<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] First, you need that $x \geq 0$. Write $x=a+ \alpha$ where $\left\lfloor \sqrt x \right \rfloor=a$ and $\alpha <1$. Take $n^2 \leq a < (n+1)^{2}$ . Now, you need to prove that $n= \left \lfloor \sqrt a \right \rfloor = \left \lfloor \sqrt {a+ \alpha} \right \rfloor$. $\left \lfloor \sqrt {a+ \alpha} \right \rfloor \leq \left \lfloor \sqrt {(n+1)^2 -1 + \alpha} \right \rfloor <\left \lfloor \sqrt {(n+1)^2} \right \rfloor =n+1$. This implies $\left \lfloor \sqrt {a+ \alpha} \right \rfloor=n $ what you needed
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