1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...

Push the natnumbers of the first column (without queue) into the field of fractions and store the hit fraction always there where the natnumber has come from. Try to push the natnumbers such that all matrix positions are occupied by them. That is best done by creating a pattern like

1, 2, 4, ...
3, 5, 8, ...
6, 9, 13, ...
.... ,

According to this simple rule it is impossible, in eternity, to remove a fraction from the matrix or to attach a natnumber to a fraction.

You have won as soon as you understand that the last sentence is true.

Congrats, WM

On Thursday, October 19, 2023 at 3:04:16 PM UTC+2, WM wrote:
> 1, 1/2, 1/3, 1/4, ...
> 2, 2/2, 2/3, 2/4, ...
> 3, 3/2, 3/3, 3/4, ...
> 4, 4/2, 4/3, 4/4, ...
> 5, 5/2, 5/3, 5/4, ...
>
> Push the natnumbers of the first column (without queue) into the field of fractions and store the hit fraction always there where the natnumber has come from. Try to push the natnumbers such that all matrix positions are occupied by them. That is best done by creating a pattern like
>
> 1, 2, 4, ...
> 3, 5, 8, ...
> 6, 9, 13, ...
> ... ,
>
> According to this simple rule it is impossible, in eternity, to remove a fraction from the matrix or to attach a natnumber to a fraction.
>
> You have won as soon as you understand that the last sentence is true.
>
> Congrats, WM

So what's the problem? How many times do you need people to explain
the error in your reasoning?
You cannot count infinitely many objects in finitely many steps.

Spelling error: the word you are looking for is spelled 'billiards' in English.

Reading comprehension and writing skills: unsatisfactory
Mathematical skills: nonexistent

JVR schrieb am Donnerstag, 19. Oktober 2023 um 19:05:26 UTC+2:
> On Thursday, October 19, 2023 at 3:04:16 PM UTC+2, WM wrote:

> > According to this simple rule it is impossible, in eternity, to remove a fraction from the matrix or to attach a natnumber to a fraction.

> You cannot count infinitely many objects in finitely many steps.

First: I need not count infinitely many elements. I have proved that every issued index fails to index a fraction, may this happen finitely or infinitely often (see first paragraph).

Second: I index as many objects as Cantor does by using the same infinite sequence as he does. He claims to index all fractions.

Regards, WM

On Thursday, October 19, 2023 at 8:22:30 PM UTC+2, WM wrote:
> JVR schrieb am Donnerstag, 19. Oktober 2023 um 19:05:26 UTC+2:
> > On Thursday, October 19, 2023 at 3:04:16 PM UTC+2, WM wrote:
>
> > > According to this simple rule it is impossible, in eternity, to remove a fraction from the matrix or to attach a natnumber to a fraction.
> > You cannot count infinitely many objects in finitely many steps.
> First: I need not count infinitely many elements. I have proved that every issued index fails to index a fraction, may this happen finitely or infinitely often (see first paragraph).
>
> Second: I index as many objects as Cantor does by using the same infinite sequence as he does. He claims to index all fractions.
>
> Regards, WM

Cantor's enumeration is not a step-by-step procedure, it is a function.
Logically the enumeration of Q is a function that maps N <-> Q,
just as sin(x) is a function that maps (0, pi/2) <-> (0,1). From the latter
you should be able to see that your X's and O's are not going to get
you anywhere. And if that doesn't convince you, the fact that there
are biunique mappings from (0,1) to (0,1)x(0,1) should.

You are clearly unable to understand set theory and point set topology
as they are in use today and have been in use for 100+ years.

Crank Professor Wolfgang Mückenheim, aka WM wrote:
> [...]
> Second: I index as many objects as Cantor does by using the same infinite sequence as he does. He claims to index all fractions.

"It's possible to fail so it is impossible to succeed", what kind
of logic is that, Crank Professor Wolfgang Mückenheim, from Hochschule
Augsburg?

JVR schrieb am Donnerstag, 19. Oktober 2023 um 20:52:18 UTC+2:

> Cantor's enumeration is not a step-by-step procedure,

Enumeration is always a step-by-step procedure: 1, 2, 3, and so on. step by step.

> Logically the enumeration of Q is a function that maps N <-> Q,
> just as sin(x) is a function that maps (0, pi/2) <-> (0,1).

No, the first one is quantized, the second one is not.

If you deny that the step-by-step procedure produces exactly the function k = (m + n - 1)(m + n - 2)/2 + m, then you should be able to find a k that is not attached by my matrices to the same field (m, n) of the matrix

1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
...

as by Cantor's function. But you cannot. My matrices reproduce Cantor's formula precisely because they have been deigned that way.

