Let \(b\) be a real number. In the expression \(b^{r}\) , \(b\) is called the base and \(r\) is called the exponent .

• If \(r=n\) where \(n\) is a positive integer, then \[b^{n}=\underbrace{b\cdot b\cdot\cdots\cdot b}_{n\ \text{times}}\]
• If \(r=0\) , we define \[b^{0}=1\quad\text{if }b\neq0\] \(0^{0}\) is not defined.
• If \(r=1/n\) where \(n\) is a positive integer, then \(b^{1/n}\) (also denoted by \(\sqrt[n]{b}\) ), ¹ called the \(n\) th root of \(b\) , is a real number \(u\) such that \(u^{n}=b\) . \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\sqrt[n]{b}=b^{1/n}=u\quad\text{means}\quad u^{n}=b}\] We need to consider two cases
  • Case 1: \(n\) is a positive odd integer. There is exactly one \(n\) th root for each real \(b\) .
  • Case 2: \(n\) is a positive even integer. If \(b<0\) , there is no real \(u\) such that \(u^{n}=b\) because \(u^{n}\geq0\) for all real \(u\) and positive even integer \(n\) (for example \((-2)^{4}=32>0\) ). Because \(n\) is even, \((-u)^{n}=u^{n}\) ; therefore, for each \(b>0\) , there will be two \(n\) th roots, \(u\) and \(-u\) . However, the symbol \(\boldsymbol{b^{1/n}}\) or \(\boldsymbol{\sqrt[n]{b}}\) is reserved for the positive \(n\) th root .
• Remark that
  • \(\sqrt[n]{b}\) is positive if \(b>0.\)
  • \(\sqrt[n]{b}\) is negative if \(b<0\) and \(n\) is odd.
  • \(\sqrt[n]{b}\) is imaginary (= not real) if \(b<0\) and \(n\) is even. (See Section: Principle Square of Negative Numbers )

• Properties of roots . We can show that the \(n\) th roots have the following properties. Let \(a\) and \(b\) be two real numbers and \(m\) and \(n\) be two integers. We assume that all the given roots exist.

  • \((ab)^{1/n}=a^{1/n}b^{1/n}\) (equivalently \(\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}\) )
  • \(\left(\dfrac{a}{b}\right)^{1/n}=\dfrac{a^{1/n}}{b^{1/n}}\) (equivalently \(\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}\) )
  • \(\left(a^{1/n}\right)^{1/m}=a^{1/(mn)}\) (equivalently \(\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}\) )
  • \(\left(a^{n}\right)^{1/n}=a\) (equivalently \(\sqrt[n]{a^{n}}=a\) ) if \(n\) is odd
  • \(\left(a^{n}\right)^{1/n}=|a|\) (equivalently \(\sqrt[n]{a^{n}}=|a|\) ) if \(n\) is even

  Note that when \(n\) is even and both \(a<0\) and \(b<0\) , then \(\sqrt[n]{ab}\) and \(\sqrt[n]{a/b}\) exist (they are real numbers) because \(ab>0\) and \(a/b>0\) , but \(\sqrt[n]{a}\) and \(\sqrt[n]{b}\) do not exist (in fact, they are imaginary).

• If \(r=m/n\) where \(m\) and \(n\) are positive integers and \(m/n\) is in lowest terms, \(b^{m/n}\) is defined to be \((b^{1/n})^{m}\) ; that is, the \(n\) th root of the \(m\) th power of \(b\) . \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{b^{m/n}=\left(b^{m}\right)^{1/n}}\] We can verify that also \[b^{m/n}=\left(b^{1/n}\right)^{m}.\] If \(n\) is even, we require that \(b\geq0\) .
• If \(r=\alpha\) where \(\alpha\) is an irrational number and \(b>0\) , the approximate value of \(b^{\alpha}\) is obtained by expressing \(\alpha\) approximately as a fraction. For example, to calculate \(b^{\sqrt{2}}\) for any given \(b>0\) , we can replace \(\sqrt{2}\) by enough digits of its decimal expansion. For instance \(3^{\sqrt{2}}\) is approximately equal to \(3^{1.41}\) and with better accuracy equal to \(3^{1.4142}\) . Because we can write \(1.41\) as \(141/100\) , \(3^{1.41}\) is raising \(3\) to a fraction \(m/n\) discussed above; the same applies to \(3^{1.4142}\) .
• Note that when \(\alpha\) is irrational, \(b^{\alpha}\) is imaginary for \(b<0\) .
• If \(b\neq0\) , we define \(b^{-r}\) to be \(\dfrac{1}{b^{r}}\) whenever \(b^{r}\) is defined. For example, \[\begin{align} &b^{-\frac{1}{2}}=\frac{1}{b^{1/2}}=\frac{1}{\sqrt{b}},\\ &b^{-\frac{5}{3}}=\frac{1}{b^{5/3}}=\frac{1}{\sqrt[3]{b^{5}}}=\frac{1}{\sqrt[3]{b^{3}b^{2}}}=\frac{1}{\sqrt[3]{b^{3}}\sqrt[3]{b^{2}}}=\frac{1}{b\sqrt[3]{b^{2}}} \end{align}\] 

If \(b>0\) , \(c>0\) and \(r\) and \(s\) are two real numbers, we can prove

1. \(b^{r}b^{s}=b^{r+s}\)
2. \(\dfrac{b^{r}}{b^{s}}=b^{r-s}\)
3. \((b^{r})^{s}=b^{rs}\)
4. \((bc)^{r}=b^{r}c^{r}\)
5. \(\left(\dfrac{b}{c}\right)^{r}=\dfrac{b^{r}}{c^{r}}\)