Non computable numbers are normals?

Question

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) ?

score 6 · Accepted Answer · answered Sep 03 '15 at 19:05

6

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.

answered Sep 03 '15 at 19:05

André Nicolas

507,029

Any deeper reason why you changed $2\to 8$? – Hagen von Eitzen Sep 03 '15 at 19:11
I was thinking the same thing. – Matt Samuel Sep 03 '15 at 19:12
1

No good reason. I was a bit worried about someone at first sight thinking I was doubling the digits. – André Nicolas Sep 03 '15 at 20:19

score 2 · Answer 2 · answered Sep 03 '15 at 19:05

2

Let $A\subset\mathbb N$ a non-recursive set. Then $\sum_{k\in A}10^{-k}$ uses only digits $0$ and $1$

answered Sep 03 '15 at 19:05

Hagen von Eitzen

374,180

Non computable numbers are normals?

2 Answers2