Why do you need to compute the n-th decimal of a number for this proof to work ? let Ω be a Chaitin Omega number it's decimal are non-computable, yet for every x, Ω + x is a perfectly valid number.

Then `floor(Ω.10^(n))-10.floor(Ω.10^n-1)` is also a number (a natural one) and actually it's the n-th decimal of Ω. It cannot be computed, but it still exist no matter what.