Exponents Radicals

MTH-4105-1 C1-C4 ExponentsRadicals_Layout 1 10-10-18 13:55 Page 1

MTH-4105-1

xponents and Radicals E

MTH-4105-1

EXPONENTS AND RADICALS

Société de formation à distance SOf AD des commissions scolaires du Québec Project Coordinator: Jean-Paul Groleau

Author: Nicole Perreault

Update: Éric Lacroix

Content Revision: Jean-Paul Groleau Line Régis

Translator: Claudia de Fulviis

Linguistic Revision: Johanne St-Martin

Electronic publishing: Productions P.P.I. inc.

Cover Page: Daniel Rémy

Printed: 2005

Reprint: 2006

All rights for translation and adaptation, in whole or in part, reserved for all countries. Any reproduction by mechanical or electronic means, including microreproduction, is forbidden without the written permission of a duly authorized representative of the Société de formation à distance des commissions scolaires du Québec (SOFAD).

Legal Deposit: 2005 Bibliothèque et Archives nationales du Québec Bibliothèques et Archives Canada ISBN 978-2-89493-283-4 050331 MTH-4105-1 Exponents and Radicals

TABLE OF CONTENTS

Introduction to the Program Flowchart ...... 0.4 The Program Flowchart ...... 0.5 How to Use this Guide ...... 0.6 General Introduction...... 0.9 Intermediate and Terminal Objectives of the Module ...... 0.10 Diagnostic Test on the Prerequisites ...... 0.13 Answer Key for the Diagnostic Test on the Prerequisites ...... 0.17 Analysis of the Diagnostic Test Results...... 0.19 Information for Distance Education Students...... 0.21

UNITS

1. The Laws of Exponents ...... 1.1 2. Simplifying Algebraic or Numerical Expressions Written in Exponential Form...... 2.1 3. Converting an Expression Containing a Radical to Exponential Form and Vice Versa ...... 3.1 4. The Sum, Difference, Product and Quotient of Numerical Expressions Containing Square Roots ...... 4.1 5. Operations on Polynomials Containing Square Roots ...... 5.1

Final Review ...... 6.1 Answer Key for the Final Review...... 6.4 Terminal Objectives ...... 6.5 Self-Evaluation Test...... 6.7 Answer Key for the Self-Evaluation Test ...... 6.15 Answer Key for the Self-Evaluation Test Results ...... 6.19 Final Evaluation...... 6.20 Answer Key for the Exercises ...... 6.21 Glossary ...... 6.59 List of Symbols ...... 6.63 Bibliography ...... 6.64

Review Activities ...... 7.1

INTRODUCTION TO THE PROGRAM FLOWCHART

Welcome to the World of Mathematics!

This mathematics program has been developed for the adult students of the Adult Education Services of school boards and distance education. The learning activities have been designed for individualized learning. If you encounter difficulties, do not hesitate to consult your teacher or to telephone the resource person assigned to you. The following flowchart shows where this module fits into the overall program. It allows you to see how far you have progressed and how much you still have to do to achieve your vocational goal. There are several possible paths you can take, depending on your chosen goal.

The first path consists of modules MTH-3003-2 (MTH-314) and MTH-4104-2 (MTH-416), and leads to a Diploma of Vocational Studies (DVS).

The second path consists of modules MTH-4109-1 (MTH-426), MTH-4111-2 (MTH-436) and MTH-5104-1 (MTH-514), and leads to a Secondary School Diploma (SSD), which allows you to enroll in certain Cegep-level programs that do not call for a knowledge of advanced mathematics.

The third path consists of modules MTH-5109-1 (MTH-526) and MTH-5111-2 (MTH-536), and leads to Cegep programs that call for a solid knowledge of mathematics in addition to other abiliies.

If this is your first contact with this mathematics program, consult the flowchart on the next page and then read the section “How to Use This Guide.” Otherwise, go directly to the section entitled “General Introduction.” Enjoy your work!

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THE PROGRAM FLOWCHART CEGEP

MTH-5112-1 Logic

MTH-5111-2 Complement and Synthesis II MTH-536 MTH-5110-1 Introduction to Vectors

MTH-5109-1 Geometry IV

MTH-5108-1 Trigonometric Functions and Equations MTH-526 Exponential and Logarithmic Functions MTH-5107-1 and Equations MTH-5104-1 Optimization II MTH-5106-1 Real Functions and Equations MTH-514 MTH-5103-1 Probability II MTH-5105-1 Conics

MTH-5102-1 Statistics III MTH-5101-1 Optimization I

MTH-4111-2 Complement and Synthesis I Trades MTH-436 MTH-4110-1 The Four Operations on DVS Algebraic Fractions

MTH-4109-1 Sets, Relations and Functions MTH-4108-1 Quadratic Functions MTH-426 MTH-4107-1 Straight Lines II MTH-4106-1 Factoring and Algebraic Functions MTH-4105-1 Exponents and Radicals You ar e here

MTH-4104-2 Statistics II

MTH-4103-1 Trigonometry I MTH-416 MTH-4102-1 Geometry III MTH-4101-2 Equations and Inequalities II

MTH-3003-2 Straight Lines I

MTH-314 MTH-3002-2 Geometry II MTH-3001-2 The Four Operations on Polynomials

MTH-2008-2 Statistics and Probabilities I MTH-216 MTH-2007-2 Geometry I MTH-2006-2 Equations and Inequalities I

MTH-1007-2 Decimals and Percent MTH-1006-2 The Four Operations on Fractions MTH-116 MTH-1005-2 The Four Operations on Integers 25 hours = 1 credit

50 hours = 2 credits

HOW TO USE THIS GUIDE

Hi! My name is Monica and I have been You’ll see that with this method, math is I’m Andy. asked to tell you about this math module. a real breeze! What’s your name?

My results on the test Whether you are ... you have probably taken a indicate that I should begin registered at an placement test which tells you with this module. adult education exactly which module you center or pur- should start with. suing distance education, ...

Now, the module you have in your ... the entry activity, which By carefully correcting this test using the hand is divided into three contains the test on the corresponding answer key, and record- sections. The first section is... prerequisites. ing your results on the analysis sheet ...

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... you can tell if you’re well enough And if I’m not, if I need a little In that case, before you start the prepared to do all the activities in the activities in the module, the results review before moving on, what analysis chart refers you to a review module. happens then? activity near the end of the module.

Good!

In this way, I can be sure I have all the prerequisites START for starting. The starting line shows where the Exactly! The second section contains the learning activities. It’s learning activities the main part of the module. begin.

The little white question mark indicates the questions ? for which answers are given in the text. The target precedes the objective to be met.

The memo pad signals a brief reminder of concepts which you have already studied.

The boldface question mark indicates practice exercises which allow you to try out what ? you have just learned. Look closely at the box to the right. It explains the symbols used to identify the various activities. The calculator symbol reminds you that you will need to use your calculator.

The sheaf of wheat indicates a review designed to ? reinforce what you have just learned. A row of sheaves near the end of the module indicates the final review, which helps you to interrelate all the learning activities in the module.

FINISH

Lastly, the finish line indicates that it is time to go on to the self-evaluation test to verify how well you have understood the learning activities.

