Rational Numbers form Ordered Integral Domain

Theorem

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

$\blacksquare$

It remains to specify a property $P$ on $\Q$ such that:

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

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

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

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 $P$ if and only if 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:

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

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

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.

$\blacksquare$

1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $9$