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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.


  • Reciprocals
  • Negative Exponents
  • Working with Negative Exponents



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.

4 1 4 = 1. This means that 4 and 1 4 are reciprocals .

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6 1 6 = 1. Hence, 6 and 1 6 are reciprocals .

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2 1 2 = 1. Hence, 2 and 1 2 are reciprocals .

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a 1 a = 1. Hence, a and 1 a are reciprocals if a 0.

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x 1 x = 1. Hence, x and 1 x are reciprocals if x 0.

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x 3 1 x 3 = 1. Hence, x 3 and 1 x 3 are reciprocals if x 0.

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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 0 .

x 3 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 1 x 3 = 1 . (See problem 6 above.) Thus, x 3 and 1 x 3 are also reciprocals.

Then, since x 3 and 1 x 3 are both reciprocals of x 3 and a real number can have only one reciprocal, it must be that x 3 = 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 = 1 x n

Sample set a

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

( 3 a ) 6 = 1 ( 3 a ) 6

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( 5 x 1 ) 24 = 1 ( 5 x 1 ) 24

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( k + 2 z ) ( 8 ) = ( k + 2 z ) 8

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Practice set a

Write each of the following using only positive exponents.

( x y ) 4

1 ( x y ) 4

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( a + 2 b ) 12

1 ( a + 2 b ) 12

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( m n ) ( 4 )

( m n ) 4

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It is important to note that a n is not necessarily a negative number. For example,

3 2 = 1 3 2 = 1 9 3 2 9

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.

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 = y 5 x 2

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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 = a 9 b 3

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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 . a 4 b 2 c 6 = a 4 b 2 c 6

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Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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I'm not sure why it wrote it the other way
I got X =-6
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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