There are also many fun things in this module. For example, A “Did you know that...”? Must I memorize what the sage says? when you see the drawing of a sage, it introduces a “Did you know that...” Yes, for example, short tidbits No, it’s not part of the learn- on the history of mathematics ing activity. It’s just there to and fun puzzles. They are in- give you a breather. teresting and relieve tension at the same time.

It’s the same for the “math whiz” pages, which are designed espe- They are so stimulating that And the whole module has cially for those who love math. even if you don’t have to do been arranged to make them, you’ll still want to. learning easier.

For example. words in bold- ... statements in boxes are important The third section contains the final re- face italics appear in the points to remember, like definitions, for- view, which interrelates the different glossary at the end of the mulas and rules. I’m telling you, the for- parts of the module. module... mat makes everything much easier.

Great!

There is also a self-evaluation Thanks, Monica, you’ve been a Later ... test and answer key. They tell big help. you if you’re ready for the final I’m glad! Now, This is great! I never thought that I would evaluation. I’ve got to run. like mathematics as much as this! See you!

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GENERAL INTRODUCTION

EXPONENTS AND RADICALS: A WORLD TO DISCOVER

In science, it is common to use numbers expressed in exponential form or as radicals. Chemists use numbers expressed in scientific notation to calculate the equilibrium constant of an equation or to determine the degree of acidity of a solution. Physicists make frequent use of radicals in mechanics or nuclear science, and most of the constants they use in calculations are expressed in scientific notation.

In this module, you will learn the laws of exponents and the rules for applying them. These laws will allow you to simplify algebraic or numerical expressions 2 1 2 × such as a a 2 . a–2

You will also learn the laws that apply to radicals. These laws will come in handy for reducing polynomials such as 52 to simplest form. 2+ 3

You are already familiar with rational numbers (natural numbers, integers, decimals and mixed numbers). In this module, you will learn about irrational numbers which, historically speaking, had difficulty taking hold in mathematics. You will also learn to rationalize the denominator of an expression, in other words, to make it rational.

To achieve the terminal objectives of this module, you will be required to solve algebraic and numerical expressions in exponential form by applying the laws of exponents. You will also be required to perform the four mathematical operations on polynomials containing square roots, by following the rules of priority of operations.

INTERMEDIATE AND TERMINAL OBJECTIVES OF THE MODULE

Module MTH-4105-1 consists of five units and requires 25 hours of study distributed as follows. The terminal objectives appear in boldface.

Objectives Number of hours* % (Evaluation)

1 and 2 13 55%

3 4 15%

4 and 5 7 30%

* One hour is allocated for the final evaluation.

1. The laws of exponents

To calculate the numerical or algebraic value of a mathematical expression containing powers of numbers or variables by applying one or more exponents laws. The exponent laws are the following: am × an = am + n m a = am – n an a–m = 1 am ao = 1 (am)n = am × n (abc)m = ambmcm a m am b = bm

where a, b, c are rational numbers or variables and m and n are rational numbers.

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2. Simplifying algebraic or numerical expressions written in exponential form

To reduce to simplest form an algebraic or numerical expression in exponential form by applying the priority of operations and one or more of the following exponent laws: am × an = am + n

m a = am – n a n a–m = 1 a m ao = 1 (am)n = am × n (abc)m = ambmcm

m m a = a b b m

where a, b, c are rational numbers or variables and m and n are rational numbers.

3. Converting an expression containing a radical to exponential form and vice versa

To convert a numerical or algebraic expression containing a radical to an exponential expression with the lowest possible base.

The given expression is of the form am • n b p , where a and b are variables or positive rational numbers that are powers of like bases; n and p are natural numbers; and m is a rational number.

4. The sum, difference, product and quotient of expressions containing square roots

To calculate the sum, difference, product and quotient of an arithmetic expression containing a maximum of four radicals with the same root index. When calculating a quotient, the denominator must be rationalized, if necessary.

5. Operations on polynomials containing square roots

To perform operations on a numerical expression containing a maximum of two polynomials and three square roots, and then reducing it to simplest form.

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DIAGNOSTIC TEST ON THE PREREQUISITES

Instructions

1. Answer as many questions as you can.

2. You may use a calculator.

3. Write your answers on the test paper.

4. Do not waste any time. If you cannot answer a question, go on to the next one immediately.

5. When you have answered as many questions as you can, correct your answers using the answer key which follows the diagnostic test.

6. To be considered correct, your answers must be identical to those in the answer key. In addition, the various steps in your solution should be equivalent to those shown in the answer key.

7. Copy your results onto the chart which follows the answer key. This chart gives an analysis of the diagnostic test results.

8. Do only the review activities that apply to your incorrect answers.

9. If all your answers are correct, you may begin working on this module.

1. Perform the following operations.

a) 3 × 1 = ...... 2 4

b) 8.02 × 4.1 = ......

c) 5.12 ÷ 0.2 = ......

d) – 7 ÷ – 5 = ...... 8 2

2. Find all the factors of the numbers below.

a) 16: ...... b) 21: ...... c) 32: ...... d) 144: ......

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3. Perform the following operations.

a) 2 + 7 = ...... 3 8

b) –3.56 – 2.49 = ......

4. Subtract the following polynomials. The resulting algebraic expression must be in simplest form. Show all the steps in your solution.

2y – (8x – 7y – 3x)

......

5. Perform the required operations on the following polynomials. The resulting algebraic expression must be in simplest form. Show all the steps in your solution.

a) 2a(7a – 5) b) (5x + 4)(3x – 2)

......

c) (5x – 3)(5x + 3) d) (3y – 4)2

......

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ANSWER KEY FOR THE DIAGNOSTIC TEST ON THE PREREQUISITES

1. a) 3 × 1 = 3 2 4 8

b) 8.02 × 4.1 = 32.882

c) 5.12 ÷ 0.2 = 25.6

d) – 7 ÷ – 5 =–7 × – 2 = 14 = 7 8 2 8 5 40 20

2. a) 16: 1, 2, 4, 8, 16 b) 21: 1, 3, 7, 21 c) 32: 1, 2, 4, 8, 16, 32 d) 144: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144

3. a) 2 + 7 = 16 + 21 = 37 or 1 13 b) –3.56 – 2.49 = –6.05 3 8 24 24 24 24

4. 2y – (8x – 7y – 3x) 2y – (5x – 7y) 2y – 5x + 7y 9y – 5x

5. a) 2a(7a – 5) b) (5x + 4)(3x – 2) (2a × 7a) + (2a × (– 5)) 5x(3x – 2) + 4(3x – 2) 14a2 – 10a 5x × 3x + 5x × (–2) + 4 × 3x + 4 × (–2) 15x2 – 10x + 12x – 8 15x2 + 2x – 8

c) (5x – 3)(5x + 3) d) (3y – 4)2 5x(5x + 3) – 3(5x + 3) (3y – 4)(3y – 4) 5x × 5x + 5x × 3 – 3 × 5x – 3 × 33y(3y – 4) – 4(3y – 4) 25x2 + 15x – 15x – 9 3y × 3y + 3y × (–4) – 4 × 3y + (–4)(–4) 25x2 – 9 9y2 – 12y – 12y + 16 9y2 – 24y + 16

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ANALYSIS OF THE DIAGNOSTIC TEST RESULTS

Answers Review Questions Correct Incorrect Section Page Before Going on to 1. a) 7.1 7.4 Unit 1 b) 7.1 7.4 Unit 1 c) 7.1 7.4 Unit 1 d) 7.1 7.4 Unit 1 2. a) 7.2 7.13 Unit 4 b) 7.2 7.13 Unit 4 c) 7.2 7.13 Unit 4 d) 7.2 7.13 Unit 4 3. a) 7.3 7.18 Unit 4 b) 7.3 7.18 Unit 4 4. 7.4 7.28 Unit 4 5. a) 7.5 7.31 Unit 5 b) 7.5 7.31 Unit 5 c) 7.5 7.31 Unit 5 d) 7.5 7.31 Unit 5

• If all your answers are correct, you may begin working on this module.

