Algebra 1 - Herman Chernoff

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Algebra I for Students Comfortable with Arithmetic

c2001 Herman Chernoff

February 6, 2002


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Chapter 1

Introduction

Why should you learn Algebra I? I could elaborate on many reasons whya knowledge of Mathematics beyond Arithmetic is of interest and of valueto people in the twenty-first century, but for the sake of brevity, I wont.It is enough to say that a high school curriculum demands it. What arethe basic concepts to master? They are the ability to translate quantitativeproblems of the real world into symbolic statements and to manipulate thosestatements to arrive at satisfactory answers.

As the title indicates, this book is written for students who are comfort-able with Arithmetic. One problem with current books on Algebra I, is thatin order to be adopted by state agencies they must be suitable for students

of all levels of attainment, including some who are still unsure of themselveswith simple problems in Arithmetic. As a result, these books tend to befull of detail and very large and heavy. To encourage the practice that isrequired to fix the basic ideas, the students are asked to do many exercises,which are essentially repetitions of the text, and not very challenging.

Assuming that the readers of this book are quite comfortable with Arith-metic, and willing to do some experimenting, permits bypassing some of thetedious details in text and exercises. This does not mean that the studentcan avoid the necessary practice. It is hoped that he or she will get thatpractice in working on more demanding problems. The student should beprepared to use lots of scratch paper and graph paper in reading this text

and in doing the problems. The custom will be to have problem sets withthree different types of problems. There will be exercises. labeled Ex, to testwhether the reader has understood the text well enough to solve a slightlymodified variation of the material in the text. There will be problems, la-beled Pr, which demand a more comprehensive knowledge of the past text

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4 CHAPTER 1. INTRODUCTION

and some imagination. A few of them are quite difficult. There are not

many problems, and the reader reader should try all of them. If one seemstoo difficult or complicated, he or she should not get hung up on it, but goon with the book after a valiant try. It might be rewarding to return toit after completing the text. Then there will be some puzzles, labeled Pu,which are almost impossible without an illuminating insight.

The object of the problems is to provide an opportunity to experimentwith alternative approaches. It is expected that this experimentation willprovide the necessary practice to develop the basic underlying techniquesof the subject without boring repetition. On the other hand one shouldbe prepared to deal with getting wrong answers. Perhaps more importantthan doing the problems correctly in the first place, is the task of finding outwhere one went wrong, how to correct for it and to avoid similar errors later.The ability to recognize when an answer is wrong by testing it in variousways is also a useful talent. Finally, even when an answer is correct, areview of the work will often indicate that the reasoning went in unnecessarycircles instead of a direct straight line. The task of refining a solution tocut out unnecessary digressions often leads to a better understanding of theessentials of the argument. Incidentally, mathematicians tend to refine theirwork so much that the readers lose the excitement involved in tracking thetrail of the hunt, where strange ideas led to hints of how to proceed next toget to the ultimate objective.

There are not many figures in this text, but the reader should be preparedto use up lots of graph paper, drawing many figures to help his understanding

of the text, examples and problems.Chapter 10 is a brief summary of the main ideas of the book. In Chapter

11, we shall present the answers to the exercises and problems. Some ofthem will be detailed solutions of the problems. Such solutions are notunique, and if the reader has another approach, that will not necessarilybe better or worse. The solutions presented will usually aim to be concise.That does not necessarily mean they will be as insightful as they could be.Finally there will be another section with hints for the puzzles. While itsometimes took me a couple of days to solve some of these puzzles, theyusually became rather easy once the hint occurred to me. I have a Ph.D.in Applied Mathematics and have been a professor of Mathematics and of

Statistics, so dont be surprised if some of them defy your efforts withoutthe hint.

In summary, this should be a short book, freely available on the worldwide web, and its mastery should not take more than a couple of months bythe student willing to spend about 4 hours a week. It would be appreciated


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if you spread the news about the availability of the book to friends near and

far. It is a good idea to collaborate with one or two others of comparableability. You should not be in the position where your collaborator doesall the heavy lifting or where you do all of it. That would put one ofyou into too passive a mode to get what you should out of the work. Anoccasional hint from a better trained friend is ok, but major dependencewill be counter productive. In the near future you may write to me withcomplaints, suggestions for improvement (which will become my property),and questions. I plan to spend a couple of hours a week on dealing withsuch comments. My e-mail address is [email protected] and I hopeto use the comments to improve the book. In the event that this bookbecomes very popular, I might plan to publish it as a small soft cover textto be sold at a modest price for students who dont want to download itfrom the computer.

Finally, a warning. This book is not appropriate for students who arenot comfortable with Arithmetic. On the other hand, the fact that a studenthas the ability to master Algebra I does not automatically imply that thisbook will be right for him or her. The approach that brings out the best ina student differs very much from one student to another. It is possible thata reasonably gifted student will not find this presentation suitable. If I knewhow to describe the characteristics that determine whether a student will findthis text useful, I would do so, but I dont. I do feel that the willingness tomake some effort and to learn from ones mistakes are important ingredientsfor a successful use of this presentation.


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6 CHAPTER 1. INTRODUCTION


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Chapter 2

Arithmetic

2.1 Introduction

How well do we understand Arithmetic? With the availability of electroniccalculators, we no longer need to be able to add, subtract, multiply, or divide.On the other hand, the ability to perform these operations is closely relatedto a basic understanding. Moreover the ability to perform these operationsaccurately and quickly requires a knowledge of the 10 10 addition andmultiplication tables.

I dont recommend feats of memory in this text. Memorizing rules forsolving problems is usually a way of avoiding understanding. Without un-

derstanding, great feats of memory are required to handle a limited class ofproblems, and there is no ability to handle new types of problems. However,memorizing the multiplication and addition tables should not be a majortask, and studying them can even be informative. I strongly recommend it,although I must admit that a friend who had a Ph.D. in Physics had to usehis fingers to add 8 and 5.

In this chapter we review Arithmetic and emphasize a couple of pointsthat are so well accepted that they are possibly not as well appreciated asthey should be. For example the fact that 35 = 53 is seldom questioned,but perhaps it should be, because it means that 5+5+5 = 3+3+3+3+3 andwhy is that obviously the case? The usual demonstration is to construct a

rectangle of 3 lines of five points each to represent 3 5. Turning thisrectangle around, we have 5 lines of 3 points each without changing thetotal number of points. See Figure 2.1.1.

Readers who find this chapter difficult are probably not ready for therest of this text. Incidentally, the example discussed above is an example

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8 CHAPTER 2. ARITHMETIC

of a general rule that goes under the name of the commutative property of

multiplication. To forget this name is no disaster as long as you can applythe principle. Chapter 10 contains a list of such principles, major results,and notation for reference when the text uses them. I doubt that you willhave many occasions to use this list.

Figure 2.1.1

2.2 Integers

The basic rules for Arithmetic with integers (whole numbers) are taught

in the early grades. The reader should test his ability to add a list of 4three digit numbers with many large digits. Then subtraction of severalexamples of 6 digit numbers, also with some large digits, should be tried.Then multiplication of a couple of 3 digit numbers should be tried. Finallywe are ready for long division. Multiply 439 by 6347 to get a product. Nowsee if dividing the product by 439 retrieves 6347. You can do several otherexamples, using a calculator to save time in checking your reults.

If you have trouble at this stage, it is likely that this text is not appro-priate for you.

2.3 Fractions

If a lawyer wants to divide 7 cents among 3 people, he will have a problem.The integer 7 is not evenly divisible by 3, and he will have a remainder of 1which he can dispose of by claiming it as his fee. Fat chance that a lawyerwould be satisfied with that fee. On the other hand if a cook has 7 pies to


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2.3. FRACTIONS 9

be divided among three diners, he can give each two pies, and then divide

the remaining one into 3 equal parts.Each of the small parts of the seventh pie is called one third of the pie.Two of these parts would be called two thirds of the pie. If each of the sevenpies were divided into thirds, there would be 21 thirds, and each of the dinerswould have seven thirds. We can regard six thirds as the equivalent of twowhole pies. In Arithmetic we write the solution of the cooks problem as 7/3or 2 1/3. The first form is a representation of the solution in fractional formwhich is given by one integer over another. Once we introduce fractions,then we have a solution of the problem of dividing one integer, the dividendor numerator by another, the divisor, or demominator even when the firstinteger is not an exact multiple of the second.

Having introduced fractions, we have to consider the arithmetic of frac-tions. First we should notice that the fraction 7/3 is no different than 14/6.In other words, if the pies were all cut into six equal pieces of 1/6 each, therewould be 6 7 = 42 pieces which would divide among 3 people equally ifeach took 14 pieces, i.e. 14/6. Thus a given fraction can be represented inmany ways where two representations are equivalent. For example 14/6 and21/9 are equal to each other. Essentially a fraction does not change value ifboth numerator and denominator are multiplied by the same integer. Both14/6 and 21/9 can be obtained from 7/3 in this way.

