Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Corollary-question:
Does the number of nines grow
when in 0.999 the decimal point is shifted
by one or more position?
WM used his keyboard to write :
Let the infinite sequence 0.999... be multiplied by 10. Does the number of >> nines grow?
No, both sequences are infinite.
Corollary-question: Does the number of nines grow when in 0.999 the decimal >> point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
On 6/25/2024 1:18 PM, WM wrote:
Let the infinite sequence 0.999... be multiplied by 10. Does the number
of nines grow?
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
Huh? .999 * 10 = 9.99
The number of nines is still three.
On 6/25/2024 4:18 PM, WM wrote:
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1
tl;dr
No.
Nuance:
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.
The positions in 0.999... correspond to
numbers in well.ordered inductive ℕᴬ⤾⁺¹₀ᐣ⤓
Le 26/06/2024 à 00:11, Jim Burns a écrit :
On 6/25/2024 4:18 PM, WM wrote:
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1
tl;dr
No.
Is the set of natural indices complete
such that
no natural number can be added?
Nuance:
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.
The positions in 0.999... correspond to
numbers in well.ordered inductive ℕᴬ⤾⁺¹₀ᐣ⤓
And they are fixed.
Therefore your answer is correct:No.
Therefore
9.999... has one 9 less [one 9 fewer]
after the decimal point than 0.999... .
On 6/26/2024 7:20 AM, Jim Burns wrote:
Inductive:
for each 9 there is its successor.9
There isn't one 9 fewer.
I still wonder why WM thinks
there is one 9 "fewer"... Strange one!
Therefore,
once again,
'humongous' and 'infinite' are different.
Indeed! :^)
Ross Finlayson has brought this to us :
On 06/25/2024 04:17 PM, FromTheRafters wrote:
WM used his keyboard to write :
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
No, both sequences are infinite.
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
What does Simon Stevin say?
The number of nines left of the radix grows, ....
Though, one might aver it's the "count" of nines,
it's also its number.
Counting and numbering are two different things,
though they're often conflated, not to be confused.
Numbers "have" a number and "make" a count.
...000099.999...
There are countably many nines on either side of the radix point.
WM has brought this to us :
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal expansion >>> representation?
You must understand that no natural number can be added to the complete set.
Of course not, a set contains *all* of its elements.
Therefore no index can be added here. 9.999... has as many nines as 0.999...
.
9.999... is not a set.
After the decimal point, there is a difference. In completed infinity.
You are
On 6/26/2024 3:15 AM, WM wrote:
Is the set of natural indices complete
such that
no natural number can be added?
You'd do better at being understood if
you said what distinguishes 'natural number'
from 'natural index'
The well.ordered inductive natural.number.indices
are complete
Nuance:
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.
If, by 'fixed', you mean that
each is not anything other than itself,
then yes.
Therefore
9.999... has one 9 less [one 9 fewer]
after the decimal point than 0.999... .
No.
"Move" the decimal point.
0.9⃒9999... and 09.9⃒9999...
0.99⃒999... and 09.99⃒999...
0.999⃒99... and 09.999⃒99...
0.9999⃒9... and 09.9999⃒9...
0.99999⃒... and 09.99999⃒...
...
There isn't one 9 fewer.
On 6/26/2024 12:15 AM, WM wrote:
9.999... has one 9 less after the decimal point than 0.999... .
NO! Think of: .(9) * 10 = 9.(9) = 10
10*0.999...999 < 9.999...999
9*0.999...999 < 9
0.999... < 1
There are still infinite nines... :^)
WM thinks an infiniteᵂᴹ number is very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹ
Le 26/06/2024 à 16:20, Jim Burns a écrit :
On 6/26/2024 3:15 AM, WM wrote:
Le 26/06/2024 à 00:11, Jim Burns a écrit :
On 6/25/2024 4:18 PM, WM wrote:
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1
tl;dr
No.
The well.ordered inductive natural.number.indices
are complete
Therefore the indices of the nines of 0.999...
are the complete set ℕ.
When they are shifted to 9.999..., none is added.
One is missing at the right of the decimal point.
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
No.
Therefore
9.999... has one 9 less [one 9 fewer]
after the decimal point than 0.999... .
No.
You believe in the magic of matheology.
Try to think.
"Countable" is an unsharp notion,
"Countable" is an unsharp notion,
too unsharp to measure the fact that
{1, 2, 3, ...} has one number less than
{0, 1, 2, 3, ...}
although this is obvious:
ℕ is a proper subsets of ℕ_0.
Is the set of natural indices complete
such that
no natural number can be added?
You'd do better at being understood if
you said what distinguishes 'natural number'
from 'natural index'
if you draw a distinction,
if being understood is something you want.
Why? There is nothing different.
The set of numbers and of indices is ℕ.
On 6/27/2024 8:03 AM, WM wrote:
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Try to think.
"Countable" is an unsharp notion,
Consider instead the notion '1.to.1 function'
Because ℕ is not larger than ℕ
the cardinality of ℕ cannot grow by 1
but
there are sets which are
ℕ with 1 element inserted or deleted.
Those sets have the same cardinality as ℕ
The cardinality didn't grow by 1
The set of numbers and of indices is ℕ.
WM explained :
If the set ℕ is actually complete, then the set ℕ_0 has one element more.
If by 'complete' you mean something entirely different from what it
means to most of the rest of the posters here.
The set of naturals is
'complete' in your sense because it is defined such that it contains
*all* of its elements. All sets are 'complete' in that sense.
WM submitted this idea :
Therefore ℕ_0 as a proper superset of ℕ
has one elements more than ℕ. Infinity does not make them equal.
Yes it does, cardinal arithmetic works differently in the two realms of finite and infinite.
on 6/27/2024, WM supposed :
The number of nines left of the radix grows, ....
And the number right of the radix point decreases.
No, it does not!
On 6/27/2024 5:09 AM, WM wrote:
Le 26/06/2024 à 20:46, "Chris M. Thomasson" a écrit :
On 6/26/2024 12:15 AM, WM wrote:
9.999... has one 9 less after the decimal point than 0.999... .
NO! Think of: .(9) * 10 = 9.(9) = 10
Wrong.
10*0.999...999 = 9.999...990
Where is that last zero coming from?
Le 27/06/2024 à 14:33, FromTheRafters a écrit :
WM submitted this idea :
Therefore ℕ_0 as a proper superset of ℕ has one elements more than ℕ. >>> Infinity does not make them equal.
Yes it does, cardinal arithmetic works differently in the two realms
of finite and infinite.
Cardinality is nonsense. Set differences are established by their
elements. ℕ_0 differs from ℕ by one element. Do you want to contradict? Don't make a fool of yourself!
Regards, WM
Le 27/06/2024 à 18:30, Jim Burns a écrit :
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
Le 26/06/2024 à 23:55, Jim Burns a écrit :What does „complete” mean?
WM thinks an infiniteᵂᴹ number is very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹNo, I assume that sets are complete. Therefore ℕ_0 as a proper superset
of ℕ has one elements more than ℕ. Infinity does not make them equal.
Le 27/06/2024 à 14:41, FromTheRafters a écrit :
on 6/27/2024, WM supposed :
The complete set of indices, when attached to the nines, can grow whenNo, it does not!And the number right of the radix point decreases.The number of nines left of the radix grows, ....
only a point of fly droppings is moved?
As was saidd, Cardinatality of infinte sets work differently than that
of finite sets.
On 6/27/2024 2:37 PM, WM wrote:
Le 27/06/2024 à 18:30, Jim Burns a écrit :
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
Sets can have elements inserted,
which makes them different sets.
The effect on size of inserting 1
is not the same for all sets.
The size of the set ℕ of all finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
cannot grow by 1
It is an
infiniteⁿᵒᵗᐧᵂᴹ.size.which.cannot.grow.by.1
The effect on size of inserting 1
is not the same for all sets.
Am Thu, 27 Jun 2024 12:15:30 +0000 schrieb WM:
Le 26/06/2024 à 23:55, Jim Burns a écrit :What does „complete” mean?
WM thinks an infiniteᵂᴹ number isNo, I assume that sets are complete. Therefore ℕ_0 as a proper superset
very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹ
of ℕ has one elements more than ℕ. Infinity does not make them equal.
With which numbers do you describe the sizes of N and N_0?
Am Thu, 27 Jun 2024 19:55:41 +0000 schrieb WM:
Le 27/06/2024 à 14:41, FromTheRafters a écrit :
on 6/27/2024, WM supposed :The complete set of indices, when attached to the nines, can grow when
No, it does not!And the number right of the radix point decreases.The number of nines left of the radix grows, ....
only a point of fly droppings is moved?
It does not grow. It stays the same.
There is always an infinite number of nines in Say:
9.(9) = 10
Le 27/06/2024 à 22:20, "Chris M. Thomasson" a écrit :No, they both have infinitely many 9s. It does not matter how they
There is always an infinite number of nines in Say:Is in 9.999... one 9 more than in 0.999... when 9.999... has been
9.(9) = 10
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
Le 28/06/2024 à 10:35, joes a écrit :It stays infinite. What does your notation mean? You cannot add digits „after” infinitely many; there is no after. You cannot count to or past infinity with naturals, as ω is not itself in N.
Am Thu, 27 Jun 2024 19:55:41 +0000 schrieb WM:
Therefore 10*0.999...999 = 9.999...990.The complete set of indices, when attached to the nines, can grow whenIt does not grow. It stays the same.
only a point of fly droppings is moved?
Le 28/06/2024 à 10:38, joes a écrit :Infinity does not have a predecessor like finite numbers.
Am Thu, 27 Jun 2024 12:15:30 +0000 schrieb WM:
Le 26/06/2024 à 23:55, Jim Burns a écrit :
WM thinks an infinite number isNo, I assume that sets are complete. Therefore ℕ_0 as a proper
very.large.but.finite
superset of ℕ has one elements more than ℕ. Infinity does not make
them equal.
Duh, the set of all natural numbers N contains all of them.What does „complete” mean?It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
It means that the subtraction of the complete set leaves {0, 1, 2, 3,There is not, since there are infinitely many of them.
..., ω} \ ℕ = {0, ω}.
It means that in {0, 1, 2, 3, ..., ω} before ω there is a natural
number.
Not their elements. I was asking for their number, how manyWith which numbers do you describe the sizes of N and N_0?Most of them are dark and cannot be used as individuals.
Le 28/06/2024 à 06:31, Jim Burns a écrit :So what is the size of N\{2}?
On 6/27/2024 2:37 PM, WM wrote:
Le 27/06/2024 à 18:30, Jim Burns a écrit :
Cardinalities which can grow by 1 are finite. Cardinalities whichCardinalities are useless. Sets can grow by 1 element.
cannot grow by 1 are infinite.
What are you replacing it with?Sets can have elements inserted, which makes them different sets.It is changing the infinite set but not its cardinality. Therefore cardinality is useless for my proof.
The effect on size of inserting 1 is not the same for all sets.
The size of N, containing all finite numbers, is itself infinite.The size of the set ℕ of all finite.sizes.which.can.grow.by.1Your discussion is off topic.
cannot grow by 1 It is an infinite.size.which.cannot.grow.by.1
You can’t remove from the right, since there is no end. You can onlyThe effect on size of inserting 1 is not the same for all sets.The effect of removing a nine from 0.999... is changing its value.
When in 0.999... the decimal point is shifted, the number of nines
remains constant. That has nothing to do with cardinalities.
Am Fri, 28 Jun 2024 13:55:34 +0000 schrieb WM:
Le 28/06/2024 à 10:38, joes a écrit :Infinity does not have a predecessor like finite numbers.
Am Thu, 27 Jun 2024 12:15:30 +0000 schrieb WM:
Le 26/06/2024 à 23:55, Jim Burns a écrit :
WM thinks an infinite number isNo, I assume that sets are complete. Therefore ℕ_0 as a proper
very.large.but.finite
superset of ℕ has one elements more than ℕ. Infinity does not make >>>> them equal.
Duh, the set of all natural numbers N contains all of them.What does „complete” mean?It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
It means that the subtraction of the complete set leaves {0, 1, 2, 3,There is not, since there are infinitely many of them.
..., ω} \ ℕ = {0, ω}.
It means that in {0, 1, 2, 3, ..., ω} before ω there is a natural
number.
Not their elements. I was asking for their number, how manyWith which numbers do you describe the sizes of N and N_0?Most of them are dark and cannot be used as individuals.
of them there are.
Am Fri, 28 Jun 2024 13:59:02 +0000 schrieb WM:
Le 27/06/2024 à 22:20, "Chris M. Thomasson" a écrit :No, they both have infinitely many 9s. It does not matter how they
There is always an infinite number of nines in Say:Is in 9.999... one 9 more than in 0.999... when 9.999... has been
9.(9) = 10
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
were „produced”.
They are the same number 10.
Am Fri, 28 Jun 2024 13:51:48 +0000 schrieb WM:
Le 28/06/2024 à 10:35, joes a écrit :
Am Thu, 27 Jun 2024 19:55:41 +0000 schrieb WM:
It stays infinite. What does your notation mean?Therefore 10*0.999...999 = 9.999...990.The complete set of indices, when attached to the nines, can grow when >>>> only a point of fly droppings is moved?It does not grow. It stays the same.
You cannot add digits
„after” infinitely many; there is no after.
You cannot count to or past
infinity with naturals,
as ω is not itself in N.
Am Fri, 28 Jun 2024 13:50:06 +0000 schrieb WM:
Le 28/06/2024 à 06:31, Jim Burns a écrit :So what is the size of N\{2}?
On 6/27/2024 2:37 PM, WM wrote:
Le 27/06/2024 à 18:30, Jim Burns a écrit :
Cardinalities which can grow by 1 are finite. Cardinalities whichCardinalities are useless. Sets can grow by 1 element.
cannot grow by 1 are infinite.
What are you replacing it with?Sets can have elements inserted, which makes them different sets.It is changing the infinite set but not its cardinality. Therefore
The effect on size of inserting 1 is not the same for all sets.
cardinality is useless for my proof.
The size of N, containing all finite numbers, is itself infinite.The size of the set ℕ of all finite.sizes.which.can.grow.by.1Your discussion is off topic.
cannot grow by 1 It is an infinite.size.which.cannot.grow.by.1
The effect on size of inserting 1 is not the same for all sets.The effect of removing a nine from 0.999... is changing its value.
When in 0.999... the decimal point is shifted, the number of nines
remains constant. That has nothing to do with cardinalities.
You can’t remove from the right, since there is no end.
You can only
remove from the left (dividing by 10)
joes was thinking very hard :
Am Fri, 28 Jun 2024 13:55:34 +0000 schrieb WM:
Duh, the set of all natural numbers N contains all of them.What does „complete” mean?It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Not their elements. I was asking for their number, how manyWith which numbers do you describe the sizes of N and N_0?Most of them are dark and cannot be used as individuals.
of them there are.
There's a number of them, however, how many there are is not a number.
Le 28/06/2024 à 06:31, Jim Burns a écrit :
On 6/27/2024 2:37 PM, WM wrote:
Le 27/06/2024 à 18:30, Jim Burns a écrit :
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
Sets can have elements inserted,
which makes them different sets.
The effect on size of inserting 1
is not the same for all sets.
It is changing the infinite set
but not its cardinality.
Therefore
cardinality is useless for my proof.
WM pretended :
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not changed >> by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different numbers of >> nines is stupid.
The number of nines is not a number.
On 6/28/2024 6:41 AM, WM wrote:
Le 28/06/2024 à 02:03, Richard Damon a écrit :
As was saidd, Cardinatality of infinte sets work differently than that
of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
Huh? Sure it can. What is:
.(428571)
? It is a non-terminating digit sequence in base ten.
I thought .999... = 1, ....
Ross Finlayson pretended :
You know, or it "goes" to.
Yes, but the number of nines in the sequence after the radix point is countably infinite rather than finite, hence NaN in this context.
On 6/28/2024 9:50 AM, WM wrote:
It is changing the infinite set
but not its cardinality.
A set with a cardinal.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
cannot change without its cardinality changing.
Therefore
cardinality is useless for my proof.
For any difficulty which cardinality presents,
not.saying 'cardinal' not.resolves the difficulty.
On 06/28/2024 02:01 PM, FromTheRafters wrote:
The number of nines is not a number.
I thought .999... = 1, ....
You know, or it "goes" to.
This recalls the "First of Zen Koans" bit again,
Two Buddhist priests observe a flag in the wind.
The first says, "the wind, moves, the flag".
The second says, "ah, that flag, moves, in the wind."
A third says "it is your mind that moves".
However,
we can reason about Avogadroᴬᵛᵒᵍᵃᵈʳᵒ 9s
by finite not.first.false claim.sequence
without going to them.
It's the same for infinitely.many 9s in that
we can't go to them, but
we can reason about them.
But
'infinite' is different from 'humongous' and
different conclusions get concluded.
Le 28/06/2024 à 23:01, FromTheRafters a écrit :Which it hasn’t.
