Integer Multiplication has Zero

Theorem
The set of integers under multiplication $\left({\Z, \times}\right)$ has a zero element, which is $0$.

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.

From the method of construction, $\left[\!\left[{c, c}\right]\!\right]$, where $c$ is any element of the natural numbers $\N$, is the identity of $\left({\Z, +}\right)$.

To ease the algebra, we will take $\left[\!\left[{0, 0}\right]\!\right]$ as a canonical instance of this equivalence class.

We need to show that:

$\forall a, b, c \in \N: \left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{0, 0}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right] \times \left[\!\left[{a, b}\right]\!\right]$.

From Natural Numbers form Commutative Semiring, we can take it for granted that addition and multiplication are commutative on the natural numbers $\N$.