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Topic: ? 533 Proof
Replies: 2   Last Post: Aug 4, 2014 1:17 PM

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Virgil

Posts: 1,458
Registered: 6/8/11
Re: ? 533 Proof
Posted: Aug 4, 2014 1:17 PM
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In article <e6127db8-e978-46b9-803f-0c6cc932ebac@googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

> On Sunday, 3 August 2014 23:48:21 UTC+2, Zeit Geist wrote:
>

> > You have Shown this For Sets of All Finite Cardinalities.
> >
> > You have Not Shown that this is True of N itself.

>
> Nobody can show anything for N itself.


Outside of WM's worthless world of WMytheology, one can easily prove
that N is a well-ordered set.

So that WM's claim that no one can prove anything for N itself is proven
FALSE!
> But you cannot show any proof that proves the
> bijection for N itself, can you?


Yes I can provide such a proof, though it is not mine:

Each member of Q is of form m/n, with m an integer,
n a natural, and with no common factor greater than 1.

Order them by increasing values of |m|+n and for equal values
of |m|+n by increasing values of m.

This is a well-ordering of Q with a first rational, 0/1, and for each
rational a unique successor rational, and none left out.

Thus each rational is now enumerated by the natural number
marking its position in that well-ordering,
and Mueckenheim is proved WRONG!.

>
> Show a proof of bijection N and Q according to your delusions.


See above!
> >
> > Now, let P(x) be "The Set x doe Not Enumerate Q."
> >
> > You Prove "For All n e N, P(n)" and Erroneously Conclude "P(N)".
> >
> > What gives you the Logical Validity to do this?

>
> I do not claim that N enumerates Q.


But the above proof shows it does enumerate when Q is well-ordered as
described!



> I claim that all natural numbers cannot
> enumerate all rational numbers.


The above proof proves otherwise!

> Set theory "proves" that all natural numbers
> can enumerate all rational numbers.


Proper mathematics proves it!

> > > > We all know that there are enumerations of the rationals (in set
> > > > theory,
> > > > not WMaths -- even you know that)


> > > You all are taking the first few numbers for "all". That is rather naive.

> > Nope, ALL.

> How do you do so?

By well-ordering Q as described above.
>
> > Each and Every of the Rationals in Q is made to Correspond to Exactly One
> > Natural in N.

>
> How is that accomplished?


By well-ordering Q as described above.

> Show it!!!

Shown above, unless WM snips it!




> Further unfounded assertions deleted.
> Here is sci.math and not sci.belief in delusions.


Until WM can prove that the well-ordering of Q described above is NOT a
well-ordering of Q, the bijection between N and Q, that WM denies, will
remain proved to exist!

And WM will remain wrong! Again!! As usual!!!
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)



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