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Topic: ? 533 Proof
Replies: 6   Last Post: Aug 1, 2014 2:06 PM

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 Virgil Posts: 10,821 Registered: 6/8/11
Re: ? 533 Proof
Posted: Jul 31, 2014 10:45 AM

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

To well-order 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 well-ordering 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 well-order 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 well-ordering of Q proves that possibility is actuality!

Despite WM's desperate ignoring of that proof.
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

Date Subject Author
7/31/14 Virgil
7/31/14 Virgil
8/1/14 Virgil