Ring of Integers has no Zero Divisors

Theorem
The integers have no zero divisors:
 * $\forall x, y, \in \Z: x \times y = 0 \implies x = 0 \lor y = 0$

This can equivalently be expressed:
 * $\forall x, y, \in \Z: x \ne 0 \land y \ne 0 \implies x \times y \ne 0$

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:


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

From Natural Numbers form Commutative Semiring, we can take it for granted that:
 * addition and multiplication are commutative and associative on the natural numbers $\N$;
 * natural number multiplication is distributive over natural number addition.

So:

We have to be careful here, and bear in mind that $a, b, c, d$ are natural numbers, and we have not defined (and, at this stage, will not define) subtraction on such entities.

, suppose that $\left[\!\left[{c, d}\right]\!\right] \ne \left[\!\left[{0, 0}\right]\!\right]$.

Then $c \ne d$.

, suppose also that $c > d$.

From Ordering in terms of Addition, $\exists p \in \N: d + p = c$ where $p > 0$.

Then:

Similarly for when $c < d$.

Thus:
 * $\left[\!\left[{c, d}\right]\!\right] \ne \left[\!\left[{0, 0}\right]\!\right] \implies \left[\!\left[{a, b}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right]$

A similar argument shows that:
 * $\left[\!\left[{a, b}\right]\!\right] \ne \left[\!\left[{0, 0}\right]\!\right] \implies \left[\!\left[{c, d}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right]$

The equivalence between the two forms of the statement of this theorem follows from De Morgan's Laws: Conjunction of Negations and the Rule of Transposition.