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# Algebraic Reasoning for Teaching Math MATH 135

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Lecture Notes for Math 135 Algebraic Reasoning for Teaching Mathematics Steffen Lempp DEPARTMENT OF MATHEMATICS UNIVERSITY OF WISCONSIN MADISON WI 53706 1388 USA E maz39l address lempp mathwisc edu URL httpwwwmathwisc edulempp Key words and phrases algebra equation function inequality linear function quadratic function exponential function student misconception middle school mathematics These are the lecture notes for the course Math 135 Algebraic Reasoning for Teaching Mathematics77 taught at UW Madison in spring 2008 The preparation of these lecture notes was partially supported by a faculty development grant of the College of Letters and Science and by summer support by the School of Education both of the University of Wisconsin Madison Feedback comments and corrections are welcome This text is currently in a constant state of change so please excuse what I am sure will be numerous errors and typosl Thanks to Dan McGinn and my Math 135 students in spring 2008 Rachel Burgan Erika Calhoon Lindsay Feest Amanda Greisch Celia Hagar Brittany Hughes and Shannon Osgar for helpful comments and corrections 2008 Steffen Lempp University of Wisconsin Madison This material may only be reproduced and distributed for cost of reproduction and distribution for educationalnon pro t purposes This book is currently available for free download from http wwwmath wise edulemppteaCh135 pdf Any dissemination of this material in modi ed form requires the authors written approval Contents Preface Introduction to the Instructor Introduction to the Student Chapter 1 Numbers and Equations Review and Basics 11 Arithmetic in the Whole Numbers and Beyond A Review 111 Addition and Multiplication 112 Subtraction and Division 12 Letters and Algebraic Expressions 121 Letters in Algebra 1211 Letters as Variables 1212 Letters as Unknowns 1213 Letters as Parameters 1214 Letters as Constants 122 Algebraic Expressions 13 Equations and Identities 14 Manipulating Algebraic Expressions 141 Using the de nition of subtraction 142 Combining Like Terms77 143 Removing Parentheses 15 Conventions in Algebra 151 Notation for Sets7 and Some Special Sets of Numbers 152 Order of Operations and Use of Parentheses 16 Algebra as Undoing Expressions77 Chapter 2 Linear Equations 21 Ratio and Proportional Reasoning 22 Velocity and Rate of Change 23 Linear Expressions 24 Line Equations 241 Graphing a Linear Equation 242 Different Forms of Equations for Lines iii vi viii iv CONTENTS 243 Varying the Parameters in a Line Equation 41 25 Solving a Linear Equation in One Variable 43 26 Solving Simultaneous Linear Equations in Two or More Variables 52 Chapter 3 Order and Linear lnequalities 59 31 Ordering the Numbers 59 32 Order and the Arithmetical Operations 61 33 Linear Inequalities in One Unknown 64 34 Estimation and Approximation 69 Chapter 4 The Concept of a Function7 and Functions Closely Related to Linear Functions 72 41 The Concept of a Function 72 42 Range of a Function7 Onto and One to One Functions7 and lnverse of a Function 77 421 Range of a Function7 and Onto Functions 78 422 171 Functions 78 423 lnverse of a Function 79 43 Some Functions Closely Related to Linear Functions 82 431 Step Functions 83 432 Piecewise Linear Functions 85 433 The Absolute Value Function 86 434 Direct and lnverse Variation 89 Chapter 5 Quadratic Functions7 Equations and lnequalities 92 51 Introduction to Quadratic Functions 92 52 Solving Quadratic Equations 98 521 Reviewing the Binomial Laws 98 522 Solving Quadratic Equations by Factoring 99 523 Solving Quadratic Equations by Taking the Square Root100 524 Solving Quadratic Equations by Completing the Square 102 525 Solving Quadratic Equations by the Quadratic Formula 104 526 The number of real solutions of a quadratic equation 105 53 A Digression into Square Roots and the Complex Numbers 109 531 Square Roots 109 532 The Number 239 and the Complex Numbers 111 54 Graphing Quadratic Functions and Solving Quadratic Equations Graphically 113 55 Quadratic lnequalities 121 CONTENTS Chapter 6 Exponential and Logarithmic Functions7 Equations and Inequalities Exponentiation A Review of the De nition Motivation for Exponentiation The Exponential Functions Properties of Exponential Functions Another Application of Exponential Functions The Logarithmic Functions and Solving Exponential Equations De nition and Properties of Logarithmic Functions Applications of Logarithmic Functions The slide rule Radioactive Decay Earthquakes and the Richter Scale Sound Volume and Decibels 61 62 63 631 632 64 641 642 6421 6422 6423 6424 Bibliography Index 124 124 128 131 131 135 138 138 143 143 144 145 146 148 149 Preface The human mind has never invented a labor saving device equal to algebra J Willard Gibbs The word algebra has several meanings in our society Historically7 the word derives from the book Al Kitab al Jabr wa l Muqabala meaning The Compendious Book on Calculation by Completion and Balancing written by the Persian mathematician Muhammad ibn Musa al Khwarizmi approx 780 850 CE7 which was the rst book dealing systematically with solving linear and quadratic equations7 based on earlier work by Greek and Indian mathematicians Curiously7 the authors last name is also the source for our word al gorithm 7 meaning a concise list of instructions7 such as in a computer program In school mathematics7 algebra often called elementary 7 inter mediate 7 high school 7 or college algebra follows the study of arithmetic Whereas arithmetic deals with numbers and operations7 algebra generalizes this from computing with concrete numbers to reasoning with unknown numbers variables 7 usually denoted by letters using equations7 functions7 etc In upper level college algebra and beyond7 algebra takes on yet another meaning which will not be the subject of this text There7 algebra now also often called higher algebra 7 abstract algebra 7 or modern algebra abstracts from the study of numbers to the study of abstract objects which behave like numbers in a very broad sense the complex numbers are a very simple pun intended example of such a structure It is the school mathematics meaning of the word algebra which will be the topic of this text7 and in which meaning we will from now on exclusively use this word The history of algebra started with this meaning of the word7 from the Babylonians around 1600 BCE to the Greeks7 Indians and Chi nese around 500 BCE to 500 CE to the Persians and Indians around 800 CE to 1100 CE to the Italians and French in the 16th vi PREFACE vii century Until the Renaissance most of the attention in algebra cen tered around solving equations in one variable How to solve quadratic equations by completing the square was already known to the Babyloni ans the Italian mathematicians Scipione del Ferro and Niccolo Fontana Tartaglia independently and the Italian mathematician Lodovico Fer rari gave the rst general solution to the cubic and quartic equation respectively1 The Italian mathematician Gerolamo Cardano was the rst to publish the general solutions to cubic and quartic equations and the formula for the cubic equation bears his name to this day The French mathematician Francois Viete 1540 1603 more com monly known by his Latinized name Vieta is credited with the rst attempt at giving the modern notation for algebra we use today before him very cumbersome notation was used The 18th and 19th century then saw the birth of modern algebra in the other sense mentioned above which also led to much more general techniques for solving equa tions including the proof by the Norwegian mathematician Niels Hen rik Abel in 1824 that there is no general formula to solve a quintic equation For a brief history of algebra see the Wikipedia entry on algebra at httpenwikipediaorgwikiAlgebraHistory 1Here a quadratic equation is an equation involving numbers I and 12 a cubic equation is one also involving 13 a quartic equation is one additionally involving 14 and a quintic equation mentioned in the following paragraph also involves z i Introduction to the Instructor These lecture notes rst review the laws of arithmetic and discuss the role of letters in algebra and then focus on linear quadratic and ex ponential equations inequalities and functions But rather than simply reviewing the algebra your students will have already learned in high school these notes go beyond and study in depth the concepts un derlying algebra emphasizing the fact that there are very few basic underlying idea in algebra which explain77 everything else there is to know about algebra These ideas center around 0 the rules of arithmetic more precisely the ordered eld ax ioms which carry over from numbers to general algebraic expressions and o the rules for manipulating equations and inequalities These basic underlying facts are contained in the few Propositions sprinkled throughout the lecture notes These notes however attempt to not just cover77 the material of algebra but to put it into the right context for teaching algebra by focusing on how real life problems lead to algebraic problems multi ple abstract representations of the same mathematical problem and typical student misconceptions and errors and their likely underlying causes These notes just provide a bare bones guide through an algebra course for future middle school teachers It is important to combine them with a lot of in class discussion of the topics and especially with actual algebra problems from school books in order to generate these discussions For these we recommend using the following Singapore math schoolbooks 0 Primary Math Textbook 5A US Edition 0 Primary Math Textbook 6A US Edition 0 New Elementary Math Textbook 1 Syllabus D 0 New Elementary Math Textbook 2 Syllabus D 0 New Elementary Math Textbook 3A Syllabus D and 0 New Syllabus Additional Mathematics Textbook viii INTRODUCTION TO THE INSTRUCTOR ix Only a few pages of the rst and last of these books need to be used the other four books should be used more extensively The course web page httpwwwmathwiscedulemppteaCh135html gives an idea of how to integrate these various components We nally refer to Milgrarn 27 Chapter 8 and Wu 6 for similar treatments of algebra7 and to Papick 3 and SzydlikKoker 5 for very different approaches A very careful and thorough introduction to algebra can also be found in Cross Introduction to the Student These notes are designed to make you a better teacher of mathe matics in middle school The crucial role of algebra as a stepping stone to higher mathe matics has been documented numerous times7 and the mathematics curricula of our nations schools have come to re ect this by placing more and more emphasis on algebra At the same time7 it has be come clear that many of our nations middle school teachers lack much of the content knowledge to be effective teachers of mathematics and especially of algebra See the recent report ofthe National Mathemat ics Advisory Panel available at httpwwwedgovaboutbdscomm 1JLstmathpanelreportfinal reportpdf7 especially its chapter 4 on algebra After some review of the laws of arithmetic and the order of oper ations7 and looking more closely at role of letters in algebra7 we will mainly study linear7 quadratic and exponential equations7 inequalities and functions But rather than simply reviewing the algebra you have already learned in high school7 we will go beyond and study in depth the concepts underlying algebra7 emphasizing the following topics problem solving modeling real life problems word prob lems as mathematical problems and then interpreting the mathematical solution back into the real world context proofs making mathematically grounded arguments about U V l and nln rirm o analyzing student solutions examining the rationale behind middle school and high school students7 mathematical work and how it connects to prior mathematical understanding and future mathematical concepts analyzing the strengths and weaknesses of a range of solution strategies including standard techniques and recognizing and identifying common student misconceptions modeling exible use of multiple representations such as graphs7 tables7 and equations including different forms7 and x INTRODUCTION TO THE STUDENT xi how to transition back and forth between them and using functions to model real world phenomena 0 symbolic pro ciency solving equations and inequalities sim plifying expressions factoring etc These lecture notes try to make the point that school algebra centers around a few basic underlying concepts which explain all or at least most of school algebra These are 0 the rules of arithmetic for numbers including those for the ordering of numbers and how they naturally carry over from numbers to general algebraic expressions and o the rules for manipulating equations and inequalities and how they allow one to solve equations and inequalities We will generally label these basic facts Propositions to highlight their central importance Many if not most of the student misconcep tions and errors you will encounter in your teaching career will come from a lack of understanding of these basic rules and how they natu rally explain algebra CHAPTER 1 Numbers and Equations Review and Basics 11 Arithmetic in the Whole Numbers and Beyond A Review 111 Addition and Multiplication Very early on in learning arithmetic7 we realize that there are certain rules which make com puting simpler and vastly decrease the need for memorizing number facts For example7 we see that 11 2772 12 606 13 3X55gtlt3 14 6gtlt00 15 3x13 For a more elaborate example7 we also see that 16 2gtlt392gtlt32gtlt9 ls there something more general going on Of course7 you know the answer is yes7 and we could write down many more such examples However7 this would be rather tiring7 and we certainly cannot write down all such examples since there are in nitely many so we7d rather nd a better way to state these as general rules In ancient and medieval times7 before modern algebraic notation was invented7 such rules were stated in words eg7 the general rule for 11 would have been stated as The sum of any two numbers is equal to the sum of the two numbers in the reverse order Obviously7 this is a rather clumsy way to state such a simple rule7 and furthermore7 when writing down rules this way7 it is not always easy to write out the rule unambiguously Pause for a moment and think how you would write down the general rule for 16 And again7 you know7 of course7 how to write down these rules in general We use letters instead of numbers and think of the letters as arbitrary 1 2 1 NUMBERS AND EQUATIONS REVIEW AND BASICS numbers So the above examples turn into the following general rules 17 abba 18 a0a 19 agtltbbgtlta 110 agtlt00 111 agtlt1a 112 agtltbcagtltbagtltc Two related issues arise now 1 For which numbers 17 b7 and 0 do these rules hold 2 How do we know these rules are true for all numbers 17 b7 and 0 ln elementary school7 these rules can be proved visually for whole numbers by coming up with models for addition and multiplication A common model for the addition of whole numbers a and b is to draw a many objects eg7 apples in a row7 followed by b many objects in the same row7 and then counting how many objects you have in total Similarly7 a common model for the multiplication of whole numbers a and b is to draw a rectangular array of a many rows of b many objects eg7 apples7 and then counting how many objects you have in total All the above rules are now easily visualized Eg7 see Figure 11 for a visual proof of 11 and Figure 12 for a visual proof of 13 2 7 FIGURE 11 A visual proof that 2 7 7 2 An easy extension of these arguments works for fractions and in deed for positive real numbers if we go from counting objects to work ing with points on the positive number line On the positive number line starting at a point identi ed with the number 07 mark off a unit interval from 0 to a new point called 1 Repeatedly marking off more unit intervals gives us points we identify with the whole numbers 27 37 4 etc For a whole number n gt 07 we can subdivide the unit interval into 71 many subintervals of equal length ie7 of length Now again7 11 ARITHMETIC IN THE WHOLE NUMBERS AND BEYOND A REVIEW 3 0 0 00 30 50 0000 K J 0 5 0 L J 3 FIGURE 12 A visual proof that 3 gtlt 5 5 gtlt 3 starting at 07 we repeatedly mark off intervals of this length and get points we identify with the fractions 5 g 7 etc We also identify these points with the directed arrows77 from 0 to that point so7 eg7 is represented both by the point labeled in Figure 137 and by the arrow from 0 to the point labeled We can now model addition of fractions and indeed for positive real numbers a and b by marking off an interval of length a from 07 followed by an interval of length b from the point marked a The resulting point can be identi ed with the sum 1 b Similarly7 for multiplication7 we can draw two positive number lines7 a horizontal one going to the right and a vertical one going up7 both starting at the same point Now the rectangle bounded below by the interval on the horizontal number line from 0 to 17 and bounded on the left by the interval on the vertical number line from 0 to b7 has area a gtlt b and so can be identi ed with this product Once we have these models of addition and multiplication of fractions or7 more generally7 positive real numbers7 we can again easily verify the above rules 177112 Draw your own pictures Two much more subtle rules about addition and multiplication so subtle that you may not even think about them any more are the following 113 abcabc 114 agtltbgtltcagtltbgtltc It is due to these rules that we can write a b c and a gtlt b gtlt c unambiguously without giving it much thought Again draw a picture 4 1 NUMBERS AND EQUATIONS REVIEW AND BASICS for each to convince yourself that they are true For 114 you will have to go to a three dimensional array Exercise 11 Give pictures for each of 18 and 1107114 in the whole numbers 112 Subtraction and Division At rst students are given problems of the type 2 5 7 or 3 gtlt 4 7 asking them to ll in the answer A natural extension of this is to then assign problems of the form 2 7 7 or 3 gtlt 7 12 asking them to ll in the missing summand or factor respectively Let7s look at the problem of the missing summand rst Solving 2 7 7 corresponds eg to the following word problem Ann has two apples Her mother gives her some more apples and now she has seven apples How many apples did her mother give her Students will quickly learn that this requires the operation of sub traction Ann7s mother gave her 772 that is 5 apples More generally the solution to a 7 c is c 7 1 But then a problem arises What if a gt 071 Obviously you now have to go from the objects model to a more abstract model eg the money model with debt If you have 82 and spend 85 then you are 83 in debt ie you have 73 dollars It is a good idea to emphasize the distinction in the use of the symbol 7 in these two cases by reading 2 7 5 as two minus ve but 73 as negative three So allowing subtraction for arbitrary numbers forces us to allow a new kind of numbers the negative numbers So we proceed from the whole numbers 012 to the integers 0 1 712 72 You will need to allow your students to adjust to this change since now your answer to a question like Can I subtract a larger number from a smaller number will change possibly confusing your students Of course one should never answer no to this question without some quali cation some students will know about the negative numbers at an early age especially in Wisconsin in the winterl But now the correct answer should be it depends namely it depends on which number system we work in At this point you will want to extend the number line to the left also see Figure 13 it can also be used to include fractions and negative fractions But this transition from the whole numbers to the integers imme diately raises three new questions about the arithmetical operations 1I canlt resist at this point to tell you my favorite German math teacher joke What does a math teacher think if two students are in the class room and ve students leave 7 If three students come back the room is emptyl 11 ARITHMETIC IN THE WHOLE NUMBERS AND BEYOND A REVIEW 5 FIGURE 13 The number line 1 How should we de ne a b and a 7 b when one or both of a and b are negative What about multiplication 2 How do we know that the rules 177114 hold for addition and multiplication of arbitrary positive or negative numbers 3 Do these rules also hold for subtraction and multiplication of arbitrary positive or negative numbers Of course7 you already know the answer to item But how would you justify it to your students Maybe there is another way to de ne addition with negative numbers The answer is twofold First of all7 one can justify the common de nition of addition of integers with a real life model such as the money and debt model or with the number line which is really just an abstraction from a real life model The justi cation for subtraction and multiplication is a bit more delicate and may not appeal to your students as much But one can also justify the common de nition of addition and multiplication of integers by the fact that we want the rules 177114 to be true for addition and multiplication of integers also So7 in a sense7 the rules in item 2 above really are rules we impose rather than rules we can prove 7 unless you are satis ed with the justi cation for item 1 by the models for addition and multiplication So what about subtraction We rst de ne 7b as the integer which when added to b gives 0 115 b7b 0 This is now a new rule about addition7 and 7b is called the ad ditive inverse of b Note that there is lots of room for confusion here 7b or simply 7b for short is sometimes called the negative of b7 a term I would avoid7 since 7b need not be negative In fact7 7b is positive if b is negative 7b is 0 if b is 0 and 7b is negative if b is positive This will confuse quite a few of your students through high school and even college7 so be sure to emphasize this early on We can then de ne subtraction via addition a 7 b is de ned as a 7b for any integers a and b Note that this works well with real life models of subtraction Adding a capital of 5 ie7 a debt of 85 to a capital of 82 is the same as subtracting 5 from 82 6 1 NUMBERS AND EQUATIONS REVIEW AND BASICS This then also gives us an answer for item 3 above We dont have to worry about new rules for subtraction since we can use the old rules for addition It is fairly clear that agtltb7cagtltb7agtltc since this is simply a gtlt b7c agtlt b7agtltc and since 7a gtlt c a gtlt 70 On the other hand in general a 7 b b 7 a is false since it would mean a 7b 71 b Similarly a 7 b 7 c a 7 b 7 c is false How would that be rewritten Let7s move on to division Again division rst comes up when a student transitions from solving 3 gtlt 4 i to solving 3 gtlt 7 12 and solving for the missing factor corresponds eg to the following word problem Ann7s mother has twelve apples If she wants to give each of her three children the same number of apples how many will each child get Again solving this problem amounts to dividing The solution to 3gtlt7 12 is 1273 or 4 More generally the solution to agtlt7 c is 07a But a similar problem as for subtraction arises Here a may not evenly divide into c eg in the above problem when Ann7s mother has 13 apples Here is where it becomes really confusing to your students There are two different ways in which school mathematics solves this problem division with remainder and extending to fractions In the above example this would mean that Ann7s mother could either give four apples each to her three children and keep a remainder of one apple or that she could cut one apple into thirds and give each child 4 apples2 Since both of these solutions are acceptable you have to make it clear to your students which solution you are aiming for Generally in the higher grades you will want the second solution but division with remainder never disappears completely from your students7 mathematics ln algebra and calculus there is division of polynomials with remainder as we will see late in this course In any case we will for now focus on extending our numbers so that we can always divide c 7 a evenly Unless a 0 Then it 2There is one other model of division Annls mother has 12 apples and wants to give three apples to each of her children How many children does she have If you now increase the number of apples to 13 then only one kind of division makes sense since cutting up children is generally frowned upon So be careful before you start a problem and want to vary it laterl 11 ARITHMETIC IN THE WHOLE NUMBERS AND BEYOND A REVIEW 7 makes no sense since it would amount to solving the original problem 0 gtlt 7 c and there will either be no solution if c 31 0 or any number is a solution if c 0 so 0 0 never makes sense We proceed from the whole numbers to fractions which will always be positive or 0 or more generally from the integers to the rational numbers which may also be negative The number line from Figure 13 still serves as a good visual model But this transition again raises a number of questions 1 How should we de ne the operations of addition subtraction multiplication and division for fractions or more generally for rational numbers 2 How do we know that the rules 177115 still hold 3 Do these rules also hold for division In answering item 1 we need to be able to come up with real life examples to justify the correct77 way to perform the arithmetical operations on fractions and more generally rational numbers See Exercise 13 below But again one can also justify this by wanting the rules 177115 to still hold for these numbers addressing item An elegant way to think about division is to rst de ne the multi plicative inverse of a nonzero number a as that number denoted by i which when multiplied by a gives 1 1 116 a gtlt a 7 1 This is now a new rule for multiplication and it works not only when a is a positive whole number but indeed for any nonzero rational number And again this removes the need for new rules for division since we can reduce division to multiplication by de ning a b as a gtlt See Exercise 14 We summarize all the rules about the arithmetical operations of number which we have discovered in the following Proposition 12 The following rules hold for all real numbers a b and c commutative law of addition a b b a associative law of addition a b c a b 0 law of the additive identity a 0 a law of the additive inverse a 7a 0 8 1 NUMBERS AND EQUATIONS REVIEW AND BASICS commutative law of multiplication a gtlt b b gtlt a associative law of multiplication a gtlt b gtlt c a gtlt b gtlt 0 law of the multiplicative identity 1 gtlt 1 a 1 law of the multiplicative inverse a gtlt 7 1 when a 31 0 a distributive law a gtlt b e a gtlt b a gtlt 0 Exercise 13 Give several word problems for the division both for partitive division as in the apple example in the main text above and for measurement division as in the footnote above Exercise 14 Explain which of the rules 197112 and 114 carry over to division if they dont carry over are there corresponding rules for division 12 Letters and Algebraic Expressions 121 Letters in Algebra Arithmetic only deals with numbers and numeiieal empressions ie expressions involving numbers and the arithmetical operations addition subtraction multiplication division later on also roots exponentiation and logarithms Each such ex pression has a de nite value or is unde ned for some reason such as division by 0 or taking the square root of a negative number Word problems often can be reduced to evaluating numerical expressions eg Ann has four apples Her friend Mary gives her three more apples each day for a ve day week How many apples does she have at the end of the week 77 Clearly this gives rise to the numerical expression 4 5 gtlt 3 and evaluating this numerical expression then gives the solution Ann has 19 apples at the end of the week77 One extension of this problems is to leave some numbers eg the number of apples Ann has at the beginning and the number of apples she receives each day unspeci ed In this case we typically use letters to denote these unspeci ed quantities and it is crucial that these letters are clearly and unambiguously speci ed So lets set a number of apples Ann has at the beginning of the week b number of apples Ann receives from Mary each day Now the word problem becomes 12 LETTERS AND ALGEBRAIC EXPRESSIONS 9 Ann has a apples Her friend Mary gives her b more apples each day for a ve day week How many apples does she have at the end of the week This gives rise to the more general expression a 5 gtlt b7 and the solution becomes Ann has a 5 gtlt b apples at the end of the week Of course7 this expression can be rewritten7 but we cannot simply write it as a number any more since we dont know a and b We will cover rewriting such expressions7 one of the key features of algebra7 later in the course For now7 let7s concentrate on the use of letters in algebra We can basically distinguish four different uses of letters 1211 Letters as Variables This is how letters were used in the example above Since we didnt know the number of apples Ann had at the beginning or that she received each day from Mary7 we used the letters a and b for them7 respectively It doesnt really matter which letters we use7 as long as we consistently use the same letter for the same quantity7 use different letters for different quantities7 and clearly state how each letter is used It is often helpful to use mnemonic ie7 easy to remember letters Eg7 in the above example7 we could have used b for the number of apples Ann has at the beginning of the week7 and d for the number of apples Ann receives daily Later on7 we will distinguish between independent and dependent variables In the above example7 a and b are independent since neither depends on any other quantity in the example7 but if we set c a 5 gtlt b for the total number of apples at the end ofthe week7 then c is a dependent variable7 since it depends on a and b In the 5th grade Singapore math book Primary Mathematics Text book 5A 