Rational Numbers form Ordered Integral Domain

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Theorem

This follows directly from Field of Rational Numbers: the set of rational numbers $\Q$ forms a totally ordered field under addition and multiplication: $\left({\Q, +, \times, \le}\right)$.

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

We have that the rational numbers $\left({\Q, +, \times}\right)$ form an integral domain. What is needed now is to specify a property $P$ on $\Q$ such that:

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

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

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

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

Let us define the property $P$ on $\Q$ as:

$\forall a \in \Q: P \left({a}\right) \iff a = \dfrac p q: P' \left({p}\right), P' \left({q}\right)$

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

Now let $a = \dfrac p q$ and $b = \dfrac r s$ such that $P \left({a}\right)$ and $P \left({b}\right)$.

Then by definition of rational addition:

$\displaystyle \frac p q + \frac r s = \frac {p s + r q} {q s}$

$\displaystyle \frac p q \times \frac r s = \frac {p r} {q s}$

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

It can be seen that if $P \left({a}\right)$ then $\neg P \left({-a}\right)$ and vice versa.

Also we note that $\neg P \left({0}\right)$ and of course $\neg P \left({-0}\right)$.

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

Hence the result.

$\blacksquare$