<s>[INST] Proof: $ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $. I need some help with the following proof:
$ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $.
I got:
(1) $[ \sqrt{x} ] \le \sqrt{x} < [\sqrt{x}] + 1 $ (by definition?).
(2) $[ \sqrt{x} ]^2 \le x < ([\sqrt{x}] + 1)^2 $.
(3) $[ \sqrt{x} ]^2 \le [x] \le x < ([\sqrt{x}] + 1)^2$
??
 [/INST] Note that every nonnegative number $x$ is between the squares of two consecutive nonnegative integers. Say $n^2 \leq x < (n+1)^2$. Then $n \leq \sqrt{x} < n + 1$, so $\lfloor \sqrt{x} \rfloor = n$ in this situation.
So you just have to make sure the left hand side is also $n$. Since $n^2$ is an integer, one has
$n^2 \leq \lfloor x \rfloor \leq x$, so $n^2 \leq \lfloor x \rfloor < (n+1)^2$. So exactly as above one has $\lfloor \sqrt{\lfloor x \rfloor} \rfloor = n$ as well.
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