Necessary and sufficient condition that $ \lceil \sqrt { \lfloor x \rfloor } \rceil = \lceil \sqrt { x } \rceil $ I am stumped at the following paragraph, which comes from Concrete Mathematics, Chapter 3, Section 2, Page 73:

What is a necessary and sufficient condition that $ \lceil \sqrt { \lfloor x \rfloor } \rceil = \lceil \sqrt { x } \rceil $? We have observed that equality holds when $ x = 3.142 $ but not when $ x = 1.618 $; further experimentation shows that it fails also when $ x $ is between $ 9 $ and $ 10 $. Oho. Yes. We see that bad cases occur whenever $ m^2 \lt x \lt m^2 + 1 $, since this gives $ m $ on the left and $ m + 1 $ on the right. In all other cases where $ \sqrt { x } $ is defined, namely when $ x = 0 $ or $ m^2 + 1 \le x \le (m + 1)^2 $, we get equality. The following statement is therefore necessary and sufficient for equality: Either $ x $ is an integer or $ \sqrt { \lfloor x \rfloor } $ isn't.

Starting from the bold sentence, I cannot follow it. The book says that when $ m^2 \lt x \lt m^2 + 1 $, the equation does not hold. But how did the authors come to this conclusion, by reasoning rather than just by observing some particular cases? And how can I derive the final conclusion (The necessary and sufficient condition is that either $ x $ is an integer or $ \sqrt { \lfloor x \rfloor } $ isn't.) from it?
 A: 
The book says that when $ m^2 \lt x \lt m^2 + 1 $, the equation does not hold. But how did the authors come to this conclusion, by reasoning rather than just by observing some particular cases? 

For $ m^2 \lt x \lt m^2 + 1 $ where $m$ is a non-negative integer, since  $m\lt \sqrt x\lt\sqrt{m^2+1}\le m+1$ we get
$$\left\lceil\sqrt{\lfloor x\rfloor}\right\rceil=m\lt m+1=\lceil\sqrt x\rceil$$
so the equality does not hold.

And how can I derive the final conclusion (The necessary and sufficient condition is that either $ x $ is an integer or $ \sqrt { \lfloor x \rfloor } $ isn't.) from it?

We may suppose that $m^2\le x\lt (m+1)^2$ separating it into three cases :
Case 1 : $x$ is an integer. We easily see that the equality holds.
Case 2 :  $m^2\lt x\lt m^2+1$. We've already seen that the equality does not hold.
Case 3 : $m^2+1\le x\lt (m+1)^2$ but $x$ is not an integer
There is no integer $n$ such that $m^2+1\le n^2\lt (m+1)^2$, so in this case, $\sqrt{\lfloor x\rfloor}$ is not an integer.
So, we get
$$m\lt \sqrt{m^2+1}\le \sqrt{\lfloor x\rfloor }\le \sqrt{(m+1)^2-1}\lt m+1\implies \left\lceil\sqrt{\lfloor x\rfloor }\right\rceil=m+1=\lceil\sqrt{x}\rceil$$
So, in this case, the equality holds.
Therefore, the necessary and sufficient condition is that either $x$ is an integer or $\sqrt{\lfloor x\rfloor}$ isn't.