7 variables are introduced visually as bars in bar diagrams rather than as letters7 one of the best known features of Singapore mathematics It is only in the next grade7 in the Singapore math book Primary Mathematics Textbook 6A 7 that variables are used as letters instead7 and its easy to see how one can translate the bar diagram pic ture into one or more equations involving variables7 where the variables correspond to bars of unknown length in the bar diagram For example7 in exercise 1 on page 23 of the Primary Mathematics Textbook 5A 7 Ryan and Juan share 8410 between them7 and Ryan receives 8100 more than Juan The problem asks you to nd Juan7s share See Figure 14 If we call Juan7s share x we can easily translate the bar diagram into an equation x 100 z 410 10 1 NUMBERS AND EQUATIONS REVIEW AND BASICS Rya n 410 Juan l l A J 7 100 FIGURE 14 What is Juan7s share 1212 Letters as Unknowns This is probably the rst way in which letters are introduced in school mathematics and is very sim ilar to the previous use as variables Now however we dont think of the letter as denoting an unspeci ed quantity but as a speci ed quantity which we dont know and usually would like to nd Eg Ann has four apples Her friend Mary gives her some more apples each day for a ve day week the same number each day How many apples does Mary give Ann each day of the week if Ann has 19 apples at the end of the week7 In this case we can express the number of apples Ann has at the end of the week in two distinct ways as the numerical expression 19 and as the expression 4 5 gtlt b where b is the number of apples she receives each day So to solve the problem we need to solve the equation 4 5 gtlt b 19 where the letter b denotes the unknown quantity For a single unknown the letter z is usually used but this is merely a convention 1213 Letters as Parameters This is probably the trickiest bit about the use of letters in algebra since the role of a letter as a pa rameter depends on the context Let7s look again at a variation of the above word problem Ann has a apples Her friend Mary gives her b more apples each day for a ve day week At the end of the week she has 19 apples77 This is not really a word problem yet since I havent told you what I am asking for In fact I could ask for two different quantities I could ask 1 for the number of apples Ann had at the beginning or 2 for the number of apples Mary gave her each day Of course as stated you cant give me a numerical answer but you can give me an expression in each case 12 LETTERS AND ALGEBRAIC EXPRESSIONS 11 For 1 you could tell me that Ann had 19 7 5 gtlt b many apples at the beginning where b is the number of apples Mary gives her each day So in this case we treat a as an unknown and b as a parameter and we express the solution ie in our case a in terms of the parameter b Why on earth would I want to know this Well I can now solve the problem as soon as you tell me b and I can solve it very quickly if you want to know a for different values of b So at the beginning Ann has four apples if Mary gives her three each day but Ann has nine apples if Mary gives her only two each day On the other hand for 2 you could tell me that Mary gives Ann 195 many apples each day where a is the number of apples Ann has at the beginning So in this case the roles of a and b are reversed Now a is the parameter and b is the unknown And the solution b is given in terms of a It is this switch in the roles of letters as unknowns or parameters which is very confusing to students so when you reach this stage you need to make this very clear and explicit Note that the distinction between a letter as a variable and a letter as a parameter is somewhat uid but roughly the following A variable is a number we don7t know and want to leave unspeci ed A parameter is a speci c number which we know but which we have not yet speci ed So in item 1 above we can think of b as a speci c number of apples Mary gives Ann every day Then b is a parameter3 Or we can think of b as an unspeci ed number for which we want to study how a 19 7 5 gtlt b as a dependent variable depends on the independent variable b Thus the distinction of the role of a letter as a parameter or as a variable very much depends on the context 1214 Letters as Constants This last use of letters is rather ob vious in general There are certain letters which are used for xed numbers The most common one is 7139 de ned as the ratio of the cir cumference of a circle divided by its diameter This is an irrational number and thus a non repeating decimal and there is no way to write it down easily using integers fractions or even roots The values 314 272 or 314159 are only approximations to its true value Another well known constant in calculus is the number 6 the in nite sum 1 1 1 1 112123 which is again an irrational number and equals approximately 2718 It is very useful in calculus but not really before A third well known 3Big I could be some really complicated number which if plugged in now would really make our computation very complicate i 12 1 NUMBERS AND EQUATIONS REVIEW AND BASICS number of the complex number 239 the square root of 71 This number is used to allow us to take the square root of negative numbers and is thus not a real number ie a number which we can write as a decimal or which we can identify with a point on the number line In physics etc there are other constants such as g for the gravitational constant The only thing to remember about these constants is that one should not use the same letters for variables unknowns or param eters when these constants are present in the context of a problem And the letter 7139 should probably never ever be used in any meaning other than the usual one 122 Algebraic Expressions Above we de ned a numerical expression as an expression involving numbers and the arithmetical op erations addition subtraction multiplication division later on also roots exponentiation and logarithms An algebraic eapressz39orz is an expression involving both numbers and letters as well the arithmetical operations We think ofthe algebraic expression as a numerical expres sion where not all numbers involved are speci ed and thus expressed as letters First a short aside Constants take sort of an intermediate role They should really be viewed as actual numbers but since they are generally not expressible as rational numbers or repeating decimals we often leave them as letters so the circumference of a circle of radius 2 units is normally given as 47139 the exact answer rather than 1257 say which is only an approximate answer Algebraic expressions can be evaluated by replacing each letter by a number so that the same letter is replaced by the same number everywhere in the expression whereas different letters may but dont have to be replaced by different numbers Eg in the example of the last subsection we evaluated 19 7 5 gtlt b when b 3 to get a value of 4 and when b 2 to get a value of 9 More importantly algebraic expressions can be rewritten as alge braic expressions which take the same value no matter what numbers replace the letters in them For this we usually use the rules given in Proposition 12 which hold for all numbers and thus also for all algebraic expressions For example 19 can be rewritten as Q 7 a 5 5 using the distributive law and the de nition of subtraction and divi sion Learning how to rewrite algebraic expressions is one of the most important topics of algebra and requires a lot of practice 14 MANIPULATING ALGEBRAIC EXPRESSIONS 13 13 Equations and Identities Equality is one of the basic notions of algebra expressing the iden tity of two expressions It is rst introduced in the role of Evaluate 2 5 7 In this role it corresponds to the way a pocket calculator would understand it namely Add 2 and 5 and give the answer rather than the role in which it is normally used in algebra namely simply as a statement that two expressions are equal This can easily lead to problems Solving 2 7 7 does not involve addition but subtraction or at rst possibly guessing the right answer and then adding 2 to check if the sum is 7 What is often confusing is that there are two types of equations Identities are equations which are always true no matter by what num bers we replace the letters as long as the expressions on both sides are still de ned The rules in Proposition 12 are examples of identities and many more can be derived from them eg a 7 b c a 7 b 7 c etc Any equation which is not an identity is thus an equation which is false when each letter is replaced by some number and both sides are still de nedl So eg 1 is an identity since for all nonzero x this equation is true and for z 0 the left hand side is unde ned On the other hand z 7 1 2x 7 3 is not an identity since eg when we replace x by 1 it becomes the equation 1 7 1 2 7 3 which is false Of course if we had replaced x by 2 then we would have gotten the true equation 2 7 1 4 7 3 but that is irrelevant here The equation x 7 1 2x 7 3 is only an identity if it is true for all values of x ie for all possible replacements of z making both sides de ned 14 Manipulating Algebraic Expressions In early algebra students are often asked to simplify complicated algebraic expressions ie to give an algebraic expression identical to the original one using identities in each step Here we call two al gebraic expressions A and B identical if no matter which numbers replace the letters in them A and B evaluate to the same number Usually such tasks appeal the identities from Proposition 12 as well as the de nition of subtraction and division But the exact way in which these steps can be justi ed can be obscure at times so we will go over the two most error prone steps in these simpli cations here and justify them by our identities and de nitions Of course l7m not trying to advocate that manipulating algebraic expressions should be taught the way I present this on the next page or so but for you as a teacher it should be valuable to know why one can perform these simpli cations 14 1 NUMBERS AND EQUATIONS REVIEW AND BASICS and that they are merely consequences of our identities in arithmetic and our de nition of subtraction But rst let7s prove a little fact about subtraction 141 Using the de nition of subtraction ln subsection 112 we de ned subtraction by addition namely a 7 b a 7b where 7b was the number such that b 7b 0 Note that this de nition can only be used after you have introduced negative numbers Before then you have to model subtraction with real life situations like taking away a part from a whole Mary had ve apples and gave two apples to Ann How many did she have left In this subsection we want to show that for any number a 117 7a 71a To show this we add both 7a and 71a to a 118 a 7a 0 by de nition 119 a71a1a71a by 111 1 i1a by 112 0a by arithmetic 0 by 110 Now we have to appeal to intuition but we could also show formally using our identities that there is exactly one number b which when added to 1 gives 0 and so that number b equals both 7a and 71a by de nition and 118 This gives the desired identity 117 142 Combining Like Terms Let7s now start on the rst of the two common steps in simplifying algebraic expressions combining like terms Here is an example 120 4z5y73z76yz7y Sounds pretty straightforward It is but when the expressions become more complicated it can be confusing And most importantly how can we justify it We rst appeal to our identities 17 and 113 4z5y73x76y4735y76y Now we have the like terms77 adjacent and can use the distributive law 121 4x73z5y76y473m576y1z71yziy 15 CONVENTIONS IN ALGEBRA 15 Of course7 the last step uses our identities 18 and 117 But the main step in combining like terms consists in the use ofthe distributive law 112 in the rst equality sign of 121 143 Removing Parentheses The other common step in sim plifying algebraic expressions is removing parentheses when adding or subtracting several terms Here are two typical examples 122 a7bc7da7bc7d 123 a7b7c7da7b7cd The rst identity 122 is pretty simple to justify using the de ni tion of subtraction and our identities 17 and 113 a7bc7d a7bc7d a7bc7d a7bc7d The second identity 123 is one of the most common sources of errors in algebra it requires lots of practice7 and a good justi cation Here it is a 7 b 7 c 7 d a i b H0 idl a i b l1C 1dl a 7 b 71c 712d by the distributive law a 7 b 70 1d by 117 and arithmetic a7b7cd by 177 113 and 18 As you can see7 this justi cation7 even for a small expression like a 7 b 7 c 7 d 7 can become quite tricky7 and so it is no wonder that removing parentheses after a minus sign is so hard for students by de nition of subtraction by 117 15 Conventions in Algebra There are a number of conventions about sets and operations which will make algebra simpler once we have mastered them7 since they allow us to write statements more precisely7 more succinctly7 and with less clutter All notation and conventions introduced in this section should be introduced in school only gradually and will require repeated practice 151 Notation for Sets and Some Special Sets of Num bers One of the fundamental notion of mathematics is that of a set Loosely speaking7 a set is a well de ned collection of objects Note that I am not actually carefully de ning what a set is here7 and for good reason The notion of a set cannot be de ned precisely7 we can 16 1 NUMBERS AND EQUATIONS REVIEW AND BASICS only state properties of sets In any case7 let7s x some notation about sets We will restrict ourselves to sets of numbers in the follow ing Not all of the notation introduced here will be used right away so consider this subsection more of a reference for later on The simplest example of a set is a nite set7 and we can simply list all its elements So the set containing exactly the numbers 07 27 and 5 would be written as 07 25 A special case of a nite set is the empty set containing no elements at all7 usually written as or Q We write n E S to denote that a number n is a member7 or an element7 of a set S ln nite or very large sets cannot be written this way For very large sets7 this is at least not practical In that case7 we will resort to using dots and write7 eg7 017 100 for the set of whole numbers less than or equal to 100 For the in nite set of all whole numbers7 we can use notation like 0127 7 etc Some sets of numbers are so common that one xes standard nota tion We denote o by N the set of all whole numbers 012 o by Z the set of all integers 0171272 o by Q the set of all rational numbers ie7 all numbers of the form where ab 6 Z and b 31 07 this set is also the set of all real numbers which can be written as repeating decimals o by R the set of all real numbers ie7 all numbers which can be written as decimals7 possibly non repeating and o by C the set of all sampler numbers We will introduce this set in detail later on Here7 N stands for natural number7 which is what mathematicians usually call a whole number Z stands for the German word Zahl meaning number and Q stands for the word quotient Now that we have all this7 we can use the set builder notation l to de ne new sets Eg7 the set of all positive integers can be written as n E Z l n gt 0 and the set of all real numbers less than 2 can be written as r E R l r lt 2 Furthermore7 later on7 it will be very convenient to have notation for intervals of real numbers7 ie7 sets of real numbers less than7 greater than7 or between xed real numbers Given real numbers a and b7 we ab forthe set rERlaSer ab for the set rERlaltrltb a7 for the set r E R l a S r and 00 foob for the set r E R l r S b 15 CONVENTIONS IN ALGEBRA 17 The sets ab ab aoo and 7oob are de ned similarly Note that the symbols 00 and 70077 do not denote real numbers but are merely used for convenience Therefore there can be no sets of the form moi wool 7001 or 70012 Finally we will sometimes use convenient notation to combine sets Given two sets A and B of numbers we let oAUB n l n E Aorn E B bethe um39zm ofthe setsA and B o A B n l n E A and n E B be the intersection of the sets A and B and o A7B n l n E A and 71 B be the dz erence of the sets A and B For example if A 012 and B 024 then AUB 0124 A B 0 2 A 7 B 1 152 Order of Operations and Use of Parentheses While the last subsection may have been rough going for you very quickly de ning a lot of notation which may be unfamiliar to you this sub section will be fairly smooth sailing We need to de ne for numerical or algebraic expressions in what order the operations are to be per formed and most of this will already be familiar to you from school even though you may never have thought about it this carefully and explicitly Note that everything we de ne in this subsection is only convention ie we arbitrarily de ne the order of operations a cer tain way to make things more convenient we could have de ned it completely differently but its certainly more convenient to have con ventions everyone agrees with So suppose you want to write an expression denoting the sum of the difference of a and the product of 4 and b and the difference of 2 and the quotient of 5 and 0 Note that its hard for anyone to even comprehend what I just wrote here so we desperately need a better way to write expressions In any case we could simply write this as 124 a 7 4 X b 2 7 5 7 0 since the rst rule on the order of operations is certainly that anything inside parentheses must be evaluated rst and if parentheses are nested one inside the other then we start from the inside out But this expression is ridiculously redundant of course as you have surely noticed we have literally cluttered it with many unnecessary 18 1 NUMBERS AND EQUATIONS REVIEW AND BASICS parentheses So we use the second rule on the order of operations All multiplications and divisions are performed before any additions and subtractions This allows us to rewrite the above expression much more simply as 125 a74gtltb275c We now come to the third rule on the order of operations Additions and subtractions without parentheses are performed from left to right and similarly multiplications and divisions are performed from left to right This allows us to rewrite our expression even more simply as u2 ai4xb275c Pause for a second here and convince yourself that we have actually used one of our rules on arithmetic to save parentheses speci cally we have used the identity 127 a74gtltb275ca74gtltb275c Now we introduce two conventions 0 In order to avoid confusion between the multiplication sym bol gtlt and the letter z especially when hand writtenl we drop the symbol gtlt unless we multiply two numbers in that latter case we use the symbol instead So eg 4 gtlt b is rewritten as 4b and 4 gtlt 5 is rewritten as 45 lnstead ofthe division symbol we use the fraction notation So eg 5 b is rewritten as g and 5 7 a 2 b is rewrit ten as Note here that the fraction notation also saves on parentheses If A and B are two complicated algebraic expres sions then 2 is an abbreviated form of A B ie the operations inside the expressions A and B are performed rst and then we divide the value of the former by the value of the latter Using these conventions we arrive at the nal simpli ed form of the expression 124 as 5 u a74b27E without any parentheses at all Later on when we introduce roots and exponentiation we also have to agree on the order of operations there For roots we simply indicate the order of operations by the length of the horizontal bar in the root symbol W and think of it as 5 So eg x23 means that we rst take the square root of 2 and then add 3 whereas V2 3 means that 16 ALGEBRA AS UNDOING EXPRESSIONS 19 we rst add 2 and 3 and then take the square root For exponentiation a fancy word for taking the power of we agree that the power is taken before any other operation is performed unless parentheses are involved So eg a b2 means that we rst square b and then add 1 whereas a b2 means that we rst add a and b and then square A word on how to read these aloud The rst is usually read as a plus b squared whereas the latter is usually read as a plus b quantity squared We summarize the rules on the order of operations as follows 1 Any operation inside parentheses is performed before any op eration outside them If nested parentheses are involved we perform the operations from the inside out Any fraction 3 for expressions A and B is taken to be A Any root A for an expression A is taken to be A 2 If parentheses do not determine the order between operations then we rst take powers then perform any multiplications and divisions in order from left to right and then any additions and subtractions in order from left to right Clearly these conventions require quite a bit of practice before a student becomes pro cient at them 16 Algebra as Undoing Expressions We have now set up some of the foundations of algebra We have discussed some important properties of arithmetical operations which hold for numbers and thus must carry over to arbitrary algebraic ex pressions We have isolated some of the different ways in which letters are used in algebra We have talked about algebraic expressions and equations We have gone over the rules which govern the evaluation of numerical expressions and therefore also algebraic expressions And we have set up some notation for sets of numbers Earlier we talked about algebra as arithmetic with letters There is one more way in which one can look at algebra namely as the art of undoing expressions Here is an example Ann has four apples Her friend Mary gives her three more apples each day for a ve day week How many apples does she have at the end of the week This is not an algebra but an arithmetic problem The answer is simply given by the numerical expression 4 5 3 apples ie 19 apples The corresponding pure arithmetic problem is 4 5 3 i Now lets turn things around Instead of asking for the answer 19 20 1 NUMBERS AND EQUATIONS REVIEW AND BASICS lets give the answer and ask for one of the things in the question let7s solve 4 5 7 19 The corresponding word problem now becomes Ann has four apples Her friend Mary gives her the same number of apples each day for a ve day week At the end of the week7 she has 19 apples How many apples did Mary give Ann each day Why do I call this undoing an expression Let b be the number of apples Mary gives Ann each day7 and c the number of apples Ann has at the end Then we have the equation 4 5 19 c The rst problem gives b and asks for c which amounts to simply evaluating the expression 4 5 b when b 3 The second problem gives 0 and asks for b This is very different It amounts to solving the equation 4 5 b 197 ie7 undoing rather than evaluating the expression 4 5 b to nd out for which b it evaluates to 19 In this case7 the undoing goes in two steps 45b19 turnsinto 5b1974 5b1974 turnsinto b19745 Now we have undone the expression 4 51 and reduced the solution to an arithmetic problem7 namely7 evaluating the numerical expression 19 7 4 5 Many algebra problems we will encounter later on will be of this nature And7 of course7 we will encounter situations later on where it is much harder to undo an expression7 where there may be no solution at all7 or where there are several or possibly in nitely many solutions So lets get started CHAPTER 2 Linear Equations 21 Ratio and Proportional Reasoning Probably the rst time a student encounters linearity is via the concept of ratio in a problem such as this one John has twelve apples which are either green or red For every green apple he has three red apples How many red apples does he have So the problem states explicitly that the ratio of the number of green apples to the number of red apples is 1 3 And one can then solve the problem by nding out that John has nine red apples so in unsimpli ed form the ratio of the number of green apples to the number of red apples is 3 9 Ratio can be a confusing concept since it can be expressed in a number of different and different looking ways 1 For every green apple John has three red apples 2 The ratio of the number of green apples to the number of red apples is 1 3 3 Out of every four apples three apples are red 4 The ratio of the number of red apples to the number of all apples is 3 4 5 Three quarters of Johns apples are red 6 75 of Johns apples are red Here statements 1 and 2 are basically the same just written in different notation However statements 1 and 3 look quite differ ent since the former expresses a ratio between parts green apples and red apples whereas the latter expresses a ratio between a part and the whole red apples and all apples Statements 4 5 and 6 formalize the ratio in statement 3 more and more abstractly statement 4 in terms of ratio similar to statement 2 statement 5 in terms of fractions still basically the same as 3 and statement 6 in terms of percentage which now takes the whole the unit not as four as in all previous statements but as 100 Figure 21 illustrates this ratio graphically 21 22 2 LINEAR EQUATIONS J J green apples red apples FIGURE 21 Ratio of the number of green apples to the number of red apples Note that the ratio notation can be taken further by relating not just two quantities but several Eg7 let7s modify the ratio problem to read John has 18 apples7 which are either green7 red7 or yellow For every green apple7 he has three red apples and two yellow apples How many red apples does he have77 Now the ratio of the number of green apples to the number of red apples to the number of yellow apples is given as 1 3 2 and we can still solve for the number of red apples as being 9 The concept of ratio directly leads to the concept of proportional reasoning Let7s look at the following table for the original version of our problem H 2 3 4 5 71 number of red apples 6 9 12 15 371 number of all apples 4 8 12 16 20 471 number of green apples C40 TABLE 21 The number of green and red apples vs all apples Each column contains numbers which are in ratio 1 3 4 The last column states this fact for a general number n of green apples Given any entry of the table7 we can easily compute the other two entries of that column Also note that7 going from one column to the next7 the numbers in each row increase by a xed amount for each row7 namely7 by 17 37 and 47 respectively On the other hand7 the ratio is usually changed when we add the same number to the entries of a column Suppose7 in our original word problem7 John receives two more green apples and two more red apples7 so that he now has ve green apples 22 VELOCITY AND RATE OF CHANGE 23 and eleven red apples Then the new ratio of the number of green apples to the number of red apples is 5 11 which is di erent from the old ratio 1 3 Similar examples can be found in many other contexts eg in recipes which call for the ingredients to be in a certain ratio such as two cups of our to one cup of sugar etc Percentage from the Latin per centum ie per one hundred is a special way to express a ratio where one thinks of one quantity in a ratio as the whole77 and sets that number equal to 100 the other quantities are then expressed in proportion to that number Eg in our rst word problem the ratio of the number of green apples to the number of red apples to the number of all apples is 25 75 100 and so we can say that there are 25 green apples and 75 red apples Usually the whole77 is the largest of the numbers and the others are considered parts77 of it but this need not always be the case Exercise 21 Think of several everyday examples of ratio and per centage where the largest number is not considered as 100 22 Velocity and Rate of Change Another common linear77 phenomenon is velocity A train moves at a constant velocity of 120 kmh How far will the train travel in half an hour In two hours In 3 hours77 Again we can make up a table time traveled in hours 0 1 ll 2131351 1 2 distance traveled in km 0 l 60 120 l 240 l 360 l 420 TABLE 22 Time and distance traveled by a train of constant velocity 120 kmh First a brief comment Why didn7t I just call 120 kmh the speed of the train rather than its velocity There is a small but important difference between these and well get to it in more detail later on For now lets just remember that velocity can be negative when the train moves backward whereas speed is always positive or zero when the train is not moving So velocity takes into account the direction in one dimension say along a rail line whereas speed does not In some sense the above velocity problem is again a ratio problem since time and distance traveled are in ratio 1 120 But that is not how 24 2 LINEAR EQUATIONS we normally think of this even though mathematically speaking it is exactly the samel The reason for this is twofold For one time and distance are measured in dz erent units But that alone doesnt really make it different since eg in a recipe we might use the ratio of one cup of sugar to one teaspoon of salt What makes time and distance really different is that they measure dz erent physical quantities So instead of considering this a ratio we normally think of it as a rate or more precisely a rate of change and express this rate of change as a quotient 2 1 1 At distance veoc1yi time Now Table 22 tells us that this quotient is xed for any time However since time and distance are measured not only in whole numbers we dont have the property of Table 21 any more that as we move along a row the numbers change by a xed amount since we have deliberately added some half hour intervals A quotient as in 21 is just one example of a rate of change And of course a rate of change need not be constant at all The train might speed up or slow down For now however we will concentrate on the case of a constant rate of change Examples of a constant rate of change abound in everyday situations 1 the cost of loose fruit and vegetables at the grocery store is computed by the weight say by the kilogram so there is a xed cost per kilogram 2 the cost of electricity you use is computed by the kWh kilo watt hour so there is a xed cost per hour 3 the water volume of a swimming pool usually increases at a constant rate when it is lled letting the time start when the pool starts lling up etc In each case we can identify a constant quotient Let7s make this relationship more formal Usually there is one independent quantity often called s and one dependent quantity often called y