3.6 Negative exponents

Basic mathematics review Page 1 / 2

This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.Objectives of this module: understand the concepts of reciprocals and negative exponents, be able to work with negative exponents.

Overview

Reciprocals
Negative Exponents
Working with Negative Exponents

Reciprocals

Two real numbers are said to be reciprocals of each other if their product is 1. Every nonzero real number has exactly one reciprocal, as shown in the examples below. Zero has no reciprocal.

$\begin{array}{l} 4 \cdot \frac{1}{4} = 1. & This means that 4 and \frac{1}{4} are reciprocals . \end{array}$

$\begin{array}{l} 6 \cdot \frac{1}{6} = 1. & Hence, 6 and \frac{1}{6} are reciprocals . \end{array}$

$\begin{array}{l} - 2 \cdot \frac{- 1}{2} = 1. & Hence, - 2 and - \frac{1}{2} are reciprocals . \end{array}$

$\begin{array}{l} a \cdot \frac{1}{a} = 1. & Hence, a and \frac{1}{a} are reciprocals if a \neq 0. \end{array}$

$\begin{array}{l} x \cdot \frac{1}{x} = 1. & Hence, x and \frac{1}{x} are reciprocals if x \neq 0. \end{array}$

$\begin{array}{l} x^{3} \cdot \frac{1}{x^{3}} = 1. & Hence, x^{3} and \frac{1}{x^{3}} are reciprocals if x \neq 0. \end{array}$

Negative exponents

We can use the idea of reciprocals to find a meaning for negative exponents.

Consider the product of $x^{3}$ and $x^{- 3}$ . Assume $x \neq 0$ .

$x^{3} \cdot x^{- 3} = x^{3 + (- 3)} = x^{0} = 1$

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

$x^{- n} = \frac{1}{x^{n}}$

Sample set a

Write each of the following so that only positive exponents appear.

$x^{- 6} = \frac{1}{x^{6}}$

$a^{- 1} = \frac{1}{a^{1}} = \frac{1}{a}$

$7^{- 2} = \frac{1}{7^{2}} = \frac{1}{49}$

${(3 a)}^{- 6} = \frac{1}{{(3 a)}^{6}}$

${(5 x - 1)}^{- 24} = \frac{1}{{(5 x - 1)}^{24}}$

${(k + 2 z)}^{- (- 8)} = {(k + 2 z)}^{8}$

Practice set a

Write each of the following using only positive exponents.

$y^{- 5}$

$\frac{1}{y^{5}}$

$m^{- 2}$

$\frac{1}{m^{2}}$

$3^{- 2}$

$\frac{1}{9}$

$5^{- 1}$

$\frac{1}{5}$

$2^{- 4}$

$\frac{1}{16}$

${(x y)}^{- 4}$

$\frac{1}{{(x y)}^{4}}$

${(a + 2 b)}^{- 12}$

$\frac{1}{{(a + 2 b)}^{12}}$

${(m - n)}^{- (- 4)}$

${(m - n)}^{4}$

Caution

It is important to note that

a^{- n}

is not necessarily a negative number. For example,

$\begin{array}{l} 3^{- 2} = \frac{1}{3^{2}} = \frac{1}{9} & 3^{- 2} \neq - 9 \end{array}$

Working with negative exponents

The problems of Sample Set A suggest the following rule for working with exponents:

Moving factors up and down

In a fraction, a factor can be moved from the numerator to the denominator or from the denominator to the numerator by changing the sign of the exponent.

Sample set b

Write each of the following so that only positive exponents appear.

$\begin{array}{l} x^{- 2} y^{5} . & The f a c t o r x^{- 2} can be moved from the numerator to the \\ denominator by changing the exponent - 2 to + 2. \\ x^{- 2} y^{5} = \frac{y^{5}}{x^{2}} \end{array}$

$\begin{array}{l} a^{9} b^{- 3} . & The f a c t o r b^{- 3} can be moved from the numerator to the \\ denominator by changing the exponent - 3 to + 3. \\ a^{9} b^{- 3} = \frac{a^{9}}{b^{3}} \end{array}$

$\begin{array}{l} \frac{a^{4} b^{2}}{c^{- 6}} . & This fraction can be written without any negative exponents \\ by moving the f a c t o r c^{- 6} into the numerator . \\ We must change the - 6 to + 6 to make the move legitimate . \\ \frac{a^{4} b^{2}}{c^{- 6}} = a^{4} b^{2} c^{6} \end{array}$

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Source: OpenStax, Basic mathematics review. OpenStax CNX. Jun 06, 2012 Download for free at http://cnx.org/content/col11427/1.2

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