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Algebra I for Students Comfortable with Arithmetic c 2001 Herman Chernoff February 6, 2002 2 Chapter 1 Introduction Why should you learn Algebra I? I could elaborate on many reasons why a knowledge of Mathematics beyond Arithmetic is of interest and of value to people in the twenty-first century, but for the sake of brevity, I won’t. It is enough to say that a high school curriculum demands it. What are the basic concepts to master? They are the ability to translate quantitative problems of the real world into symbolic statements and to manipulate those statements to arrive at satisfactory answers. As the title indicates, this book is written for students who are comfortable with Arithmetic. One problem with current books on Algebra I, is that in 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 themselves with simple problems in Arithmetic. As a result, these books tend to be full of detail and very large and heavy. To encourage the practice that is required 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 Arithmetic, and willing to do some experimenting, permits bypassing some of the tedious details in text and exercises. This does not mean that the student can avoid the necessary practice. It is hoped that he or she will get that practice in working on more demanding problems. The student should be prepared 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 with three different types of problems. There will be exercises. labeled Ex, to test whether the reader has understood the text well enough to solve a slightly modified variation of the material in the text. There will be problems, labeled Pr, which demand a more comprehensive knowledge of the past text 3 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 seems too difficult or complicated, he or she should not get hung up on it, but go on with the book after a valiant try. It might be rewarding to return to it 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 experiment with alternative approaches. It is expected that this experimentation will provide the necessary practice to fix the basic underlying techniques of the subject without boring repetition. On the other hand one should be prepared to deal with getting wrong answers. Perhaps more important than doing the problems correctly in the first place, is the task of finding out where 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 various ways is also a useful talent. Finally, even when an answer is correct, a review of the work will often indicate that the reasoning went in unnecessary circles instead of a direct straight line. The task of refining a solution to cut out unnecessary digressions often leads to a better understanding of the essentials of the argument. Incidentally, mathematicians tend to refine their work so much that the readers lose the excitement involved in tracking the trail of the hunt, where strange ideas led to hints of how to proceed next to get to the ultimate objective. There are not many figures in this text, but the reader should be prepared to use up lots of graph paper, drawing many figures to help his understanding of the text, examples and problems. In Appendix A, we shall present the answers to the exercises and problems. Some of them will be detailed solutions of the problems. Such solutions are not necessarily unique, and if the reader has another approach, that will not necessarily be 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 it sometimes took me a couple of days to solve some of these puzzles, they usually 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 don’t be surprised if some of them defy your efforts without the hint. In summary, this should be a short book, freely available on the world wide web, and its mastery should not take more than a couple of months by the student willing to spend about 4 hours a week. It would be appreciated if you spread the news about the availability of the book to friends near and 5 far. It is a good idea to collaborate with one or two others of comparable ability. You should not be in the position where your collaborator does all the heavy lifting or where you do all of it. That would put one of you into too passive a mode to get what you should out of the work. An occasional hint from a better trained friend is ok, but major dependence will be counter productive. In the near future you may write to me with complaints, suggestions for improvement (which will become my property), and questions. I plan to spend a couple of hours a week on dealing with such comments. My e-mail address is [email protected] and I hope to use the comments to improve the book. In the event that this book becomes very popular, I might plan to publish it as a small soft cover text to be sold at a modest price for students who don’t want to download it from the computer. Finally, a warning. This book is not appropriate for students who are not comfortable with Arithmetic. On the other hand, the fact that a student has the ability to master Algebra I does not automatically imply that this book will be right for him or her. The approach that brings out the best in a student differs very much from one student to another. It is possible that a reasonably gifted student will not find this presentation suitable. If I knew how to describe the characteristics that determine whether a student will find this text useful, I would do so, but I don’t. I do feel that the willingness to make some effort and to learn from one’s mistakes are important ingredients for a successful use of this presentation. 