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Mathematics LibreTexts

2.6: Powers with Arbitrary Real Exponents. Irrationals

  • Page ID
    19033
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    In complete fields, one can define \(a^{r}\) for any \(a>0\) and \(r \in E^{1}\) (for \(r \in N,\) see §§5-6, Example \((\mathrm{f}) ) .\) First of all, we have the following theorem.

    Theorem \(\PageIndex{1}\)

    Given \(a \geq 0\) in a complete field \(F,\) and a natural number \(n \in E^{1}\), there always is a unique element \(p \in F, p \geq 0,\) such that

    \[p^{n}=a.\]

    It is called the \(n\) th root of \(a,\) denoted

    \[\sqrt[n]{a} \text{ or } a^{1 / n}.\]

    (Note that \(\sqrt[n]{a} \geq 0,\) by definition.)

    Proof

    A direct proof, from the completeness axiom, is sketched in Problems 1 and 2 below. We shall give a simpler proof in Chapter 4,§9, Example (a). At present, we omit it and temporarily take Theorem 1 for granted. Hence we obtain the following result.

    Theorem \(\PageIndex{2}\)

    Every complete field \(F\) (such as \(E^{1}\)) has irrational elements, i.e., elements that are not rational.

    In particular, \(\sqrt{2}\) is irrational.

    Proof

    By Theorem 1, \(F\) has the element

    \[p=\sqrt{2} \text{ with } p^{2}=2\]

    Seeking a contradiction, suppose \(\sqrt{2}\) is rational, i.e.,

    \[\sqrt{2}=\frac{m}{n}\]

    for some \(m, n \in N\) in lowest terms (see §7, final note).

    Then \(m\) and \(n\) are not both even (otherwise, reduction by 2 would yield a smaller \(n ) .\) From \(m / n=\sqrt{2},\) we obtain

    \[m^{2}=2 n^{2};\]

    so \(m^{2}\) is even.

    Only even elements have even squares, however. Thus \(m\) itself must be even; i.e., \(m=2 r\) for some \(r \in N .\) It follows that

    \[4 r^{2}=m^{2}=2 n^{2}, \text{ i.e., } 2 r^{2}=n^{2}\]

    and, by the same argument, \(n\) must be even.

    This contradicts the fact that \(m\) and \(n\) are not both even, and this contradiction shows that \(\sqrt{2}\) must be irrational. \(\square\)

    Note 1. Similarly, one can prove the irrationality of \(\sqrt{a}\) where \(a \in N\) and \(a\) is not the square of a natural. See Problem 3 below for a hint.

    Note 2. Theorem 2 shows that the field \(R\) of all rationals is not complete (for it contains no irrationals), even though it is Archimedean (see Problem 6 ). Thus the Archimedean property does not impleteness (but see Theorem 1 of §10).

    Next, we define \(a^{r}\) for any rational number \(r>0\).

    Definition

    Given \(a \geq 0\) in a complete field \(F,\) and a rational number

    \[r=\frac{m}{n} \quad\left(m, n \in N \subseteq E^{1}\right)\]

    we define

    \[a^{r}=\sqrt[n]{a^{m}}.\]

    Here we must clarify two facts.

    (1) If \(n=1,\) we have

    \[a^{r}=a^{m / 1}=\sqrt[1]{a^{m}}=a^{m}.\]

    If \(m=1,\) we get

    \[a^{r}=a^{1 / n}=\sqrt[n]{a}.\]

    Thus Definition 1 agrees with our previous definitions of \(a^{m}\) and \(\sqrt[n]{a}\) \((m, n \in N) .\)

    (2) If \(r\) is written as a fraction in two different ways,

    \[r=\frac{m}{n}=\frac{p}{q},\]

    then, as is easily seen,

    \[\sqrt[n]{a^{m}}=\sqrt[q]{a^{p}}=a^{r},\]

    and so our definition is unambiguous (independent of the particular representation of \(r ) .\)

    Indeed,

    \[\frac{m}{n}=\frac{p}{q} \text{ implies } m q=n p,\]

    whence

    \[a^{m q}=a^{p n},\]

    i.e.,

    \[\left(a^{m}\right)^{q}=\left(a^{p}\right)^{n};\]

    cf. §§5-6, Problem 6.

