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.