WM pretended :
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of >>> elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not
changed by more than 0.
They are the same.Or this is not the case.
But assuming completeness and simultaneously claiming different
numbers of nines is stupid.
So what? Every index is increased by 1, blah blah PA.The number of nines is not a number.The set of natural indices is a set.
Shifting it to the left shifts every index and every indexed nine to the left.
Le 29/06/2024 à 15:24, Jim Burns a écrit :Then how do you measure infinite sets?
On 6/28/2024 9:50 AM, WM wrote:
Cardinality is irrelevant.It is changing the infinite set but not its cardinality.A set with a cardinal.growable.by.1 (finite)
cannot change without its cardinality changing.
Working as designed. How do you distinguish them?Cardinality does not present a difficulty. It is simply unable toTherefore cardinality is useless for my proof.For any difficulty which cardinality presents,
not saying 'cardinal' not.resolves the difficulty.
distinguish |ℕ| and |ℕ_0|.
All nines of 0.999... are from the sequence 0.9, 0.09, 0.009, ... NoneThat’s one way to construct it. Of course none of the infinite series individually reaches the limit (otherwise the following ones :-P would go
of the ℵo nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1. ℵo nines fail to make 0.999... = 1.
Le 28/06/2024 à 19:16, joes a écrit :Why is that the same as ℕ\{1}? They are not supersets of each other.
Am Fri, 28 Jun 2024 13:50:06 +0000 schrieb WM:It is |ℕ| - 1. That was easy.
Le 28/06/2024 à 06:31, Jim Burns a écrit :So what is the size of N\{2}?
On 6/27/2024 2:37 PM, WM wrote:
Le 27/06/2024 à 18:30, Jim Burns a écrit :
Cardinalities which can grow by 1 are finite. Cardinalities whichCardinalities are useless. Sets can grow by 1 element.
cannot grow by 1 are infinite.
Which you define how? Starting maybe with |ℕ|. What numbers do you use?Number of elements.What are you replacing it with?Sets can have elements inserted, which makes them different sets.It is changing the infinite set but not its cardinality. Therefore
The effect on size of inserting 1 is not the same for all sets.
cardinality is useless for my proof.
Which is as a concrete number…? [not OT?]Yes, it is |ℕ|.The size of N, containing all finite numbers, is itself infinite.The size of the set ℕ of all finite sizes which can grow by 1 cannot >>>> grow by 1. It is an infinite size which cannot grow by 1.
Only when replacing with another digit.The effect of removing a nine from 0.999... is changing its value.
Yes it does; that is how WE define the number of nines.When in 0.999... the decimal point is shifted, the number of nines
remains constant. That has nothing to do with cardinalities.
Or equivalently, add 9.You can’t remove from the right, since there is no end.If all are a complete set, then we can shift all by one position to the
left.
Where do the superfluous digits go: 0.(9)9 ?You can only remove from the left (dividing by 10)Removing from the right is done by multiplying by 10.
Le 28/06/2024 à 19:23, joes a écrit :It does.
Am Fri, 28 Jun 2024 13:55:34 +0000 schrieb WM:
Le 28/06/2024 à 10:38, joes a écrit :
Am Thu, 27 Jun 2024 12:15:30 +0000 schrieb WM:
Le 26/06/2024 à 23:55, Jim Burns a écrit :
WM thinks an infinite number is very.large.but.finiteNo, I assume that sets are complete. Therefore ℕ_0 as a proper
superset of ℕ has one elements more than ℕ. Infinity does not make >>>>> them equal.
What is immediately before ω?This is misleadingly notated, implying a predecessor. You mean ω u ℕ,
Infinity does not have a predecessor like finite numbers.
Duh, the set of all natural numbers N contains all of them.What does „complete” mean?It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
You are confusing the infinitely long, linearly scaled natural numberLinearity excludes more than one at a position. Immediately before ωIt means that the subtraction of the complete set leaves {0, 1, 2, 3,There is not, since there are infinitely many of them.
..., ω} \ ℕ = {0, ω}.
It means that in {0, 1, 2, 3, ..., ω} before ω there is a natural
number.
there is at most one natural number.
And that is…? You gave only a relation.|N| + 1 = |N_0|Not their elements. I was asking for their number, how many of themWith which numbers do you describe the sizes of N and N_0?Most of them are dark and cannot be used as individuals.
there are.
Le 29/06/2024 à 00:23, "Chris M. Thomasson" a écrit :Which number does it describe?
On 6/28/2024 6:41 AM, WM wrote:That is a formula determining an infinite digit sequence
Le 28/06/2024 à 02:03, Richard Damon a écrit :
As was saidd, Cardinatality of infinte sets work differently than
that of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
Huh? Sure it can. What is: .(428571)?
Then all real numbers are formulas.It is a non-terminating digit sequence in base ten.No. Like "0.111..." it is a formula.
Formulas determine sequences. The other way round is not possible.
Le 29/06/2024 à 15:24, Jim Burns a écrit :
On 6/28/2024 9:50 AM, WM wrote:
It is changing the infinite set
but not its cardinality.
A set with a cardinal.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
cannot change without its cardinality changing.
Cardinality is irrelevant.
Therefore cardinality is useless for my proof.
For any difficulty which cardinality presents,
not.saying 'cardinal' not.resolves the difficulty.
Cardinality does not present a difficulty.
It is simply unable to distinguish |ℕ| and |ℕ_0|.
All nines of 0.999... are from
the sequence 0.9, 0.09, 0.009, ...
None of the ℵo nines makes
its partial sum 0,9, 0.99, 0.999, ... equal to 1.
ℵo nines fail to make 0.999... = 1.
Le 29/06/2024 à 20:01, Jim Burns a écrit :
However,
we can reason about Avogadroᴬᵛᵒᵍᵃᵈʳᵒ 9s
by finite not.first.false claim.sequence
without going to them.
We can reason about ℵo nines,
all missing the limit.
We can reason about ℵo nines,
all missing the limit.
It's the same for infinitely.many 9s in that
we can't go to them, but
we can reason about them.
But most matheologians don't understand that
the sequence 0.9, 0.09, 0.009, ... contains
ℵo nines without containing the limit 0.
But
'infinite' is different from 'humongous' and
different conclusions get concluded.
If you think straight,
then only one conclusion follows:
0.999... < 1.
And Bob does not disappear.
On 6/29/2024 2:14 PM, WM wrote:
If you think straight,
then only one conclusion follows:
0.999... < 1.
The values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
0.999... ≠ 0.9 < 1
0.999... ≠ 0.99 < 1
0.999... ≠ 0.999 < 1
0.999... ≠ 0.9999 < 1
0.999... ≠ 0.99999 < 1
...
0.999... = 1
On 6/29/2024 10:01 AM, WM wrote:
Le 29/06/2024 à 00:23, "Chris M. Thomasson" a écrit :
On 6/28/2024 6:41 AM, WM wrote:That is a formula determining an infinite digit sequence
Le 28/06/2024 à 02:03, Richard Damon a écrit :
As was saidd, Cardinatality of infinte sets work differently than
that of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
Huh? Sure it can. What is:
.(428571) >
Try 3/7 represented in base 10 decimal notation ;^)
? It is a non-terminating digit sequence in base ten.
No. Like "0.111..." it is a formula.
Formulas determine sequences. The other way round is not possible.
Here is a simple recursive formula that creates infinite 9's during iteration:
|ℕ| = |ℕ∪{Bob]|
exists f: ℕ∪{Bob} → ℕ : bijection
Bob ∉ f(ℕ∪{Bob})
And Bob disappears.
Yes it does, the sequence of partial sums approaches/converges to one.
This *complete* ordered field of reals guarantees cauchy sequence convergence.
See how real numbers are defined.
WM pretended :
The set of natural indices is a set.
Shifting it to the left shifts every index and every indexed nine to the
left.
That still doesn't affect the fact that aleph_null is not a real
number.
The values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
Am Sat, 29 Jun 2024 17:18:26 +0000 schrieb WM:
All nines of 0.999... are from the sequence 0.9, 0.09, 0.009, ... NoneThat’s one way to construct it. Of course none of the infinite series individually reaches the limit (otherwise the following ones :-P would go above it?). Together they do. It is reached after exactly Aleph0 steps.
of the ℵo nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1.
ℵo nines fail to make 0.999... = 1.
Am Sat, 29 Jun 2024 17:01:10 +0000 schrieb WM:
Like "0.111..." it is a formula.
Formulas determine sequences. The other way round is not possible.
Then all real numbers are formulas.
Am Fri, 28 Jun 2024 18:25:57 +0000 schrieb WM:
It does.ℕ_0 as a proper
superset of ℕ has one elements more than ℕ. Infinity does not make >>>>>> them equal.
Of course none of the infinite series
individually reaches the limit (otherwise the following ones :-P would go above it?). Together they do.
On 06/30/2024 02:55 AM, FromTheRafters wrote:
This *complete* ordered field of reals guarantees
cauchy sequence convergence.
See how real numbers are defined.
It's axiomatic,
and about the usual open topology.
There are others, ....
Le 29/06/2024 à 22:46, joes a écrit :Their cardinality, of course.
Am Fri, 28 Jun 2024 18:25:57 +0000 schrieb WM:
It does.ℕ_0 as a proper superset of ℕ has one elements more than ℕ. >>>>>>> Infinity does not make them equal.
Your world is not mine and is not maths.
Le 29/06/2024 à 22:25, joes a écrit :Which is to say, after all of the steps.
Am Sat, 29 Jun 2024 17:18:26 +0000 schrieb WM:
All nines of 0.999... are from the sequence 0.9, 0.09, 0.009, ... NoneThat’s one way to construct it. Of course none of the infinite series
of the ℵo nines makes its partial sum 0,9, 0.99, 0.999, ... equal to
1. ℵo nines fail to make 0.999... = 1.
individually reaches the limit (otherwise the following ones :-P would
go above it?). Together they do. It is reached after exactly Aleph0
steps.
Are there ℵo partial sums 0.9, 0.99, 0.999, ... ?That is the limit, yes.
If yes they can be combined to 0.999... .
Is the limit one of them?
Le 30/06/2024 à 00:48, Jim Burns a écrit :The limit is the Aleph-zeroth, if you will.
The values of infinite.length decimals are assigned by a different
method from how the values of finite.length decimals are assigned.
Are there ℵo partial sums 0.9, 0.99, 0.999, ... ?
It the limit one of them?
Le 30/06/2024 à 12:29, Jim Burns a écrit :
On 6/29/2024 2:14 PM, WM wrote:
If you think straight,
then only one conclusion follows:
0.999... < 1.
The values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
0.999... ≠ 0.9 < 1
0.999... ≠ 0.99 < 1
0.999... ≠ 0.999 < 1
0.999... ≠ 0.9999 < 1
0.999... ≠ 0.99999 < 1
...
If you use only definable length,
then always ℵo terms are missing.
All finite indices guarantee finite length.
0.999... = 0.999... < 1
0.999... = 1
That is wrong.
If you insert parentheses, nothing changes
..((((0,9)9)9)9)... = 0,999...
0.9 + 0.09 + 0.009 + ...
contains ℵo terms with ℵo nines,
all together smaller than 1.
Le 30/06/2024 à 12:02, FromTheRafters a écrit :Which it doesn’t. There is no end to it.
WM pretended :
But the set of indices is invariable. It cannot even expand by oneThe set of natural indices is a set.That still doesn't affect the fact that aleph_null is not a real
Shifting it to the left shifts every index and every indexed nine to
the left.
number.
index.
Le 30/06/2024 à 12:29, Jim Burns a écrit :Did you mean: /finite/ length ?
The values of infinite length decimals are assigned by a different
method from how the values of finite length decimals are assigned.
0.999... ≠ 0.99999 < 1 ...
If you use only definable length, then always ℵo terms are missing.
All finite indices guarantee finite length.
0.9 + 0.09 + 0.009 + ... contains ℵo terms with ℵo nines, all together smaller than 1.We are not talking about the terms, but their limit, which is all of them /taken together/.
Le 29/06/2024 à 23:18, joes a écrit :
Am Sat, 29 Jun 2024 17:01:10 +0000 schrieb WM:
Like "0.111..." it is a formula.
Formulas determine sequences.
The other way round is not possible.
Then all real numbers are formulas.
Yes, you got it!
Let the infinite sequence 0.999...
Le 26/06/2024 à 23:55, Jim Burns a écrit :
WM thinks an infiniteᵂᴹ number is
very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹ
No,
I assume that sets are complete.
Therefore ℕ_0 as a proper superset of ℕ has
one elements more than ℕ.
Infinity does not make them equal.
On 06/30/2024 08:44 AM, Jim Burns wrote:
On 6/30/2024 10:48 AM, Ross Finlayson wrote:
On 06/30/2024 02:55 AM, FromTheRafters wrote:
This *complete* ordered field of reals guarantees
cauchy sequence convergence.
See how real numbers are defined.
It's axiomatic,
and about the usual open topology.
There are others, ....
...but not in that discussion.
Yes,
we CAN discuss things other than the real numbers.
However,
if we ARE discussing the real numbers,
then we AREN'T doing that.
Define the Dedekind.complete real numbers.
Prove the Intermediate Value Theorem.
Counter.propose(?) the rational numbers,
for which the Intermediate Value Theorem is false.
So what?
The rational numbers aren't the real numbers.
We can still apply the Intermediate Value Theorem
to the real numbers,
which is all anyone has claimed.
Well, iota-values are defined and
satisfy making for the IVT
which results the FTC's,
Fundamental Theorems of Calculus.
For the past several years
a Mikhail Katz has been working on
rehabilitating infinitesimals,
and it reminds
me of a story where an educator surveyed an introductory
class whether .999... was the same, or different, than 1.0,
and at least according to their thought processes,
it was about 50/50.
On 06/30/2024 06:38 PM, Jim Burns wrote:
On 6/30/2024 5:05 PM, Ross Finlayson wrote:
Well, iota-values are defined and
satisfy making for the IVT
which results the FTC's,
Fundamental Theorems of Calculus.
If I use the usual definitions for
the limit of a sequence of sets
for your iota.values,
they do not satisfy the Intermediate Value Theorem.
I understand your iota.values to be the limit
n/d: 0≤n≤d: d → ∞
For n/d: 0≤n≤d I read {0/d,1/d,...,d/d}
For lim[d → ∞] I read ⋂[0<dᵢ<∞] ⋃[dᵢ<d<∞]
Is that what you mean? You (RF) don't say.
⋂[0<dᵢ<∞] ⋃[dᵢ<d<∞] {0/d,1/d,...,d/d}
does not satisfy the Intermediate Value Theorem.
Yes it does, the iota-values result that they do
make for the IVT,
Am Sun, 30 Jun 2024 15:07:41 +0000 schrieb WM:
Le 30/06/2024 à 00:48, Jim Burns a écrit :The limit is the Aleph-zeroth, if you will.
The values of infinite.length decimals are assigned by a different
method from how the values of finite.length decimals are assigned.
Are there ℵo partial sums 0.9, 0.99, 0.999, ... ?
It the limit one of them?
1 is near almost.all (all.but.finitely.many) of
0.9 0.99 0.999 0.9999 0.99999 ...
for any sense > 0 of 'near'
For that reason,
we assign 1 to 0.999...
The values of infinite.length decimals
are assigned by a method different from how
the values of finite.length decimals are assigned.
..((((0,9)9)9)9)... = 0,999...
0.9 + 0.09 + 0.009 + ...
contains ℵo terms with ℵo nines,
all together smaller than 1.
β is the least.upper.bound of the terms
0.9 0.99 0.999 0.9999 0.99999 ...
least.upper.bound β is not less than 1
| Assume otherwise.
Am Sun, 30 Jun 2024 14:51:42 +0000 schrieb WM:
Le 30/06/2024 à 12:29, Jim Burns a écrit :
Did you mean: /finite/ length ?The values of infinite length decimals are assigned by a different
method from how the values of finite length decimals are assigned.
0.999... ≠ 0.99999 < 1 ...
If you use only definable length, then always ℵo terms are missing.
All finite indices guarantee finite length.
0.(9) is not finitely long though, so there are no „missing” terms
0.9 + 0.09 + 0.009 + ... contains ℵo terms with ℵo nines, all together >> smaller than 1.
We are not talking about the terms, but their limit,
which is all of them
/taken together/.
On 6/30/2024 11:15 AM, WM wrote:
Le 29/06/2024 à 23:18, joes a écrit :
Am Sat, 29 Jun 2024 17:01:10 +0000 schrieb WM:
Like "0.111..." it is a formula.
Formulas determine sequences.
The other way round is not possible.
Then all real numbers are formulas.
Yes, you got it!
Then there are more formulas than there are formulas.
So, there is a point there.
So, there is a formula there.
But there isn't an indexed.formula there.
So, all the indexed.formulas
(finite strings, finite alphabet)
aren't all the formulas.
Somehow.
