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I am working on the problem:

Let $m$ and $n$ be integers in the ring of integers. Show that $m\mathbb Z$ contains $n\mathbb Z$ if and only if $m$ divides $n$.

It's an if and only if proof so two directions to show. Start with the fact $n\mathbb Z$ is a subset of $m\mathbb Z$ and want to show $n=mx$ for some $x \in \mathbb Z$. Help to proceed?

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If $n\mathbb{Z}\subset m\mathbb{Z}$ then $n\in m\mathbb{Z}$ so $n=md$ for some $d\in\mathbb{Z}$.

If $n=md$ for some $d\in\mathbb{Z}$ and $k\in n\mathbb{Z}$ then $k=ne=mde\in m\mathbb{Z}$ for some $e\in\mathbb{Z}$, so $n\mathbb{Z}\subset m\mathbb{Z}$.

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Hint: $n\in n\Bbb Z\subseteq m\Bbb Z$.

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$m\mathbb{Z} \subset n\mathbb{Z}$ iff $ m \in n \mathbb{Z}$ iff $ m = nk $ for some integer $k$ iff $ n \mid m$

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