It is easy to add 4/3 and 7/3 to get 11/3. But, to add 4/3 and 7/5 isnot so easy. On the other hand 4/3 = 20/15, multiplying numerator anddenominator by 5, and 7/5 = 21/15. Now the two fractions have a common

denominator, and we can add them to get 41/15 = 2 11/15. To add 5/6 and11/4 we can convert these fractions to 20/24 and 66/24 resulting in a sumof 86/24 = 43/12. Here we multiplied the numerator and denominator ofeach fraction by the denominator of the other fraction, thereby obtaining, ascommon denominator, the product of the two denominators. This operationcan be unnecessarily tedious, involving huge denominators if we have to addseveral fractions. That effort can be reduced somewhat by noticing that 6factors into 2 3 and 4 into 2 2. We can get a common denominator bytaking 2 2 3. Then we can multiply the numerator and denominator of5/6 by 2 and those of 11/4 by 3, getting 10/12 + 33/12 = 43/12. In thisexample 12 is the least common denominator. While it is not necessary to

get the least common denominator, it is often a good idea to get as small acommon denominator as is readily available. The same appproach works insubtracting fractions.

How about multiplication? It is no surprise that 27/3 is equal to 14/3.Nor should it be surprising that 1/3 of 7/2 is 7/6. After all 7/2 = 21/6 and


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one third of 21 is 7. In short, to multiply two fractions, we multiply the

numerators and the denominators. Often the result can be simplified bydividing the numerator and denominator by a common factor. For example3/7 2/9 = 6/63 = 2/21. It is customary to take advantage offactoringbe-fore doing all the multiplications. Factoring the 9 in the second denominatorinto 3 3 permits us to cancel the 3 in the first numerator with one of thethrees in the seond denominator. While this device is trivial in this example,it can result in substantial time savings in more complicated examples.

It remains to deal with the division of one fraction by another. To divide3/4 by 2/5, the rule is to multiply 3/4 by 5/2. In other words we multiplythe dividend by the fraction obtained from the divisor by interchanging thenumerator and denominator of the divisor. What is the rationale for this?Division is called the inverse operation of multiplication just as subtractionis the inverse operation of addition. When we divide 18 by 2, we essentiallyask what multiplied by 2 will give us 18. What multiplied by 2/5 willgive us 3/4? Here we have 2/5 (5/2 3/4) will give us 3/4 because the2/5 5/2 = 1. Clearly the rule presented works.

To indicate a mastery of the arithmetic of fractions, you should be ableto do the following exercise.

Ex 2.3.1 Express 23 +415 310 divided by 73 + 512 + 118 as a fraction.

Pu 2.3.2. How do two boys divide a pie fairly so that no one could blameanyone but himself if he gets less than his fair share? The standard answeris to have one split the pie and the other to choose which part he prefers.It isnt so easy to find a resolution for this problem when there are three ormore boys to divide the pie. How do 3 or more boys divide the pie?

2.4 Real Numbers

In Algebra we shall deal with real numbers. The set of real numbers corre-sponds to the distances along a horizontal line, where one point labeled 0 isthe origin. See Figure 2.4.1. A length of one is marked off to the right of0. It corresponds to the integer one or the fraction 1/1. An equal distanceto the right of that point is another corresponding to 2 = 2 /1. The positiveintegers 3,4, etc. correspond to the points which are three, and four times as

far from 0 as is 1, etc. The negative integers correspond to equally distantpoints to the left of 0. The rational numbers are the fractions with integernumerators and denominators, and correspond to some of the points on theline. For example 1/3 is the point one third of the way from 0 to 1. Similarly1/3 corresponds to the point one third os the way from 0 to -1.


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2.4. REAL NUMBERS 11

Why have we introduced negative numbers and how do we ordinarily

interpret them? Of course we can use the interpretation of position alongthe line. But there are other rationales for introducing negative numbers.Suppose that we count how much money a man has. If he has $100 and buysan item for $200, then he is in debt $100 and it makes sense to say that hehas $100. Then we might reasonably say that 30 + 40 200 + 50 = 80 isthe amount a gambler won at the races if his profits were 30,40 and 50 andhis loss was 200. The result -80 would indicate a resulting profit of -80, orequivalently, a loss of 80. One way to do the arithmetic of adding positiveand negative numbers is to add all the positive entries and all the negativeentries separately, and then to subtract the sum of the negatives from thesum of the positives. Another interpretation of the above sum is that onemoves a point 30 units to the right of the origin, then 40 more to the right,then 200 units to the left, and finally 50 units to the right, resulting in amovement of 80 units to the left.

Figure 2.4.1

o o

-3 -2 -1 0 1 2 3 4

Let me explain the rationale for the following rule. The product of two

positive or two negative numbers is positive. The product of a positive anda negative number is negative.

It is sensible to regard 2 times -3 or 2 3 = 6 using the aboveinterpretations of negative numbers. But what sort of interpretation wouldmake sense of3 2? Consider a sheet of graph paper which is a page with


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a grid of equally spaced and intersecting horizontal and vertical lines. See

Figure 2.4.2. Mark one of the intersections as the origin. For every pointon the sheet, we have a horizontal and vertical distance to the origin. Thehorizontal distance is given a sign of plus or minus depending on whether it isto the right or left of the origin and is called the abscissaor, more commonly,the x-value. The vertical distance is given a sign of plus or minus dependingon whether it is above or below the origin, and is called the ordinate or,more commonly, the y-value. Take the line which goes through the originand rises 2 units for every unit to the right of the origin. When the x-valueis 3 the y-value is 3 times 2 or 3 2 = 6. When the x-value is 2 the y-valueis 2 2 = 4. When the x-value is 1, the y-value is 1 2 = 2, and when thex-value is 0, the y-value is 0 2 = 0. When the x-value is -1, the y-valueis -2 which can be reasonably interpreted as

1

2. Similarly when the

x-value is -3, the y-value can be regarded as 3 2 = 6. this is especiallycomforting since it implies that 32 = 23 = 6, and the commutativelaw of Section 2.1 applies in this case involving a negative number.

Figure 2.4.2

-3 -2 -1 0 1 2 3

-6

-4

-2

0

2

4

6

x

y

How about the product of two negative numbers? Take the line whichgoes through the origin and goes down 2 units for every unit to the right ofthe origin. When the x-value is 3, the y-value will be -6. For this interpreta-tion it makes sense to regard 3 2 = 6. Considering what happens to the


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2.5. DECIMALS 13

second line when the x-value is -3, it also makes sense to regard 32 = 6.Thus multiplying a negative number by a positive number results in a neg-ative product, and multiplying two negative numbers results in a positiveproduct.

While we have shown where the integers and fractions appear on thehorizontal line, one may wonder whether these represent all the points onthe line. It is apparent that we can find fractions or rational numbers inevery interval on the real line, no matter how small that interval is. Howeverwe shall later show that these do not cover all points. In fact the square rootof 2, the number which when multiplied by itself yields 2, is not one of therational values represented by a fraction with an integer numerator and aninteger denominator.

2.5 Decimals

The Romans were outstanding engineers, many of whose accomplishmentscontinue to amaze us after 2,000 years. Brilliant engineers are not stoppedby a poor system of counting, but such a system can slow down progress, andit can frustrate some of the less brilliant practitioners. The Roman numbersystem did not make arithmetic easy. Until recently, the English countingsystem for money was unnecessarily complicated. A shilling was 12 pence,and a pound was 20 shillings and a guinea was 21 shillings. This systemmade sense for specialists where, for example, a dozen eggs at three pence

an egg was obviously 3 shillings. It did not make arithmetic easy, and theEnglish have since gone to a decimal system for money. The Americans stilluse a non-decimal system for weights and distances, and the whole worldcontinues to use a non-decimal system for time. Most of the civilized worldis moving toward decimal systems for money, weights (grams), and distances(meters). Thus a thousand grams is a kilogram and a thousand meters akilometer.

While the decimal system is not essential, it is convenient and doesfacilitate the use of arithmetic in many practical situations. So we shall saya few words about decimals. The amount $20.64 represents twenty dollarsand 64 cents. Actually a cent stands for one hundreth of a dollar. One way

to look at the amount above is to regard it as 2,064 cents or hundreths of adollar. Similarly 3.476 may be regarded as the fraction 3, 476/1, 000. Whendoing arithmetic with decimals, we are essentially dealing with our familiarintegers, except that the units may be tenths or hundreths or even smallerportions of one. Any real number can be arbitrarily well approximated by a


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decimal with enough positions after the decimal point. Interestingly enough

some of the simple rationals, as for example, 1/3 can not be representedexactly by a decimal with a finite number of positions after the decimalpoint.

I shall assume that the reader has been properly trained in the arithmeticof decimals and needs no further explanations.Ex 2.5.1 (a) Add 2.6; 9.35; 15.16; -22.1; -3.124.(b) Subtract -3.124 from -2.1.(c) Multiply 26.2 by -3.71; -3.124 by -2.1.(d) Divide 1,864.874 by 63.8. Carry 3 places after the decimal point.


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Chapter 3

Algebra

3.1 Two Examples

The main point of this text is to demonstrate how the essentials of Algebra 1consists of translating problems into algebraic expressions and manipulatingthose expressions to derive meaningful answers to the questions raised. Weillustrate with two examples, one of which is very simple, and the otherwhich would be difficult if it were not the case that the answers were integersand could be found by trial and error. In the discussion we will see how aconvenient notation is used before it is completely explained.

The solution of the first example is made relatively easy by the use of

Algebra, although it could be solved without the explicit use of that tool.Example 3.1.1 Simon is upset because he invested some money in a stockand lost $100. He doesnt mind so much the money loss, but feels that therewas some funny business. The stock lost 20% in the first month. But thenit gained 20% in the second month, when he needed the cash and had to sellout. Under these circumstances he feels that he should have come out even,(not counting the brokers charges of $20). But he lost $100. My problemis to find out what was the original value of the stock that he bought.

