Integer Multiplication is Closed

Theorem
The set of integers is closed under multiplication:
 * $\forall a, b \in \Z: a \times b \in \Z$

Proof
Let us define $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$ as in the formal definition of integers.

That is, $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.

$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\left({x_1, y_1}\right) \boxminus \left({x_2, y_2}\right) \iff x_1 + y_2 = x_2 + y_1$.

In order to streamline the notation, we will use $\left[\!\left[{a, b}\right]\!\right]$ to mean $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$, as suggested.

Integer multiplication is defined as:


 * $\forall a, b, c, d \in \N: \left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{c, d}\right]\!\right] = \left[\!\left[{ac + bd, ad + bc}\right]\!\right]$

We have that:
 * $\forall a, b, c, d \in \N: \left[\!\left[{a, b}\right]\!\right] \in \Z, \left[\!\left[{c, d}\right]\!\right] \in \Z$

Also:
 * $\forall a, b, c, d \in \N: \left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{c, d}\right]\!\right] = \left[\!\left[{ac + bd, ad + bc}\right]\!\right]$

But:
 * $ac + bd \in \N, ad + bc \in \N$

So:
 * $\forall a, b, c, d \in \N: \left[\!\left[{ac + bd, ad + bc}\right]\!\right] \in \Z$

Therefore integer multiplication is closed.