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: § 534 Finis
Replies: 30   Last Post: Feb 22, 2015 8:14 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Virgil

Posts: 10,821
Registered: 6/8/11
Re: � 534 Finis
Posted: Aug 9, 2014 8:11 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

In article <1e0fd76d-1bae-4651-b327-ad9537494dc7@googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

> On Saturday, 9 August 2014 00:14:28 UTC+2, Zeit Geist wrote:
> > On Friday, August 8, 2014 2:07:28 PM UTC-7, muec...@rz.fh-augsburg.de
> > wrote:


> > > I have a proof about all natural numbers.

It is only about EACH natural number,
which WM acknowledges is quite different.

> > In that you Show something for Each Set of Cardinality n e N.

Not even thatn, only for FISONs!
>

> > You show Nothing for the Set N.
>
> I show it for all natural numbers.


Only for each, not for all.

> > > It applies to all natural numbers.

Only for each, not for all.

> > P(x) is True For All x e N, does Not imply P(N) is Necessarily True.
>
> Nobody is interested in P(N).


Everyone other than WM is!
> >
> > For you, P(y) is a Set of Cardinality y can Not Exhaust The Set Q.
> >

> All natural numbers cannot exhaust all rational numbers.

N enueates W when Q has been well-ordered SA dollows:
Once again, since WM is having so much trouble understanding it:

Each member of Q has UNIQUE representation as m/n, with m being an
integer, n being a positive integer, and with m and n having no common
factor greater than 1.

Order Q by increasing values of abs(m)+n, and within equal values of
abs(m)+n by increasing values of m.

Note that for positive m, m/1 has successor -(m+1)/1.
For any other form, m/n will have successor of form
(m + k)/(n - k) for some natural k with 0 < k < n.

This is a well-ordering of Q with a first rational, 0/1, and for each
rational a uniquely defined successor rational, and with no rationals
left out.

Thus each rational is now enumerated by the natural number marking its
position in the above well-ordering, everywhere outside of WM's
worthless world of WMytheology.

> > > It shows that "all natural numbers" is a non-existing notion in
> > > mathematics.


The SET of all natural numbers certainly is an existing notion
everywhere outside of WM's worthless world of WMytheology.

> Most set theorists call it N.
> >
> >
> >
> > BTW, have you Figured out how to Wriggle Out of the Fact that your System
> > shows that the Set N, which contains More Elements than Any Natural Number,
> > is Finite?

>
> It is potentially infinite. That means it is not actually infinity.


Outside of WM's worthless world of WMytheology, N is actually
infinite in the sense that one can inject N into a proper subset of N.
E,g., f: N -> N: n -> 2*n is such a function.

> what I have proven (and known for a long time already)
Is mostly false in proper mathematics.
>
> Regards, WM

--
Virgil
"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.