The other day I got to thinking about the decimal expansion of √2, and I stumbled upon a somewhat embarrassing problem.

There cannot be only one digit that occurs infinitely often in the decimal expansion of √2, because otherwise it would be rational (e.g. √2=1.41421356237…11111111… 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. √2=1.41421356237…12112111211112…?

**Answer**

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.

**Attribution***Source : Link , Question Author : Mankind , Answer Author : Wojowu*