Proof that a composite number has a prime factor at most its square root

Proof

1. Let $n$ be a composite number. Then $n > 1$ and $n$ is not a prime number.
2. So $n$ has a positive divisor $a$ with $a \neq 1$ and $a \neq n$. Write $n = a b$ with an integer $b$.
3. Since $n > 0$ and $a > 0$, $b \ge 1$, so $a \le n$. Hence $1 < a < n$.
4. Then $b = n / a$ satisfies $1 < b < n$. If $b = 1$ then $a = n$, and if $b = n$ then $a = 1$; both are excluded.
5. By swapping $a$ and $b$ if needed, we may assume $a \le b$. Then $a^2 \le a b = n$.
6. By the existence of a prime divisor, since $a \ge 2$, there is a prime $p$ with $p \mid a$.
7. Since $p$ is a positive divisor of the positive integer $a$, $p \le a$. So $p^2 \le a^2 \le n$.
8. Since $p \mid a$ and $a \mid n$, $p \mid n$. So $p$ is a prime factor of $n$ with $p^2 \le n$, that is, $p \le \sqrt{n}$.