Let $b\geq 2$ be an integer. A real $r \in [0,1]$ is said to be normal with respect to $b$ if every finite string made from the elements $\{0,\ldots,b-1\}$ appears in the $b$-ary expansion of $r$.

Are there integers $b, b'\geq 2$ as well as a real number $r\in[0,1]$ such that $r$ is normal with respect to $b$, but not with respect to $b'$?

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    $\begingroup$ See my 5 July 2002 sci.math post Numbers normal to one base but not to another base. (Note: In that post I seem to have reversed the definitions of multiplicatively dependent and multiplicatively independent.) $\endgroup$ Jan 21 at 10:42
  • $\begingroup$ The answer, with references, is on Wikipedia. $\endgroup$
    – Wojowu
    Jan 21 at 11:12
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    $\begingroup$ This is nonstandard terminology. Usually, a number is normal to base $b$ if any string $w\in\{0,\dots,b-1\}^n$ appears in the base-$b$ expansion of the number with asymptotic density $b^{-n}$; this is much stronger than the mere fact that each string appears in the expansion. $\endgroup$ Jan 21 at 11:13
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    $\begingroup$ Apparently numbers in which every sequence appears are called disjunctive or rish in the given base. $\endgroup$
    – Wojowu
    Jan 21 at 11:14
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    $\begingroup$ I didn't notice until seeing Emil Jeřábek's and Wojowu's comments (just now) that you are incorrectly defining what "normal numbers" are. To follow-up on their comments, for more about this larger collection of numbers see this 19 February 2003 sci.math post and this 5 January 2012 MSE answer and this 9 April 2018 MSE answer. $\endgroup$ Jan 21 at 16:14

1 Answer 1


A 1960 paper by Wolfgang Schmidt states the following:

We write $r \sim s$, if there exist integers $n$, $m$ with $r^n = s^m$. Otherwise, we put $r \not\sim s$.

In this paper we solve the following problem. Under what conditions on $r$, $s$ is every number $\xi$ which is normal to base $r$ also normal to base $s$? The answer is given by


A) Assume $r \sim s$. Then any number normal to base $r$ is normal to base $s$.
B) If $r \not\sim s$, then the set of numbers $ξ$ which are normal to base $r$ but not even simply normal to base $s$ has the power of the continuum.

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    $\begingroup$ While the result is probably true regardless, note that the OP is using a nonstandard definition of normality which does not agree with Schmidt’s paper. $\endgroup$ Jan 21 at 11:10

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