# 2.3: Integers and Rationals

- Page ID
- 19030

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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.