On 6/27/2024 8:15 AM, WM wrote:
I assume that sets are complete.
I (JB) think that what you (WM) call 'logic' is that
each nonempty set has a trichotomous order such that
each nonempty subset holds a first and a last,
whether its first or last is visibleᵂᴹ or darkᵂᴹ
Is that what you're saying?
Am 25.06.2024 um 22:18 schrieb WM:
Let the infinite sequence 0.999...
0.999... ist keine "infinite sequence", sondern eine Zahl,
Le 30/06/2024 à 20:45, Moebius a écrit :
Am 25.06.2024 um 22:18 schrieb WM:
Let the infinite sequence 0.999...
0.999... ist keine "infinite sequence", sondern eine Zahl,
Wrong.
0.999... is <dumschwatz>
On 7/1/2024 9:15 AM, WM wrote:
Le 30/06/2024 à 20:45, Moebius a écrit :.(9) is just <etc.>
Am 25.06.2024 um 22:18 schrieb WM:
Let the infinite sequence 0.999...
0.999... ist keine "infinite sequence", sondern eine Zahl,
Wrong. 0.999... is an abbreviation of ...(((0.9)9)9)... = 0.9, 0.99,
0.999, ... converging to but never reaching 1.
Le 30/06/2024 à 22:59, Jim Burns a écrit :
On 6/27/2024 8:15 AM, WM wrote:
I assume that sets are complete.
I (JB) think that what you (WM) call 'logic' is that
each nonempty set has a trichotomous order such that
each nonempty subset holds a first and a last,
whether its first or last is visibleᵂᴹ or darkᵂᴹ
Is that what you're saying?
Yes, but
only under the pr4condition of completed infinity.
https://en.wikipedia.org/wiki/Finite_set
| Necessary and sufficient conditions for finiteness
| (Paul Stäckel)
| S can be given a total ordering which is
| well-ordered both forwards and backwards.
| That is, every non-empty subset of S has
| both a least and a greatest element in the subset.
What [ordinal number] is immediately before ω?
What [ordinal number] is immediately before ω?
Am 28.06.2024 um 20:25 schrieb WM:
What [ordinal number] is immediately before ω?
Keine, Mückenheim.
WIE OFT SOLL MAN DIR DAS NOCH SAGEN? Du scheinst geistig-mental sogar
noch schlechter aufgestellt zu sein, als Biden.
What [ordinal number] is immediately before ω?
Siehe oben.
{1, 2, 3, ...} has one number less than {0, 1, 2, 3, ...} although this <bla>
Le 29/06/2024 à 03:05, Ross Finlayson a écrit :
I thought .999... = 1, ....That is wrong.
Am 01.07.2024 um 18:15 schrieb WM:
Le 30/06/2024 à 20:45, Moebius a écrit :
Am 25.06.2024 um 22:18 schrieb WM:
Let the infinite sequence 0.999...
0.999... ist keine "infinite sequence", sondern eine Zahl,
Wrong.
Nein, nicht wrong, Du hirnloser Depp.
0.999... e IR, daher ist 0,999... eine reelle Zahl. Es gilt sogar
0,999... = 1 (siehe https://simple.wikipedia.org/wiki/0.999...),
0.999... ist also identisch mit der reellen Zahl 1.
0.999... is <dumschwatz>
Cardinality is nonsense.
Le 30/06/2024 à 11:55, FromTheRafters a écrit :
See how real numbers are defined.
I have taught it for 30 years.
The set of natural indices is a set.
On 06/30/2024 09:22 PM, Jim Burns wrote:
On 6/30/2024 11:32 PM, Ross Finlayson wrote:
On 06/30/2024 06:38 PM, Jim Burns wrote:
On 6/30/2024 5:05 PM, Ross Finlayson wrote:
Well, iota-values are defined and
satisfy making for the IVT
which results the FTC's,
Fundamental Theorems of Calculus.
If I use the usual definitions for
the limit of a sequence of sets
for your iota.values,
they do not satisfy the Intermediate Value Theorem.
I understand your iota.values to be the limit
n/d: 0≤n≤d: d → ∞
For n/d: 0≤n≤d I read {0/d,1/d,...,d/d}
For lim[d → ∞] I read ⋂[0<dᵢ<∞] ⋃[dᵢ<d<∞]
Is that what you mean? You (RF) don't say.
⋂[0<dᵢ<∞] ⋃[dᵢ<d<∞] {0/d,1/d,...,d/d}
does not satisfy the Intermediate Value Theorem.
Yes it does, the iota-values result that they do
make for the IVT,
Tell me what you are talking about.
I understand your iota.values to be the limit
n/d: 0≤n≤d: d → ∞
For n/d: 0≤n≤d I read {0/d,1/d,...,d/d}
For lim[d → ∞] I read ⋂[0<dᵢ<∞] ⋃[dᵢ<d<∞]
Is that what you mean? You (RF) don't say.
⋂[0<dᵢ<∞] ⋃[dᵢ<d<∞] {0/d,1/d,...,d/d}
does not satisfy the Intermediate Value Theorem.
There must be satisfied "extent density completeness
measure" to satisfy the IVT, though one may aver that
"extent density completeness" would suffice.
So, iota-values or
ran(EF) of the natural/unit equivalency function,
or sweep, has "extent density completeness measure",
thus the IVT follows.
It's sort of irrelevant what I intend
as I don't see value in nominalist fictionalism,
what it is is what it is, what it is.
It's not really any of the initial approximations,
this limit, this infinite limit, this continuum limit.
It's an _infinite_ limit.
On 7/1/2024 3:38 PM, Moebius wrote:
Am 01.07.2024 um 20:32 schrieb Jim Burns:
https://en.wikipedia.org/wiki/Finite_set
| Necessary and sufficient conditions for finiteness
| (Paul Stäckel)
| S can be given a total ordering which is
| well-ordered both forwards and backwards.
| That is, every non-empty subset of S has
| both a least and a greatest element in the subset.
Makes much sense.
Thank you for bringing this to our attention (again).
https://en.wikipedia.org/wiki/Paul_St%C3%A4ckel
https://de.wikipedia.org/wiki/Paul_St%C3%A4ckel
"In the area of prime number theory, he used
the term /twin prime/
(in its German form, "Primzahlzwilling")
for the first time." :-)
"Paul Gustav Samuel Stäckel
(20 August 1862, Berlin – 12 December 1919, Heidelberg)"
Perhaps WM has not yet caught up
to this late.breaking development.
Give him another century.
Am 01.07.2024 um 20:32 schrieb Jim Burns:
https://en.wikipedia.org/wiki/Finite_set
| Necessary and sufficient conditions for finiteness
| (Paul Stäckel)
| S can be given a total ordering which is
| well-ordered both forwards and backwards.
| That is, every non-empty subset of S has
| both a least and a greatest element in the subset.
Makes much sense.
Thank you for bringing this to our attention (again).
https://en.wikipedia.org/wiki/Paul_St%C3%A4ckel https://de.wikipedia.org/wiki/Paul_St%C3%A4ckel
"In the area of prime number theory, he used
the term /twin prime/
(in its German form, "Primzahlzwilling")
for the first time." :-)
Am 02.07.2024 um 03:22 schrieb Jim Burns:
On 7/1/2024 3:38 PM, Moebius wrote:
Am 01.07.2024 um 20:32 schrieb Jim Burns:
https://en.wikipedia.org/wiki/Finite_set
| Necessary and sufficient conditions for finiteness
| (Paul Stäckel)
| S can be given a total ordering which is
| well-ordered both forwards and backwards.
| That is, every non-empty subset of S has
| both a least and a greatest element in the subset.
Makes much sense.
Thank you for bringing this to our attention (again).
https://en.wikipedia.org/wiki/Paul_St%C3%A4ckel
https://de.wikipedia.org/wiki/Paul_St%C3%A4ckel
"In the area of prime number theory, he used
the term /twin prime/
(in its German form, "Primzahlzwilling")
for the first time." :-)
"Paul Gustav Samuel Stäckel
(20 August 1862, Berlin – 12 December 1919, Heidelberg)"
Perhaps WM has not yet caught up
to this late.breaking development.
Give him another century.
Remember that WM is a physicist.
His knowledge of math is rather superficial.
On 07/01/2024 06:10 PM, Jim Burns wrote:
What that means is that
I think
theory is a strong mathematical platonism,
it matters what is _attained_ to,
or that to which we _attain_,
the "true" objects of
a universe of mathematical objects,
"a" universe, then with regards to descriptions
there's that
the "extent density completeness measure" provide
"extent density completeness"
which you would agree that
"extent density completeness" makes for
satisfying the IVT.
Then, as I mentioned,
there's a theory,
in all the universe of theories,
all the abstract and contingent and fanciful and
practical and otherwise,
one of which is "the true theory",
that among those,
there's one where it appears that
"it is so" is an axiom.
So, given that
you won't accept that via inspection,
that a least-upper-bound is given and
also a sigma algebra is given,
given that extent and density are givens,
then,
given that it's axiomatic,
and, doesn't contradict the ordinary
because
it just makes for the "only-diagonal" contra
the "anti-diagonal",
then, how's that.
Good sir, ....
On 06/30/2024 09:22 PM, Jim Burns wrote:
[...]
It's not really any of the initial approximations,
this limit, this infinite limit, this continuum limit.
It's an _infinite_ limit.
On 7/1/2024 7:42 PM, Ross Finlayson wrote:
On 06/30/2024 09:22 PM, Jim Burns wrote:
[...]
It's not really any of the initial approximations,
this limit, this infinite limit, this continuum limit.
It's an _infinite_ limit.
Again, despite the name,
the continuum limit and the continuum are different.
In the continuum limit,
the least.upper.bound of neighbor.distances is 0
In the continuum,
in each nonempty split F ᴬ<ᴬ H
foresplit F holds a last or hindsplit H holds a first.
The rationals are an example of
the continuum limit which isn't the continuum.
A non-terminating digit sequence does not determine a real number.
A non-terminating digit sequence does not determine a real number.
in {0, 1, 2, 3, ..., ω} before ω there is a natural number.
Le 28/06/2024 à 02:03, Richard Damon a écrit :
[...]
A non-terminating digit sequence
does not determine a real number.
Therefore,
A non.terminating decimal determines
exactly one real number (line.point).
But it is impossible to <bla bla bla>
On 6/28/2024 9:41 AM, WM wrote:
Le 28/06/2024 à 02:03, Richard Damon a écrit :
[...]
A non-terminating digit sequence
does not determine a real number.
A non.terminating decimal determines
no less than one real number (line.point) and
no more than one real number (line.point).
Am 28.06.2024 um 15:55 schrieb WM:
in {0, 1, 2, 3, ..., ω} before ω there is a natural number.
Ja, "before ω" (im Sinne von kleiner) sind unendlich viele natural
numbers, nämlich die natürlichen Zahlen 0, 1, 2, 3, ... usw.
Es gibt aber keine natürliche Zahl die "unmittelbar vor" ω steht.
Zu jeder
natürlichen Zahl n gibt es eine natürliche Zahl m (beispielsweise n+1),
so dass n < m < ω gilt.
Am 28.06.2024 um 15:41 schrieb WM:
A non-terminating digit sequence does not determine a real number.
Natürlich tut sie das
Die Folge (d_1, d_2, d_3, ...) mit d_i e {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
für alle i e IN "determiniert" die reelle Zahl SUM_(n=1..oo) d_n/10^n.
Le 03/07/2024 à 19:09, Moebius a écrit :
Am 28.06.2024 um 15:55 schrieb WM:
in {0, 1, 2, 3, ..., ω} before ω there is a natural number.
Ja, "before ω" (im Sinne von kleiner) sind unendlich viele natural
numbers, nämlich die natürlichen Zahlen 0, 1, 2, 3, ... usw.
Es gibt aber keine natürliche Zahl die "unmittelbar vor" ω steht.
You <bla>
Am 03.07.2024 um 22:07 schrieb WM:
Le 03/07/2024 à 19:01, Moebius a écrit :
Am 28.06.2024 um 15:41 schrieb WM:Gib mal eine konkret an.
A non-terminating digit sequence does not determine a real number.
Natürlich tut sie das
Die Folge (d_1, d_2, d_3, ...) mit d_i e {0, 1, 2, 3, 4, 5, 6, 7, 8,
9} für alle i e IN "determiniert" die reelle Zahl SUM_(n=1..oo) d_n/10^n. >>>
Da wäre z. B. die Folge (d_n)_(n e IN), die durch d_i = 3 für alle i e
IN definiert ist.
Diese Folge "determiniert" die reelle Zahl
SUM_(n=1..oo) d_n/10^n = 1/3.
Le 03/07/2024 à 19:01, Moebius a écrit :
Am 28.06.2024 um 15:41 schrieb WM:
A non-terminating digit sequence does not determine a real number.
Natürlich tut sie das
Die Folge (d_1, d_2, d_3, ...) mit d_i e {0, 1, 2, 3, 4, 5, 6, 7, 8,Gib mal eine konkret an.
9} für alle i e IN "determiniert" die reelle Zahl SUM_(n=1..oo) d_n/10^n. >>
Le 03/07/2024 à 22:46, Moebius a écrit :
Am 03.07.2024 um 22:07 schrieb WM:
Le 03/07/2024 à 19:01, Moebius a écrit :
Am 28.06.2024 um 15:41 schrieb WM:Gib mal eine konkret an.
A non-terminating digit sequence does not determine a real number. >>>>Natürlich tut sie das
Die Folge (d_1, d_2, d_3, ...) mit d_i e {0, 1, 2, 3, 4, 5, 6, 7, 8,
9} für alle i e IN "determiniert" die reelle Zahl SUM_(n=1..oo)
d_n/10^n.
Da wäre z. B. die Folge (d_n)_(n e IN), die durch d_i = 3 für alle i e
IN definiert ist.
That is
Am 03.07.2024 um 23:10 schrieb WM:
Le 03/07/2024 à 22:46, Moebius a écrit :
Am 03.07.2024 um 22:07 schrieb WM:
Le 03/07/2024 à 19:01, Moebius a écrit :
Am 28.06.2024 um 15:41 schrieb WM:Gib mal eine konkret an.
A non-terminating digit sequence does not determine a real number. >>>>>Natürlich tut sie das
Die Folge (d_1, d_2, d_3, ...) mit d_i e {0, 1, 2, 3, 4, 5, 6, 7,
8, 9} für alle i e IN "determiniert" die reelle Zahl SUM_(n=1..oo)
d_n/10^n.
Da wäre z. B. die Folge (d_n)_(n e IN), die durch d_i = 3 für alle i
e IN definiert ist.
That is
Mückenheim, Du bist für jede Art von Mathematik zu dumm und zu blöde. Deine Psychose und (neuerdings) Demenz machen es praktisch UNMÖGLICH,
mit Dir über irgend ein mathematisches Thema zu sprechen.
Und tschüss.
∀x > 0: NUF(x) = ℵo is true.
between every x [in IR] and 0 there [...] lie ℵo unit
fractions and ℵo finite distances between them.
Le 03/07/2024 à 20:24, Jim Burns a écrit :
On 6/28/2024 9:41 AM, WM wrote:
A non-terminating digit sequence
does not determine a real number.
A non.terminating decimal determines
no less than one real number (line.point) and
no more than one real number (line.point).
But it is impossible to
construct or define or find or recognize or communicate
a non-terminating decimal without using a finite formula,
and be it as simple as "0.111...".
because it cannot be handled with eniugh precision.
Always almost all digits are dark.
Always almost all digits are dark.
Le 03/07/2024 à 19:09, Moebius a écrit :There is no such thing as „visibility”.
Am 28.06.2024 um 15:55 schrieb WM:
in {0, 1, 2, 3, ..., ω} before ω there is a natural number.
Ja, "before ω" (im Sinne von kleiner) sind unendlich viele natural
numbers, nämlich die natürlichen Zahlen 0, 1, 2, 3, ... usw.
Es gibt aber keine natürliche Zahl die "unmittelbar vor" ω steht.
You cannot recognize it. It is existing but dark.
Zu jeder natürlichen Zahl n gibt es eine natürliche Zahl mFor every visible natural number this is true.
(beispielsweise n+1),
so dass n < m < ω gilt.
Like ∀x > 0: NUF(x) = ℵo is true. But this truth proves the existence of dark numbers, because between every such x and 0 there must lie ℵo unit fractions and ℵo finite distances between them.Why? That distance can be divided infinitely.
Wir definieren eine "non-terminating digit sequence" the following way:
Wir betrachten die Folge (d_n)_(n e IN), die durch d_i = 3 für alle i e
IN definiert ist.
Diese Folge "determiniert" die reelle Zahl SUM_(n=1..oo) d_n/10^n = 1/3.
Le 03/07/2024 à 23:58, Moebius a écrit :
Am 03.07.2024 um 22:05 schrieb WM:
∀x > 0: NUF(x) = ℵo is true.
