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Re: Matheology § 258
Posted:
May 2, 2013 5:16 PM


WM <mueckenh@rz.fhaugsburg.de> writes:
> On 2 Mai, 16:03, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote: >> WM <mueck...@rz.fhaugsburg.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 antidiagonal, the FIS of which are >> >> > always in the next line. So the antidiagonal 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