depending on the independent quantity In the train example z is the time trav eled and y is the distance traveled In the above examples 173 z is the weight the electricity used more precisely the electrical energy in kWh and the time since the pool starts lling up respectively And y is the cost in both 1 and and the water volume respectively Since we are assuming that the ratio or quotient between these two quantities z and y is constant we7ll denote this ratio by k for now 22 VELOCITY AND RATE OF CHANGE 25 we thus have 9 22 k 7 for any value of z and any corresponding y In the train example k is the velocity in the other three examples it is the cost per kilogram the cost per kWh and the increase in water volume per time unit say per minute respectively We can now rewrite 22 as 23 y k w to make the dependence of y on x more explicit For our train example switching to the more common notation of t for time 1 for velocity and s for distance we arrive at 24 s 1 t The equation 24 is just one way to represent this situation An other representation is via a table as in Table 22 A third representa tion is visual via a graph as in Figure 22 500 v 400 300 200 1 oo o t o 05 1 15 2 25 a 35 4 FIGURE 22 Time It and distance 5 1 t traveled by a train of constant velocity 1 120 kmh What are the advantages and disadvantages of these three repre sentations 0 An equation lets us quickly compute the exact value of the dis tance s for any time t As we will see later an equation is also 26 2 LINEAR EQUATIONS handy when you need a description of how the distance varies with time inside another7 more complicated computation On the other hand7 an equation is not so easy to visualize with out some practice A table lets us quickly look up the distance 5 for the particular values of t listed in the table But it has us guessing the distance for other values of t In fact7 strictly speaking7 we cant really tell the distance for other values of t at all we are really just guessing them since we cant tell from a table how the distance really varies as time varies o A graph is a great visual aid and lets us see at a glance how the distance varies with time The drawbacks are twofold Its not really possible to read from a graph the precise distance at a given time t and this is impossible for values oft off the graph And its hard to work with the graph when you have to use these values inside another7 more complicated computa tion Graphs become a bit more complicated when they can no longer be drawn without lifting the pencil off the page7 ie7 when the value of 5 suddenly jumps at some value of t well talk about such graphs in more detail later Since tables can only tell us the distance for nitely many values of t they are generally not used as much in algebra There are excep tions7 however eg7 you may know that the values of the trigonometric functions at speci c multiples of 7139 are handy to have around7 and even to have memorized On the other hand7 it7s generally not hard but only tedious at worst to generate a table from an equation7 say From now on7 we will mostly concentrate on equations and graphs One of the main tasks in algebra will be to transfer back and forth between these two representations7 a task which requires lots and lots of practice A graphing calculator can be of great help in visualizing the graphs of more complicated equations but for simpler equations7 you7 and hopefully later on your students7 should be able to visualize their graphs mentally and sketch them quickly 23 Linear Expressions We now slightly generalize from the above examples of ratio and proportion Here is an example A pay phone charges 506 for the rst minute and 10C for each additional minute of a long distance call Let7s look at a table of how much a phone call costs There are four things to note right away in Table 23 23 LINEAR EXPRESSIONS 27 numberofminutes 1121314151 6 7 sl 9110 costincents l50l60l70l80l90l100l110l120l130l140l TABLE 23 Cost of a pay phone call 1 We still have the property as in Table 21 that each row in creases by a xed amount by 1 and 10 respectively 2 We no longer have the properties from Tables 21 and 22 that each column has numbers in the same ratio Eg the ratios 1 50 and 2 60 from the second and third column are different 3 As in Table 21 but unlike in Table 22 it doesnt make real world77 sense to look at the cost of a phone call lasting 0 min utes 4 As in Table 21 but unlike in Table 22 it doesnt make sense to look at the cost of a phone call lasting a number of minutes which is not a whole number such as half a minute We7re assuming here that the phone company charges the same for a fractional77 minute ie a time interval lasting less than a minute as for a whole minute So eg a phone call lasting 2 minutes and 1 second and another call lasting 2 minutes and 59 seconds will both be billed as a 3 minute call by the phone company As you can see there are a whole lot of new issues coming up and we need to separate them out one by one We7ll defer the issue of whole number minutes in item 4 until Section 431 we7ll assume for now that we just round up77 and so consider only whole numbers The issue in item 3 is fairly easy to resolve if we just look for the pattern77 in Table 23 We could simply assume that the phone company charges 406 for a 0 minute phone call This is of course not how this works out in the real world but it makes the table follow the pattern noted in item So we are down to considering the issues from items 1 and Let7s look at the graph for the cost of the phone call see Figure 23 Note that in light of item 3 above I didn7t draw this graph as a line but only used dots for positive whole number values of the number of minutes What we would like of course is an equation describing this graph Let7s denote by t the number of minutes of the phone call and by c the cost of the phone call in cents Using the information from the original problem we notice that t7 1 is the number of additional minutes and 28 2 LINEAR EQUATIONS 14039 o 120 n 100 I FIGURE 23 Cost of a pay phone call so the cost of a phone call can be calculated by the equation 25 05010t71 where 506 is the cost of the rst minute and 106 is the cost for each additional rninute Alternatively7 using the trick77 from above and letting 406 be the cost77 of a 0 minute phone call7 we notice that if we subtract this amount from all the other possible costs7 then we do indeed get a xed ratio7 narnely7 1 107 for all the columns in Table 23 This reasoning quickly leads to the equation 26 c 10t 40 where 406 is the cost77 of a 0 minute call and 106 is the cost per additional77 rninute It is also not hard to see that equations 25 and 26 give the same cost using our rules of arithmetic It should be clear that rnanipulating expressions such as 10t 40 or 50 10t 7 1 is very important to algebra So lets look at letters and algebraic expressions in more detail Here is the way Singapore rnath books in Primary Mathernatics 6A rst introduce letters as variables Think of the letter m7 say7 as the number of rnarbles in a bag where we assume that each bag con tains the sarne7 but unknown7 number of rnarbles Now it makes sense 30 2 LINEAR EQUATIONS 2 4 Line Equations The last section should have convinced you that equations of the form 28 y mz b and their graphs are very important in algebra so we will study them mostly in the abstract in this section with some asides about appli cations We have changed notation slightly here to conform with the usual conventions on these equations We have called the independent variable z the dependent variable y and we have denoted the two pa rameters by m and b As usual for parameters we think of them as xed numbers An equation as in 28 is often called a linear equa tion but it is not the most general form of a linear equation as we will see later on 241 Graphing a Linear Equation Before we can formally de ne the graph of a linear equation we need to formally de ne the two dimensional coordinate system which connects algebra with ge ometry or more precisely coordinate geometry A coordinate system consists of a horizontal number line usually called the x acsis if z is the independent variable and a vertical number line usually called the y acsis if y is the dependent variable So both axes are two way in nite lines called the coordinate ames Of course if the variables are denoted by different letters we also use these letters to denote the cor responding axes The point where the two axes intersect is called the origin and usually denoted by the letter 0 See Figure 25 We also introduce the set of ordered pairs of reals zy by let ting x and y vary over the real numbers and by de ning two pairs of reals zhyl and z2y2 to be equal if both 1 2 and y1 yg We now identify each pair of reals z y with the point P in the coordinate plane obtained by moving from the origin x units to the right or if z lt 0 7z units to the left and y units up or if y lt 0 7y units down We observe that for each pair of reals there is exactly one such point in the coordinate plane Similarly for each point P in the coordinate plane there is exactly one pair of reals found as follows Draw a horizontal and a vertical line through the point P let x be the number on the z axis viewed as a number line at which the ver tical line intersects the z axis and let y be the number on the y axis viewed as a number line at which the horizontal line intersects the y axis Then the point P corresponds to the pair of reals We call x the x eoordinate of P and y the y eoordinate of P Figure 25 illustrates this for the point P 23 The vertical line through P 24 LINE EQUATIONS 31 P2 3 o oo0 FIGURE 25 The coordinate system with two points labeled intersects the a axis at 27 and the horizontal line through P intersects the y axis at 3 Now the graph of the equation y mzb from 28 consists of the set of all points Ly in the real plane satisfying this equation For an exarnple7 see Figure 26 for the graph of the equation y x We want to argue that the equation 28 always describes a line in the plane For this7 we rst compute the ratio of the change of y to the change of x Fix any two distinct points A zhyl and B x2y2 on the graph Since A and B satisfy the equation 287 we have 29 yl mm b and y2 mm b If 1 2 then7 by 297 yl yg and so A B7 contrary to our assumption Thus 1 31 x2 which will allow us below to divide by 2 7 1 2 2 LmEAR mummus mumze Thelmey We now compute me nun 12 me mm enhe Mange me We eehmee as 71 m2b New meme ms nun does hm depend m me We 0 we mm A and B m we ywhcnhe Equation 23 and m am always equalslhe 1 ma a m Now let s Pick three damn mm A B and B m we mph and me they axe axxmged m we alder mm Jan to new and um m gt o 37 that 39 and h be dame A m the mammals Plane harder cshqwthaltheywhdoursq crm me we ed to imw um three Paints axe cn me Fe 2 m heywhm thesquu m s A B and 539 he on alme So we we m know um yen u we M the mum CC Jespemvely helhepomlwhezelhevenmdhnelhmu hB 2 A um Emunous 31 new A G4 sen1 I39heslwpenf uelmey egaenea mun 2 7 ened with two d ezenl tnang Bu hespeetwety and the horizontal hne througu A hnteheet then we can tnnn twn mangee AABC and AAB C Hehe the Picture m e2neenneet ThehnnteAcandceauydnhennnnehne Shae the edee Ac and Ac axe hmh hnnanma 12 12212112 to the 273x15 Now the meme calculaunn yxelds for the lengthe d the sdes of the tnaneiee that Toquot cA cA hennnhe the angee 2c and 2c axe herthneht aneiee andthne the mangee AABC and AAB C39 axe enuax shee the sdes Ac and Ac 2121de md hysmdanlye the angee LBACmd LB AC enmede the sdes AB and AB aae Paxa el aeweu Ihuethe hnnte A md B mlh ywhm lhesqualm 23 dcmdeedhenn three methane our Mme m gure 27 We call the qunthent m the stop dlbhneyvenbytbequalmnymb Emcee 2 2 How the the hmtwe m memes 2 e and 27 change when m he neeatwe2 What hWPEm when m e m owe enuax 12me that the exam orm ahne m each case Now that we have found zvmd Werwelumdlh haxameteh me what ahnut the haxameteh b2 Settmee a we nhtnn mm the enua Mon 23 that the hnnt at he nn our hne Ths he where the hne 4 arm the teams Ihezetme b 15 called the 17mm n he hue and the equatm y m b 15 ca ed the stopehaempz ltwmnf the hhe PIGsz I39hes wpenfthelmey 22 aemea shed wnh two mew thahgee at the emame h the statemeht we tust tame ahme he the A hmvemcax hhe h the cmzdmz39e 1 ven by m y mb w u talk aheht the Spemdcase wecahmh 09217me It must mlezxcl the it eh me real number I m mhez the 1 AB axe 122121121 Fax heme lets aeehhe that hmh B that t h 39axegtbetthatthehmhtssahd lle Ax the alele sh ecmdzzwpmmscmdcsamp 228 RE on me through A ahdthe vezttea hhee shee an three sdes of the mange AABC and AAB C39 axe hahwtee haxaneL these thaws axe smuax and en B W CA m h the 1mm Me See The hhhee that the Met g the rm dzpendnnthecknxae n h mm B mdwie demethsqumxem 24 LINE EQUATIONS 35 by m If we now let B Ly be any point on the line to the right of A then y 7 b x 7 0 m which can be rewritten as yibmd and so y 7nd b We have thus shown the following Theorem 23 I The graph of any equation y ind b is a non vertical line with slope in intersecting the y accis at the point 07 b 2 Any non vertical line Z is the graph of an equation of the form y 7nd b where b is determined by the y coordinate of the point where the line Z intersects the y adis and m is the quo tient 12 11 2 1 for any two distinct points B zhyl and B zgy2 on the line Z Exercise 24 Strictly speaking7 we have shown part 2 of the above theorem only for the case when the line is increasing and then only in the case where the points B and B lie to the right of the point A 0J9 Finish the proof by analyzing the cases where the line is decreasing77 or horizontal7 and where one or both of the points B and B lie to the left of the point A 07 b 242 Different Forms of Equations for Lines We have al ready seen one form of an equation for a non vertical line7 namely7 the slope intercept form y 7nd b where m is called the slope2 and b the y intercept7 ie7 the y coordinate of the point at which the line intersects the y axis So we can easily write down the equation of a non vertical line if we are given its slope and y intercept ln reality7 however7 we are much more likely to be given two points on the line7 both in geometry or in real life applications such as the following ZThe Singapore math books call the slope of a line its gradient In US text books7 this term is normally used in a different but similar meaning in calculus 36 2 LINEAR EQUATIONS A car travels at constant speed At 830 am it crosses a state line where the mile markers start at 0 At 1010 am it passes mile marker 100 So letting t be the number of hours since midnight and s the dis tance from the state line to be taken as negative before the car crosses the state line and converting times to fractions the problem tells us that the graph of the cars motion which is a line since the car travels at constant speed contains the two points 8 0 and 10100 So how do we obtain an equation for a non vertical line if we only know two distinct points on it The answer is of course that the two points tell us the slope First remember the mantra which really just says that the y change rise is proportional ie in a xed ratio to the z change run The word rise may be somewhat confusing since it may be negative ie y may decrease as x increases but then the slope is simply negative As a third case y may just be constant as x changes in which case the slope is 0 and the graph of the equation is a horizontal line Let7s call the two given points on the line P0 and P1 and assume that their coordinates are zoy0 and shyl Then the slope can be computed as 210 m 1 T 0 1 7 0 Similarly given any other point P my on this line we have y yo i 7 0 211 m The equation 211 is called the point slope form of a line equation and can with after little bit of algebra also be written as y yom0 or y ms imxo yo Of course in this last form it is basically the same as the slope intercept form except that the parameter b is now more complicated But the point slope form which is useful if we are given the slope and just some point P0 zoy0 on a line isn7t really what we were after at rst when we asked for a line equation given two distinct points on a non vertical line However it is now easy to combine 24 LINE EQUATIONS 37 equations 210 and 211 to obtain the two point form of a line equa tion 212 y 90 91 yo 70 1 70 Again with a little bit of algebra this can be rewritten as 91 yo yiyo70 1 0 or in slope intercept form as 91 90 11 110 y 7 70yo 1 7 0 1 7 0 And again of course in this last form this is basically the same as the slope intercept form except that now both parameters m and b are very complicated Note that in equations 2107212 not only the letter m but also the letters 0 1 yo and y1 are parameters and only the letters z and y are variables something which many students will nd highly mystifying The above three forms of the line equation all apply exactly to all non vertical lines There is a fourth and more general form of the line equation which also applies to vertical lines and not surprisingly is called the general form of the line equation 213 AzByO0 Here A B and C are parameters ie xed numbers and z and y are variables namely the a and y coordinates of points in the coordinate plane We will tacitly assume that at least one of A and B is nonzero since otherwise the equation 213 degenerates to the trivial equation 0 0 which is either true or false independent of the values of the variables x and y and thus does not describe a line The equation 0 0 either describes the whole plane if the parameter 0 is zero or the empty set otherwise3 On the other hand if at least one of A and B is nonzero then the equation 213 really does describe a line as we will now see We consider four cases for the general line equation 213 sepa rately Here the third case will not exclude the rst two cases 3Note that in this equation C 0 the letter C is the parameter and not the unknownl The unknowns or variables in C 0 are I and yl 38 2 LINEAR EQUATIONS Case A 0 By our tacit assumption we must then have B 31 0 and so the equation reads By C 0 and can be rewritten as C y B This is just a special case of the slope intercept form of a line with slope 0 and so describes a horizontal line which crosses the y axis at the point 0 7 Note that equation 214 does not depend on the value of the variable z as is to be expected of a horizontal line the equation simply xes a particular value of the variable y namely 7 214 Case B 0 By our tacit assumption we must then have A 31 0 and so the equation reads Ax C 0 and can be rewritten as C 215 z 7 A This equation does not depend on the value of the variable y and thus describes a vertical line which crosses the z axis at the point 7 0 the equation simply xes a particular value of the variable x namely 7 Note that equation 215 cannot be rewritten in slope intercept form or either of the other two forms introduced earlier in this section Case C 0 First note that in this case one of A or B may also equal 0 so this case does not exclude either of the rst two special cases However since the equation now reads Ax By 0 we immediately see that the origin 00 is a point on the line If A or B also equals 0 then by the previous cases the equation describes the z axis or the y axis as a line respectively If neither A nor B equals 0 then the equation can be rewritten as A 216 y i B x Since this line crosses both the z axis and the y axis at the origin both the y intercept de ned before as the y coordinate of the point where a line crosses the y axis and the z intercept de ned now as the z coordinate of the point where a line crosses the z axis equals 0 And the slope of the line clearly equals m 7E Case A B C 31 0 In that case the equation 213 can be rewritten as A C if z 7 i y B B and so has nonzero slope m 7 Also since the parameter C does not equal 0 the line does not cross the axes at the origin Rather by 2 4 LINE EQUATIONS 39 setting a O and and solving for y and then setting y O and solving r 3 respectively it is not hard to see that the line contains the two distinct points 7 O and 17 on the use and yraxis respectively o t e reintercept of the line is 7 and its yrintercept is 7 Since we can multiply the equation 2 13 by any nonzero number without a ecting its truth we may assume having multiplied equation 2 13 by the appropriate number that the parameter 0 ter renaming of parameters equals 71 Then equation 2 13 reads with the new values for our parameters A B an O A By 71 0 or A3 By 1 Now the use and yrintercepts are L and respectively De ning new ar meters a and 17 to e ual i and respectively we can thus rewrite equation 2 13 still assuming that the parameter 0 equals 71 as I y 2 17 a Z7 7 1 This is another normal form77 of an equation of a line not containing the origin It is sometimes called the intercept farm of such a line See Figure 2 9 for the example 7 l FIGURE 29 The line 12 1 with reintercept a 72 and yrintercept l7 1 Finally let s look at the five di erent forms of a line equation in terms of a realrlife example In each case a car will cross a state line where the mile markers start at O and travel at a constant speed We use the independent variable t for the time elapsed in hours since 4O 2 LINEAR EQUATIONS noon7 and the dependent variable 5 for the distance traveled in miles starting at the state line taking 5 to be negative before the car crosses the state line In all versions of this exarnple7 we will get an equation for the same line7 see Figure 210 25039 sraxls P1 25 240 Po1150 Bo90 o539 o39o539 1 391539 2 15 39 1 A 15 0 FIGURE 210 The line 5 60t90 describing the travel of a car Slope intercept form A car travels at a constant speed of 60mph At noon7 it has traveled 90 miles past the state line77 Slope m 607 s intercept b 907 equation 5 601 90 Point slope form A car travels at a constant speed of 60mph At 1 prn7 it has traveled 150 miles past the state line77 Slope m 607 point P0 11507 equation 5 7 150 60 or 5715060t71 Two point form A car travels at a constant speed At 1 prn7 it has traveled 150 miles past the state line at 230 prn7 it 24 LINE EQUATIONS 41 has traveled 240 miles past the state line Points P0 1150 and P1 257 2407 equation 5715072407150 15072407150t 1 til 7 25710 7 2571 Intercept form and general form A car travels at a constant speed At 1030 am7 it crosses the state line at noon7 it has traveled 90 miles past the state line t intercept a 715 s intercept b 907 equation in intercept form t s i 1 715 90 7 and after clearing denominators by multiplying by 790 in general form 60t78900 Although the different phrasings of this problem may seem some what contrived7 they show that each form of a line equation can be modeled by a real life example7 which is what I was trying to con vince you of Exercise 25 Find a very differen real life example for the equation y 72x 7 2 and explain how this problem would be phrased in terms of the four different forms of the line equation slope intercept7 point slope7 two point7 and intercept form 243 Varying the Parameters in a Line Equation Let7s now see what happens when we vary the parameters Most of this should be clear from the visual or physical interpretation of the parameters in the different forms of the line equation covered in the previous subsection In the slope intercept form of a line equation 218 ymzb what happens if we vary the slope m and the y intercept b See Figure 211 for examples Here are some fairly obvious remarks about the slope m o If m is positive7 then the line described by 218 goes up or is increasing 7 ie7 y increases as x increases o If m is negative7 then the line described by 218 goes down or is decreasing 7 ie7 y decreases as x increases to LINEAR EQUATIONS FIGURE 211 Examples of varying slope and y intercept of a line o If m is 0 then the line described by 218 is horizontal ie y remains constant as x varies 0 As m increases and assuming it is always positive the line described by 218 becomes steeper 0 As m decreases and assuming it is always positive the line described by 218 becomes atter or less steep Similarly here are some remarks about the y intercept b o If b is positive then the line described by 218 crosses the y axis above the origin o If b is negative then the line described by 218 crosses the y axis below the origin o If b is 0 then the line described by 218 crosses the y axis at the origin 25 SOLVING A LINEAR EQUATION IN ONE VARIABLE 43 0 As b increases or decreases7 the point at which the line de scribed by 218 crosses the y axis moves up or down respectively Similar comments can be made about the and y intercepts in the intercept form of the line equation 9 1 b 2 7 a 25 Solving a Linear Equation in One Variable After all this long preparation let7s now solve linear equations We7ll again start with two examples A pool is being lled with water at a constant rate of 50 Zmin liters per minute7 starting at noon If the pool already contained 1000 liters of water at noon7 how much water will be in the pool after 30 minutes A pool is being lled with water at a constant rate of 50 Zmin7 starting at noon If the pool already contained 1000 liters of water at noon7 what time will it be when the pool contains 2500 liters We can see that the time t in minutes since noon and the water volume V in the pool in liters are related in both examples by the equation 219 V 50t 1000 since the slope is the rate of 50 Zmin7 and the V intercept is the 1000 liters of water already in the pool at noon Now the rst example above is just an arithmetic problem We replace t by 30 and obtain the answer V 50 30 1000 2500 in liters On the other hand7 the second problem gives V 2500 and asks for a solution of the equation 220 50t1000 2500 This requires undoing the operation of computing the volume from the time as explained in section 167 ie7 it requires rewriting the equation by isolating t We undo 220 in two steps 50t 1000 2500 turns into 501 15007 and 50t 1500 turns into t 30 This then gives the answer It will be 1230 pm when the pool con tains 2500 liters 44 2 LINEAR EQUATIONS More precisely what we are saying is that the following three equa tions are equivalent namely have the same solutions in t 221 50t1000 2500 222 50t 1500 223 t 30 But why do we know that they are equivalent Notice that we really have to show two things 1 Any solution to equation 221 is also a solution to equa tion 223 and 2 any solution to equation 223 is also a solution to equa tion 221 Let7s address item 1 rst Equation 222 is obtained from equa tion 221 by subtracting 1000 on both sides But if we replace t in equation 221 by any number which makes both sides of the equa tion equal then this equality remains true if we subtract 1000 from both sides After all the two sides of equation 221 represent the same number before the subtraction so they must still represent the same number after the subtraction The same is true in going from equation 222 to equation 223 If we replace t in equation 222 by any number which makes both sides of the equation equal then this equality remains true after we divide by 50 on both sides The argument for item 2 is similar but with a crucial difference We rst have to show that if we replace t in equation 223 by any number which makes both sides of the equation equal then this equal ity is true before we divide both sides by 50 There is only one way to show this Note that equation 222 can be obtained from equa tion 223 by multiplying both sides by 501 Now we can argue as for item 1 If we replace t in equation 223 by any number which makes both sides of the equation equal then this equality remains true if we multiply both sides by 50 and the resulting equation is equation 222 A similar argument shows that any solution to equation 222 is a so lution to equation 221 since equation 221 can be obtained from equation 222 by adding 1000 to both sides of the equation Both steps of the argument for item 2 can be viewed as undoing the corresponding steps of the argument for item This leads us to two crucial observations about solving equations in general Proposition 26 1 Any solution to an equationA B where A and B are algebraic empressions remains a solution if we add 25 SOLVING A LINEAR EQUATION IN ONE VARIABLE 45 or subtract the same algebraic eppression on both sides of the equation and vice uersa any solution to the equation after adding or subtracting the same algebraic eppression is also a solution of the original equation A B 2 Any solution to an equation A B where A and B are alge braic eccpressions remains a solution if we multiply or diuide both sides of the equation by the same nonzero algebraic ecc pression and vice uersa any solution to the equation after multiplying or dividing by the same nonzero algebraic empres sion is also a solution of the original equation A B Here an algebraic eppression is nonzero if it cannot equal 0 no mat ter what numbers replace the letters in the eppression Here are two ne points which you can ignore on a rst reading but which are important in general Remark 27 There is small caueat here If the algebraic eppression C in Proposition 26 which we add subtract multiply or divide by on both sides of an equation cannot be computed for all values of its uariables then our proposition only holds for those values at which 0 can be computed There is also an eactension of Proposition 26 If the algebraic eppression 0 equals 0 for some but not all replacements of letters in C by numbers then Proposition 26 remains true for those values at which 0 does not equal 0 Let7s rst look at the caveat Here are two very simple examples I If we add i to both sides of the equation x 0 say then the new equation zi obviously does not have the same solutions as x 0 but the proposition still holds for all values of x at which i is de ned namely whenever z is nonzero Both equations z 0 and z i i are false for all nonzero z as the proposition states 2 If we add V5 to both sides of the equation x 71 say then the new equation x V 71 obviously does not have the same solutions as x 71 but the proposition still holds for all values of x at which is de ned namely whenever z is non negative Both z 71 and z f 71 V are false for all non negative x as the proposition states As for the extension of Proposition 26 2 here is a