$\lfloor \sqrt{\lceil x \rceil} \rfloor = \lfloor \sqrt{x} \rfloor, \forall x \in \mathbb{R}$ 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.

The result,  as it is written now, is not true. There are many counterexamples. Let $x=8.5$. Then the right-hand side is $2$, while the left-hand side is $3$, since $\lceil 8.5\rceil=9$.
