.
Deine Definition von NUF(x) warI recognized lately that you use the wrong definition of NUF.
bisher: "die Anzahl (Kardinalzahl) aller Stammbrüche, die kleiner-gleich
x sind".
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.
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.
You (WM) have mad [sic!] an unreliable quantifier shift.
Am 04.08.2024 um 17:53 schrieb Jim Burns:
You (WM) have mad [sic!] an unreliable quantifier shift.
You (WM) are mad from an unreliable quantifier shift.
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,
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.
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
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
Only you (WM) think that ∃u ∀x>0: u < x
follows from ∀x>0 ∃u: u < x,
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.
ℵ₀ unit fractions occupy ℵ₀*2^ℵ₀ points.
Not even one unit fraction can be
smaller than all positive x.
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
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.
The latter is close to [...]:
There are NUF(x) [u ∈ ⅟ℕ:] u < x.
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
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.
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 claim concerns only definable x.
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.
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".
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 > 0 ⇒ NUFᵈᵉᶠ(x) = ℵ₀
and, implicitly quantified, as
x > 0 ⇒ NUFᵈᵉᶠ(x) = ℵ₀
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?
Impossible. [...] Not even one unit fraction can be smaller than all positive x.
Huh?
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...
See?
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
...
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
...
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,
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 (*).
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.
∃^ℵ₀ u ∈ ⅟ℕ: ∀x > 0: u < x (false) .
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 .
Btw. we might implement this function (in Python) the followimg way:
def next_unit_fraction(n, m):
return (1, m//n + 1)
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)
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.
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).
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
On 8/5/2024 3:04 PM, WM wrote:
For every number of unit fractions
NUF(x) gives the smallest interval (0, x).
Le 04/08/2024 à 21:18, Richard Damon a écrit :
On 8/4/24 11:23 AM, WM wrote:Always.
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.
Regards, WM
On 8/5/2024 3:04 PM, WM wrote:
NUF(1) = ℵ₀
NUF(x) = ℵ₀
For each x > 0
⅟ℕᵈᵉᶠ∩(0,x) is not finite.
On 8/5/24 2:58 PM, WM wrote:
Always.If there are all, then there is half of all.
Not with infinities.
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.
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
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).
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.
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).
There are infinite[ely many] even[ numbers] and there are infinite[ly many] odd[ numbers].
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.
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.
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.
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. :-)
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!
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.
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.
(We could however subdivide the distance
between u_3 and u_4.)
| 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.
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
So 0 is an odd even number? :-)
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? :-)
I say even?
Well, both (odd and even) at the same time?
This makes me think of signed zero...
+0
-0
and just, 0?
0 is just zero?
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?
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
Le 06/08/2024 à 03:35, Richard Damon a écrit :
On 8/5/24 2:58 PM, WM wrote:
Always.If there are all, then there is half of all.
Not with infinities.
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
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.
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 ?
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.
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.
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
But we have
proof INVNUF(3) does not exist.
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?
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?
On 8/7/2024 6:05 AM, WM wrote:
Peano is not valid for all dark numbers.
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.
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?
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
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)
(1) (1/n - 1/(n+1))_(n e IN)
is a null sequence
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
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.
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]
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.
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.
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.
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.
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 [starting with NUF(0) = 0].
you (WM) find silence with regard to
the rest of your argument
more advantageous, apparently.
There is no rest.
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).
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)."
Exists x with x = INVNUF(1).
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]
Infinite is beyond all finites, even big.finites.
Infinite does not have finiteness.properties.
There exists x = INVNUF(1).
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) >
...
Exists x with x = INVNUF(1).
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?
Exists x with x = INVNUF(1).
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.
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.
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...
you [WM] have a HYPER finite mind that simply cannot deal with
infinity in any way, shape or from in the first place...
The main point is that there is no smallest unit fraction
Sorry for any confusion. ;^o
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 <.
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.
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_ ?
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.
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)! :-)
⎛ <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_ ?
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.
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.
On 8/8/2024 4:59 PM, Moebius wrote:
Am 08.08.2024 um 22:36 schrieb Chris M. Thomasson:Yup. Instead if using first, I should say largest... Any better?
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 <.
Surely, a lawyer wouldn't think that
"Boom! Here's the conclusion"
is an _argument_ ?
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.
and he thinks that
a set ordered with two ends is
more complete than
the same set with one or zero ends.
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.
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?
and [WM] thinks that
a set ordered with two ends is
more complete than [...]
The set of unit fractions has only one end.
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
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
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, [...]
See: https://www.youtube.com/watch?v=VScSEXRwUqQ
See: https://www.youtube.com/watch?v=VScSEXRwUqQ
Please study it closely!
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,
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?
What causes an exception: nₓ ∈ ℕ without ⅟(nₓ+1) ?
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.
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.
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.
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
Each positive unit fraction is not
the first positive unit fraction.