very simple ex ample The equation z I obviously does not have the same solutions as the equations 2 z obtained by multiplying both sides by However if we only look at the solutions to both equations where z 46 2 LINEAR EQUATIONS is nonzero then they do indeed have the same solutions namely only z 1 Now for the proof of Proposition 26 Adding to subtracting from multiplying by a nonzero expression or dividing by a nonzero expres sion both sides of an equation preserves all solutions of the original equation and since these steps can be reversed by subtracting adding dividing and multiplying respectively we also dont get any extra solutions Notice that the crucial part in Proposition 26 is the vice versa part Not only do we preserve a solution by adding subtracting multiplying or dividing both sides of an equation we also dont get any new solution Of course in part 2 of Proposition 26 we de nitely need the restriction that we dont multiply or divide by 0 or an expression which could equal 0 Dividing by 0 is not allowed anyway and multiplying an equation A B by 0 leads to the equation 0 0 which almost always has more solutions than the original equation namely any number is a solution to 0 0 A more sophisticated ex ample and an example of a very common error is to divide both sides of the equation 2 x by z to arrive at the equation x 1 Clearly z may be zero so this would not be allowed under Proposition 26 and it is also easy to see that Proposition 26 fails in this example The equation 2 x has two solutions namely 0 and 1 whereas the equation x 1 only has the solution 1 However as noted above the extension of our proposition still allows us to state that the nonzero solutions of 2 z and z 1 agree More generally why am 1 putting so much emphasis on the vice versa part7 After all its obvious that you shouldnt multiply both sides of an equation by 0 while solving an equation The reason is that this vice versa part will become much more subtle later on and leads to frequent mistakes For example if we square an equation A B then any solution to the original equation A B is still a solution to the equation A2 B2 after squaring but not vice versa as you can see from the simple example z 1 It has only one solution namely 1 but the equation x2 1 ie after squaring has two solutions namely 1 and 71 So as we extend Proposition 26 later on eg in Proposi tion 53 to allow more operations on equations we always have to make sure that the vice versa part also holds since it can fail in subtle ways Let7s now vary the pool problem a bit to arrive at the general form of a linear equation 25 SOLVING A LINEAR EQUATION IN ONE VARIABLE 47 A pool is being lled at a constant rate of 50 Zmin starting at noon at which time it already contains 1000 liters of water If that same pool had contained 1600 liters of water at noon and had been lled at the rate of only 30 Zmin starting at noon it would have been full at the same time At what time will the pool be full How much water can the pool hold777 If we still let t be the time in minutes since noon the expression 50t 1000 describes the water volume under the rst scenario while the expression 30t 1600 describes it under the second scenario The problem tells us that the pool is full and so contains the same volume of water at the same time It so we arrive at the equation 224 501 1000 301 1600 Now we can use Proposition 26 three times to isolate77 t Equa tion 224 is equivalent to 20t1000 1600 by subtracting 30t from both sides Next we subtract 1000 from both sides to arrive at the equivalent equation 20t 600 Finally we divide by 20 on both sides to arrive at the solution It 30 which yields the answer The pool will full at 1230 pm77 A small further computation shows that the pool can hold 2500 liters since 50 30 1000 30 30 1600 2500 There is one other way to solve our pool problems namely graph ically Let7s return to the rst equation we tried to solve 220 501 1000 2500 Consider the equation describing the water volume in terms of the time 219 V 501 1000 We can view solving 220 as determining the t coordinate of the point P 30 2500 where the line described by equation 219 inter sects the horizontal line V 2500 See Figure 212 Let7s turn to the other equation we solved in this section 224 501 1000 301 1600 48 2 LINEAR EQUATIONS P 30 25001 V2500 1500 V50t100 100 P0 o 1000 FIGURE 212 Solving the equation 50t 1000 2500 graphically Again7 there is a nice graphical solution We can describe the water volume in terms of the time for the two different rates as follows 225 V 50t1000 for the rate of 50 Zmin 226 V 30t 1600 for the rate of 30 Zmin The problem now asked for the time at which the volume is the same7 ie7 to determine the t coordinate of the point P 302500 where the two lines described by the equations 225 and 226 intersect See Figure 213 A general linear equation has the form moi b0 mm b1 and can be viewed graphically as nding the a coordinate of the point at which the lines described by 228 y mox be and y mg b1 25 SOLVING A LINEAR EQUATION IN ONE VARIABLE 49 P 30 2500 V30t1500 P1 o 1600 V50t100 100 P0 o 1000 FIGURE 213 Solving the equation 50t 1000 30t 1600 graphically intersect as7 eg7 in Figure 2127 where one line is horizontal7 or Fig ure 2137 where neither line is horizontal Algebraically speaking7 we rst subtract up to two times to get all the x on one side77 and all the constants on the other side In order to see what might happen7 its easier to View this problem graphically as nding the s coordinate of the point where the two non vertical lines described by the equations in 228 intersect There are now three cases for the two lines described by the equations in 228 Case 1 The lines intersect in a single point P7 say Then the two lines must have different slope7 ie7 m0 31 m1 In that case7 there is exactly one solution to the equation 2277 namely the s coordinate 50 2 LINEAR EQUATIONS of P Algebraically we can nd this value in three steps m0 7 m1 z be b1 by subtracting mlz m0 7 m1 z b1 7 be by subtracting b0 b 7 b x by dividing by me 7 m1 m0 m1 Here the last step is possible since m0 31 m1 and thus m0 7 m1 31 0 A real life example of such a situation would be the following Two trains leave the same city each heading in the same direction at a constant speed of 80kmh and 120kmh respectively At noon the rst train is at a distance of 140km from the city and the second train at a distance of 100km At what time will the second train pass the rst It is easy to see that solving this problem amounts to solving the equation 80f 140 120t 100 and after some algebra one obtains the solution It 1 where the time t is measured in hours since noon This solution is shown in Figure 214 as the point P at the intersection of the lines 1 and 0 Case 2 The lines are parallel and have no point in common Then the two lines must have the same slope ie m0 m1 but different y intercept ie b0 31 b1 In that case there is no solution to the equation 227 Algebraically we can see this in three steps m0 7 m1 z be b1 by subtracting mlx m0 7 m1 m b1 7 be by subtracting b0 0 b1 7 be since m0 ml The last equation is always false and thus has no solution x A similar real life example of such a situation would be the following Two trains leave the same city both heading in the same direction at a constant speed of 120kmh At noon the rst train is at a distance of 160km from the city and the second train at a distance of 100km At what time will the second train pass the rst It is easy to see that solving this problem amounts to solving the equation 120t 160 120t 100 which has no solution The sec ond train will never pass the rst This situation is also shown in Figure 214 The two lines 2 and 0 do not intersect Case 3 The lines are parallel and in fact coincide Then the two lines must have the same slope and the same y intercept ie m0 2 5 soume A mama gemnon m om mums menu 2 1A Solvmgthe three Lums pxoblems gaphxcauy m and bu m In that case every value of z 15 a solution to the equauon 2 27 Algebxaacauy we can see ths agam m Lhreesteps mu 7 m z bu b by subtuctmg mg m 7 m z m e bu by subcxaamg 00 smoemmandbum The last equauon ls always true and so my wine of z 15 a solution A smile somewhat noneenacal zeaere example orsuda a smuauon would be the following Two txuns leave the same city both heading m the some direction at a constant speed of 120kmh At 52 2 LINEAR EQUATIONS noon7 both trains are at a distance of 100km from the city At what time will one train pass the other It is easy to see that solving this problem amounts to solving the equation 120t 100 120t1007 and that the trains will always be at the same distance from the city This situation is shown in Figure 214 The travel of both trains is described by the same line 0 26 Solving Simultaneous Linear Equations in Two or More Variables Lets start again with an example 17m thinking of two numbers The larger of the two numbers is less than twice the smaller number by 3 The sum of the two numbers is 18 What are my numbers77 Now it is7 of course7 possible to solve this problem using only what we already know7 namely7 linear equations in one variable lf 1 call the smaller number a then the larger number can be expressed as 2x 7 3 since l7m told that the larger of the two numbers is less than twice the smaller number by 3 The other piece of information then tells us about the sum of the two numbers7 namely7 that z 2x 7 3 18 which can be rewritten as 3x 73 18 and solved to z 7 as discussed in the previous section But it may7 and will7 not always be so easy to express one unknown quantity in terms of another so an alternative and more general approach is to call the two numbers z and y and lets say z is the smaller of the two Now we can translate the information from the problem into two equations y 2x 7 3 Q29 y18 We can rewrite these equations into two line equations in general form as 72x y 3 0 Q3 xy7180 Solving these equations now means that we have to nd all the values for the unknowns z and y which make both equations true There are a number of different methods to solve such simultaneous linear equations as well call them7 ie7 several equations mentioning several unknowns but where none of the unknowns is raised to any 26 SOLVING LINEAR EQUATIONS IN TWO OR MORE VARIABLES 53 power7 or occurs in a denominator or in a product with another un known so that each equation is equivalent to a line equation in general form Geornetrically7 solving two simultaneous linear equations in two un knowns means that we need to nd the points of intersection of two lines in the plane If the lines are not parallel7 there will be exactly one point of intersection Figure 215 shows the two lines from our example problem and the point of intersection P 711 o 39 391 392 39 5 39 394 39 395 39 39 391 39 5 FIGURE 215 Solving the two equations 2x y 3 0 and z y 7 18 0 graphically Algebraically7 there are a number of different ways to solve simulta neous linear equations7 and experience will tell you and your students which method is more ef cient in a particular situation Eg7 starting from 2297 it would be easiest to use the substitution method We already know that y 2x 7 37 so we could simply replace y in the other equation x y 18 by 2x 7 3 and obtain the single equation 54 2 LINEAR EQUATIONS z 2x 7 3 18 in a single unknown which can then be solved as in the previous section This is essentially how we solved this problem at rst at the beginning of the current section We could also have started from 230 say and found out from the equation x y 18 that z 18 7 y by subtracting y on both sides now we can replace x by 18 7 y in the other equation 72x y 3 0 to arrive at the equation 7218 7 y y 3 0 in a single variable which can then be simpli ed to 3y 7 33 0 and solved for y as in the previous section namely to y 11 Clearly the second way of using the substitution method is somewhat more cumbersome here but the guiding principle is always the same Rewrite one of the equations so that one variable occurs alone on one side and not at all on the other side and then substitute it in the other equation to arrive at an equation in a single variable Once we have found the value of one unknown its easy to nd the value for the other unknown Eg in the substitution of the previous paragraph we obtain z 7 and then compute y 2773 11 In the substitution of the current paragraph we obtain y 11 and then compute z 18 711 7 Starting from 230 there is a very different way to solve the equa tions called the elimination method The idea is that the equation x y 7 18 0 is equivalent to 2x 2y 7 36 0 by multiplying both sides of the equation by 2 Now we can add this new equation 2x 2y 7 36 0 and the other equation 72x y 3 0 and arrive at 2 2y 7 36 7295y 3 0 0 After simplifying to 3y 7 33 0 we see that we have eliminated one unknown namely x and so have an equation we can again solve by the method of the previous section Alternatively we could have subtracted the rst equation 72xy 3 0 in 230 from the second equation m y 7 18 0 to arrive at xy718 772xy3 070 and after simplifying to 3x721 0 we have again eliminated one unknown namely y this time and so have an equation we can again solve by the method of the previous section In either way of eliminating one unknown we can solve the result ing equation in one unknown and then replace the solution for this unknown in one of the original equations to now solve for the other unknown Eg in the last paragraph once we have solved the new equation 3x 7 21 to z 7 we can solve for y by solving 7 y 718 0 and arrive at y 11 Note that the elimination method introduces a new way to rewrite equations In Proposition 26 we had discussed rewriting a single equa tion by adding or subtracting the same expression to or from both sides 26 SOLVING LINEAR EQUATIONS IN TWO OR MORE VARIABLES 55 or by multiplying or dividing both sides by the same nonzero expres sion on both sides The elimination method requires a new operation on two simultaneous equations Proposition 28 1 Any solutions to two simultaneous equa tions A B and C D where A B C and D are alge braic empressions remain solutions if we add or subtract one equation to or from the other ie the simultaneous solu tions ofA B and C D are still simultaneous solutions of the simultaneous equations A C B D and C D and also simultaneous solutions of the simultaneous equations A 7 C B 7 D and C D Similarly for later reference ifC and D are nonzero eccpressions then any solutions to two simultaneous equations A B and C D where A B C and D are algebraic eccpressions remain solutions if we multi ply or divide one equation by the other ie the simultaneous solutions ofA B and C D are still simultaneous solutions of the simultaneous equations AC BD and C D and also simultaneous solutions of the simultaneous equations g and C D 2 Conversely any simultaneous solutions of the equations A C BD andC D orA7C B7D andC D respec tively are also simultaneous solutions of the original equations A B and C D Similarly again for later reference ifC and D are nonzero eccpressions then any simultaneous solu tions of the equations AC BD and C D or g g and C D respectively are also simultaneous solutions of the original equations A B and C D The proof of the part I of Proposition 28 is simple If we replace the variables in A B C and D by any xed numbers then the truth of the equations A B and C D with the variables replaced by these xed numbers implies the truth of the equations A C B D A70 B7D AC BD and g For part 2 of Proposition 28 we have to work just a little harder Eg for the case AC BD and C D we use the fact that we can reverse the operation of adding equations by subtracting From A C B D and C D we obtain A C 7 C B D 7 D which simpli es to the desired equation A B Similarly for the case AC BD and C D we use the fact that we can reverse the operation of multiplying equations by dividing From AC BD and C D we obtain B753 which simpli es to the desired equation A B given that C and D are nonzero expressions 56 2 LINEAR EQUATIONS Let7s now return to our geometric intuition for solving two simul taneous linear equations in two unknowns again to see what else can happen We had said that geometrically solving two simultaneous linear equations in two unknowns amounts to nding the points of intersection of the two lines described by the two equations This is easy by either of the two algebraic methods substitution method or elimination method described above as long as the two lines are not parallel and therefore intersect in exactly one point Let7s vary our original problem a bit as follows 17m thinking of two numbers The sum of the two numbers is 12 The sum of twice each of the two numbers exceeds 24 by 3 What are my numbers77 We can quickly translate this problem into the following two simul taneous equations 231 z y 12 2x 2y 7 3 24 If we tried to solve 231 by the substitution method we could rewrite the rst equation as x 12 7 y and thus the second equa tion as 212 7 y 2y 7 3 24 But this last equation simpli es to 24 7 3 24 which is always false Since this last equation has no si multaneous solutions z and y neither do the original equations 231 by Proposition 28 Or we could have tried to solve 231 by the elimination method multiplying the rst equation by 2 to obtain 2x 2y 24 and then subtracting that from the second equation to arrive at 2z2y 7 2 2y 7 3 24 7 24 which simpli es to 3 0 again an equation which is always false And this implies that the original equations 231 do not have any simultaneous solutions again by Proposition 28 So what happened If we rewrite 231 as line equations in slope intercept form y712 232 y 7s 135 we quickly see that these lines are parallel and do not intersect so it is also geometrically clear that they cannot have any simultaneous solutions Finally it is possible that the two lines from two simultaneous linear equations coincide in which case there are in nitely many simultaneous solutions Here is a somewhat non sensical example 26 SOLVING LINEAR EQUATIONS IN TWO OR MORE VARIABLES 57 Pm thinking of two numbers The sum of the two numbers is 12 The sum of twice each of the two numbers equals 24 What are my numbers This problem can be expressed by the two simultaneous equations z y 12 2s 2y 24 But these equations represent the same line7 with the equation y is 12 in slope intercept form7 so any solution x is possible as long as y is then chosen so as to satisfy y is 12 233 Finally7 the above carries over to simultaneous linear equations in more than two unknowns Here is an example Richard spends 35 on two books7 a CD7 and a DVD Susan can only afford to pay 25 for a CD and a DVD And Tom pays just 20 for a book and a DVD Assuming that each book costs the same7 each CD costs the same7 and each DVD costs the same7 what is the price of each item Let7s denote the price of a book by b7 the price of a CD by c and the price of a DVD by d Then we arrive at the equations 2bcd35 234 cd25 bd20 representing in turn the total cost of Richard7s7 Susan7s7 and Tom7s purchases7 respectively It is possible to visualize a solution to this problem in three dimensional coordinate space Each linear equation then represents a plane in three dimensional space7 and a solution is given by any point at which all three planes intersect Obviously7 this is rather hard to draw or even visualize7 so school mathematics generally approaches this kind of problem only algebraically by one of our two methods7 the substitution method or the elimination method7 or a combination of the two Here are two of several possible ways to solve 234 algebraically 1 The second and third equation of 234 allow us to express both c and b in terms of d by subtracting d on both sides of each equation7 arriving at 0257d b207d A D V 2 LINEAR EQUATIONS Now we can replace 0 and b in the rst equation of 234 and obtain the following equation in only one unknown 2207d257dd35 This equation sirnpli es to 65 7 2d 35 and gives us the solution d 15 as in the previous section Now we can also compute 0257d2571510 b207d207155 We can start by subtracting the second equation of 234 from the rst and arrive at the equation 2bcd 7 cd 35725 which simpli es to the following equation in only one unknown 2b 10 This gives us the solution b 5 Now we can use the third equation of 234 to compute d207b207515 and then use the second equation of 234 to compute 0257d2571510 In either case7 we see that a book costs 85 a CD 8107 and a DVD 815 CHAPTER 3 Order and Linear Inequalities 31 Ordering the Numbers All the number systems we will consider here except for the com plex numbers which we will touch upon only brie y in subsection 532 carry a natural order which formalizes the very intuitive idea of more and less This concept is introduced very early for the whole numbers If Mary has seven apples and John has ve then Mary has more apples than John Using like denominators one can carry over this concept to fractions and using place value to positive decimals On the other hand the concept of more and less is somewhat counterintuitive for negative numbers and doesnt always conform with everyday language If Susan owes 5 and Bob owes 88 then Bob owes more than Susan but if one considers their savings and allows this to be negative in the case of debt then Susan has more savings than Bob A better model for negative numbers and their order is temperature 720 is less than 710 and also less than 10 even though 20 is bigger than 10 Another intuitive way to think about the order of all numbers positive negative or 0 is on the number line Being less than a number means being to the left of the point on the number line representing that number Such concepts are expressed in mathematics by inequalities which is somewhat of a misnomer We express that two numbers say 2 and 3 are unequal by writing 2 31 3 which is literally an inequality but one usually understands an inequality to be a statement involving one of the symbols lt gt g or 2 We will start by formally using only lt and we will consider a gt b to be merely an abbreviation for b lt a Here are some basic properties of the ordering lt for all numbers a b and e irre exivity a lt a is false transitivity a lt b and b lt 0 implies a lt e comparability a lt b or a b or a gt b So the rst property irre exivity states that no number is less than itself thus the name of the property re exive means referring 59 6O 3 ORDER AND LINEAR INEQUALITIES to itself The second property transitivity states that from two facts one can conclude a third lf 1 know that b is greater than a but less than c then certainly a is less than c The third property comparability states that whenever two numbers a and b are not equal then one must be less than the other What is a bit confusing is that there is an alternative way to state properties of the ordering of the numbers using the symbol 3 Again a 2 b is simply an abbreviation for b S a so formally we dont use 2 at all Now there are two ways to deal with the symbol 3 On the one hand we could simply consider a S b as an abbreviation by letting 31 a S b abbreviate a lt b or a b and so the symbol 3 gives no additional meaning which we couldn7t express before On the other hand we could replace the symbol lt by the symbol 3 and only deal with the latter Then the basic properties of the ordering under 3 become the following for all numbers a b and c re exivity a S a transitivity a S b and b S 0 implies a S c antisymmetry a S b and b S a implies a b comparability a S b or a 2 b So now the rst property re exivity states that every number is less than or equal to itself thus again the name of the property The second property transitivity states that from two facts one can con clude a third lf 1 know that b is greater than or equal to a but less than or equal to c then certainly a is less than or equal to c The third property antisymmetry states that if a number is both less than or equal to and greater than or equal to another number then the two numbers are indeed equal The last property comparability states that for any two numbers one is less than or equal the other or vice versa Now assuming that we have only introduced the symbol 3 so far but not lt we can think of lt as an abbreviation by letting 32 a lt b abbreviate a S b and a 31 b So in summary it doesnt really matter which of the two symbols lt or 3 we introduce rst we can simply de ne the other from it And in fact the properties of S can be proved from the properties of lt and vice versa 32 ORDER AND THE ARITHMETICAL OPERATIONS 61 Exercise 31 1 Show that assuming the three properties of lt and the de nition of S by 31 we can prove the four prop erties of g 2 Conversely show that assuming the four properties of S and the de nition of lt by 32 we can prove the three properties of lt Two more remarks are in order In light of the comparability prop erties we dont need any symbols like or since we can rewrite egabasa2bieasbgaanda basagtbieasblta Finally there are many different ways to express order in English using words like exceed not etc Exercise 32 For each of the statements a lt b a S b a gt b and a 2 b nd at least four dz erent ways to express these in the English language for quantities a and b 32 Order and the Arithmetical Operations So far we have considered the ordering of the numbers and the arithmetical operations separately But they are closely connected in deed as we will now explore in detail These connections will be es sential in solving inequalities later on First of all recall the following trivial properties linking equality of numbers and the arithmetical operations These were crucial in establishing Propositions 26 and 28 33 abimpliesacbc 34 abimpliesaicbic 35 c 74 0 and a b implies ac be A AAA V b 36 c0andabimpliesgi c 0 In fact not only do these properties hold but its also not hard to see that their converses hold ie the right hand side of each statement im plies the left hand side always assuming that c 74 0 in 35 and 36 This is since we can undo the addition in 33 using 34 the sub traction in 34 using 33 the multiplication in 35 using 36 and the division in 36 using 35 62 3 ORDER AND LINEAR INEQUALITIES There are now the following similar properties linking the order lt of the numbers and the arithmetical operations 37 altbirnpliesacltbc 38 altbirnpliesaicltbic 39 c gt 0 and a lt b irnplies ac lt bc b 310 c gt 0 and a lt b irnplies lt 2 Why do properties 377310 hold This is easy and intuitive to see for properties 37738 Adding or subtracting c to both sides of an inequality does not change its truth Similarly for properties 397 310 rnultiplying or dividing both sides of an inequality by a positive number 0 does not change its truth There are also more visual proofs The implications 37 and 38 hold since we can think of them as amounting to shifting77 a and b on the number line to the right or left by an equal distance given by c The implications 39 and 310 hold since we can think of them as amounting to scaling a and b on the number line by the positive factor or divisor 0 Again we have the converses of these properties the right to left direction of these irnplications undoing77 the operations as for 337 36 still under the assumption that c gt 0 in 39 and 310 We can also state similar properties and their converses using 3 except that we still assume 0 gt 0 in 313 and 314 311 agbirnpliesac bc 312 agbirnpliesaicgbic 313 c gt 0 and a S b irnplies ac S be b 314 c gt 0 and a S b irnplies S E And again we have the converses of these properties the right to left direction of these irnplications undoing77 the operations as for 337 36 still under the assumption that c gt 0 in 313 and 314 But this leads to an obvious question and one which is at the root of many errors when solving inequalities Why do we need the restriction that c gt 0 in properties 397310 and 31373147 We start with the following easy but crucial observations 315 a lt b irnplies 7 a gt 7b 316 a S b irnplies 7 a 2 7b 32 ORDER AND THE ARITHMETICAL OPERATIONS 63 Note that 315 and 316 hold not only for positive numbers a and b7 but for 0 and negative numbers as well In order to see why 315 holds7 we distinguish three cases Case 0 S a lt b Then b lies to the right of a on the number line7 and both are 2 0 Since taking the negative of a number means to re ect the points on the number line about 07 7b will now lie to the left of a on the number line7 and both are S 0 See Figure 31 I I I I I b a O a b FIGURE 31 The proof of 315 when 0 S a lt b Case a lt b S 0 Then b lies to the right of a on the number line7 and both are S 0 Since taking the negative of a number means to re ect the points on the number line about 07 7b will now lie to the left of 7a on the number line7 and both are 2 0 See Figure 32 FIGURE 32 The proof of 315 when a lt b S 0 Case a lt 0 lt b Then b lies to the right of 07 and a to the left of 07 on the number line Since taking the negative of a number means to re ect the points on the number line about 07 7b will now lie to the left of 07 and 7a to the right of 07 on the number line See Figure 33 FIGURE 33 The proof of 315 when a lt 0 lt b In each case7 we have established 7b lt 7a as required for 315 The proof for 316 is similar Using 3157 we can now prove versions of 39 and 310 as well as 313 and 314 when 0 lt 0 64 3 ORDER AND LINEAR INEQUALITIES 317 c lt 0 and a lt b implies ac gt bc b 318 c lt 0 and a lt b implies E gt 7 c c 319 c lt 0 and a S b implies ac 2 bc b 320 c lt 0 and a S b implies 2 2 Each of 3177320 follows from the corresponding version when c gt 0 by applying 315 or 316 Eg 317 is shown as follows assuming c lt 0 a lt b implies a7c lt b7c 321 implies a717c gt b717c implies ac gt bc The rst implication follows by 39 using that 7c gt 0 and the second by 315 using that 7a7c a717c while the third just evaluates 717c c A more visual proof of 317 closely following 321 goes as fol lows Multiplying an inequality a lt b by a negative number c amounts to rst scaling a and b on the number line by the positive number 7c as in the rst line of 321 followed by a re ection of the number line about the number 0 as in the second line of 321 Again we have the converses of 3177320 ie the right to left77 implications also hold always assuming that c lt 0 since we can undo the multiplication or division by c by