An unexplained condition on $a$ in a proof on the primes?

Question

Lemma

A positive integer $n$ is a prime if $(n,p) = 1$ for every prime integer $p \leq \sqrt{n}$

Proof in my text

Let $(n,p) = 1$ for every prime $p \leq \sqrt{n} \:$. Suppose $n$ is not a prime, we may write $n=ab$ with $1 < a \leq b \:$ , then $a \leq \sqrt {n} \:$. Any prime $p$ dividing $a$ also divides $n$ and we have $p \leq a \leq \sqrt {n} \:$, contradicting $(n,p) = 1$. Hence $n$ is a prime.

My question

I follow the proof, except the part where it is stated that $a \leq \sqrt {n} \:$. I just don't understand how this condition follows from $n = ab \rightarrow 1 < a \leq b$

Answer 1

If you don't understand a condition, it is generally a good idea to assume its opposite and see what goes wrong. In this case, $a > \sqrt n$ implies $b \ge a > \sqrt n$, hence $n = ab > n$.

Hope that helps,

Answer 2

Note that if $a > \sqrt n$, then since $ a \leq b$: $$a \cdot b > \sqrt n \cdot \sqrt n = n$$ meaning $ab > n$, which is in contradiction with the original statement $n = ab$.