What causes an exception: nₓ ∈ ℕ:
⅟nₓ > 0 without ⅟(nₓ+1) > 0 ?
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
Le 11/08/2024 à 19:56, Jim Burns a écrit :asserts 1/(n+1) exists, '∀n ∈ ℕ'
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
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.
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.
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]
∀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.
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.
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 .
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
∀n ∈ ℕ: 1/n > 1/(n+1)
If 1/(n+1) exists. [WM]
Uhhhh..... WHAT!?
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 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.
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
In der Mathematik ist es so:
Es gibt das allgemein akzeptierte Peano-Axiom:
An e IN: s(n) e IN.
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.
On 8/13/2024 10:27 AM, WM wrote:
the smallest unit fractions
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.
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.
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.
resulting in a real coordinate x with NUF(x) = 1.
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.
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.
But increase by more than 1 is excluded by
the gaps between unit fractions.
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.
But increase by more than 1 is excluded by
the gaps between unit fractions.
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.
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.
It depends on context. Are we going from 1/1 all the way down, or from
0 all the way up to 1/1.
0 is not a unit fraction, which means there is no smallest one... Fair enough?
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?
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?
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.
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.
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.
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
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!
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.
resulting in a real coordinate x with NUF(x) = 1.
INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
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.
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:
In fact, Ax > 0: NUF(x) = aleph_0, while NUF(0) = 0.
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.
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.
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.
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.
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.
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.
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.
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).
On 8/14/2024 8:23 AM, WM wrote:
No.
Is "No" an argument?
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.
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
INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
On 8/14/2024 8:23 AM, WM wrote:
Therefore there is no step at 0.
There is a step at 0.
However,
0 is not a unit.fractionᵈᵉᶠ.
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.
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,
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,∞).
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?)
The first point with unit fractions is x = INVNUF(1).</WM>
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?
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.
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
I've just proved INVNUF(1) not.exists?
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
I've just proved INVNUF(1) not.exists?
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
I've just proved INVNUF(1) not.exists?
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.
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.)
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.
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.)
INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
NUF(INVNUF(1)) > 1
Contradiction.
Our problem, as a community, is that this crook is actually teaching
in an academic institution.
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.)
INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋
NUF(INVNUF(1)) > 1
Contradiction.
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...
Am 15.08.2024 um 17:53 schrieb Jim Burns:
I've just proved INVNUF(1) not.exists?
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".
Defining ⅟𝔊 to be the smallest unit fraction
isn't a claim that ⅟𝔊 exists.
'⅟𝔊' can be used to derive contradictions,
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.
The smallest point of (0, oo) belongs to (0, oo).
It cannot be seen. It is dark.
Translate ¬∃ᴿx(x = 1/0) to
¬∃ᴿx: 0⋅x = 1
Prove that.
Proof by contradiction (RRA).
Translate ¬∃ᴿx(x = 1/0) to
¬∃ᴿx: 0⋅x = 1
Prove that.
Proof by contradiction (RRA).
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,
Translate ¬∃ᴿx(x = 1/0) to
¬∃ᴿx: 0⋅x = 1
Prove that.
Proof by contradiction (RRA).
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).
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.
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.
That's a hill which I doubt we need to climb.
A definition isn't a claim something exists.Yeah, if you define a "property", say,
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
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!
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.
That's a hill which I doubt we need to climb.
Want to be the the next crank in sci.math? (After all
the other cranks have left?)
Half or more of my proofs to WM say
"Assume otherwise... However... Contradiction."
But maybe I'm crazy.
[nonsense deleted]
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]
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?
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 [...]
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.
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 [...]
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.
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 [...]
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]
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
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.
√2 is irrational.
⎛ Assume otherwise.
⎜ Assume p₃,q₃ ∈ ℕ₁: p₃⋅p₃ = 2⋅q₃⋅q₃:
⎝ Contradiction.
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?
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.
√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.
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
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
On 8/15/2024 9:52 AM, WM wrote:
⎜ Assume NUF(x) = 0 and x > 0
Our problem, as a community, is not proving Wolfgang Mückenheim wrong.
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!
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
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.
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.
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
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
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.
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?
On 8/17/2024 9:28 AM, WM wrote:
SBZ(x) starts with 0 at 0 and increases,
but [bla bla bla].
Therefore there is a smallest unit fractions and vice versa a greatest
natnumber.
What can't you understand?
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,
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
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.
Dark unit fractions have a smallest element.
No smaller unit fractions is existing
No smaller unit fractions is existing
because no larger natnumber is existing.
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."
WM: "Dark unit fractions have a smallest element."
WM says a lot of things.
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
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.
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.
On 8/19/2024 7:40 AM, WM wrote:
Dark unit fractions have a smallest element.
Darkᵂᴹ unit.fractions are positive. Correct?
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.
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.
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,
But, you can't just claim sommething is there that isn't.
The Number system is DERIVED from rules.
