On 10 Apr., 14:50, Dan <dan.ms.ch...@gmail.com> wrote: > > That situation is better described as: > > "Therefore we have a contradiction". > > We have a contradiction only if we believe with religious fervor that > "it should" be on the list.
No, we can prove that for all n: (d_1, ..., d_n) is infinitely often repeated in the sequence of all rational numbers - with mathematical precision. What do you not understand with this result?
> > Your conclusion is wrong. Something is not in the list for some k and > > for all n in the list: > > It always fascinating how one can manage to string up words together > in sentences that are grammatically correct but don't hold a shred of > meaning, semantically speaking .
Yes I know. Many set theorists do so when claiming that an irrational number in decimal representation "is more" than all its finite initial segments, since this "more" is required to argue that d "is more" than all its digits d_n. Or do you think that for all n: d_n is more than for all n: (d_1, d_2, d_3, ..., d_n) namely d_1, d_2, d_3, ... ?
If you tend to think so, remember that *for all* Cantor-arguments the digit substitution a_nn --> d_n takes place at a finite index n. All these substitutions belong to a finite initial segment of lines 1 to n of the list, namely to the decimal representation of a terminating decimal d_1, d_2, d_3, ..., d_n.
The "beyond" is not included in Cantor's argument. This easily explained by the observation that *for all n* we have 1/n =/= 2/n, but *n the limit* the inequality fails: lim 1/n = lim 2/n for n --> oo. So even if the "limit beyond all finite n" is considered, Cantor is not saved.