To understand why he lost $100, forget the brokers charges and supposethat Simon had invested $1,000. When the stock lost 20%, the value wentdown to $800. When it regained 20%, that 20% of 800 or 160, increased

the value to $960 and he lost $40 because 20% of the reduced amount isless that 20% of the original amount. Thus for every $1,000 invested he lost$40. If he had invested 100/40 = 2.5 times the $1,000, or $2,500 we wouldhave accounted for the loss of $100, not counting the brokers fee of $20. Toaccount for that we must assume that the stock only lost $80 and that the

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16 CHAPTER 3. ALGEBRA

original value of the stock purchased was $2,000.

This problem is not difficult, but slightly involved. It would become moredifficult if the brokers fees were not fixed at $20, but used a rate dependingon the amount being traded.

How would we deal with this problem using Algebra? The main toolis to represent an unknown quantity by a symbol, typically a letter of thealphabet, and to do the arithmetic with that symbol. We illustrate. In thiscase the unknown quantity, the original value of the stock, (in dollars), isdesignated by x. When its value is decreased by 20%, its value shrinks to80% of x represented by .8 x. We find it convenient to suppress the timessign when there is no ambiguity, and we have the reduced value of .8x, whichwhen increased by 20% becomes (1.2).8x. Here we have used parenthesesabout 1.2 to separate 1.2 from .8 and to avoid the ambiguity of 1.2.8. Weexpress the statement that the loss of $100 is the original value minus thefinal value plus the fee by the equation

x (1.2).8x + 20 = 100

or

x .96x + 20 = 100.04x + 20 = 100

.04x = 80

x = 80/.04

x = 2, 000.

In ordinary usage these 6 steps would have been abbreviated consider-ably and I would have written, after the first equation, the third and sixthequations. For people familiar with algebraic manipulations, it is easy toskip many steps, but I notice, as I grow older, that skipping leads to errors.

Let me review these steps in detail. The first step, that of writing downthe first equation, is fundamental, and involves translating the basic factsor statement into an equation. The remainder consists of algebraic manip-ulations of the underlying equation to yield an expression for the desiredquantity.

In the second equation, we converted (1.2).8x to .96x. In the thirdequation, we used the fact that x .96x is .04x. In the next equation weeliminated the 20 on the left hand side of the equation by subtracting 20from both sides. This equation tells us that a multiple of x is 80, in whichcase x is 80 divided by that multiple. In summary, we solved the original


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3.1. TWO EXAMPLES 17

equation by combining all multiples of x on one side of the equation and the

other quantities on the other side.Some of these steps involve notions which are a little more sophisticatedthan may seem necessary to the typical student who is used to doing arith-metic. These are explained later.

Now I would like to introduce a more complicated example.Example 3.1.2 Bill and John are two men, whose ages sum up to 91. Bill istwice as old as John was when Bill was as old as John is now. How old arethey now?

Let b and j stand for their current ages, and b and j for their formerages. Then

b + j = 91

b

= jb = 2j.

are obvious equations to fit this problem. We wish to find b and j, butwe have only three equation involving four unknowns, and it is typicallydifficult to solve for more unknowns than we have equations. Is there ahidden relationship in our problem statement or is this problem unsolvable?

There is another relationship, not overtly stated, but implicit in thestatement. The time interval between now and then is the same for Bill andJohn. That means that

b b = j j.

Note that we could have introduced another unknown, t for time interval,and written the two equations b = b + t and j = j + t. If we did that, wewould have 5 equations in five unknowns. As it is, we have four equationsinvolving four unknowns which should be plenty.

One way to handle this example is to use each equation to eliminateone of the unknowns. We illustrate. Since b = j, we can eliminate thatequation and replace every b by j in the remaining equations. Thus thelast equation becomes b j = j j, which can be rewritten as b = 2j jby adding j on both sides. Substituting for b in the first and third of theoriginal equations, they now become

2j j + j = 912j j

= 2j

.Rewriting these two equations in j and j, we have

3j j = 912j = 3j.


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The last equation can be written j = (2/3)j. Susbstituting in the other

equation3j (2/3)j = 91

which leaves us with one equation and one unknown. This kind of reductioncould have been done using one equation at a time in a different order. Nowwe have

(7/3)j = 91

or

j =91

7/3= 91(3/7) = 39.

Using the first equation, we have b = 91

j = 52. These represent the

solution to our problem. As a matter of curiosity, what are b and j?Obviously b = j = 39 and j = (2/3)j = 26, and indeed, Bill is twice asold as John was, and also the elapsed time is t = b b = 13. Incidentally,it always pays to check out the solution to see that it satisfies the originalequations and the statement of the problem.

When I was a student, we were given certain types of special problemsand methods of solving these problems and expected to memorize thosemethods. Such a method is called an algorithm. In my opinion this may bea good policy for finding solutions to many examples of the same problemdiffering only in the input numbers, but it is a very poor way to foster theunderstanding which provides the flexibility to handle different variations of

these problems. There are very few algorithms worth memorizing. Thesedays they can be handled by computers, and putting numbers into a com-puter algorithm is usually a poor way to learn something worthwhile besidesthe answers to those specific problems.

Knowing that the answers to the last problem were integers, we couldhave avoided algebra and used trial and error. We could have tried everypossible value of b from 46 to 91 till we found out where b and j = 91 bsolved the problem. But if we were not told that the answers were integers,that method would not work, although we could develop some idea for whatwould constitute an approximate answer to the problem. A clever use ofgraph paper could be helpful.

Pr 3.1.1 Suppose that in the first example, the broker charged 1% as thetransaction cost for buying and for selling the stock. If the original purchasewas for $2,000 worth of stock, he would get $20 for that purchase and more,later, when he sold the stock. Granted that the stock went down 20% andthen came up 20%, and Simon lost $200 counting the brokers fees, what


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3.2. NOTATION 19

was the value of the stock he originally bought? (Not how much he spent

originally.)Pr 3.1.2 It takes Bill 3 days to clear an acre. It takes John 5 days to dothe same task. How long will it take for the two, working together at theirnatural rates, to clear 2 acres? Note: rate time = amount of work.Ex 3.1.3 In celebration of Johns birthday he receives a large gift and hisbrother Bill gets a smaller one. The two boys are naughty and play cardsfor money with Harry. Together Bill and John start out with $100. Aftertwo hours, John has lost half his money and Bill has doubled his, and nowthey have the same amount. What was the size of Johns gift?

Ex 3.1.4 John has $10 more than Bill. He earns $2 an hour and Bill earns$10 per hour. In 5 hours, Bill has twice as much as John. How much do

they have then?

3.2 Notation

Up to now we have illustrated how the elementary symbolic methods ofAlgebra can be used to solve problems that may be difficult to handle with-out these methods. In the process we have cut a few corners in the hopethat the exposition was elementary enough to be understood without beingcompletely precise. We now take the opportunity to go into some of theneglected details in notation and basic laws.

First we can use symbols, for example alphabetic letters, to represent

real numbers. Sometimes we may use subscripts as in x1 and x2 to representdifferent numbers which are similar in some sense. It is customary, but notessential, to use the later symbols of the alphabet to represent unknown realnumbers or variables and earlier symbols such as a,b,c to represent knownnumbers which are temporarily unspecified.

Fortran is an early computer language which also uses symbols to repre-sent real numbers. The rules for Fortran are different from those for Algebrabecause Fortran was compelled to use the characters available on a keyboardwith limited ability to use different sizes and subscripts. While a human canlive with a certain amount of ambiguity, where context makes things clear,the computer was not so flexible in the early days. On the other hand the

human doing Algebra requires displays which are easily interpreted or rela-tively transparent to the observer, whereas the computer, as distinct fromthe computer programmer, is not much concerned with such visual trans-parency. This digression may help explain some of the following text.

In Fortran, computational efficiency made it valuable to distinguish be-


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tween integers and real numbers. If a symbol started with an i,j,k,l,m,n

it was supposed to represent an integer, and otherwise a real number, unlessthere was a declaration to the contrary. In Algebra, there is no such rule,but the symbols i,j,k,l,m,n are often reserved for integers.

In Fortran, multiplication is represented by an asterisk which is readilyavailable on the keyboard. In Algebra, visualization is frequently enhancedby omitting the multiplication sign, which can also be represented by themid-level dot, as in xy = x y. Parentheses are used in both Fortran andAlgebra to group real numbers represented by a complex process, such asin x(y + z). Whereas Fortran would use parentheses within parenthesesfor something like x(y + (z + w)), the writing in Algebra is liable to usebrackets and braces or parentheses of different sizes to enhance the readersunderstanding of what is going on, for example, x[y + (z + w)]. Fortran cannot do that because the keyboard does not allow different size parentheses,and brackets and braces have special meanings in Fortran.

When we represent a relationship between x and y as in the graph ofthe line in Figure 2.4.2, we are dealing with a function. We say that y =f(x) = 2x + 3 represents y as a function f ofx since this equation assigns avalue of y for each of a set of values of x. This introduces a certain amountof ambiguity in our notation, for it could be interpreted as the product of fand x. Unfortunately, this notational ambiguity is prevalent, and we shallhave to learn to live with it, and to make some declaration where the contextfails to make the proper interpretation quite obvious. Note that the abovefunction is a special case of y = ax + b where a and b are specified to be 2

and 3 respectively.We will use |x| to indicate the absolute value of x , that is the distance

from x to 0. Thus |4| = |4| = 4. The exponentialnotation uses superscriptsto express xn as the product of x with itself n times where n is a positiveinteger. We will expand on this concept in Chapter 4.

