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Show that ∀a, b, c ∈ Z, a|b ∨ a|c ⇒ a|bc.

How would I prove the following in discrete math?

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closed as off-topic by Shaun, mrtaurho, Joshua Mundinger, Lee David Chung Lin, dantopa Apr 16 at 21:04

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    $\begingroup$ What have you tried? Hint: $a\,|\,b\implies b=ad$ for some $d\in \mathbb Z$. $\endgroup$ – lulu Apr 16 at 20:02
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    $\begingroup$ So apparently (based on your comment to Peter Foreman's answer) you don't understand the notation used in the question. That's an honest matter. However, the fact that instead of asking what the notation means, you're just typing in the question and hoping someone will do your homework for you so you don't have to learn anything yourself, that is worth a downvote. $\endgroup$ – Henning Makholm Apr 16 at 20:16
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You don't even need that $a|c$ for this to be true. If $a|b$ then there exists $d\in\mathbb{Z}$ such that $b=ad$ hence $bc=abd$. This implies that $a|bc$ as $bc=a(bd)$ so as there exists $bd\in\mathbb{Z}$ such that $bc=a(bd)$ we must have $a|bc$.

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  • $\begingroup$ a|c means the difference of a and c, right? $\endgroup$ – user664799 Apr 16 at 20:12
  • $\begingroup$ No it means that $a$ divides $c$. $\endgroup$ – Peter Foreman Apr 16 at 20:12
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This is evidently a HW question and I assume your teacher won't like this, but anyways I will leave a small bug that needs touch up in here.

Proof:

without loss of generality, a|b.

Hence $\exists p\in\mathcal{Z}$ S.T.$b = ap$. Thus $bc/a = apc/a = pc$ which is in $\mathcal{Z}$ since the set of $\mathcal{Z}$ is closed under multiplication.

QED

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Since $bc=cb$, assume without loss of generality that $a|b$. Then, by definition of divisibility, we have $b=ad$, for some $d\in \Bbb Z$. Consider the product $bc$. Can you continue from here?

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