# 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 $\struct {\Z, +, \times}$ 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 $\struct {\N_{> 0}, +, \times}$ is a (commutative) semiring.

So it follows by definition of semiring, in particular the fact that $+$ and $\times$ on $\N_{> 0}$ are closed, that:

- $\forall a, b \in \N_{> 0}: a + b \in \N_{> 0}$

- $\forall a, b \in \N_{> 0}: a \times b \in \N_{> 0}$

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

- $\forall x \in \Z: \map P x \iff x \in \N_{> 0}$

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

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

Follows directly from the fact that $P \left({x}\right) \iff x \in \N_{> 0}$.

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

Follows directly from the fact that $P \left({x}\right) \iff x \in \N_{> 0}$.

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

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 $\struct {\Z, +, \times}$ forms an ordered integral domain.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $9$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers