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Topic: Matheology § 201
Replies: 32   Last Post: Jan 28, 2013 2:26 PM

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 Virgil Posts: 8,833 Registered: 1/6/11
Re: Matheology � 201
Posted: Jan 27, 2013 5:42 PM

In article
WM <mueckenh@rz.fh-augsburg.de> 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.

For all n in |N, (n+1)^2 = n^2 + 2*n + 1.
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