Exercise 1.3.24 (if $n$ is composite, it must have a prime factor $p \leq \sqrt{n}$.)

Proof. Since $n>1$ is composite, then $n=\mathit{ab}$ for some integers $a,b$ such that $1<a\le b<n$. Then $a^2\le \mathit{ab}=n$, thus $a\le \sqrt{n}$. Since $a>1$, $a$ has some prime factor $p$, and $p\le a\le \sqrt{n}$.

If $n$ is composite, it must have a prime factor $p\le \sqrt{n}$. □