2
$\begingroup$

This is my attempt to prove the following (excerpt from the exercises in my textbook):

Let $p$ be an integer other than $0, \pm 1$ with this property: Whenever $b$ and $c$ are integers such that $p \vert bc$, then $p\vert b$ or $p\vert c$. Prove that $p$ is prime.

Here is my attempt:

If $d\vert p$ then we have $p=dt$ for some $t\in \mathbb Z$. This implies that $p\vert d$ or $p\vert t$. If $p\vert d$ then $\vert p \vert \leq \vert d \vert$ and since we have assumed $d \vert p$ we have $\vert d \vert \leq \vert p \vert$. This implies that $\vert d \vert = \vert p \vert$ so $d= \pm p$.

If $p$ does not divide $d$ then $p \vert t$ which implies $\vert p \vert \leq \vert t \vert$. Together with $p=dt$ this implies that $d=\pm 1$. This shows that the only possible divisors of $p$ are $\pm 1$ and $\pm p$, meaning that $p$ is prime.

I would greatly appreciate any constructive feedback/criticism on my attempted proof.

$\endgroup$
1
  • $\begingroup$ thanks, I'll edit $\endgroup$ Mar 8 '19 at 1:08
0
$\begingroup$

I have trouble understanding your logic, but here is my version. Assume on the contrary that $p$ is not prime, then $p=rs$, now p|rs implies $p|r$ or $p|s$, but both are false.

$\endgroup$
3
  • $\begingroup$ I believe if $p$ is not prime, then $p=rs$ does not imply $p\vert r$ or $p\vert s$ consider $p=12$, $r=3$ and $s=4$. Then we have $12=3 \cdot 4$, but $12$ does not divide $3$ and $12$ does not divide $4$. $\endgroup$ Mar 8 '19 at 11:41
  • $\begingroup$ I said "if p=rs, then ..." by your assumption. $\endgroup$ Mar 8 '19 at 12:38
  • $\begingroup$ I am using proof by contradiction . $\endgroup$ Mar 8 '19 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.