Problem: Let $\sqrt{3}=1.b_1b_2b_3\dots_{(2)}$ be the binary representation
of $\sqrt 3$. Prove that for any positive integer $n$, at least one of
the digits $b_n, b_{n+1}, \dots, b_{2n}$ equals 1.

Solution: Assume the contrary, so that for some integer $k$ we have
\[k < 2^{n-1} \sqrt 3 < k + \frac{1}{2^{n+1}}. \]
Squaring gives
\begin{align*}
k^2 < 3 \cdot 2^{2n-2} &< k^2 + \frac{k}{2^n} + \frac{1}{2^{2n+2}} \\
&\le k^2 + \frac{2^{n-1} \sqrt 3}{2^n} + \frac{1}{2^{2n+2}} \\
&= k^2 + \frac{\sqrt3}{2} + \frac{1}{2^{2n+2}} \\
&\le k^2 + \frac{\sqrt3}{2} + \frac{1}{16} \\
&< k^2+1
\end{align*}
and this is a contradiction.