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So I'm unsure as to how to prove this:

If $a \mid b$ and $b \mid c$, then $a \mid (b+c)$.

I'm aware of the divisibility properties such as: if $a \mid b$, then $b=ak$ for some integer $k$.

I also know the Transitivity of Divisibility: Let $a$, $b$, and $c$ be integers. If $a \mid b$ and $b \mid c$, then $a \mid c$.

Any help as to how to approach this implication, or even some hints would be really appreciated.

Thanks.

Ken
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E 4 6
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    So you know that $a\mid b$ and $a \mid c$. Now what can you say about $b + c = ak + c = \ldots$? – Arthur Sep 10 '14 at 23:15
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    I think I've got it.. $b+c=b+bk$ so $b+c=b(k+1)$ -> $b+c=al(k+1)$ and since $l(k+1)$ is an integer, $a|(b+c)$ – E 4 6 Sep 10 '14 at 23:21
  • Yes, that's it. I was thinking of using $a\mid c$ to make it into $b + c = ak + al = a(k+l)$ (this would be a different $l$ to yours), but doing it your way means you don't have to directly use the transitivity. Incidently, you've more or less proven transitivity in there somewhere. – Arthur Sep 10 '14 at 23:28

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As you said, $a|c$, that means $c = ap$ for some integer $p$. Then:

$$b+c = ak+ap = a(k+p)$$

As both $k,p$ are integers, $k+p$ is an integer. By definition of divisibility, $a|b+c$

Rono
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