Consider the decimal expansion of $ \sqrt 2$ in base $2$: $ \sqrt 2=1+\displaystyle \sum_{k\ge1} r_k 2^{-k}, r_k=0, 1. $ Let us prove the following fact: for every $n\ge1$, there is at least $n\le k\le 2n$ s.t $r_k=1.$
(Note that the proof is inspired from an existing one regarding $\sqrt 3$, but I lost the link}).
Assume that $r_k=0 $ for $ k=n,n+1,..,2n$. Then $2^{n-1} \sqrt 2=m+\epsilon, $ where $m\in\mathbf{N}$ and $0<\epsilon \le 2^{-n-1}$. It follows that $ 2^{2n-1}=m^2+2 m \epsilon+\epsilon^2.$ Moreover, we have $m< 2^{n-1} \sqrt 2,$ thus $0<2 m \epsilon+\epsilon^2< 2^{-1}\sqrt 2 + 2^{-2n-2}<1. $ Hence $2^{2n-1}=m^2$ which is impossible.
I wonder if the following result is also true: for every $n\ge1$ large enough, there is at least $n\le k\le 2n$ s.t $r_k=0.$