• For each incorrect answer, find the related section listed in the Review column. Do the review activities for that section before beginning the unit listed in the right-hand column under the heading Before Going on to.

MTH-4105-1 Exponents and Radicals

INFORMATION FOR DISTANCE EDUCATION STUDENTS

You now have the learning material for MTH-4105-1 and the relevant homework assignments. Enclosed with this package is a letter of introduction from your tutor, indicating the various ways in which you can communicate with him or her (e.g. by letter or telephone), as well as the times when he or she is available. Your tutor will correct your work and help you with your studies. Do not hesitate to make use of his or her services if you have any questions.

DEVELOPING EFFECTIVE STUDY HABITS

Learning by correspondence is a process which offers considerable flexibility, but which also requires active involvement on your part. It demands regular study and sustained effort. Efficient study habits will simplify your task. To ensure effective and continuous progress in your studies, it is strongly recommended that you:

• draw up a study timetable that takes your work habits into account and is compatible with your leisure and other activities;

• develop a habit of regular and concentrated study.

The following guidelines concerning theory, examples, exercises and assignments are designed to help you succeed in this mathematics course.

Theory

To make sure you grasp the theoretical concepts thoroughly:

1. Read the lesson carefully and underline the important points.

2. Memorize the definitions, formulas and procedures used to solve a given problem; this will make the lesson much easier to understand.

3. At the end of the assignment, make a note of any points that you do not understand using the sheets provided for this purpose. Your tutor will then be able to give you pertinent explanations.

4. Try to continue studying even if you run into a problem. However, if a major difficulty hinders your progress, contact your tutor before handing in your assignment, using the procedures outlined in the letter of introduction.

Examples

The examples given throughout the course are applications of the theory you are studying. They illustrate the steps involved in doing the exercises. Carefully study the solutions given in the examples and redo the examples yourself before starting the exercises.

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Exercises

The exercises in each unit are generally modeled on the examples provided. Here are a few suggestions to help you complete these exercises.

1. Write up your solutions, using the examples in the unit as models. It is important not to refer to the answer key found on the coloured pages at the back of the module until you have completed the exercises.

2. Compare your solutions with those in the answer key only after having done all the exercises. Careful! Examine the steps in your solutions carefully, even if your answers are correct.

3. If you find a mistake in your answer or solution, review the concepts that you did not understand, as well as the pertinent examples. Then redo the exercise.

4. Make sure you have successfully completed all the exercises in a unit before moving on to the next one.

Homework Assignments

Module MTH-4105-1 comprises three homework assignments. The first page of each assignment indicates the units to which the questions refer. The assignments are designed to evaluate how well you have understood the material studied. They also provide a means of communicating with your tutor.

When you have understood the material and have successfully completed the pertinent exercises, do the corresponding assignment right away. Here are a few suggestions:

1. Do a rough draft first, and then, if necessary, revise your solutions before writing out a clean copy of your answer.

2. Copy out your final answers or solutions in the blank spaces of the document to be sent to your tutor. It is best to use a pencil.

3. Include a clear and detailed solution with the answer if the problem involves several steps.

4. Mail only one homework assignment at a time. After correcting the assignment, your tutor will return it to you.

In the section “Student’s Questions,” write any questions which you wish to have answered by your tutor. He or she will give you advice and guide you in your studies, if necessary.

In this course Homework Assignment 1 is based on units 1 and 2. Homework Assignment 2 is based on units 3 to 5. Homework Assignment 3 is based on units 1 to 5.

CERTIFICATION

When you have completed all your work, and provided you have maintained an average of at least 60%, you will be eligible to write the examination for this course.

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START

UNIT 1

THE LAWS OF EXPONENTS

1.1 LEARNING CONTEXT

Numbers That Defy the Imagination

While reading an article in a newspaper or science magazine, or a chemistry or physics textbook, you may notice that the author used “astronomical” or “infinitesimal” numbers to describe various facts.

Here are some examples.

• In 2002, the federal government debt was $421 000 000 000 (421 billion), which is equivalent to $14 000 per inhabitant, regardless of age.

• The distance between our planet and the sun is 150 000 000 km, whereas Alpha Centauri is located 40 400 000 000 000 000 000 km away from the sun.

• The human body contains about 15 000 000 000 000 blood corpuscles, also called red blood cells; each of these cells has a diameter of 0.007 5 mm. .

• Each cubic centimetre of air (the equivalent of a thimbleful) we breathe contains 27 000 000 000 000 000 000 molecules.

• The HTLV-III retrovirus ( a type of virus), the transmitting agent for AIDS, has a diameter of 0.000 042 5 mm.

Fortunately, there is a method for writing such large numbers. This method is called scientific notation. The table below shows the above numbers written in scientific notation.

$ $421 000 000 000 = 4.21 × $1011

150 000 000 km = 1.5 × 108 km

13 CH3 15 000 000 000 000 = 1.5 × 10 OH 0.007 5 mm = 7.5 × 10-3 mm

27 000 000 000 000 000 000 = 2.7 × 1019

0.000 042 5 mm 4.25 × 10-5 mm

Figure 1.1 From the extremely large to the extremely small

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Later in this unit we will look at how to express a number in scientific notation. For the time being, let’s spend some time examining exponents, which are an essential part of this type of notation.

An exponent is a numerical or algebraic expression that indicates how many times a quantity is to be multiplied by itself. Thus, in the expression 52 = 5 × 5, the exponent is 2.

We know that 22 (2 squared) = 4 because 2 × 2 = 4. We also know that 23 (2 cubed) = 8 because 2 × 2 × 2 = 8. In the expression 32 = 9, the base is 3 and the exponent is 2.

In an algebraic or numerical expression, the base is the number or variable with an exponent.

To calculate the power of a number, we apply the definition of exponents.

☞ an = a × a × ... × a

n times { Note

The base can be represented by any type of number: a natural number ( ), an integer ( ), a fraction or a decimal ( ).

Example 1

3 2 = 2 × 2 × 2 = 8 5 5 5 5 125 (3.5)2 = (3.5) × (3.5) = 12.25

To achieve the objective of this unit, you must be able to solve various forms of exponential expressions by applying the laws of exponents.

? What is the value of (–2)2? ......

? What is the value of (–2)3? ......

Your answers to the above questions should be 4 and –8, since (–2)2 = (–2) × (–2) = 4 and (–2)3 = (–2) × (–2) × (–2) = 4 × (–2) = –8.

Law of signs in multiplication (+) × (+) = + (–) × (–) = + (+) × (–) = – (–) × (+) = –

Remember to pay particular attention to parentheses and signs when solving expressions.