6 CHAPTER 1. INTRODUCTION Chapter 2 Arithmetic 2.1 Introduction How well do we understand Arithmetic? With the availability of electronic calculators, 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 related to a basic understanding. Moreover the ability to perform these operations accurately and quickly requires a knowledge of the 10 × 10 addition and multiplication tables. I don’t recommend feats of memory in this text. Memorizing rules for solving problems is usually a way of avoiding understanding. Without understanding, great feats of memory are required to handle a limited class of problems, and there is no ability to handle new types of problems. However, memorizing the multiplication and addition tables should not be a major task, 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 use his fingers to add 8 and 5. In this chapter we review Arithmetic and emphasize a couple of points that are so well accepted that they are possibly not as well appreciated as they should be. For example the fact that 3×5 = 5×3 is seldom questioned, but perhaps it should be, because it means that 5+5+5 = 3+3+3+3+3 and why 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 this rectangle around, we have 5 lines of 3 points each without changing the total number of points. See Figure 2.1.1. Readers who find this chapter difficult are probably not ready for the rest of this text. Incidentally, the example discussed above is an example 7 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 apply the principle. Chapter 10 contains a list of such principles, major results, and notation for reference when the text uses them. I doubt that you will have 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 4 three digit numbers with many large digits. Then subtraction of several examples of 6 digit numbers, also with some large digits, should be tried. Then multiplication of a couple of 3 digit numbers should be tried. Finally we are ready for long division. Multiply 439 by 6347 to get a product. Now see if dividing the product by 439 retrieves 6347. You can do several other examples, 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 appropriate 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 1 which he can dispose of by claiming it as his fee. Fat chance that a lawyer would be satisfied with that fee. On the other hand if a cook has 7 pies to 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 seven pies were divided into thirds, there would be 21 thirds, and each of the diners would have seven thirds. We can regard six thirds as the equivalent of two whole pies. In Arithmetic we write the solution of the cook’s problem as 7/3 or 2 1/3. The first form is a representation of the solution in fractional form which is given by one integer over another. Once we introduce fractions, then we have a solution of the problem of dividing one integer, the dividend or numerator by another, the divisor, or demominator even when the first integer is not an exact multiple of the second. Having introduced fractions, we have to consider the arithmetic of fractions. 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, there would be 6 × 7 = 42 pieces which would divide among 3 people equally if each took 14 pieces, i.e. 14/6. Thus a given fraction can be represented in many ways where two representations are equivalent. For example 14/6 and 21/9 are equal to each other. Essentially a fraction does not change value if both numerator and denominator are multiplied by the same integer. Both 14/6 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 is not so easy. On the other hand 4/3 = 20/15, multiplying numerator and denominator 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 and 11/4 we can convert these fractions to 20/24 and 66/24 resulting in a sum of 86/24 = 43/12. Here we multiplied the numerator and denominator of each fraction by the denominator of the other fraction, thereby obtaining, as common denominator, the product of the two denominators. This operation can be unnecessarily tedious, involving huge denominators if we have to add several fractions. That effort can be reduced somewhat by noticing that 6 factors into 2 × 3 and 4 into 2 × 2. We can get a common denominator by taking 2 × 2 × 3. Then we can multiply the numerator and denominator of 5/6 by 2 and those of 11/4 by 3, getting 10/12 + 33/12 = 43/12. In this example 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 a common denominator as is readily available. The same appproach works in subtracting fractions. How about multiplication? It is no surprise that 2 × 7/3 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 10 CHAPTER 2. ARITHMETIC 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 by dividing the numerator and denominator by a common factor. For example 3/7 × 2/9 = 6/63 = 2/21. It is customary to take advantage of factoring before doing all the multiplications. Factoring the 9 in the second denominator into 3 × 3 permits us to cancel the 3 in the first numerator with one of the threes 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 divide 3/4 by 2/5, the rule is to multiply 3/4 by 5/2. In other words we multiply the dividend by the fraction obtained from the divisor by interchanging the numerator and denominator of the divisor. What is the rationale for this? Division is called the inverse operation of multiplication just as subtraction is the inverse operation of addition. When we divide 18 by 2, we essentially ask what multiplied by 2 will give us 18. What multiplied by 2/5 will give us 3/4? Here we have 2/5 × (5/2 × 3/4) will give us 3/4 because the 2/5 × 5/2 = 1. Clearly the rule presented works. To indicate a mastery of the arithmetic of fractions, you should be able to do the following exercise. 