Right, if our domain of discourse is IR (i.e. if x ranges over the
elements in IR).
This means:
between every x [in IR] and 0 there [...] lie ℵo unit fractions and
ℵo finite distances between them.
Exactly!
Why do you adhere to the [...] idea that every x > 0 is larger than [the sum of] ℵo finite distances?
Hinweis: Für jedes x e IR mit x > 0 gibt es (abzählbar) unendlich<bla>
viele Stammbrüche s, so dass s < x ist.
Am 03.07.2024 um 22:05 schrieb WM:
∀x > 0: NUF(x) = ℵo is true.
Right, if our domain of discourse is IR (i.e. if x ranges over the
elements in IR).
This means:
between every x [in IR] and 0 there [...] lie ℵo unit
fractions and ℵo finite distances between them.
Exactly!
Hinweis: Für jedes x e IR mit x > 0 gibt es (abzählbar) unendlich viele Stammbrüche s, so dass s < x ist.
On 7/3/2024 4:00 PM, WM wrote:
But it is impossible to
construct or define or find or recognize or communicate
a non-terminating decimal without using a finite formula,
and be it as simple as "0.111...".
Answer the questions
"What is a non.terminating.decimalⁿᵒᵗᐧᵂᴹ?" and
"What is a line.pointⁿᵒᵗᐧᵂᴹ?"
and one can prove from very reasonable assumptions
that non.terminating.decimalsⁿᵒᵗᐧᵂᴹ _exist_
and line.pointsⁿᵒᵗᐧᵂᴹ _exist_
and
a non.terminating.decimalⁿᵒᵗᐧᵂᴹ determines
no less than one real number (line.pointⁿᵒᵗᐧᵂᴹ) and
no more than one real number (line.pointⁿᵒᵗᐧᵂᴹ).
If it is impossible to constructᵂᴹ or defineᵂᴹ or findᵂᴹ or
recognizeᵂᴹ or communicateᵂᴹ a non.terminating.decimalⁿᵒᵗᐧᵂᴹ
because it cannot be handled with enough precision.
Then it _exists_ and is unable to be
handled.with.enough.precisionᵂᴹ.
Always almost all digits are dark.
All.nonempty.digit.setsⁿᵒᵗᐧᵂᴹ hold a first member. All.digitsⁿᵒᵗᐧᵂᴹ have a first.after.digit
Am 04.07.2024 um 15:19 schrieb WM:
Le 03/07/2024 à 23:58, Moebius a écrit :
Why do you adhere to the [...] idea that every x > 0 is larger than [the sum of]
ℵo finite distances?
Because it is true?
Am Wed, 03 Jul 2024 20:05:48 +0000 schrieb WM:
There is no such thing as „visibility”.so dass n < m < ω gilt.For every visible natural number this is true.
Like ∀x > 0: NUF(x) = ℵo is true. But this truth proves the existence of >> dark numbers, because between every such x and 0 there must lie ℵo unitWhy? That distance can be divided infinitely.
fractions and ℵo finite distances between them.
Am Wed, 03 Jul 2024 20:05:48 +0000 schrieb WM:
Le 03/07/2024 à 19:09, Moebius a écrit :
Am 28.06.2024 um 15:55 schrieb WM:
in {0, 1, 2, 3, ..., ω} before ω there is a natural number.
Ja, "before ω" (im Sinne von kleiner) sind unendlich viele natural
numbers, nämlich die natürlichen Zahlen 0, 1, 2, 3, ... usw.
Es gibt aber keine natürliche Zahl die "unmittelbar vor" ω [ist].
It is existing but <bla>
Zu jeder natürlichen Zahl n gibt es eine natürliche Zahl m
(beispielsweise n+1), so dass n < m < ω gilt.
For every <bla bla bla>
There is no such thing as „visibility”.
On 25/06/2024 21:18, WM wrote:
A more interesting question; suppose a set containing an infinite number
of 9's.
Now copy that set and add a 2.
Is the second set bigger than the first set?
Let the infinite sequence 0.999... be multiplied by 10. Does the number
of nines grow?
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
Regards, WM
Le 04/07/2024 à 13:00, joes a écrit :
There is no such thing as „visibility”.
There is invisibility
For all visible/choosable n: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Le 04/07/2024 à 15:49, Moebius a écrit :
Am 04.07.2024 um 15:19 schrieb WM:
Le 03/07/2024 à 23:58, Moebius a écrit :
Why do you adhere to the [...] idea that every x > 0 is larger than
[the sum of] ℵo finite distances?
Because it is true?
<bla bla bla> Useless to discuss with you further.
Am 04.07.2024 um 15:59 schrieb WM:
Le 04/07/2024 à 15:49, Moebius a écrit :
Am 04.07.2024 um 15:19 schrieb WM:
Le 03/07/2024 à 23:58, Moebius a écrit :
Why do you adhere to the [...] idea that every x > 0 is larger than
[the sum of] ℵo finite distances?
Because it is true?
<bla bla bla> Useless to discuss with you further.
Da gibt's auch nicht zu "diskutieren", Mückenheim.
Hinweis: Für jedes x e IR mit x > 0 gibt es (abzählbar) unendlich viele Stammbrüche s, so dass s < x ist. Die Summe der (abzählbar) unendlich vielen Abstände zwischen diesen Stammbrüchen ist dann natürlich < x.
Am 04.07.2024 um 16:07 schrieb WM:
For all visible/choosable n: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Mal davon abgesehen, dass niemand weiß, was Deine "visible/choosable n"
sein sollen (im Gegensatz zu den anderen natürlichen Zahlen), gilt:
Am 04.07.2024 um 16:30 schrieb Moebius:
Da gibt's auch nicht zu "diskutieren",
Le 04/07/2024 à 16:45, Moebius a écrit :
Am 04.07.2024 um 16:07 schrieb WM:
For all visible/choosable n: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Mal davon abgesehen, dass niemand weiß, was Deine "visible/choosable
n" sein sollen (im Gegensatz zu den anderen natürlichen Zahlen), gilt:
Es bedeutet, dass man alle Zahlen von 1 bis n angeben kann.
Auf diese endlich vielen folgen noch unendlich viele, von denen man
unendlich viele nicht angeben kann.
Le 04/07/2024 à 16:45, Moebius a écrit :n" sein sollen (im Gegensatz zu den anderen natürlichen Zahlen), gilt:
Am 04.07.2024 um 16:07 schrieb WM:
For all visible/choosable n: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Mal davon abgesehen, dass niemand weiß, was Deine "visible/choosable
Es bedeutet, dass man alle Zahlen von 1 bis n angeben kann.unendlich viele nicht angeben kann.
Auf diese endlich vielen folgen noch unendlich viele, von denen man
die Verteilung der ersten Stammbrüche
Am 04.07.2024 um 17:09 schrieb WM:
Le 04/07/2024 à 16:45, Moebius a écrit :
Am 04.07.2024 um 16:07 schrieb WM:
n" sein sollen (im Gegensatz zu den anderen natürlichen Zahlen), gilt:For all visible/choosable n: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Mal davon abgesehen, dass niemand weiß, was Deine "visible/choosable
Es bedeutet, dass man alle Zahlen von 1 bis n angeben kann.unendlich viele nicht angeben kann.
Auf diese endlich vielen folgen noch unendlich viele, von denen man
Aus dem nämlichen Grund hat man sog. Variablen eingeführt
Diese erlauben uns auch Aussage über alle Element einer sehr großen endlichen oder unendlichen Menge zu tätigen.
Le 04/07/2024 à 17:22, Moebius a écrit :
Am 04.07.2024 um 17:09 schrieb WM:
Le 04/07/2024 à 16:45, Moebius a écrit :"visible/choosable n" sein sollen (im Gegensatz zu den anderen
Am 04.07.2024 um 16:07 schrieb WM:;
For all visible/choosable n: |ℕ \ {1, 2, 3, ..., n}| = ℵo.;
Mal davon abgesehen, dass niemand weiß, was Deine
natürlichen Zahlen), gilt:
;unendlich viele nicht angeben kann.
Es bedeutet, dass man alle Zahlen von 1 bis n angeben kann.
Auf diese endlich vielen folgen noch unendlich viele, von denen man
Aus dem nämlichen Grund hat man sog. Variablen eingeführt
Ja, natürlich. n ist gerade so eine. Damit kann man beweisen, dass fast
alle natürlichen Zahlen nicht visible sind. Nur endlich viele sind
visible, unendlich viele sind es nicht.
Diese erlauben uns auch Aussage über alle Element einer sehr großen
endlichen oder unendlichen Menge zu tätigen.
Auf diese Weise beweist man: Fast alle natürlichen Zahlen sind dunkel.
Le 03/07/2024 à 23:58, Moebius a écrit :That depends on the size of those distances.
Am 03.07.2024 um 22:05 schrieb WM:
∀x > 0: NUF(x) = ℵo is true.Right, if our domain of discourse is IR (i.e. if x ranges over the
elements in IR). This means:
Why do you adhere to the silly idea that every x > 0 is larger than ℵo finite distances?between every x [in IR] and 0 there [...] lie ℵo unit fractions and ℵo >>> finite distances between them.Exactly!
Note that you are wrong. Not every x > 0 is larger than ℵo finite distances.
Which ones aren’t?Hinweis: Für jedes x e IR mit x > 0 gibt es (abzählbar) unendlich vieleOnly for x larger than ℵo finite distances.
Stammbrüche s, so dass s < x ist.
Le 04/07/2024 à 05:20, Jim Burns a écrit :
On 7/3/2024 4:00 PM, WM wrote:
Always almost all digits are dark.
All.nonempty.digit.setsⁿᵒᵗᐧᵂᴹ hold a first member.
All.digitsⁿᵒᵗᐧᵂᴹ have a first.after.digit
Most are unknown.
All.nonempty.digit.setsⁿᵒᵗᐧᵂᴹ hold a first member.
All.digitsⁿᵒᵗᐧᵂᴹ have a first.after.digit and a last.before.digit,
except the first.digit, which has only a first.after.digit.
Most are unknown.
But it is impossible to
construct or define or find or recognize or communicate
a non-terminating decimal without using a finite formula,
and be it as simple as "0.111...".
Answer the questions
"What is a non.terminating.decimalⁿᵒᵗᐧᵂᴹ?" and
"What is a line.pointⁿᵒᵗᐧᵂᴹ?"
and one can prove from very reasonable assumptions
that non.terminating.decimalsⁿᵒᵗᐧᵂᴹ _exist_
and line.pointsⁿᵒᵗᐧᵂᴹ _exist_
and
a non.terminating.decimalⁿᵒᵗᐧᵂᴹ determines
no less than one real number (line.pointⁿᵒᵗᐧᵂᴹ) and
no more than one real number (line.pointⁿᵒᵗᐧᵂᴹ).
Yes, the line point is the limit of an infinite sequence.
Example: 0.999... --> 1.
But the sequence cannot be given other than
by its defining formula like "0.999...".
because it cannot be handled with enough precision.
Then it _exists_ and is unable to be
handled.with.enough.precisionᵂᴹ.
It is handled by a formula,
sometimes even with absolute precision like
0.999... --> 1.
It is handled by a formula,
sometimes even with absolute precision like
0.999... --> 1.
Peter Fairbrother formulated on Thursday :
On 25/06/2024 21:18, WM wrote:
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
Regards, WM
A more interesting question; suppose a set containing an infinite
number of 9's. Now copy that set and add a 2.
Is the second set bigger than the first set?
Peter Fairbrother
Lrf, pneqvanyvgl bar naq pneqvanyvgl gjb.
I was just thinking that infinite is infinite,
there is an infinite number of natural numbers,
there are an infinite amount of [real] numbers between say, .0000001 and .00000001
On 7/4/2024 1:01 PM, FromTheRafters wrote:
Providing one infinite set and the other
are both countable or both uncountable.
If one set is size Aleph_zero
and the other is 2^Aleph_zero
then they are not the same size.
I was just thinking that infinite is infinite,
Am Thu, 04 Jul 2024 13:19:35 +0000 schrieb WM:
Not every x > 0 is larger than ℵo finiteThat depends on the size of those distances.
distances.
Which ones aren’t?Hinweis: Für jedes x e IR mit x > 0 gibt es (abzählbar) unendlich viele >>> Stammbrüche s, so dass s < x ist.Only for x larger than ℵo finite distances.
Am 04.07.2024 um 17:16 schrieb WM:
die Verteilung der ersten Stammbrüche
es gibt keine "ersten Stammbrüche". Zu _jedem_ Stammbruch
gibt es (abzählbar) unendlich viele kleinere Stammbrüche.
Am 04.07.2024 um 22:23 schrieb Chris M. Thomasson:
there are an infinite amount of [real] numbers between say, .0000001 and
.00000001
_Uncountably_ infinitely many.
On 7/4/2024 9:49 AM, WM wrote:
Yes, the line point is the limit of an infinite sequence.
Example: 0.999... --> 1.
But the sequence cannot be given other than
by its defining formula like "0.999...".
Givingᵂᴹ the sequence isn't needed in order for us
to know there is only one line.point.
Chris M. Thomasson laid this down on his screen :
On 7/4/2024 9:58 AM, FromTheRafters wrote:
Peter Fairbrother formulated on Thursday :
On 25/06/2024 21:18, WM wrote:
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
Regards, WM
A more interesting question; suppose a set containing an infinite
number of 9's. Now copy that set and add a 2.
Is the second set bigger than the first set?
Peter Fairbrother
Lrf, pneqvanyvgl bar naq pneqvanyvgl gjb.
Different set with an infinite number of elements. Their sizes are the
same for infinity = infinity... :^)
Providing one infinite set and the other are both countable or both uncountable. If one set is size Aleph_zero and the other is 2^Aleph_zero
then they are not the same size.
Le 04/07/2024 à 18:39, joes a écrit :It is infinite for every x>0 and zero at 0. There is no first point.
Am Thu, 04 Jul 2024 13:19:35 +0000 schrieb WM:Wrong, and easily proved wrong: NUF(x), the number of unit fractions
Not every x > 0 is larger than ℵo finite distances.That depends on the size of those distances.
between 0 and x increases from 0 to more. It cannot grow by more than 1
at any real point x because at any x only one unit fraction can exist.
Which is?Any x within the first distance.Which ones aren’t?Hinweis: Für jedes x e IR mit x > 0 gibt es (abzählbar) unendlichOnly for x larger than ℵo finite distances.
viele Stammbrüche s, so dass s < x ist.
Note that without a first distance ℵo distances cannot exist.Why not? You’re counting from the wrong end.
After serious thinking Peter Fairbrother wrote :[..]
On 04/07/2024 21:01, FromTheRafters wrote:
Chris M. Thomasson laid this down on his screen :
Providing one infinite set and the other are both countable or both
uncountable. If one set is size Aleph_zero and the other is
2^Aleph_zero then they are not the same size.
I imagined the 9's as being the digits of 0.99... so
distinguishable and countably infinite.
There is nothing wrong with a sequence having duplicate members. Sets
(ZFC) don't have duplicates though.
A difference between the idea of 'same' and 'equal size'. 'Same' if each
can be a subset/superset of the other (matching elements) and 'equal
size' in terms of cardinality (pairing elements).
Does saying that the reals are infinitely denser than the naturals make
any sense?
As in there are more reals than naturals even if they both
are infinite.
On 7/4/2024 1:30 PM, Moebius wrote:
Am 04.07.2024 um 22:23 schrieb Chris M. Thomasson:
I was just thinking that infinite is infinite,
No, it isn't.
For some reason I like to think of the density of infinity. The Natural numbers are not dense at all when compared to the reals...
there is an infinite number of natural numbers,
Right, _countably_ infinitely many.
there are an infinite amount of [real] numbers between say, .0000001
and .00000001
_Uncountably_ infinitely many.
Okay. I get a little confused by that sometimes. Trying to count the
reals is not possible because of all those infinite infinities that are embedded in them...
However The naturals have no infinities between say,
1 and 2. Make any sense to you?
Le 04/07/2024 à 17:25, Moebius a écrit :
Am 04.07.2024 um 17:16 schrieb WM:
die Verteilung der ersten Stammbrüche
es gibt keine "ersten Stammbrüche". Zu _jedem_ Stammbruch gibt es
(abzählbar) unendlich viele kleinere Stammbrüche.
Wrong,
Does saying that the reals are infinitely denser than the naturalsmake any sense?
As in there are more reals than naturals even if they both are infinite.
Le 04/07/2024 à 22:30, Moebius a écrit :Nope. Und jetzt halt endlich mal die Fresse, Du Depp!
Am 04.07.2024 um 22:23 schrieb Chris M. Thomasson:
Wrongthere are an infinite amount of real numbers between say, .0000001
and .00000001
_Uncountably_ infinitely many.
Peter Fairbrother was thinking very hard :
On 05/07/2024 10:59, FromTheRafters wrote:
After serious thinking Peter Fairbrother wrote :[..]