Example 2

23 = 2 × 2 × 2 = 8. The base is 2 and the exponent is 3. –23 = –(2 × 2 × 2) = –8. The base is 2 and the exponent is 3. (–2)3 = (–2) × (–2) × (–2) = –8. The base is –2 and the exponent is 3.

? Calculate: 34 = ...... –34 = ...... (–3)4 = ......

You probably obtained 81, –81 and 81 respectively, since:

34 = 3 × 3 × 3 × 3 = 81 –34 = –( 3 × 3 × 3 × 3) = –81 (–3)4 = (–3) × (–3) × (–3) × (–3) = 81

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You no doubt noticed that the last result is positive even though the base is negative. Indeed, any negative base with an even-numbered exponent yields a positive result; however, a negative base with an odd-numbered exponent yields a negative result.

• Any power with a positive base is positive. • Any power with a negative base is positive if the exponent is an even number. • Any power with a negative base is negative if the exponent is an odd number.

Now, try your hand at the following exercises to check your understanding of the above concepts.

Exercise 1.1

1. Calculate the following expressions.

4 a) 23 = ...... b) 2 = ...... 3

c) 13 = ...... d) (–1.4)2 = ......

e) (–5)3 = ......

2. Is the value of the following expressions positive or negative?

3 a) 1 : ...... b) (–0.5)2: ...... 2

c) (–0.35)3: ...... d) (–1.12)4: ......

e) –1.124: ......

Now let’s look at the laws of exponents. The first two laws were covered in a previous course; however, we will review them here and examine them in more depth.

First Law of Exponents

When two or more powers with like bases are multiplied, the product has the same base and the exponents are added. In general: am × an = am + n

Example 3

Calculate t3 × t4.

We know that t3 = t × t × t and t4 = t × t × t × t. Thus, t3 × t4 = t × t × t × t × t × t × t = t7.

By applying the first law of exponents, we get: t3 × t4 = t3 + 4 = t7.

Example 4

Calculate 22 × 23 × 2. 22 × 23 × 2 = 22 × 23 × 21 = 22 + 3 + 1 = 26 = 64

Example 5

Calculate (–y)5 × (–y)8. (–y)5 × (–y)8 = (–y)5 + 8 = (–y)13 = –y13

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1 1 ? What is the product of x 2 × x 3 ? ......

The first law of exponents always applies, regardless of the nature of the 1 1 1 + 1 5 exponents. Thus, x 2 × x 3 = x 2 3 = x 6 .

Example 6

Calculate –3y4 × 2y3.

First, we multiply the numerical coefficients of the terms. Then, we apply the first law of exponents. –3y4 × 2y3 = –6y4y3 = –6y4 + 3 = –6y7

Example 7

Calculate (–2)3 × 25.

Since the bases are not identical, we must first determine the sign of (–2)3. We can then apply the first law of exponents. (–2)3 × 25 = –23 × 25 = –28 = –256

Second Law of Exponents

When two powers with like bases are divided, the quotient has the same base and the exponents are subtracted. In general: m a = am – n, a ≠ 0 an

Example 8

5 Calculate a . a3

5 × × × × × a = a a a a a = a a = a2 a3 a × a × a 1

5 If we apply the second law of exponents we get: a = a5 – 3 = a2. a3

3 ? What is the quotient of x ? ...... x 4

2y –2 ? What is the quotient of ? ...... 4y 3

2 a 3 ? Calculate the quotient of 1 ...... a 4 2 ? Calculate the quotient of 6 ...... 6 2 3y 4 ? Calculate the quotient of ...... 6y 3

You probably obtained the following answers:

5 1 1 –5 1 , y , a 12 , 1 and y . x 2 2

The last quotient is calculated by first simplifying the numerical coefficients. 3y 4 1y 4 We get: = . We then apply the second law of exponents: 6y 3 2y 3 1y 4 1y 4–3 1y y = = or or 1 y . 2y 3 2 2 2 2

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Until now, we have used numbers that are relatively simple to calculate. However, this is not always the case in mathematics or science. These subjects often involve operations on very small or very large numbers, making manual calculations long and tedious. Fortunately, we have calculators to make our task easier. In the following examples, you will learn how to use a calculator to determine the numerical value of exponential expressions.

Example 9

37 × 32 = 37 + 2 = 39

To determine the value of this expression, key in: 3 yx 9 = 19683

Therefore: 39= 19 683

Example 10

(–2)14 ( 2 +/– ) yx 1 4 = 16 384

Example 11

–214

We know that the value is negative. We key in: 2 yx 1 4 = 16 384 and write –16 384.

N.B. Be careful if you are using a calculator to find –214, as there are two types of calculators.

• Conventional calculator: 2 +/– yx 1 4 = .

• Direct algebraic logic (DAL) calculator: +/– 2 yx 1 4 = .

Until now, we have looked at bases with positive exponents. But what happens when the exponent is negative? In your opinion, what is the value of the expression a–2? This leads us to the third law of exponents.

Third Law of Exponents

When the base has a negative exponent, the base is inverted and the exponent becomes positive.

a–m = 1 where m ∈ + am

We can now answer the above question. a–2 = 1 a2

This law stems from the second law of exponents.

3 For example, a = a3 – 5 = a–2. a5 3 × × But a = a a a = 1 = 1 a5 a × a × a × a × a a × a a2 Therefore, a–2 = 1 . a2

? Calculate 3–2...... Your answer should be 1 , since 3–2 = 1 = 1 . 9 32 9

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–4 Calculate 1 ...... ? 3 –4 4 4 Your result should be 81, since 1 = 1 = 3 = 3 1 1 3 3 × 3 × 3 × 3 = 81 = 81. 1 1 1 1 1

Example 12

–3 Convert – 1 . 4

The base is – 1 and its reciprocal is – 4 . 4 1 –3 3 Thus – 1 =–4 =–4 × – 4 × – 4 =–64 = –64. 4 1 1 1 1 1

Example 13

–3 Convert 2 . 5

The base is 2 and its reciprocal is 5 . 5 2 –3 3 Thus 2 = 5 = 5 × 5 × 5 = 125 . 5 2 2 2 2 8

Example 14

Convert 5a–2.

1 The base is a and its reciprocal is a .

Thus, 5a–2 = 5 × 1 = 5 . a2 a2

N.B. Here, the base with the negative exponent is a. The number 5 is the numerical coefficient. It therefore remains in the numerator position.

? Can you identify which of the following statements is true?

–1 a) 3–1 = –3: ...... b) (–2)–1 = 2: ...... c) 7 =–3 : ...... 3 7

–2 –3 d) – 5 =–16 : ...... e) – 2 =–125 : ...... f) (–1)–1 = 1: ...... 4 25 5 8

If you answered that only statement e) is true, you’re right! Indeed,

–1 a) 3–1 = 1 b) (–2)–1 = – 1 c) 7 = 3 3 2 3 7

–2 2 –3 3 d) – 5 =–4 = 16 e) – 2 =–5 =–125 f) (–1)–1 = –1 4 5 25 5 2 8

The first two laws of exponents remain valid when the expressions involved have negative exponents.

