Is every irrational number normal in at least one base? 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?
 A: 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.
