In article <email@example.com>, firstname.lastname@example.org wrote:
> On Thursday, 7 August 2014 00:14:25 UTC+2, Zeit Geist wrote: > > > > > > > > > Up to every desired n and q_n. Alas they belong to a finite initial > > > > > segment. Infinitely many follow. The infinite is never completed. > > > > > > You try to Show a Contradiction in Set Theory > > > > > I have shown a contradiction between set theory and mathematics. > > > > No! You just Think you do. > > I think the following: For all natural numbers I have shown that uncounted > rationals remain.
No, only for each natural number And since in WM's world "all" is different from "each", he is now disproved!
And equally many uncounted naturals ALSO remain after each natural.
At least when Q is well-ordered 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 them by increasing values of abs(m)+n, and within equal values of abs(m)+n by increasing values of m, if any.
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, at least everywhere outside of WM's worthless world of WMytheology. -- Virgil "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)