Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

2.3: Integers and Rationals

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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 \(q=1\)

    \[R \supseteq J \supset N\]

    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 \(ab\).


    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 \(F.)\)


    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,

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



    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.

    • Was this article helpful?