Example 15

–3 y Calculate . y –7

–3 y According to the second law of exponents, = y–3 – (–7) = y–3 + 7 = y4. y –7

Example 16

Calculate a × a–3.

According to the first law of exponents, a × a–3 = a1 × a–3 = a1 + (– 3) = a–2 = 1 . a2

We have seen what to do with negative exponents. But what about the exponent 0? In your opinion, what is the value of 50? To find out, let’s look at the fourth law of exponents.

1.12 © SOFAD MTH-4105-1 Exponents and Radicals

Fourth Law of Exponents

Any expression other than 0 that has the exponent 0 is equal to 1. a0 = 1 where a ≠ 0

We can now answer the above question. 50 = 1

In fact, this law stems from the second law of exponents.

3 For example, 5 = 53 – 3 = 50. 53

3 × × But 5 = 5 5 5 = 125 = 1, therefore 50 = 1. 53 5 × 5 × 5 125

It’s now time to apply what you have learned. The following exercises will allow you to practice working with the first four laws of exponents.

Exercise 1.2

1. Express the product of the following powers in exponential form and calculate the value of the resulting expression, if necessary. Your answers should not contain any negative exponents.

a) 53 × 52 = ......

b) 32 × 3–1 = ......

3 –5 c) 7 × 7 = ...... 4 4

1 d) x 4 × x 2 = ......

e) 2y–2 × 3y3 = ......

1 5 f) (–4) 3 × (–4) 3 = ......

g) (–2.5)3 × 2.5 = ......

h) (–3.25) × (–3.25)2 = ......

i) –0.2x3 × 0.5x2 = ......

j) 5x3 × x–7 × 2x = ......

–4 k) 1 x 2 × – 1 x 3 × 2x = ...... 3 2

2. Express the quotients of the following powers in exponential form.

3 a) 3 = ...... 32

(– x)4 b) = ...... (– x)2

3 4a 2 c) 1 = ...... 2a 2

d) –5t8 ÷ t6 = ......

4 4 e) 10 y 5 ÷ 5y 5 = ......

3 f) 3.5a = ...... 0.7a–3

g) 1.5x2 ÷ (–0.3x) = ......

1.14 © SOFAD MTH-4105-1 Exponents and Radicals

3 h) –3a = ...... 6a3

i) –33y8 ÷ 11y9 = ......

j) 2 a6 ÷ – 1 a4 = ...... 3 2

Did you know that…

… the number 10 is the base of the “decimal” number system? This means that all possible numbers can be written with only 10 digits. Why was this base chosen rather than another? It appears that the reason is one of convenience. Almost all historians agree that this system was adopted because humans have 10 fingers.

The Celts used a base 20 number system, very likely because they used their fingers and toes to count. Traces of this number system were found in the old British monetary system, in which 1 pound equalled 20 shillings.

Base 5 is also widely used. In many languages the words “five” and “hand” are similar or have a common root. For example, “pentcha” means “hand” in Persian and “pantche” means “five” in Sanskrit.

However, bases 5, 10 and 20 were not necessarily used worldwide. Certain tribes in Africa, South America and Australia used the base 2 (binary) system while others used the base 3 (tertiary) system because their calculations were based on the 3 phalanxes of the fingers opposite the thumb.

Base 4, while more rare, is still used by the Yiki Indians of California who count by using the spaces between the fingers of one hand.

The most complicated number system is unquestionably that of the Mesopotamian civilization. It is called the sexagesimal system because its base is 60. Sixty different signs are required to represent the numbers from 0 to 59. Despite its complexity, this system is still used to measure the radian of an angle in degrees or the passing of time. There are 60 minutes in an hour and 60 seconds in a minute. It is obvious that if your watch reads 5:07:09, then 5 hours, 7 minutes and 9 seconds have elapsed. What is not so obvious is that in Mesopotamian notation, 5, 7, 9 mean: 5 × 602 + 7 × 601 + 9 × 600 or 18 429 in our decimal system. If 5 hours, 7 minutes and 9 seconds have gone by, then 18 429 seconds have elapsed, starting from midnight.

SCIENTIFIC NOTATION

Before we examine the last three laws of exponents, let’s look at how to convert a decimal number to scientific notation and vice versa.

• When we multiply a decimal number by a power of 10 (i.e. 10, 100, 1 000, 10 000, and so on), the decimal point is moved to the right by as many places as there are zeros in the power of 10. Thus,–3.29 × 10 = –32.9 41.7 × 100 = 4 170

1.16 © SOFAD MTH-4105-1 Exponents and Radicals

• When we divide a decimal number by a power of 10, the decimal point is moved to the left by as many places as there are zeros in the power of 10.

Thus,5.187 ÷ 1 000 = 0.005 187 8 ÷ 100 = 0.08

N.B. Zeros must be added where there are no digits.

Use your calculator to find 616.

6 yx 1 6 = 2.821109907 12

N.B. Some calculators cannot display more than 8 digits show a result of

2.8211099 12.

You may have guessed what these results mean. It is the answer written in scientific notation. The number 2.821109907 12 stands for 2.821 109 907 × 1012. With this notation, the calculator displays an approximation of the exact result, which could not be displayed otherwise. The exact result is: 2 821 109 907 456.

N.B. To display 2.821109907 × 1012 directly on your calculator, key in 2.821109907 and press EXP 1 2 .

In scientific notation, a number is always expressed in the form a × 10n, where n is an integer and a is a decimal, such that 1 ≤ a < 10.

Let’s now look at how to convert a number expressed in scientific notation to a decimal.

Example 17

Convert 7.547 8 × 102 to a decimal.

You need only move the decimal point two places to the right since the base 10 exponent is 2. 7.547 8 × 102 = 754.78

Indeed, 7.547 8 × 102 = 7.5478 × 100 = 754.78.

What happens if the base 10 exponent is negative?

Example 18

Convert 5.789 × 10–4 to a decimal.

Simply move the decimal point four places to the left since the base 10 exponent is –4. 0 0005.789 × 10–4 = 0.000 578 9

Indeed, 5.789 × 10–4 = 5.789 × 1 = 5.789 × 1 = 5.789 ÷ 10 000 = 10 4 10 000 0.000 578 9.

1.18 © SOFAD MTH-4105-1 Exponents and Radicals

Let a be a decimal greater than or equal to 1 such that 1 ≤ a < 10, and let n be a natural number. The number a × 10n can be converted to a decimal by moving the decimal point in a to the right by n places. The number a × 10–n can be converted to a decimal by moving the decimal point in a to the left by n places.

? Canada’s debt in 2002 can be expressed as $4.21 × 1011. Can you convert it to a decimal?

......

If your answer is $421 000 000 000, you’re right!

The decimal point must be moved 11 places to the right by adding as many zeros as necessary. Do you find it more encouraging to have a debt of $4.21 × 1011 or a debt of $421 000 000 000?

? The HTLV-III retrovirus has a diameter of 4.25 × 10–5 mm. Can you write this as a decimal?

......

If your answer is 0.000 042 5 mm, you’re right!

The decimal point must be moved 5 places to the left by adding as many zeros as necessary.

Now, let’s do the reverse and express a decimal in scientific notation.

Example 19

Express 7 547.8 in scientific notation.

