(i) Show that for real $x ≥ 0$ the following is false: $√⌊x⌋ = ⌊√x⌋$.
(ii) Prove that for real $x ≥ 0$ we have $⌊√⌊x⌋⌋ = ⌊√x⌋$.
(i) We show this with a counter-example.
We choose $x=2$,
$\sqrt{\lfloor 2 \rfloor} = \sqrt{2} = 1.4142 \ldots$
$\lfloor \sqrt{2} \rfloor = \lfloor 1.4142 \ldots \rfloor = 1$
So $x=2$ shows that $\sqrt{\lfloor x \rfloor} = \lfloor \sqrt{x} \rfloor$ is not always true.
(ii) If $d$ is an integer such that
$$ d \le \sqrt{x} < d + 1 $$
then $d = \lfloor \sqrt{x} \rfloor$, by definition of the floor function.
Since squaring positive real numbers is monotonically increasing, we can square over this inequality
$$ d^2 \le x < (d + 1)^2 $$
Since $d^2$ is an integer, we can say
$$d^2 \le \lfloor x \rfloor \tag{i}$$
Also, since $\lfloor x \rfloor \le x$ and $x < (d + 1)^2$ we have
$$ \lfloor x \rfloor < (d+1)^2 \tag{ii}$$
Combining (i) and (ii)
$$d^2 \le \lfloor x \rfloor < (d+1)^2 $$
We can take the square root since it is a monotonically increasing function
$$d \le \sqrt{\lfloor x \rfloor} < d+1 $$
This is means $d = \lfloor \sqrt{\lfloor x \rfloor} \rfloor$.
So $d = \lfloor \sqrt{x} \rfloor$, and $d = \lfloor \sqrt{\lfloor x \rfloor} \rfloor$, which means we have shown
$$ \lfloor \sqrt{\lfloor x \rfloor} \rfloor = \lfloor \sqrt{x} \rfloor$$