On 04/07/2024 21:01, FromTheRafters wrote:
Chris M. Thomasson laid this down on his screen :
Providing one infinite set and the other are both countable or both
uncountable. If one set is size Aleph_zero and the other is
2^Aleph_zero then they are not the same size.
I imagined the 9's as being the digits of 0.99... so distinguishable
and countably infinite.
There is nothing wrong with a sequence having duplicate members. Sets
(ZFC) don't have duplicates though.
Oh dear. Let's forget about the nines (which are not duplicates, the
nth 9 is different from the (n+1)th 9).
One set is all the natural numbers, the second is that plus an orange.
OK? Both countable infinities, no duplicates.
Yes, this is exactly the infinite cardinal arithmetic WM doesn't grasp.
Or rather, say, he objects to its validity.
[...]
A difference between the idea of 'same' and 'equal size'. 'Same' if
each can be a subset/superset of the other (matching elements) and
'equal size' in terms of cardinality (pairing elements).
Maybe. but I'm not concerned about "same" here. only size.
In that case you need not have any concern about what the elements are,
only cardinality.
And the elements are pairable (if an orange can be paired to a
number), but one set has an orange in it and the other doesn't.
That doesn't matter, it's another element even if it's a fish.
So, pairing (and cardinality) don't really work for sizing these sets;
in everyday terms, the second is bigger.
Pairing does work,
On 05/07/2024 15:33, FromTheRafters wrote:
Peter Fairbrother was thinking very hard :
One set is all the natural numbers, the second is that plus an
orange. OK? Both countable infinities, no duplicates.
Yes, this is exactly the infinite cardinal arithmetic WM doesn't
grasp. Or rather, say, he objects to its validity.
So, pairing (and cardinality) don't really work for sizing these
sets;
in everyday terms, the second is bigger.
Pairing does work,
Eh? I don't follow that. AFAICT every element in one set can be paired
with an element of the other, so if that is a definition of being the
same size, they are the same size.
But one set contains an orange more than the other one does,*)
Le 04/07/2024 à 20:52, Jim Burns a écrit :
On 7/4/2024 9:49 AM, WM wrote:
Yes, the line point is the limit of an infinite sequence.
Example: 0.999... --> 1.
But the sequence cannot be given other than
by its defining formula like "0.999...".
Givingᵂᴹ the sequence isn't needed in order for us
to know there is only one line.point.
Who doubted that?
But it is impossible to
construct or define or find or recognize or communicate
a non-terminating decimal without using a finite formula,
and be it as simple as "0.111...".
Yes, the line point is the limit of an infinite sequence.
Example: 0.999... --> 1.
But the sequence cannot be given
other than by its defining formula like "0.999...".
It is not impossible to construct every desired digit
of the sequence.
But it is impossible to do so without a finite formula.
And it is impossible to desire all digits.
ℵo are always missing.
But uncountability could only be established by
completed infinite sequences, not by finite formulas.
But uncountability could only be established by
completed infinite sequences, not by finite formulas.
On 05/07/2024 10:59, FromTheRafters wrote:
A difference between the idea of 'same' and 'equal size'.
'Same' if each can be a subset/superset of the other (matching elements)
and 'equal size' in terms of cardinality (pairing elements).
Maybe. but I'm not concerned about "same" here. only size.
And the elements are pairable (if an orange can be paired to a number),
but one set has an orange in it and the other doesn't.
So, pairing (and cardinality) don't really work for sizing these sets;
in everyday terms, the second is bigger.
That would require complete infinite sequences which cannot be given because the not given part is always larger, namely infinite.
The other way around.
It's set.inclusion which stops working as a guide to size.
Am 05.07.2024 um 20:08 schrieb Jim Burns:
The other way around.
It's set.inclusion which stops working as a guide to size.
Right. How would we be able to compare, say, the sets {1, 2, 3, ...} and
{-1, -2, -3, ...} concerning "size" by relying on "set inclusion"? Or,
say, {1, 2, 3, ...} and {1.5, 2.5, 3.5, ...} etc.
Or even {1, 2, orange} and {1, 2, 3}.
The other way around.
It's set.inclusion which stops working as a guide to size.
On 07/05/2024 07:12 PM, Moebius wrote:? smaller than the superset,
[...]
I'm reminded many years ago,
when studying size relations in sets,
that one rule that arrived was that
a proper subset, had a size relation,
and was told that
it was not so,
while, still it was written how
it was so.
Then, Fred Katz pointed me to
his Ph.D. from M.I.T. and OUTPACING,
showing that it was a formal result that
it was so.
So,
the "conclusion", seems to be, "not a conclusion",
for all the "considerations",
their conclusions, together.
Then another one was asymptotic density and
the size relation of sets not just being ordered
but also having a rational value,
this was the "half of the integers are even".
It involves a bit of book-keeping, yet,
it is possible to keep these various notions,
while still there's cardinality sort of in the middle,
where of course on the other side of
these refinements of the notion of
the relation of size in infinite sets of
numbers their spaces their elements then
there's an entire absolute of "ubiquitous ordinals",
that have the infinite sets as of a "size".
So, when you mean cardinal, say cardinal.
There are other notions of
"size", and "measure", and, "number".
Le 04/07/2024 à 17:08, Moebius a écrit :Du zählst vom falschen Ende. Es gibt keine Stammbrüche, die kleiner als
Am 04.07.2024 um 16:30 schrieb Moebius:
Doch, die Frage, wie Du Dir die Verteilung der ersten Stammbrüche, also derjenigen, die kleiner als jeder, den Du angeben kannst, sind,Da gibt's auch nicht zu "diskutieren",
vorstellst, bleibt.
Le 04/07/2024 à 17:08, Moebius a écrit :
Am 04.07.2024 um 16:30 schrieb Moebius:
Da gibt's auch nicht zu "diskutieren",
Doch, die Frage, wie Du Dir
die Verteilung der ersten Stammbrüche,
also derjenigen, die kleiner als jeder,
den Du angeben kannst, sind, vorstellst, bleibt.
(Off limits for native English speakers.)
On 7/4/2024 11:16 AM, WM wrote:
Le 04/07/2024 à 17:08, Moebius a écrit :
Am 04.07.2024 um 16:30 schrieb Moebius:
Da gibt's auch nicht zu "diskutieren",
Doch, die Frage, wie Du Dir
die Verteilung der ersten Stammbrüche,
also derjenigen, die kleiner als jeder,
den Du angeben kannst, sind, vorstellst, bleibt.
(Off limits for native English speakers.)
⎛ ∀j ∈ ℕ₁:
⎜ ∃u ∈ ⅟ℕ: u = ⅟j ∧
⎜ ¬∃v ∈ ⅟ℕ: ⅟j = v ≠ u
⎜
⎜ ∀u ∈ ⅟ℕ:
⎜ ∃j ∈ ℕ₁: j = ⅟u ∧
⎜ ¬∃k ∈ ℕ₁: ⅟u = k ≠ j
⎜
⎜ ∀j,k ∈ ℕ₁:
⎜ ∀u,v ∈ ⅟ℕ: u = ⅟j ∧ v = ⅟k ⟹
⎝ j < k ⇔ u > v
⎛ ∀A ⊆ ℕ₁: A ≠ {} ⟹
⎜ ∃j ∈ A: ∀k ∈ A: j ≤ k
⎜
⎜ ∀B ⊆ ⅟ℕ: B ≠ {} ⟹
⎝ ∃v ∈ B ∀u ∈ B: v ≥ u
⎛ ∀j ∈ ℕ₁: ℕ₁\{1} ∋ j⁺¹ > j
⎜
⎝ ∀u ∈ ⅟ℕ: ⅟ℕ\{1} ∋ ⅟(⅟u)⁺¹ < u
⎛ ∀k ∈ ℕ₁\{1}: ℕ₁ ∋ k⁻¹ < k
⎜
⎝ ∀v ∈ ⅟ℕ\{1}: ⅟ℕ ∋ ⅟(⅟v)⁻¹ > v
On 7/4/2024 11:16 AM, WM wrote:The question is only this: Are there more than one unit fractions in the
Le 04/07/2024 à 17:08, Moebius a écrit :
Am 04.07.2024 um 16:30 schrieb Moebius:
Da gibt's auch nicht zu "diskutieren",
Doch, die Frage, wie Du Dir
die Verteilung der ersten Stammbrüche,
also derjenigen, die kleiner als jeder,
den Du angeben kannst, sind, vorstellst, bleibt.
(Off limits for native English speakers.)
⎛ ∀j ∈ ℕ₁:
⎜ ∃u ∈ ⅟ℕ: u = ⅟j ∧
⎜ ¬∃v ∈ ⅟ℕ: ⅟j = v ≠ u
⎜
Am Thu, 04 Jul 2024 15:16:23 +0000 schrieb WM:
Le 04/07/2024 à 17:08, Moebius a écrit :Du zählst vom falschen Ende.
Am 04.07.2024 um 16:30 schrieb Moebius:Doch, die Frage, wie Du Dir die Verteilung der ersten Stammbrüche, also
Da gibt's auch nicht zu "diskutieren",
derjenigen, die kleiner als jeder, den Du angeben kannst, sind,
vorstellst, bleibt.
Es gibt keine Stammbrüche, die kleiner als
alle anderen sind;
On 7/5/2024 3:53 AM, WM wrote:
But uncountability could only be established by
completed infinite sequences, not by finite formulas.
Uncountability has been established for {y:N₁→10:¬∃0…}
It has been, historically.
It has been, here, for you (WM).
It could be established again, for you,
Am 05.07.2024 um 09:39 schrieb WM:
Le 04/07/2024 à 17:25, Moebius a écrit :
Am 04.07.2024 um 17:16 schrieb WM:
die Verteilung der ersten Stammbrüche
es gibt keine "ersten Stammbrüche". Zu _jedem_ Stammbruch gibt es
(abzählbar) unendlich viele kleinere Stammbrüche.
Wrong,
Nope.
On 7/7/2024 1:19 PM, WM wrote:
Le 05/07/2024 à 19:09, Jim Burns a écrit :
On 7/5/2024 3:53 AM, WM wrote:
But uncountability could only be established by completed infinite
sequences, not by finite formulas.
Uncountability has been established for {y:N₁→10:¬∃0…}
It has been, historically.
It has been, here, for you (WM).
It could be established again, for you,
It is impossible to completely write an infinite digit sequence.
An infinite digit sequence cannot be determined by writing all its terms.
An infinite digit sequence can only be determined by a finite expression.
There are only countably many finite expression.
What generated this?
.(121232)
?
On 7/5/2024 6:26 AM, Moebius wrote:
Still, the rational numbers are countable! (Not enough "infinite
infinities embedded in them"!)
[...] How many embedded infinite infinities are
"needed" _before_ it can be deemed uncountable?
Say between 0 and 1.
There seems to be an infinite number of rationals that can fill in the
"gap", so to speak.
0 + (1/8 + 1/8 + 1/4 + 1/2) = 1
0 + (1/16 + 1/16 + 1/8 + 1/4 + 1/2) = 1
0 + (1/16 + 1/16 + 1/8 + 2/3 + 1/3 - 1/4) = 1
These are all rational, right?
:^) Are the complex numbers more "dense" than the reals?
For every real we can use it wrt a non-zero y-axis, or imaginary axis if
you will. Fair enough?
Le 07/07/2024 à 19:39, Jim Burns a écrit :
On 7/4/2024 11:16 AM, WM wrote:
Le 04/07/2024 à 17:08, Moebius a écrit :
Am 04.07.2024 um 16:30 schrieb Moebius:
Da gibt's auch nicht zu "diskutieren",
Doch, die Frage, wie Du Dir
die Verteilung der ersten Stammbrüche,
also derjenigen, die kleiner als jeder,
den Du angeben kannst, sind, vorstellst, bleibt.
(Off limits for native English speakers.)
⎛ ∀j ∈ ℕ₁:
⎜ ∃u ∈ ⅟ℕ: u = ⅟j ∧
⎜ ¬∃v ∈ ⅟ℕ: ⅟j = v ≠ u
⎜
The question is only this:
Are there more than one unit fractions in
the point where NUF(x) changes from 0 to more,
or is there only one unit fraction?
The question is only this:
Are there more than one unit fractions in
the point where NUF(x) changes from 0 to more,
or is there only one unit fraction?
Am 07.07.2024 um 22:04 schrieb WM:
Are there more than one unit fractions in the point where NUF(x)
changes from 0 to more, or
Mückenheim, Ihr dummes Gerede tut allein schon beim Lesen weh.
Grundsätzlich kann sich eine Funktion in einem Punkt nicht ändern, Depp.
f(x_0) hat (falls x_0 im Definitionsbereich der Funktion f liegt) einen bestimmten Wert und damit hat es sich. Da ändert sich nichts.
In Bezug auf NUF ist es so, dass NUF(0) = 0 ist und NUF(x) = aleph_0 für alle x e IR, x > 0.
Es gibt also keinen spezifische Punkt wo NUF "sich von 0 auf aleph_0 ändert", weil es solche Punkte gar nicht gibt.
Du bist einfach für jede Art von Mathematik zu doof und zu blöde.
Are there more than one unit fractions in the
point where NUF(x) changes from 0 to more, or
On 7/7/2024 4:04 PM, WM wrote:
The question is only this:
Are there more than one unit fractions in
the point where NUF(x) changes from 0 to more,
or is there only one unit fraction?
The complete distribution of unit fractions,
NUF(0) = 0
NUF(x) changes at 0
There is no unit fraction at 0
The question is only this:
Are there more than one unit fractions in
the point where NUF(x) changes from 0 to more,
or is there only one unit fraction?
The third choice: none.
On 7/7/2024 1:27 PM, WM wrote:
There are only countably many finite expression.
What generated this?
.(121232)
?
The finite formula ".(121232)" generated this.
Did you know that a rational generated it?
(4) Es gilt also IN DER MENGENLEHRE für alle x e IR, x > 0: |{m e {1/n :
n e IN} : m <= x}| = aleph_0.
(5) Mit der Definition NUF(x) := |{m e {1/n : n e IN} : m <= x}| folgt
also für alle x e IR, x > 0: NUF(x) = aleph_0.
Am 07.07.2024 um 22:04 schrieb WM:
Grundsätzlich kann sich eine Funktion in einem Punkt nicht ändern
In Bezug auf NUF ist es so, dass NUF(0) = 0 ist und NUF(x) = aleph_0 für alle x e IR, x > 0.
Es gibt also keinen spezifische Punkt wo NUF "sich von 0 auf aleph_0 ändert", weil es solche Punkte gar nicht gibt.
Le 08/07/2024 à 04:52, Moebius a écrit :
(4) Es gilt also IN DER MENGENLEHRE für alle x e IR, x > 0: |{m e {1/n
: n e IN} : m <= x}| = aleph_0.
(5) Mit der Definition NUF(x) := |{m e {1/n : n e IN} : m <= x}| folgt
also für alle x e IR, x > 0: NUF(x) = aleph_0.
ℵo unit fractions occupy at least ℵo different points on the positive real axis.
Hence NUF(x) = ℵo cannot be true for x less than ℵo points, that is for
a set of ℵo positive x.
Le 08/07/2024 à 04:35, Moebius a écrit :
In Bezug auf NUF ist es so, dass NUF(0) = 0 ist und NUF(x) = aleph_0
für alle x e IR, x > 0.
That is
Es gibt also keinen spezifische Punkt, wo NUF "sich von 0 auf aleph_0
ändert", weil es solche Punkte gar nicht gibt.
Le 08/07/2024 à 04:35, Moebius a écrit :WTF does that even mean. Since when is this a question about measure?
Am 07.07.2024 um 22:04 schrieb WM:
In Bezug auf NUF ist es so, dass NUF(0) = 0 ist und NUF(x) = aleph_0That is nonsense. NUF(x) cannot be ℵo unless x consists of ℵo points.
für alle x e IR, x > 0.
Am 08.07.2024 um 16:12 schrieb WM:
Le 08/07/2024 à 04:52, Moebius a écrit :
(4) Es gilt also IN DER MENGENLEHRE für alle x e IR, x > 0: |{m e
{1/n : n e IN} : m <= x}| = aleph_0.
(5) Mit der Definition NUF(x) := |{m e {1/n : n e IN} : m <= x}|
folgt also für alle x e IR, x > 0: NUF(x) = aleph_0.
ℵo unit fractions occupy at least ℵo different points on the positive
real axis.
Nicht nur "at least", sondern "genau", Mückenheim.
Hence NUF(x) = ℵo cannot be true for x less than ℵo points, that is
for a set of ℵo positive x.
Huh?! Du redest wirres Zeug, Mann. :-)
NUF(x) = ℵo ist für alle x e IR, x > 0 wahr. Also für überabzählbar viele Punkte.
Wir brauchen NUF aber gar nicht auf der Definitionsmenge IR zu
definieren; auch Q ginge als Definitionsmenge:
NUF(x) := |{m e {1/n : n e IN} : m <= x}| (x e Q).
