Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

 Messages: [ Previous | Next ]
 mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Matheology § 201
Posted: Jan 27, 2013 4:39 PM

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.

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.

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.

Regards, WM