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



Virgil
Posts:
10,821
Registered:
6/8/11


Re: ? 533 Proof
Posted:
Aug 4, 2014 1:17 PM


In article <e6127db8e97846b9803f0c6cc932ebac@googlegroups.com>, mueckenh@rz.fhaugsburg.de 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 wellordered set.
So that WM's claim that no one can prove anything for N itself is proven FALSE! > 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 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!.
> > 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 wellordered as described!
> 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 wellordering 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 wellordering 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 wellordering of Q described above is NOT a wellordering of Q, the bijection between N and Q, that WM denies, will remain proved to exist!
And WM will remain wrong! Again!! As usual!!!  Virgil "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)



