Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.


Virgil
Posts:
3,665
Registered:
6/8/11


Re: ? 533 Proof
Posted:
Aug 4, 2014 1:27 PM


In article <db0da8bf3e3b4792adc6e8284c6d8a1a@googlegroups.com>, mueckenh@rz.fhaugsburg.de wrote:
> On Monday, 4 August 2014 18:37:58 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. All proofs in set theory apply all > > > natural numbers only. > > > > Now, what on Earth are you Talking about? > > > > We can Prove lots of Theorems about N. > > Always you need the elements, at least for proving bijections.
One can prove a bijection between N and Q by proving Q can be wellordered which it can be. > > > > > > > Only a Fool would Think Otherwise! > > > > > > > > > > You seem to think otherwise. But you cannot show any proof that proves > > > the bijection for N itself, can you? (Further you cannot show an > > > uncountability proof that does not construct a defined element, although > > > you boasted you knew such.) > > > > > First, there are Many Enumations of Q, or at least Q+. For instance Take > > the SternBrocot Tree > > A tree? Do you use a mollusc called tree? Or do you use its nodes and path? > It is simply foolish to claim any bijection without considering the elements.
Any two sets which are wellorderable with unique nonsuccessor and every element a predecessor biject with each other. N is natually such set. Q becomes such a set when properly reordered: 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 wellordering 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 wellordering, and Mueckenheim is proved wrong AGAIN!.
> You need all natural numbers to prove a bijection NQ.
The normal wellordeing of N provides them.
> > > > Show a proof of bijection N and Q according to your delusions. > > > Done
And done!. > > You are either very stupid or very, very stupid. In both cases it is > appropriate to stop this nonsense.
Since WM is the only one producing any nonsense, he is the only one able to stop it. > > Regards, WM  Virgil "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)