Auch in diesem Fall gilt: NUF(0) = 0 und für alle x e IR, x > 0: NUF(x)
= aleph_0.
Das ist alles ziemlich trivial.
Am Mon, 08 Jul 2024 13:58:06 +0000 schrieb WM:
Le 08/07/2024 à 04:35, Moebius a écrit :
Am 07.07.2024 um 22:04 schrieb WM:
In Bezug auf NUF ist es so, dass NUF(0) = 0 ist und NUF(x) = aleph_0
für alle x e IR, x > 0.
That is nonsense. NUF(x) cannot be ℵo unless x consists of ℵo points.
WTF does that even mean. Since when is this a question about measure?
Le 08/07/2024 à 04:35, Moebius a écrit :für alle x e IR, x > 0.
In Bezug auf NUF ist es so, dass NUF(0) = 0 ist und NUF(x) = aleph_0
That is
aleph_0 ändert", weil es solche Punkte gar nicht gibt.Es gibt also keinen spezifische Punkt, wo NUF "sich von 0 auf
Le 08/07/2024 à 01:57, Jim Burns a écrit :
On 7/7/2024 4:04 PM, WM wrote:
The question is only this:
Are there more than one unit fractions in
the point where NUF(x) changes from 0 to more,
or is there only one unit fraction?
The complete distribution of unit fractions,
does not answer this question.
NUF(0) = 0
NUF(x) changes at 0
There is no unit fraction at 0
Therefore it does not change at 0.
Am Mon, 08 Jul 2024 13:58:06 +0000 schrieb WM:
Le 08/07/2024 à 04:35, Moebius a écrit :
Am 07.07.2024 um 22:04 schrieb WM:
WTF does that even mean. Since when is this a question about measure?In Bezug auf NUF ist es so, dass NUF(0) = 0 ist und NUF(x) = aleph_0That is nonsense. NUF(x) cannot be ℵo unless x consists of ℵo points.
für alle x e IR, x > 0.
On 7/8/2024 9:49 AM, WM wrote:
the answer to your question is the third option: none.
There is no point x > 0: NUF(x) < ℵ₀
There is no point x < 0: NUF(x) > 0
NUF(x) changes "at" 0.
Therefore it does not change at 0.
Functions do not change at single points.
A change needs be _with respect to_ something,
Consider
Am 08.07.2024 um 16:12 schrieb WM:
Le 08/07/2024 à 04:52, Moebius a écrit :
(4) Es gilt also IN DER MENGENLEHRE für alle x e IR, x > 0: |{m e {1/n
: n e IN} : m <= x}| = aleph_0.
(5) Mit der Definition NUF(x) := |{m e {1/n : n e IN} : m <= x}| folgt
also für alle x e IR, x > 0: NUF(x) = aleph_0.
ℵo unit fractions occupy at least ℵo different points on the positive
real axis.
Nicht nur "at least", sondern "genau"
Hence NUF(x) = ℵo cannot be true for x less than ℵo points, that is for >> a set of ℵo positive x.
Huh?!
NUF(x) = ℵo ist für alle x e IR, x > 0 wahr.
Auch in diesem Fall gilt: NUF(0) = 0 und für alle x e IR, x > 0: NUF(x)
= aleph_0.
Das ist alles ziemlich trivial.
Le 08/07/2024 à 19:33, Jim Burns a écrit :Which ten.
On 7/8/2024 9:49 AM, WM wrote:
There is no point x > 0: NUF(x) < ℵ₀That is wrong because 10 unit fractions and their finite distances
occupy a part of the positive axis which has a finite positive measure. Therefore there exist x > 0: NUF(x) < 11. in order to accumulate ℵ₀ unit fractions, at least 10 must exist at the beginning. Are you unable to understand that?
NUF changes at 0 from 0 to omega.There is no point x < 0: NUF(x) > 0 NUF(x) changes "at" 0.The function f(x) = [x] changes at 1 from 0 to 1, at 2 from 1 to 2, and
so on.
The function NUF(x) changes at unit fractions by 1, but not before, not
at 0.
And NUF changes like the sign function.You are wrong. The function f(x) = [x] changes at the points 1, 2, 3,Therefore it does not change at 0.Functions do not change at single points.
...
Le 08/07/2024 à 16:35, Moebius a écrit :That is as precise as it gets.
Am 08.07.2024 um 16:12 schrieb WM:There is no precision, no "genau" with ℵo.
Le 08/07/2024 à 04:52, Moebius a écrit :Nicht nur "at least", sondern "genau"
ℵo unit fractions occupy at least ℵo different points on the positive >>> real axis.
Grammar?Not for the first ℵo points and their intervals required that the numberHence NUF(x) = ℵo cannot be true for x less than ℵo points, that isNUF(x) = ℵo ist für alle x e IR, x > 0 wahr.
for a set of ℵo positive x.
of unit fractions can grow to ℵo.
There are no "first points"!Auch in diesem Fall gilt: NUF(0) = 0 und für alle x e IR, x > 0: NUF(x)You cannot draw the conclusion: If X unit fractions occupy X points,
= aleph_0. Das ist alles ziemlich trivial.
then there cannot exist more than X unit fractions at the first X
points.
Am Mon, 08 Jul 2024 19:57:33 +0000 schrieb WM:
Le 08/07/2024 à 19:33, Jim Burns a écrit :
On 7/8/2024 9:49 AM, WM wrote:
Which ten.There is no point x > 0: NUF(x) < ℵ₀That is wrong because 10 unit fractions and their finite distances
occupy a part of the positive axis which has a finite positive measure.
Therefore there exist x > 0: NUF(x) < 11. in order to accumulate ℵ₀ unit >> fractions, at least 10 must exist at the beginning. Are you unable to
understand that?
NUF changes at 0 from 0 to omega.There is no point x < 0: NUF(x) > 0 NUF(x) changes "at" 0.The function f(x) = [x] changes at 1 from 0 to 1, at 2 from 1 to 2, and
so on.
The function NUF(x) changes at unit fractions by 1, but not before, not
at 0.
Are all mathematicians [...] stupid?
Ob alle Mathematiker zu blöde sind, um das zu verstehen?
A person who believes in loss by exchange and NUF changing at 0 is
not of interest to me.
Le 05/07/2024 à 15:54, Moebius a écrit :
Am 05.07.2024 um 09:39 schrieb WM:
Le 04/07/2024 à 17:25, Moebius a écrit :
Am 04.07.2024 um 17:16 schrieb WM:;
;
die Verteilung der ersten Stammbrüche;
es gibt keine "ersten Stammbrüche". Zu _jedem_ Stammbruch gibt es
(abzählbar) unendlich viele kleinere Stammbrüche.
Wrong,
Nope.
In sufficient argument.
Le 08/07/2024 à 19:33, Jim Burns a écrit :
A change needs be _with respect to_ something,
yes, to the value befor that point.
Consider
No.
A person who believes in
loss by exchange and
NUF changing at 0
is not of interest to me.
Am 08.07.2024 um 22:47 schrieb WM:
Are all mathematicians [...] stupid?
Certainly not, but
On 7/8/2024 3:57 PM, WM wrote:
Le 08/07/2024 à 19:33, Jim Burns a écrit :
A change needs be _with respect to_ something,
yes, to the value befor that point.
Yes,
for example, floor(0) with respect to floor(0-ε)
Or,
yes, to the value after that point
A person who believes in
loss by exchange and
NUF changing at 0
_near_ 0
Does each nonempty set S of unit.fractionsᵂᴹ
hold a largest S.element?
Does each unit.fractionᵂᴹ u (including 1)
have a next.smaller unit.fractionᵂᴹ ⅟(1+⅟u) ?
Does each unit.fractionᵂᴹ v (excluding 1)
have a next.larger unit.fractionᵂᴹ ⅟(-1+⅟v) ?
Or is what you're talking about irrelevant to
what you're saying?
Am 07.07.2024 um 22:24 schrieb WM:
Le 05/07/2024 à 15:54, Moebius a écrit :
Am 05.07.2024 um 09:39 schrieb WM:
Le 04/07/2024 à 17:25, Moebius a écrit :
Am 04.07.2024 um 17:16 schrieb WM:;
;
die Verteilung der ersten Stammbrüche;
es gibt keine "ersten Stammbrüche". Zu _jedem_ Stammbruch gibt es
(abzählbar) unendlich viele kleinere Stammbrüche.
Wrong,
Nope.
Insufficient argument.
Hinweis: Wenn s ein Stammbruch ist, dann ist 1/(1/s + 1) ein Stammbruch,
der kleiner ist als s.
Le 08/07/2024 à 23:12, Moebius a écrit :What does this mean? It should read "smaller than ANY".
Am 08.07.2024 um 22:47 schrieb WM:
Are all mathematicians [...] stupid?Certainly not, but
many don't understand that ℵo unit fractions cannot occupy a distance smaller than all positive distances. Can you?
Le 09/07/2024 à 02:15, Jim Burns a écrit :Why this direction of less/greater than?
On 7/8/2024 3:57 PM, WM wrote:No, my definition _for this concrete case_ is this: A change of NUF(x) happens at point x if for all y < x NUF(x) > NUF(y).
Le 08/07/2024 à 19:33, Jim Burns a écrit :Yes, for example, floor(0) with respect to floor(0-ε)
A change needs be _with respect to_ something,yes, to the value befor that point.
Or, yes, to the value after that point
Why should there be a smallest element?Does each nonempty set S of unit.fractionsᵂᴹ hold a largest element?A largest and a smallest. Alas the smallest can only be found if it is
not dark.
"Named", which is all of them.Does each unit.fractionᵂᴹ u (including 1)Obviously not, as I have demonstrated irrefutably (refuted only by
have a next.smaller unit.fractionᵂᴹ ⅟(1+⅟u) ?
people who cannot think clear enough. But every unit fraction that can
be named has a next smaller unit fraction.
The only step is at 0.Does each unit.fractionᵂᴹ v (excluding 1)Yes, but for all dark unit fractions this cannot be found. Every unit fraction excluding 1/1 that can be named has a next larger unit
have a next.larger unit.fractionᵂᴹ ⅟(-1+⅟v) ?
fraction.
Or is what you're talking about irrelevant to what you're saying?Relevant is this and only this: NUF(0) = 0, and the first step happens
at x > 0. Like every step it is a step by 1.
Le 09/07/2024 à 03:04, Moebius a écrit :
Am 07.07.2024 um 22:24 schrieb WM:
Le 05/07/2024 à 15:54, Moebius a écrit :
Am 05.07.2024 um 09:39 schrieb WM:
Le 04/07/2024 à 17:25, Moebius a écrit :
Am 04.07.2024 um 17:16 schrieb WM:;
;
die Verteilung der ersten Stammbrüche;
es gibt keine "ersten Stammbrüche". Zu _jedem_ Stammbruch gibt es >>>> >> (abzählbar) unendlich viele kleinere Stammbrüche.
Wrong,
Nope.
Insufficient argument.
Hinweis: Wenn s ein Stammbruch ist, dann ist 1/(1/s + 1) ein
Stammbruch, der kleiner ist als s.
Not true for the smallest unit fraction.
Am Tue, 09 Jul 2024 11:49:21 +0000 schrieb WM:
many don't understand that ℵo unit fractions cannot occupy a distance
smaller than all positive distances. Can you?
What does this mean? It should read "smaller than ANY".
Am Tue, 09 Jul 2024 11:49:21 +0000 schrieb WM:
many don't understand that ℵo unit fractions cannot occupy a distanceWhat does this mean? It should read "smaller than ANY".
smaller than all positive distances. Can you?
Am Tue, 09 Jul 2024 12:08:46 +0000 schrieb WM:
Relevant is this and only this: NUF(0) = 0, and the first step happensThe only step is at 0.
at x > 0. Like every step it is a step by 1.
Am 09.07.2024 um 15:23 schrieb joes:
Am Tue, 09 Jul 2024 11:49:21 +0000 schrieb WM:
many don't understand that ℵo unit fractions cannot occupy a distance
smaller than all positive distances. Can you?
What does this mean? It should read "smaller than ANY".
Es bedeutet, dass Mückenheim (in diesem Kontext) nicht zwischen AxEy und EyAx unterscheiden kann.
Either there is a first unit fraction or this is not the case.
If it is not the case, then NUF(x) increases by more than 1,
say by X, at that x where it is leaving 0.
But then there must exist an x <bla bla bla>
Le 09/07/2024 à 02:15, Jim Burns a écrit :
On 7/8/2024 3:57 PM, WM wrote:
Le 08/07/2024 à 19:33, Jim Burns a écrit :
A change needs be _with respect to_ something,
yes, to the value befor that point.
Yes,
for example, floor(0) with respect to floor(0-ε)
Or,
yes, to the value after that point
for example, ceiling(0) with respect to ceiling(0+ε)
No, my definition _for this concrete case_ is this:
A change of NUF(x) happens at point x if
for all y < x NUF(x) > NUF(y).
A person who believes in
loss by exchange and
NUF changing at 0
_near_ 0
So the change does not happen at zero.
Does each nonempty set S of unit.fractionsᵂᴹ
hold a largest S.element?
A largest and a smallest.
Alas the smallest can only be found if it is not dark.
Does each unit.fractionᵂᴹ u (including 1)
have a next.smaller unit.fractionᵂᴹ ⅟(1+⅟u) ?
Obviously not,
as I have demonstrated irrefutably
(refuted only by people who cannot think clear enough.
But every unit fraction that can be named
has a next smaller unit fraction.
Does each unit.fractionᵂᴹ v (excluding 1)
have a next.larger unit.fractionᵂᴹ ⅟(-1+⅟v) ?
Yes, but
for all dark unit fractions this cannot be found.
Every unit fraction excluding 1/1 that can be named
has a next larger unit fraction.
Or is what you're talking about irrelevant to
what you're saying?
Relevant is this and only this:
NUF(0) = 0,
and the first step happens at x > 0.
Like every step it is a step by 1.
On 7/9/2024 3:11 AM, FromTheRafters wrote:
Does that mean there are as many rationals as there are reals?
On 7/9/2024 9:53 AM, WM wrote:
Le 09/07/2024 à 16:35, Moebius a écrit :
Am 09.07.2024 um 15:23 schrieb joes:
Am Tue, 09 Jul 2024 11:49:21 +0000 schrieb WM:
many don't understand that ℵo unit fractions cannot occupy a distance >>>>> smaller than all positive distances. Can you?
What does this mean? It should read "smaller than ANY".
Es bedeutet, dass Mückenheim (in diesem Kontext) nicht zwischen AxEy
und EyAx unterscheiden kann.
That is nonsense. Either there is a first unit fraction or this is not
the case.
There is a first unit fraction at 1/1. However, there is no last unit fraction... :^)
On 7/9/2024 5:11 AM, WM wrote:
Le 09/07/2024 à 03:04, Moebius a écrit :
Am 07.07.2024 um 22:24 schrieb WM:
Le 05/07/2024 à 15:54, Moebius a écrit :
Am 05.07.2024 um 09:39 schrieb WM:
Le 04/07/2024 à 17:25, Moebius a écrit :
Am 04.07.2024 um 17:16 schrieb WM:;
;
die Verteilung der ersten Stammbrüche;
es gibt keine "ersten Stammbrüche". Zu _jedem_ Stammbruch gibt es >>>>> >> (abzählbar) unendlich viele kleinere Stammbrüche.
Wrong,
Nope.
Insufficient argument.
Hinweis: Wenn s ein Stammbruch ist, dann ist 1/(1/s + 1) ein
Stammbruch, der kleiner ist als s.
Not true for the smallest unit fraction.
There is NO smallest unit fraction. Dark or not. :^)
On 7/9/2024 5:11 AM, WM wrote:
Le 09/07/2024 à 03:04, Moebius a écrit :
Am 07.07.2024 um 22:24 schrieb WM:
Le 05/07/2024 à 15:54, Moebius a écrit :
Am 05.07.2024 um 09:39 schrieb WM:
Le 04/07/2024 à 17:25, Moebius a écrit :
Am 04.07.2024 um 17:16 schrieb WM:
die Verteilung der ersten Stammbrüche
es gibt keine "ersten Stammbrüche". Zu _jedem_ Stammbruch gibt es
(abzählbar) unendlich viele kleinere Stammbrüche.
Wrong,
Nope.
Insufficient argument.
Hinweis: Wenn s ein Stammbruch ist, dann ist 1/(1/s + 1) ein
Stammbruch, der kleiner ist als s.
Not true for the smallest unit fraction.
There is NO smallest unit fraction. Dark or not. :^)
On 7/9/2024 2:07 PM, Moebius wrote:What is "up to infinite precision" and "represent"?
Am 09.07.2024 um 22:10 schrieb Chris M. Thomasson:
On 7/9/2024 3:11 AM, FromTheRafters wrote:
Does that mean there are as many rationals as there are reals?
I already told you that: The set of rational numbers is countable
infinite while the set of real numbers is _uncountable_.