1. Place the decimal point after the first non-zero digit. 7 547.8

2. Count the number of places you moved the decimal point: 3 places to the left. 7.547 8

3. Write 3 as the base 10 exponent to show the number of places the decimal point was moved. 7.547 8 × 103

And that’s pretty much all there is to scientific notation! The number 7 547.8 is equal to 7.547 8 × 103 since 7 547.8 × 1 000 = 7.547 8 × 103.

Example 20

Convert 0.000 647 to scientific notation.

1. Place the decimal point after the first non-zero digit. 0.000 6 47

2. Count the number of places you moved the decimal point: 4 places to the right. 0 000 6.47

3. Write –4 as the base 10 exponent to show the number of places the decimal point was moved. 6.47 × 10–4

1.20 © SOFAD MTH-4105-1 Exponents and Radicals

Simple, isn’t it? The number 0.000 647 is indeed equal to 6.47 × 10–4 since

0.000 647 = 6.47 = 6.47 × 1 = 6.47 × 1 = 6.47 × 10–4. 10 000 10 000 10 4

? Canada’s debt in 2002 was $421 000 000 000. Can you express this in scientific notation?

......

If you answered 4.21 × 1011, you’re right!

You need only move the decimal point 11 places to the left (4.21 000 000 000) and then write 11 as the base 10 exponent: 4.21 × 1011.

This is indeed the amount of Canada’s debt since: $421 000 000 000 = 4.21 × $100 000 000 000 = $4.21 × 1011.

? Now try converting the diameter of the HTLV-III retrovirus to scientific notation: 0.000 042 5 mm?

......

If you answered 4.25 × 10–5, you’re right!

You simply move the decimal point 5 places to the right (0000 04.25) and then write –5 as the base 10 exponent: 4.25 × 10–5.

This is, in fact, the diameter of the HTLV-III retrovirus since 0.000 042 5 mm =

4.25 mm = 4.25 × 1 mm = 4.25 × 1 mm = 4.25 × 10–5 mm. 100 000 100 000 10 5

? Can you now explain why your calculator displays 2.821109907 12 when you calculate 616?

......

In fact, 2.821109907 12 stands for 2.821109907 × 1012. When we convert this result to a decimal, we get 2 821 109 907 000, since the decimal point was moved 12 places to the right. This result is an approximation of the exact result: 2 821 109 907 456.

N.B. This integer has 13 digits since we moved the decimal point 12 places to the right.

To express a number in scientific notation: 1. Place the decimal to the right of the first non-zero digit. 2. Count the number of places the decimal point was moved. 3. Write this number as the base 10 exponent: a) this exponent is positive if the decimal point was moved to the left; b) this exponent is negative if the decimal point was moved to the right.

Exercise 1.3

1. Convert the following numbers to scientific notation.

a) 142 857 = ...... b) 1 230 000 000 = ......

c) 0.054 = ...... d) 0.007 21 = ......

1.22 © SOFAD MTH-4105-1 Exponents and Radicals

2. Convert the following numbers to decimals.

a) 5.1 × 105 = ...... b) 7.654 3 × 103 = ......

c) 8.193 × 10–4 =...... d) 4 × 10–7 =......

3. Which number is larger: 0.000 007 2 or 7.1 × 10–5?

......

4. The speed of sound in dry air at standard temperature and pressure is about 331.4 m/s. Convert this number to scientific notation.

......

5. The speed of light in a vacuum is about 2.997 924 6 × 108 m/s. Convert this number to a decimal.

......

6. Express 532 in scientific notation by keeping three digits after the decimal point.

......

7. How many digits are in 2131?

......

Now let’s look at the last three laws of exponents.

You are now ready to tackle a greater level of difficulty. Problem-solving in mathematics or science sometimes requires mastery of another law of exponents. For example, how would you solve an expression of the form (x2)3?

We know that x2 = x × x.

If this expression is itself cubed (power of three), then we get: x2 × x2 × x2 (x × x) × (x × x) × (x × x) which is equivalent to x6. Therefore, (x2)3 = x2 × 3 = x6

We have just discovered the fifth law of exponents.

Fifth Law of Exponents

To raise a power m with base a to a power n, multiply the exponents. In general: (am)n = am × n

Example 21

(24)3 = 24 × 3 = 212

To determine the value of this expression, key in: 2 yx 1 2 = 4096

Therefore: (24)3 = 212 = 4 096

1.24 © SOFAD MTH-4105-1 Exponents and Radicals

Example 22

–3(a2)6 = –3a2 × 6 = –3a12

☞ Here, the negative sign applies only to the numerical coefficient of base a. We must therefore keep the negative sign throughout the calculations.

Example 23

((–3)2)–2 = (32)–2 = (3)2 × (– 2) = (3)–4 = 1 = 1 34 81

☞ Here, base –3 is negative. As the power of a negative base is positive when the exponent is an even number, the result is therefore positive.

• The power of a positive base is positive. • The power of a negative base is positive if the exponent is an even number. • The power of a negative base is negative if the exponent is an odd number.

Example 24

–(32)2 = –(32 × 2) = –(34) = –81

☞ Here the base is positive. Be careful: –(32)2 ≠ [(–3)2]2. The base is positive in the first expression, and negative in the second.

To assign an exponent to several variables or numbers, place these between parentheses. Otherwise, the exponent will apply only to the closest variable or number.

? What is the value of (xyz)2? ......

? Solve (a2y3)2 ......

These problems are slightly more complex. But don’t worry, solving them is child’s play – or almost!

To determine the value of the first expression:

(xyz)2 = (x × y × z) × (x × y × z) = x2 y2 z2

The second problem is solved as follows: (a2y3)2 = (a × a × y × y × y) × (a × a × y × y × y) = a4y6

As you may already have guessed, there is yet another law of exponents that can be used to solve this type of problem.

Sixth Law of Exponents

To raise a product in exponential form to a power n, multiply the exponents of each of the factors by n. In general: (abc)m = am bm cm where a, b et c represent exponential expressions.

Example 25

2 2 2 1 2 2 2 1 3a b 2 = 3 × (a ) × b 2 • 6th law of exponents

2 2 1 2 × 1 × 2 3a b 2 = 3 × a2 2 × b 2 • 5th law of exponents

2 2 1 3a b 2 = 9 × a4 × b

2 2 1 3a b 2 = 9a4b

1.26 © SOFAD MTH-4105-1 Exponents and Radicals

Example 26

(x2y–2)3 = (x2)3 × (y–2)3 • 6th law of exponents (x2y–2)3 = x2 × 3 × y–2 × 3 • 5th law of exponents (x2y–2)3 = x6 × y–6

6 (x2y–2)3 = x • 3rd law of exponents y 6

N.B. In future, we will apply the fifth and sixth laws at the same time. We will therefore go directly from (x2y–2)3 to x2 × 3 × y–2 × 3.

Example 27

(–x2 y)3 = (–1)3 × x2 × 3 × y1 × 3 = –x6 y3

Example 28

(–a3 b2)2 = (–1)2 × a3 × 2 × b2 × 2 = a6 b4

☞ Be careful!

Always remember that an expression of the type (2abc)2 stands for (2 × a × b × c)2. By applying the sixth law of exponents, we get (22 × a2 × b2 × c2). The expression 2(abc)2 is equivalent to 2 × a2 × b2 × c2. Watch out for the parentheses!

