Integers form Totally Ordered Ring

Theorem
The structure $$\left({\mathbb{Z}, +, \times; \le}\right)$$ is a totally ordered ring.

Proof

 * From Integers form Integral Domain, $$\left({\mathbb{Z}, +, \times}\right)$$ is an integral domain, which is a commutative ring.


 * From Integer Addition forms Totally Ordered Group, $$\left({\mathbb{Z}, +; \le}\right)$$ is a totally ordered group.


 * We need to show that $$\le$$ is a compatible ordering on $$\mathbb{Z}$$. That is, that:


 * 1) $$\le$$ is compatible with $$+$$;
 * 2) $$\forall x, y \in \mathbb{Z}: 0 \le x, 0 \le y \Longrightarrow 0 \le x \times y$$.

The first one follows from the fact that $$\left({\mathbb{Z}, +; \le}\right)$$ is a totally ordered group.

Now, from Multiplicative Ordering on Integers, we have:

$$\forall z, x, y \in \mathbb{Z}, 0 < y: z \le x \iff y \times z \le y \times x$$

Let $$z = 0, 0 \le x$$.

Then $$0 = x \times z \le x \times y = y \times x$$.

So if $$y \ne 0$$, the condition $$0 \le x, 0 \le y \Longrightarrow 0 \le x \times y$$ holds.

Now if $$y = 0$$, we have $$x \times y = 0$$ and thus $$0 \le x \times y$$.

Thus $$\le$$ is compatible with $$\mathbb{Z}$$, and the result follows.