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


Virgil
Posts:
6,853
Registered:
6/8/11


Re: ? 533 Proof
Posted:
Jul 31, 2014 10:45 AM


In article <353e990f74d344969faf0766f55c1bc6@googlegroups.com>, mueckenh@rz.fhaugsburg.de wrote:
> On Wednesday, 30 July 2014 22:42:57 UTC+2, Zeit Geist wrote: > > > > > Of course every n is finite. Every FIS is finite. > > > > > But we have "forall n in N" here. > > > > > No, you Have "forall n in N, if n* = { m e N  m <= n } then Something is > > True.". > > > > Subtle Difference from "forall n e N, Something is True.". > > Subtle is the Lord. But not you. > > I prove "forall n e N" n leaves infinitely many rationals without index.
But for all q in Q, the following well ordering of Q gives each qin Q a natural number positon in that wellordering!
To wellorder the positive rationals. Write each one as p/q where naturals p and q have no common factor (other than 1). Then order them by ascending value of (p+q), and within each set of p+q values, order by ascending p.
Thus for ALL positive rationals one has a wellordering with 1/1 being first and for each positive rational, a unique immediate successor rational: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1...
One can then wellorder the whole set of rationals, Q, by putting zero at the start and interleaving each negative after its corresponding positive. > > > > This what you Fail to See. > > And what you fail to show. For what n is my claim wrong? If you cannot show > one, then my claim is true for all
But that claim does not override the above proof that each rational can be indexed by a different natural
> > Your set N is of no interest for me or people who are interested whether it > is possible to index all rationals.
But the above wellordering of Q proves that possibility is actuality!
Despite WM's desperate ignoring of that proof.  Virgil "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)