We often wish to use inequalities. Here the useful symbols are >, ,


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3.3. BASIC LAWS 21

3.3 Basic Laws

Real numbers have some properties inherited from the integers and fractions,that are so common that they are ordinarily used without special attentionpaid to them. The commutative laws state that for all real x and y,

x + y = y + x

andxy = yx.

Thus the order in which you add or multiply two reals is unimportant.Implicit in these laws is the fact that addition and multiplication are dyadicoperations involving two elements. When we add three terms, x, y and z,

it is, in principle, necessary to state the order in which the additions takeplace. Thus (x + y) + z would use the parentheses to indicate that first weadd x and y, and that sum is then added to z. The parentheses are used toseparate units that must be defined or calculated before further operationstake place.

Abstractly, there is no prior reason for a dyadic operation to be suchthat (x y) z is the same as x (y z). However the associative lawsof addition and multiplication, two special dyadic operations, state that forall real numbers x, y and z,

(x + y) + z = x + (y + z)

and(xy)z = x(yz).

By combining the associative and commutative laws, it can be seenthat the order in which several items are added is irrelevant and that theparentheses are unnecessary in representing such a sum. For example,(x + y) + z = x + (z + y) = y + (z + x), and these can be repesented byx + y + z without danger of confusion. The same applies to multiplication.

It is often neglected to point out that changing the sign of a real numberis considered to be a unitary operation (involving one real number). Thusconverting 6 to -6 is such an operation. One consequence, typically ignored

because it is so simple is that 6 + (8) = 6 8. How are these two sidesconsidered so different that one must write such an equation? The left handside is the sum of two reals, one of which is negative. The right hand side isthe difference of two positive reals. That these are equal is often acceptedas too obvious to mention.


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Another important property of real numbers, that deserves mention is

the distributive law. This states that for all real x, y and z,

x(y + z) = xy + xz.

For example 3 times 5 + 3 times 8 is equal to 3 times 13. That does not seemto need much explanation for integers, and it holds true for real numberstoo. Nevertheless it is an important property, and it has been used in ourinformal treatment of the examples of Section 3.1. For example we claimedthat x .96x which is the same as 1 x .96x is equal to .04x, because,1 x .96 x = x 1 x .96 = x(1 .96) = x .04 = .04x.

The dyadic operations, multiplication and addition, have inverse oper-ations, subtraction and division. The difference y

x is that number z for

which x + z = y. Similarly, the quotient, z = y/x is that number z for whichxz = y. While subtraction always yields a unique solution, that is not thecase for division. If x = 0 and y = 0, there is no real number z for whichxz = y. Ifx = 0 and y = 0, then any real number z will do. For this reason,it is important to avoid dividing by zero. One can easily generate apparentparadoxes by using division by an expression which may be zero. We shallillustrate later.

Both addition and multiplication have identity elements. The real num-ber 0 is the identity for addition since x + 0 = x, and it has no influence onthe sum. Similarly the real number 1 is the identity for multiplication sincex

1 = x. The unitary inverse operations are negative,

x, for addition since

x + (x) = 0, and reciprocal, (1/x), for multiplication since x 1/x = 1 forx = 0. These convert a real number to the number which when added ormultiplied yields the identity element.

In other contexts, the properties mentioned here, or the lack of thembecome important. We often use these properties without paying muchattention to them, and when we need to point them out, we will do so.

We review these laws briefly, mentioning only that it is no crime to forgetthe names and that they can be extended in obvious ways to include morevariables.

x + y = y + x

xy = yx

(x + y) + z = x + (y + z)

(xy)z = x(yz)

x(y + z) = xy + xz


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3.3. BASIC LAWS 23

One application of these laws is

(x + y)(z + w) = (x + y)z + (x + y)w

= z(x + y) + w(x + y)

= (zx + zy) + (wx + wy)

= xz + yz + xw + yw

First of all, we used the custom of not repeating the left hand side in eachstep. Second it took us four steps to do what we would ordinarily do inone step. Finally, we actually skipped a few of the obvious steps in thefinal conclusion. Going from the next to last to the last step we used acombination of the associative and commutative properties that technically

might have required about 3 additional steps. These are trivial steps and wemay feel free to add or multiply any number of reals in any order we wishbecause of the laws mentioned.

Finally we present the apparent paradox formerly promised.

(x + 2)(x 2) = x2 + 2x 2x 4= x2 4

Now let x = 2. Then x2 = 2x, and x2 4 = 2x 4 = 2(x 2). It followsthat

(x + 2)(x

2) = 2(x

2).

Dividing both sides by x 2, we have x + 2 = 2 or x = 0. But x = 2. Howdid we get this contradiction? The simple explanation is that we dividedboth sides of the equation by x 2 which is equal to zero, and that is liableto give us any result.Pr 3.3.1 Use the commutative and associative laws to show that ( x+y)+z =y + (z + x).Pr 3.3.2 Show that (x + y)(x y) = x2 y2.Pr 3.3.3 Two dyadic operations are defined below. For each, does the opera-tion have the commutative property? If yes, show why. If no, give a counterexample.

(a)x y = x(y + 1).(b)x y = |x/y y/x| for positive x and y.Pr 3.3.4 Win.com has $100,000 in the bank. The amount it spends thismonth is $30,000, and is increasing by $5,000 per month. The incomeWin.com brings in this month is $10,000, and is increasing by $7,000 per


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month. Will Win.com go bankrupt? If not, when will it start making a

profit? If yes, how much more money does it need in the bank to survive?Pr 3.3.5 Conservation of energy is an important principle in Physics. Anobject of mass m and velocity v has a kinetic energy e which is half theproduct of the mass and the square of the velocity. Translate that statementinto an equation.

Pr 3.3.6 Each town in Massachusetts is expected to aim at the goal of achiev-ing 10% affordable housing. As long as the town has not achieved this goaldevelopers of apartments are entitled to special privileges if they devote25% of their new apartments to affordable housing. If a town currently hasa proportion p < .10 of affordable housing and a current housing stock ofx units, find a formula for the number of additional units, y, built by suchdevelopers, required to achieve the towns goal.

Pr 3.3.7 Casting out nines is a method of partially checking addition andmultiplication that used to be popular many years ago. For each positiveinteger i there are remainders r and r after i and the sum, s, of the digitsthat make up the decimal representation of i are divided by 9.(a) Show that if i1 i2 is divisible by 9, i1 and i2 have the same remaindersafter division by 9.(b) Show that i and s have the same remainder, and hence r = r.The method of casting out nines consists of checking the remainder ofi1+i2+ +in against that ofs1+s2+ +sn. For example 37+438+7853 = 8328.The remainders on the left are 1, 6, 5 whose sum is 12 with remainder 3.The remainder on the right is that of 8+3+2+8 = 21 which is also 3. The

same rule applies to multiplication.Show that if i1 and i2 have remainders r1 and r2,(c) The remainder of i1 + i2 is the same as that of r1 + r2, and(d) The remainder of i1i2 is the same as that of r1r2.

3.4 Inequalities

Dealing with inequalities can be a little tricky. The user of Algebra wouuldbe well advised to test his methods when working with inequalities by tryingthem on a variety of simple examples. I admit that I often do that instead

of relying on my memory for special variations.The basic rules can be summarized briefly. We say that x > y when

x y is positive, i.e. x y > 0. The basic rules are simple and they follow.If x > 0 and y 0 then x + y > 0.If x > 0 and y > 0 then xy > 0.


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3.4. INEQUALITIES 25

If x > 0 and y < 0 then xy < 0.

Using these basic rules, it follows, (see Pr3.4.1), thatx1 > y1 and x2 y2 imply x1 + x2 > y1 + y2,x > y and z > 0 imply xz > yz, andx > y and z < 0 imply xz < yz.

There are additional variations and extensions which follow from these.We shall not use inequalities very often in this text.Pr 3.4.1 Establish the last three consequences of the basic rules.Pr 3.4.2 (a) Show that if x > y > 0 and z > 0, then z/x < z/y.(b) Show that if x > 0 > y and z > 0, then z/x > z/y.Ex 3.4.3 What is the minimum possible value of y = (x 3)2 + 4?


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27/65

Chapter 4

Exponents

4.1 Introduction

In the preceeding chapter we had occasion to use the exponential notation x2

to replace the clumsy expression xx to represent the product ofx with itself.Thus x x is represented by x2 and x x x by x3. In general, the expressionxn represents the product of x with itself n times. Thus x xn = xn+1.Notice that we used the dot to represent multiplication here. Otherwise,if we had written xxn, someone might have interpreted it as the symbolxx multiplied by itself n times. We try, not always successfully, to avoidambiguities like that by tending to avoid the use of a pair of characters to

represent one variable. The dot, parentheses, brackets and braces are alsouseful when we detect possible sources of misunderstanding. Incidentallythe use of subscripts is also helpful. The symbol n in the expression xn iscalled the exponent.

4.2 Algebra of Integer Exponents

When we are dealing with positive integer exponents, it is obvious thatxn xm = xn+m and that (xn)m = xnm. Also, ifm and n are positive integerswith m greater than n, indicated by the symbol m > n, then x

m

/xn

= xmn

.Suppose however that m is less than n, that is m < n. Then how should weinterpret xmn and how should we interpret x with a negative exponent? Itseems very reasonable to take xn = 1/xn if x = 0.Pr 4.2.1 Which is larger, (1 + 1/4)4 or (1 + 1/2)2 ?