You argument is tantamount to claiming that the set of natural numbers is not given by the step-by-step formula {1, 2, 3, ...}.

Of course you must not cut this sequence at any point. That would make it differ from ℕ. But the complete sequence is ℕ as much as the complete sequence of my matrices reproduces Cantor's indexing. My way however shows, that not all positions of the matrix get indexed by Cantor because the fractions remain without index in every case. That's the billiards-effect.

Regards, WM

Python schrieb am Freitag, 20. Oktober 2023 um 00:23:31 UTC+2:
> Professor Wolfgang Mückenheim, aka WM wrote:

> > Second: I index as many objects as Cantor does by using the same infinite sequence as he does. He claims to index all fractions.
> "It's possible to fail so it is impossible to succeed", what kind
> of logic is that,

Cantor fails to index all positions of the matrix. I succeed in proving that because the fractions will never get an index but they all stay at positions of the matrix.

Regards, WM

On 10/20/2023 1:05 PM, WM wrote:
> Python schrieb am Freitag, 20. Oktober 2023 um 00:23:31 UTC+2:
>> Professor Wolfgang Mückenheim, aka WM wrote:
>
>>> Second: I index as many objects as Cantor does by using the same infinite sequence as he does. He claims to index all fractions.
>> "It's possible to fail so it is impossible to succeed", what kind
>> of logic is that,
>
> Cantor fails to index all positions of the matrix. I succeed in proving that because the fractions will never get an index but they all stay at positions of the matrix.

lol. Are you familiar with Cantor pairing?

on 10/20/2023, WM supposed :
> JVR schrieb am Donnerstag, 19. Oktober 2023 um 20:52:18 UTC+2:
>
>> Cantor's enumeration is not a step-by-step procedure,
>
> Enumeration is always a step-by-step procedure: 1, 2, 3, and so on. step by
> step.

[...]

https://en.wikipedia.org/wiki/Enumeration#Countable_vs._uncountable

On Friday, October 20, 2023 at 10:02:45 PM UTC+2, WM wrote:

> If you deny that the step-by-step procedure produces <bla>

A bijektion is not a "step-by-step procedure", you psychotic asshole full of shit!

On Friday, October 20, 2023 at 10:05:34 PM UTC+2, WM wrote:

> Cantor fails to index all positions of the matrix.

No, he doesn't.

Fritz Feldhase schrieb am Samstag, 21. Oktober 2023 um 14:47:08 UTC+2:
> On Friday, October 20, 2023 at 10:02:45 PM UTC+2, WM wrote:
>
> > If you deny that the step-by-step procedure produces <bla>
>
> A bijektion is not a "step-by-step procedure",

Counting is step-by-step procedure.

But my infinite sequence of matrices represents every pair (k, (m, n)) of the bijection. Could you find one that is not described by my matrices (which simultaneously prove that no fraction is indexed ever) then you had won. But you can't.

Regards, WM

On Saturday, October 21, 2023 at 6:57:49 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Samstag, 21. Oktober 2023 um 14:47:08 UTC+2:
> > On Friday, October 20, 2023 at 10:02:45 PM UTC+2, WM wrote:
> > >
> > > If you deny that the step-by-step procedure produces <bla>
> > >
> > A bijektion is not a "step-by-step procedure",
> >
> Counting is step-by-step procedure.

Indeed! But a bijektion is not based on counting, you psychotic asshole full of shit.

Mückenheim, geh doch bitte mal zum Psychiater und lass Dich behandeln!

On Saturday, October 21, 2023 at 7:06:04 PM UTC+2, WM wrote:
> FromTheRafters schrieb am Samstag, 21. Oktober 2023 um 13:44:13 UTC+2:
> >
> > https://en.wikipedia.org/wiki/Enumeration#Countable_vs._uncountable
> >
> They write there: An enumeration e of a set S with domain IN induces a well-order

Ach?

> When the first billiards are executed,

Wie meinen, Mückenheim? Gehts noch? Piep-Piep?

Wann gehst Du endlich mal zum Psychiater, um Dich behandeln zu lassen?

Fritz Feldhase schrieb am Samstag, 21. Oktober 2023 um 23:22:21 UTC+2:
> On Saturday, October 21, 2023 at 6:57:49 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Samstag, 21. Oktober 2023 um 14:47:08 UTC+2:
> > > On Friday, October 20, 2023 at 10:02:45 PM UTC+2, WM wrote:
> > > >
> > > > If you deny that the step-by-step procedure produces <bla>
> > > >
> > > A bijektion is not a "step-by-step procedure",
> > >
> > Counting is step-by-step procedure.
> Indeed! But a bijektion is not based on counting,

A bijection with ℕ is counting. Did Dedekind use a bijection by closed formula?

