Uirgil
Posts:
185
Registered:
4/18/12


Re: ? 533 Proof
Posted:
Aug 4, 2014 7:55 PM


In article <70758aefd7bc4f5894ed1e2d026e591b@googlegroups.com>, mueckenh@rz.fhaugsburg.de wrote:
> On Monday, 4 August 2014 22:02:36 UTC+2, Martin Shobe wrote: > > > > > The lemma of my proof holds for every n that is mapped on an a/1. That > > > was intended and makes the theorem true. In fact the theorem is true for > > > every n. The proof could be reformulated to make also the lemma true for > > > every n. But why should it? There is every k in N of not enumerated > > > rationals. > > > > > > > > You're babbling. Try again, but make it coherent this time. > > Your discovery does not spoil the result of the proof. But here is the > corrected proof of the lemma: > > Proof: Let a/1 be the largest fraction indexed by n.
But there is no largest fraction in Q (or smallest fraction other than 0/1) indexed by some n in N. in fact in the following wellordering of Q, there is no fraction at all without its n in N:
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!.
> > You said, "If my proof concerning all finite initial segments is not > > accepted by you as sufficient, then you must believe that there is > > something in N that isn't in at least one initial segment. Or have you > > a tertium datur?"
The subset of even naturals is NOT contained in any finite initial segment of N > > > > > > > > Your proof is not sufficient since I found a flaw in it. > > > This flaw has been removed.
Not hardly! > > > That does not > > mean that I have to believe that there is something in N in the lemma that > > isn't in > > at least one finite initial segment.
But you can belie it, since , for example, the subset of even naturals is NOT contained in any finite initial segment of N but is contained in N. > > In that case my proof shows that no natural number and no set of natural > numbers can index all rational numbers.
And the following proof shows that WM's "proof" is invalid anywhere outside of WM's worthless world of WMytheology:
Each member of the set of all rational numbers, Q, is of form m/n, with m an integer, n a natural, and with no common factor greater than 1.
Reorder them by increasing values of m+n and for equal values of m+n by increasing values of m.
This new ordering is a wellordering of Q with a first rational, 0/1, and for each rational a unique successor rational, and none omitted.
Thus each rational is now enumerated by the natural number marking its position in that wellordering, and both Mueckenheim and his worthless world of WMytheology are proved totally wrong!

