Show $a_1 \mathbb Z · a_2 \mathbb Z$ = $(a_1 · a_2)$ · $\mathbb Z$ 
Let $a_1$, $a_2$ ∈ $\mathbb Z$, Show:
$a_1 \mathbb Z · a_2 \mathbb Z$ = $(a_1 · a_2)$ · $\mathbb Z$
where $c\mathbb Z = \{c · n : n ∈ \mathbb Z\}$

My attempt:
Let $x$ ∈ $a_1 \mathbb Z$ then $x$ can be written as $x = a_1.n$ for some $n$ ∈ $\mathbb Z$
Let $y$ ∈ $a_2 \mathbb Z$ then $y$ can be written as $y = a_2.m$ for some $m$ ∈ $\mathbb Z$
Let $z$ ∈ $(a_1.a_2) \mathbb Z$ then $z$ can be written as $z = (a_1.a_2).l$ for some $l$ ∈ $\mathbb Z$
$x.y = a_1.n . a_2.m$ = $a_1. a_2. n. m$ since multiplication is commutative in $\mathbb Z$
But since $l$ ∈  $\mathbb Z$, it can be written as a linear combination, say $l = n.m$ so $z = (a_1.a_2).l = a_1. a_2. n. m = x.y$, and hence $x.y=z$ and thus $x.y$ ∈  $(a_1 · a_2)$ · $\mathbb Z$, and therefore $a_1 \mathbb Z · a_2 \mathbb Z$ = $(a_1 · a_2)$ · $\mathbb Z$
Is my attempt correct?
 A: The proof is lacking a bit of clarity in its presentation. You start by defining an arbitrary element, $x \in a_1 \Bbb{Z}$, then an unrelated arbitrary element $y \in a_2 \Bbb{Z}$, and finally, a third arbitrary element $z \in (a_1 a_2) \Bbb{Z}$, seemingly unrelated to both $x$ and $y$. In the final paragraph, you seem to change $z$ to now relate to $x$ and $y$, and in particular, so that it simplifies to $xy$.
It would be better if you declared $x$ and $y$ (and deduced the existence of $m$ and $n$ as you have), then let $z = (a_1a_2)(mn)$. You can conclude that $z \in (a_1a_2)\Bbb{Z}$, and that $z = xy$. This means $xy \in (a_1a_2)\Bbb{Z}$, as required.
This way, you don't have to pretend $z$ is arbitrary initially, and you only need to define it once it becomes useful to your proof. You could even skip defining $z$, and note that $xy = (a_1a_2)(mn) \in (a_1a_2)\Bbb{Z}$.
Also, bear in mind that all you've proven is $(a_1 \Bbb{Z})(a_2 \Bbb{Z}) \subseteq (a_1a_2) \Bbb{Z}$. You also need a separate argument for the converse direction. That is, you need to be able to start with some arbitrary $z \in (a_1a_2) \Bbb{Z}$ and construct elements of $x \in a_1 \Bbb{Z}$ and $y \in a_2 \Bbb{Z}$, such that $z = xy$. Hint: $1$ is a perfectly fine integer.
A: Show two inclusions in a systematic way:
Let $z \in a \Bbb Z \cdot b \Bbb Z$. This means that we can write $z=xy$ for some $x \in a \Bbb Z $ and $y \in \Bbb Z $. By definition we then know there is some $n \in \Bbb Z $ and some $m \in \Bbb Z $ such that $x=an$ and $y=bm$. 
Then we can conclude that (using standard commutative ring properties of $\Bbb Z $)
$$z=xy=(an)(bm)=(ab)(nm)$$ and because $nm \in \Bbb Z $ we conclude that $z \in (ab) \Bbb Z $, showing one inclusion.
Now let $z \in (ab) \Bbb Z $ be given. By definition this means that $z=(ab)n$ for some $n \in \Bbb Z $. Again using simple ring facts:
$$z=(ab)n= (an)(b1) \in a \Bbb Z \cdot b \Bbb Z $$
where the last step is by definitions again: $an \in a \Bbb Z $, $b1 \in b \Bbb Z $ and so their product is in the product of the sets.
Both inclusions being shown, equality is shown too. The only "clever(ish)" bit is that we use $1$ as an integer in the reverse inclusion. The rest should write itself.