dividing or multiplying by This uses that c lt 0 implies lt 0 which follows by dividing both sides of c lt 0 by twice 33 Linear Inequalities in One Unknown Equipped with the properties from the last section it is now easy to solve linear inequalities in one unknown ie inequalities where both sides are linear expressions Lets start with an example Little Red Riding Hood falls asleep in the forest 5km away from her grandmother7s house The next morn ing she walks toward her grandmother7s house at a constant speed of 3kmh for a while If she is now at most 2km away from her grandmother7s house how long will she have walked 33 LINEAR INEQUALITIES IN ONE UNKNOWN 65 Lets denote the time in hours from when she starts walking by t and the distance from her grandmother7s house in kilometers by 5 Then we have the following equation for time and distance 322 s 73t 5 The problem tells us that this quantity is at most 2 at the end of the walk leading to the inequality 731 5 S 2 Now we can rst subtract 5 from both sides using 312 to obtain 731 S 73 and then divide by 73 using 320 to arrive at the solution 323 t 2 1 Thus she will have walked at least one hour Note that solving linear inequalities is thus very similar to solving linear equations the one difference being that whenever we multiply or divide by a negative number we have to invert the inequality sign This is re ected in the following proposition which closely follows Proposi tion 26 Proposition 33 1 Any solution to an inequality A lt B where A and B are algebraic eccpressions remains a solution if we add or subtract the same algebraic eppression on both sides of the inequality and vice versa any solution to the inequality af ter adding or subtracting the same algebraic eppression is also a solution of the original inequality A lt B 2 Any solution to an inequality A lt B where A and B are alge braic eccpressions remains a solution if we multiply or divide both sides of the inequality by the same positive algebraic ecc pression and vice versa any solution to the inequality after multiplying or dividing by the same positive algebraic empres sion is also a solution of the original inequality A lt B Here an algebraic eppression is positive if it is positive no matter what numbers replace the letters in the eppression 3 Any solution to an inequality A lt B where A and B are al gebraic eccpressions is a solution of the inequality AC gt BC or g gt g obtained by multiplying or dividing respectively both sides of the inequality by the same negative algebraic ecc pression and switching the inequality sign and vice versa any solution to the inequality AC gt BC or gt after multi plying or dividing respectively by the same negative algebraic 66 3 ORDER AND LINEAR INEQUALITIES eppression and switching the inequality sign is also a solution of the original inequality A lt B Here an algebraic empres sion is negative if it is negative no matter what numbers re place the letters in the eppression These results also hold with lt and gt switched with lt and gt re placed by S and 2 respectively and with lt and gt replaced by 2 and g respectively The proof of Proposition 33 is basically the same as for Proposi tion 26 Adding or subtracting an algebraic expression on both sides of an inequality or multiplying and dividing both sides of an inequal ity by a nonzero algebraic expression does not affect the solutions by the results of Section 32 and the fact that all these operations on in equalities can be undone except for the caveat that rnultiplication or division by a negative number ips the inequality syrnbol Here is now a slightly more complicated exarnple Little Red Riding Hood falls asleep in the forest 5km away from her grandrnother7s house The next morn ing she starts walking toward her grandrnother7s house at a constant speed of 3kmh When she starts walk ing the big bad wolf starts along the same trail at a constant speed of 6kmh from a distance of 8km from the grandrnother7s house How long will Little Red Riding Hood be safe from the wolf Let7s denote the time in hours from when she starts walking by t and the distance from her grandrnother7s house in kilometers by SE and SW for Little Red Riding Hood and the big bad wolf respectively Then we have the following equation relating time and distance for Little Red Riding Hood 513 73t 5 And for the big bad wolf the equation relating time and distance reads 5W 76t 8 The problem tells us that we want to nd the times t when 5R lt 5W since Little Red Riding Hood is only safe from the wolf as long as she is closer to her grandrnother7s house than the wolf This leads to the inequality 324 73t5 lt 76t8 33 LINEAR INEQUALITIES IN ONE UNKNOWN Now we can rst subtract 5 from both sides using 38 to obtain 731 lt 761 3 then add 6t to both sides using 37 to obtain 3t lt 3 and nally divide by 3 using 310 to arrive at the solution 325 t lt 1 Thus she will be safe at any time less than one hour There are two other nice ways in addition to 323 and 325 to represent the solutions to inequalities like 322 and 324 0 We can represent the solutions as intervals using the notation frorn subsection 151 The solutions to 322 are all t 2 17 and so the solution set can be written as 17 oo Sirnilarly7 the solutions to 324 are all t lt 17 and so the solution set can be written as 7001 We can represent the solutions on the number line7 by let ting the points corresponding to solutions be denoted by the thicker part of the number line7 and by using solid dots for interval endpoints which are a solution7 and circles for interval endpoints which are not a solution The solutions to 322 and 324 on the number line are now pictured in Figures 34 and 357 respectively FIGURE 34 The solution set of 322 on the number line Here is one nal exarnple7 illustrating how more than one inequality for a quantity works Little Red Riding Hood falls asleep in the forest7 5km away from her grandrnother7s house The next rnorn ing7 she walks toward her grandrnother7s house at a constant speed of 3krnh for some time How long 68 3 ORDER AND LINEAR INEQUALITIES l w N o N 1 FIGURE 35 The solution set of 324 on the number line will she have walked when she is between 3km and 2km from her grandmother7s house The equation relating time and distance for Little Red Riding Hood still reads 5 731 5 where s is the distance from grandmother7s house in kilometers and t the time from when she starts walking in hours The information from the problem now tells us that the distance is between 2 and 3 leading to the inequalities 326 2 S 731 5 and 3 2 731 5 which is usually written more compactly as 327 2 S 731 5 S 3 Note that this compact notation only works since the inequalities in 326 are linked by the word and In general you can abbreviate the statement A S B and B S C by A S B S C and similarly forAltBltCAZBZCandAgtBgtC Moreaboutthisina minute We can perform the same operations on the inequality 327 to solve it First subtract 5 on all sides to obtain 73 S 731 3 i2 and then divide by 73 on all sides to arrive at the solution 2 328 1 2 t 2 g So Little Red Riding Hood walks between 40 minutes and one hour Again we can represent the solution set by an interval namely 1 or on the number line as in Figure 36 34 ESTIMATION AND APPROXIMATION 69 FIGURE 36 The solution set of 327 on the number line Note that in this last example7 the set of t which are not a solution can be described by the statement 2 gt 73t5 or3 lt 73t5 which is frequently but incorrectly abbreviated as 3 lt 731 5 lt 2 This last statement makes no sense at all since it would imply that 3 lt 2 The real77 problem is that one cannot write the set of non77 solutions of 327 as a single interval rather7 it consists of two intervals7 namely7 700 and 100 Using our set notation from subsection 1517 we can also write the set of non77 solutions of 327 as foo7 U Loo We will return to more complicated inequalities later on and then discuss possible solution sets of inequalities in more detail For now7 we turn to one common application of inequalities estimation and approximation 34 Estimation and Approximation In a purely mathematical setting7 the number 2 means exactly that Its the number 27 no more and no less And writing it differently7 eg7 as 20 or 2007 does not change that fact But in the real world 7 we cant measure quantities emactly we can only measure them approcm39mately And there often is a convention that 2 means 2 rounded to the nearest integer ie7 2 denotes a quan tity x which satis es the inequality 15 S m lt 257 whereas 20 means 2 rounded to the nearest tenth ie7 20 denotes a quantity x which sat is es the inequality 195 S x lt 2057 and 200 means 2 rounded to the nearest hundredth ie7 200 denotes a quantity x which satis es the inequality 1995 S x lt 2005 A better and more precise way to write these would be to write 2 i 5 2 i 05 and 2 i 005 respectively One advantage is that it allows us to write things like 2 i 1 ie7 2 rounded to the closest 7O 3 ORDER AND LINEAR INEQUALITIES fth77 or more precisely a quantity x which satis es the inequality 19 S x lt 21 Another advantage is that when reading 120 we cant really be sure whether this means 120 rounded to the closest integer77 ie a quantity x which satis es the inequality 1195 3 z lt 1205 or 120 rounded to the closest ten77 ie a quantity x which satis es the inequality 115 S x lt 125 whereas reading it as 120 i 5 or 120 i 5 respectively clari es which of these two we mean There are now some quirks about estimation and approximation which are frequently swept under the rug in school mathematics but which are important to keep in mind for you as a teacher and for your students to at least be aware of especially in the age of ubiquitous pocket calculators which can spew out a number with eight signi cant digits regardless of how imprecise the input was We7ll restrict ourselves here to just one example with two parts Given a rectangle of length 84cm and width 52cm what are 1 its perimeter and 2 its area Now suppose that length and width are only given to the nearest tenth what are the possible values for 1 its perimeter and 2 its width Your rst inclination may be to reach for a pocket calculator and to compute the perimeter as p284252272 and the area as A 8452 4368 so purely mathematically the perimeter is exactly 272cm and the area exactly 4368cm2 But in real life the length l measuring 84cm and the width w measuring 52cm means that 835 S l lt 845 and 515 S w lt 525 For the perimeter this implies that 2728352515 2l2wlt28452525274 and so p 272 i 2 in our earlier notation Similarly for the area 430025 835515 S l w lt 845 525 443625 and so the area can vary almost all the way between 43cm2 and 444cm21 Using the i notation here is very messy And in general its quite hard to give the exact bounds We therefore see that especially computing the area as exactly 4368cm2 is highly misleading The general rule of thumb is that one 34 ESTIMATION AND APPROXIMATION 71 should never give more signi cant digits for the output than for the in puts Under this rule ofthurnb7 we would have computed the perimeter as 27cm and the area as 44crn2 rounding each to two signi cant dig its7 which is not quite accurate by our cornputations above since the latter suggests that 265 S p lt 275 and 435 S A lt 4457 but at least it is in the right ballpark77 for the accuracy of the answer CHAPTER 4 The Concept of a Function and Functions Closely Related to Linear Functions 41 The Concept of a Function In Chapter 2 on linear equations we already saw examples of how one quantity let7s denote it by y varies depending on another quantity which we will call x for now In particular the equation for a non vertical line can always be written in the form 41 ymzb for some xed parameters in and b In such a situation we call x the independent variable and y the dependent Udiidble since we consider how y varies as varies First note that in the particularly simple case of 41 and as long as m 74 0 the choice of which variable to call dependent and which independent is somewhat arbitrary from a mathematical point of view since 41 can be rewritten as y b m in However the real life77 context might make one choice more natural77 than the other Say z might be the time elapsed since a train left the station at a constant speed and y might be the distance of the train from the station It would then be more natural to consider the distance as time varies rather than the time as distance varies since time cant be made exible77 whereas the distance of the train can vary in a different pattern if the speed equals a different constant or is no longer constant at all Putting this minor point aside let7s concentrate on the concept of how a quantity y varies depending on how another quantity z varies Mathematics usually expresses such a situation by the concept of a function 42 1 Wt First of all note that this notation may already cause some confu sion f does not mean that we multiply quantity f by quantity x 72 41 THE CONCEPT OF A FUNCTION 73 but rather that as a computer scientist might put it we apply some rule f to the input z in order to determine the output y The set of possible values for the quantity z is called the domain of the function it is determined either mathematically by when the equation y fx makes sense or is de ned or by the real life context as to when it makes sense for the real life quantity x The critical feature of a function is however the fact that for each argu ment z in the domain of the function there is emaetly one value for y such that y fx holds as already suggested by the way we write this But note that we could have written the relation between the quantities z and y differently and it wouldnt have been so easy to check whether the relation de nes a function Eg the relation 43 x i y 0 can be expressed as s y2 and so we can think of it as describing a function for the quantity x as y varies The fact that I switched the roles of z and y here is deliberate The choice of what to call the variables is also completely arbitrary and one should not always choose the same letter for the independent and the dependent variable respectively On the other hand 43 does in general not describe y as a function of x Eg setting z 1 both y 1 and y 71 satisfy 43 However we can resolve this problem by restricting the allowable values of y If we specify in addition to 43 that y 2 0 then 43 can be rewritten as y xE and y is a function of x All this should convince you that when a relationship between quantities is speci ed determining whether or not this relationship can be described as one quantity being a function of another may depend on additional constraints being imposed and sometimes this is needed in order to obtain a function And of course the real life context may impose such or other constraints So lets look at the concept of a function more abstractly1 well also introduce some more handy notation going along with functions 1In these lecture notes welll restrict ourselves to functions from numbers to numbers More generally one can consider functions from any set of objects to any other set of objects Eigi one could de ne a function with domain the set of all Americans which on the input person returns the value eligible to vote or not eligible to vote As long as we assume that for each American exactly one of eligible to vote or not eligible to vote is true this would also be a function albeit admittedly much more abstract 74 4 FUNCTIONS First of all let7s x two sets of numbers X and Y we7ll call them the domain and the codomain2 of our function f We then use 44 fiX Y to denote that f is a function with domain X and codomain Y Next for each argument z ie each number x in the domain X we have to specify emaetly one number y called the value of f at m in the codomain Y well then denote this number y as To expand on our previous notation we could write 4 5 f X a Y 39 l x H M So in this notation the function f 2x 1 could be written as f R a R z gt gt 2x 1 but we could also restrict the domain to the set of positive real numbers say and write f000gtR zgt gt2x1 or to the set of whole numbers and write fZaZ zgt gt2x1 Note in this notation that the two arrows are slightly different In the second line we use the special arrow gt gt to indicate that z is mapped to77 y so a function is sometimes also called a map or a mapping Often in school mathematics functions are given by simple formu las such as f 2z1 or f x2 But they do not have to be so simple Here are three perfectly reasonable77 functions for which there 2While the notion of a domain is completely standard the notion of a codomain is not and often either no name is used for it at all or it is sometimes called the range which however con icts with our later notationi 41 THE CONCEPT OF A FUNCTION 75 are no easy formulas 4396 f1 iszZ 0 otherwise x if z E Q 47 ya 0 otherwise 48 h the largest integer n S x Note that in each of 46748 we clearly have a function in the sense in which we de ned it at the beginning of the chapter with domain the set of all real numbers For each argument x there is exactly one value y satisfying each of the three statements respectively Another way to represent a function apart from a formula or a description of a function as above is in the form of a table Table 41 shows some values for the functions f g and h The problem with a table is obviously that it can only show the values of a function at a very small number of values and in the case of the function g the table would not distinguish between 9 and the function m s for example if we hadn7t added the irrational number xi to our table in spite of the fact that it looks a bit out of place z 1 05 0 051 fz 1 01 01 0 g 1 05 0 05 1 0 hz 1 05 0 0 1 1 TABLE 41 A table for the functions described in 46748 A third and much more useful way to represent a function is in the form of a graph First of all the graph of a function is the set of all points x in the two dimensional plane where z varies over all arguments z in the domain of the function Figures 41 and 42 show the graphs of f and h The graph of 9 cannot be represented so easily Since every irrational number x is arbitrarily close to a rational number x eg the decimal representation of z truncated at any xed number of digits we have f 0 and fz x for arguments z and x which can be arbitrarily close together and that is not really possible to draw At the beginning of the section we de ned a function as map ping each argument z in the domain X to exactly one value y in the 76 4 FUNCTIONS A D A 1 To 1 FIGURE 41 The graph of the function f from 46 Mb NU 2 y 1 o A X 2 1 n Y 2 0 0 2 FIGURE 42 The graph of the function greatest integer function h from 48 codornain Y There is a Visual way to check this property whether any given picture curve in the plane is the graph of a function To check whether for any argurnent x in the domain X7 there is only one value y we perform the Vertical Line Test77 by drawing a vertical 42 RANGE ONTO AND 171 FUNCTIONS AND INVERSE 77 line at z and checking how often the curve intersects the vertical line If the line and the curve intersect only once for each argument x then the curve passes the Vertical Line Test and is the graph of a function This is because this unique point of intersection Ly7 say7 tells us that the value of the function f at z is y See Figure 43 for an example of a curve described by the equation x yz which does not pass the Vertical Line Test The line given by z 1 intersects the curve at the points 11 and 171 Algebraically7 its also easy to see that the curve has two parts7 one described by y V the upper half of the curve 7 the other described by y ixE the lower half of the curve and both halves individually pass the Vertical Line Test7 as they should7 since they are the graphs of the two functions given by y and y 7 x with domain the set of nonnegative real numbers x1 11 FIGURE 43 The Vertical Line Test for the curve z y2 with the line z 1 42 Range of a Function Onto and Oneto One Functions and Inverse of a Function We now discuss some properties of functions7 leading up to the concept of undoing or inverting a function ie7 given the value y nding the argument Of course7 this may not always be possible7 as we will see 78 4 FUNCTIONS 421 Range of a Function and Onto Functions First of all let the range of a function f X a Y be the set of all y in the codomain Y of f which are of the form f for some argument z in the domain X of f we will denote the range of f by ranf Similarly we will denote the domain of f by domf Sometimes such as in the example of the function f R a Rx gt gt 2x 1 the range of f is the same as the codomain This is because any y in the codomain R can be written as 2x 1 by simply setting z On the other hand the range may consist of only a single number such as in the case of the function f R a Rx gt gt 1 Here only the number 1 is ofthe form fx and in this case we can choose z arbitrarily and will always obtain the value 1 In general the range is a subset ofthe codomain ie by de nition ofthe codomain of a function f X a Y any number of the form f for some z in the domain X must be in the codomain Y There are many other possibilities for the range of a function Eg in the examples 46748 we have that the range of f is 0 1 the range of g is Q and the range of h is Z This is easy to see from Figures 41 and 42 for the functions f and h For the function 9 we distinguish two cases for gx If x is a rational number then so is g whichjust equals z and any rational number y satis es y gy so all rational numbers must be in the domain of 9 On the other hand if z is irrational then g equals 0 which is rational so the range of g is exactly the set of all rational numbers In general determining the range of a function is a very complicated problem We call a function f X a Y onto or onto Y to be more precise if the codomain and the range of the function f are the same Eg as remarked above the function f R a Rx gt gt 2x 1 is onto while the function f R a Rm gt gt 1 is not None of the functions in the examples 46748 is onto as we saw above 422 171 Functions Next we call a function f X a Y me to zme or 171 for short if for each value y in the codomain Y there is at most one x in the domain X with y We can rephrase this in several equivalent ways 1 The function f is 171 if for each value y in the range of f there is exactly one x in the domain X with y 2 The function f is 171 if for all y in the codomain Y whenever we have y f fx for arguments z and x in the domain X then actually z x 3 A visual way of checking whether a function is 171 is the Hor izontal Line Test Draw a horizontal line at the value y and 42 RANGE ONTO AND 171 FUNCTIONS AND INVERSE 79 check how often the graph of the function intersects the ver tical line If the line and the graph intersect at most once for each value y then the graph passes the Horizontal Line Test and is the graph of a 171 function This is because that point of intersection zy say tells us that the value of the function f equals y at at most one value x See Figure 44 for the example of the function f R a R z gt gt 2 which fails the Horizontal Line Test eg at the line y 1 and thus is not 171 If we were to restrict the function f to the domain 000 ie the right half of the graph or to the domain foo 0 ie the left half of the graph then the function would pass the Horizontal Line Test and thus would be 171 yxquot2 11 y1 11 1 FIGURE 44 The Horizontal Line Test for the curve y 2 with the line y 1 423 Inverse of a Function Next we will investigate when we can undo or invert a function f X a Y in the sense that we can nd a function g Y a X such that the statements y fz and z 9y are equivalent If such a function 9 exists we will call it the inverse function of f and denote it by f 13 3Note that the exponent 71 here does not denote that we take the 70th power ie f 1z does not mean l Later on in trigonometry when we 80 4 FUNCTIONS What are the conditions now under which the inverse function of a function f X a Y exists Note that we must ensure two things 1 For each y in the codomain Y of f there is an argument z in the domain X of f such that y fz and 2 for each y in the codomain Y of f there is not more than one argument z in the domain X of f such that y Now observe that condition 1 is exactly the condition that f is onto and condition 2 is the condition that f is 1711 These two conditions can be combined into one condition 3 For each y in the codomain Y of f there is emactly one argu ment z in the domain X of f such that y It should be clear that condition 3 is exactly the condition on the ex istence of an inverse function which we are looking for Condition 1 is necessary since otherwise for some y we will not be able to de ne f 1y Condition 2 is necessary as well since otherwise for some y we will not know which value x to choose for f 1y On the other hand condition 3 is also suf cient to ensure the existence of an in verse function f 1 Y a X Given y in Y condition 3 ensures the existence of exactly one x in X such that fx y and so it is both natural and inevitable to de ne f 1y x There is also a very nice visual way to nd the graph of the inverse function of a function f Recall that the graph of a function f X a Y is the set of all pairs Ly such that y fx whereas the graph of the inverse function f 1 Y a X is the set of all pairs yz such that z f 1y ie such that y So the graph of f 1 is obtained from the graph of f by simply interchanging the coordinates of each point my Graphically this corresponds to re ecting the graph of f along the main diagonal77 described by y x but of course this works only after we rename the variables from x f 1y to y f 1z Figures 45 and 46 show two examples We conclude with another example of our concepts for functions from this section Let X 012 and Y 123 be two nite sets of numbers Let f and g be two functions from X to Y de ned commonly write expressions like sin2 I in place of sin z2 this notation can lead to serious ambiguities since we wonlt know how to interpret sin 1 1 unless the context clari es this and 1 always recommend to use the special notation arcsinz for the inverse of the sine function in place of sin 1 I but welll leave these intricacies aside or now 81 42 RANGE ONTO AND 171 FUNCTIONS AND INVERSE 2 y3x3 39main diagonal39 x Y 3 FIGURE 45 The graphs of the function y 3x 7 3 and its inverse function y ix 1 39main diagonal39 x y FIGURE 46 The graphs of the function y 3 and its inverse function y W 82 4 FUNCTIONS by f0 1 90 1 f0 3 91 3 f2 3 92 2 We can represent f and g visually as shown in Figure 47 f 9 FIGURE 47 The nite functions f and g from X 012 to Y 123 It is now easy to see that f is neither onto nor 171 f is not onto since 2 is in the codornain but not in the range of f and it is not 171 since f1 f2 It is therefore impossible to nd an inverse function for f Given the value 2 in the codornain Y of f7 we cant nd any x in the domain X of f with f 2 furtherrnore7 for the value 3 in the codornain Y of f7 there are two different z in the domain X of f with fx 3 So it is not possible to nd exactly one value for f 1y for either y 2 or y 3 On the other hand7 for each y in Y7 there is exactly one x in X such that g y so we can de ne g l Y a X with and 9 thus has an inverse function 43 Some Functions Closely Related to Linear Functions So far7 we have mainly dealt with one kind of function7 linear func tions which can be expressed as fx ms b for xed pararneters m 43 SOME FUNCTIONS CLOSELY RELATED TO LINEAR FUNCTIONS 83 and b In this section7 we will encounter some functions closely related to linear functions 431 Step Functions These are functions whose graph is hor izontal except at certain points where it jumps77 from one value to another We already saw an example of a step function in Figure 42 it is usually denoted by y m de ned by 48 the largest integer n S x and called the greatest integer function sometimes also the oor func tion There is a similar function called the least integer function or sometimes the ceiling function it is de ned by 49 the smallest integer n 2 z Its graph is shown in Figure 48 23y 0 10 2 o 1 392 0 1 2 FIGURE 48 The graph of the least integer function from 49 These two functions are particularly useful to computer scientists7 who use them quite frequently But they come up in real life77 situa tions quite early in school mathematics in fact7 we have already seen an example which really requires such a function if done carefully 84 4 FUNCTIONS A pay phone charges 506 for the rst minute and 10C for each additional minute of a long distance call77 Before we had restricted the domain of the function describing the cost of the phone call to the set of positive integers so we de ned this function as fil23gtR n 1 gt 1071 40 where n is the length of the phone call in minutes and fn is its cost in cents and its graph was given in Figure 23 But this is really cheating since phone calls can be of length any positive real at least in principle and then the phone company rounds up the length of the phone call to the next integer So the cost function is really given by g 0 00 a R z 1 gt 10 40 The graph of the real77 cost function g is given in Figure 49 410 411 15a 9 0 0 0 0 100 0 0 0 0 0 5MP o o 1 2 a 4 s a 7 a 9 10 FIGURE 49 The cost of a phone call from 49 What are the pros and cons of describing the cost function accu rately as in 411 rather than just as f 10x 40 as in 26 Certainly the former is more accurate it is the actual cost function as computed by the phone company On the other hand f 10x 40 is a close approximation which is much easier to work with Unlike g f is 171 and onto and thus has an inverse function so we can easily compute the length of a 90 cent phone call say using f or rather f l If we tried this for an 85 cent phone call say we7d still get an answer 43 SOME FUNCTIONS CLOSELY RELATED TO LINEAR FUNCTIONS 85 using f l namely 45 minutes and even though the answer would be wrong there is no phone call costing exactly 85 centsl it would tell us that the phone call would be around 4 or 5 minutes So the upshot is that a linear function is sometimes just a useful approximation to the real function such as a step function which is much harder to deal with School mathematics is often happy