Floor of square root

Let $x \in \mathbb{R}$ be a positive real number. Prove that

$\left \lfloor \sqrt{\left \lfloor x \right \rfloor} \right \rfloor = \left \lfloor \sqrt{x} \right \rfloor$

Solution

We have successively:

$\begin{align*} n = \left \lfloor \sqrt{x} \right \rfloor & \Rightarrow n \leq \sqrt{x} < n+1 \\ &\Rightarrow n^2 \leq x < (n+1)^2 \\ &\Rightarrow n^2 \leq \left \lfloor x \right \rfloor < (n+1) ^2\\ &\Rightarrow n \leq \sqrt{\left \lfloor x \right \rfloor} < n+1 \\ &\Rightarrow n = \left \lfloor \sqrt{\left \lfloor x \right \rfloor} \right \rfloor \end{align*}$