The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: ? 533 Proof
Replies: 2   Last Post: Aug 4, 2014 1:17 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View  

Posts: 10,821
Registered: 6/8/11
Re: ? 533 Proof
Posted: Aug 4, 2014 1:17 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

In article <>, wrote:

> On Sunday, 3 August 2014 23:48:21 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.

Outside of WM's worthless world of WMytheology, one can easily prove
that N is a well-ordered set.

So that WM's claim that no one can prove anything for N itself is proven
> But you cannot show any proof that proves the
> bijection for N itself, can you?

Yes I can provide such a proof, though it is not mine:

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!.

> Show a proof of bijection N and Q according to your delusions.

See above!
> >
> > Now, let P(x) be "The Set x doe Not Enumerate Q."
> >
> > You Prove "For All n e N, P(n)" and Erroneously Conclude "P(N)".
> >
> > What gives you the Logical Validity to do this?

> I do not claim that N enumerates Q.

But the above proof shows it does enumerate when Q is well-ordered as

> I claim that all natural numbers cannot
> enumerate all rational numbers.

The above proof proves otherwise!

> Set theory "proves" that all natural numbers
> can enumerate all rational numbers.

Proper mathematics proves it!

> > > > We all know that there are enumerations of the rationals (in set
> > > > theory,
> > > > not WMaths -- even you know that)

> > > You all are taking the first few numbers for "all". That is rather naive.

> > Nope, ALL.

> How do you do so?

By well-ordering Q as described above.
> > Each and Every of the Rationals in Q is made to Correspond to Exactly One
> > Natural in N.

> How is that accomplished?

By well-ordering Q as described above.

> Show it!!!

Shown above, unless WM snips it!

> Further unfounded assertions deleted.
> Here is sci.math and not sci.belief in delusions.

Until WM can prove that the well-ordering of Q described above is NOT a
well-ordering of Q, the bijection between N and Q, that WM denies, will
remain proved to exist!

And WM will remain wrong! Again!! As usual!!!
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.