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.
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.
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.
Are you about to teach this at Technische Hochschule Augsburg
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.
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
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.
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.
Are you about to teach this at Technische Hochschule Augsburg
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.
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?
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.
It starts from 0.
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.
Dark unit fractions cannot be defined.
The GLB that can be defined is 0.
And 0 is glb.⅟ℕᵈᵉᶠ not.in ⅟ℕᵈᵉᶠ
It starts from 0.
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
On 8/20/2024 9:07 AM, WM wrote:
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
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.
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'
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,
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
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
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
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.
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 exists at x > 0 because NUF(0) = 0.
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.
Le 20/08/2024 à 22:05, Jim Burns a écrit :
No min.⅟ℕᵈᵉᶠ exists.
The reason is potential infinity.
But dark unit fractions are assumed to be
actually infinite.
[...] actually infinite.
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
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.
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.
WM presented the following explanation :
Since no unit fraction is below or at zero, the end is before. What else?
No end, Duh!!
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.
There is no set ℕᵈᵉᶠ because of lacking completeness. Sets of set theory
are complete.
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.
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
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.
Since no unit fraction is below or at zero, the end is before. What else?
It exists at x > 0 because NUF(0) = 0.
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,
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.
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.
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
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.
Le 22/08/2024 à 16:22, FromTheRafters a écrit :
WM used his keyboard to write :
Logic is finite.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.
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.
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.
Logic is finite.
All finiteⁿᵒᵗᐧᵂᴹ ordered sets _and their subsets_
are 2.ended, except for {}.
Le 22/08/2024 à 16:22, FromTheRafters a écrit :
WM used his keyboard to write :
Logic is finite.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.
Regards, WM
I guess by your finite logic, Achilles can't pass the tortoise,
On 8/23/24 10:49 AM, WM wrote:
NUF(x) increases from 0 to more. But it cannot increase by more than 1And, since there is no finite value of x that makes NUF(x) equal to 1,
without shzowing a constant level.
since for every finite x, their exists an x/2
Just because you have verbally defined a function doesn't mean it exists
and has proper values.
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
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 thanAnd, since there is no finite value of x that makes NUF(x) equal to 1,
1 without shzowing a constant level.
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
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.
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.
¬∃ᴿx>0: NUF(x) = 1
Why do you say that it is "1000" states that are dark?
Your final answer is basically just admitting that your logic can't
supply the needed properties of the Natural Numbers.
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.
The function exists if actual infinity exists.
The function does not exist if only potential infinity exists.
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.
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?
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
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
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
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.
¬∃ᴿx>0: NUF(x) = 1
Then NUF(x) does not exist
Well NUF(x) does not exist, but that doesn't say that infinity is not
actual,
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.
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.
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.
An actually.infiniteᵂᴹ set is
a not.potentially.infiniteᵂᴹ set with
a potentially.infiniteᵂᴹ subset.
¬∃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,
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.
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?
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.
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?
Georg Cantor [...]
"Further it is useful to have
a symbol expressing the absence of points. [...]
I propose a very conservative answer:
that we accept at least
the empty set existsᴲ,
Does it?
Le 28/08/2024 à 15:25, WM a écrit :
Gottlob Frege shared his opinion:
things,"If, according to our previous use of the word, a class consists of
whenis a collection, a collective union of them, then it must disappear
thenthese things disappear. If we burn down all the trees of a forest,
Hildesheim (1990) p. 195]we burn down the forest. Thus an empty class cannot exist."
[G. Frege: "Kleine Schriften", I. Agelelli (ed.), 2nd ed., Olms,
Anyhow, "zero things" means "no things". So we can
safely say <bla bla bla>
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?
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?
each non.empty S ⊆ B is 2.ended.
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,
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
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.
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
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
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.
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.
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.
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
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.
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.
On 8/29/24 9:26 AM, WM wrote:
Le 28/08/2024 à 22:35, Moebius a écrit :No, for all x > 0,
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.
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
On 8/29/24 9:27 AM, WM wrote:
Le 29/08/2024 à 01:48, Richard Damon a écrit :Because you can't start at an end that isn't there.
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?
Try to tell me the actual number you are going to start at.
Logic that depends on the existance of something that doesn't exist is
just broken.
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 :No, for all x > 0,
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.
Easier to answer: Are there two unit fractions lesorequal than all unit fractions?
Regards, WM
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 :Because you can't start at an end that isn't there.
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?
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
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.
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.
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.
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.
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.
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.
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.
But its not a unit fraction. If it was, then 1/0 would
be defined as a unit fraction, but
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 :1/0 is NOT a unit fraction damn it! wow.
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. >>>
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.
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? ;^)
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.
On 8/31/2024 8:27 PM, Moebius wrote:
Am 01.09.2024 um 04:54 schrieb Chris M. Thomasson:How does that fit with WM who thinks there is a smallest unit fraction
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.
to start counting from?
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