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So I gotta prove that if $p\mid b$ and $q\nmid b$, then $pq\nmid b$, where $p,q$ are primes and $b\in\mathbb{Z}$ . The thing it seems really simple, but I coudn't do it.

So I have $p\mid b$, ie, $b=kp$. And $q\nmid b$, ie, $b=qs+r$. Which gives me $kp=qs+r$. And then, I got nothing.

Thought about isolating $q$ on the last equation, but got anywhere too.

Can someone give me a hint/direction?

Bill Dubuque
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geep
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    Hint : Prove by contraposition that $q \not\mid b \Longrightarrow pq \not\mid b$, and conclude. – TheSilverDoe Jun 21 '22 at 21:07
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    They do not need to be prime. If $pq|b$ then ...? – Henry Jun 21 '22 at 22:02
  • $pq\mid b\Rightarrow q\mid b$ by transitivity of "divides". This is a dupe so please delete the question once all is clear. – Bill Dubuque Jun 22 '22 at 01:59
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    Primes have nothing to do with it and $p$ can be any integer in the universe. If $q\not \mid b$ then $pq\not \mid b$ no matter what $p$ is. Why? Because if $pq\mid b$ then both $q\mid b$ and $p\mid b$ but you were told explicitelythat $q$ doesn't divide $b$. So no multiple of $q$ can divide $b$. – fleablood Jun 22 '22 at 02:30

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We’ll use proof by contradiction.
Suppose, by way of contradiction, that $pq|b$. Then, $\exists$ some integer m such that $$b=mpq$$ This can be recast as $$b=(mp)q=aq$$ whre a is another integer. But $b=aq$ implies that $q|b$, which contradicts the hypothesis. Hence our assumption that $pq|b$ is false. Hence proved.
NOTE that $p|b$ is irrelevant here because we haven’t used it at all in our argument. Thus, the conclusion is that $pq|b \implies p|b$ as well as $q|b$.

insipidintegrator
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