On 16 Jun., 11:48, "|-|ercules" <radgray...@yahoo.com> wrote: > "WM" <mueck...@rz.fh-augsburg.de> wrote ... > > > > > By induction we prove: There is no initial segment of the (ANTI)diagonal > > that is not as a line in the list. > > Right, therefore the anti-diagonal does not contain any pattern of digits > that are not computable. > Sorry, you misquoted me. I wrote: By induction we prove: There is no initial segment of the diagonal that is not as a line in the list. And there is no part of the diagonal that is not in one single line of the list. But I have to excuse because I wrote somewhat unclear.
The meaning is: 1) Every initial segment of the decimal expansion of pi is in at least one line of your list 3. 3.1 3.14 3.141 ... What we can finde in the diagonal (not the anti-diagonal), namely 3.141 and so on, exactly that can be found in one line. This is obvious by construction of the list.
2) Every part of the diagonal is in at least one line. That means, every part is in one single line, or there are parts that are in different lines but not in one and the same.
The latter proposition can be excluded. If there are more than one lines that contain parts of pi, then it can be proved, be induction, that two of them contain the same as one of them. This can be extended to three lines and four lines and so on for every n lines.
Hence we prove that all of pi, that is contained in at least one of the finite lines of your list, is contained in one single line.
Conclusion: Either the complete diagonal pi does not exist, or it exists also in one and the same single line. As the latter is wrong, so is the former. There does not exist an actually infinite sequence. Actually infinite mean finished infinite. That is nonsense.
(The proof has the same status as Cantor's diagonal proof. Also his proof is valid only for all finite n. In a similar way we find above: For all finite n: All of pi that is in the list up to line n, is in a single line of the list.)