Regards, WM

On Sunday, October 22, 2023 at 11:52:39 AM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Samstag, 21. Oktober 2023 um 23:22:21 UTC+2:
> > On Saturday, October 21, 2023 at 6:57:49 PM UTC+2, WM wrote:
> > > Fritz Feldhase schrieb am Samstag, 21. Oktober 2023 um 14:47:08 UTC+2:
> > > > On Friday, October 20, 2023 at 10:02:45 PM UTC+2, WM wrote:
> > > > >
> > > > > If you deny that the step-by-step procedure produces <bla>
> > > > >
> > > > A bijektion is not a "step-by-step procedure",
> > > >
> > > Counting is step-by-step procedure.
> > >
> > Indeed! But a bijektion is not based on counting,
> >
> A bijection with ℕ is counting.

This statement does not make any sense.

On Sunday, October 22, 2023 at 11:52:39 AM UTC+2, WM wrote:

> Did Dedekind [prove the existence of] a bijection by [presenting a] closed formula?

Das ist nicht immer nötig, Mückenheim. Man kann Abzählbarkeit auch auf andere Weise beweisen.

Tipp: Die Vereinigung abzählbar unendlich vieler abzählbar unendlicher Mengen ist abzählbar unendlich.

On 10/20/2023 1:26 PM, Chris M. Thomasson wrote:
> On 10/20/2023 1:05 PM, WM wrote:
>> Python schrieb am Freitag, 20. Oktober 2023 um 00:23:31 UTC+2:
>>> Professor Wolfgang Mückenheim, aka WM wrote:
>>
>>>> Second: I index as many objects as Cantor does by using the same
>>>> infinite sequence as he does. He claims to index all fractions.
>>> "It's possible to fail so it is impossible to succeed", what kind
>>> of logic is that,
>>
>> Cantor fails to index all positions of the matrix. I succeed in
>> proving that because the fractions will never get an index but they
>> all stay at positions of the matrix.
>
> lol. Are you familiar with Cantor pairing?
>

Fwiw, here is some experimental music I made for fun using Cantor pairing:

https://youtu.be/XkwgJt5bxKI

On 10/21/2023 5:47 AM, Fritz Feldhase wrote:
> On Friday, October 20, 2023 at 10:05:34 PM UTC+2, WM wrote:
>
>> Cantor fails to index all positions of the matrix.
>
> No, he doesn't.

Indeed.

On Sunday, October 22, 2023 at 3:50:28 AM UTC-7, Fritz Feldhase wrote:
> On Sunday, October 22, 2023 at 11:52:39 AM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Samstag, 21. Oktober 2023 um 23:22:21 UTC+2:
> > > On Saturday, October 21, 2023 at 6:57:49 PM UTC+2, WM wrote:
> > > > Fritz Feldhase schrieb am Samstag, 21. Oktober 2023 um 14:47:08 UTC+2:
> > > > > On Friday, October 20, 2023 at 10:02:45 PM UTC+2, WM wrote:
> > > > > >
> > > > > > If you deny that the step-by-step procedure produces <bla>
> > > > > >
> > > > > A bijektion is not a "step-by-step procedure",
> > > > >
> > > > Counting is step-by-step procedure.
> > > >
> > > Indeed! But a bijektion is not based on counting,
> > >
> > A bijection with ℕ is counting.
> This statement does not make any sense.

Calculating or counting can't reach the unlimited.
You would be doing them forever instead...

Fritz Feldhase schrieb am Sonntag, 22. Oktober 2023 um 12:50:28 UTC+2:
> On Sunday, October 22, 2023 at 11:52:39 AM UTC+2, WM wrote:

> > A bijection with ℕ is counting.
> This statement does not make any sense.

Try better. Read Cantor.

Dabei nenne ich zwei wohlgeordnete Mengen von demselben Typus und schreibe ihnen gleiche Anzahl zu, wenn sie sich unter Wahrung der festgesetzten Rangordnung ihrer Elemente gegenseitig eindeutig aufeinander abbilden, oder wie man sich gewöhnlich ausdrückt, aufeinander abzählen lassen.

