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題目

Prove that $\lfloor\sqrt{\lfloor{x}\rfloor}\rfloor = \lfloor\sqrt{x}\rfloor$ for any real number $x$.

解答

It's clearly that there exist a positive integer n, s.t.

$\begin{aligned} & n^2\leq\lfloor{x}\rfloor\leq x < {(n+1)}^2\\ \Rightarrow & n\leq \sqrt{\lfloor{x}\rfloor} \leq \sqrt{x} < n+1\\ \Rightarrow & n\leq \lfloor\sqrt{\lfloor{x}\rfloor}\rfloor \leq \lfloor\sqrt{x}\rfloor < n+1 \end{aligned}$

Hence $n = \lfloor\sqrt{\lfloor{x}\rfloor}\rfloor = \lfloor\sqrt{x}\rfloor$