Integers form Ordered Integral Domain

Theorem

The integers $\Z$ form an ordered integral domain under addition and multiplication.

Proof

From Integers form Integral Domain we have that $\left({\Z, +, \times}\right)$ forms an integral domain.

From Natural Numbers are Non-Negative Integers we have that the set $\N$ can be considered as a subset of the integers.

Then we have that $\left({\N^*, +, \times}\right)$ is a (commutative) semiring.

So it follows by definition of semiring, in particular the fact that $+$ and $\times$ on $\N^*$ are closed, that:

$\forall a, b \in \N^*: a + b \in \N^*$

$\forall a, b \in \N^*: a \times b \in \N^*$

It follows that we can define a property $P$ on $\Z$ such that:

$\forall x \in \Z: P \left({x}\right) \iff x \in \N^*$

Checking that $P$ fulfils the conditions for it to be the positivity property:

$(1): \quad \forall a, b \in \Z: P \left({a}\right) \land P \left({b}\right) \implies P \left({a + b}\right)$

Follows directly from the fact that $P \left({x}\right) \iff x \in \N^*$.

$(2): \quad \forall a, b \in \Z: P \left({a}\right) \land P \left({b}\right) \implies P \left({a \times b}\right)$

Follows directly from the fact that $P \left({x}\right) \iff x \in \N^*$.

$(3): \quad \forall a \in \Z: P \left({a}\right) \lor P \left({-a}\right) \lor a = 0$

This follows from the definition of the integers as the inverse completion of the natural numbers.

If $a \in \Z - \N$ then $\exists b \in \N: a + b = 0$ and so $a = -b$.

Hence the conditions are fulfilled and $\left({\Z, +, \times}\right)$ forms an ordered integral domain.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 2.7$: Example $9$