4 3 5 1 − 10 divided by 73 + 12 + 18 as a fraction. Ex 2.3.1 Express 23 + 15 Pu 2.3.2. How do two boys divide a pie fairly so that no one could blame anyone but himself if he gets less than his fair share? The standard answer is to have one split the pie and the other to choose which part he prefers. It isn’t so easy to find a resolution for this problem when there are three or more 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 corresponds to the distances along a horizontal line, where one point labeled 0 is the origin. See Figure 2.4.1. A length of one is marked off to the right of 0. It corresponds to the integer one or the fraction 1/1. An equal distance to the right of that point is another corresponding to 2 = 2/1. The positive integers 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 distant points to the left of 0. The rational numbers are the fractions with integer numerators and denominators, and correspond to some of the points on the line. For example 1/3 is the point one third of the way from 0 to 1. Similarly −1/3 corresponds to the point one third os the way from 0 to -1. 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 along the 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 buys an item for $200, then he is in debt $100 and it makes sense to say that he has −$100. Then we might reasonably say that 30 + 40 − 200 + 50 = −80 is the amount a gambler won at the races if his profits were 30,40 and 50 and his loss was 200. The result -80 would indicate a resulting profit of -80, or equivalently, a loss of 80. One way to do the arithmetic of adding positive and negative numbers is to add all the positive entries and all the negative entries separately, and then to subtract the sum of the negatives from the sum of the positives. Another interpretation of the above sum is that one moves 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 a movement of 80 units to the left. Figure 2.4.1 • • • o•o • • • • -3 -2 -1 2 3 4 0 1 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 and a negative number is negative. It is sensible to regard 2 times -3 or 2 × −3 = −6 using the above interpretations of negative numbers. But what sort of interpretation would make sense of −3 × 2? Consider a sheet of graph paper which is a page with 12 CHAPTER 2. ARITHMETIC 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 point on the sheet, we have a horizontal and vertical distance to the origin. The horizontal distance is given a sign of plus or minus depending on whether it is to the right or left of the origin and is called the abscissa or, more commonly, the x-value. The vertical distance is given a sign of plus or minus depending on 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 origin and rises 2 units for every unit to the right of the origin. When the x-value is 3 the y-value is 3 times 2 or 3 × 2 = 6. When the x-value is 2 the y-value is 2 × 2 = 4. When the x-value is 1, the y-value is 1 × 2 = 2, and when the x-value is 0, the y-value is 0 × 2 = 0. When the x-value is -1, the y-value is -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 especially comforting since it implies that −3×2 = 2×−3 = −6, and the commutative law of Section 2.1 applies in this case involving a negative number. Figure 2.4.2 6 • • • 4 • y • 2 • • 0 • -2 • -4 -6 • x • • • -3 -2 -1 0 1 2 3 How about the product of two negative numbers? Take the line which goes through the origin and goes down 2 units for every unit to the right of the origin. When the x-value is 3, the y-value will be -6. For this interpretation it makes sense to regard 3 × −2 = −6. Considering what happens to the 2.5. DECIMALS 13 second line when the x-value is -3, it also makes sense to regard −3×−2 = 6. Thus multiplying a negative number by a positive number results in a negative product, and multiplying two negative numbers results in a positive product. While we have shown where the integers and fractions appear on the horizontal line, one may wonder whether these represent all the points on the line. It is apparent that we can find fractions or rational numbers in every interval on the real line, no matter how small that interval is. However we shall later show that these do not cover all points. In fact the square root of 2, the number which when multiplied by itself yields 2, is not one of the rational values represented by a fraction with an integer numerator and an integer denominator. 