Example 29

(2x3 3y–2 4z2)–2 = 2–2 × x3 × (–2) × 3–2 × y–2 × (–2) × 4–2 × z2 × (–2)

(2x3 3y–2 4z2)–2 = 1 × x –6× 1 × y 4 × 1 × z –4 22 32 4 2

(2x3 3y–2 4z2)–2 = 1 × 1 × 1 × y 4 × 1 × 1 4 x 6 9 16 z 4 4 y (2x3 3y–2 4z2)–2 = 576x 6z 4

You will now learn a useful trick for simplifying some of your calculations. In science and mathematics, we must often simplify numerical or algebraic expressions. Thus (22 × 8)2 = 22 × 2 × 8. So far, no problem. However, 82 can also be written (23)2 because 8 = 2 × 2 × 2. The purpose of this type of conversion is to obtain a single base in order to make your calculations easier. Therefore 22 × 2 × 82 = 24 × (23)2 = 24 × 26 = 210.

Convert each of the following expressions to expressions with a single base.

? (3 × 272)3 ......

? (16 × 82)2 ......

The first expression can be converted to [3 × (33)2]3 because 27 = 3 × 3 × 3 = 33. [3 × (33)2]3 = (3 × 36)3 = 33 × 318 = 321 or (3 × 36)3 = (37)3 = 321.

To simplify the second expression, you must convert it using base 2. (16 × 82)2 = [24 × (23)2]2 = (24 × 26)2 = 28 × 212 = 220 or (24 × 26)2 = (210)2 = 220.

As you can see, it’s quite simple. Pay close attention to the parentheses and the signs of the bases, the numerical coefficients and the exponents, and presto! Now let’s see if you have understood the concepts you have learned so far.

Exercise 1.4

1. Simplify the following expressions.

a) (x2)3 = ...... b) (y–3)–2 = ......

1 c) 3(a6)2 = ...... d) –( x 3) 3 = ......

e) (–y2)2 = ......

2. Calculate the value of the following expressions by applying the appropriate law of exponents.

a) (–23)–1 = ...... b) (–42)3 = ......

1 4 c) (3–4)–1 = ...... d) 4 4 = ......

1 –2 e) 2 2 = ......

3. The following products are in exponential form. Raise them to the required power.

1 a) (x2y3)2 = ...... b) –2( a8b4) 2 =......

c) m3n 6 p3 3 = ...... d) (a–1b)–3 = ......

e) (x–1yz2)–2 = ......

4. Calculate the value of the following expressions by applying the appropriate law of exponents.

a) (42 × 33)2 = ......

1 b) (–1.5)2 × (–2.2)4 2 = ......

c) (–34 × 52)2 = ......

–2 –1 1 d) 35 × 7 2 = ......

5. Simplify and convert the following expressions to exponential form.

a) (22 × 4)3 =......

b) (33 × 92)–1 = ......

c) (4 × 162 × 33)2 = ......

3 2 d) 1 × 1 = ...... 2 8

1 1 1 2 e) 8 2 × 16 3 = ......

There is one last case involving both a quotient and exponents. In your opinion, 3 2 what is the quotient of a ? To answer this question, you need to know the b3 seventh and last law of exponents.

Seventh Law of Exponents

To raise a quotient of two exponential expressions to a power n, multiply the exponents of each expression by n. In general:

m m a = a where b ≠ 0 b bm

Determine the quotient of each of the following expressions.

3 ? x = ...... y 2

2 2 ? 3a = ...... 4b3

To obtain the first result, proceed as follows:

3 3 3 x = x = x y 2 y 2 × 3 y 6

To determine the second quotient, proceed as follows:

2 2 2 × 2 × 2 4 3a = 3 a = 9a 4b3 4 2 × b3 × 2 16b6

Obviously, we can always make things more difficult! Here are some examples.

Example 30

–3 9 –2 –2× (–3) 6 6 y x = x = x = x = x 6 × = x6y9 y 3 y 3 × (–3) y –9 1 1 y 9

Example 31

3 2 2 × 3 6 6 6 a = a = a = a =–a (because the exponent is an odd (–b)3 (–b)3 × 3 (–b)9 – b9 b9 number)

Since there’s nothing like practice, try your hand at a few exercises involving the seventh law of exponents.

Exercise 1.5

1. The quotients below are in exponential form. Raise them to the required power.

2 4 2 a) m = ...... b) x = ...... n y 3

3 –2 –2 1 y a 2 c) 3 = ...... d) 1 = ...... x b 4 3 4 e) x = ...... (– y)3

2. Calculate the following expressions by applying the appropriate exponent law of exponents.

3 a) 3 = ...... 2

3 (–6) 3 b) = ...... (–4)2

–2 2 c) 3 = ...... 4 3

2 (–2)2 d) – = ...... (–5)3 1 4 2 e) 2.5 = ...... 0.52

Before you continue, here is a summary of the laws of exponents and the rules for applying them.

Laws of Exponents

1. am × an = am + n

m 2. a = am – n an m 3. a–m = 1 where 1 a am

4. a0 = 1 where a ≠ 0

5. (am)n = am × n

6. (abc)m = ambmcm

m m 7. a = a where b ≠ 0 b bm

Rules for Applying the Laws of Exponents

1. The power of a positive base is positive. 2. The power of a negative base is positive if the exponent is an even number. 3. The power of a negative base is negative if the exponent is an odd number. 4. When a base has a negative exponent, invert the base and change the sign of the exponent to obtain an equivalent expression in which the exponent is positive. 5. When an expression preceded by a negative sign has an exponent, each term must be raised to this power, including the digit –1.

It can never be repeated enough: signs merit special attention. For example, they can radically change the result of a chemistry experiment or a financial statement! The law of signs, while simple, applies and will always apply regardless of the complexity of the problems to be solved. Speaking of signs…

Did you know that…

…around the year 500 CE, a bank employee of East Indian origin came up with the ingenious idea of using a sign to distinguish between assets and liabilities in his clients’ accounts.

All the rules regarding how these numbers are handled were established and codified by the mathematicians Brahmagupta in the VII century and Bhaskara in the XII century. East Indians, lovers of poetry and mathematics, translated the law of signs into sayings. These sayings are still used today:

“The friends of my friends are my friends.” “The enemies of my friends are my enemies.” “The friends of my enemies are my enemies.” “The enemies of my enemies are my friends.”

Substitute the + sign for the word “friends” and the – sign for the word “enemies” and you get the current law of signs for multiplication and division.

At the beginning of this unit, we studied the sign of the power of a number. We will now go a little further in our analysis by describing the value of am depending on the sign and value of a as well as that of its exponent m.

Example 32

Describe the sign and the value of am knowing that –1 < a < 0 and m is a positive even integer.

2 An example of this is – 4 . 5

2 – 4 =–4 × – 4 = 16 5 5 5 25

16 has a positive sign and is between 0 and 1. 25 We can therefore conclude that am has a positive sign and 0 < am < 1.

Example 33

Let’s describe the sign and the value of am knowing that a < –1 and m is an odd negative integer.

–3 An example of this situation is – 5 . 4

–3 3 According to the third law of exponents, – 5 =–4 . 4 5 3 – 4 =–4 × – 4 × – 4 =– 64 5 5 5 5 125

– 64 has a negative sign and is between –1 and 0. 125 We can therefore deduce that am has a negative sign and that –1 < am < 0.

