# If normal with respect to prime base then normal for all bases

I tried to find it on internet but couldn't so m asking this here. I want to ask if a number is normal with respect to all prime number base then do we know that it is normal with respect to any base. Obviously we know that if a number is normal with respect to base b then for any $b^k$ it is normal.

Thanks

• This question is not appropriate for this site, which is for research mathematics. Math.stackexchange would be a better fit. – Noah Schweber Jan 24 '17 at 18:05
• I beg to differ – this is definitely a question of research mathematics. Wolfgang Schmidt, Über die Normalität von Zahlen zu verschiedenen Basen, Acta Arith. 7 1961/1962 299–309, MR0140482 (25 #3902), according to the review by N. G. de Bruijn, proved this: Let the set of all integers $>1$ be split into two disjoint classes R and S, such that for every $n>1$ all powers of $n$ belong to the same class as $n$ itself. Then there exists a continuum of numbers $\xi$ which all have the property that $\xi$ is normal with respect to every $r\in R$ and non-normal with respect to every $s\in S$. – Gerry Myerson Jan 24 '17 at 22:26
• This should be an answer, surely? – Adam P. Goucher Jan 27 '17 at 8:05
• @Adam, done – I had to wait until the question got undeleted and reopened. – Gerry Myerson Jan 27 '17 at 12:26

Gerry Myerson already answered in the comments, but let me be a little more explicit. Take any number $m\in\mathbb{N}$ with at least two prime factors (for example $6$). Then construct a Cantor set $C=C_m$ similar to the ternary Cantor set, except that instead of keeping two intervals of relative length $1/3$ we keep two intervals of relative length $1/m$ at each step. Alternatively, $C_m$ is the set of points in $[0,1]$ whose base $m$ expansion has only digits $0, m-1$. Clearly, no point in $C$ is remotely normal to base $m$.
On the other hand, $C$ carries a natural measure $\mu$, which can be defined by the property that is assigns the same mass $2^{-k}$ to all $2^k$ intervals in the $k$-th stage of the construction ($\mu$ is also a multiple of Hausdorff measure of the appropriate dimension on $C$).
It is then known that $\mu$-almost all points are normal to any base $p$ such that $\log p/\log m$ is irrational. In particular, this is true for all prime numbers $p$. So $\mu$-almost all points are counterexamples to the question.
The proof of this essentially goes back to Cassels [Cassels, J. W. S. On a problem of Steinhaus about normal numbers. Colloq. Math. 7 1959 95--101], who proved the special case $m=3$, by looking at the Fourier transform of $\mu$ along sequences $p^n \xi$. I am quite sure the proof works for any $m$. See also our paper [Hochman, Michael; Shmerkin, Pablo. Equidistribution from fractal measures. Invent. Math. 202 (2015), no. 1, 427--479.] for more discussion and generalizations.
Wolfgang Schmidt, Über die Normalität von Zahlen zu verschiedenen Basen, Acta Arith. 7 1961/1962 299–309, MR0140482 (25 #3902), according to the review by N. G. de Bruijn, proved this: Let the set of all integers $>1$ be split into two disjoint classes $R$ and $S$, such that for every $n>1$ all powers of $n$ belong to the same class as $n$ itself. Then there exists a continuum of numbers $\xi$ which all have the property that $\xi$ is normal with respect to every $r\in R$ and non-normal with respect to every $s\in S$.
In particular, let $R$ be the set of all primes and prime powers, $S$ the set of all integers with two or more distinct prime factors. Then there is a continuum (that is, an uncountable infinity) of numbers normal to every prime base and not normal to any base that isn't prime.