27


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28 CHAPTER 4. EXPONENTS

4.3 Square Roots

The square root of a positive real number is that number which when multi-plied by itself gives the original number. In algebraic notation, we can writethat y is a square root of the positive number x if x = y2. For each positivenumber x, there are two square roots. For example the square roots of 4are 2 and -2, which we can write as 2. For x = 0, there is only one squareroot, namely zero. Generally speaking, when we refer to the square root ofa number greater than 0, we will mean the positive square root. If we donot wish to specify the sign, we will refer to a square root.

There are two ways used to indicate the square root. These are

xand x1/2. The second representation is natural, considering our algebra ofexponents, for then (x1/2)2 would be expected to be x21/2 or x1 = x.

4.4 Rational Exponents

Once we introduce the exponent 1/2, it makes sense to see whether we canreasonably define expressions with other fractional or rational exponents. Itis easy to consider that x1/3 is a number y so that y cubed or y3 = x. Thenx2/3 would stand for the square ofx1/3. Thus it makes sense to define xm/n

as ym where yn = x. In those cases where there is more than one possiblevalue of xm/n we may have to indicate whether we want only the positivevalue. In those cases where n is odd, we can deal with both positive and

negative values of x.When the exponent is 0, we get the value one except for the case where

x = 0. The expression 00 does not have a clear meaning. Otherwise thealgebraic rules in Section 4.2 apply to rational exponents.

In fact these definitions for rational exponents can be extended to applyto real exponents also, but we will avoid that extension in this text, and willalways assume that the exponents are rational. In the meantime, it is easyto establish the rules

xyxz = xy+z

and

(xy)z = xyz

for rational y and z and positive x. Also, ifx and y are positive, and z isrational, then

xzyz = (xy)z .


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4.5. SQUARE ROOTS AGAIN 29

Ex 4.4.1 The software firm Google is named after the googol which was

defined to be 10100

, a very large number. The exponent can be written as102. How does a googol compare with (1010)2. Note that 1010 is ten billion.Pr 4.4.2 Logarithms were introduced to simplify approximations for complexcalculations. If x = 10y, we say that y is the logarithm of x, and we writey = log(x). Show that(a) log(10x) = log(x) + 1,(b) log(x1x2) = log(x1) + log(x2) and(c) log(xz) = z log(x).

4.5 Square Roots Again

There is an algorithm for calculating square roots that I learned when Istudied Algebra. It is rather complicated to describe, and it makes moresense to use a calculator to do that calculation. However there is an inter-esting alternative method for calculating square roots in an iterative fashionwhich I will describe here. Given a positive number x, let y0 be a roughapproximation to the square root ofx. For example we might take for x = 2,the value of y0 = 1 which is not an especially good rough approximation.The next approximation is obtained by using the formula

yn+1 = .5(yn + x/yn)

Thus we have y1 = 3/2 = 1.5, y2 = 17/12 = 1.416667, and y3 = 1.414215 . . .,

whereas 2 = 1.414213 . . .. Try this algorithm with a different value of y0and possibly a different value of x.

We will now demonstrate that the square root of 2 is not rational, that itcan not be expressed as the ratio of two integers. Suppose that it is rational.Then we can write

2 = m/n

where m and n are positive integers. If they have a common factor, wecan divide by that factor, and so we may assume that they do not have acommon factor. But then

m2 = 2n2

which means that m must be even, say m = 2k where k is an integer. Thenm2 = 4k2 = 2n2 and

n2 = 2k2

from which it follows that n is also even, contradicting our assumption thatm and n have no common factor. This contradiction establishes that

2 is


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30 CHAPTER 4. EXPONENTS

not rational. Interestingly enough it can be shown that, in a certain special

sense, there are more irrational real numbers than there are real ones. Thissense is quite special, since there are infinitely many rational numbers.While it is obvious that there is no real solution to the equation y2 = 1,

there is an interesting branch of Mathematics which deals with an extensionof real numbers, called complex numbers, for which that equation has twosolutions called imaginary. We will not discuss complex numbers in thistext, even though they are especially useful in the study of Electricity andAerodynamics.Pr 4.5.1 Find the square root of (x + y)2 4xy.Pr 4.5.2 Let en = y

2n x be a measure of the error in estimating

x using

yn. Show that en+1 = e2n/4y

2n for the iteration rule of this section. Show

that for yn > 0, en+1

0 and for en

0, en+1

en/4.


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Chapter 5

Arithmetic and Geometric

Progressions

5.1 Introduction

There is a theory that the professions that people choose are much influencedby a combination of special talents and psychological needs. Personally,I have felt that one of the psychological factors that plays a role in thedecision to become a mathematician is an underlying belief in magic whichis seldom articulated. I suspect that it is this belief which leads one to make

conjectures for no other reason than that they look good. These conjecturesplay an important role in the success of a mathematicians work.

For example, since 1 = 12, 1+3 = 22, 1+3+5 = 32, 1+3+ 5+7 = 42 and1 + 3 + 5 + 7 + 9 = 52, it is natural for one who notices this and believes inmagic to conjecture that this relation, indicating that the sum of the first nodd numbers is n2 is true for all integer values of n. In fact this conjectureis valid, but a practice of leaping to such conclusions inevitably leads tofalse ones. To compensate, mathematicians have had to develop a highlyrigorous approach to checking or proving their conjectures. Thus, in whatseems paradoxical, a belief in magic leads to a habit of very cold hard logicalproof. Nevertheless, good mathematicians have always depended on flights

of fancy to raise conjectures, some of which have subsequently been soundlyestablished.

Let us return to the example of the sum of the first n odd integers. Foreach positive integer i, the i-th even number is 2i and the i-th odd numberis 2i 1. Each odd number differs from the previous one by 2. Thus we are

31


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32 CHAPTER 5. ARITHMETIC AND GEOMETRIC PROGRESSIONS

interested in the sum of the first n odd numbers which may be written as

s = 1 + 3 + 5 + + (2n 3) + (2n 1).

Writing the sum of the terms in reverse order, we have

s = (2n 1) + (2n 3) + + 5 + 3 + 1

If we add the corresponding terms in the two equations, we see that on theright hand side of the first equation, each term increases by 2, while for thesecond equation each term decreases by 2. Thus each of the n sums remainsfixed at 1 + (2n 1) = 2n. We have n such sums, and hence

2s = n 2n = 2n2

and our conjectured claim that s = n2 has been demonstrated.

5.2 Arithmetic Progressions

The argument we have used can easily be generalized to get a result, whichis not as appealing to me, but very useful. We say that a sequence ofreal numbers, a1, a2, , , an form an arithmetic progressionif each numberdiffers from the preceeding one by a fixed amount d. Then a2 = a1 + d, a3 =a2 + d , . . .. In general, an = an1 + d = a1 + (n 1)d. Examples are 1,3,5,7and 1,2,3,4 and 2,0,-2,-4,-6. The sum of an arithmetic progression is

s = a1 + a2 + + ans = an + an1 + + a1.

Adding, we have the same sum a1 + an, n times and dividing by 2, we have

s = n(a1 + an)/2 = n[2a1 + (n 1)d]/2. (5.1)

The middle expression in the above equations may be described as n timeshalf the sum of the first and last terms of the arithmetic progression. When

applied to the special example of the sum of the first n integers, we havethe conclusion that the sum satisfies s = n(n + 1)/2 which is often useful toknow. Check it out for n = 5.

Ex 5.2.1 For each integer n let an = an1 + 2 and a1 = 5. Find the sums = a1 + a2 + + a100.


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5.3. GEOMETRIC PROGRESSIONS. 33

5.3 Geometric Progressions.

A geometric progression is a sequence where each term is a multiple of thepreceeding one by the same multiple or factor, r, which stands for the wordratio. Thus 1,2,4,8,16, and 3,1,1/3,1/9, and 2,-2,2,-2,2 are geometric pro-gressions corresponding to the ratios 2, 1/3, and -1 respectively. We shallnow derive an expression for the sum of a geometric progression. Notice thatif a1, a2, a3, , an form a geometric progression, an = ran1 = r ran2 =r2an2. It is easy to see that an = r

n1a1 for all positive integers n. Thenthe sum of a geometric progression can be written

s = a1 + ra1 + r2a1 + + rn2a1 + rn1a1

andrs = ra1 + r

2a1 + r3a1 + + rn1a1 + rna1.

These expressions for s and rs sum the same terms with the exception ofthe first term of s and the last term of rs. Subtracting we have

(r 1)s = (rn 1)a1

and, if r = 1,s = a1

rn 1r 1 = a1

1 rn1 r . (5.2)

The choice of which equality is more convenient to use depends on

whether r is greater or less than one in magnitude.For example 1 + 2 + 4 + + 2n1 = 2n1 and 1+1/2 + 1/4 + + 2(n1) = (12n)/(11/2) =2 ( 12)n1.

This last example raises an interesting issue on which we will rarelyelaborate in this text, with the exception of the next section. That issue isone of limits.Ex 5.3.1 What is the sum s of a geometric progression when r = 1 and (a)when n is odd, and (b) when n is even.Ex 5.3.2 Suppose an = 2an1 and a1 = 3. What is the sum s = a1 + a2 + + a7?Pu 5.3.3 Suppose an = 2an1 +1 and a1 = 1. Find an expression for an and

for sn = a1 + + an.Pr 5.3.4 The bank lends money at an annual interest rate of 6% compoundedmonthly. That means that each month .5% is added to the principal or theprincipal is multiplied by 1.005. I borrow $100,000.(a) How much will I owe the bank at the end of 2 years?