? Can you describe the sign and value of am knowing that –1 < a < 0 and that m is an odd negative integer?

......

If you answered am < –1, you’re right! –3 A numerical example of this situation is – 4 . 5 –3 3 According to the third law of exponents, – 4 =–5 . 5 4 3 – 5 =–5 × – 5 × – 5 =–125 and – 125 is smaller than –1. 4 4 4 4 64 64 We can therefore conclude that am < –1.

This method of solving problems is relatively simple; however, sometimes it is necessary to generalize. There are several ways of doing this.

We know that any base with an even-numbered positive exponent yields a positive result. Thus, if a > 1 or a < –1, then am > 1. For example, 52 = 25 and (–5)2 = 25. Likewise, if 0 < a < 1 or –1 < a < 0, then a < am < 1. For example, 2 2 1 = 1 and – 1 = 1 . 5 25 5 25

This, of course, is a fairly easy example.

? In your opinion, what is the value of am if m > 0 and odd and that a > 1 or that a < –1?

......

? Now what is the value of am if m > 0 and odd and that 0 < a < 1 or that –1 < a < 0?

......

If you answered am > 1 or am < –1 to the first question, you’re right. Indeed, 53 = 125 or (–5)3 = –125. Likewise, 0 < am < 1 or –1 < am < 0 in the second case. 3 3 You will be convinced of this if you consider that 1 = 1 and – 1 =– 1 . 5 125 5 125

Is everything clear up to now? Yes? Then we can go one step further and look at what happens when m < 0.

? What do we know when m < 0? In other words, what rule do we follow?

......

We must invert the base and then use the same reasoning as for the preceding cases.

Let m be an even-numbered negative exponent. If a > 1 or a < –1, then 0 < am < 1. If you are not convinced, then consider this: 5–2 = 1 = 1 and 52 25 (–5)2 = 1 = 1 . (–5)2 25

However, if m is an odd-numbered negative exponent, 0 < am < 1 or –1 < am < 0: 5–3 = 1 = 1 and (–5)3 = 1 =– 1 . 53 125 (–5)3 125

? What is the value of am if m < 0 and even and if 0 < a < 1 or –1 < a < 0?

......

? And if m is an odd number?

......

Answer

–2 2 –2 2 In the first case, am > 1 1 = 5 =25 and – 1 =–5 =25 , while 5 1 5 1

–3 3 in the second case am > 1 or am < –1 1 = 5 = 125 and 5 1

–3 – 1 =(–5)3 = –125 . 5

Until now, we have proved each statement with a numerical example. Try doing the following exercise without looking at the above examples.

Exercise 1.6

1. Describe the sign and value of am knowing that:

a) a < –1 and m is an even positive integer......

b) a < –1 and m is an even negative integer......

c) –1 < a < 0 and m is an odd negative integer......

d) 0 < a < 1 and m is an odd negative integer......

e) a < –1 and m is an odd positive integer......

The time has come to put what you have learned to the test. The following practice exercises will tell you whether you have understood the concepts in this unit. If you have difficulty doing the exercises, reread the relevant passages. You should have a firm grasp of the content of this unit since it forms the basis for the material in later units.

? 1.2 PRACTICE EXERCISES

1. Determine the numerical or algebraic value of the following expressions by applying the appropriate law of exponents. N.B. Round off your answers to the nearest thousandth.

a) 0.7–2 =......

b) –3y2 × y–1 = ......

1 2 c) 2 2 × –(2) = ......

d) 3y2 ÷ 2y = ......

2 e) –8.1c =...... 0.9c 2

1 2 f) (–5) 2 = ......

1 g) (x 2 ) 3 = ......

2 4 h) 3 = ...... 23

i) (1.43 × 2.2–1)2 = ......

–2 1 j) m3n 2 p2 = ......

k) –1.6–3 = ......

l) (–3)–2 = ......

m) 3x2 × 2x–1 = ......

n) 2 y 2 ÷ 1 y 3 =...... 3 6

o) 1 y –2× 1 y 2 = ...... 3 8

–4 p) 3a = ...... 2a–5

q) 6x3 ÷ 2x3 =......

2 –2 r) 1.2 =...... 0.43

s) 1.5 × 103 × 0.4 × 10–2 =......

–1 3 2 4 t) 1 × 1 × 1 = ...... 3 2 4

–2 –3 1 u) 8 × 2 2 = ......

2. a) In 1988, the federal government debt was about $310 000 000 000. Express this number in scientific notation.

......

b) The mass of the moon is about 7.36 × 1022 kg. Express this number as a decimal.

......

c) The mass of an electron at rest is 0.000 000 000 000 000 000 000 000 000 000 911 kg. Express this number in scientific notation.

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d) How many digits are in 345.

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3. What is the sign and the value of am, knowing that:

a) a > 1 and m is an even negative integer......

b) –1 < a < 0 and m is an odd positive integer......

4. Which of the following expressions is equal to (5x6)–2?

A) 25 ...... B) 5 ...... C) 25x4...... D) 1 ...... E) 5x4...... x 2 x 6 25x 12

12 F) x ...... G) 1 ...... H) 1 ...... I) 1 ...... 25 5x 12 5x 4 25x 4

1.3 REVIEW ACTIVITIES

1. In your own words, describe the seven laws of exponents and give an example of each one.

1......

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Example:......

2......

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Example:......

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Example:......

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Example:......

5......

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Example:......

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Example:......

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Example:......

2. Complete the following sentences by inserting the missing terms or expressions in the blanks.

a) The power of a positive base is ...... The power of a negative base is ...... if the exponent is an even number. The power of a ...... base is negative if the exponent is an odd number.

b) When a base has a negative exponent, simply ...... the base and change the ...... of the exponent to obtain an equivalent expression in which the exponent is ......

c) Let a be a decimal greater than or equal to 1 and smaller than 10 and let n be a natural number. The number a × 10n is converted to a decimal by moving the decimal point in a to the ...... by n places. The number a × 10–n is converted to a decimal by moving the decimal point in a to the ...... by n places.

d) To express a number in scientific notation: 1. Place the decimal point to the ...... of the first non-zero digit. 2. Count the number of places the decimal point was moved. 3. Write this number as the base 10 exponent: a) this exponent is ...... if the decimal point was moved to the left; b) this exponent is ...... if the decimal point was moved to the right.

e) Let exponent m be a positive integer: 1. if a > 1, then am ...... 1; 2. if ...... , then 0 < am < 1.

1.4 THE MATH WHIZ PAGE

Spotlight on Scientific Notation

Convert the numbers in the problems below to scientific notation and perform the calculations using the same notation. Watch those units!

1. Each second, Niagara Falls releases more than 840 000 000 000 drops of water. Each of these drops of water contains 1 700 000 000 000 000 000 000 molecules. Calculate the number of water molecules that fall each second.

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2. The earth’s circumference is approximately 40 000 km.

a) If sound travels at a speed of 340 m/s, how much time will it take a sound to circle the earth?

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b) The speed of light is 300 000 km/s. How many times will light circle the earth in one minute?

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c) We say that an airplane flies at Mach 2 when its speed is double that of sound. How much time would it take such an airplane to circle the earth?

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