Is every irrational number normal in at least one base?

Question

It is not known if $\pi$ and $\sqrt{2}$ are normal in base 10. But is every irrational number normal in at least one base?

Answer 1

Comment expanded into an answer to stop this from unanswered.

The answer is NO.

Absolutely abnormal irrational numbers (i.e. one that is not normal in every base $b \ge 2$) do exist.
A web search return this paper by Greg Martin.

To construct an example, the paper first define a sequence of integers

$$d_2 = 2^2,\; d_3 = 3^2,\; d_4 = 4^3,\; d_5 = 5^{16},\; d_6 = 6^{30517578125},\; \ldots$$

with the recursive rule:

$$d_j = j^{d_{j-1}/(j-1)} \quad\text{ for } j \ge 3.$$

One then define another sequence of rational numbers

$$\alpha_k = \prod_{j=2}^k \left( 1 - \frac{1}{d_j}\right)$$

and the limit

$$\alpha = \lim_{k\to\infty} \alpha_k \approx 0.6562499999956991 \underbrace{99999\ldots99999}_{23747291559\;\text{9s}} 8528404201690728\ldots$$

will be an example for a abnormal irrational number. In fact, by construction, this number is a Liouville number and hence transcendental.

Despite its horrible looking, the basic idea behind this construct is not that complicated.

For any base $b \ge 2$, we knew how to construct an irrational that is non-normal for that base. For example, for base $10$, the Liouville number defined by

$$\beta = \sum_{n=1}^\infty 10^{-n!} \approx 0.11000100000000000000000100\ldots$$

is non-normal. It is easy to verify $\beta$ is non-normal in base $10$ because each successive rational approximation of $\beta$, $\sum_{n=1}^k 10^{-n!}$, are $10$-adic fractions. The problem is we cannot rule out the possibility that this number $\beta$ is normal in some other base. To address this issue, the paper construct the Liouville number $\alpha$ above as one whose successive rational approximations are $b$-adic fractions with $b$ varying!

For more details how this is done and a rigorous proof that $\alpha$ is absolutely abnormal, please consult the paper mentioned above.