2.5 Decimals The Romans were outstanding engineers, many of whose accomplishments continue to amaze us after 2,000 years. Brilliant engineers are not stopped by a poor system of counting, but such a system can slow down progress, and it can frustrate some of the less brilliant practitioners. The Roman number system did not make arithmetic easy. Until recently, the English counting system for money was unnecessarily complicated. A shilling was 12 pence, and a pound was 20 shillings and a guinea was 21 shillings. This system made 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 the English have since gone to a decimal system for money. The Americans still use a non-decimal system for weights and distances, and the whole world continues to use a non-decimal system for time. Most of the civilized world is moving toward decimal systems for money, weights (grams), and distances (meters). Thus a thousand grams is a kilogram and a thousand meters a kilometer. While the decimal system is not essential, it is convenient and does facilitate the use of arithmetic in many practical situations. So we shall say a few words about decimals. The amount $20.64 represents twenty dollars and 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 a dollar. Similarly 3.476 may be regarded as the fraction 3, 476/1, 000. When doing arithmetic with decimals, we are essentially dealing with our familiar integers, except that the units may be tenths or hundreths or even smaller portions of one. Any real number can be arbitrarily well approximated by a 14 CHAPTER 2. ARITHMETIC decimal with enough positions after the decimal point. Interestingly enough some of the simple rationals, as for example, 1/3 can not be represented exactly by a decimal with a finite number of positions after the decimal point. I shall assume that the reader has been properly trained in the arithmetic of 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. Chapter 3 Algebra 3.1 Two Examples The main point of this text is to demonstrate how the essentials of Algebra 1 consists of translating problems into algebraic expressions and manipulating those expressions to derive meaningful answers to the questions raised. We illustrate with two examples, one of which is very simple, and the other which would be difficult if it were not the case that the answers were integers and could be found by trial and error. In the discussion we will see how a convenient 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 stock and lost $100. He doesn’t mind so much the money loss, but feels that there was some funny business. The stock lost 20% in the first month. But then it gained 20% in the second month, when he needed the cash and had to sell out. Under these circumstances he feels that he should have come out even, (not counting the broker’s charges of $20). But he lost $100. My problem is to find out what was the original value of the stock that he bought. To understand why he lost $100, forget the broker’s charges and suppose that Simon had invested $1,000. When the stock lost 20%, the value went down 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 is less 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 would have accounted for the loss of $100, not counting the broker’s fee of $20. To account for that we must assume that the stock only lost $80 and that the 15 16 CHAPTER 3. ALGEBRA original value of the stock purchased was $2,000. This problem is not difficult, but slightly involved. It would become more difficult if the broker’s fees were not fixed at $20, but used a rate depending on the amount being traded. How would we deal with this problem using Algebra? The main tool is to represent an unknown quantity by a symbol, typically a letter of the alphabet, and to do the arithmetic with that symbol. We illustrate. In this case the unknown quantity, the original value of the stock, (in dollars), is designated by x. When its value is decreased by 20%, its value shrinks to 80% of x represented by .8 × x. We find it convenient to suppress the times sign when there is no ambiguity, and we have the reduced value of .8x, which when increased by 20% becomes (1.2).8x. Here we have used parentheses about 1.2 to separate 1.2 from .8 and to avoid the ambiguity of 1.2.8. We express the statement that the loss of $100 is the original value minus the final 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 considerably and I would have written, after the first equation, the third and sixth equations. For people familiar with algebraic manipulations, it is easy to skip 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 down the first equation, is fundamental, and involves translating the basic facts or statement into an equation. The remainder consists of algebraic manipulations of the underlying equation to yield an expression for the desired quantity. In the second equation, we converted (1.2).8x to .96x. In the third equation, we used the fact that x − .96x is .04x. In the next equation we eliminated the 20 on the left hand side of the equation by subtracting 20 from both sides. This equation tells us that a multiple of x is 80, in which case x is 80 divided by that multiple. In summary, we solved the original 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 sophisticated than may seem necessary to the typical student who is used to doing arithmetic. 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 is twice as old as John was when Bill was as old as John is now. How old are they now? Let b and j stand for their current ages, and b∗ and j ∗ for their former ages. Then b + j = 91 b∗ = j b = 2j ∗ . are obvious equations to fit this problem. We wish to find b and j, but we have only three equation involving four unknowns, and it is typically difficult to solve for more unknowns than we have equations. Is there a hidden relationship in our problem statement or is this problem unsolvable? There is another relationship, not overtly stated, but implicit in the statement. The time interval between now and then is the same for Bill and John. 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, we would have 5 equations in five unknowns. As it is, we have four equations involving four unknowns which should be plenty. One way to handle this example is to use each equation to eliminate one of the unknowns. We illustrate. Since b∗ = j, we can eliminate that equation and replace every b∗ by j in the remaining equations. Thus the last equation becomes b − j = j − j ∗ , which can be rewritten as b = 2j − j ∗ by adding j on both sides. Substituting for b in the first and third of the original equations, they now become 2j − j ∗ + j = 91 2j − j ∗ = 2j ∗ . Rewriting these two equations in j and j ∗ , we have 3j − j ∗ = 91 2j = 3j ∗ . 18 CHAPTER 3. ALGEBRA The last equation can be written j ∗ = (2/3)j. Susbstituting in the other equation 3j − (2/3)j = 91 which leaves us with one equation and one unknown. This kind of reduction could have been done using one equation at a time in a different order. Now we have (7/3)j = 91 or j= 91 = 91(3/7) = 39. 7/3 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 as old 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 original equations and the statement of the problem. When I was a student, we were given certain types of special problems and methods of solving these problems and expected to memorize those methods. Such a method is called an algorithm. In my opinion this may be a good policy for finding solutions to many examples of the same problem differing only in the input numbers, but it is a very poor way to foster the understanding which provides the flexibility to handle different variations of these problems. There are very few algorithms worth memorizing. These days they can be handled by computers, and putting numbers into a computer algorithm is usually a poor way to learn something worthwhile besides the answers to those specific problems. Knowing that the answers to the last problem were integers, we could have avoided algebra and used trial and error. We could have tried every possible value of b from 46 to 91 till we found out where b and j = 91 − b solved 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 what would constitute an approximate answer to the problem. A clever use of graph paper could be helpful. Pr 3.1.1 Suppose that in the first example, the broker charged 1% as the transaction cost for buying and for selling the stock. If the original purchase was 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% and then came up 20%, and Simon lost $200 counting the broker’s fees, what 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 do the same task. How long will it take for the two, working together at their natural rates, to clear 2 acres? Note: rate × time = amount of work. Ex 3.1.3 In celebration of John’s birthday he receives a large gift and his brother Bill gets a smaller one. The two boys are naughty and play cards for money with Harry. Together Bill and John start out with $100. After two hours, John has lost half his money and Bill has doubled his, and now they have the same amount. What was the size of John’s 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 of Algebra can be used to solve problems that may be difficult to handle without these methods. In the process we have cut a few corners in the hope that the exposition was elementary enough to be understood without being completely precise. We now take the opportunity to go into some of the neglected 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 represent different numbers which are similar in some sense. It is customary, but not essential, to use the later symbols of the alphabet to represent unknown real numbers or variables and earlier symbols such as a, b, c to represent known numbers which are temporarily unspecified. Fortran is an early computer language which also uses symbols to represent real numbers. The rules for Fortran are different from those for Algebra because Fortran was compelled to use the characters available on a keyboard with limited ability to use different sizes and subscripts. While a human can live 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 relatively transparent to the observer, whereas the computer, as distinct from the computer programmer, is not much concerned with such visual transparency. This digression may help explain some of the following text. In Fortran, computational efficiency made it valuable to distinguish be- 20 CHAPTER 3. ALGEBRA 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, unless there 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 readily available on the keyboard. In Algebra, visualization is frequently enhanced by omitting the multiplication sign, which can also be represented by the mid-level dot, as in xy = x · y. Parentheses are used in both Fortran and Algebra to group real numbers represented by a complex process, such as in x(y + z). Whereas Fortran would use parentheses within parentheses for something like x(y + (z + w)), the writing in Algebra is liable to use brackets and braces or parentheses of different sizes to enhance the