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Topic: ? 533 Proof
Replies: 46   Last Post: Aug 4, 2014 8:39 PM

 Messages: [ Previous | Next ]
 Uirgil Posts: 185 Registered: 4/18/12
Re: ? 533 Proof
Posted: Aug 4, 2014 7:55 PM

mueckenh@rz.fh-augsburg.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 well-ordering 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 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!.

> > 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.

Re-order them by increasing values of |m|+n and for equal values of
|m|+n by increasing values of m.

This new ordering is a well-ordering 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 well-ordering, and both Mueckenheim and his worthless
world of WMytheology are proved totally wrong!

Date Subject Author
7/30/14 Ben Bacarisse
7/31/14 mueckenh@rz.fh-augsburg.de
7/31/14 Michael Klemm
7/31/14 Tanu R.
7/31/14 Virgil
8/1/14 mueckenh@rz.fh-augsburg.de
8/1/14 Martin Shobe
8/1/14 mueckenh@rz.fh-augsburg.de
8/1/14 Virgil
8/1/14 Virgil
8/1/14 Virgil
8/1/14 YBM
8/2/14 Martin Shobe
8/2/14 mueckenh@rz.fh-augsburg.de
8/2/14 Tanu R.
8/2/14 Virgil
8/2/14 Virgil
8/2/14 Martin Shobe
8/3/14 none3
8/3/14 none3
8/3/14 mueckenh@rz.fh-augsburg.de
8/3/14 Virgil
8/3/14 Tanu R.
8/4/14 Martin Shobe
8/4/14 mueckenh@rz.fh-augsburg.de
8/4/14 Martin Shobe
8/4/14 Tanu R.
8/4/14 Virgil
8/4/14 Martin Shobe
8/4/14 mueckenh@rz.fh-augsburg.de
8/4/14 Tanu R.
8/4/14 Martin Shobe
8/4/14 Virgil
8/4/14 Uirgil
8/4/14 Virgil
8/4/14 Tanu R.
8/4/14 Virgil
8/3/14 Virgil
8/1/14 Virgil
7/31/14 Virgil