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tech / sci.math / Re: A game like billards

SubjectAuthor
* A game like billardsWM
+* Re: A game like billardsJVR
|`* Re: A game like billardsWM
| +* Re: A game like billardsJVR
| |`* Re: A game like billardsWM
| | +* Re: A game like billardsFromTheRafters
| | |`- Re: A game like billardsFritz Feldhase
| | `* Re: A game like billardsFritz Feldhase
| |  `* Re: A game like billardsWM
| |   `* Re: A game like billardsFritz Feldhase
| |    `* Re: A game like billardsWM
| |     +* Re: A game like billardsFritz Feldhase
| |     |+* Re: A game like billardsmitchr...@gmail.com
| |     ||`* Re: A game like billardsWM
| |     || `- Re: A game like billardsFritz Feldhase
| |     |`* Re: A game like billardsWM
| |     | +* Re: A game like billardsDieter Heidorn
| |     | |`* Re: A game like billardsWM
| |     | | `* Re: A game like billardsDieter Heidorn
| |     | |  `* Re: A game like billardsWM
| |     | |   `* Re: A game like billardsDieter Heidorn
| |     | |    `* Re: A game like billardsWM
| |     | |     `* Re: A game like billardsDieter Heidorn
| |     | |      `* Re: A game like billardsWM
| |     | |       `* Re: A game like billardsDieter Heidorn
| |     | |        +* Re: A game like billardsWM
| |     | |        |`* Re: A game like billardsDieter Heidorn
| |     | |        | +* Re: A game like billardsWM
| |     | |        | |`* Re: A game like billardsDieter Heidorn
| |     | |        | | `- Re: A game like billardsWM
| |     | |        | `* Re: A game like billardsWM
| |     | |        |  `* Re: A game like billardsDieter Heidorn
| |     | |        |   `- Re: A game like billardsWM
| |     | |        +- Re: A game like billardsFritz Feldhase
| |     | |        `- Re: A game like billardsWM
| |     | `* Re: A game like billardsWM
| |     |  +- Re: A game like billardsFromTheRafters
| |     |  +- Re: A game like billardsJim Burns
| |     |  +* Re: A game like billardsWilliam
| |     |  |`* Re: A game like billardsWM
| |     |  | `* Re: A game like billardsWilliam
| |     |  |  +* Re: A game like billardsWilliam
| |     |  |  |+* Re: A game like billardsWM
| |     |  |  ||+* Re: A game like billardsFromTheRafters
| |     |  |  |||`- Re: A game like billardsFredJeffries
| |     |  |  ||+- Re: A game like billardsChris M. Thomasson
| |     |  |  ||`- Re: A game like billardsChris M. Thomasson
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+* Re: A game like billardsWM
| |     |  |  ||`* Re: A game like billardsJim Burns
| |     |  |  || +* Re: A game like billardsWM
| |     |  |  || |`* Re: A game like billardsJim Burns
| |     |  |  || | `* Re: A game like billardsWM
| |     |  |  || |  +- Re: A game like billardsChris M. Thomasson
| |     |  |  || |  `* Re: A game like billardsJim Burns
| |     |  |  || |   `* Re: A game like billardsWM
| |     |  |  || |    +* Re: A game like billardsJim Burns
| |     |  |  || |    |`* Re: A game like billardsWM
| |     |  |  || |    | +* Re: A game like billardsJim Burns
| |     |  |  || |    | |`* Re: A game like billardsWM
| |     |  |  || |    | | +- Re: A game like billardsChris M. Thomasson
| |     |  |  || |    | | `* Re: A game like billardsJim Burns
| |     |  |  || |    | |  +* Re: A game like billardsWM
| |     |  |  || |    | |  |`* Re: A game like billardsJim Burns
| |     |  |  || |    | |  | +* Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  | |`- Re: A game like billardsJim Burns
| |     |  |  || |    | |  | `* Re: A game like billardsWM
| |     |  |  || |    | |  |  `* Re: A game like billardsJim Burns
| |     |  |  || |    | |  |   `* Re: A game like billardsWM
| |     |  |  || |    | |  |    `- Re: A game like billardsJim Burns
| |     |  |  || |    | |  +* Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  |`* Re: A game like billardsJim Burns
| |     |  |  || |    | |  | +- Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  | +- Re: A game like billardsWM
| |     |  |  || |    | |  | +* Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  | |`* Re: A game like billardsJim Burns
| |     |  |  || |    | |  | | +- Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  | | +* Re: A game like billardsWM
| |     |  |  || |    | |  | | |`* Re: A game like billardsJim Burns
| |     |  |  || |    | |  | | | +* Re: A game like billardsWM
| |     |  |  || |    | |  | | | |+* Re: A game like billardsJim Burns
| |     |  |  || |    | |  | | | ||`- Re: A game like billardsFredJeffries
| |     |  |  || |    | |  | | | |+- Re: A game like billardsFromTheRafters
| |     |  |  || |    | |  | | | |`- Re: A game like billardsJim Burns
| |     |  |  || |    | |  | | | +- Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  | | | +- Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  | | | `- Re: A game like billardsWolfgang
| |     |  |  || |    | |  | | +- Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  | | `* Re: A game like billardsWM
| |     |  |  || |    | |  | +- Re: A game like billardsWM
| |     |  |  || |    | |  | `- Re: A game like billardsFritz Feldhase
| |     |  |  || |    | |  +- Re: A game like billardsWM
| |     |  |  || |    | |  `* Re: A game like billardsFritz Feldhase
| |     |  |  || |    | `- Re: A game like billardsChris M. Thomasson
| |     |  |  || |    `- Students: You can only understand calculus from me. No one else canEram semper recta
| |     |  |  || `- Re: A game like billardsFritz Feldhase
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+* Re: A game like billardsWM
| |     |  |  |+* Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+* Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+* Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsFritz Feldhase
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+* Re: A game like billardsFritz Feldhase
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+* Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsFritz Feldhase
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsFredJeffries
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWM
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsFredJeffries
| |     |  |  |+* Re: A game like billardsWolfgang
| |     |  |  |+- Re: A game like billardsWolfgang
| |     |  |  |+* So it is my last post hereAmine Moulay Ramdane
| |     |  |  |+- Re: A game like billardsFritz Feldhase
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsFredJeffries
| |     |  |  |+- Re: A game like billardsFredJeffries
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+- Re: A game like billardsWolfgang
| |     |  |  |+- Re: A game like billardsFritz Feldhase
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |+* Re: A game like billardsGanzhinterseher
| |     |  |  |+* Re: A game like billardsGanzhinterseher
| |     |  |  |+* Re: A game like billardsGanzhinterseher
| |     |  |  |+- Re: A game like billardsFredJeffries
| |     |  |  |+- Re: A game like billardsFredJeffries
| |     |  |  |+- Re: A game like billardsWilliam
| |     |  |  |`* Re: A game like billardsTransfinity
| |     |  |  `* How would a singularity explode...?mitchr...@gmail.com
| |     |  +- Re: A game like billardsFritz Feldhase
| |     |  +- Re: A game like billardsFritz Feldhase
| |     |  +- Re: A game like billardsFritz Feldhase
| |     |  `* Re: A game like billardsFritz Feldhase
| |     +* Re: A game like billardsFritz Feldhase
| |     +* Re: A game like billardsJim Burns
| |     `- Re: A game like billardsJim Burns
| `* Re: A game like billardsPython
+- Re: A game like billardsChris M. Thomasson
`- Re: A game like billardsAdam Polak

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Re: A game like billards

<68b91404-311b-2146-a6f4-9d6dea4d9ddd@att.net>

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From: james.g....@att.net (Jim Burns)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Wed, 25 Oct 2023 11:06:44 -0400
Organization: A noiseless patient Spider
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 by: Jim Burns - Wed, 25 Oct 2023 15:06 UTC

On 10/25/2023 6:30 AM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 24. Oktober 2023 um 17:47:27 UTC+2:

>> B does not have two ends.
>> Procedures have two ends.
>
> No.

