In article <firstname.lastname@example.org>, WM <email@example.com> wrote: > Am Samstag, 7. Dezember 2013 01:37:34 UTC+1 schrieb Zeit Geist: > > On Friday, December 6, 2013 2:21:02 PM UTC-7, WM wrote: > > > Am Freitag, 6. Dezember 2013 18:59:16 UTC+1 schrieb Zeit Geist: > > > Let d differ from r, then there is a first digit where both differ. Let d > > > differ from p, q, r, then there is a first digit where d differs from all > > > three. Let d differ from all rationals, then there is a first digít where > > > d differs from all rationals.
> > That is incorrect.
> That is what you say when you claim that d differs from all rationals.
No! That may be what WM claims, but no one else hs to, because it is not true.
Quite the contrary, one would have to say that even though d does differ from all rationals there need not be any first digit of d by which it differs from more than finitely many them. For example if for each n d differs from the nth digit of the nth rational but no prior digit of the nth rational it will still differ from all rationals but never have any finite position by which it does so.
> > > You say for every rational q, there is a digit, where q differs from d. > > > You intentionally forget: For every rational q which is followed by > > > infinitely many other rationals, there is a first digit, where q differs > > > from d.
That does not imply any of the many false claims WM is making in trying to deny the truth.
> You asked at which point your quantifyer confusion appears. There are many > points one of them is this:
Before WM can justify his claims that that the rationals, |Q, are not well-orderable so as to be order-isomorphic to (|N, <), he will have to show why the following is not just such an ordering of |Q:
Consider this well-ordering of the rationals: Each rational, n/d, is represented by the quotient of an integer numerator, n, and a natural number denominator, d, with no common integer divisors greater than 1 then define a new ordering on the rationals so that n1/d1 > n2/d2 if and only if either | n1 | + d1 < | n2 | + d2 or both | n1 | + d1 = | n2 | + d2 and n1 < n2.
Then the set of all rationals reordered as above described is order-isomorphic to the naturally well-ordered set of naturals, producing a natural bijection between |Q and |N. --