The modern notion is bijection.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 22. Oktober 2023 um 13:01:08 UTC+2:
> On Sunday, October 22, 2023 at 11:52:39 AM UTC+2, WM wrote:
> > Did Dedekind [prove the existence of] a bijection by [presenting a] closed formula?
>
> Das ist nicht immer nötig, Mückenheim. Man kann Abzählbarkeit auch auf andere Weise beweisen.
>
> Tipp: Die Vereinigung abzählbar unendlich vieler abzählbar unendlicher Mengen ist abzählbar unendlich.

Die Vereinigung aller Umordnungen mittels der Matrizen A_i ist eine unendliche Menge. Alle Indizes k, die Cantor in

1, 2, 4, ...
3, 5, 8, ...
6, 9, 13, ...
....

verteilt, werden auch durch die Matrixfolge (A_i) mit Matrixfeldern gepaart: (k, (m, n)).

Gruß, WM

mitchr...@gmail.com schrieb am Montag, 23. Oktober 2023 um 03:30:24 UTC+2:

> Calculating or counting can't reach the unlimited.
> You would be doing them forever instead...

Indeed, but Cantor claims "The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)] He calls this conting the fractions.

Regards, WM

On 10/22/2023 5:52 AM, WM wrote:
> Fritz Feldhase schrieb am Samstag,
> 21. Oktober 2023 um 23:22:21 UTC+2:
>> On Saturday, October 21, 2023
>> at 6:57:49 PM UTC+2, WM wrote:
>>> Fritz Feldhase schrieb am Samstag,
>>> 21. Oktober 2023 um 14:47:08 UTC+2:
>>>> On Friday, October 20, 2023
>>>> at 10:02:45 PM UTC+2, WM wrote:

>>>>> If you deny that
>>>>> the step-by-step procedure
>>>>> produces <bla>
>>>>
>>>> A bijektion is not
>>>> a "step-by-step procedure",
>>>
>>> Counting is step-by-step procedure.
>>
>> Indeed!
>> But a bijektion is not based on counting,
>
> A bijection with ℕ is counting.

A bijection B of set A with ℕ
is a set B ⊆ ℕ×A of pairs ⟨n,x⟩ for
each n ∈ ℕ and each x ∈ A
one match each in B and no others

Set B bijects ℕ and A ⟺
∀n ∈ ℕ, ∀x ∈ A:
∃⟨n,x⟩ ∈ B:
¬∃⟨n,x′⟩ ∈ B: x′ ≠ x ∧
¬∃⟨n′,x⟩ ∈ B: n′ ≠ n

> Did Dedekind use
> a bijection by closed formula?

_Whatever you call a bijection_
can't change that:
for B = {⟨n,p/q⟩| n = (p+q-1)(p+q-2)/2+p }
∀n ∈ ℕ, ∀p/q ∈ ℕ×ℕ:
∃⟨n,p/q⟩ ∈ B:
¬∃⟨n,p′/q′⟩ ∈ B: p/q′ ≠ p/q ∧
¬∃⟨n′,p/q⟩ ∈ B: n′ ≠ n

It is a radically pointless goal which
you are pursuing.
Even if, miraculously, you convinced us
to call _something else_ "bijection",
that miracle would leave the circumstances of
_your actual objection_ untouched.

On Monday, October 23, 2023 at 9:34:14 AM UTC+2, WM wrote:

> Die Vereinigung aller Umordnungen mittels der Matrizen A_i ist eine unendliche Menge.

Mumbo-jumbo.

Definiere erst mal die von Dir verwendeten Begriffe, Mückenheim.

"[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving
precise definitions, his fundamental misunderstanding of elementary mathematical
concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

Du vereinigst also irgendetwas und heraus kommt eine unendliche Menge, ja und?

> Alle Indizes k, die Cantor in
>
> 1, 2, 4, ...
> 3, 5, 8, ...
> 6, 9, 13, ...
> :
>
> verteilt, werden auch durch die Matrixfolge (A_i) mit Matrixfeldern gepaart: (k, (m, n)).

Ja, Deine Fettecke wächst von Matrix zu Matrix, und jetzt?

Hast Du damit die Mengenlehre widerlegt? Schon wieder? :-)

WAS GENAU soll denn Deine lächerliche Folge von Matrizen zeigen? Dass Deine Fettecke(n) niemals den gesamten verfügbaren "Platz" (also die gesamte Matrix) ausfüllen wird/werden? Geschenkt. Wahrlich eine großartige Erkenntnis. :-)

Zur Erinnerung: An e IN {1, ..., n} =/= IN.Aber U_(n e IN) {1, ..., n} = IN.