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$\begin{array}{ll}4\cdot \frac{1}{4}=1.\hfill & \text{This}\text{\hspace{0.17em}}\text{means}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{1}{4}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{reciprocals}.\hfill \end{array}$
$\begin{array}{ll}6\cdot \frac{1}{6}=1.\hfill & \text{Hence,}\text{\hspace{0.17em}}6\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{1}{6}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{reciprocals}.\hfill \end{array}$
$\begin{array}{ll}-2\cdot \frac{-1}{2}=1.\hfill & \text{Hence,}\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}-\frac{1}{2}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{reciprocals}.\hfill \end{array}$
$\begin{array}{ll}a\cdot \frac{1}{a}=1.\hfill & \text{Hence,}\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{1}{a}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{reciprocals}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}a\ne 0.\hfill \end{array}$
$\begin{array}{ll}x\cdot \frac{1}{x}=1.\hfill & \text{Hence,}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{1}{x}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{reciprocals}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\ne 0.\hfill \end{array}$
$\begin{array}{ll}{x}^{3}\cdot \frac{1}{{x}^{3}}=1.\hfill & \text{Hence,}\text{\hspace{0.17em}}{x}^{3}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{1}{{x}^{3}}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{reciprocals}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\ne 0.\hfill \end{array}$
We can use the idea of reciprocals to find a meaning for negative exponents.
Consider the product of ${x}^{3}$ and ${x}^{-3}$ . Assume $x\ne 0$ .
${x}^{3}\cdot {x}^{-3}={x}^{3+(-3)}={x}^{0}=1$
Thus, since the product of ${x}^{3}$ and ${x}^{-3}$ is 1, ${x}^{3}$ and ${x}^{-3}$ must be reciprocals.
We also know that ${x}^{3}\cdot \frac{1}{{x}^{3}}=1$ . (See problem 6 above.) Thus, ${x}^{3}$ and $\frac{1}{{x}^{3}}$ are also reciprocals.
Then, since ${x}^{-3}$ and $\frac{1}{{x}^{3}}$ are both reciprocals of ${x}^{3}$ and a real number can have only one reciprocal, it must be that ${x}^{-3}=\frac{1}{{x}^{3}}$ .
We have used $-3$ as the exponent, but the process works as well for all other negative integers. We make the following definition.
If $n$ is any natural number and $x$ is any nonzero real number, then
${x}^{-n}=\frac{1}{{x}^{n}}$
Write each of the following so that only positive exponents appear.
${x}^{-6}=\frac{1}{{x}^{6}}$
${a}^{-1}=\frac{1}{{a}^{1}}=\frac{1}{a}$
${7}^{-2}=\frac{1}{{7}^{2}}=\frac{1}{49}$
${(3a)}^{-6}=\frac{1}{{(3a)}^{6}}$
${(5x-1)}^{-24}=\frac{1}{{(5x-1)}^{24}}$
${(k+2z)}^{-(-8)}={(k+2z)}^{8}$
Write each of the following using only positive exponents.
$\begin{array}{ll}{3}^{-2}=\frac{1}{{3}^{2}}=\frac{1}{9}\hfill & {3}^{-2}\ne -9\hfill \end{array}$
The problems of Sample Set A suggest the following rule for working with exponents:
Write each of the following so that only positive exponents appear.
$\begin{array}{ll}{x}^{-2}{y}^{5}.\hfill & \text{The}\text{\hspace{0.17em}}factor\text{\hspace{0.17em}}{x}^{-2}\text{\hspace{0.17em}}\text{can}\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}\text{moved}\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numerator}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{the}\hfill \\ \hfill & \text{denominator}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{changing}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{exponent}\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}+2.\hfill \\ {x}^{-2}{y}^{5}=\frac{{y}^{5}}{{x}^{2}}\hfill & \hfill \end{array}$
$\begin{array}{ll}{a}^{9}{b}^{-3}.\hfill & \text{The}\text{\hspace{0.17em}}factor\text{\hspace{0.17em}}{b}^{-3}\text{\hspace{0.17em}}\text{can}\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}\text{moved}\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numerator}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{the}\hfill \\ \hfill & \text{denominator}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{changing}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{exponent}\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}+3.\hfill \\ {a}^{9}{b}^{-3}=\frac{{a}^{9}}{{b}^{3}}\hfill & \hfill \end{array}$
$\begin{array}{ll}\frac{{a}^{4}{b}^{2}}{{c}^{-6}}.\hfill & \text{This}\text{\hspace{0.17em}}\text{fraction}\text{\hspace{0.17em}}\text{can}\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}\text{written}\text{\hspace{0.17em}}\text{without}\text{\hspace{0.17em}}\text{any}\text{\hspace{0.17em}}\text{negative}\text{\hspace{0.17em}}\text{exponents}\hfill \\ \hfill & \text{by}\text{\hspace{0.17em}}\text{moving}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}factor\text{\hspace{0.17em}}{c}^{-6}\text{\hspace{0.17em}}\text{into}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numerator}\text{.}\hfill \\ \hfill & \text{We}\text{\hspace{0.17em}}\text{must}\text{\hspace{0.17em}}\text{change}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}-6\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}+6\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{make}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{move}\text{\hspace{0.17em}}\text{legitimate}\text{.}\hfill \\ \frac{{a}^{4}{b}^{2}}{{c}^{-6}}={a}^{4}{b}^{2}{c}^{6}\hfill & \hfill \end{array}$
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