Integers form Integral Domain

Theorem
The integers form an integral domain under addition and multiplication.

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 $$\left({x_1, y_1}\right) \boxminus \left({x_2, y_2}\right) \iff x_1 + y_2 = x_2 + y_1$$.

We have that $$\Z = \left\{{\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus: a, b \in \N}\right\}$$.


 * First we establish that the system of integers is a ring.


 * 1) The algebraic structure $$\left({\Z, +}\right)$$ is an abelian group.
 * 2) The algebraic structure $$\left({\Z, \times}\right)$$ is a monoid and therefore a semigroup.
 * 3) Integer Multiplication Distributes over Addition.

Thus all the ring axioms are fulfilled, and $$\left({\Z, +, \times}\right)$$ is a ring.

By Integer Multiplication has a Zero, the zero is $$\left[\!\left[{\left({0, 0}\right)}\right]\!\right]_\boxminus$$.


 * Next we show that the additional properties are fulfilled for $$\left({\Z, +, \times}\right)$$ to be an integral domain.


 * 1) $$\left({\Z, +, \times}\right)$$ is a commutative ring as Integer Multiplication is Commutative.
 * 2) $$\left({\Z, +, \times}\right)$$ has a unity, and the unity is $\left[\!\left[{\left({1, 0}\right)}\right]\!\right]_\boxminus$.
 * 3) $\left({\Z, +, \times}\right)$ has no divisors of zero.

Thus the algebraic structure $$\left({\Z, +, \times}\right)$$ is an integral domain, whose zero is $$\left[\!\left[{\left({0, 0}\right)}\right]\!\right]_\boxminus$$ and whose unity is $$\left[\!\left[{\left({1, 0}\right)}\right]\!\right]_\boxminus$$.