The Indescribable Random Numbers
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From: Jacko on 10 Aug 2010 18:16 On 10 Aug, 20:57, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Jacko <jackokr...(a)gmail.com> writes: > > I don't think so. Unless you have a full proof that there are only a > > finite number of numbers below infinity, and not some messed up proof > > by negation with a not(red) = blue => no green. > > Why should I have such a proof? > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechen kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus Because the table is diagonalized making another real, hence it was not complete, as assumed. There would be no reason why the created real would not be added to the table ad infinitum. Like counting them starting from a table of 1 real. :-) You'd need the proof to demonstrate that the table was complete, and could have no more things added.
From: Jacko on 10 Aug 2010 18:18 On 10 Aug, 21:05, Shubee <e.shu...(a)gmail.com> wrote: > On Aug 10, 2:33 pm, Jacko <jackokr...(a)gmail.com> wrote: > > > A Cantor is 'Jazz Singer' recountor of the torah. Cardinality is > > related to Cardinal. A sin is a religious moral judgement. The 1984 UK > > Data Protection Act referes to storing false or misleading information > > of a personal nature and having it used against the individual the > > information is about. > > > Does this imply that quoting Cantor proved X and sullying his name, > > when he is not about to retort and consider the option of withdrawing > > the 'proof' is an offence under the act? > > Wouldn't it be more interesting to investigate the hopelessly > indescribable real numbers, i.e., the ethereal ones that can't be > defined with a finite number of words, numbers or mathematical > symbols? 'Can't be' is not the same as 'have not currently been'. Can't be requires a logically sound proof.
From: Aatu Koskensilta on 10 Aug 2010 18:22 Jacko <jackokring(a)gmail.com> writes: > Because the table is diagonalized making another real, hence it was > not complete, as assumed. There's no need to make this assumption, since it is in fact not used in the proof at all; there's no reason we should present the diagonal argument as an indirect proof. It's a curious phenomenon that so many otherwise sensible people are inclined to turn perfectly fine direct proofs into indirect ones for no apparent reason. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechen kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Chip Eastham on 10 Aug 2010 18:49 On Aug 10, 1:38 pm, Jacko <jackokr...(a)gmail.com> wrote: [snip] > Cantor's diagonalization argument just prooves that > unordered sets are not countable, it does not proove > that an ordered countable set of the > reals can not be made. Here's an unordered set: {x,y}. Exactly why do you claim this is not countable? One application of Cantor's diagonalization argument is that the reals are not countable. Diagonalization arguments find a place in many other proofs, such as the nonexistence of a Turing machine that solves the halting problem, Godel's incompleteness theorem, the existence of sets that are recursively enumerable but not recursive, etc. I hope this helps. Chip
From: Jacko on 10 Aug 2010 20:25
On 10 Aug, 23:49, Chip Eastham <hardm...(a)gmail.com> wrote: > On Aug 10, 1:38 pm, Jacko <jackokr...(a)gmail.com> wrote: > > [snip] > > > Cantor's diagonalization argument just prooves that > > unordered sets are not countable, it does not proove > > that an ordered countable set of the > > reals can not be made. > > Here's an unordered set: {x,y}. > > Exactly why do you claim this is not countable? Some disordered sets are not countable. This does not mean that the is no order which is countable. English is not logic, so why do people insist on assuming it is complete? > One application of Cantor's diagonalization argument > is that the reals are not countable. Diagonalization > arguments find a place in many other proofs, such as > the nonexistence of a Turing machine that solves the > halting problem, Godel's incompleteness theorem, the > existence of sets that are recursively enumerable > but not recursive, etc. It doesn't the reals are countable as the reals form an isomorphism with the rationals, and the zig-zag argument applies. 0.0 0.1 1.0 2.0 1.1 0.2 0.3 ... 3.0 4.0 ... 0.4 etc. The number 1234.5678 for example is isomorphic to 1234/8765 in the 2D zig zag table of rationals. > I hope this helps. Suprisingly it doesn't. All proofs by negation of assumption are only sound if the assumption and the counter assumption are form a pair. The creation of a real not in the infinite Cantor basket of reals, means either the basket is full hence finite and a hyper-real exists, or the basket is not full as in the meaning of ALL reals are within. You can stick your head up your asymptote as much as you choose, but it won't change the fact that the reals are countable. |