P is the ordered set of steps sⱼ of a process.

Steps sₛ‖sₑ exist as start‖end of P
and,
for each split F ᣔ≪ᣔ H of P
some sᵢ‖sᵢ₊₁ exist as end-of-F‖start-of-H

> The procedure accomplishes,
> according to Cantor,
> "that every element of the set stands at
> a definite position of this sequence".

There aren't two ends.

Each fraction p/q stands at
the definite position n
n = (s-1)(s-2)/2+p
s = p+q

Each position n is where
the definite fraction p/q stands.
p = n-(s-1)(s-2)/2
q = s-p
s = ⌈(((8⋅n+1)¹ᐟ²+1)/2)⌉

And there aren't two ends.

>> B is not a procedure, however,
>> counting to any ⟨n,xₙ⟩ in B
>> has two ends, and is a procedure.
>
> Not according to Cantor:
> "such that
> every element of the set stands at
> a definite position of this sequence".

According to arithmetic,
there aren't two ends.

Each {p}×ℕ is followed by {p⁺¹}×ℕ

Each ℕ×{q} is followed by ℕ×{q⁺¹}

Each p/q is followed by
p⁺¹/q⁻¹ if q ≠ 1
1/p⁺¹ if q = 1

A second end is not followed, thus
no p/q no ℕ×{q} no {p}×ℕ
is a second end.

Your choices are
to reject two-ended-ness or
to reject arithmetic.

Re: A game like billards

<uhbb0t$r2a9$1@dont-email.me>

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From: FTR...@nomail.afraid.org (FromTheRafters)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Wed, 25 Oct 2023 11:08:10 -0400
Organization: Peripheral Visions
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 by: FromTheRafters - Wed, 25 Oct 2023 15:08 UTC

WM presented the following explanation :
> Fritz Feldhase schrieb am Dienstag, 24. Oktober 2023 um 21:26:05 UTC+2:
>> On Tuesday, October 24, 2023 at 1:52:19 PM UTC+2, WM wrote:
>>> Fritz Feldhase schrieb am Montag, 23. Oktober 2023 um 18:02:11 UTC+2:
>>>>
>>>> 2. A bijection is not a process.
>> A BIJECTION IS NOT A PROCESS.
>
> A bijection with |N can be executed as a process.

But that doesn't mean it *is* a process.

Re: A game like billards

<uhbb8k$r4ei$1@dont-email.me>

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https://www.novabbs.com/tech/article-flat.php?id=151219&group=sci.math#151219

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Path: i2pn2.org!i2pn.org!eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: FTR...@nomail.afraid.org (FromTheRafters)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Wed, 25 Oct 2023 11:12:17 -0400
Organization: Peripheral Visions
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 by: FromTheRafters - Wed, 25 Oct 2023 15:12 UTC

It happens that WM formulated :
> Fritz Feldhase schrieb am Dienstag, 24. Oktober 2023 um 21:26:05 UTC+2:
>> On Tuesday, October 24, 2023 at 1:52:19 PM UTC+2, WM wrote:
>>> Fritz Feldhase schrieb am Montag, 23. Oktober 2023 um 18:02:11 UTC+2:
>>>>
>>>> 2. A bijection is not a process.
>> A BIJECTION IS NOT A PROCESS.
>
> A bijection with |N can be executed as a process. "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)]
>>
>>> Every pair of a bijection with IN can be counted to. This [counting] is a
>>> process [in each case].
>>
>> YES, EVER PAIR OF A BIJECTION CAN BE COUNTED TO. THIS COUNTING IS A PROCESS
>> (in each case).
>>
>> BUT NO PAIR OF A BIJECTION IS THIS BIJECTION, _THE BIJECTION CANNOT BE
>> COUNTED TO_.
>
> Irrelevant. We count to every place of the matrix, we do not count to the
> matrix.
>>
>> *MOREOVER*: THE BIJECTION AS WELL AS EACH AND EVERY PAIR OF THE BIJECTION IS
>> NOT A PROCESS.
>
> Each and every pair is: "The infinite sequence thus defined [...]

Each and every pair is a very short sequence.

Re: A game like billards

<71b0241b-e335-4fe4-bad8-8ebb7dc61eecn@googlegroups.com>

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Subject: Re: A game like billards
From: wpihug...@gmail.com (William)
Injection-Date: Wed, 25 Oct 2023 18:19:37 +0000
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 by: William - Wed, 25 Oct 2023 18:19 UTC

On Wednesday, October 25, 2023 at 11:56:04 AM UTC-3, WM wrote:
> William schrieb am Mittwoch, 25. Oktober 2023 um 14:27:06 UTC+2:
> > On Wednesday, October 25, 2023 at 6:57:34 AM UTC-3, WM wrote:
> > > William schrieb am Dienstag, 24. Oktober 2023 um 20:13:54 UTC+2:
> > > > A bijection with |N, B. is a set
> > > > > Every pair of a bijection with |N can be counted to.
> > > > Indeed, every element of B can be "counted to".
> > > > B cannot be "counted to".
> > > B is countable
> > So What? B cannot be "counted to"
> Who wants to count to [B]?
Anyone who wants to claim that B is the result of a step.

> Every pair (n, x) of a bijection that cannot be proven to be the result of step n with n the last element of a FISON,

There is no such pair. The fact that you cannot count to B does not imply there is such a pair.

Re: A game like billards

<kpt8r0Fk2dbU1@mid.individual.net>

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From: d.heid...@t-online.de (Dieter Heidorn)
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Subject: Re: A game like billards
Date: Wed, 25 Oct 2023 21:32:49 +0200
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 by: Dieter Heidorn - Wed, 25 Oct 2023 19:32 UTC

WM schrieb:
> Dieter Heidorn schrieb am Dienstag, 24. Oktober 2023 um 17:49:00 UTC+2:
>> WM schrieb:
>>
>>>>> Man *kann* jedes Paar konstruieren. Was nicht konstruiert werden kann, gehört nicht zur Abzählung. Bist Du gegenteiliger Meinung?
>>>> Selbstverständlich.
>>>
>>> O, Du proklamierst dunkle Zahlen.
>> Selbstverständlich nicht. Ich "proklamiere" nur mathematisch existente
>> Objekte.
> Was nicht konstruiert werden kann, gehört nicht zur Abzählung. Mathematisch existente
> Objekte, hier also Paare, können konstruiert werden. "such that every element of the set stands at a definite position of this sequence".
>

Und an welcher Position einer Folge sie stehen, hängt von der
verwendeten Zuordnungsvorschrift ab. Dir scheint Cantors Aussage
offensichtlich nicht zu gefallen, da sie deine Schwurbelei vom
"schrittweisen konstruieren" zurückweist:

|"[Es] wird bewiesen, daß äquivalente Mengen immer eine und dieselbe
| Mächtigkeit oder Kardinalzahl haben und daß auch umgekehrt Mengen
| von derselben Kardinalzahl äquivalent sind. [...]
| Die Kenntnis nur eines /Zuordnungsgesetzes/ für zwei Mengen M und M_1
| /genügt/, um die Äquivalenz derselben zu konstatieren; doch gibt es
| immer viele, im allgemeinen sogar unzählig viele Zuordnungsgesetze,
| durch welche zwei äquivalente in gegenseitig eindeutige und
| vollständige Beziehung zueinander gebracht werden können."
(Cantor: Gesammelte Abhandlungen, S.413)

Kein Wunder also, dass du es immer wieder fort lässt...