Again: One of my math professors once tried to express this state of
affairs the following way: "There are (in a certain sense) much more
real numbers than rational numbers."
Strange that any real can be represented by a rational up to infinite precision...
On 7/9/2024 2:07 PM, Moebius wrote:
Am 09.07.2024 um 22:10 schrieb Chris M. Thomasson:
On 7/9/2024 3:11 AM, FromTheRafters wrote:
Does that mean there are as many rationals as there are reals?
I already told you that: The set of rational numbers is countable
infinite while the set of real numbers is _uncountable_.
Again: One of my math professors once tried to express this state of
affairs the following way: "There are (in a certain sense) much more
real numbers than rational numbers."
Strange that any real can be represented by a rational up to infinite precision...
There cannot be ℵ₀ without first 10.
Am 08.07.2024 um 22:47 schrieb WM:
There cannot be ℵ₀ without first 10.
There cannot be ℵ₀ rational numbers > 0 without first 10?
Which are these numbers? Pleeeze tell us!
The first 10 Mückenheim-darkies?
We might call them WM_1, WM_2, .., WM_10.
But shouldn't there be a rational number WM_0 := WM_1/2 with WM_0 < WM_1 (since WM_1/2 < WM_1)?
Am 09.07.2024 um 18:53 schrieb WM:
Either there is a first unit fraction or this is not the case.
It is not the case.
Hint: If s is a unit fraction, 1/(1/s + 1) is a smaller one.
If it is not the case, then NUF(x) increases by more than 1,
say by X, at that x where it is leaving 0.
1. NUF does not "increase" but "jump".
2. "That x where it is leaving 0" does not exist. :-)
Hint: NUF(x) = 0 for all x e IR, x <= 0 and NUF(x) = aleph_0 for all x e
IR, x > 0.
But then there must exist an x <bla bla bla>
Hint: For all x e IR, x > 0 there are infinitely many unit fractions
which are smaller than x.
On 7/9/2024 8:08 AM, WM wrote:
A change of NUF(x) happens at point x if
for all y < x NUF(x) > NUF(y).
Then
Relevant is this and only this:
NUF(0) = 0,
and the first step happens at x > 0.
Like every step it is a step by 1.
If NUF(x) = 0 at x > 0
then
Contradiction.
So how can there be "more" real numbers than rational numbers? :-P
This is one of WM's arguments to show that card(IR) and card(Q) are the
same (iirc).
Am 10.07.2024 um 01:23 schrieb Moebius:
Am 08.07.2024 um 22:47 schrieb WM:
There cannot be ℵ₀ without first 10.
There cannot be ℵ₀ rational numbers > 0 without first 10?
Which are these numbers? Pleeeze tell us!
Le 09/07/2024 à 20:51, Jim Burns a écrit :
Contradiction.
Not in dark numbers.
Le 10/07/2024 à 03:02, Moebius a écrit :
Am 10.07.2024 um 01:23 schrieb Moebius:
Am 08.07.2024 um 22:47 schrieb WM:
There cannot be ℵ₀ without first 10.
There cannot be ℵ₀ rational numbers > 0 without first 10?
Which are these numbers? Pleeeze tell us!
They are dark.
On 7/10/2024 12:02 PM, WM wrote:
Le 09/07/2024 à 20:51, Jim Burns a écrit :
On 7/9/2024 8:08 AM, WM wrote:
A change of NUF(x) happens at point x if
for all y < x NUF(x) > NUF(y).
Then ceiling(x) _doesn't_ changeᵂᴹ "at" 0
Then what ever. It is my definition.
Very much like Humpty Dunpty and "glory".
[1]
When you say that
NUF(x) doesn't changeᵂᴹ at 0
you mean that
NUF(x) doesn't changeᵂᴹ like ceiling(x) doesn't.
However,
your argument depends upon
it being unclear what you (WM) mean.
Not the best argument.
Relevant is this and only this:
NUF(0) = 0,
and the first step happens at x > 0.
Like every step it is a step by 1.
If NUF(x) = 0 at x > 0
then
Contradiction.
Not in dark numbers.
For the unit.fractionsⁿᵒᵗᐧᵂᴹ.set ⅟ℕⁿᵒᵗᐧᵂᴹ glb.⅟ℕⁿᵒᵗᐧᵂᴹ > 0 is contradictory.
Each nonempty ⅟ℕⁿᵒᵗᐧᵂᴹ.subset S
holds a largest S.element.
Each ⅟ℕⁿᵒᵗᐧᵂᴹ.element u (including 1) has
a next.smaller ⅟ℕⁿᵒᵗᐧᵂᴹ.element ⅟(1+⅟u)
Each ⅟ℕⁿᵒᵗᐧᵂᴹ.element v (excluding 1) has
a next.larger ⅟ℕⁿᵒᵗᐧᵂᴹ.element ⅟(-1+⅟v)
However,
your argument depends upon
it being unclear what you (WM) mean.
Not the best argument.
Le 09/07/2024 à 20:51, Jim Burns a écrit :
On 7/9/2024 8:08 AM, WM wrote:
A change of NUF(x) happens at point x if
for all y < x NUF(x) > NUF(y).
Then ceiling(x) _doesn't_ changeᵂᴹ "at" 0
Then what ever. It is my definition.
Relevant is this and only this:
NUF(0) = 0,
and the first step happens at x > 0.
Like every step it is a step by 1.
If NUF(x) = 0 at x > 0
then
Contradiction.
Not in dark numbers.
Am 10.07.2024 um 19:58 schrieb Jim Burns:
On 7/10/2024 12:02 PM, WM wrote:
your argument depends upon
it being unclear what you (WM) mean.
Not the best argument.
I beg to differ. It's the best (type of) argument WM will ever have.
On 7/10/2024 12:02 PM, WM wrote:
your argument depends upon
it being unclear what you (WM) mean.
Not the best argument.
On 7/9/2024 2:49 PM, Moebius wrote:
I guess you might have a _sequence_ of rational numbers in mind, say,Well, basically, I was thinking that for any element of:
(1, 1.4, 1.41, 1.414, ...).
So we might say that this SEQUENCE represents the real number sqrt(2)
- in a certain sense. :-P
Actually, its limit is sqrt(2).
(1, 1.4, 1.41, 1.414, ...)
there is a rational that can represent it.
So, it kind of makes my brain want to bleed from time to time, shit happens! Uggg.
Taken to infinity, there are rationals that can represent [...] sqrt 2:
(1, 1.4, 1.41, 1.414, ...)
However, there is no single rational that equals sqrt 2.
Humm... Fair enough?
Think of the whole as simply, sqrt(2)
The finite parts are the step wise construction of the whole, can be something akin to:
(1, 1.4, 1.41, 1.414, ...)
Is this a decent line of thought?
(1, 1.4, 1.41, 1.414, ...)
For each element there is a rational that can represent it.
No single rational can represent the whole...
However, a real can represent [...] the whole...
Fair enough?
Or am I drifting off deeper into WM land?
Math SHOULD BE fun! (imho)
Am 11.07.2024 um 00:55 schrieb Moebius:
Math SHOULD BE fun! (imho)
Hmmm... It's quite clear that WM doesn't have ANY fun with math, I'd say.
Am 11.07.2024 um 01:00 schrieb Moebius:
Am 11.07.2024 um 00:55 schrieb Moebius:
Math SHOULD BE fun! (imho)
Hmmm... It's quite clear that WM doesn't have ANY fun with math, I'd say.
Actually, he doesn't DO any math.
Am 11.07.2024 um 01:00 schrieb Moebius:say.
Am 11.07.2024 um 00:55 schrieb Moebius:
Math SHOULD BE fun! (imho)
Hmmm... It's quite clear that WM doesn't have ANY fun with math, I'd
Actually, he doesn't DO any math.
On 7/10/2024 3:55 PM, Moebius wrote:
Am 11.07.2024 um 00:26 schrieb Chris M. Thomasson:
(1, 1.4, 1.41, 1.414, ...)
For each element [term] [in (1, 1.4, 1.41, 1.414, ...)] there is a rational that can represent it.
Well, A = A? Agree!
Should I add in the word finite here:
For each _finite_ element there is a rational that can represent it.
Any clearer, or does it add mud to the clear waters?
No single rational can represent the whole...
Right!
Otherwise the question would be: Which one? :-)
However, a real can represent [...] the whole...
Exactly! (!!!)
Fair enough?
Sure!
On 7/10/2024 4:29 PM, Moebius wrote:
Am 11.07.2024 um 01:02 schrieb Moebius:
Am 11.07.2024 um 01:00 schrieb Moebius:Most cranks don't to any (real) math, they just "criticizes" (sort of)
Am 11.07.2024 um 00:55 schrieb Moebius:
Math SHOULD BE fun! (imho)
Hmmm... It's quite clear that WM doesn't have ANY fun with math, I'd
say.
Actually, he doesn't DO any math.
things they don't really understand.
If I am having a hard time understanding something, I ask/read around until
I can finally grasp the underlying meaning of the problem to a point where
I can code up a working solution. It has certainly tended worked for me in the past.
On 7/10/2024 6:55 PM, Moebius wrote:
Am 11.07.2024 um 00:26 schrieb Chris M. Thomasson:
(1, 1.4, 1.41, 1.414, ...)
For each element there is a rational that can represent it.
Well, A = A? Agree!
Just to keep things from becoming too clear...
WM has disputed A = A for darkᵂᴹ numbers A
To be precise,
he has disputed _saying_ 'A = A'
...which I guess indicates that he _agrees_ with it
and dislikes how it messes up his "proofs".
Am 11.07.2024 um 00:26 schrieb Chris M. Thomasson:
(1, 1.4, 1.41, 1.414, ...)
For each element there is a rational that can represent it.
Well, A = A? Agree!
On 7/10/2024 5:32 PM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:[...]
For example, with these tiles, the sequence 1, 2, 1, 3 produces
aa bb aa abb
aab ba aab b
r[0] = aa
r[1] = bb
r[3] = abb
r[4] = aab
r[5] = ba
r[6] = b
Where:
r[0] r[1] r[0] r[3]
r[4] r[5] r[4] r[6]
Is a legit mapping?
On 7/10/2024 5:32 PM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
On 7/10/2024 4:29 PM, Moebius wrote:Here's and interesting problem to code up. The input is a finite array
Am 11.07.2024 um 01:02 schrieb Moebius:
Am 11.07.2024 um 01:00 schrieb Moebius:Most cranks don't to any (real) math, they just "criticizes" (sort of) >>>> things they don't really understand.
Am 11.07.2024 um 00:55 schrieb Moebius:
Math SHOULD BE fun! (imho)
Hmmm... It's quite clear that WM doesn't have ANY fun with math, I'd >>>>>> say.
Actually, he doesn't DO any math.
If I am having a hard time understanding something, I ask/read around until >>> I can finally grasp the underlying meaning of the problem to a point where >>> I can code up a working solution. It has certainly tended worked for me in >>> the past.
of n pairs of strings, so there are 2n strings in all. They are often
thought of as strings on the top and bottom of a collection of tiles or
dominos, but that just help visualise the problem. An example with n=3
might be
aa bb abb
aab ba b
Just making up an example:
ab ab ab = ababab
aba ba b = ababab
Would that work?
Does each string need to be unique?
On 7/10/2024 5:32 PM, Ben Bacarisse wrote:
"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:[...]
Just making up a quick example, I need to work on something else right
now... Sorry Ben.
On 7/10/2024 6:55 PM, Moebius wrote:
Am 11.07.2024 um 00:26 schrieb Chris M. Thomasson:
(1, 1.4, 1.41, 1.414, ...)
For each element there is a rational that can represent it.
Well, A = A? Agree!
Just to keep things from becoming too clear...
WM has disputed A = A for darkᵂᴹ numbers A
To be precise,
he has disputed _saying_ 'A = A'
...which I guess indicates that he _agrees_ with it
and dislikes how it messes up his "proofs".
On 7/10/2024 12:02 PM, WM wrote:
When you say that
NUF(x) doesn't changeᵂᴹ at 0
you mean that
your argument depends upon
it being unclear what you (WM) mean.
Le 11/07/2024 à 02:52, Jim Burns a écrit :
On 7/10/2024 6:55 PM, Moebius wrote:
Am 11.07.2024 um 00:26 schrieb Chris M. Thomasson:
(1, 1.4, 1.41, 1.414, ...)
For each element
there is a rational that can represent it.
Well, A = A? Agree!
Just to keep things from becoming too clear...
WM has disputed A = A for darkᵂᴹ numbers A
To be precise,
he has disputed _saying_ 'A = A'
Dark numbers are equal to themselves.
We cannot prove it,
because we cannot know them.
But it has to be assumed in any meaningful theory.
Further
it happens to be true for
every dark number that becomes visible.
Dark numbers are equal to themselves.
...which I guess indicates that he _agrees_ with it
and dislikes how it messes up his "proofs".
says a person who believes that
by exchanging two elements
one of them can be lost.
Le 10/07/2024 à 19:58, Jim Burns a écrit :
On 7/10/2024 12:02 PM, WM wrote:
When you say that
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0, namely NUF(0) = 0.
your argument depends upon it being unclear what you (WM) mean.
That <bla bla bla>
You (WM) are confused.You think?
Le 10/07/2024 à 19:58, Jim Burns a écrit :
On 7/10/2024 12:02 PM, WM wrote:
When you say that
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0, namely NUF(0) = 0.
your argument depends upon
it being unclear what you (WM) mean.
That <bla bla bla>
On 7/12/2024 1:04 PM, WM wrote:
says a person who believes that
by exchanging two elements
one of them can be lost.
I have never claimed ℵ₀ = 2
The set ℕ⁺ᴮᵒᵇ of
cardinals which, when augmented, are bigger plus Bob
has the same cardinality ℵ₀ as
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0, namely NUF(0) = 0.
But it changes its value from aleph_0 for each and every x > 0 to 0 at x
= 0,
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that;
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0, namely NUF(0) = 0.
But it changes its value from aleph_0 for each and every x > 0 to 0 at
x = 0,
That claim is wrong,
Le 12/07/2024 à 22:23, Jim Burns a écrit :
On 7/12/2024 1:04 PM, WM wrote:
says a person who believes that
by exchanging two elements
one of them can be lost.
I have never claimed ℵ₀ = 2
You have claimed that
by exchanging X and O an O can disappear,
in fact
infinitely many can disappear by pure exchange.
The set ℕ⁺ᴮᵒᵇ of
cardinals which, when augmented, are bigger plus Bob
has the same cardinality ℵ₀ as
Cardinality is nonsense as my proof shows.
There is no counting but only exchanging.
You cannot consistently argue with
rules which are disproved.
Cardinality is
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that;
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0, namely NUF(0) = 0.
But it changes its value from aleph_0 for each and every x > 0 to 0 at
x = 0,
That claim is wrong,
Nope.
On 7/13/2024 11:43 AM, WM wrote:
Le 12/07/2024 à 22:23, Jim Burns a écrit :
On 7/12/2024 1:04 PM, WM wrote:
says a person who believes that
by exchanging two elements
one of them can be lost.
I have never claimed ℵ₀ = 2
You have claimed that
by exchanging X and O an O can disappear,
Yes.
in fact
infinitely many can disappear by pure exchange.
Yes.
What we also which understand and you (WM) don't is that
ℕ₁⁺ᴮᵒᵇ and ℕ₁×ℕ₁ aren't finiteⁿᵒᵗᐧᵂᴹ sets,
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that;
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0, namely NUF(0) = 0.
But it changes its value from aleph_0 for each and every x > 0 to 0
at x = 0,
That claim is wrong,
Nope.
That claim is wrong,
Am 14.07.2024 um 14:58 schrieb WM:
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that;
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0, namely NUF(0) = 0.
But it changes its value from aleph_0 for each and every x > 0 to 0
at x = 0,
That claim is wrong,
Nope.
That claim is wrong,
Nope.
Le 13/07/2024 à 23:55, Jim Burns a écrit :
On 7/13/2024 11:43 AM, WM wrote:
Le 12/07/2024 à 22:23, Jim Burns a écrit :
On 7/12/2024 1:04 PM, WM wrote:
says a person who believes that
by exchanging two elements
one of them can be lost.
I have never claimed ℵ₀ = 2
You have claimed that
by exchanging X and O an O can disappear,
Yes.
That disqualifies you from any serious discussion.
in fact infinitely many can disappear
by pure exchange.
Yes.
Laughable.
Are you aware that nobody followed you?
What we also [understand which] you (WM) don't is that
ℕ₁⁺ᴮᵒᵇ and ℕ₁×ℕ₁ aren't finiteⁿᵒᵗᐧᵂᴹ sets,
Logic remains valid for all correct mathematics,
finite and infinite.
Are you aware that nobody followed you?
For each and every x e IR, x > 0: NUF(x) = aleph_0
Am 14.07.2024 um 14:58 schrieb WM:
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that;
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0, namely NUF(0) = 0.
