Composite Number has Prime Factor not Greater Than its Square Root

Theorem
Let $$n \in \mathbb{Z}$$ and $$n = p_1 \times p_2 \times...\times p_j$$, where $$p_1,...,p_j \in \mathbb{P}$$ are prime factors of $$n$$. Then, $$\exists p_i \in \mathbb{P}$$, such that $$p_i \le \sqrt{n}$$.

That is, if $$n \in \mathbb{Z}$$ is composite, then $$n$$ has a prime factor $$p \le \sqrt{n}$$.

Proof

 * Since $$n$$ is composite, let $$n=ab$$ where $$a,b \in \mathbb{Z}$$ and $$1 \sqrt{n}$$, then $$b \ge a > \sqrt{n}$$.
 * However, if $$b \ge a > \sqrt{n}$$ is true, then $$n=ab > \sqrt{n}\sqrt{n} > n$$.
 * This is clearly a contradiction.

Therefore, any prime factor of $$a$$ must be less than or equal to $$\sqrt{n}$$.