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Topic: Matheology § 258
Replies: 4   Last Post: May 3, 2013 3:56 PM

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Alan Smaill

Posts: 748
Registered: 1/29/05
Re: Matheology § 258
Posted: May 2, 2013 5:16 PM
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WM <mueckenh@rz.fh-augsburg.de> writes:

> On 2 Mai, 16:03, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>> > On 1 Mai, 23:31, Dan <dan.ms.ch...@gmail.com> wrote:
>> >> > Yes, that is true. But (and please read this very attentively!):
>> >> > Cantor's argument requires the existence of the complete sequence
>> >> > 0.111.... in digits:

>> >> > You can see this easily here:
>> >> > The list
>> >> > 0.0
>> >> > 0.1
>> >> > 0.11
>> >> > 0.111
>> >> > ...

>> >> > when replacing 0 by 1 has an anti-diagonal, the FIS of which are
>> >> > always in the next line. So the anti-diagonal is not different from
>> >> > all lines, unless it has an infinite sequence of 1's. But, as we just
>> >> > saw, this is impossible.

>> >> I see no significant difference between referring to a mathematical
>> >> object by a formula and referring to it by 'writing it down' .

>> > But Cantor's argument is invalid, in this special case, unless it can
>> > produce 0.111... with actually infinitely many 1's, i.e. more than
>> > every finite number of 1's.

>> > It does not matter whether 1/9 exists as a fraction or whether it
>> > exísts in the ternary system as 0.01. In order to differ from every
>> > entry of my list Cantor's argument needs to produce, digit by digit,
>> > the infinite sequence. And that does not exist.

>> Not at all;
>> you accept that for any naturals n,m, (n/m)^2 =/= 2,
>> and that because you reason that any particular choice
>> leads to a contradiction.  You do not worry in that situation
>> that you need to check infinitely many cases.

> I have not to check infinitely many cases. I have to check exactly one
> case. I assume no common divisor and prove a common divisor.

One general case, which is sufficient to show the result in general.

>> Just reason in the same way here.

> Cantor needs the digits - all.

Simply repeating your mantra endlessly does not make it true.

In the diagonal argument, you check exactly one case:
assume the antidiagonal is in the list, and show the antidigonal
is not in the list.

Do you really believe that the *proof* of the diagonal argument
rests on individual examination of every possible position
in the list? For sqrt(2), you are happy to reach a comclusion
which applies to every pair of naturals n,m, without doing
such a case by case analysis. Why should this case be different?

Why in earth would you treat these two situations differently?
-- other than looking for any arbitrary excuse to deny
a conclusion you find unpalatable, of course.

> Regards, WM

Alan Smaill

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