to just deal with the approximation but you as teachers should be aware of this distinction and the reasons why one uses approximations since it may confuse some of your students and you should then be able to give at least a roughly correct answer 432 Piecewise Linear Functions Another very common kind of function one encounters in real life problems is a piecewise lin ear function The graph of such a function is not a line but consists of several line segments and is bent at the end points of these line segments without any jumps however Here is an example John swims for half an hour at a constant speed of 2kmh then jogs for an hour at a constant speed of 8kmh and nally bikes for half an hour at a constant speed of 20kmh Describe the distance traveled by John in terms of the time elapsed Let7s let t in hours be the time elapsed since John starts and s in km his distance traveled since then Then the graph of s as a function of t is as shown in Figure 410 Let7s rst describe this graph verbally John rst swims from time zero hours to time one half hour at a speed of 2kmh covering a dis tance of one kilometer Then he jogs for one hour from time one half hour to time one and a half hours at a speed of 8kmh covering an additional eight kilometers for a total distance of nine kilometers for now Finally he bikes for half an hour from time one and a half hours to time two hours at a speed of 20kmh covering an additional ten kilometers for nal total distance of nineteen kilometers Let7s translate this into an algebraic description of our function 2t ifogtgg 412 5t 18t7 ifggt 1 920t71 if1 t 2 So in the intervals 0 i E 1 and 1 2 s behaves like the linear functions 5 2t 5 1 8t 7 and s 9 20t71 respectively as can be seen eg from the point slope formula of a linear function as A Marlow mu 4 10 The distance s meta by thh gram of Lhe hme telzpsad mdascanbeseenalw nmhhegzphshnwnmngureAlO NokehhaL aLLhepmm wecmcnmpukesbnthasi gmdasl8r sh h but does hen Jum quot 3L g and we axe all lt1 mLhe xst mundane 2 1 as 5 200 e 1 speeds at Lhevazmus hms cmespmds m hhe slaps mime m us seynems h hhe yayh 4 a a The Absolute Value Elmth The ehshlme veluehm mm s avery mpemhc special case of e piecewise m me e h hhe numbez hhe Very mgemlh dun hhe As you pmbzbly shewh knew hhe ehshlme value of e s dammed by M The yayh of Lhe zbmluke value fumhnn 5 shown h n A 11 The mml d mmnn of Lhe abxoiu m of e s yven by A 13 43 SOME FUNCTIONS CLOSELY RELATED TO LINEAR FUNCTIONS 87 X lel 2 TI FIGURE 411 The graph of the absolute value function y 1961 This simple and elegant de nition is the source of endless confu sion If x is negative then is is positive so the second case of 413 always produces a positive value Therefore the absolute value of a number is always non negative one of the most important facts about the absolute value 414 2 0 for all z Before we discuss an example of how the absolute value function is used let7s state and discuss three important rules about this function which hold for all real numbers a and all non negative real numbers 7 415 lal r exactly when a r or a 7r 416 lal lt r exactly when 7 r lt a lt r 417 lal gt r exactly when a lt 7r or a gt 7 Why do properties 4157417 hold true This is easy to see intuitively77 from the intuitive de nition of as the distance77 of z from 0 The two numbers exactly at distance r from 0 are r and 7r If r 0 there is only one number here of course The expression lal lt 7 states that a is at a distance less than r from 0 which clearly holds exactly for the numbers in the interval 7717quot This interval is empty if r 0 as there are no numbers at a distance less than 0 from 0 Finally the expression lal gt 7 covers all the cases not included in the rst two cases ie all numbers in the intervals foo 7r and 7 oo 88 4 FUNCTIONS A second more visual way to see why properties 4157417 hold true is to look at the graph in Figure 411 Eg the expression lal lt 7 means that we need to look at all values lt r on the y axis and the arguments z for which lt 7 correspond to the points on the line segments from 7737 to the origin and from the origin to 737 not including the points 7737 and 737 themselves A third more algebraic way to see why properties 4157417 hold true is to analyze the statements on the left hand sides in light of the formal de nition of in 413 Eg lal 7 means that either a 2 0 and so lal a forcing a r or that a lt 0 and so lal 7a forcing 7a 7 this gives exactly the two cases on the right hand side of 415 Similar arguments can be made for 4167417 The main use of the absolute value lies in the fact that it does not only allow us to measure the distance of a number from 0 on the number line but indeed the distance on the number line between any two numbers Namely the distance between numbers a and b is given by la 7 bl The properties 4157417 now immediately translate to properties of the distance between numbers a and b for any non negative real number 7quot 418 laibl7 exactlywhenabrorab7r 419 laibl ltrexactly whenb7rltaltbr 420 laiblgtrexactly whenaltb7roragtbr since the right hand sides of 4187420 are equivalent to the right hand sides of 4157417 when we replace a by a 7 b A very common application of the absolute value is in estimating the accuracy of a measurement We already saw that in real life77 applications a length Z of 10 meters say really means a length of 10mi5m ie that 421 95 3 Z 3 105 We actually wrote 95 3 Z lt 105 before but lets ignore that ne point for the moment in applications it makes little difference An other way to express 421 would be lZ 7 10 S 5 as can be seen from 4187419 Note that the negation of this state ment ie that the true length Z differs from 10m by more than half a 43 SOME FUNCTIONS CLOSELY RELATED TO LINEAR FUNCTIONS 89 meter is expressed by ll 7 10 gt 5 or equivalently by l lt 95 or Z gt 105 434 Direct and Inverse Variation We already saw in Chap ter 2 that often the ratio ie the quotient between two quantities z and y say is a constant which we7ll call k 9 422 k 7 lt gt 5 This can be rewritten as 423 y kx In this case we say that y is proportional to z or that y varies directly with x An example of this from section 22 was constant speed i as the ratio of distance 5 traveled over time t 424 i 7 lt gt At other times we have that the product between two quantities z and y say is a constant which we7ll call k 425 k yp This can again be rewritten as k 426 7 y 96 In this case we say that y is inversely proportional to z or that y varies inversely with x The speed equation 424 also gives us an example of this if instead of constant speed we assume that the distance is constant and we want to study how the speed varies as we vary the time Eg you may consider the travel over a xed distance and thus consider the speed as depending on the time it takes you to travel this distance Figures 412 and 413 show the graphs of the functions in 423 and 426 for k 1 Sometimes we want to study the relationship between more than two quantities and then the concepts of direct and inverse variation are particularly useful The so called Combined Gas Law77 from physics describing properties of gas in a sealed container provides such an example PV 427 o 90 4 FUNCTIONS 439 y 3 2 yx 1 o x 4 4 2 391 a 1 2 a i 1 2 4 4 FIGURE 412 The graph of the function y z Here7 P is the pressure in pascal7 say7 V the volume in mil7 say7 T the absolute temperature in degrees Kelvin7 and C is a constant which is xed as long as we only consider a xed amount of a particular kind of gas The Combined Gas Law can now be rewritten7 and stated as saying the following OT 1 P 7 The pressure varies directly with the temperature at constant volume7 and inversely with the volume at con stant temperature 2 V 7 The volume varies directly with the temperature at constant pressure7 and inversely with the pressure at con stant temperature 7 e empera ure varies 1rec y w1 e pressure 3 T 0 Th t t d tl thth at constant volume7 and also directly with the volume at constant pressure u sown mono CLch mm To um mom 9 mamas I39heyayhn hefumtmy CHAPTER 5 Quadratic Functions Equations and Inequalities 51 Introduction to Quadratic Functions So far7 the only functions we have considered are functions which are linear functions7 or at least fairly close to linear functions We will now turn to quadratic functions Lets start with an example Consider the sequence of shapes in Figure 51 Ell Fl I l n2 n3 n4 FIGURE 51 A sequence of shapes There are several ways in which we could think of the n lst shape being generated from the nth shape here are two de nitions leading to the same sequence 1 The nlst shape is generated from the nth shape by adding a new row of squares below the last row of the nth shape7 extending that last row one extra square to the left and one extra square to the right 2 The n lst shape is generated from the nth shape by adding two new squares to the right of each row of the nth shape and then adding one new square above the middle of the new top row Whichever way you de ne the sequence of shapes7 it is important to de ne it precisely Simply letting the student guess the pattern is highly misleading7 since there are many dz erent ways to continue the pattern Eg7 we could have a pattern which stabilizes at the 4th shape7 ie7 the nth shape for all n gt 4 is the same as the 4th shape 92 51 INTRODUCTION TO QUADRATIC FUNCTIONS 93 Or we could have a pattern which oscillates between the 1st and the 4th shape ie the 5th shape is the same as the 3rd the 6th shape is the same as the 2nd the 7th shape is the same as the 1st the 8th shape is the same as the 2nd again etc There is no reason why one pattern should be considered more reasonable than the others This is why a careful de nition of the intended pattern is essential Now lets look at the perimeter and the area of the nth shape in Figure 51 as de ned above assuming that the side length of each square is one unit length and each square has area one unit square Table 51 below shows these for the rst ve shapes number of shape 1 2 3 4 5 perimeter of shape 4 10 16 22 28 area of shape 1 4 9 16 25 TABLE 51 Perirneter and area of the shapes from Figure 51 We will now see that perimeter and area show a very different pat tern Table 51 suggests that the perimeter increases by a xed amount namely 6 each time We can also see this more precisely and for general n from our de nitions of the shapes Eg from the above de nition 1 for the shapes we see that each time we add to the pattern we move the bottom edge down one unit and then we add a square on each side of the new bottom row resulting in an increase in perirneter by six edges of these two squares We can also see this from the above de nition 2 for the shapes Each time we add to the pattern we move all the edges on the right over by two units and then we add four edges at the top including for the new top square and two edges at the bottom resulting in an increase in perirneter by six edges Therefore the perimeter let7s call it of the nth shape is a linear function of n with slope 6 and the graph containing the point 14 since the 1st shape has perirneter 4 By the point slope form for a line equationsee 211 we arrive at the formula 51 Pn 6n146n72 Here is another way of looking at nding the expression for this function Let7s de ne a new function called the rst dz erence func tion by 52 13171 Pn 1 7 Pn 94 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES So this function P1 tells us not the function P itself but by how much the function P increases from n to n 1 As we showed above this increase is constant namely we have 53 P101 6 for all whole numbers n gt 0 Note that 53 together with 52 does determine the function P eg the function P n 6n also satis es both 53 and 52 But if we also specify that 54 P1 4 then we can compute the function P at all whole numbers n gt 0 by starting with n 1 in 54 and then repeatedly using that 53 and 52 give us Pn 1 Pn 13171 Pn 6 to compute P2 P3 etc1 We note that any function P de ned on the positive whole num bers with constant rst difference function P1 must be linear Suppose that P1n c for all n gt 0 Then we can compute P2 P1 c P3 P2 c P1 2c P4 P3c P13c PoiPn71cP1n71c On the other hand the increase in area from one shape to the next is not constant The increase from the 1st to the 2nd shape is 3 from the 2nd to the 3rd is 5 from the 3rd to the 4th is 7 etc So instead of the increase being constant it appears from Table 51 that the increase increases by a constant namely 2 We can also see this more precisely from our de nitions of the shapes Eg from the above de nition 1 for the shapes we see that each time we add to the pattern we add two more squares to the shape than we did the last time since the new bottom row is two units wider than the previous bottom row And we can see this from Figure 52 The squares marked new77 in regular font are the same number as was added at the previous step the two squares marked new 7 in slanted font are added in addition to the 1Such a de nition of a function is sometimes called a de nition by recursion 51 INTRODUCTION TO QUADRATIC FUNCTIONS 95 other new squares so we do indeed add two more squares at each step than we did at the previous step new new new new new new new new D new new l l newW I l new new n1 n2 n3 n4 FIGURE 52 Adding new squares in our sequence of shapes This results in the following equation for the rst difference func tion A1 of the area A 55 A101 An 1 7 An 2n 1 So by the argument on the previous page the rst difference function A1n An I 7 An is a linear function since it increases by 2 each time namely the so called sccond di crcncc function A201 A1n 1 7 A1n is constant namely 2 Whenever the second difference function is constant the rst dif ference function must be linear as we saw before and whenever the rst difference function is linear then the function is quadratic as we will now show First of all a quadratic function is a function of the form 56 An 1712 bn c 2If you happen to know some calculus note the analogy We consider here so called discrete functions ie functions de ned on the positive integers say in an area called discrete mathematics For such functions the rst difference function is the analogue of the rst derivative of a continuous function and the second difference function is the analogue of a continuous function What we just said was that a discrete function is linear exactly when its rst difference function is constant and that a discrete function is quadratic exactly when its second difference function is constant Of course the derivatives do not exist for all continuous functions but the difference functions of discrete functions always exist 96 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES for parameters a b and c Now note that 56 satis es A1 An 1 7 An an12bn1c 7an2bnc 57 an 12 7 anz bn 1 7 bn c 7 c an22ana7an2 bnb7bn 2an a b and so A1 is a linear function of n But we have An 1 An An 1 7 An A1n and so having A1 and a formula for A1 completely determines the function A Comparing 55 and 57 we now have 2n12anab and so 2 2a 1 a b which solves to 1 b 0 Now we still have to nd 0 for 56 Setting n 1 we also have A1 1 12 c and therefore 0 0 Thus the area of the nth shape in Figure 51 is 58 An n2 Of course you probably guessed the formula for the area all along from Table 51 However I hope you found the formal way of deriving the formula above instructive since it works for any sequence of values with constant second difference we7ll work out a second example below to show how this works in a more complicated case But lets rst prove the formula 58 for the area ofthe nth shape geometrically Figure 53 shows how to move some of the old squares to new places in such a way that the nth shape becomes a square of side length n and thus of area n2 Now as promised let7s slightly vary the problem Consider the shapes in Figure 54 It is now no longer so easy to guess a formula for 51 INTRODUCTION TO QUADRATIC FUNCTIONS 97 old old old n1 n2 n3 n4 FIGURE 53 Computing the area of the nth shape in our sequence of shapes from Figure 51 n1 n2 n3 n4 FIGURE 54 A new sequence of shapes the area of the nth shape But its not hard to see that the areas of these shapes satisfy 59 An 1 7 An 71 1 and so the second difference is constant 1 Comparing 59 with 57 above7 we arrive at n12anab and so 1 2a 1ab which solves to 1 a 7 2 51 2 n Now we still need to nd 0 Setting 1 gives A11 12 c 98 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES and therefore again 0 0 Thus the area ofthe nth shape in Figure 51 is 510 An i712 in Exercise 51 Find a formula for the perimeter of the shapes in Fig ure 547 and prove both it and the area formula 510 geometrically in at least two different ways 52 Solving Quadratic Equations We call an equation a quadratic equation in an unknown a if it is of the form 511 a2bc0 for some parameters a7 b7 0 in R We will also assume from now on that a 31 0 since otherwise 511 reduces to a linear equation Note that a more general equation a2bxca2bzc can easily be transformed into the format 511 by subtracting the right hand side to the left so we can restrict ourselves in general to equations of the form 511 We will now consider increasingly sophisticated and more compli cated techniques to solve a quadratic equation7 starting with a review of the binomial laws7 which will prove very useful 521 Reviewing the Binomial Laws The binomial laws for algebraic expressions a7 b and 0 state that 512 ab2 a22abb2 513 aw ah 2abb2 514 abaib azib2 and are simple but very useful consequences of the laws of arithmetic from Chapter 1 We will prove them in two different ways7 algebraically from the laws of arithmetic7 and geometrically using models for addition and multiplication Lets start with an algebraic proof of 512 a b2 a b a b by de nition of squaring aa b ba b by distributive law a2 ab ba b2 by distributive law again a2 2ab b2 by commassocdistrib laws 52 SOLVING QUADRATIC EQUATIONS 99 For a geometric proof of 512 for ab gt 07 refer to Figure 55 The area of the big square7 which has side length ab is ab27 and is the sum of the areas of the two smaller squares and the two rectangles7 which have areas 12 bz7 ba and ab respectively The binomial law 512 now follows immediately b ba b2 a a2 ab a b FIGURE 55 A geometric proof for ab gt 0 showing that a b2 a2 2ab b2 Exercise 52 Give similar algebraic and geometric proofs of the bi nomials laws 5137514 For the geometric proofs7 assume that 0 lt b lt a to make geometric sense Explain why this assumption is desirable for the geometric proofs 522 Solving Quadratic Equations by Factoring Lets start with an example 515 22730 Since 2 2x 7 3 x 7 1z 3 by using the distributive property repeatedly7 this can be rewritten as 516 x71x3 0 Of course7 its not always so easy to see this immediately7 and in fact it is often impossible to see such a factoring of the left hand side of a quadratic equation at all There is one useful fact from advanced algebras which often comes in handy Suppose we have an equation of 3Here I am using the word algebra not in the sense of school algebra7 but in the sense in which research mathematicians use it The proof of this fact is typically covered in an advanced undergraduate course on Modern Algebra 100 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES the form 2 bx c 0 ie the parameter 1 equals 1 or put differ ently x2 is not preceded by a number and the parameters b and c are integers then all the solutions are positive or negative integers which divide the parameter c or else necessarily involve square roots So in the above example 515 we know right away without even factoring that the solutions to 515 either involve square roots or are among the numbers 1 713 73 which are the only integers dividing 3 Now we use a simple fact about the real numbers Multiplying two nonzero numbers always produces a nonzero number 517 A7 0andB7 0impliesAB7 0 Equivalently if a product of two numbers is zero then one of the factors has to be zero 518 AB0impliesA00rB0 Both 517 and 518 hold for all numbers A and B And since the converse of these is trivial a product where one factor is zero must be zero we actually have 519 A B 0 holds exactly when A 0 or B 0 Now lets apply 519 to 516 1x73 0 holds exactlywhenx100r x730 which shows that the equation 515 has exactly the two solutions 71 and 3 More generally and again using 519 the solutions to a qua dratic equation of the form zAzB 0 are 7A and 7B Of course if A B then there is only one solution here Lets talk about another very simple way to solve quadratic equa tions of another special form next 523 Solving Quadratic Equations by Taking the Square Root Let7s again start with an example Consider the quadratic equa tion 2 2x 3 Since 2 2x 1 x 12 this can be rewritten as x 1 4 after adding 1 to both sides Of course its not always so easy to see this immediately but we will learn a technique in subsection 524 which shows that this can always be done The point of this rewriting 52 SOLVING QUADRATIC EQUATIONS 101 is that now we can take the square root77 on both sides and we arrive at 21 i2 which is an abbreviation for z 1 2 or z 1 72 Dont forget the i here More about that in a minute Subtracting 1 on both sides then yields z 71 i 2 which is an abbreviation for z 1 or z 73 Clearly taking the square root77 here did the trick but why was that justi ed This step and the i part of it is a frequent source of student errors so lets look at it more closely What we really did was to take the following steps each resulting in an equivalent statement 520 z 12 4 521 la 112 1212 522 195 11 121 523 z1i2 iez12orx172 Let7s analyze each step carefully Going from 520 to 521 uses that 4 22 that 2 2 and the identity A2 1A1 which holds for all algebraic expressions A and follows immediately from the de nition 413 of the absolute value Going from 521 to 522 is a new operation on equations for us If both sides of an equation A B for algebraic expressions A and B are non negative then the equations A B and A2 B2 have the same solutions Finally the step from 522 to 523 follows immediately from 415 Since this sequence of three steps is used quite frequently we state it as a proposition Proposition 53 The solutions to the equation A2 B2 for algebraic expressions A and B are the same as the solutions to the statement A B 07quot A 7B The proof of Proposition 53 is the same the proof for the special case above The following four statements are equivalent by the same 102 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES reasoning as for 5207523 A2B2 lAl2lBl2 lAl lBl ABorA7B Proposition 53 now gives us a general procedure to solve quadratic equations of the form x A2 B for any numbers A and B But there is a small wrinkle we havent considered yet What if B is negative Since the square x A2 is always non negative7 there cannot be any solution at least in the real numbers in this case So we7ll assume in addition that B 2 0 Then the solutions to the quadratic equation x A2 B are x 7Aandxi 7A Of course7 there will be two different solutions if B is positive7 and only one solution namely7 7A if B 0 This shows a particularly simple way to solve quadratic equations7 and well see in subsection 524 that we can always reduce the solution of a quadratic equation to this case 524 Solving Quadratic Equations by Completing the Square ln subsection 5227 we saw that sometimes a quadratic equa tion x2 bx c 0 can be solved very ef ciently7 namely7 whenever we see77 how to factor the polynomial77 x2 bx c as a product x Ax B for some numbers A and B But if the solutions to a quadratic equation are too complicated eg7 if they involve square roots7 then we need another way to solve the equation7 which will al ways lead to a solution Subsection 523 already gave a hint as to how we might proceed First of all7 we can always reduce solving an arbitrary quadratic equation to an equation of the form5 524 x2pxq0 egx22x730 simply by dividing the original quadratic equation by the number be fore77 x2 which is usually called the coe cient of 2 Next7 we can rewrite 524 as 525 x2 p iq eg7 2 2x 773 3 4ln subsection 5327 welll brie y cover the complex numbers in which we can solve the equation I A2 B even when E is negative 5We will always accompany our general considerations in this subsection by an example to make things clearer 52 SOLVING QUADRATIC EQUATIONS 103 by subtracting q on both sides Now we can add 102 to both sides of 525 to arrive at 526 x2pxip2ip27q eg22x1134 and we are in the case of subsection 5237 since we can now rewrite 526 as 2 527 z p i122 7 q es7 96 12 4 See Figure 56 for a geometric picture of how completing the square works for the left hand side of the equation The area corresponding to pm is split in half each half is put along a side of a square of area 2 Now7 in order to get a square again7 we need to add a square of area 411102 in the corner7 to arrive at a square of area x 10 Of course7 this picture assumes p gt 0 but a similar picture can be drawn for the case p lt 0 p pz I 12p 12px 14p2 x x2 x 12px X x 12p FIGURE 56 Completing the square geometrically As in subsection 5237 we can now solve 524 as 528 z piip2q egx1i2 and so 529 x7pi1ip27q egx71i2 The main trick in this way of solVing the quadratic equation is in going from 525 to 526 this step is often calling completing the 104 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES square since we are adding just the right number in order to make the left hand side into an expression which can be written as a square as shown geometrically in Figure 56 525 Solving Quadratic Equations by the Quadratic For mula The format of 529 above already suggests that one could solve a general quadratic equation with parameters in place of num bers by simply nding a general formula which gives the solutions to the equation Such a formula does indeed exist and we will now deduce it in its most general form 529 is the formula for the special form of the quadratic equation in 524 ie when the coef cient of 2 is 1 We start with the general quadratic equation6 530 ax2bzc 0 eg 2x275 0 and still assume that a 31 0 First we can rewrite 530 as 531 of in 70 eg 2 7s 75 by subtracting c on both sides and then as b c 532 2 Ex 78 eg 2 x 7 by dividing by a on both sides b 2 Now we can add to both sides of 532 to arrive at 533 225 b 2 b 2 0 527455 x is 7 7 7 7 7 2a 2a 2a 1 4a 7 7 2 7 2 5 49740 9 egv 952 f 595 1 1 5 16 E and we are in the case of subsection 523 since we can now rewrite 533 as b 2 b2 7 4ac 2 534 z E W eg z E 1 As in subsection 523 we can now solve 530 as b Mlin 535 zi2 m egz ig a 1 6Again we will accompany our general considerations in this subsection by an example to make things clearer 52 SOLVING QUADRATIC EQUATIONS 105 and so arrive at the quadratic formula b i xb2 7 4ac 2a 2a 5 eg z 7E i ie z 71 or z 75 536 z i for the solution to the general quadratic equation 530 axz bx c 0 We see from the quadratic formula 536 that the quantity b2 7 4ac is crucial it is called the discriminant of the quadratic equation and determines the number of solutions in the real numbers If the dis criminant is positive then 536 gives two solutions If the discrimi nant is 0 then 536 gives only one solution namely Finally if the discriminant is negative then 536 gives no solution which makes sense if we look back at 534 If the discriminant is negative then the right hand side of 534 is negative since the denominator of the right hand side is a square and thus positive whereas the left hand side of 534 is a square and thus non negative To summarize there are different ways to solve a quadratic equa tion and there is not one best way While the technique of com pleting the square and the quadratic formula will always work the techniques we discussed earlier are sometimes more ef cient Further more the quadratic formula is rather complicated and its easy to memorize it wrong so at the very least one should remember along with the quadratic formula also how to derive it so that one can quickly reconstruct it 526 The number of real solutions of a quadratic equa tion We already saw in the last two subsections that a quadratic equation 530 axz bx c 0 has at most two solutions in the real numbers and we will see this again graphically in section 54 Algebraically this fact depended on Proposition 53 which states that for all algebraic expressions A and B the equation A2 B2 holds exactly when A B or A 7B holds There is another proof of the fact that any quadratic equation can have at most two real solutions which uses factoring along the lines of subsection 522 Lets start with an example Consider the equation 537 z23z 1 0 106 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES From the quadratic forrnula7 we can see that one solution of 537 is 7 Of course7 the other solution from the formula is 7 7 x but lets ignore that for now We will now divide the expression z23x1 by the expression x 7 ie7 the expression z 7 solution 7 for a reason well see later7 similarly to the way long division works for whole nurnbers7 except that now the digits of our expressions are not numbers between 0 and 9 as in usual long division but expressions of the form azz bx or c where 17 b and c are numbers This procedure is called polynomial division and works in our example as shown in Table 52 z x2 3x 1 7 z 3 m 1 7 same 1 0 TABLE 52 Dividing 2 3x 1 7 x g 7 by polynornial division Let7s review this process of polynomial division carefully We want to divide x2 3x 1 by z 7 Just as in long division7 we do it digit by digit except that instead of digits 7 we now use the coef cients ln 2 3x 1 7 the coef cient 1 of x2 corresponds to the hundreds digit 7 the coef cient 3 of z corresponds to the tens digit 7 and the constant coef cient 1 corresponds to the ones digit in z the coef cient 1 of z corresponds to the tens digit 7 and the constant coef cient 3 corresponds to the ones digit We rst check how often z goes into x2 and nd that it goes z times So we rnultiply z by z and subtract the result from x2 3x 17 leaving us with 7 We now take down the coef cient 1 and check again how often z goes into 7 we nd it goes 7 tirnes 7 rnultiply x 3 by 7 and subtract the result from 7 17 leaving us with rernainder 0 So x2 3x 1 divides evenly into z ie7 there is no rernainder7 and the