# Non computable numbers are normals?

We know that almost all real number are normal and almost all real number are non computable. This does not suffice to deduce that all non computable numbers are normals but , intuitively (??) this seems reasonable. There is some proof ( or disproof) ?

Disproof: The number with decimal expansion $0.a_1a_2a_3\dots$ is non-computable if and only if the number $0.a_18a_28a_38\dots$ is non-computable. But the second number is definitely not normal.
• Any deeper reason why you changed $2\to 8$? – Hagen von Eitzen Sep 3 '15 at 19:11
Let $A\subset\mathbb N$ a non-recursive set. Then $\sum_{k\in A}10^{-k}$ uses only digits $0$ and $1$