> On Monday, 18 August 2014 19:39:18 UTC+2, Zeit Geist wrote: <snip> >> This Shows NOTHING about the Non-Existence of a Bijection from N to Q. > > It shows that there are not enough naturals to index all > rationals.
You agree that, given S(q) = 1 / (2 floor(q) - r + 1) and B(n) = S^n(1),
Q+ = image(B) B(p) = B(q) iff p = q for all n c N, exists q c Q+, q = B(n)
> More is not intended.
More than the above is not required. You can't express your apparent disagreement with set theory with anything like the clarity with which you've expressed your agreement with it.