Originally Posted by

**emakarov** I agree with the reply above:But the short answer to "Why 3?" questions is, because it is true (namely, if n >= 5, then n >= 3). It is one thing to find that some step in a proof is incorrect and another to not understand the author's intention behind a step. The former obviously invalidates the whole proof; the latter happens all the time and is normal. Hopefully, when you reach the end of the proof, you will understand why 3 and not 2 or 4 is chosen. In fact, you write that you see "that this gets the answer to the needed 2^(n+1) > n^2 + 2n + 1 = (n+1)^2".

Similarly, if n >= 5, then n > 1, so 2n + n > 2n + 1. To be extremely formal, you start with a true inequality n > 1 (which follows from n >= 5) and add 2n to both sides.

This is precisely how the proof proceeds, it just justifies n^2 > 2n + 1 in more detail.

You probably mean $\displaystyle n^2>3n=2n+n$.

I would write, Since $\displaystyle n > 1$ and, as we have shown, $\displaystyle n^2>2n+n$, we have $\displaystyle n^2 > 2n + 1$. (The fact that $\displaystyle n>1$ does not imply that $\displaystyle n^2>2n+n$.)