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I'm having trouble proving this one. I know its true. Any ideas? Here is what I have so far:

If $a\mid b$, then there exists an integer $q_1$ such that $b = aq_1$.

If $a\mid c$, then there exists an integer $q_2$ such that $c = aq_2$.

I know the next part is gonna be like: Therefore, $c-b=aq_2-aq_1$.

I'm just a little lost at this point.

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If $b = aq$ and $c = aq_2$, then

$$c - b = aq_2 - aq = a(q_2 - q)$$

So $c$ is a multiple of $a$.

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Hint you are done! $a\mid a(q_2-q)=(c-b)$, isn't it?

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If $a|b$ and $a|c$, then $a|(c-b)$.

Since $a|b$ we know that $b=ax$ where $x\in\mathbb{Z}$ and since $a|c$ we know that $c=ay$ where $y\in\mathbb{Z}$. Consider $c-b=ay-ax=a(y-x)$. We know that $x,y\in\mathbb{Z}$ and so $y-x\in\mathbb{Z}$. Thus it follows that $a|(c-b)$.

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