1. ## Prime Theorem-Proof?

P\a

2. Originally Posted by thegarden
Proof required for this theorem:
Let p be prime, p|ab implies p|a or p|b.
Thanks.
Surely p isn't a prime then?

3. I think we can assume p is prime

4. If p is prime, then p does not divide a. And p does not divide b. By definition of being prime. (Unless a and b are 1 and p).
So proving that p divides a or b would be rather difficult, no?

Are you sure you've got the question right?

a and b are integers, I'm assuming...

5. Originally Posted by thegarden
Proof required for this theorem:
Let p be prime, p|ab implies p|a or p|b.
Thanks.
This theorem is correct. (" $p|a$" means that $p$ divides $a$, or that $a$ is a multiple of $p$, not the contrary)

It results from Gauss Theorem: We assume that $p|ab$. If $p|a$, we are done. Suppose $p\not|a$. Then $a$ and $p$ are relatively prime.
(If $d$ divides $a$ and $p$, then $d=1$ or $d=p$ because $p$ is prime, hence necessarily $d=1$ since $p\not|a$; hence $1$ is the only positive common divisor of $a$ and $p$)
Because $p|ab$, you deduce from Gauss Theorem that $p|b$. qed