Rational Numbers form Ordered Integral Domain

Theorem
The rational numbers $\Q$ form an ordered integral domain under addition and multiplication.

Proof
This follows directly from Rational Numbers form Totally Ordered Field:

The set of rational numbers $\Q$ forms a totally ordered field under addition and multiplication: $\struct {\Q, +, \times, \le}$.

However, it is useful to demonstrate this directly from the definition of the ordered integral domain.

We have that the rational numbers $\struct {\Q, +, \times}$ form an integral domain.

What is needed now is to specify a property $P$ on $\Q$ such that:


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


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


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

We have that the integers $\struct {\Z, +, \times}$ form an ordered integral domain.

Let $P'$ be the (strict) positivity property on $\struct {\Z, +, \times}$.

Let us define the property $P$ on $\Q$ as:
 * $\forall a \in \Q: \map P a \iff a = \dfrac p q: \map {P'} p, \map {P'} q$

That is, an element $a = \dfrac p q$ has $P$ both $p$ and $q$ have the (strict) positivity property in $\Z$.

Now let $a = \dfrac p q$ and $b = \dfrac r s$ such that $\map P a$ and $\map P b$.

Then by definition of rational addition:


 * $\dfrac p q + \dfrac r s = \dfrac {p s + r q} {q s}$

and rational multiplication:


 * $\dfrac p q \times \dfrac r s = \dfrac {p r} {q s}$

It can be seen from the definition of (strict) positivity $P'$ on $\Z$ that $\map P {a + b}$ and $\map P {a \times b}$.

It can be seen that if $\map P a$ then $\neg \map P {-a}$ and vice versa.

Also we note that $\neg \map P 0$ and of course $\neg \map P {-0}$.

So the property $P$ we defined fulfils the criteria for the (strict) positivity property.

Hence the result.