readers understanding of what is going on, for example, x[y + (z + w)]. Fortran can not 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 of the 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 of x since this equation assigns a value of y for each of a set of values of x. This introduces a certain amount of ambiguity in our notation, for it could be interpreted as the product of f and x. Unfortunately, this notational ambiguity is prevalent, and we shall have to learn to live with it, and to make some declaration where the context fails to make the proper interpretation quite obvious. Note that the above function 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 exponential notation uses superscripts to express xn as the product of x with itself n times where n is a positive integer. We will expand on this concept in Chapter 4. We often wish to use inequalities. Here the useful symbols are >, ≥, <, ≤, 6=, which stand for greater than, greater than or equal to , less than, less than or equal to, and not equal to, respectively. For example x ≥ y, means x is greater than or equal to y, while x 6= y means that x and y are unequal. Mathematicians have several favorite symbols to represent special numbers. Thus they use the greek letter π = 3.14159... to indicate the famous pi which is important in the relationships among the diameter, perimeter and area of a circle. The letter e is often used for a special value, 2.71828..., important to mathematicians. In a later chapter we shall introduce the Greek capital sigma, Σ, to represent a sum. 3.3. BASIC LAWS 3.3 21 Basic Laws Real numbers have some properties inherited from the integers and fractions, that are so common that they are ordinarily used without special attention paid to them. The commutative laws state that for all real x and y, x+y =y+x and xy = 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 dyadic operations 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 take place. Thus (x + y) + z would use the parentheses to indicate that first we add x and y, and that sum is then added to z. The parentheses are used to separate units that must be defined or calculated before further operations take place. Abstractly, there is no prior reason for a dyadic operation to be such that (x y) z is the same as x (y z). However the associative laws of addition and multiplication, two special dyadic operations, state that for all 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 seen that the order in which several items are added is irrelevant and that the parentheses are unnecessary in representing such a sum. For example, (x + y) + z = x + (z + y) = y + (z + x), and these can be repesented by x + 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 number is considered to be a unitary operation (involving one real number). Thus converting 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 sides considered so different that one must write such an equation? The left hand side is the sum of two reals, one of which is negative. The right hand side is the difference of two positive reals. That these are equal is often accepted as too obvious to mention. 22 CHAPTER 3. ALGEBRA 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 seem to need much explanation for integers, and it holds true for real numbers too. Nevertheless it is an important property, and it has been used in our informal treatment of the examples of Section 3.1. For example we claimed that 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 operations, 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 which xz = y. While subtraction always yields a unique solution, that is not the case for division. If x = 0 and y 6= 0, there is no real number z for which xz = y. If x = 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 apparent paradoxes by using division by an expression which may be zero. We shall illustrate later. Both addition and multiplication have identity elements. The real number 0 is the identity for addition since x + 0 = x, and it has no influence on the sum. Similarly the real number 1 is the identity for multiplication since x · 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 for x 6= 0. These convert a real number to the number which when added or multiplied yields the identity element. In other contexts, the properties mentioned here, or the lack of them become important. We often use these properties without paying much attention 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 forget the names and that they can be extended in obvious ways to include more variables. x+y = y+x xy = yx (x + y) + z = x + (y + z) (xy)z = x(yz) x(y + z) = xy + xz 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 each step. Second it took us four steps to do what we would ordinarily do in one step. Finally, we actually skipped a few of the obvious steps in the final conclusion. Going from the next to last to the last step we used a combination of the associative and commutative properties that technically might have required about 3 additional steps. These are trivial steps and we may feel free to add or multiply any number of reals in any order we wish because 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 follows that (x + 2)(x − 2) = 2(x − 2). Dividing both sides by x − 2, we have x + 2 = 2 or x = 0. But x = 2. How did we get this contradiction? The simple