Question:
$\lfloor \sqrt{\lceil x \rceil} \rfloor = \lfloor \sqrt{x} \rfloor, \forall x \in \mathbb{R}$
My Attempt:
Let $a = \lfloor \sqrt{\lceil x \rceil} \rfloor$
$$a \leq \sqrt{\lceil x \rceil} < a + 1\\ a^2 \leq \lceil x \rceil < (a+1)^2$$
Since $a^2 = \lceil x \rceil \Rightarrow a^2 \leq \lceil x \rceil < a^2+1$, It follows
$$a^2 \leq x < (a+1)^2\\ a \leq \sqrt{x} < a+1$$
It follows,
$$a = \lfloor \sqrt{x} \rfloor$$
Since equal (=), is also an equivalence $\iff$, I reckon I should also prove the converse, but it is easy to see from here. I'm new to Discrete Math and floor/ceil functions, thus I need someone to confirm if my proof is right.
Note: There's also a known inequality (I've seen from one of the posts here), which is not in my book $k \geq \lceil r \rceil \iff k \geq r$ and $k \leq \lfloor r \rfloor \iff k \leq r$, I rcekon I could also use this one, but since it's not in my book I'm abstaining from using it.