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

(1) Natural Powers

  If $r=n$ where $n$ is a positive integer, then\[b^{n}=\underbrace{b\cdot b\cdot\cdots\cdot b}_{n\ \text{times}}\]

(2) Zero Power

If $r=$ 0, we define

\[b^{0}=1\quad\text{if }b\neq 0\]

00 is not defined.

(3) Rational Powers

If $r=1/n$ where $n$ is a positive integer, then $b^{1/n}$ (also denoted by $\sqrt[n]{b}$), called the $n$th root of $b$, is a real number $u$ such that $u^{n}=b$. \[ \bbox[#F2F2F2,5px,border:2px solid black]{\sqrt[n]{b}=b^{1/n}=u\qquad\text{means}\qquad u^{n}=b}\]

We need to consider two cases.

    • Case 1:  n is a positive odd integer. There is exactly one nth root for each real b.
    • Case 2: $n$ is a positive even integer:

Case 2a: If  b < 0 , there is no nth root. 

[If we multiply a positive or negative number by itself an even number of times the result will be positive (for example, (-2)4=32>0). So there cannot be a real number u such that un=b as un ≥ 0 for all real u and positive even integer n.

Case 2b: If b > 0, there are two nth roots (one positive and one negative). 

However, the symbol  b1/n or $\boldsymbol{\sqrt[n]{b}}$ is reserved for the positive $n$th root.
[There are two nth roots because (-u)n=un when n is even; that is, for each b>0 there will be two nth roots, u and -u.]  


    • $\sqrt[n]{b}>0$ if $b>0$
    • $\sqrt[n]{b}<0$ if $b<0$ and $n$ is odd
    • $\sqrt[n]{b}$ is imaginary ( = not real) if $b<0$ and $n$ is even.


Properties of the nth roots

Let a and b be two real numbers and m and n be two integers. We can show that the nth roots have the following properties.

(a) $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$             (equivalently $(ab)^{1/n}=a^{1/n}b^{1/n}$)

(b) $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$             (equivalently  $\left(\dfrac{a}{b}\right)^{1/n}=\dfrac{a^{1/n}}{b^{1/n}}$)

(c) $\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$          (equivalently $\left(a^{1/n}\right)^{1/m}=a^{1/(mn)}$)

(d) $\displaystyle{\left(a^n \right )^{1/n}=\sqrt[n]{a^n}=\left\{\begin{matrix}
a & & (\text{if } n \text{ is odd})\\
|a|& & (\text{if } n \text{ 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 $b$.

\[ \bbox[#F2F2F2,5px,border:2px solid black]{b^{m/n}=\left(b^{m}\right)^{1/n}}\]

We can verify that also \[b^{m/n}=\left(b^{1/n}\right)^{m}.\]

If n is even, we require that b > 0.

(4) Irrational Powers

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, \[3^{\sqrt{2}}\approx  3^{1.41}\] and with better accuracy \[3^{\sqrt{2}}\approx 3^{1.4142}.\]

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 .

(5) Negative Powers

For any real number $r$, if $b\neq $ 0, we define $b^{-r}$ to be $\dfrac{1}{b^r}$ whenever $b^r$ is defined.

\[ \bbox[#F2F2F2,5px,border:2px solid black]{b^{-r}=\frac{1}{b^r}}\]

For example, 

\[b^{-\frac{1}{2}}=\frac{1}{b^{1/2}}=\frac{1}{\sqrt{b}},\quad b^{-\frac{5}{3}}=\frac{1}{b^{5/3}}=\frac{1}{\sqrt[3]{b^{5}}}=\frac{1}{\sqrt[3]{b^{3}b^{2}}}=\frac{1}{\sqrt[3]{b^{3}}\sqrt[3]{b^{2}}}=\frac{1}{b\sqrt[3]{b^{2}}}\]

General Properties

If $b>0$, $c>0$, and $r$ and $s$ are two real numbers, we can prove: 

(a)    $b^{r}b^{s}=b^{r+s}$

(b)    $\dfrac{b^{r}}{b^{s}}=b^{r-s}$

(c)    $(b^{r})^{s}=b^{rs}$

(d)    $(bc)^{r}=b^{r}c^{r}$

(e)    $\left(\dfrac{b}{c}\right)^{r}=\dfrac{b^{r}}{c^{r}}$