Prove that if $n=ab$, $a$ and $b$ are positive integers, then $a \leq \sqrt{n}$ or $b \leq \sqrt{n}$
Method chosen: Proof by contradiction.
suppose $n=ab$, $a$ and $b$ are positive integers, then $a>\sqrt{n}$ and $b>\sqrt{n}$
Let $n=30$ and $a=5$, $b=6$.
$5 >\sqrt{30}$ and $6>\sqrt{30}$.
Contradiction, $5 > \sqrt{30}$ is false.
Therefore it is not the case $a>\sqrt{n}$ and $b>\sqrt{n}$, so it is true that $a \leq \sqrt{n}$ or $b \leq \sqrt{n}$
Did I do this right? Any help would be appreciated.