>> Bijektionen haben - wie alle Funktionen - eine Definitionsmenge
>> und eine Wertemenge. Die Definition einer Bijektion zwischen zwei Mengen
>> stellt einen Zusammenhang zwischen zwei definierten Mengen her - und
>> dunkel ist daran nicht das Geringste.
>
> Was nicht konstruiert werden kann, gehört nicht zur Abzählung. Das hast Du abgelehnt.

Abgelehnt habe ich deine irrige Vorstellung, dass man bei einer
Bijektion sämtliche Paare "schrittweise konstruieren" müsse, um die
Bijektion zu "überprüfen". Eine Bijektion ist kein "Prozess".
Damit lehne ich ebenfalls deine idiotische Aussage ab, dass - da man
dies bei unendlichen Mengen nicht durchführen kann - es "dunkle Zahlen"
geben müsse.
In der Mathematik gilt: Eine Bijektion zwischen zwei unendlichen Mengen
ist vollständig beschrieben durch Angabe von Definitionsmenge,
Wertemenge und Funktionsvorschrift ("Zuordnungsgesetz").

>> Damit ist die "Konstruktion" aller Paare angegeben. Die Cantorsche
>> Paarungsfunktion ist ein Beispiel für eine solche Bijektion.
>
> Und die führe ich aus, so dass alle O in der Matrix bleiben.

Du führst die Cantorsche Paarungsfunktion nicht aus, sondern führst in
einer Darstellung der Definitionsmenge Verschiebungen und Vertauschungen
der Elemente aus, und hältst es für für eine sensationelle Entdeckung,
dass dabei keine Darstellung der Wertemenge der Funktion entsteht.
Aber wenn man - so wie du - Bijektionen und unendliche Mengen nicht
versteht, dann kommt eben so ein Unfug heraus.

>Damit ist die Existenz nicht nummerierbarer Felder und damit auch die Unvollständigkeit von Cantors "Bijektion" gezeigt.

Ganz gewiss nicht, da dein Unfug nichts mit der Cantorschen
Paarungsfunktion zu tun hat.

>>> "Werden nun die Zahlen p/q in einer solchen Reihenfolge gedacht, [...] so kommt jede Zahl p/q an eine ganz bestimmte Stelle einer einfach unendlichen Reihe," [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) S. 126] Natürlich kann man jede ganz bestimmte Stelle bestimmen. Sie ist ja schon bestimmt.
>> Solche Zitatschnipsel besagen gar nichts.
>> Dieser Satz steht in folgendem
>> Zusammenhang:

.... den du wohlweislich unterschlagen hast. Fakten und klare Aussagen
sind eben nicht so dein Ding...

Cantor betrachtet hier positive Brüche p/q mit den Eigenschaften

* p und q sind relativ prim,
* p/q ∈ [0,1]

und stellt eine Zuordnung her:

* p/q -> p + q = N .

die _nicht injektiv_ ist - und damit ist sie natürlich auch nicht
bijektiv.

>> * p/q -> p + q = N .
>>
>> die _nicht injektiv_ ist - und damit ist sie natürlich auch nicht
>> bijektiv.
>>
>> Ein Zusammenhang mit der bijektiven Cantor-Funktion zur Abzählung der
>> Menge der positiven Brüche besteht nicht.
>
> Er besteht natürlich,

* Eine nicht injektive Zuordnung steht in keinem Zusammenhang mit der
bijektiven Cantorschen Paarungsfunktion zur Abzählung der Menge 𝔹+
der positiven Brüche.

* Eine Zuordnung, deren Definitionsmenge die rationalen Zahlen p/q
(p und q relativ prim) im Intervall [0,1] ist, steht nicht im
Zusammenhang mit der bijektiven Cantor-Funktion zur Abzählung der
Menge 𝔹+ der positiven Brüche, deren Definitionsmenge die positiven
Brüche m/n, interpretiert als Paar (m,n), ist.

> "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)]

Dein extrem verkürztes Zitieren grenzt langsam an Fälschung.

In dem Brief an Lipschitz gibt Cantor den Anfang einer Aufzählung der
positiven rationalen Zahlen an, wie zu erkennen ist, wenn man etwas mehr
zitiert:

|"Gestatten Sie mir bei dieser Gelegenheit, Ihnen eine specielle Frage
| vorzulegen. Folgende Reihe von rationalen Zahlen erscheint mir sehr
| merkwürdig:
| | 1/1; 1/2, 2/1; 1/3, 3/1; 1/4, 2/3, 3/2, 4/1; 1/5, 5/1;
| 1/6, 2/5, 3/4, 4/3, 5/2, 6/1; 1/7, 3/5, 5/3, 7/1;
| 1/8, 2/7, 4/5, 5/4, 7/2, 8/1; 1/9, 3/7, 7/3, 9/1; etc.
| | Das Gesetz dieser Reihe ist ein höchst einfaches: Sie sehen, dass die
| Reihe nach gewissen Abschnitten fortschreitet, von denen jeder
| zwischen zwei ;; eingeschlossen ist. Der erste Abschnitt enthält
| φ(2) = 1, der zweite φ(3) = 2, der (n-1)-te Abschnitt φ(n) Zahlen, wo
| φ(n) die Anzahl aller relativen Primzahlen zu n, die kleiner als n
| sind, bestimmt.
| Innerhalb des (n-1)ten Abschnittes bilden die Zähler der rationalen
| Zahlen die aufsteigende Reihe der φ(n) Zahlen rel. prim zu n und
| kleiner als n, die Nenner die absteigende Reihe derselben φ(n) Zahlen.
| Die so definirte unendliche Reihe hat nun das merkwürdige an sich,
| sämmtliche positiven rationalen Zahlen und jede von ihnen nur einmal
| an einer bestimmten Stelle zu enthalten. Bezeichnet man die Glieder
| jener Reihe mit
| | F(1), F(2), F(3), ... , F(ν), ...
| so daß:
| F(1) = 1; F(2) = 1/2; F(3) = 2/1; F(4) = 1/3; F(5) = 3/1; u.s.w.
| so ist F(ν) eine zahlentheoretische Function, welche wenn ν alle
| positiven ganzen Zahlen durchläuft, ihrerseits alle positiven
| rationalen Zahlenwerthe und jeden nur einmal annimmt.
| Liesse sich nicht mit den Mitteln der analytischen Zahlentheorie
| (Ausdrucksweise von Mertens) ein analytischer Ausdruck für die
| Function F(ν) finden?"

