"WM" <firstname.lastname@example.org> wrote in message news:email@example.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. >
> And if r is not computable, then it is impossible to prove > disagreement with any r'. >
Computing the diagonal number is actually very easy.