2.3: Integers and Rationals

All natural elements of a field $F,$ their additive inverses, and 0 are called the integral elements of $F,$ briefly integers.

An element $x \in F$ is said to be rational iff $x=\frac{p}{q}$ for some integers $p$ and $q$ $(q \neq 0) ; x$ is irrational iff it is not rational.

We denote by $J$ the set of all integers, and by $R$ the set of all rationals, in $F .$ Every integer $p$ is also a rational since $p$ can be written as $p / q$ with $q=1$
Thus

$$R \supseteq J \supset N$$

In an ordered field,

$$N=\{x \in J | x>0\} .(\mathrm{Why} ?)$$

Theorem $\PageIndex{1}$

If $a$ and $b$ are integers (or rationals) in $F,$ so are $a+b$ and $ab$.

Proof

For integers, this follows from Examples (a) and (d) in Section 2; one only has to distinguish three cases:

(i) $a, b \in N$;

(ii) $-a \in N, b \in N$;

(iii) $a \in N,-b \in N$.

The details are left to the reader (see Basic Concepts of Mathematics, Chapter $2, § 7,$ Theorem 1$)$.

Now let $a$ and $b$ be rationals, say,

$$a=\frac{p}{q} \text{ and } b=\frac{r}{s}$$

where $q s \neq 0 ;$ and $q s$ and $p r$ are $i n t e g e r s$ by the first part of the proof (since $p, q, r, s \in J ) .$

$$a \pm b=\frac{p s \pm q r}{q s} \text{ and } a b=\frac{p r}{q s}$$

where $q s \neq 0 ;$ and $q s$ and $p r$ are integers by the first part of the proof (since $p, q, r, s \in J )$.

Thus $a \pm b$ and $a b$ are fractions with integral numerators and denominators. Hence, by definition, $a \pm b \in R$ and $a b \in R .$ $\square$

Theorem $\PageIndex{2}$

In any field $F,$ the set $R$ of all rationals is a field itself, under the operations defined in $F,$ with the same neutral elements 0 and 1. Moreover, $R$ is an ordered field if $F$ is. (We call $R$ the rational subfield of $F.)$

Proof

We have to check that $R$ satisfies the field axioms.

The closure law 1 follows from Theorem 1.

Axioms 2, 3, and 6 hold for rationals because they hold for all elements of $F ;$ similarly for Axioms 7 to 9 if $F$ is ordered.

Axiom 4 holds in $R$ because the neutral elements 0 and 1 belong to $R ;$ indeed, they are integers, hence certainly rationals.

To verify Axiom 5, we must show that $-x$ and $x^{-1}$ belong to $R$ if $x$ does. If, however,

$$x=\frac{p}{q} \quad(p, q \in J, q \neq 0)$$

then

$$-x=\frac{-p}{q}$$

where again $-p \in J$ by the definition of $J ;$ thus $-x \in R$.

If, in addition, $x \neq 0,$ then $p \neq 0,$ and

$$x=\frac{p}{q} \text{ implies } x^{-1}=\frac{q}{p} .(\mathrm{Why} ?)$$

Thus $x^{-1} \in R$. $\square$

Note. The representation

$$x=\frac{p}{q} \quad(p, q \in J)$$

is not unique in general; in an ordered field, however, we can always choose $q>0,$ i.e., $q \in N($ take $p \leq 0$ if $x \leq 0)$.

Among all such $q$ there is a least one by Theorem 2 of $\$ 85-6 .$ If $x=p / q$, with this minimal $q \in N,$ we say that the rational $x$ is given in lowest terms.