    By definition, however,

    \[\left(\sqrt[n]{a^{m}}\right)^{n}=a^{m} \text{ and } \left(\sqrt[q]{a^{p}}\right)^{q}=a^{p}.\]

    Substituting this in \(\left(a^{m}\right)^{q}=\left(a^{p}\right)^{n},\) we get

    \[\left(\sqrt[n]{a^{m}}\right)^{n q}=\left(\sqrt[q]{a^{p}}\right)^{n q},\]

    whence

    \[\sqrt[n]{a^{m}}=\sqrt[q]{a^{p}}.\]

    Thus Definition 1 is valid, indeed.

    By using the results of Problems 4 and 6 of §§5-6, the reader will easily obtain analogous formulas for powers with positive rational exponents, namely,

    \[\begin{aligned} a^{r} a^{s} &=a^{r+s} ;\left(a^{r}\right)^{s}=a^{r s} ;(a b)^{r}=a^{r} b^{r} ; a^{r}<a^{s} \text { if } 0<a<1 \text { and } r>s \\ a &<b \text { iff } a^{r}<b^{r}(a, b, r>0) ; a^{r}>a^{s} \text { if } a>1 \text { and } r>s ; 1^{r}=1 \end{aligned}\]

    Henceforth we assume these formulas known, for rational \(r, s>0\).

    Next, we define \(a^{r}\) for any real \(r>0\) and any element \(a>1\) in a complete field \(F .\)

    Let \(A_{a r}\) denote the set of all members of \(F\) of the form \(a^{x},\) with \(x \in R\) and \(0<x \leq r ;\) i.e.,

    \[A_{ar}=\{a^{x} | 0<x \leq r, x \text{ rational}\}.\]

    By the density of rationals in \(E^{1}\) (Theorem 3 of §10), such rationals \(x\) do exist; thus \(A_{\text {ar}} \neq \emptyset\).

    Moreover, \(A_{a r}\) is right bounded in \(F .\) Indeed, fix any rational number \(y>r\). By the formulas in \((1),\) we have, for any positive rational \(x \leq r\),

    \[a^{y}=a^{x+(y-x)}=a^{x} a^{y-x}>a^{x}\]

    since \(a>1\) and \(y-x>0\) implies

    \[a^{y-x}>1.\]

    Thus \(a^{y}\) is an upper bound of all \(a^{x}\) in \(A_{a r}\).

    Hence, by the assumed completeness of \(F,\) sup \(A_{ar}\) exists. So we may define

    \[a^{r}=\sup A_{a r}.\]

    We also put

    \[a^{-r}=\frac{1}{a^{r}}.\]

    If \(0<a<1\) (so that \(\frac{1}{a}>1 ),\) we put

    \[a^{r}=\left(\frac{1}{a}\right)^{-r} \text{ and } a^{-r}=\frac{1}{a^{r}},\]

    where

    \[\left(\frac{1}{a}\right)^{r}=\sup A_{1 / a, r},\]

    as above.

    Summing up, we have the following definitions.

    Definition

    Given \(a>0\) in a complete field \(F\), and \(r \in E^{1}\), we define the following.(i) If \(r>0\) and \(a>1\), then

    \[a^{r}=\sup A_{a r}=\sup \left\{a^{x} | 0<x \leq r, x \text { rational }\right\}\]

    (ii) If \(r>0\) and \(0<a<1\), then \(a^{r}=\frac{1}{(1 / a)^{r}}\), also written \((1 / a)^{-r}.\)

    (iii) \(a^{-r}=1 / a^{r}\). (This defines powers with negative exponents as well.)

    We also define \(0^{r}=0\) for any real \(r>0\), and \(a^{0}=1\) for any \(a \in F\), \(a \neq 0\); \(0^{0}\) remains undefined.

    The power \(a^{r}\) is also defined if \(a<0\) and \(r\) is a rational \(\frac{m}{n}\) with \(n\) because \(a^{r}=\sqrt[n]{a^{m}}\) has sense in this case. (Why?) This does not work for other values of \(r\). Therefore, in general, we assume \(a>0\).

    Again, it is easy to show that the formulas in (1) remain also valid for powers with real exponents (see Problems } 8-13 below), provided \(F\) is complete.