quotient is z 7 We can write this also 52 SOLVING QUADRATIC EQUATIONS 107 as z23z1 3 1 x 777g or News 2 2 z23z1 So why all this work We can now use 518 The solutions z to the equation 537 x23z 1 0 are exactly the solutions to the statement 3 1 3 1 wlt22g700rlt2 270 Thus equation 537 has exactly two solutions namely 7 7 and 7 is So what have we accomplished We started with a quadratic equa tion and one solution for it We then used polynomial division and computed another solution and we also saw that these two solutions are the only ones The point of all this is that this can be done in general leading us not only to a way of nding one solution of a quadratic equation from another but more importantly a proof of the following Proposition 54 Any quadratic equation a2bc 0 with a 31 0 has at most two real solutions In fact it has at most two eomplew solutions The parenthetical statement above can easily be seen once we have covered the complex numbers so lets concentrate on the case for the real numbers First of all we may assume by dividing both sides by a that a 1 so lets only consider a quadratic equation of the form 538 z2pzq 0 Furthermore if this equation has no real solution at all then were done So lets x one real solution and call it we We will now attempt polynomial division of 2 px q by z 7 x0 and see what happens as shown in Table 53 Lets go over this very abstract polynomial division in detail We want to divide 2 10 q by z 7 x0 Again just as in long division we do so digit by digit except that instead of digits we now use the coef cients In x2 px q the coef cient 1 of x2 corresponds to the hundreds digit the coef cient p of z corresponds to the tens 108 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES I p 960 x 7 zo z px q 7 2 7 ms 10 o q i 10 95095 i 10 950 10 o o 9 TABLE 53 Dividing 2 10 q 7 x 7 x0 by poly nomial division digit 7 and the constant coef cient q corresponds to the ones digit in z 7 0 the coef cient 1 of z corresponds to the tens digit 7 and the constant coef cient 7 corresponds to the ones digit We rst check how often z goes into77 x2 and nd that it goes z times So we multiply z 7 x0 by z and subtract the result from 2 10 q leaving us with p zox We now take down77 the coef cient q and check again how often z goes into77 Pxox we nd it goes p x0 tirnes 7 rnultiply z 7 0 by p 0 and subtract the result from p x0x q7 leaving us with rernainder p x0z0 q So 2 10 q divides into z 7 0 with rernainder p x0x0 q7 and the quotient is z p 0 We can write this also as 2 539 f 1 f q x p 950 p 0 q or i 7 0 i 7 0 540 962 p96 q 96 p 96096 7 x0 p atom q Note that 540 is obtained from 539 simply by multiplying by z 7 0 on both sides Now lets use the fact that 0 is a solution of 2 px q 0 Set z zo Then the left hand side of 540 is equal to 0 But the left half of the right hand side of 540 is also equal to 0 for z 0 so this forces the remainder p x0z0 q to be 0 as well7 ie7 x2 10 q divides into z 7 0 evenly without rernainderl This simpli es 540 to 962 1096 q 95 10 95095 i 950 and so7 by 5187 the solutions of x2 p q 0 are the solutions to the statement px0 00rs7z00 ie7 the solutions are 0 and 7p 0 So there are at most two solutions as claimed As a by product7 we have also found a formula 53 SQUARE ROOTS AND THE COMPLEX NUMBERS 109 which allows us to compute one solution of a quadratic equation from the other Note that 0 may equal 71 0 so there may be only one single solution 53 A Digression into Square Roots and the Complex Numbers 531 Square Roots As you can see from the quadratic formula solving quadratic equations can necessarily involve square roots so lets brie y review them rst By de nition a square root of a real number a is a solution to the equation 541 2 a If a 0 then x 0 is a unique solution to this equation If a is positive then there are two solutions namely a positive number which we denote by 5 and its additive inverse i The existence of such a number 5 is actually a very deep fact about the real numbers which we cannot prove here In order to make the square root unique for all a 2 0 we de ne the non negative solution to the equation 2 a to be the77 square root of a and denote it by 5 Note that this is somewhat arbitrary it is simply a de nition of what we mean by the77 square root of a non negative number well get to what a square root of a negative real number could be later in subsection 532 when we discuss the complex numbers Let7s now state some rules about square roots The simplest ones are about multiplication and division with square roots For all non negative real numbers a and b we have 542 5 w ab V5 9 543 W 4 b where we assume b 31 0 for 543 These rules follow immediately from the de nition of square roots Let 3 the unique non negative solution x to the equation 2 a l the unique non negative solution y to the equation yz b xab the unique non negative solution 2 to the equation 22 ab a a the unique non negative solution w to the equation w2 3 110 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES Then 542 simply states that 544 zy z and 542 states that 2 545 y w Since all the numbers involved are non negative we can apply Propo sition 53 Squaring equations 544 and 545 we nd that they are equivalent to 2 2 2 d 2 2 z y w an yz respectively both of which are trivially true There is one other useful rule about square roots which connects them to absolute value For my real number a ie also for a negative real number a we have 546 V52 la This rather surprising and at rst unexpected fact can be shown directly from the de nition of square roots and absolute value distin guishing the two cases in the de nition of absolute value If a 2 0 then xCTZ is the unique non negative solution x to the equation 22 a2 in the unknown 2 with parameter a but 2 a is a non negative so lution to this equation and since the non negative solution is unique we must have x a lal On the other hand if a lt 0 then again xCTZ is the unique non negative solution x to the equation 22 a2 in the unknown 2 with parameter 1 but 2 7a is a non negative solution to this equation since a2 7602 and since the non negative solution is unique we must have x 7a lal We7ll conclude this subsection with two non rules which fre quently creep into school mathematics They are the following two statements The statement a b 3 W is generally false The statement a 7 f 7 is generally false Once you realize the above formulas are false its very easy to see why Just plug in almost any numbers and check your answer Remember Showing that a general rule7 fails is much easier than showing that it holds For a general rule to hold you must check it in all7 cases usually by some mathematical proof in order to show that it fails you need just one counterexample 53 SQUARE ROOTS AND THE COMPLEX NUMBERS 111 532 The Number 239 and the Complex Numbers We now return to the question of how one might de ne the square root of a negative real number a Our rst attempt would be to de ne it as a solution x to the equation x2 a But there is an obvious problem for any real number a x2 is non negative and thus cannot equal a So there cannot be a real square root of a But we can extend our number system again Think back to the time when you rst learned about the negative numbers Back then it was impossible to solve an equation like z3 2 since that would be the solution to a non sensical problem like Mary receives three apples and now has two apples How many apples did she have before 7 In this context introducing negative numbers makes no sense Mary cannot have had 71 apples before But if we expand our model of what numbers mean to notions like temperature the equation x 3 2 suddenly makes sense The temperature rose three degrees and is now at two degrees What was the temperature before7 We are in a similar situation with solving an equation like 2 71 If we insist on staying within the real numbers there is no solution The dif culty is that there is no simple and visual model for the com plex numbers The usefulness of the complex numbers can be seen re ally only in higher mathematics after most of calculus in upper level undergraduate mathematics and beyond and in fairly sophisticated science and engineering applications Thus it is far from obvious why one would want to introduce the complex numbers except for the fact that it enables one to solve some equations In order to introduce the complex numbers lets start with the sim plest quadratic equation which cannot be solved in the real numbers namely x2 71 We introduce a new number 239 called the imagi nary unit and de ne it to be a solution of the equation x2 71 ie we de ne that 2392 71 Assuming that all the rules of arithmetic we have for the real numbers also hold for the complex numbers we see that 702 2392 71 and so the equation x2 71 has two solu tions 239 and 7239 Furthermore we can factor the expression 2 1 as x 239 x 7 239 Let7s now move on to solving a slightly more general equation namely 2 7a for positive a Set b ya Since we assumed all the rules of arithmetic for the complex numbers we have bt2 m2 14 7a 112 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES and 4m2 422392 a71 7a Thus the equation 2 7a for positive a has the two solutions in and 7b239 While one does not in general de ne one of M or 7M to be the square root of 7a both in and 7b239 serve as a good substitute for the square root of 7a Whenever we write expressions like ixja it doesnt matter anyhow which of in or 7bz39 we use for H Finally we7ll de ne a complcd number to be any number of the form a M where a and b are arbitrary real numbers Addition subtraction and multiplication are de ned on the complex numbers just as expected 547 a in c dz a c b dz39 548 a b2 7 c dz a 7 c b 7 dz39 549 a bz39c dz39 ac 7 bd ad bcz39 Here the formula 549 for multiplication just uses repeated applica tion of the distributive law for the left hand side plus the fact that 2392 71 Division is somewhat more complicated since one has make the denominator a real number this involves a slight trick and some messy computations a M a b c 7 dz39 cd cdz39c 7dz39 53950 ac bd bc 7 adz39 c2 d2 ac bd bc 7 ad 2 2 2 Z c d c d2 Note that the last line of 550 is well de ned as long as cd 74 0 ie as long as at least one of c and d is nonzero making c2 d2 nonzero So for c dz39 74 0 the formula 550 gives us again an expression of the form A Bz39 for real numbers A and B ie a complex number in our sense It is also possible to solve quadratic equations with complcd parameters a b and c but that would require the de nition of square roots of complex numbers which is beyond the scope of these notes We now return to the quadratic formula 2 7 536 z 43 4 2a 2a for the solution to the general quadratic equation 530 axz in c 0 54 GRAPHING QUADRATIC FUNCTIONS 113 where 17 b and c are now arbitrary real numbers with a 31 0 The quadratic formula then gives us one or two complex solutions in all cases namely7 one solution if b2 746w 07 and two solutions otherwise We conclude with an example Consider the quadratic equation 551 z22z2 0 Here7 a 1 and b c 2 So the solutions to 551 are 2 22 i 8 x if 7 2 2 1 i 7V4 2 i 1 i 2239 7 2 71 it We can check that these two solutions really work 71 2271 2172 71722 2 171722722 0 and 71i 2271i 212 717272 2 171722272z390 54 Graphing Quadratic Functions and Solving Quadratic Equations Graphically Solving a quadratic equation of bx c 0 corresponds7 of course7 to undoing77 the function f of bx c ie7 to nding all arguments x at which the function f takes the value 0 We had seen this before in section 25 with solving a linear equations cw b 07 which corresponded to nding all x at which the function 92 ax b takes the value 0 As in the case of linear equations7 one can always nd the solutions to quadratic equations by graphing at least approximately but the graphs of quadratic functions are naturally a bit more complicated Lets start with an example from physics An arrow is shot straight up from the ground with an initial velocity of 00 50msec If the height of the arrow in meters at time t after being red7 in seconds is described by l 552 s Dot 7 Egt2 114 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES neglecting friction7 where g 98msec2 is the earth7s gravity constant7 how high will the arrow go7 and when will it hit the ground again77 1 2 m s voi 2 gt FIGURE 57 The motion of an arrow The motion of the arrow is described approximately by the graph of the function 553 5t 50t i 49 shown in Figure 57 We only show the graph for values of t in the interval 0102 since the graph appears to be below the t axis outside this interval7 which would correspond to the arrow being below the ground7 contradicting the real life77 context It is now not hard to check that the arrow is at maximum height at just over ve seconds7 when it reaches a height of approximately 130 meters Furthermore7 the arrow appears to hit the ground after just over ten seconds Could we have computed this without a picture of the graph Of course7 but that requires a bit of thinking and computation Let7s address the second question from the problem rst When does the arrow hit the ground again This corresponds to a solution of the quadratic equation 50t i 49t2 0 54 GRAPHING QUADRATIC FUNCTIONS 115 since the arrow hits the ground when 5 equals 0 again We can solve this equation by factoring 749 50t 491mi 0 49 and so 5 equals 0 at time t 0 when the arrow is red and at time x x 102 The latter time 102 seconds after being red must be the approximate time when the arrow hits the ground again The rst question How high will the arrow go7 and when will it be at maximum height is harder to answer7 The solution uses our technique of completing the square from subsection 524 5 50t i 4912 749 t2 i gt 2 2 554 749 t2 7 gt g 7 g gt 2 49ltt2tlt5gt We 749 t 7 g g Since 749 7 2 S 0 for all t we can now see that the height 5 cannot exceed x 1275 meters7 but that that height is indeed the maximum height and is achieved at g m 51 seconds Now lets do the above in the abstract Consider a quadratic func tion 555 f azz bx c where a 31 0 We can rewrite this7 completing the square7 as b f a x2 Ex c 7 2 b b 2 b 2 T a z a 2a 2a C 2 b b 2 52 a x is 7 c i 7 1 2a 4a b 2 52 altz 07 andso 7Calculus was invented to solve problems like this7 but in our easy case7 we donlt need itl 556 116 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES and so 557 Setting 54754475 to g y 0 3 equation 557 simpli es to 558 fx 7 yo az 7 02 or fx az 7 x02 yo The graph of the quadratic function is called a parabola and the point V with coordinates zoy0 is called the vertem of the parabola In the arrow example 553 above the vertex is approximately the point 511275 We are still assuming that a 31 0 and now need to distinguish two cases before we can discuss the graph of a quadratic function in more detail Case 1 a gt 0 Then az 7 02 y0 2 yo for all a and so the minimal 559 value of f is yo which is achieved only at z 0 The equation describes a parabola which opens upward In order to solve the quadratic equation az 7x02 y0 0 we need to distinguish three subcases Case 1a yo lt 0 Then the equation 559 has two real solutions z0i 7170 V a since the graph of the function f az 7 02 yo intersects the a axis at the two points 0 70 0 and 0 7 7330 0 The function achieves its minimum value at the vertex 0 yo See Figure 58 for an example with 0 2 yo 71 and so the vertex is V 271 the graph of f intersects the a axis at 10 and 30 Case 1b yo 0 Then the equation 559 has only one real solu tion 0 since the graph of the function f az 7 02 yo intersects the a axis only at the vertex 0 0 at which the function also achieves its minimum value See Fig ure 59 for an example with x0 2 yo 0 so the vertex 54 GRAPHING QUADRATIC FUNCTIONS 117 fixix 4x3x3 1 v21 FIGURE 58 The graph of the quadratic function fx z274z3 is V 27 07 which is also the point where the graph of f intersects the z axis Case 1c yo gt 0 Then the equation 559 has no real solutions since f ax7x02yo 2 gt 0 for all t The function still achieves its minimum value at the vertex xmyo See Figure 510 for an example with 0 27 yo 17 so the vertex is V 217 and the graph of f does not intersect the z axis Case 2 a lt 0 Then a 7x02y0 3 yo for all L and so the maximal value of f is yo which is achieved only at z 0 The equation describes a parabola which opens downward In order to solve the quadratic equation 559 az 7 x02 yo 0 we again need to distinguish three subcases 118 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES 2 2 lezx 4lt4x 2I FIGURE 59 The graph of the quadratic function fx z274z4 4 3 2 2 2 figt0x 4x5x 1 1 1 V2 1 o I I I I I 0 1 2 3 4 5 FIGURE 510 The graph of the quadratic function fx x274x5 Case 2a 54 GRAPHING QUADRATIC FUNCTIONS 119 yo gt 0 Then the equation 559 has two real solutions z0i 710 V a since the graph of the function f az 7 02 yo intersects the s axis at the two points z0 7 0 and 07 7 0 The function achieves its maximum value at the vertex mayo See Figure 511 for an example with 0 72 yo 17 and so the vertex is V 721 the graph of f intersects the s axis at 17 0 and 30 1 V2 1 o P10 Q3390 o 392 394 1 2 2 fxx 4x 3 3 FIGURE 511 The graph of the quadratic function f 96 7z24z73 Case 2b yo 0 Then the equation 559 has only one real solu Case 2c tion i 0 since the graph of the function f az 7 02 yo intersects the s axis only at the vertex 07 07 at which it also achieves its maximum value yo lt 0 Then the equation 559 has no real solutions since f ax7x02yo Eye lt 0 120 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES for all t The function still achieves its maximum value at the vertex x07y0 Finally7 let7s study the effect of varying the three parameters 17 0 and yo in the equations of a quadratic function 558 fx 7 yo az 7 x02 or fx az 7 x02 yo Since the vertex of the parabola described by these equations is always the point x07y07 the effect of changing 0 or yo is clear It shifts the whole parabola right or left when changing 0 or up or down when changing yo respectively The effect of changing a is a bit more subtle We saw already that if a gt 0 then the parabola opens upward 7 and that it if a lt 0 then the parabola opens downward The effect of increasing or decreasing lal is now to make the parabola steeper or less steep Compare Figure 58 where a 1 with Figure 512 where a 4 4 3 2 2 fx4x 15x154x2 1 2 1 0 P150 Q250 0 1 2 3 4 5 6 1 V21 FIGURE 512 The graph of the quadratic function fx 4x2 716 15 55 QUADRATIC INEQUALITIES 121 55 Quadratic Inequalities We conclude this chapter by considering quadratic inequalities Let7s vary the arrow problem from section 54 a bit An arrow is shot straight up from the ground with an initial velocity of 00 50msec If the height of the arrow in meters at time t after being red7 in seconds is described by l 552 s Dot 7 Egtz neglecting friction7 where g 98msec2 is the earth7s gravity constant7 when will the arrow be at least 50 meters above the ground We already saw that the motion of the arrow is described approxi mately by the graph of the function 553 5a 50f i 49t2 shown in Figure 57 So answering the above problem amounts to solving the quadratic inequality 560 50f i 49t2 2 50 First note a common mistake We cannot simply rewrite this inequal ity as 7491 50f 7 50 2 0 and use the quadratic formula to 7 2 7 7 7 get It 2 W this solution doesnt even make any sense since it speci es t to be 2 some number i some other number lnstead7 we can either use the graph of the function or use the technique of completing the square For a graphical solution7 we simply look at the graph in Figure 57 and note that the graph is at or above the line 5 50 approximately between times t 1 seconds and t 9 seconds For an exact solution7 we use the completed square version of equation 553 2 2 554 s 749 t i g 122 5 QUADRATIC FUNCTIONS EQUATIONS AND INEQUALITIES We can now solve our inequality in a few steps as follows 561 749 t 7 g 375 2 50 2 2 562 749 7 2 5 7 563 52lt w 39 49 749 49 492 25 252 4 49 50 564 tii 31 49 492 25 252 4 49 50 25 252 4 49 50 565 fin K7 1 l 49 492 49 492 which gives the approximate solution 1124 S t S 908 Looking over the steps in 5617565 we see that the key steps are in going from 563 to 564 and from 564 to 565 The second of these two steps involves statements 415 and 416 about the absolute value namely combining the two that we have 566 lal S r exactly when 7 r S a S r The rst of these two steps going from 563 to 564 involves a modi cation of Proposition 53 to the case of inequalities Proposition 55 Let A be an algebraic eapression and r 2 0 a num beiquot Then I The solutions to the equation A2 r are the same as the solutions to the statement A 77 07quot A ixT 2 The solutions to the inequality A2 lt r are the same as the solutions to the statement i lt A lt W7 3 The solutions to the equation A2 gt r are the same as the solutions to the statement Alt707quotAgt7 55 QUADRATIC INEQUALITIES 123 Proposition follow immediately from 4157417 and from the fact that for all non negative numbers a and b 567 a lt b exactly when a2 lt b2 We now see that the key step above7 going from 563 to 5647 follows immediately from Proposition 55 It is also not hard to see that we could now carry out a general analy sis and solution for quadratic inequalities of the form of bx c S 07 lt 07 2 0 or gt 07 respectively7 in the style of section 52 for an algebraic solution or section 54 for a graphical solution CHAPTER 6 Exponential and Logarithmic Functions Equations and Inequalities In this last chapter we7ll cover the exponential functions and their inverse functions the logarithmic functions 61 Exponentiation A Review of the De nition There is one common arithmetical operation from school mathemat ics which we have mostly ignored so far exponentiation or taking the power This operation is initially introduced as repeated multiplica tion just as multiplication in the whole numbers can be viewed as repeated addition The actual de nition of exponentiation is quite tricky and we will present it here very carefully in a number of steps Given any number a and any nonzero whole number m we de ne the mth power Ufa to be m q maaaa 61 1 Here and from now on we will call a the base and m the exponent or the power1 We read 1 as a to the mth power or a to the mth for short For m 2 we read a2 as a squared or the square of a and for m 3 we read a3 as a cubed or the cube of a This de nition of exponentiation as repeated multiplication gives us immediately the following rules of exponentiation for now only for nonzero whole numbers m and n 62 am a amn 63 Zi am if m gt n 64 trn am 65 a bm am b 1In the Singapore math books the exponent is also called the index plural indices 124 61 EXPONENTIATION A REVIEW OF THE DEFINITION 125 To see that eg 62 is true simply observe that m 71 PEPh ama aaaaaaaa mn PW aaaa amn And to see that 64 is true we check n m m m m mn zz z a aaaaaaaaaaaa39Haaaa mn PH aaaa Exercise 61 How you would prove 63 and 65 this way Why does it make sense to impose the restriction m gt n in 63 for now So 62765 follow immediately from our previous rules for mul tiplication Wouldn7t it be nice to have these rules hold not just for nonzero whole numbers as exponents Of course you know that these rules are still true right Well there is a small problem We havent even de ned what 1 means unless m is a nonzero whole number We will do this in a sequence of small steps each time justifying our de nition for more and more possible exponents From now on we will assume that the base a is nonzero This will make sense in a moment when we see how division by 1 becomes involved in our de nition Now we can de ne exponentiation for an exponent which is 0 or a negative integer as follows 66 a0 1 I 67 am f for any negative integer m a m First note here that a in 67 has already been de ned in 61 since 7m is a nonzero whole number if m is a negative integer More importantly note that we are forced to de ne exponentiation this way if we want the analogue of 63 to still hold 0 a E 1 a a 1 a 7 7 1 7quot 1 7quot 126 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS What about rational exponents For this7 we need to restrict the base 1 even further We7ll assume from now on that a is not only nonzero but indeed a positive real2 Now we7ll de ne exponentiation for rational exponents as follows for any integer n and any positive integer p Of course7 this immediately raises another problem We haven7t de ned yet what FE means for a positive real number a and a positive integer p But as for the case p 27 ie7 the square root7 we can de ne the pth root of a simply as the unique positive solution3 to the equation x17 1 Again note also that the de nition in 69 already requires the de nition in 68 And again7 more importantly7 note that we are forced to de ne exponentiation this way if we want the analogue of 64 to still hold But there are two hidden problems with these de nitions 1 It is possible that the rational number for an integer n and a positive integer p happens to equal an integer m7 say Now we need to show that the de nition in 69 coincides with the de nition from 61 and 66767 In that case7 we have m g and so 71 mp Now we can argue as follows 2We impose this restriction since welll be taking roots in a moment It is actually possible to de ne am in a meaningful way for a negative real and some rational exponents m7 but the rules for when this is possible are rather complicated7 so welll ignore this case here 3The fact that such a solution always exists and is unique is hard to show and beyond the scope of these notes so welll simply assume it For an odd positive integer p one can also de ne 5 for negative a this way but welll not do so here 61 EXPONENTIATION A REVIEW OF THE DEFINITION 127 using 64 for integer exponents onlyl i W 2 Similarly given a rational number 7 it can be written in many different ways as a fraction So in order to show that a is uniquely de ned we really need to show that whenever 1 g for integers n and n and for positive integers p and p then B n a aZ397 under the de nition in 69 This proof is similar to the one in 1 but quite a bit more tedious since it involves many cases so well skip it here Finally can we de ne a for irrational exponent r The answer is a quali ed yes quali ed because in order to make this de nition pre cise one really needs calculus which we clearly dont want to assume here Instead lets give the intuition For an irrational number 7 and arbitrary positive real number a we de ne a as being more and more closely approximated by 1 7 where q is a rational number close to r The fact that this de nition works is a miracle we have to accept here since proving it would take us far beyond the scope of these notes Finally looking back we really need to return to checking our rules of exponentiation 62765 for arbitrary exponents We state them again here for reference as a proposition 4Strictly speaking what we are doing here is merely what you have already done for the other arithmetical operations like addition and multiplication for arbitrary real numbers Eigi the sum 7 0 7 1 of two irrational numbers 7 0 and 7 1 say is approximated more and more closely by the sum q0 ql of rational numbers go and q1 close to 7 0 and 7 1 respectively 128 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Proposition 62 The following exponential rules hold for all positive real numbers a and b and for all real numbers r and s 610 a a5 a a7 611 7 H lt gt a a 612 a5 a 613 11 a b If both r and s are integers then these rules hold for arbitrary nonzero real numbers a and b The formal proof of Proposition 62 is quite tedious7 so well skip it here and simply accept the exponential rules as given from now on There is one special case of 611 which comes up frequently7 namely7 the case when r 07 so well state this as a separate formula 1 1 5 1 1 4 614 0 9 i 7 or equivalently at 7 a F a 62 Motivation for Exponentiation So far7 we have only discussed the de nition of exponentiation in fair detail and have left the motivation rather weak Exponentiation is repeated multiplication7 just like multiplication is repeated addition Of course7 this motivation is not really satisfying one could now go on and talk about repeated exponentiation 7 but there is a really good reason why almost no one does There are essentially no real life phenomena which can be described by repeated exponentiation 7 but there are quite a few which can be described by exponentiation Let7s look at the following two examples with which Id like to contrast linear vs exponential functions A sh in Ann7s aquarium is giving birth to babies Every hour7 three new babies are born If she has ve sh at the beginning7 how many sh will she have after n hours The number of bacteria in a lab dish doubles every hour If there are one hundred bacteria at the begin ning7 how many will there be after n hours Look at the rst problem rst Table 61 shows the number of sh during the rst few hours We see right away that the number of sh increases each hour by a xed amount7 namely7 by 3 Therefore the number F of sh after n 62 MOTIVATION FOR EXPONENTIATION 129 2 3 4 5 nurnber of hours 0 1 11 14 17 20 number of sh 5 8 TABLE 61 The number of sh after 71 hours hours is given by the linear function 3n 5 or more awkwardly for reasons which will become apparent in a few paragraphs as 615 53n Note that this function really only makes sense for whole number values of n it doesnt make sense to talk about there being 95 sh after 15 hours But lets ignore this side issue for now Now lets look at the second problem Table 62 shows the number of bacteria during the rst few hours nurnberofhours 0 1 2 3 4 5 number of bacteria 100 200 400 800 1600 3200 TABLE 62 The number of bacteria after 71 hours We see right away that the number of bacteria does not increase by a xed amount each hour In fact the increase itself increases rather drarnatically Frorn hour 0 to hour 1 the increase is by only 100 but from hour 4 to hour 5 the increase is already by 1600 In fact this leads us to the following observation If we denote the number of bacteria after 71 hours by Bn and consider the rst difference function77 as de ned in section 51 narnely B101 Bn 1 7 Bn then it turns out that in fact B101 Bn ie the rst difference function B1 equals the original function B