On 15 Jun., 14:45, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote: > "WM" <mueck...@rz.fh-augsburg.de> wrote in message > > news:62ae795b-1d43-4e1f-8633-e5e2475851aa@x21g2000yqa.googlegroups.com... > > > On 15 Jun., 12:26, stevendaryl3...@yahoo.com (Daryl McCullough) wrote: > > >> (B) There exists a real number r, > >> Forall computable reals r', > >> there exists a natural number n > >> such that r' and r disagree at the nth decimal place. > > > In what form does r exist, unless it is computable too? > > Of course its computable. > > You compute it by changing the nth digit of the nth number on the list to a > 7, unless it is already a 7 in which case you make it an 8. > > Sounds pretty easy to compute, I would have thought. I reckon I could code > it in about 2 minutes. > > > But if r is computable, then this theorem shows that the computable > > numbers are uncountable. Contradiction. > > Huh? > > > And if r is not computable, then it is impossible to prove > > disagreement with any r'. > > Computing the diagonal number is actually very easy.
Therefore the diagonal number is a computable real. Then Cantor's proof works exclusively on a countable set, namely the set of computable reals.
And it shows that this set is uncountable. This result is wrong.
