Integers form Integral Domain

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

We have:

$$\mathbb{Z} = \left\{{\left[\left[{a, b}\right]\right]_\boxminus: a, b \in \mathbb{N}}\right\}$$

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

Proof

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

* Integer Multiplication is Closed; * Integer Multiplication is Associative;
 * 1) The algebraic structure $$\left({\mathbb{Z}, +}\right)$$ is an abelian group.
 * 2) The algebraic structure $$\left({\mathbb{Z}, \times}\right)$$ is a semigroup:
 * 1) Integer Multiplication Distributes over Addition

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


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


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

I want to structure the sections in the first block so that it goes: 1., 2., *, *, 3. But I haven't worked out how.