Auch hier betrachtet Cantor positive Brüche p/q, bei denen p und q
relativ prim sind, also die Menge ℚ+. Die von ihm gesuchte Funktion F
würde eine Abzählung dieser Brüche darstellen:

F: ℕ → ℚ+ .

Natürlich kann man auch hier wieder zur Veranschaulichung mit
Matrixdarstellungen M(m,n) arbeiten wie bei der Cantorschen
Paarungsfunktion zur Abzählung der Menge 𝔹+ der positiven Brüche.
Wie dort liest man die Indizes m,n der Zeilen und Spalten als Zähler und
Nenner der vollständig gekürzten positiven Brüche, also der positiven
rationalen Zahlen m/n.

M(m,n):

1 2 3 4 5 6 7 8 9 n
|---|----|----|----|----|----|----|----|----|---
|
1- 1/1 1/2 1/3 1/4 1/5 1/6 1/7 1/8 1/9 ...
|
2- 2/1 - 2/3 - 2/5 - 2/7 - 2/9 ...
|
3- 3/1 3/2 - 3/4 3/5 - 3/7 3/8 3/9 ...
|
4- 4/1 - 4/3 - 4/5 - 4/7 - 4/9 ...
|
5- 5/1 5/2 5/3 5/4 - 5/6 5/7 5/8 5/9 ...
|
6- 6/1 - - - 6/5 - 6/7 - - ...
|
7- 7/1 7/2 7/3 7/4 7/5 7/6 - 7/8 7/9 ...
|
8- 8/1 - 8/3 - 8/5 - 8/7 - 8/9 ...
|
9- 9/1 9/2 - 9/4 9/5 - 9/7 9/8 - ...
|
| ...
|
m

Diese Matrix stellt einen Ausschnitt aus der Wertemenge der gesuchten
Zuordnung F: ℕ → ℚ+ dar.

In einer zweiten Matrix L(m,n) kann man die Positionsnummern der
rationalen Zahlen in Cantors angefangener Reihe eintragen:

1/1; 1/2, 2/1; 1/3, 3/1; 1/4, 2/3, 3/2, 4/1; 1/5, 5/1;
1 2 3 4 5 6 7 8 9 10 11

1/6, 2/5, 3/4, 4/3, 5/2, 6/1; 1/7, 3/5, 5/3, 7/1;
12 13 14 15 16 17 18 19 20 21

1/8, 2/7, 4/5, 5/4, 7/2, 8/1; 1/9, 3/7, 7/3, 9/1;
22 23 24 25 26 27 28 29 30 31

L(m,n):

1 2 3 4 5 6 7 8 9 n
|---|----|----|----|----|----|----|----|----|---
|
1- 1 2 4 6 10 12 18 22 28 ...
|
2- 3 - 7 - 13 - 23 - ...
|
3- 5 8 - 14 19 - 29 ...
|
4- 9 - 15 - 24 - - ...
|
5- 11 16 20 25 - ...
|
6- 17 - - - - - - ...
|
7- 21 26 30 ...
|
8- 27 - - - ...
|
9- 31 - - - ...
|
| ...
|
m


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Re: A game like billards

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Subject: Re: A game like billards
From: askaske...@gmail.com (WM)
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 by: WM - Thu, 26 Oct 2023 16:12 UTC

Jim Burns schrieb am Mittwoch, 25. Oktober 2023 um 17:06:55 UTC+2:
> On 10/25/2023 6:30 AM, WM wrote:

> > The procedure accomplishes,
> > according to Cantor,
> > "that every element of the set stands at
> > a definite position of this sequence".
> There aren't two ends.

There is a state where no fraction is enumerated, and there is a state wher all fractions are enumerated, according to Cantor. These are two ends. (Same with unit fractions: The two ends are 1 and 0.)
>
> And there aren't two ends.

Then all O remain forever in the matrix.

> > Not according to Cantor:
> > "such that
> > every element of the set stands at
> > a definite position of this sequence".
> According to arithmetic,
> there aren't two ends.

That's why arithmetic is incompatible with actual infinity. Arithmetic is compatible with constructivism: "(1) Start with I. (2) When x is reached, add I. [...] These rules supply a constructive definition of numbers (namely their scheme of construction). Now we can immediately say that according to these rules infinitely many numbers are possible. One has to be aware of the fact that here only the possibility is asserted – and this is secured by the rule itself." [Paul Lorenzen: "Das Aktual-Unendliche in der Mathematik", Philosophia naturalis 4 (1957) 3-11]

Regards, WM

Re: A game like billards

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Subject: Re: A game like billards
From: askaske...@gmail.com (WM)
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 by: WM - Thu, 26 Oct 2023 16:15 UTC

FromTheRafters schrieb am Mittwoch, 25. Oktober 2023 um 17:08:24 UTC+2:
> WM presented the following explanation :
> > Fritz Feldhase schrieb am Dienstag, 24. Oktober 2023 um 21:26:05 UTC+2:
> >> On Tuesday, October 24, 2023 at 1:52:19 PM UTC+2, WM wrote:
> >>> Fritz Feldhase schrieb am Montag, 23. Oktober 2023 um 18:02:11 UTC+2:
> >>>>
> >>>> 2. A bijection is not a process.
> >> A BIJECTION IS NOT A PROCESS.
> >
> > A bijection with |N can be executed as a process.

According to Cantor. Of course not in reality.

> But that doesn't mean it *is* a process.

A process is what can be executed as a process.

Regards, WM

Re: A game like billards

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Subject: Re: A game like billards
From: askaske...@gmail.com (WM)
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 by: WM - Thu, 26 Oct 2023 16:21 UTC

William schrieb am Mittwoch, 25. Oktober 2023 um 20:19:42 UTC+2:
> On Wednesday, October 25, 2023 at 11:56:04 AM UTC-3, WM wrote:

> > Every pair (n, x) of a bijection that cannot be proven to be the result of step n with n the last element of a FISON,

> There is no such pair.

Because dark elements cannot be paired. But there are many dark natnumbers and fractions which cannot be paired. The O indicate not paired fractions and remain in all possible steps and do not vanish in the limit.

Regards, WM

Re: A game like billards

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Subject: Re: A game like billards
From: askaske...@gmail.com (WM)
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 by: WM - Thu, 26 Oct 2023 16:33 UTC

Dieter Heidorn schrieb am Mittwoch, 25. Oktober 2023 um 21:32:59 UTC+2:
> WM schrieb:
> > Dieter Heidorn schrieb am Dienstag, 24. Oktober 2023 um 17:49:00 UTC+2:
> >> WM schrieb:
> >>
> >>>>> Man *kann* jedes Paar konstruieren. Was nicht konstruiert werden kann, gehört nicht zur Abzählung. Bist Du gegenteiliger Meinung?
> >>>> Selbstverständlich.
> >>>
> >>> O, Du proklamierst dunkle Zahlen.
> >> Selbstverständlich nicht. Ich "proklamiere" nur mathematisch existente
> >> Objekte.
> > Was nicht konstruiert werden kann, gehört nicht zur Abzählung.. Mathematisch existente
> > Objekte, hier also Paare, können konstruiert werden. "such that every element of the set stands at a definite position of this sequence".
> >
> Und an welcher Position einer Folge sie stehen, hängt von der
> verwendeten Zuordnungsvorschrift ab.