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The present value of x, paid in n months is x(1.005)n.

(b) If I pay $4,000 per month starting at the end of the month, what is thepresent value of how much I have paid at the end of 2 years?(c) What will be my remaining principal (what I still owe) at the end of 2years?

5.4 Infinite Geometric Progressions

.In the last example, we notice that if n becomes large, the last term in

the sum becomes small and gets arbitrarily close to zero for n large enough.In that case we can regard the sum of the progression 1 + 1/2 + 1/4 + with infinitely many terms, as the limiting value of the finite sums or simply2.

The infinite geometric progression with initial term a and ratio r where|r| < 1 is easily seen to have the sum

s = a/(1 r).

Thus 1+1/3 + 1/9 + = 3/2 while 1+1/4 + 1/16 + = 4/3. We formerlypointed out that the decimal expansion of 1/3 does not terminate. It consistsof an infinite sequence of threes. In mathematical terms, it can be regardedas 3/10 + 3/100 + 3/1000 + . By the rules of sums of infinite geometricprogressions this sum should be (3/10)/(1

1/10) = 3/9 = 1/3 which is a

verification of what we expected. On the other hand the decimal .9999 . . .should equal (9/10)/(1 1/10) = 1 This has the strange interpretation thatthe infinite decimal consisting of all nines is equivalent to 1.

In our problem sets we explore the fact that every rational number canbe expressed in decimal notation as either having a finite number of nonzerodigits after the decimal point, or having a repeating sequence of digits suchas .25343434 . . . which is the sum .25 + .0034 (1 + .01 + .0001 + ) =.25 + .0034(1 1/100) which is obviously rational. Thus the fact that such arepeating sequence is rational is not surprising. What is a little less obviousis that every rational can be so expressed.Pu 5.4.1 Show that every rational number can be represented by a decimal

with a repeating part.Pu 5.4.2 There is an anecdote about Von Neumann, a famous mathematicianand physicist. He was given a puzzle, and the method he used to solve itwould determine whether he was primarily a physicist or mathematician.Two trains head toward each other at 50 miles per hour and 30 miles per


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5.5. SUMMATION NOTATION 35

hour respectively. They start out 80 miles apart. A fly flies from one train

to the other at 100 miles per hour, and instantly reverses to fly to the firsttrain and continues to shuttle between them until it is crushed. How far hasthe fly flown?

Von Neumann answered instantly. He was told that he was a physicistbecause the mathematician would have figured out how far the fly flew eachtime till he was crushed, and added the infinite series. A physicist wouldhave used a much faster and simpler calculation. When this was explainedto him, he admitted that he had added the infinite series. The simplercalculation is a puzzle for the reader. The infinite series is a problem whichis complicated and difficult, but not very tricky. Try both approaches if youcan think of the physicists. In any case, do the infinite series.

5.5 Summation Notation

I have tried to minimize the amount of algebraic notation used in this text.However good notation plays a useful role in making more transparent whatmay seem very complicated otherwise. The down side is that it takes practiceto become familiar with a great deal of notation, and the usefulness declinesif there is little opportunity to use it. The standard notation for sumsinvolves the greek letter capital sigma which follows, . The sum of asequence of terms such as a5 + a6 + + a15 is represented by capital sigmawith the lower and upper limits below and above the sigma. That is

a5 + a6 + + a15 =15i=5

ai.

In this notation the character i seems to come out of nowhere. What is itdoing? It is called a dummy index. The right hand side is saying that weare adding all the numbers ai where the symbol i varies in steps of 1 from5 to 15. It is a dummy because the right hand side would be the same ifwe changed i to j or any other symbol that had not previously been givena meaning. The sigma notation is useful, because the alternative we havebeen using is sometimes clumsy and misleading. I often like the alternative,

but it is embarassing to write a1 + a2 + a3 + + an1 + an when n is oneor two.

One should keep in mind that the usual convention for the use of sigmahas the upper index greater than or equal to the lower index. Otherwisethere is some confusion. Using this notation, the result for arithmetic and


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geometric progressions can be written in the following equations.

ni=1

a + (i 1)d = n[2a + (n 1)d]/2n

i=1

ari1 = a(1 rn)/(1 r)


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Chapter 6

Polynomials

6.1 Introduction

Consider the relation

y = P1(x) = ax + b.

As we saw, this can be represented on a graph by a straight line, and it istherefore called a linear function. On the other hand

y = P2(x) = ax2 + bx + c

involves the second power of x and is called a quadratic relation if a = 0.

Figure 6.1.1

-1 0 1 2 3 4

10

-5

0

5

10

15

x

y

More generally, we have an n-th degree polynomial which has the form

y = Pn(x) = a0xn + a1x

n1 + + an1x + an

37


38/65

38 CHAPTER 6. POLYNOMIALS

when a0 = 0, and n is a positive integer. If we plot y = x3 3x2 + x 7, wesee that y becomes large for large x and highly negative for large negativevalues of x. (See Figure 6.1.1.) As x increases y rises from highly negativevalues to a local peak and decreases to a local valley before rising again,crossing the x-axis near x = 3.3. There are a variety of curves that canbe represented by a third degree polynomial. We shall concentrate on thelinear and quadratic cases.

The form y = ax + b yields a straight line which intercepts the y-axisat x = 0 and y = b. The usual convention is to designate that point as(x, y) = (0, b). As x increases by 1, y increases by the slope a ifa is positive.If a is negative the line slopes down and y decreases by |a| for each increaseof 1 in x.Ex 6.1.1 Plot y = 4x

3; y = x2

3x+4; y =

x+5; y =

x2; y = x2

6x+2

6.2 Quadratic Functions

The quadratic polynomial

y = P2(x) = ax2 + bx + c

with a = 0 is represented by a curve called a parabolawhere y rises as |x| getslarge ifa > 0 and y decreases as |x| gets large ifa < 0. This parabola eithernever crosses the x-axis (y = 0), or it crosses it twice, or simply touchesthe x-axis at one value of x. This geometric statement has an algebraic

interpretation which involves solving the equation

ax2 + bx + c = 0

for the value or values of x in terms of a, b and c.To solve the equation we use the method called completing the square.

Sincea(x + b2a)

2 = a(x2 + bax +b2

4a2 )

= ax2 + bx + b2

4a

ax2

+ bx + c = a(x +b

2a)2

+ (c b2

4a) = 0

a(x + b2a)2 = b

24ac4a

(x + b2a)2 = b

24ac4a2 .


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6.3. POLYNOMIALS AS PRODUCTS 39

Finally, if the discriminant d = b2 4ac 0,

x +b

2a=

b2 4ac2a

or

x =b

b2 4ac

2a. (6.1)

The three cases we considered were those where d < 0 (no solution),d > 0 (two solutions), and d = 0 (one solution). When d < 0 there is nosolution because there is no real square root of a negative number. Note thatif we introcuced the concept of an imaginary number i, such that i2 = 1,and complex numbers of the form u + vi where u and v are real, we could

extend the concept of solutions of quadratic equations so that there weretwo solutions when d < 0.The formula for the solution of the quadratic equation is one of the few

that I have committed to memory. I dont recommend forcing yourself tomemorize this equation, but I would suggest recalling the idea of completingthe square.Ex 6.2.1 If there is a solution, solve for x in(a) x2 x 6 = 0,(b) x2 + 8x + 16 = 0, and(c) 2x2 + 6x + 5 = 0.Pr 6.2.2 For what value of x is 3x2 + 5x 2 minimized? Hint: Complete thesquare.

6.3 Polynomials as Products

Let us address polynomials from another point of view. Consider the product

y2 = (x x1)(x x2) = x2 + bx + c

where b = (x1 + x2) and c = x1x2. Then y2 = 0 if and only x = x1 orx = x2. Similarly,

y3 = (x

x1)(x

x2)(x

x3) = x

3 + a1x2 + a2x + a3

where a1 = (x1 + x2 + x3) , a2 = x1x2 + x1x3 + x2x3, and a3 = x1x2x3.In English terms, a1 is the sum of the xi, a2 is the sum of the productstwo at a time, and a3 is the product of all three. This result generalizesfor polynomials which are the product of n factors of the form (x xi).


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A special case of interest is that ofy2 = (xx1)2 = x22xx1+x21. In thatcase there is only one solution of y2 = 0. That is x = x1. But in this case x1is considered a double solution, because (x x1) can be factored out of thepolynomial for y2 twice.. There is a sophisticated theorem in the analysisof complex variables that states that every polynomial of degree n can befactored into a0 times the product of n terms of the form (x xi), wherethe xi are complex numbers. Since we will concentrate on real numbersin this text, we shall not use that theorem directly, but it implies thatevery polynomial with real coefficients, a0, a1, , an can be factored into aproduct of terms of the form (x xi) or ((x ui)2 + v2i ), where xi,ui andvi are real. Since the latter term is always positive, the polynomial can be0 only for the values of x = xi.

6.4 Rational Functions

Just as the fraction m/n , where m and n are integers, is called a rationalnumber, the ratio of two polynomials y = P1(x)/P2(x) is called a rationalfunction. If P1 and P2 have a common factor P0(x), that can be factoredout, so usually we assume that they have no common factor.

