In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 27 Jan., 21:40, William Hughes <wpihug...@gmail.com> wrote: > > On Jan 27, 6:46 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 27 Jan., 18:32, William Hughes <wpihug...@gmail.com> wrote: > > > > > > On Jan 27, 6:05 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > <snip> > > > > > > >..the diagonal > > > > > cannot differ from all lines > > > > > (it differs from every line, though). > > > > > > The fact that the diagonal differs from every line is > > > > enough to show (induction) that the diagonal is not > > > > equal to any line in the list. > > > > > No. > > > > Let the antidiagonal be d and the nth line be l(n) > > > > We know that for each n in |N, d is not equal to l(n) > > > > You have agreed that this implies > > > > There is no m in |N such that d equals l(m) > > No. There is no m between 1 and n such that d equals I(m). > This holds for *every* n. > This holds not for *all* n.
In standard math, tertium non datur. Thus in order for f(n) not to hold for all n, there must be an n for which it does not hold. > > Note: After every natural number there are infinitely many natural > numbers. After all natural numbers there is (would be, if the notion > "all natural numbers" was meaningful) none.
Bu in this case "for all" means collectively not individually.
For all naturals numbers, n , there is a natural after n. > > Note: Also induction shows that you never have all n. They say, > induction proves something for all n. But that is inaccurate speech. > Induction proves something for every n. For instance that you never > have all n.