Hier geht es nur um eine einzige, nämlich diese:
k = (m + n - 1)(m + n - 2)/2 + m.

> Dir scheint Cantors Aussage
> offensichtlich nicht zu gefallen, da sie deine Schwurbelei vom
> "schrittweisen konstruieren" zurückweist:

Was nach dieser Zuordnungsvorschrift nicht konstruiert werden kann, ist dunkel, denn alles andere kann konstruiert werden, egal was Cantor sonst noch sagt.

> > Was nicht konstruiert werden kann, gehört nicht zur Abzählung.. Das hast Du abgelehnt.
> Abgelehnt habe ich deine irrige Vorstellung, dass man bei einer
> Bijektion sämtliche Paare "schrittweise konstruieren" müsse,

Nicht müsse, sondern könne. Was nach dieser Zuordnungsvorschrift nicht konstruiert werden kann, ist dunkel, denn alles andere kann konstruiert werden.

> In der Mathematik gilt: Eine Bijektion zwischen zwei unendlichen Mengen
> ist vollständig beschrieben durch Angabe von Definitionsmenge,
> Wertemenge und Funktionsvorschrift ("Zuordnungsgesetz").

Ja, aber was nach dieser Zuordnungsvorschrift nicht konstruiert werden kann, ist dunkel, denn alles andere kann konstruiert werden.

> >> Damit ist die "Konstruktion" aller Paare angegeben. Die Cantorsche
> >> Paarungsfunktion ist ein Beispiel für eine solche Bijektion.
> >
> > Und die führe ich aus, so dass alle O in der Matrix bleiben.
> Du führst die Cantorsche Paarungsfunktion nicht aus, sondern führst in
> einer Darstellung der Definitionsmenge Verschiebungen und Vertauschungen
> der Elemente aus,

die genau Cantors Vorschrift reproduzieren.

> >Damit ist die Existenz nicht nummerierbarer Felder und damit auch die Unvollständigkeit von Cantors "Bijektion" gezeigt.
> Ganz gewiss nicht, da dein Unfug nichts mit der Cantorschen
> Paarungsfunktion zu tun hat.

Entweder bist Du unfähig einzusehen, dass ich genau Cantors Vorschrift befolge, oder unwillig. Beides ändert nichts daran, dass Du keinen Schritt angeben kannst, der von Cantors Formel k = (m + n - 1)(m + n - 2)/2 + m abweicht.

Also finde einen solchen, oder schweige. Reine gegenteilige Behauptungen Deinerseits werde ich nicht akzeptieren.

Gruß, WM

Re: A game like billards

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Subject: Re: A game like billards
From: wpihug...@gmail.com (William)
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 by: William - Thu, 26 Oct 2023 16:37 UTC

On Thursday, October 26, 2023 at 1:21:23 PM UTC-3, WM wrote:
> William schrieb am Mittwoch, 25. Oktober 2023 um 20:19:42 UTC+2:
> > On Wednesday, October 25, 2023 at 11:56:04 AM UTC-3, WM wrote:
>
> > > Every pair (n, x) of a bijection that cannot be proven to be the result of step n with n the last element of a FISON,
>
> > There is no such pair.
> Because dark elements cannot be paired. But there are many dark natnumbers and fractions which cannot be paired. The O indicate not paired fractions and remain in all possible steps

Correct, you cannot get B using steps (you cannot "count to B"). This does not mean B does not exist.

Re: A game like billards

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Subject: Re: A game like billards
From: askaske...@gmail.com (WM)
Injection-Date: Thu, 26 Oct 2023 16:52:52 +0000
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 by: WM - Thu, 26 Oct 2023 16:52 UTC

William schrieb am Donnerstag, 26. Oktober 2023 um 18:37:09 UTC+2:
> On Thursday, October 26, 2023 at 1:21:23 PM UTC-3, WM wrote:
> > The O indicate not paired fractions and remain in all possible steps
> Correct, you cannot get B using steps (you cannot "count to B"). This does not mean B does not exist.

B, the matrix
1, 2, 4, ...
3, 5, 8, ...
6, 9,13, ...
....
containing only natural numbers does exist. It can be defined. But if you start from 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, ...
....
B cannot be reached by the game of billiards without having all fractions in the dark parts of this matrix, surrounding the natnumbers. The game of billiards however reproduces Cantor's enumeration. Defining B has nothing to do with enumerating the fractions.

Regards, WM

Re: A game like billards

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Subject: Re: A game like billards
From: fredjeff...@gmail.com (FredJeffries)
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 by: FredJeffries - Thu, 26 Oct 2023 17:07 UTC

On Thursday, October 26, 2023 at 9:12:29 AM UTC-7, WM wrote:
> Jim Burns schrieb am Mittwoch, 25. Oktober 2023 um 17:06:55 UTC+2:
> > On 10/25/2023 6:30 AM, WM wrote:
>
> > > The procedure accomplishes,
> > > according to Cantor,
> > > "that every element of the set stands at
> > > a definite position of this sequence".
> > There aren't two ends.
> There is a state where no fraction is enumerated, and there is a state wher all fractions are enumerated, according to Cantor. These are two ends. (Same with unit fractions: The two ends are 1 and 0.)
> >
> > And there aren't two ends.
> Then all O remain forever in the matrix.

That is correct

Congratulations

Re: A game like billards

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From: james.g....@att.net (Jim Burns)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Thu, 26 Oct 2023 13:47:45 -0400
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 by: Jim Burns - Thu, 26 Oct 2023 17:47 UTC

On 10/26/2023 12:12 PM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 25. Oktober 2023 um 17:06:55 UTC+2:
>> On 10/25/2023 6:30 AM, WM wrote:

>>> The procedure accomplishes,
>>> according to Cantor,
>>> "that every element of the set stands at
>>> a definite position of this sequence".
>>
>> There aren't two ends.
>
> There is a state where
> no fraction is enumerated,
> and there is a state [where]
> all fractions are enumerated, according to Cantor.
> These are two ends.

for each split F ᣔ≪ᣔ H of ordered process-set P
some sᵢ‖sᵢ₊₁ exist as end-of-F‖start-of-H

The enumeration of all the fractions
n = p+(p+q-1)(p+q-2)/2
is not a process

Either
the all-enumerated state sₑ is not-in P
or
not all splits F ᣔ≪ᣔ H have
end-of-F‖start-of-H states.

Either way, enumerating the fractions
is not a process.

That is not a proof of dark states.
Even with dark states, it is not a process.
Even without dark states, it is not a process.

It is the nature of the accessible, palpable,
olfactible states that
each is followed by another accessible, palpable,
olfactible state.

The split F ᣔ≪ᣔ H between
all accessible, palpable, olfactible states
and
anything (if anything) after
all accessible, palpable, olfactible states
cannot have a last-in-F
because
there is no last accessible, palpable,
olfactible state.

