Prove that if $n \in \Bbb{N}$ and $n > 1$ is not prime, then $\exists p$ prime such that $p \mid n$ and $p \leq \sqrt{n}$ Not really sure how to do this question

this is what I have so far

$n = a \cdot b$

$(a \leq b)$

$a &gt; \sqrt{n}$

$b &gt; \sqrt{n}$

$ab &gt; n$
You actually proved that if $n$ is not prime, it has a divisor $a&lt;\sqrt n$.

What remains to prove is that the smallest $a&gt;1$ that divides  $n$  is prime.

Indeed, suppose $a$ has a non-trivial factorisation: write $a=bc$, $\enspace1&lt;b,c&lt;a$. Then $n$ has divisors $&gt;1$ that are strictly less than $a$, so that $a$ is not the smallest divisor $&gt;1$ of $n$. Contradiction.
As $n$ is not prime it will at least have $2$ prime factors (may be repeated) and both of them can't be greater than $\sqrt n$, otherwise their product will exceed $n$. Hence at least of them will be $\leq \sqrt n$.