On 10 Apr., 13:40, Dan <dan.ms.ch...@gmail.com> wrote: > On Apr 10, 1:33 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 10 Apr., 08:59, William Hughes <wpihug...@gmail.com> wrote: > > > > Next argument.- > > > Consider a Cantor-list that contains a complete sequence (q_k) of all > > rational numbers q_k. The first n digits of the anti-diagonal d are > > d_1, d_2, d_3, ..., d_n. It can be shown for every n that the Cantor- > > list beyond line n contains infinitely many rational numbers q_k that > > have the same sequence of first n digits as the anti-diagonal d. > > Proof: There are infinitely many rationals q_k with this property. > > All are in the list by definition. At most n of them are in the first > > n lines of the list. Infinitely many must exist in the remaining part > > of the list. So we have obtained: > > For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ..., > > q_kn > > > Every property that holds for all FISs d_1, d_2, d_3, ..., d_n of d, > > also holds for all digits d_n of d that can be subject to digit- > > reversal and digit comparison. > > > They have the same sequence of first n digits, but you'll never find a > number in the list that has "all digits the same as the anti- > diagonal" .
For all n is not for all n? d has more digits than all? My proof is valid for all n. If you can name a digit d_n that is not covered, please let me know. If not, then you should think over your basis of doing mathematics. > > For any natural number k , the k'th member of your list will have its > k'th digit different from the k'th digit of the antidiagonal . > Therefore the antidiagonal itself is not in the list , even as > arbitrarily large "partial sub-sequences" of the antidiagonal may > appear on the list .- Ziti
Your conclusion is wrong. Something is not in the list for some k and for all n in the list: That situation is better described as: "Therefore we have a contradiction".