The existence or non-existence of
non-accessible non-palpable non-olfactible
states after them doesn't change them.

> (Same with unit fractions:
> The two ends are 1 and 0.)

0 is not an end of the unit fractions.
0 is a lower bound of the unit fractions,
the greatest lower bound.

In order to be an end,
0 needs to be a unit fraction,
which 0 isn't.

>> And there aren't two ends.
>
> Then all O remain forever in the matrix.

There is no process which
moves 1×1 infinitely-many
accessible, palpable, olfactible Os
in finitely-many steps.

Not because of dark Os
Because of each-not-last
accessible, palpable, olfactible Os

>>> Not according to Cantor:
>>> "such that
>>> every element of the set stands at
>>> a definite position of this sequence".
>>
>> According to arithmetic,
>> there aren't two ends.
>
> That's why arithmetic is incompatible with
> actual infinity.

Arithmetic is incompatible with
ineffably-self-equal things.
In arithmetic, _it can be said_
of each thing that it equals itself.

Your actual-infinityᵂᴹ requires
ineffably-self-equal things.

Your actually-infiniteᵂᴹ A cannot biject with
any proper subset S ≠⊂ A, |S| ≠ |A|
but
it contains a subset B ≠⊂ A which _can_
biject with a proper subset S ≠⊂ B, |S| = |B|

That's only possible if
the remainder A\B cannot biject with itself.
That requires ineffably-self-equal things.

Yes.
Arithmetic is incompatible with
your actual-infinityᵂᴹ
That's not a problem.
Being compatible would be a problem.

Re: A game like billards

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From: FTR...@nomail.afraid.org (FromTheRafters)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Thu, 26 Oct 2023 14:00:46 -0400
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 by: FromTheRafters - Thu, 26 Oct 2023 18:00 UTC

WM wrote on 10/26/2023 :
> William schrieb am Mittwoch, 25. Oktober 2023 um 20:19:42 UTC+2:
>> On Wednesday, October 25, 2023 at 11:56:04 AM UTC-3, WM wrote:
>
>>> Every pair (n, x) of a bijection that cannot be proven to be the result of
>>> step n with n the last element of a FISON,
>
>> There is no such pair.
>
> Because dark elements cannot be paired.

Of course, since we don't do pairing with things that do not exist.

If we could enumerate them, there would be a first nonexistent and a
next nonexistent thing and so on -- then we could pair them.

Re: A game like billards

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From: FTR...@nomail.afraid.org (FromTheRafters)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Thu, 26 Oct 2023 14:05:41 -0400
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 by: FromTheRafters - Thu, 26 Oct 2023 18:05 UTC

WM presented the following explanation :
> FromTheRafters schrieb am Mittwoch, 25. Oktober 2023 um 17:08:24 UTC+2:
>> WM presented the following explanation :
>>> Fritz Feldhase schrieb am Dienstag, 24. Oktober 2023 um 21:26:05 UTC+2:
>>>> On Tuesday, October 24, 2023 at 1:52:19 PM UTC+2, WM wrote:
>>>>> Fritz Feldhase schrieb am Montag, 23. Oktober 2023 um 18:02:11 UTC+2:
>>>>>>
>>>>>> 2. A bijection is not a process.
>>>> A BIJECTION IS NOT A PROCESS.
>>>
>>> A bijection with |N can be executed as a process.
>
> According to Cantor. Of course not in reality.
>
>> But that doesn't mean it *is* a process.
>
> A process is what can be executed as a process.

Explain.

Re: A game like billards

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Subject: Re: A game like billards
From: wpihug...@gmail.com (William)
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 by: William - Thu, 26 Oct 2023 18:24 UTC

On Thursday, October 26, 2023 at 1:52:57 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. Oktober 2023 um 18:37:09 UTC+2:
> > On Thursday, October 26, 2023 at 1:21:23 PM UTC-3, WM wrote:
> > > The O indicate not paired fractions and remain in all possible steps
> > Correct, you cannot get B using steps (you cannot "count to B"). This does not mean B does not exist.
> B, the matrix
> 1, 2, 4, ...
> 3, 5, 8, ...
> 6, 9,13, ...
> ...
> containing only natural numbers does exist.

Correct, a bijection between |N and |N x |N exists.

Re: A game like billards

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Subject: Re: A game like billards
From: fredjeff...@gmail.com (FredJeffries)
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 by: FredJeffries - Thu, 26 Oct 2023 18:44 UTC

On Thursday, October 26, 2023 at 11:01:03 AM UTC-7, FromTheRafters wrote:
> WM wrote on 10/26/2023 :
> > William schrieb am Mittwoch, 25. Oktober 2023 um 20:19:42 UTC+2:
> >> On Wednesday, October 25, 2023 at 11:56:04 AM UTC-3, WM wrote:
> >
> >>> Every pair (n, x) of a bijection that cannot be proven to be the result of
> >>> step n with n the last element of a FISON,
> >
> >> There is no such pair.
> >
> > Because dark elements cannot be paired.

> Of course, since we don't do pairing with things that do not exist.

Ridiculous. We do it all the time:

Juliet <--> Romeo
Ophelia <--> Hamlet
Katherine <--> Petruchio
Lena Lamont <--> Don Lockwood
....

Re: A game like billards

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From: d.heid...@t-online.de (Dieter Heidorn)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Thu, 26 Oct 2023 21:41:44 +0200
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 by: Dieter Heidorn - Thu, 26 Oct 2023 19:41 UTC

WM schrieb:
> Dieter Heidorn schrieb am Mittwoch, 25. Oktober 2023 um 21:32:59 UTC+2:
>> WM schrieb:
>>> Dieter Heidorn schrieb am Dienstag, 24. Oktober 2023 um 17:49:00 UTC+2:
>>>> WM schrieb:
>>>>
>>>>>>> Man *kann* jedes Paar konstruieren. Was nicht konstruiert werden kann, gehört nicht zur Abzählung. Bist Du gegenteiliger Meinung?
>>>>>> Selbstverständlich.
>>>>>
>>>>> O, Du proklamierst dunkle Zahlen.
>>>> Selbstverständlich nicht. Ich "proklamiere" nur mathematisch existente
>>>> Objekte.
>>> Was nicht konstruiert werden kann, gehört nicht zur Abzählung. Mathematisch existente
>>> Objekte, hier also Paare, können konstruiert werden. "such that every element of the set stands at a definite position of this sequence".
>>>
>> Und an welcher Position einer Folge sie stehen, hängt von der
>> verwendeten Zuordnungsvorschrift ab.

Wie bereits Cantor schrieb:

|"[Es] wird bewiesen, daß äquivalente Mengen immer eine und dieselbe
| Mächtigkeit oder Kardinalzahl haben und daß auch umgekehrt Mengen
| von derselben Kardinalzahl äquivalent sind. [...]
| Die Kenntnis nur eines /Zuordnungsgesetzes/ für zwei Mengen M und M_1
| /genügt/, um die Äquivalenz derselben zu konstatieren; doch gibt es
| immer viele, im allgemeinen sogar unzählig viele Zuordnungsgesetze,
| durch welche zwei äquivalente in gegenseitig eindeutige und
| vollständige Beziehung zueinander gebracht werden können."
(Cantor: Gesammelte Abhandlungen, S.413)

> Hier geht es nur um eine einzige, nämlich diese:
> k = (m + n - 1)(m + n - 2)/2 + m.

