William schrieb am Mittwoch, 22. Juni 2022 um 23:12:57 UTC+2:
> On Wednesday, June 22, 2022 at 5:15:17 PM UTC-3, WM wrote:
> not mean the bijection (a set) does not exist.
> > My argument proves the not existence of the most famous bijection.
> Nope, You demonstrate that your method does not get the bijection.

My method is Cantor's method. Only the language differs slightly. But we could even use Cantor's original language, if only the indices first were associated with the integer fractions and then taken from there.

Regards, WM

On 6/23/2022 8:12 AM, WM wrote:
> William schrieb am Mittwoch, 22. Juni 2022 um 23:12:57 UTC+2:
>> On Wednesday, June 22, 2022 at 5:15:17 PM UTC-3, WM wrote:
>> not mean the bijection (a set) does not exist.
>>> My argument proves the not existence of the most famous bijection.
>> Nope, You demonstrate that your method does not get the bijection.
>
> My method is Cantor's method.

Liar. You cannot even generate an equation.

> Only the language differs slightly.

Liar. Admit your Swaparoofest is purely deception.

> But we could even use Cantor's original language,if only the indices first were associated with the integer fractions and then taken from there.

Clearly, you will not understand Cantor.

>
> Regards, WM

the only reason you try to disprove Cantor, is because you cannot post any original math yourself.

Crank Wolfgang Mückenheim, aka WM wrote:
> William schrieb am Mittwoch, 22. Juni 2022 um 23:12:57 UTC+2:
>> On Wednesday, June 22, 2022 at 5:15:17 PM UTC-3, WM wrote:
>> not mean the bijection (a set) does not exist.
>>> My argument proves the not existence of the most famous bijection.
>> Nope, You demonstrate that your method does not get the bijection.
>
> My method is Cantor's method. Only the language differs slightly.

No, you're doing far more than that. There is an irrefutable way
to check that Cantor's argument is sound, while yours is not:
Cantor's proof can be written in a theorem solver such as Coq
or HOL (and checked), while your idiotic sophistry and shell
game cannot.

You are a kook, Wolfgang, and a shame to German Academia.

Python schrieb am Donnerstag, 23. Juni 2022 um 15:44:00 UTC+2:
> There is an irrefutable way
> to check that Cantor's argument is sound, while yours is not:
> Cantor's proof can be written in a theorem solver such as Coq
> or HOL (and checked),

There are no irrefutable ways. The basis of these problem solvers is wrong. Therefore they are as wrong as their masters.

Look: There are mostly dark numbers. They cannot be used for mappings. That is what the problem solvers and their masters did not know. Here is the proof:

Every definable endsegment is infinite. Compared to |N it has lost only few natnumbers. More cannot get lost in a definable way.
But all natnumbers have vanished in the itersection of all undefinable endsegments.

Regards, WM

WM presented the following explanation :
> William schrieb am Mittwoch, 22. Juni 2022 um 23:12:57 UTC+2:
>> On Wednesday, June 22, 2022 at 5:15:17 PM UTC-3, WM wrote:
>> not mean the bijection (a set) does not exist.
>>> My argument proves the not existence of the most famous bijection.
>> Nope, You demonstrate that your method does not get the bijection.
>
> My method is Cantor's method. Only the language differs slightly. But we
> could even use Cantor's original language, if only the indices first were
> associated with the integer fractions and then taken from there.

Which they were not, and for good reason. It simply doesn't *SHOW* that
the bijection exists when done your way. Your inability to prove the
existence of something does not mean that that something does not
exist.

FromTheRafters schrieb am Donnerstag, 23. Juni 2022 um 16:56:50 UTC+2:
> WM presented the following explanation :
> > William schrieb am Mittwoch, 22. Juni 2022 um 23:12:57 UTC+2:
> >> On Wednesday, June 22, 2022 at 5:15:17 PM UTC-3, WM wrote:
> >> not mean the bijection (a set) does not exist.
> >>> My argument proves the not existence of the most famous bijection.
> >> Nope, You demonstrate that your method does not get the bijection.
> >
> > My method is Cantor's method. Only the language differs slightly. But we
> > could even use Cantor's original language, if only the indices first were
> > associated with the integer fractions and then taken from there.
> Which they were not, and for good reason.

Can you name one?

> It simply doesn't *SHOW* that
> the bijection exists when done your way. Your inability to prove the
> existence of something does not mean that that something does not
> exist.

It seems to be detrimental first to check the number of indices by indexing the integer fractions. Strange.

Regards, WM

On Thursday, June 23, 2022 at 10:13:03 AM UTC-3, WM wrote:

> My method is Cantor's method.

Nope. Cantor defined an algorithm by which you could get rational given a natural. . Cantor does not use a series of exchanges to define a set. Let the algorithm be f. Then the bijection is { (n, f(n) ) | n element of N} . You cannot use a series of exchanges to define this set.
(You can use a series of exchanges to define the algorithm.

--
William Hughes

On Thursday, June 23, 2022 at 11:45:32 AM UTC-3, WM wrote:

> There are mostly dark numbers.

Piffle. The natural numbers are a Peano set and a Peano set does not contain "dark" numbers.

William Hughes

On Thursday, June 23, 2022 at 5:19:26 PM UTC+2, William wrote:
> On Thursday, June 23, 2022 at 10:13:03 AM UTC-3, WM wrote:
> >
> > My method is Cantor's method.
> >
> Nope. Cantor defined an algorithm [etc,]

I don't think so. Cantor defined a _function_.

See: https://de.wikipedia.org/wiki/Cantorsche_Paarungsfunktion

(The Wikipedia entry https://en.wikipedia.org/wiki/Pairing_function is misleading, sorry.)