On Thursday, 13 June 2013 18:24:14 UTC+2, Zeit Geist wrote:
> I said for each particular q e Q, you must be able to produce a natural number m_(1/2) such that at step m(1/2) we have for at least one-half of all rationals p, p < q, p is in the natural order.
Why should that be required? Is there some axiom demanding it?
> For a general well-ordering of the rationals, not in order of magnitude, we don't have to worry about, say, 1/2 of the rations less than q.
With exactly the same right you could demand that there is an n such that half of all natural numbers have been attached to rationals. And that there is an m such that 2/3 of all natural numbers have been attached, and so on. In fact that would make sense and we could trust in countability - as your naive idea shows. But that is not possible. Quite the reverse! Enumerate the rationals as far as you like. Let n be the largest natural number that ever a human being will think about. Then with 1, 2, ..., n you will have enumerated less than 10^-10000000000000000000000000000000000 of all rationals. Much less!
Your idea should show you very clearly what a nonsense notion countability is.