
# 2.3: Integers and Rationals

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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.