2
$\begingroup$

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) ?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Any deeper reason why you changed $2\to 8$? $\endgroup$ – Hagen von Eitzen Sep 3 '15 at 19:11
  • $\begingroup$ I was thinking the same thing. $\endgroup$ – Matt Samuel Sep 3 '15 at 19:12
  • 1
    $\begingroup$ No good reason. I was a bit worried about someone at first sight thinking I was doubling the digits. $\endgroup$ – André Nicolas Sep 3 '15 at 20:19
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.