Floor function identity for $[\sqrt{n^2-n}]$ From my textbook, where [n] is the greatest integer less than or equal to n.

Since $n^2-n=(n-\frac{1}{2})^2-\frac{1}{4}$, it follows, $[\sqrt{n^2-n}]=n-1$

This one feels like a great leap, I really don't see it at all and it doesn't feel like the kind of question where I can say "I tried doing ..."
 A: It's a bit of a joke in the mathematical community that the phrase "it follows" doesn't necessarily give any indication of how elementary the implication is.  Ideally, an author has an appropriate sense of their audience when they say that, but sometimes it takes a wall of algebra to justify a conclusion that an established mathematician would regard as "obvious".  As you grow as a student, you will find your sense of obviousness will expand.
In this case, one presumes that $n$ is positive integer.  Then
$$(n-1)^2=n^2-2n+1\le n^2-n<n^2$$
and since every expression there is positive we can take the square root of everything to get
$$n-1\le\sqrt{n^2-n}<n$$
And the stated implication follows immediately from the definition of the greatest integer function.

What was the author going for with the statement they cited?  I suppose they were saying that $\sqrt{n^2-n}\approx n-\frac12$ as $n$ becomes large.  Frankly, I don't think that's a particularly strong argument since you don't how a priori how the square root behaves before $n$ becomes large, but one assumes that it is "obvious" to the author.