But it changes its value from aleph_0 for each and every x > 0 to 0
at x = 0,
That claim is wrong,
Nope.
That claim is wrong,
Nope.
Le 14/07/2024 à 17:04, Moebius a écrit :As has been repeated, there is no single fixed number less than all/every
Am 14.07.2024 um 14:58 schrieb WM:That claim has been disproved. Something that is before each and every x
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that NUF(x) doesn't changeᵂᴹ at 0 you mean that >>>>>> > it does not change its value from x < 0, namely NUF(0) = 0.
But it changes its value from aleph_0 for each and every x > 0 to 0 >>>>>> at x = 0,
0 sits at zero or at the negative real line.
Am 14.07.2024 um 15:16 schrieb WM:
Are you aware that nobody followed you?
Are YOU aware that nobody follows YOU, Mückenheim?
Actually, Jim "approach" is indeed "questionable" (imho), but there is a
way to formulate it in a sensible way. And the result is: Bye Bob!
Am Mon, 15 Jul 2024 12:21:52 +0000 schrieb WM:
Le 14/07/2024 à 17:04, Moebius a écrit :As has been repeated, there is no single fixed number less than all/every other number.
Am 14.07.2024 um 14:58 schrieb WM:That claim has been disproved. Something that is before each and every x
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that NUF(x) doesn't changeᵂᴹ at 0 you mean that >>>>>>> > it does not change its value from x < 0, namely NUF(0) = 0.
But it changes its value from aleph_0 for each and every x > 0 to 0 >>>>>>> at x = 0,
0 sits at zero or at the negative real line.
However, every number has a smaller one. Do you understand
the difference?
Le 14/07/2024 à 17:04, Moebius a écrit :
Am 14.07.2024 um 14:58 schrieb WM:
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that
NUF(x) doesn't changeᵂᴹ at 0
you mean that
it does not change its value from x < 0,
namely NUF(0) = 0.
But it changes its value from aleph_0
for each and every x > 0
to 0 at x = 0,
That claim is wrong,
Nope.
That claim is wrong,
Nope.
That claim has been disproved.
Something that is before each and every x > 0
sits at zero or at the negative real line.
Le 15/07/2024 à 15:33, joes a écrit :I don't see how that follows.
Am Mon, 15 Jul 2024 12:21:52 +0000 schrieb WM:Hence your claim that for every x > 0: NUF(x) = ℵo, is blatantly wrong.
Le 14/07/2024 à 17:04, Moebius a écrit :As has been repeated, there is no single fixed number less than
Am 14.07.2024 um 14:58 schrieb WM:That claim has been disproved. Something that is before each and every
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that NUF(x) doesn't changeᵂᴹ at 0 you mean that >>>>>>>> > it does not change its value from x < 0, namely NUF(0) = 0. >>>>>>>> But it changes its value from aleph_0 for each and every x > 0 to >>>>>>>> 0 at x = 0,
x 0 sits at zero or at the negative real line.
all/every other number.
Real as in elements of R?However, every number has a smaller one. Do you understand theThe function NUF(x) counts only real points. They are separated.
difference?
Therefore there is a first unit fraction. It is an x, a single numberNon sequitur. Like Moebius said, 1/(1+1/x) is smaller.
which contradicts your claim.
Am Mon, 15 Jul 2024 14:19:36 +0000 schrieb WM:
Le 15/07/2024 à 15:33, joes a écrit :I don't see how that follows.
Am Mon, 15 Jul 2024 12:21:52 +0000 schrieb WM:Hence your claim that for every x > 0: NUF(x) = ℵo, is blatantly wrong.
Le 14/07/2024 à 17:04, Moebius a écrit :As has been repeated, there is no single fixed number less than
Am 14.07.2024 um 14:58 schrieb WM:That claim has been disproved. Something that is before each and every >>>> x 0 sits at zero or at the negative real line.
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
Le 10/07/2024 à 19:58, Jim Burns a écrit :
When you say that NUF(x) doesn't changeᵂᴹ at 0 you mean that >>>>>>>>> > it does not change its value from x < 0, namely NUF(0) = 0. >>>>>>>>> But it changes its value from aleph_0 for each and every x > 0 to >>>>>>>>> 0 at x = 0,
all/every other number.
Real as in elements of R?However, every number has a smaller one. Do you understand theThe function NUF(x) counts only real points. They are separated.
difference?
Therefore there is a first unit fraction. It is an x, a single numberNon sequitur. Like Moebius said, 1/(1+1/x) is smaller.
which contradicts your claim.
Am 15.07.2024 um 15:33 schrieb joes:
Am Mon, 15 Jul 2024 12:21:52 +0000 schrieb WM:
Le 14/07/2024 à 17:04, Moebius a écrit :
Am 14.07.2024 um 14:58 schrieb WM:That claim has been disproved. Something that is before each and every x > 0
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
; Le 10/07/2024 à 19:58, Jim Burns a écrit :
But it changes its value from aleph_0 for each and every x > 0 to 0 >>>>>>>> at x = 0,; When you say that NUF(x) doesn't changeᵂᴹ at 0 you mean that >>>>>>>> > it does not change its value from x < 0, namely NUF(0) = 0. >>>>>>>>
sits at zero or at the negative real line.
As has been repeated, there is no single fixed number less than all/every
other number. However, every number has a smaller one. Do you understand
the difference?
Wie schon ein paar Mal (also ein paar 100-mal) erwähnt, scheint
Mückenheim nicht fähig (oder gewillt) zu sein, zwischen
Ax Ey (y < x)
und
Ex Ay (y < x)
zu unterscheiden. Der Mann hat einen schweren Dachschaden, das steht fest.
Am Mon, 15 Jul 2024 12:21:52 +0000 schrieb WM:
Le 14/07/2024 à 17:04, Moebius a écrit :
Am 14.07.2024 um 14:58 schrieb WM:That claim has been disproved. Something that is before each and every x > 0 >> sits at zero or at the negative real line.
Le 13/07/2024 à 21:37, Moebius a écrit :
Am 13.07.2024 um 17:56 schrieb WM:
Le 13/07/2024 à 04:38, Moebius a écrit :
Am 12.07.2024 um 19:29 schrieb WM:
; Le 10/07/2024 à 19:58, Jim Burns a écrit :
But it changes its value from aleph_0 for each and every x > 0 to 0 >>>>>>> at x = 0,; When you say that NUF(x) doesn't changeᵂᴹ at 0 you mean that >>>>>>> > it does not change its value from x < 0, namely NUF(0) = 0. >>>>>>>
As has been repeated, there is no single fixed number less than all/every other number. However, every number has a smaller one. Do you understand
the difference?
Am 15.07.2024 23:57:09 Moebius schrieb:
Wie schon ein paar Mal (also ein paar 100-mal) erwähnt, scheint
Mückenheim nicht fähig (oder gewillt) zu sein, zwischen
Ax Ey (y < x)
und
Ex Ay (y < x)
zu unterscheiden. [...]
wenn du m i t WM nicht immer weiter erfolglos herumlabern willst.
Siehe auch: https://en.wikipedia.org/wiki/Quantifier_shift
There are infinite successors to any finite number?
Say, 5:
5 + 1
On 7/15/2024 5:41 AM, WM wrote:
Le 14/07/2024 à 17:27, Moebius a écrit :
For each and every x e IR, x > 0: NUF(x) = aleph_0
Wrong. All unit fractions are separated. Therefore there is a first
one at y. NUF(y) = 1.
There is a first unit fraction at 1/1. There is no last unit fraction.
Get over it man!
Naturals:
a = any natural number
b = a + 1 is the /successor/ of a?
;^)
going left to right, so to speak, [...]
1/1, 1/2, 1/3, 1/4, ...
Huh? WM is backwards?
0 is origin!
On 7/15/2024 6:47 PM, Moebius wrote:
Am 16.07.2024 um 03:32 schrieb Chris M. Thomasson:Actually, my real line is, say
going left to right, so to speak, [...]
1/1, 1/2, 1/3, 1/4, ...
Nope, here you are going "from right to left" (on the real line).
You see, 1/1 is larger than, say, 1/2, hence on "the real line" 1/1 is
RIGHT from 1/2. No?
Hint:
... | ... | ... | ... (real line)
0 1/2 1/1 (=1)
Huh? WM is backwards?
Actually, not. :-)
...-1...0...+1...
So, left to right with origin at zero [...] the natural numbers, starting at zero the
origin of the real line:
0, 1, 2, 3, 4, ...
0 is at origin point on my real line, the x axis so to speak. It has an origin at 0. Fair enough?
Left of zero, or origin if you will, is negative, right of zero is positive...
all of the signed imaginary parts go off axis. [...]
On 7/15/2024 6:57 PM, Moebius wrote:
See? :-P
I see that you increased the granularity from natural numbers into the unit fractions...
Wrt enumeration unit fractions I like to go from 1/1, to 1/2, to 1/3,
ect...
Is that wrong?
we have to think of a a smallest unit fraction, WM world, right?
On 7/15/2024 6:57 PM, Moebius wrote:the unit fractions...
See? 😛
I see that you increased the granularity from natural numbers into
Wrt enumeration unit fractions I like to go from 1/1, to 1/2, to 1/3,ect...
Is that wrong?
we have to think of a a smallest unit fraction, WM world, right?
Repost:
Am 16.07.2024 um 04:02 schrieb Chris M. Thomasson:
On 7/15/2024 6:57 PM, Moebius wrote:
See? 😛
I see that you increased the granularity from natural numbers into the unit fractions...
Yeah, the real numbers comprise the naturals, the integers, the unit fractons, the rational numbers, ..., you know. 🙂
Wrt enumeration unit fractions I like to go from 1/1, to 1/2, to 1/3, ect...
You may like to do that, still:
0 < ... < 1/3 < 1/2 < 1/1.
Meaning: Concening the < relation as _defined on the reals_ (as well on
the rationals) 1/3 is SMALLER than 1/2 and 1/2 is smaller than 1/1. In general: 1/(n+1) is smaller than 1/n.
Is that wrong?
Nope. You may define a SEQUENCE (of unit fractions):
(1/1, 1/2, 1/3, ...)
Here (referring to these sequence) 1/1 is "before", say, 1/2. 🙂
we have to think of a a smallest unit fraction, WM world, right?
Right. There simply is no such unit fraction because for each and every
unit fraction u: 1/(1/u + 1) is a unit fraction that is smaller than u.
∀ᴿx>0: NUF(x) = ℵ₀
Am Mon, 15 Jul 2024 14:19:36 +0000 schrieb WM:
Hence your claim that for every x > 0: NUF(x) = ℵo, is blatantly wrong.I don't see how that follows.
Therefore there is a first unit fraction. It is an x, a single numberNon sequitur. Like Moebius said, 1/(1+1/x) is smaller.
which contradicts your claim.
Am 15.07.2024 um 15:33 schrieb joes:
As has been repeated, there is no single fixed number less than all/every
other number. However, every number has a smaller one.
Wie schon ein paar Mal (also ein paar 100-mal) erwähnt, scheint
Mückenheim nicht fähig (oder gewillt) zu sein, zwischen
Ax Ey (y < x)
Am 16.07.2024 um 00:11 schrieb Tom Bola:
Am 15.07.2024 23:57:09 Moebius schrieb:
Wie schon ein paar Mal (also ein paar 100-mal) erwähnt, scheint
Mückenheim nicht fähig (oder gewillt) zu sein, zwischen
Ax Ey (y < x)
und
Ex Ay (y < x)
zu unterscheiden. [...]
Mal etwas für Kenner: Diese Unterscheidung muss man nicht nur bei einer aktualen Auffassung der Unendlichkeit (wie in der _Mengenlehre_),
sondern auch bei einer "potentialistischen" Auffassung des Unendlichen machen.
Zweifellos gibt es bei letzterer zu jedem Stammbruch, den man
konstruieren kann, einen kleineren Stammbruch, den man konstruieren kann (vgl. Streckenteilung in der Geometrie). Es gibt aber auch hier keinen Stammbruch, den man konstruieren kann, der kleiner ist als alle
Stammbrüche, die man konstruieren kann.
Le 16/07/2024 à 01:57, Moebius a écrit :
Am 16.07.2024 um 00:11 schrieb Tom Bola:
Am 15.07.2024 23:57:09 Moebius schrieb:
Wie schon ein paar Mal (also ein paar 100-mal) erwähnt, scheint
Mückenheim nicht fähig (oder gewillt) zu sein, zwischen
Ax Ey (y < x)
und
Ex Ay (y < x)
Fehler! x und y vertauscht!
Le 15/07/2024 à 23:57, Moebius a écrit :
Am 15.07.2024 um 15:33 schrieb joes:
As has been repeated, there is no single fixed number less than all/
every other number. However, every number has a smaller one.
That is
Wie schon ein paar Mal (also ein paar 100-mal) erwähnt, scheint
Mückenheim nicht fähig (oder gewillt) zu sein, zwischen
Ax Ey (y < x)
That is potential infinity.
In der
heutigen Mathematik, die auf der Mengenlehre basiert ("classical mathematics"), gilt selbstverständlich ebenfalls
Ax Ey (y < x) ,
Le 16/07/2024 à 16:18, Moebius a écrit:
In der heutigen Mathematik, die auf der Mengenlehre basiert
("classical mathematics"), gilt selbstverständlich ebenfalls
Ax Ey (y < x) ,
Die ist ja auch <bla>
Am 16.07.2024 um 16:23 schrieb WM:
Le 16/07/2024 à 16:18, Moebius a écrit:
In der heutigen Mathematik, die auf der Mengenlehre basiert
("classical mathematics"), gilt selbstverständlich ebenfalls
Ax Ey (y < x) ,
Die ist ja auch <bla>
Gut, dass wir darüber geredet haben, Mückenheim.
Du hast jetzt also verstanden, dass man zwischen
| Ax Ey (y < x)
| und
| Ey Ax (y < x)
unterscheiden muss? Insbesondere wenn der Kontext z. B. die reellen oder rationalen oder ganzen Zahlen sind?
Dann hast Du ja jetzt endlich auch joes Aussage verstanden:
"As has been repeated, there is no single fixed number less than all
numbers. However, every number has a smaller one." (joe)
In Zeichen: Ey Ax (y < x) ist falsch, aber Ax Ey (y < x) ist wahr.
Le 15/07/2024 à 23:08, joes a écrit :
Like Moebius said, 1/(1+1/x) is smaller
Only for x which can be named.
There is no smallest unit fraction! Damn it!!! :^)
Le 14/07/2024 à 17:27, Moebius a écrit :
For each and every x e IR, x > 0: NUF(x) = aleph_0
Wrong. All unit fractions are separated. Therefore there is a first one
at y. NUF(y) = 1.
Regards, WM
All unit fractions are separated. Therefore there is a first one
On 7/15/24 8:41 AM, WM wrote:
Le 14/07/2024 à 17:27, Moebius a écrit :
For each and every x e IR, x > 0: NUF(x) = aleph_0
Wrong. All unit fractions are separated. Therefore there is a first one
at y. NUF(y) = 1.
Nope.
That means there is a highest natural number n = 1/y.
but m = n+1 is also a natural number,
Your problem is you logic can't handle unbounded numbers.
Am 15.07.2024 um 14:41 schrieb WM:
All unit fractions are separated. Therefore there is a first one
All integers are separated. Therefore there is a
first one
On 7/16/2024 6:21 AM, WM wrote:
Ax Ey (y < x)
That is potential infinity. There is no smallest y.
In actual infinity there are all points including the smallest unit
fraction.
There is no smallest unit fraction!
Huh? Take any unit fraction you can think of. Now, does it perfectly
equal zero? No matter how small it is! NEVER! There is no smallest unit fraction... And for any unit fraction there is one smaller than it that
does not equal zero. Period.
On 7/17/2024 3:31 AM, WM wrote:
Le 16/07/2024 à 23:51, "Chris M. Thomasson" a écrit :
On 7/16/2024 6:21 AM, WM wrote:
Ax Ey (y < x)
That is potential infinity. There is no smallest y.
In actual infinity there are all points including the smallest unit
fraction.
There is no smallest unit fraction!
Don't claim. Explain how NUF(x) can increase from 0 to more when all
unit fractions are separated. That is the crucial argument! The result
is that mathematics has been erroneous for millenia. No dark numbers
have been recognized. But they are existing, if actual infinity is
existing.
Huh? Take any unit fraction you can think of. Now, does it perfectly
equal zero? No matter how small it is! NEVER! There is no smallest unit fraction... And for any unit fraction there is one smaller than it that
does not equal zero. Period.
| Sysop: | Keyop |
|---|---|
| Location: | Huddersfield, West Yorkshire, UK |
| Users: | 493 |
| Nodes: | 16 (2 / 14) |
| Uptime: | 23:44:18 |
| Calls: | 9,729 |
| Calls today: | 19 |
| Files: | 13,741 |
| Messages: | 6,182,405 |