Ja, das ist die bijektive Cantorsche Paarungsfunktion f, die jedem Paar
(m,n)∈ℕ×ℕ eine eindeutige natürliche Zahl k∈ℕ zuordnet. Mit

f: ℕ×ℕ→ℕ , f(m,n) = m + (m + n - 1)*(m + n - 2) = k

sind alle Paare ((m,n),k) definiert.

> Was nach dieser Zuordnungsvorschrift nicht konstruiert werden kann,
> ist dunkel, denn alles andere kann konstruiert werden

Die Paarungsfunktion ordnet _jedem_ Paar (m,n)∈ℕ×ℕ genau eine
natürliche Zahl k∈ℕ zu. Durch die Definition (nicht: "Konstruktion")
dieser Funktion sind alle Paare ((m,n),k) definiert.

>> Abgelehnt habe ich deine irrige Vorstellung, dass man bei einer
>> Bijektion sämtliche Paare "schrittweise konstruieren" müsse,
>
> Nicht müsse, sondern könne.

Auch deine Vorstellung, dass man bei einer Bijektion sämtliche Paare
"schrittweise konstruieren könne" ist im Wortsinne falsch, da es sich um
unendliche Mengen handelt. Was dagegen möglich ist, ist bei unendlichen
Mengen Bijektionen zu definieren, so dass alle Paare (d,w) von Elementen
d aus der Definitionsmenge und Elementen w aus der Wertemenge dadurch
erfasst sind. Fertig.

>> In der Mathematik gilt: Eine Bijektion zwischen zwei unendlichen Mengen
>> ist vollständig beschrieben durch Angabe von Definitionsmenge,
>> Wertemenge und Funktionsvorschrift ("Zuordnungsgesetz").
>
> Ja, aber

Nichts "aber". So ist es, und Einwände dagegen sind sinnlos.

>>>> Damit ist die "Konstruktion" aller Paare angegeben. Die Cantorsche
>>>> Paarungsfunktion ist ein Beispiel für eine solche Bijektion.
>>>
>>> Und die führe ich aus, so dass alle O in der Matrix bleiben.
>> Du führst die Cantorsche Paarungsfunktion nicht aus, sondern führst in
>> einer Darstellung der Definitionsmenge Verschiebungen und Vertauschungen
>> der Elemente aus,
>
> die genau Cantors Vorschrift reproduzieren.

Nein. Cantor gibt eine Vorschrift an, die eine Zuordnung von Paaren
(m,n)∈ℕ×ℕ zu natürlichen Zahlen k∈ℕ beschreibt.
Bildlich veranschaulicht:

f
Definitionsmenge ------------------> Wertemenge
ℕ×ℕ
(aufgefasst als Menge ℕ
positiver Brüche)

1 2 3 3 5 n |
|---|----|----|----|----|------ ..-->-1
| |..................................... |
1- 1/1 1/2 1/3 1/4 1/5 ... ..-->-2
| |................................ |
2- 2/1 2/2 2/3 2/4 2/5 ... ..-->-3
| |..................................... |
3- 3/1 3/2 3/3 3/4 3/5 ... -4
| usw. |
4- 4/1 4/2 4/3 4/4 4/5 ... -5
| |
5- 5/1 5/2 5/3 5/4 5/5 ... -6
| |
| ... |
| k
m

Da eine solche Darstellung nicht sehr übersichtlich ist, wird die
Matrixdarstellung der Wertemenge benutzt, in der die Werte k∈ℕ
in einer quadratischen Matrix eingetragen werden, und zwar an den
Stellen von Elementen (m,n), die dem jeweiligen Urbild von k
entspricht:

Wertemenge von f in Matrixform:

1 2 3 3 5 n
|---|----|----|----|----|--------
|
1- 1 2 4 7 11 ...
|
2- 3 5 8 12 17 ...
|
3- 6 9 13 18 24 ...
|
4- 10 14 19 25 32 ...
|
5- 15 20 26 33 41 ...
|
| ...
|
m

_Es wird also nach Cantor nichts verschoben_
_und auch nichts ausgetauscht_.

Du kannst dir die zweite Darstellung auch so vorstellen, dass eine nicht
transparente Folie mit der Anordnung der Werte k auf die Matrix, welche
die Definitionsmenge darstellt, gelegt wird. Damit sind _alle_ Paare
((m,n),k) der Paarungsfunktion f erfasst.

> Entweder bist Du unfähig einzusehen, dass ich genau Cantors Vorschrift befolge,

Jeder mit ein wenig Kenntnis der Materie kann einsehen, dass du etwas
veranstaltest, was mit Cantors Paarungsfunktion nichts zu tun hat.
Du produzierst reinsten Blödsinn und bist unfähig, das zu erkennen.

> dass Du keinen Schritt angeben kannst, der von Cantors Formel k = (m + n - 1)(m + n - 2)/2 + m abweicht.

Jeder einzelne "Schritt" weicht bei dir von Cantor ab:
In jedem "Schritt" erscheint bei dir eine Menge, welche nichts anderes
darstellt, als eine

_Umordnung der Definitionsmenge der Paarungsfunktion f_.
Und
_daraus kann nicht die Wertemenge von f in der Matrixform_
_hergestellt werden_.

> Also finde einen solchen,

Erledigt.

Dieter Heidorn

Re: A game like billards

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From: chris.m....@gmail.com (Chris M. Thomasson)
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Subject: Re: A game like billards
Date: Thu, 26 Oct 2023 13:26:18 -0700
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 by: Chris M. Thomasson - Thu, 26 Oct 2023 20:26 UTC

On 10/26/2023 9:21 AM, WM wrote:
> William schrieb am Mittwoch, 25. Oktober 2023 um 20:19:42 UTC+2:
>> On Wednesday, October 25, 2023 at 11:56:04 AM UTC-3, WM wrote:
>
>>> Every pair (n, x) of a bijection that cannot be proven to be the result of step n with n the last element of a FISON,
>
>> There is no such pair.
>
> Because dark elements cannot be paired.
[...]

Huh? WTF are you rambling on about? I have my doubts that you even
understand Cantor Pairing...

I have a dark number for you. I know what it is. You do not. Now in your
strange world, my number does not exist, or is _dark_. In your wild
world of the WM Moronica...

Re: A game like billards

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Subject: Re: A game like billards
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 by: Chris M. Thomasson - Thu, 26 Oct 2023 20:28 UTC

On 10/26/2023 9:21 AM, WM wrote:
> William schrieb am Mittwoch, 25. Oktober 2023 um 20:19:42 UTC+2:
>> On Wednesday, October 25, 2023 at 11:56:04 AM UTC-3, WM wrote:
>
>>> Every pair (n, x) of a bijection that cannot be proven to be the result of step n with n the last element of a FISON,
>
>> There is no such pair.
>
> Because dark elements cannot be paired. But there are many dark natnumbers and fractions which cannot be paired. The O indicate not paired fractions and remain in all possible steps and do not vanish in the limit.

