Virgil
Posts:
4,482
Registered:
1/6/11
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Re: Matheology � 201
Posted:
Jan 28, 2013 2:16 PM
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In article
<d89b0fec-b246-4f8b-bbb0-4293d5ca83ed@k4g2000yqn.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 28 Jan., 08:44, William Hughes <wpihug...@gmail.com> wrote:
> > On Jan 27, 11:23 pm, William Hughes <wpihug...@gmail.com> wrote:
> >
> >
> >
> >
> >
> > > On Jan 27, 11:16 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > On 27 Jan., 23:02, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > > On Jan 27, 10:39 pm, WM <mueck...@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.
> >
> > > > > You contradict yourself. You have agreed
> > > > > that if P(n) is true for every n then
> > > > > the is no n such that P(n) is false.
> >
> > > > Don't turn the words in my mouth.
> >
> > > You have apparently forgotten the
> > > thread in which you agreed to this.
>
> I did not agree to that,
> >
> > A simple proof.
> >
> > The statements
> >
> > i. for every natural number n, P(n) is true
> > ii. there exists a natural number m such that P(m)
> > is false
> >
> > cannot both be true at the same time.
> > If you prove that i. is true then it follows
> > that ii. is false.-
>
> You have not yet understood.
He appears to understand a great deal better than WM does.
> For every natural n, P(n) is true does
> not imply anything about the existence of P(m).
"i. for every natural number n, P(n) is true" certainly implies the
existence of a lot of P(m)'s as well as their truth.
> You have not yet
> grasped that there is no "for all".
There is outside of Wolkenmuekenheim.
>
> Of course for every n it is true that there are infinitely many
> naturals beyond.
> Of course there is no m such that there is no natural beyond.
> But if you assume that all natural exist, then there is no natural
> beyond all.
>
> You must unederstand that, therefore, there is no "all naturals".
If there is "no natural beyond all", then we already have "all naturals".
At least outside Wolkenmuekenheim
> All
> we can do is state P(n) for every n. Nevertheless it is not
> necessarily true for all n.
So WM claims he can find an instance in which
For every n in |N P(n) and for some n in |N not P(n)
Note that outside of WN's Wolkenmuekenheim
[NOT for all in |N P(n)]
if and only if
[Exists n in N such that not P(n)]
--
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