On Wednesday, 12 June 2013 18:06:40 UTC+2, Zeit Geist wrote:
> > > > And for which k do you fail to well-order by size all rationals which have (m + n) < k? > It will fail at no k, all k are natural numbers. It will fail for the set of all natural numbers.
My proof does not concern "the set of all natural numbers" whatever you may understand by and expect from that phrase. My proof concerns "all natural numbers" and "all rational numbers" without any exception.
If your "sets" are different or more, then I am not interested in such matheological notions of magic power.