WM says... > >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?
Another answer to the question is that r is *definable*.
Every real number between 0 and 1 can be represented as an infinite series of the form: sum from n=1 to infinity of a_n 2^{-n}.
So a real number r can be said to be definable by a formula Phi(x) in the language of arithmetic if
r = sum over all n such that Phi(n) of 2^{-n}
The formula Phi(x) uniquely determines the value of r.
In this sense, the antidiagonal of the list of all computable reals is definable (but not computable).