Let us show, by example, how P1 can be divided by a linear polynomialP2. The procedure is basically the same as the one used in ordinary longdivision. Let P1(x) = 6x

3 + 4x2 + 7 and P2(x) = 2x + 1. We note that,taking the lead terms, 6x3/2x = 3x2. Thus we have

6x3 + 4x2 + 7 = 3x2(2x + 1) + R1(x)

where R1(x) is the remainder (6x3 + 4x2 + 7) (6x3 + 3x2) = x2 + 7. This

step has eliminated the highest power of P1(x). Now x2/2x = x/2 and we

have

x2 + 7 = (x/2)(2x + 1) + R2(x)

where R2(x) = (x2 + 7) (x/2)(2x + 1) = x/2 + 7 is the next remainder.

Finally (x/2)/(2x) = 1/4, and

x/2 + 7 = (

1/4)(2x + 1) + R3(x)

where R3(x) = (x/2+ 7) (x/2 1/4) = 7 + 1/4 = 29/4. Thus we have,

6x3 + 4x2 + 7

2x + 1= 3x2 x/2 1/4 + 29/4

2x + 1.


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6.5. REDUCED FORM 41

Just as in ordinary long division, we indicate an algorithm which puts

the above process into a regular format.

3x2 +x/2 1/42x + 1 6x3 +4x2 +0 x +7

6x3 +3x2

x2 +0 xx2 +x/2

x/2 +7x/2 1/4

29/4More generally, if P1(x) is a polynomial of degree n in x, and P2(x) is

linear, we have a quotient Q(x) of degree n

1 and a remainder R which is

a constant. The remainder is 0 if and only if P2(x) is a factor of P1(x). Inthat case, if P2(x) = bx + c, and hence vanishes at x = x0 = c/b, so willP1. Otherwise we will have

P1(x)

P2(x)= Q(x) +

R

bx + c.

If R = 0, the expression 1/(bx + c) behaves strangely for x near x0. Toappreciate this, simply plot y = 1/(2x + 1), and note that for x slightlylarger than -1/2, y is very large, and for x slightly less than 1/2 , y ishighly negative. For x far from -1/2, y becomes small. The plot consists ofthe two branches of a curve called a hyperbola.

A similar process of division can also be carried out when the divisor isnot linear. For example, if P2(x) = 3x

2 + 6x + 5, the procedures describedabove need be modified in only the most trivial fashion. In this case theremainder may be a constant or a linear polynomial in x.Pr 6.4.1 Divide 2x4 6x2 + 3x +5 by x2 3x + 1. Find the quotient Q andthe remainder R.

6.5 Reduced Form

In the arithmetic of fractions it is often desirable to represent the fractionin reduced form where the numerator and denominator are integers with no

common factor. A similar reduction problem appears when dealing withsquare roots. For example, we can simplify

y =2 +

3

4 33 .


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If we note that (a + b(c))(a b

c) = a2 b2c, we can multiply numeratorand denominator of y by 4 + 33, to get

y =(2 +

3)(4 + 3

3)

16 27 = 17/11 (10/11)

3,

which is a simpler looking form than that offered originally.


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Chapter 7

Analytic Geometry

7.1 Introduction

Combining methods of Algebra and Geometry has often proved to be veryinsightful in both fields. We will not expand much in this area of Mathemat-ics, but the theorem of Pythagoras is useful in calculating distances betweentwo points on a graph and in appreciating some important facts.

In the following problems use is made of the fact that for 0 < w < 1,the quantity (1 w)x1 + wx2 = x1 + w(x2 x1) represents a point which isw part of the way from x1 toward x2. In particular (x1 + x2)/2 is half waybetween x1 and x2, while (2x1 + x2)/3 is only one third of the way from x1to x2.Pr 7.1.1 (a) Let 0 < w < 1. Show that

(1 w)x21 + wx22 [(1 w)x1 + wx2]2

with equality if and only if x1 = x2.(b) Interpret this result for the graph of y = x2.Pr 7.1.2 (a) Show that for positive x1 and x2

1

2

1

x1+

1

x2

1

(x1 + x2)/2

with equality if and only if x1 = x2.

(b) Let 0 < w < 1. Show that for positive x1 and x2,

(1 w) 1x1

+ w1

x2 1

(1 w)x1 + wx2 ,

with equality if and only if x1 = x2.

43


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44 CHAPTER 7. ANALYTIC GEOMETRY

7.2 Theorem of Pythagoras

The main point of this section is to announce the fact that the distance dbetween two points on a graph with coordinates (x1, y1) and (x2, y2) satisfies,

d2 = (x2 x1)2 + (y2 y1)2,

and hence the distance from (x, y) to the origin is

x2 + y2.This fact is an application of the Theorem of Pythagoras which states

that the length of the hypoteneuse of a right triangle with sides of size xand y, is

x2 + y2. This statement uses several technical terms in geome-

try, and a proof requires the use of additional technical terms like square,rectangle, perpendicular, orthogonal, right angle, and areas of triangles and

of rectangles.

Figure 7.2.1

y

x

z

x

y

Figure 7.2.1 is the basis of a proof of the theorem, which we outlinewithout most of the details that justify the statement that the areas of theouter square, each of the inner triangles, and the inner square are(x + y)2,xy/2, and z2 respectively. Totalling the areas,we have

(x + y)2 = z2 + 4xy/2

but

(x + y)2

= x2

+ 2xy + y2

,and it follows that

x2 + y2 = z2,

where x and y are the sides of the right triangle with hypoteneuse z.


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7.3. CIRCLES AND ELLIPSES 45

7.3 Circles and Ellipses

The points (x, y) for which

x2 + y2 = r2

where r > 0, are r units away from the origin, and characterize a circle ofradius r, and diameter 2r with center at the origin.

The points (x, y) for which

x2

a2+

y2

b2= 1

is called an ellipse with major and minor axes a and b. The major axiscorresponds to the larger of a and b. One characteristic of an ellipse, is thatit has two focal points and that the sum of the distances from the points onthe ellipse to the focal points is constant. Figure 7.3.1 presents a circle withradius r = 1, and an ellipse with a = 3, and b = 2. The circle is a specialor degenerate case of an ellipse where the two focal points coincide at thecenter.

Figure 7.3.1

-3 -2 -1 0 1 2 3-2

-1

0

1

2

x

y

Pr 7.3.1 Find the point on the line y = 2x + 7 which is closest to the origin,and how far it is from there.

Pr 7.3.2 Find the point on the parabola y = x2 4 closest to the origin, andhow far it is from there.


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46 CHAPTER 7. ANALYTIC GEOMETRY

Pr 7.3.3 (a) Show that ifa b and the two focal points of the ellipse aboveare at (c, 0) and (c, 0), then c = (a2 b2), and that the sum of thedistances to the focal points is 2a. (b) Establish that indeed for all pointson the elipse, the sum of the distances is 2a. This problem is not easy. Usea pencil and two tacks, holding down the ends of a cord to draw an ellipse.


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Chapter 8

Simultaneous Equations

8.1 Two Linear Equations

Example 8.1.1. Consider the two equations in which x and y appear in linearform (to the first power).

3x + 2y = 7

x + y = 9.

Is there a solution which yields values ofx and y for which both equations aresatisfied? The answer is yes. We take y = 9x from the second equation andsubstitute for that in the first equations yielding 3x +2(9 x) = x + 18 = 7from which it follows that x = 11 and hence y = 20. To check we see that3 (11) + 2 (20) = 33 + 40 = 7 and 11 + 20 = 9.

Graphically,these two equations represent straight lines in the (x, y)plane and these two lines intersect at the one point which corresponds tothe solution of the two simultaneous equations. Those equations are calledlinear equations because both x and y appear only in the first power. Is italways the case that two such equations have a unique soluton? No. Forexample, 3x + 2y = 7 and 6x + 4y = 9 represent two parallel lines whichnever intersect, and have no solution. On the other hand the equations3x + 2y = 7 and 6x + 4y = 14 both represent the same line, and thus every

point on that line satisfies the two equations. More generally, if we put theseequations in the form

a1x + a2y = a0

b1x + b2y = b0

47


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48 CHAPTER 8. SIMULTANEOUS EQUATIONS

these two equations have a unique solution if and only if the coefficients

(b1, b2) are not proportional to (a1, a2). Another equivalent form of thatcondition is that a1b2 = a2b1. If a1b2 = a2b1, then we have either nosolution or infinitely many solutions depending on whether or not the twoequations represent two parallel lines or the same line.

Another approach to solving these two equations when a1b2 = a2b1 isindicated in the next example.

Example 8.1.2. Consider the equations

2x + 3y = 11

3x + 5y = 15.

Multiply both sides of the first equation by 5 and of the second by 3 to get

10x + 15y = 55

9x + 15y = 45.

Now that the coefficients ofy in the two equations are equal we can subtractthe second equation from the first to obtain

x = 10

from which it easily follows that 3y = 11 2x or y = 3, which checks withthe original equations.

We can deal with 3 linear equations in three variables x, y, and z byeither of the two methods indicated above. We could substitute in oneequation to find an expression for one of the variables in terms of the othertwo. Substituting that expression in the other two equations, leads to twoequations in the two remaining variables. That is an extension of the substi-tution method, and we used that in Chapter 3. The analogue of the methodof equating coefficients used above, is to multiply two equations by the co-efficients which make those of one of the variables match. Subtracting theresulting two equations leads to a fourth equation with one of the three vari-ables missing. Then apply the same technique to another pair of the originalequations to equalize the coefficients of the variable that was cancelled be-

fore. Once more we get another equation in the other two variables. Eitherway we get two equations in two variables. We can use either method tosolve for one of the variables. Use the remaining equations to solve for theother variables. This may sound a bit abstract, but trying the techniqueson the problems will show how the method works.