WM: I have never seen an aphid, therefore they do not exist.

Q: I have seen a number that you have never seen before...

WM: How is that possible?

lol!

Re: A game like billards

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Thu, 26 Oct 2023 13:37:20 -0700
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 by: Chris M. Thomasson - Thu, 26 Oct 2023 20:37 UTC

On 10/26/2023 11:05 AM, FromTheRafters wrote:
> WM presented the following explanation :
>> FromTheRafters schrieb am Mittwoch, 25. Oktober 2023 um 17:08:24 UTC+2:
>>> WM presented the following explanation :
>>>> Fritz Feldhase schrieb am Dienstag, 24. Oktober 2023 um 21:26:05 UTC+2:
>>>>> On Tuesday, October 24, 2023 at 1:52:19 PM UTC+2, WM wrote:
>>>>>> Fritz Feldhase schrieb am Montag, 23. Oktober 2023 um 18:02:11 UTC+2:
>>>>>>>
>>>>>>> 2. A bijection is not a process.
>>>>> A BIJECTION IS NOT A PROCESS.
>>>>
>>>> A bijection with |N can be executed as a process.
>>
>> According to Cantor. Of course not in reality.
>>
>>> But that doesn't mean it *is* a process.
>>
>> A process is what can be executed as a process.
>
> Explain.

WM: I am a ultra hardcore finite type of person... ;^o

Re: A game like billards

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From: james.g....@att.net (Jim Burns)
Newsgroups: sci.math
Subject: Re: A game like billards
Date: Thu, 26 Oct 2023 17:40:13 -0400
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 by: Jim Burns - Thu, 26 Oct 2023 21:40 UTC

On 10/26/2023 4:37 PM, Chris M. Thomasson wrote:
> On 10/26/2023 11:05 AM, FromTheRafters wrote:
>> WM presented the following explanation :
>>> FromTheRafters schrieb am Mittwoch,
>>> 25. Oktober 2023 um 17:08:24 UTC+2:
>>>> WM presented the following explanation :
>>>>> Fritz Feldhase schrieb am Dienstag,
>>>>> 24. Oktober 2023 um 21:26:05 UTC+2:
>>>>>>> Fritz Feldhase schrieb am Montag,
>>>>>>> 23. Oktober 2023 um 18:02:11 UTC+2:

>>>>>>>> 2. A bijection is not a process.
>>>>>>
>>>>>> A BIJECTION IS NOT A PROCESS.
>>>>>
>>>>> A bijection with |N can be executed as a process.
>>>
>>> According to Cantor. Of course not in reality.
>>>
>>>> But that doesn't mean it *is* a process.
>>>
>>> A process is what can be executed as a process.
>>
>> Explain.
>
> WM:
> I am a ultra hardcore finite type of person...
> ;^o

WM:
I reject arithmetic
in order to keep two-ended-ness.

Re: A game like billards

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Subject: Re: A game like billards
From: askaske...@gmail.com (WM)
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 by: WM - Fri, 27 Oct 2023 10:10 UTC

Dieter Heidorn schrieb am Donnerstag, 26. Oktober 2023 um 21:41:56 UTC+2:
> WM schrieb:

> > Hier geht es nur um eine einzige, nämlich diese:
> > k = (m + n - 1)(m + n - 2)/2 + m.
> Ja, das ist die bijektive Cantorsche Paarungsfunktion f, die jedem Paar
> (m,n)∈ℕ×ℕ eine eindeutige natürliche Zahl k∈ℕ zuordnet.

That is wrong as the OP clearly proves.

> >> Abgelehnt habe ich deine irrige Vorstellung, dass man bei einer
> >> Bijektion sämtliche Paare "schrittweise konstruieren" müsse,
> >
> > Nicht müsse, sondern könne.
> Auch deine Vorstellung, dass man bei einer Bijektion sämtliche Paare
> "schrittweise konstruieren könne" ist im Wortsinne falsch, da es sich um
> unendliche Mengen handelt.

Every single pair that exists can be checked. What visible number cannot be checked? Only dark or not existing elements cannot be checked.

> > Entweder bist Du unfähig einzusehen, dass ich genau Cantors Vorschrift befolge,
> Jeder mit ein wenig Kenntnis der Materie kann einsehen, dass du etwas
> veranstaltest, was mit Cantors Paarungsfunktion nichts zu tun hat.

Rede kein dummes Zeug, sondern gib eine Abweichung an.
> Du produzierst reinsten Blödsinn und bist unfähig, das zu erkennen.

Sobald Du eine Abweichung namhaft machst, werde ich sie erkennen.

> > dass Du keinen Schritt angeben kannst, der von Cantors Formel k = (m + n - 1)(m + n - 2)/2 + m abweicht.

> Jeder einzelne "Schritt" weicht bei dir von Cantor ab:

Nein. Hier ist der zweite, denn der erste reproduziert die Ausgangslage:

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, ...
....

wird zu

1, 2, 1/3, 1/4, ...
1/2, 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, ...
....
Nach Cantor wird das Feld 1/2 mit 2 indiziert. Also völlige Übereinstimmung.

> In jedem "Schritt" erscheint bei dir eine Menge, welche nichts anderes
> darstellt, als eine
>
> _Umordnung der Definitionsmenge der Paarungsfunktion f_.

> Und
> _daraus kann nicht die Wertemenge von f in der Matrixform_
> _hergestellt werden_.

Welches ist denn der erste fehlende Wert?

Gruß, WM

Re: A game like billards

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Subject: Re: A game like billards
From: askaske...@gmail.com (WM)
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 by: WM - Fri, 27 Oct 2023 10:13 UTC

FredJeffries schrieb am Donnerstag, 26. Oktober 2023 um 19:08:03 UTC+2:
> On Thursday, October 26, 2023 at 9:12:29 AM UTC-7, WM wrote:
> > Jim Burns schrieb am Mittwoch, 25. Oktober 2023 um 17:06:55 UTC+2:

> > > And there aren't two ends.
> > Then all O remain forever in the matrix.
> That is correct

Of course. They indicate not indexed fractions.

Regards, WM

Re: A game like billards

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Subject: Re: A game like billards
From: askaske...@gmail.com (WM)
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 by: WM - Fri, 27 Oct 2023 10:19 UTC

Jim Burns schrieb am Donnerstag, 26. Oktober 2023 um 19:47:55 UTC+2:

> The enumeration of all the fractions
> n = p+(p+q-1)(p+q-2)/2
> is not a process

But all visible fractions can be enumerated by a process, one after the other. The reason is the existence of this sequence: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

> Either way, enumerating the fractions
> is not a process.

At which step n does the possibility of doing step n+1 fail?
>
> That is not a proof of dark states.
> Even with dark states, it is not a process.
> Even without dark states, it is not a process.

The proof of dark states is done by the O remaining in the matrix but not being visible.

> In arithmetic, _it can be said_
> of each thing that it equals itself.

But you cannot see these things if they are dark.

> Arithmetic is incompatible with
> your actual-infinityᵂᴹ

It is Cantor's actual infinity.

Regards, WM


tech / sci.math / Re: A game like billards

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