1. ## Little algebra proof

If $\displaystyle c > 0$ is an integer and not prime, then there is an integer $\displaystyle b > 0$ such that $\displaystyle b|c$ and $\displaystyle b \leq \sqrt{c}$.

Approach so far:

Contradiction, let us assume that $\displaystyle b > \sqrt{c}$.

Since $\displaystyle br = c$ for some integer $\displaystyle r$, then $\displaystyle r < c$.

I feel I can arrive at a contradiction, since $\displaystyle r \geq c$ from our initial statement, but not sure how to arrive at such.

2. Originally Posted by rowe
If $\displaystyle c > 0$ and not prime, then there is an integer $\displaystyle b > 0$ such that $\displaystyle b|c$ and $\displaystyle b \leq \sqrt{c}$.
I am sure you mean that c is a positive integer.
If $\displaystyle c$ is not prime then there are positive integers, $\displaystyle a~\&~b$ such that $\displaystyle c=a\cdot b$.
What if $\displaystyle a>\sqrt{c}~\&~b>\sqrt{c}?$

3. Then $\displaystyle ab \neq c$, but why do we assume $\displaystyle a > \sqrt{c}$?