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8.2. NONLINEAR EQUATIONS 49

Geometrically, we see that each linear equation represents a plane in

three-dimensional space. If two of the equations do not represent parallelplanes, they intersect in a straight line. If that straight line is not in theplane of the third equation, or parallel to that plane, the planes will intersectin one unique point. Otherwise we may have no solution or many possiblesolutions.

The general theory of determinants and matrices was developed to solven linear equations in n unknowns. We shall not elaborate on that theoryhere.Ex 8.1.1 Solve the following equations for x and y.

6x + 7y = 33

5x

2y = 4.

Pr 8.1.2 Solve the following equations for x, y and z.

3x 2y + z = 12x 5y = 1

4x + 3y + 2z = 4.

Pr 8.1.3 Find a solution of the two simultaneous equations below in termsof the coefficients assuming that a1b2 = a2b1.

a1x + a2y = a0

b1x + b2y = b0

8.2 Nonlinear Equations

We now consider a special example where one equation is linear, but thesecond is not.Example 8.2.1 Let

3x + 2y = 13

x2 + 2y2 = 17

What are the values of x and y?

Before solving these equations, it is well to note what they mean graph-ically. The first equation is a straight line with a y-intercept of 6.5, and thesecond equation is an ellipse which goes through the points (17, 0) and(0, 8.5) and is centered at the origin. The solution or solutions corre-spond to those points, if any, where the line intersects the ellipse.


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50 CHAPTER 8. SIMULTANEOUS EQUATIONS

We proceed algebraically. From the first equation, we have y = 6.51.5x.Substituting that in the second equation, we have x

2

+ 2(6.5 1.5x)2

=x2 + 84.5 39x + 4.5x2 = 17. Collecting terms with the same powers ofx,we have

5.5x2 39x + 67.5 = 0which is a quadratic equation, with discriminant d =

1521 1485 = 6 and

solutions (39 6)/11 or x = 3 and x = 45/11. The corresponding values ofy are y = 2 and y = 4/11.Pr 8.2.1 Solve for x and y .

2x y = 6xy = 3.

Pr 8.2.2 Find the value of b for which the line y = 2x + b is tangent to theparabola y = x2. That means that the line touches the parabola, but doesnot cross it at the point of contact. What is the point of contact?


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Chapter 9

Combinations and

Permutations

9.1 Introduction

Combinations and permutations are useful tools for the study of Probabilityinvolving equally likely events. While we wont pursue the field of Proba-bility, we will find the elementary results very useful to derive the BinomialTheorem which provides a representation of an expression for (a + b)n forinteger values of n.

9.2 A Fundamental Rule

A fundamental rule is that if one thing can be done in m ways, and a secondthing can be done in n ways for each of the first m ways, then the two thingscan be done in mn ways. We illustrate with 3 examples.

Example 9.2.1. Select a number from 1 to 4 and a color from red and black.How many possible selections of these two items are there?

According to the fundamental rule, there should be 8. If we list all pos-sibilities systematically, we get 1R,1B,2R,2B,3R,3B,4R,4B, where R standsfor red and B for black. Indeed there are 8 such possibilities.

Example 9.2.2. Select a card from a pack of 4 cards labeled 1 to 4, to bethe first choice and then a second card from among those remaining. Howmany possible selections can be made?

The possible selections of a first and second card can be labeled, 12, 13,14, 21, 23, 24, 31, 32, 34, 41, 42, 43. There are 12 possibilities, and they

51


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52 CHAPTER 9. COMBINATIONS AND PERMUTATIONS

satisfy the fundamental rule since there are 4 ways of selecting the first card,

and for each of those, there are 3 ways of selecting the second card.If repetitions had been allowed in Example 9.2.2, that is to say that wecould select the second card from the same four cards, we would have had16 possible pairs. We now consider a third example, a variation of Example9.2.2.Example 9.2.3. Suppose that we wish to select a committee of 2 individualsfrom a group of 4 people. How many possible such committees are therewhere no attention is paid to the order in which they are selected?

If we label the people from 1 to 4, and keep track of the order in whichthey were selected, we would have the previous list of 12 pairs. But eachcommittee of 2 individuals is counted twice in that list. There are only12/2=6 such committees. They are 12, 13, 14, 23, 24, 34.

9.3 Permutations

If we are given the n integers from 1 to n, any ordered arrangement of theseintegers will be called a permutation of those integers. In the same way,an ordered arrangement of any n distinct items is a permutation of theseitems. How many permutations or different arrangements can be made ofthe n integers or of any n distinct items? There are n ways of selecting thefirst item. For each of these there are (n 1) ways of selecting the seconditem. Thus there are n(n 1) ways of selecting the first pair. For each ofthese there are n

2 ways of selecting a third item. Proceeding in this way, it

is clear that altogether there are 123 (n1) n different permutations ofthe n items. For convenience we introduce the term n factorialto representthe product of the first n integers. The notation for n factorial is given bythe exclamation mark, n!. Thus 1! = 1, 2! = 2, 3! = 6, 4! = 24 and 5! = 120.It will be convenient to define 0! = 1, for then n! = n(n 1)! for all positiveintegers n.

While we have found that the number of permutations of n items is n!,we are also interested in the number of ways we can select, in order, r outof n items where r is a positive integer less than n. But our derivation ofn factorial incidentally answered this question. It is clear that the numberof such permutations of n items r at a time is given by Pn,r which is the

product of the r integers decreasing from n. That is

Pn,r = n(n 1) (n r + 1) = n!/(n r)!.

A special case is the one with which we started. That is Pn,n = n!/0! = n!


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9.4. COMBINATIONS 53

Notice that Example 8.2.2 in the preceeding section gave us P4,2 = 12.

9.4 Combinations

The term combination is used to indicate a subgroup formed from a groupof distinct individuals. In Example 9.2.3 above, a committee represented asubgroup of 2 of the 4 individuals, where the order was not important, andhence a combination of 2 out of 4. How many combinations of r out of ndistinct members of a group are there?

The amount Pn,r represented the number of arrangements of r items ofthe group ofn where order was important. But for any choice ofr members,there are r! permutations which are counted in Pn,r. Thus, the number of

combinations of n items, r at a time,is given by

Cn,r = Pn,r/r! =n!

r!(n r)!Another way to look at this quantity is as the quotient of two products.

The numerator is the product of r successive integers going down from nto, and including, n r + 1, and the denominator is the product of the firstr integers. The equation above highlights one important property of thenumber of combinations. That is, for 0 r n

Cn,r = Cn,nr.

From a certain point of view, this property is not surprising. For each

committee of r members selected, there is a corresponding committee ofn r members who were not selected.

The Pascal Triangle below is another interesting representation of com-binations.

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

Each horizontal line is formed by starting with a 1 in the middle forthe apex. Then each following line is obtained by adding the two numbersimmediately to the right and left on the preceeding line. The n-th line belowthe apex represents the number of combinations Cn,r for r = 0, 1, 2, . . . , n.To show that this is the case one must demonstrate

Cn,r = C(n1),(r1) + C(n1),r


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54 CHAPTER 9. COMBINATIONS AND PERMUTATIONS

Note that we have defined C0,0 = 1. Two properties of the Pascal Tri-

angle that are worth noting are that the sum of the elements of the n-throw is 2n, and if we alternated signs in that row, the sum would be zero.For the odd numbered rows, the latter property is not surprising becauseof the symmetry in those rows. But it is not immediately obvious that weshould expect 1 4 + 6 4 + 1 = 0 These properties are a consequence ofthe binomial theorem. described in the next section.

Combinations and permutations play an important role in probabilitytheory, on which we shall not expend much text. But combinations are alsofundamental in the binomial theorem.Pr 9.4.1 Show that the equality above for Cn,r is valid.Pr 9.4.2 I have 13 black balls and 10 red balls.(a) In how many ways can I select 6 black and 2 red balls?(b) In how many ways can I select 8 balls from the 21?

9.5 Binomial Theorem

The binomial theorem deals with an expression for (a + b)n. First we notethat

(a + b)2 = a2 + ab + ba + b2 = a2 + 2ab + b2,

Then

(a + b)3 = (a + b)(a + b)2

= (a + b)(a2

+ 2ab + b2

)= a3 + a2b + 2a2b + 2ab2 + ab2 + b3

= a3 + 3a2b + 3ab2 + b3.

Note that the coefficients of the terms in (a + b)2 and (a + b)3 are thecombinations that appear in the second and third rows below the apex ofthe Pascal Triangle. This is no accident. We shall derive

(a + b)n = Cn,0an + Cn,1a

n1b + + Cn,ianibi + + Cn,nbn.

This equation can also be represented using the summation symbol, as

(a + b)n =n

i=0

Cn,ianibi.

Before deriving this equation we may observe that taking a = b = 1,implies that the sum of the n-th row below the apex of the Pascal Triangle


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9.5. BINOMIAL THEOREM 55

is 2n. By taking a = 1, b = 1 , we obtain the fact that the sum of theitems with alternating signs is zero.There are several ways of deriving the theorem. I choose the following,which is easier to follow

Algebra 1 - Herman Chernoff

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