Proof involving primes and divisibility

156 Views Asked by At

I am in a basic proofs class, and am having trouble with the following question:

Let $a \in \mathbb{Z}$ and let $p$ and $q$ be distinct primes. Show that if $p|a$ and $q|a$, then $pq|a$.

Since we're just starting, we're really only allowed to use the properties of primes and divisibility. However, after trying to prove it directly, through the contrapositive, and through contradiction, I was unable to get to any conclusion; noticeably, I'm not sure where to use the fact that p and q are primes in the question. On a related note: when trying to prove the contrapositive, I have: $$ a \ne pq*k$$ (for any $k \in \mathbb{Z}$). Since $a \ne p(qk)$, why does this not imply that $p$ does not divide $a$ (which is what we want to complete the contrapositive)?

Any and all help is appreciated. Thank you kindly!

1

There are 1 best solutions below

13
On BEST ANSWER

How about the following. $p|a,$ so $a=kp$ for some $k\in \mathbb{N}$. Now, $q|a$, so $q|kp$. But $p$ is prime, so $q$ can not divide $p$ because $p\ne q$. Then it must be the case that $q|k$. Then $k=jq$ for some $j\in \mathbb{N}$. Combining these facts, $a=jpq$, and we see that $pq|a$.