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. To find only some rationals which differ from some digit whereas always, for every n, infinitely many remain with d_1, ..., d_n is not justifying your "all rationals are different" claim.
> > > > 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 is irrelevant.
Another important reply.
You asked at which point your quantifyer confusion appears. There are many points one of them is this:
You start with A( r e Q ) E( n e N ) ( x_n ~= r_n ) and then you switch to E( n e N ) A( r e Q ) ( x_n ~= r_n )
Otherwise you would get from A( n e N ) E( r e Q) ( x_n = r_n ) that d with has only its d_n finitely indexed cannot differ from all rationals. It is a rational.
The reason for this is not irrelevant. It is the fact that A( r e Q) are not available.
That may be irrelevant in order to do matheology but it is not irrelevant in mathematics.