
# 2.6: Powers with Arbitrary Real Exponents. Irrationals

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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.