Prove that a*b is not divisible by prime number p

Question

Can't understand if my thoughts make sense or not.

The question is the following: $ab$ is not divisible by prime $p$, if both $a$ and $b$ are not divisible by $p$.

I thought to prove it like this: For $ab$ there are 2 cases: a) divisible b) not divisible

For $a$ and $b$ there cases:

1. $a$ is divisible by $p$, $b$ is not
2. $b$ is divisible by $p$, $a$ is not
3. $a$ and $b$ are divisible by $p$
4. $a$ and $b$ are not divisible by $p$

Cases 1,2,3 correspond to the case a) it is clear probably. Then we have the only left case 4, which corresponds to the only left case b). Means that if both $a$ and $b$ are not divisible by $p$, then $ab$ is not divisible by $p$.

Answer 1

You're on the right track but you created more work for yourself by working with these cases separately. I'm assuming you are working over the integers, and the proof should follow directly from the unique factorization property. Let $a = p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}$ and $b = q_1^{f_1}q_2^{f_2}\cdots q_m^{f_m}$ and suppose that a prime $\ell$ does not divide $a$ or $b.$ Then,

$ab = p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}q_1^{f_1}q_2^{f_2}\cdots q_m^{f_m}$

where the prime factors need not be distinct (since $a$ and $b$ could have prime factors in common) but $\ell$ will not be a factor since it is not contained in any of the $p_i$ or $q_j$ for $1 \leq i \leq n, 1 \leq j \leq m.$ We conclude $\ell \not \mid ab.$