In article <8d610f94-e350-49b9-82c9-e2efc2a3b89e@fn10g2000vbb.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 7 Feb., 20:12, William Hughes <wpihug...@gmail.com> wrote: > > On Feb 7, 8:06 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 7 Feb., 19:46, William Hughes <wpihug...@gmail.com> wrote: > > > > > > Gosh, you are really running away > > > > from the fact that induction can > > > > show d is not in the list. > > > > > Induction can show that *your* d does not exist. > > > > My d? You are the one who defined d to be > > the antidiagonal of the list. > > The antidiagonal of a list is not always in the list, but the diagonal > of the list > > 1 > 11 > 111 > ... > > is with certainty in this very list - since it is nothing else but a > potentially infinite sequence of 1' and not longer than the lines.
If it is not longer than all lines, then there must be a first such line that it is not longer than.
So which is the first line that the diagonal is as short as?
No first line implies no line at all.
> > You also > > show by induction that the antidiagonal of > > a list is not in the list.
Actually, one should not speak of THE antidiagonal as there are at least as many antidiagonals as lines in the original list. > > No, that depends on the list.
> > The antidiagonal of the list
Which one? There are lots of them > > 0.0 > 0.1 > 0.11 > 0.111 > ... and none of them are in the list itself. --
