The other day I got to thinking about the decimal expansion of $\sqrt{2}$, and I stumbled upon a somewhat embarrassing problem.
There cannot be only one digit that occurs infinitely often in the decimal expansion of $\sqrt{2}$, because otherwise it would be rational (e.g. $\sqrt{2} = 1.41421356237\ldots 11111111\ldots$ is not possible).
So there must be at least two digits that occur infinitely often, but are there more? Is it possible that e.g. $\sqrt{2} = 1.41421356237\ldots 12112111211112\ldots$?
This problem is wide open. It is conjectured that every irrational algebraic number is absolutely normal (i.e. in every base, digits appear asymptoticaly with the same density). However, it is not even known whether there is any algebraic irrational with some three digits appearing infinitely many times in any base! Hence, to the best of our knowledge, every irrational algebraic number could eventually have only zeroes and ones in every base.