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 $$\N$$.

$$ $$ $$ $$