# Does normalcy in one base imply normalcy in any other base?

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 $$b'$$?

• 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.) Jan 21 at 10:42
• The answer, with references, is on Wikipedia. Jan 21 at 11:12
• 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. Jan 21 at 11:13
• Apparently numbers in which every sequence appears are called disjunctive or rish in the given base. Jan 21 at 11:14
• 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. Jan 21 at 16:14

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.