We could now try as in section 51 to take the second difference function B201 B1n 1 7 B101 but of course this will give us the same function again So not only is the function E not linear it is also not quadratic by the discussion in section 51 130 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS On the other hand you will surely have noticed de nite patterns in Tables 61 and 62 In the former the additive change is constant namely each hour the number of sh increases by a summtmd of 3 On the other hand in the latter the multiplicative change is constant namely each hour the number of bacteria increases by a factor of 2 It is then not hard to see that the function E can be described by 616 Bn 100 2 Note that there is some similarity between 615 and 616 The starting poin at n 0 is 5 and 100 respectively and shows up rst in the expression on the right hand side Next comes the op eration symbol for the way in which the increase occurs additively in 615 and multiplicatively in 616 Next comes the rate of change namely 3 in 615 for the additive change of 3 each hour and 2 in 616 for the multiplicative change of 2 each hour respectively And nally comes n for the number of times by which this increase oc curs preceded by symbol for the repeated operation namely multi plication as repeated addition in 615 and exponentiation as repeated multiplication in 616 respectively In both our sh and our bacteria example it is now easy to compute the number of sh or bacteria respectively at time 10 hours say It is 5 3 10 35 sh and 100 210 102400 bacteria respectively5 But as we mentioned already in section 25 such problems are really problems of arithmetic In order to solve them we merely have to evaluate the numerical expressions 5 3 10 and 100 210 respectively On the other hand suppose that we are given the number of sh or bacteria respectively and want to compute the number of hours after which there are this many sh or bacteria Suppose I want to know when there will be 50 sh This is now a real algebra problem and clearly amounts to solving the equation 53n50 by undoing the algebraic expression 5 3 71 to arrive step by step at the solution 371 50 7 5 50 7 5 i 3 7 15 5In the latter case of course we should really write approximately 100000 bacteria since 102400 gives a false sense of precision for our answer 63 THE EXPONENTIAL FUNCTIONS 131 So it will be after 15 hours that there will be 50 sh in the aquarium and in order to arrive at the solution we rst subtract 5 from both sides of the equation and then divide by 3 on both sides Let7s try the same for the bacteria Suppose I want to know after how many hours there will be 6400 bacteria Again this is now a real algebra problem and clearly amounts to solving the equation 100 2 6400 by undoing the algebraic expression 100 2 to arrive step by step at the solution 2 6400 64 100 n 6 So it will be after 6 hours that there will be 6400 bacteria in the lab dish However obtaining this solution is conceptually more dif cult In the rst step we simply divide both sides by 100 But the second step going from 2 64 to n 6 is a bit mysterious at this point We will learn how to solve such so called exponential equations in general in section 64 when we introduce the logarithmic functions In our particular example with small numbers we can of course just use guess and check 6 63 The Exponential Functions 631 Properties of Exponential Functions Before we pro ceed to solving exponential equations such as 2 64 in the example above lets talk about exponential functions in general and note some of their properties First of all for each possible base b gt 0 we can de ne an esponentz39al function f b which is de ned for all real numbers m7 In the case that b 1 we get the rather boring constant function f 1m so we will also assume b 74 1 for the remainder of this section We rst look at two typical examples of exponential functions the function f 2 see Figure 61 for its graph and the exponential function g see Figure 62 for its graph From these two graphs we observe some properties which we state in the following proposition We will then argue that all exponential functions look like one these two functions depending on whether b gt 1 or 0 lt b lt 1 respectively 6And I had to start with the rather arti cial number of 6400 bacteria to make the problem work out 7We explained already in section 61 why the assumption 1 gt 0 is needed 132 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 8 1 FIGURE 61 The graph of the exponential function f96 2 Proposition 63 For any base b gt 0 with b 31 1 the eaponential function h bm has the following properties I The emponentialfunetion h bm is de ned for all real num bers n ie the domain ofh is the set R of all real numbers 2 The graph crosses the y acois at the point 17 0 3 The graph lies entirely above the x ais ie bm gt 0 for all m 63 THE EXPONENTIAL FUNCTIONS 133 8 1 FIGURE 62 The graph of the exponential function 1 M i 4 Ifb gt1 then a the eccponential function h bm increases as x in creases ie ifxo lt 1 then bmo lt b b the graph increases quotwithout bound as x grows larger and larger and 134 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS c the graph approaches the d accis arbitrarily closely as x grows smaller and smaller ie as x quotgrows larger and larger in the negative direction 5 If0ltblt 1 then a the eccponential function h bm decreases as x in creases ie ifxo lt 1 then bmo gt b b the graph approaches the d accis arbitrarily closely as x grows larger and larger and c the graph increases quotwithout bound as x grows smaller and smaller ie as x quotgrows larger and larger in the negative direction 6 The eccponential function h bm is 171 7 The range of the eccponential function hd bm is the set 0 00 of all positive real numbers In summary the eccponential function hd bm is a 171 function with domain R and range 0 00 the set of all positive real numbers Let7s look at these statements one by one and try to argue that they hold true not only for the special cases b 2 and b shown in Figures 61 and 62 but indeed for all positive bases b 31 1 First of all 173 follow by analyzing the de nitions in section 61 Assume that b gt 1 Then for 4a note that do lt 1 irnplies by 611 that bml I 1 bTO b 1 O gt 1 since 1 7 x0 gt 0 and so b gt b Next 4b and 4c are somewhat harder to check In the special case b 2 we see that z lt 2 for all x more generally for arbitrary b gt 1 we have that b 7 1z lt b But now b 7 1z grows without bound as x does and so must b Finally 4c follows by 614 since as x grows smaller and smaller ie as x grows larger and larger in the negative direction 7d grows larger and larger in the positive direction but then by 614 1 bm b7m and this expression gets closer and closer to 0 since b grows without bound Now assume that 0 lt b lt 1 Then similarly for 5a note that 0 lt 1 irnplies by 611 that b bTO berm lt1 63 THE EXPONENTIAL FUNCTIONS 135 since 1 7x0 gt 0 and so bw1 lt b Next 5b and 5c can be reduced to 4c and 4b respectively 0 lt b lt 1 implies 1 lt by 39 Now apply 4c to see that as x grows larger and larger bm gets arbitrarily close to 0 Similarly apply 4b to see that as x grows smaller and smaller ie as x grows larger and larger in the negative direction bm grows larger without bound Furthermore 6 follows by 4a and 5a since x0 lt 1 implies bmo lt bm1 or bmo gt bf1 and thus bmo 31 b Finally 7 can be seen from 4b and 4c for b gt 1 and from 5b and SC for 0 lt b lt 1 respectively8 There is one issue we need to address still How does the value of the base b impact on the graph of the exponential function Figure 63 shows some examples note that for clarity we used various dotted lines for some of the graphs so you can quickly tell them apart As you can see from these graphs for b gt 1 the bigger b is the more steeply the graph of the exponential function increases On the other hand for 0 lt b lt 1 the smaller b the more steeply the graph of the exponential function decreases One nal note If you look at other algebra books or at calculus books you will nd a special exponential function being mentioned often called the exponential function This is the exponential func tion f em with special base 5 m 2718 which is sometimes called Euler s number The fact that this base is special does not become apparent until one studies calculus and using 6 as a base before is re ally more of a pain than a convenience since 6 is actually an irrational number Therefore we have chosen to ignore the base 6 here 632 Another Application of Exponential Functions We have already seen one application of exponential functions in real life namely to bacterial growth in section 62 We will brie y cover one other application here to show you that exponential functions are in deed very useful and not only in mathematics We had seen in section 62 that the multiplicative change of the number of bacteria from one hour to the next was constant After each hour the number of bacteria doubled There is a much more every day phenomenon in which such growth occurs namely in interest computations 8Technically we are cheating a bit here The problem is that we havenlt shown that there are no holes in the range of the exponential function ie that the range really is the whole interval 0 But calculus is required to actually show this carefully so we re skipping over this fine point here 136 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS FIGURE 63 The graphs of various exponential functions The setup for interest computations is typically as follows You have a certain amount of capital or debt called the principal At the end of each xed time period say7 a year7 or a rnonth7 or a quarter7 a xed percentage of the current amount is added This added amount is called the interest and the percentage is called the interest rate per time period7 eg7 per year Typically7 the interest is compounded7 ie7 63 THE EXPONENTIAL FUNCTIONS 137 the interest is added to the amount and next time interest is paid interest is also added for the previous interest For a simple example suppose I have 8100 in a savings account and the bank pays me 5 interest at the end of each year At the end of the rst year I then have 8105 namely the principal of 8100 plus 5 of it in interest at the end of the second year I have 11025 namely the previous amount of 8105 plus 5 of it in interest etc See Table 63 number of years 0 l 1 l 2 l 3 l 4 l l amount in account l 810000 l 810500 l 811025 l 811576 l 812155l TABLE 63 Savings after 71 years We can now see that my savings grow by a factor of 105 105 each year Thus the capital 1 have is described by the function 617 Cn 100 105 where n is the number of years I have saved This number 71 must be a whole number here for the problem to make sense More generally the capital or debt starting with a principal P and under an interest rate of r per time period after 71 time periods is given by the function 7quot V L 618 o P 1 7 n 100 where again the number n of time periods must be a whole number Of course 618 can also be used to solve for the principal P C 1 3 or to solve for the interest rate r of time periods 1 L PO 100 T n 1mPO 7 quot13071 It is also possible to solve for the number n of time periods but that requires the logarithmic functions to be introduced in section 64 Another application of exponential functions concerns nuclear de cay but again this will rst require logarithmic functions and will be covered in subsection 642 138 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 64 The Logarithmic Functions and Solving Exponential Equations Our goal in this section is to nd functions which allow us to solve exponential equations ie7 equations of the form 619 bm c where b and c are parameters Compare this to previous situations where we tried to solve equations of the form 620 b c or 621 b x c We could easily solve 620 and 621 by undoing the operations of addition and multiplication7 respectively zcib 70 1 In each case7 we undid the equation by using an inverse function7 namely7 subtraction to invert addition7 and division to undo mul tiplication More formally In 620 and 6217 we consider b to be xed7 so the functions we are inverting are the two functions f b z and g b x and the inverse functions are f 1y y 7 b and g 1y respectively Of course7 when inverting 9 we have to assume that b 31 0 otherwise the function g b z cannot be inverted This should lead us to the conclusion that in order to solve expo nential equations as in 6197 we need to nd an inverse function for the exponential function f b We saw already that we need to assume that b gt 0 for the exponential function to be de ned Also7 we need to assume that b 31 1 otherwise7 the exponential function f bm is a constant function and cannot be inverted So7 for the rest of this section7 we7ll assume that b gt 0 and b 31 1 641 De nition and Properties of Logarithmic Functions ln Proposition 637 we observed that for any positive base b 31 17 the exponential function is a 171 function with domain R and range 07 00 Typically7 one sets the codomain as de ned in section 41 to be the set 07 00 of positive reals so that the exponential function then becomes a 171 and onto function 622 epr R a 07 oo zgt gtb 64 LOGARITHMIC FUNCTIONS 139 Here we have introduced the special symbol expl for the exponential function with base b Now recall from subsection 423 that a function has an inverse func tion exactly when the function is 171 and onto This inverse function of the exponential function eprz bm is usually called the logarithmic function to base b it is de ned as log 000 a R 623 y gt gt the unique x such that bm y Let7s immediately return to solving exponential equations for a minute The equation 619 b c can be solved using the logarithmic function log to arrive at 624 z logbc But how do we evaluate logbc in practice with actual numbers Eg for the bacterial growth example from section 62 in order to solve 625 2 64 we have to evaluate 626 n log264 Your rst reaction may be How do I evaluate log264 on my calcu lator There is no log2 button on my calculator The only buttons 1 see are called log77 and ln And that is only if you have a scienti c calculator and even then it may not have both l7ll give two answers here First of all in simple cases you can compute logarithmic functions in your head To solve 626 you could simply recall the de nition of log264 627 log264 the unique 71 such that 2 6477 holds This means that you have to nd the right exponent n to make the statement 2 64 true A moment7s thought tells you that that expo nent must be 6 Of course this method will not always work so how do you do this with a calculator The real problem is that there is a different logarithmic function log for each base b and no calculator gives them all Your calculator could give you the option of choosing both b and z when computing logbx but l7ve never seen any calculator like that Fortunately there is a simple little trick Most scienti c calculators provide you with two special logarithmic functions namely log and ln 140 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS The rst called log is really logl0 and sometimes called the common logarithm or the decadlc logarithm lts importance is mainly historic as we will explain later on but since its somewhat easier than the other logarithm on your calculator we will mainly use it The other logarithmic function on your calculator called ln is really loge for the number 5 m 2718 mentioned at the end of section 631 and is usually called the natural logarithm which explains the symbol ln it is somewhat cumbersome and not all that useful until calculus when it becomes very important so important that most mathematicians call it the77 logarithmic function and ignore all othersl9 There is now a handy little formula which allows you to compute any logarithmic function from loglO lOEioW 628 logb 10g10b This formula is somewhat tedious to deduce so we7ll skip this here10 Incidentally the same formula also works with e in place of 10 ie we also have logb Note that formula 628 is exactly what you7re looking for with a calculator which can handle logo The right hand side only in volves computing logl0 and division Eg here is how we can com pute log264 the hard way using a calculator log1064 N 1806 7 6 log102 N 0301 7 We will see applications of the logarithmic functions in subsec tion 642 where the use of a calculator and thus the use of for mula 628 is inevitable But before we head into applications of the logarithmic functions lets look at some properties of these functions We summarize them in the following proposition which is the counterpart of Proposition 63 for the exponential functions Figure 64 shows the graphs of some logarithmic functions and of the corresponding exponential functions 10g264 9Some calculators have only one button for logarithmic functions typically called log and now you donlt necessarily know which of the two logarithmic func tions above that means so be carefull You can test them as follows If log10 computes as l on your calculator then you have loglo since 101 10 If it com putes as approximately 23 then you have loge since e23 m 272393 m 10 10But if you insist here is the deduction First of all z 0351 l 101031002 ogbw as well as I 101 310 by de nition of log and loglo respectively By 612 this gives 101 giob391 gb1 101 310b10gb 10103100 Since exp10 is 171 we obtain log10b logbz log10 and thus logbz 64 LOGARITHMIC FUNCTIONS 141 5 h FIGURE 64 The graphs of two exponential functions and the corresponding logarithmic functions note that for xed base b7 the graphs of the exponential and the loga rithrnic functions to base b are the re ections of each other along the main diagonal77 y x as mentioned in subsection 423 In order to better distinguish the graphs of the various functions in Figure 647 we have drawn some of them with dotted lines but note that in reality7 they should all be drawn with solid lines Proposition 64 For any base b gt 0 with b 7 1 the logarithmic function log7 has the following properties I The logarithmic function logbp is de ned for all positive real numbers x ie the domain of logb is the set 000 of all positive real numbers 2 The graph crosses the p accis at the point 01 142 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 3 Ifb gt1 then a the logarithmic function logbz increases as x increases ie if 0 lt 1 then logbx0 lt logbd1 b the graph increases quotwithout bound as x grows larger and larger and c the graph decreases quotwithout bound as x approaches 0 4 If0 ltblt 1 then a the logarithmic function logbd decreases as x increases ie if 0 lt 1 then logbx0 gt logbd1 b the graph decreases quotwithout bound as w grows larger and larger and c the graph increases quotwithout bound as x approaches 0 5 The logarithmic function logb is 171 6 The range of the logarithmic function logb is the set R of all real numbers In summary the logarithmic function log is a 171 function with do main 000 the set of all positive real numbers and range R The various parts of Proposition 64 can all be veri ed similarly to the way we veri ed Proposition 63 we7ll skip this here for brevity7s sake We nally note the following useful properties of the logarithmic functions which follow from properties of the corresponding exponen tial functions stated in Proposition 62 Proposition 65 The following logarithrnic rules hold for all positive real numbers a 31 1 all positive real numbers r and s and all real numbers y 629 low 5 loam lows 630 10mg log r e lows 631 10w 1 locum We will not prove these here but merely list them for referenceM 11They actually are not hard to show In each case we set r a and s ay Firstly logar s loga r logas holds since a ay awry Next loga logar 7 loga s holds since 7 ax y And finally loga ry y loga r holds since ay am 64 LOGARITHMIC FUNCTIONS 143 642 Applications of Logarithmic Functions Let7s now n ish our coverage of the logarithmic functions by looking at some appli cations The problem with these is not that they are particularly dif cult7 but that each is typically phrased quite differently7 which tends to confuse folks quite a bit 6421 The slide rule The slide rule was invented in the 1600s as a mechanical calculator and was in wide use until the 19707s7 when it was quickly crowded out by the modern pocket calculator 11 2x36 w m H quot le Bil l 390910 2 3909103 lag1O 6 1 FIGURE 65 A slide rule The basic idea of the slide rule is to take advantage of the log arithmic rules 629 and 630 in Proposition 65 in order to make approximate manual computations easier by converting multiplica tion to graphical addition 7 and division to graphical subtraction 12 Figure 65 shows a basic slide rule with a magnifying sliding cursor7 allowing you to better align numbers on the scales we will mainly use the middle two scales of it called scales C and D on our slide rule for now Note that these two scales are sliding number lines7 each from 1 to 107 but not drawn to scale like a usual number line7 but such that each number x between 1 and 10 is drawn at a distance log10x to the right of 1 Note7 therefore7 that the distance from 1 to 4 is twice the distance from 1 to 2 since log104 log1022 2log102 by 6317 and similarly the distance from 1 to 8 is three times the distance from 1 12This short subsection is not intended as a fullblown guide to using a slide rule7 but rather to give you the basic idea as to how and why a slide rule works You can practice with a virtual slide rule on your computer screen at the web site http wwwantiquark comsliderulesimn909esvirtua1n909es html 144 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS to 2 since log108 log1023 3log102 by 631 You can check this against scale L at the very bottom which is the usual number line from 00 to 10 its numbers represent the base 10 logarithms of the numbers on scale D so log101 0 log102 03 log104 06 log108 09 and log1010 1 To illustrate the use of the slide rule lets start with a simple mul tiplication problem Suppose you want to multiply 2 by 3 The slide rule would accomplish this in the following steps see Figure 65 1 Convert 2 and 3 to log102 and log103 represented by the length 2 on scale D and the length 3 on scale C respectively 2 Add these two lengths and read off their sum as the length log102 log103 log102 3 log106 on scale D us ing 629 3 Convert lOg106 back to 1010g106 6 Similarly in order to divide 6 by 3 say we would proceed as follows again refer to Figure 65 1 Convert 6 and 3 to log106 and log103 represented by the length 6 on scale D and the length 3 on scale C respectively 2 Subtract these two lengths and read off their difference as the length log106 7 log103 log10g log102 on scale D using 630 3 Convert lOg102 back to 1010g102 2 There are many other computations which can be performed on a slide rule but since pocket calculators are nowadays a much more ef cient way to perform the same computations we7ll leave it at that 6422 Radioactive Decay The next application concerns radioac tive materials As radioactive material decays its mass shrinks by a xed amount per time period which is proportional to the amount left Mathematically this is similar to the balance of an interest bearing savings account which grows by a xed amount per time period as we discussed in subsection 632 except that the amount of radioactive material decreases whereas the balance of an interest bearing savings account increases Now just to confuse you the decay of a radioactive material is typically described by giving its half life namely the time period it takes for the amount of material to decrease to half of its original amount Eg the half life of plutonium is approximately 24000 years Table 64 shows the amount left after multiples of 24000 years starting with 1kg of plutonium As you can see the mass of the plutonium shrinks in half every 24000 years as the name half life77 indicates 64 LOGARITHMIC FUNCTIONS 145 time years 0 24000 48000 72000 96000 t 1 24h00 mass kg 1 05 025 0125 00625 7 2 TABLE 64 Radioactive Decay Approximate mass of plutonium left starting with 1kg But suppose we want to describe the amount of plutonium left af ter only 1000 years Well after t many years 24 000 many multiples of the half life period 24000 years will have passed so there will be t 1 W many kg of plutonium left So in particular after 1000 1 2345300000 1 714 years there will be 9715kg 9175g of pluto nium left Now the logarithmic function also allows us to solve the inverse problem Suppose I want to know after how many years there will be 100g of plutonium left This then amounts to solving the equation 1 24h00 632 E 01 which can be undone as t 7 1 01 24 000 Og lt giving t 24 000 log 01 log01 log05 71 70301 m 24000 332 m 79700 24 000 by 628 m 24000 So it will take approximately 79700 years for 90 of the plutonium to decay and thus only 10 of it to be left 6423 Earthquakes and the Richter Scale Let7s look another ap plication of logarithms which at rst looks quite different the Richter scale for earthquakes This is done as follows One rst measures the 146 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS maximum amplitude A ie how far the ground swings from its rest ing position at a distance of 100km from the epicenter13 the amplitude is measured in am read micrometer which is equal to 0001mm or 0000001m The problem with amplitude is that one has to deal with very large numbers The amplitude of a barely noticeable earthquake is about 100nm the amplitude of a very destructive earthquake is as high as 100000000um 100m or more As you can see its not very practical to use the amplitude so one uses the decadic logarithm of the am plitude called the Richter magnitude or more correctly called local magnitude 633 ML log10A Thus a barely noticeable earthquake of amplitude 100nm has Richter magnitude ML log10100 log10102 2 On the other hand a very destructive earthquake of amplitude 100000000um has Richter magnitude ML log10100 000000 log10108 8 So as long as the amplitude is just a power of 10 the Richter magnitude will be a whole number namely the number of 07s in A when A is ex pressed in am and the Richter scale was chosen to make the Richter magnitude a number roughly between 0 and 10 One more quirk If you have an earthquake of Richter magnitude M0 say then an earth quake of Richter magnitude M0 1 has ten times the amplitude This is because the rst earthquake has amplitude A0 say where M0 log10A0 or A0 10 so the earthquake of Richter mag nitude M0 1 has amplitude A1 say where M0 1 log10A1 or A1 101l1quotJ l 10 10MO 10 A0 This explains why an earthquake of Richter magnitude M0 1 is generally much more destructive than an earthquake of Richter magnitude M0 6424 Sound Volume and Decibels The decadic logarithm ap pears similarly in the measurement of sound volume or sound inten sity One rst expresses volume in terms of power P formally de ned in physics as energy per time unit where the volume or power is expressed in watts abbreviated Again the problem is that com monly occurring sound volumes can range from 1W barely audible to 1000000000000 W one trillion watts where ear damage occurs even during short exposure So it is convenient to de ne a more practical 13There are formulas which one can use to convert the amplitude measured at a different distance from the epicenter to the amplitude one would have measured at a distance of lOOkm we will not worry about this here 64 LOGARITHMIC FUNCTIONS 147 measure of volume or sound intensity as LdB 10 log10P When expressed as LdB the unit of volume is decibelsM Using the above examples we then see that a barely audible sound of power 1W has volume LdB 10 log101 10 l0910100 decibels whereas a damaging sound of power 1000000000000 W has volume LdB 10 log101000000 000000 10 log101012 10 12 120 decibels 14There is also a unit called bel which equals 10 decibels and is now rarely used It was de ned as log10Pi In a sense bels are more similar to the de nition of the Richter magnitude but for some reason it is now common practice to use the unit decibels Bibliography 1 Gross Herbert ll Algebra by Example An Elementary Course Di Ci Heath amp Col Lexington Mass 1978 out of print 2 Milgram Ri James The Mathematics PreService Teachers Need to Know available online at ftp math stanfordedupubpapersmi1gramFIEbookpdf 2005 es pecially Chapter 8 Discussion of lssues in the Algebra Course Papick lra Ji Algebra Connections Mathematics for Middle School Teachers Prentice Hall Upper Saddle River NJ 2007 Parker Thomas Hi and Baldridge Scott J Elementary Mathematics for Teachers Sefton Ash Publishing Okernos Mich 2004 Szydlik Jennifer and Koker John Big Ideas in Mathematics for Future Mid dle Grades Teachers Big Ideas in Algebra University of WisconsinOshkosh preprinti Wu HungHsi Introduction to School Algebra DRAFT available online at http mathberke1ey edu wuA1gebrasummary pdf 2005 EE E E 148 Index Abel Niels Henrik vii absolute value 86 additive change 130 additive identity law of the 8 additive inverse 5 additive inverse law of the 8 Al Kitab al Jabr walMuqabala vi algebraic expression 12 evaluating 12 identical 13 negative 66 nonzero 45 positive 65 rewriting 12 algorithm vi argument of a function 74 associative law of addition 8 associative law of multiplication 8 bacterial growth 128 base 124 binomial laws 98 Cardano Gerolamo vii ceiling function 83 change additive 130 multiplicative 130 codomain of a function 74 coef cient 102 common logarithm 140 commutative law of addition 8 commutative law of multiplication completing the square 103 complex numbers 16 112 compounding 136 149 constant 11 converse of a statement 34 coordinate 30 coordinate axis 30 coordinate system 30 cube of a number 124 cubic equation vii decadic logarithm 140 decibels 147 decimal nonrepeating 16 repeating 16 del Ferro Scipione vii difference of sets 17 difference function rst 93 second 95 direct variation 89 direction of velocity 23 discriminant 105 distance on the number line 88 distributive law 8 domain of a function 73 74 e Euler s number 135 earthquake 146 elimination method 54 empty set 16 equality properties of 61 equation equivalent 43 Eulerls number 135 exponent 12 exponential equation 131 138 exponential function 131 exponential rules 128

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