explanation is that we divided both sides of the equation by x − 2 which is equal to zero, and that is liable to 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 − y 2 . Pr 3.3.3 Two dyadic operations are defined below. For each, does the operation have the commutative property? If yes, show why. If no, give a counter example. (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 this month is $30,000, and is increasing by $5,000 per month. The income Win.com brings in this month is $10,000, and is increasing by $7,000 per 24 CHAPTER 3. ALGEBRA 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. An object of mass m and velocity v has a kinetic energy e which is half the product of the mass and the square of the velocity. Translate that statement into an equation. Pr 3.3.6 Each town in Massachusetts is expected to aim at the goal of achieving 10% affordable housing. As long as the town has not achieved this goal developers of apartments are entitled to special privileges if they devote 25% of their new apartments to affordable housing. If a town currently has a proportion p < .10 of affordable housing and a current housing stock of x units, find a formula for the number of additional units, y, built by such developers, required to achieve the town’s goal. Pr 3.3.7 Casting out nines is a method of partially checking addition and multiplication that used to be popular many years ago. For each positive integer i there are remainders r and r∗ after i and the sum, s, of the digits that 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 remainders after 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 of i1 +i2 + · · ·+in against that of s1 +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 i1 i2 is the same as that of r1 r2 . 3.4 Inequalities Dealing with inequalities can be a little tricky. The user of Algebra wouuld be well advised to test his methods when working with inequalities by trying them 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. 3.4. INEQUALITIES 25 If x > 0 and y < 0 then xy < 0. Using these basic rules, it follows, (see Pr3.4.1), that x1 > y1 and x2 ≥ y2 imply x1 + x2 > y1 + y2 , x > y and z > 0 imply xz > yz, and x > 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? 26 CHAPTER 3. ALGEBRA 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 of x with itself. Thus x · x is represented by x2 and x · x · x by x3 . In general, the expression xn 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 symbol xx multiplied by itself n times. We try, not always successfully, to avoid ambiguities like that by tending to avoid the use of a pair of characters to represent one variable. The dot, parentheses, brackets and braces are also useful when we detect possible sources of misunderstanding. Incidentally the use of subscripts is also helpful. The symbol n in the expression xn is called the exponent. 4.2 Algebra of Integer Exponents When we are dealing with positive integer exponents, it is obvious that xn ·xm = xn+m and that (xn )m = xnm . Also, if m and n are positive integers with m greater than n, indicated by the symbol m > n, then xm /xn = xm−n . Suppose however that m is less than n, that is m < n. Then how should we interpret xm−n and how should we interpret x with a negative exponent? It seems very reasonable to take x−n = 1/xn if x 6= 0. Pr 4.2.1 Which is larger, (1 + 1/4)4 or (1 + 1/2)2 ? 27 28 4.3 CHAPTER 4. EXPONENTS Square Roots The square root of a positive real number is that number which when multiplied by itself gives the original number. In algebraic notation, we can write that y is a square root of the positive number x if x = y 2 . For each positive number x, there are two square roots. For example the square roots of 4 are 2 and -2, which we can write as ±2. For x = 0, there is only one square root, namely zero. Generally speaking, when we refer to the square root of a number greater than 0, we will mean the positive square root. If we do not wish to specify the sign, we will refer to a square root. √ There are two ways used to indicate the square root. These are x and x1/2 . The second representation is natural, considering our algebra of exponents, for then (x1/2 )2 would be expected to be x2·1/2 or x1 = x. 4.4 Rational Exponents Once we introduce the exponent 1/2, it makes sense to see whether we can reasonably define expressions with other fractional or rational exponents. It is easy to consider that x1/3 is a number y so that y cubed or y 3 = x. Then x2/3 would stand for the square of x1/3 . Thus it makes sense to define xm/n as y m where y n = x. In those cases where there is more than one possible value of xm/n we may have to indicate whether we want only the positive value. 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 the algebraic rules in Section 4.2 apply to rational exponents. In fact these definitions for rational exponents can be extended to apply to real exponents also, but we will avoid that extension in this text, and will always assume that the exponents are rational. In the meantime, it is easy to establish the rules xy xz = xy+z and (xy )z = xyz for rational y and z and positive x. Also, if x and y are positive, and z is rational, then xz y z = (xy)z . 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 as 102 . 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 complex calculations. If x = 10y , we say that y is the logarithm of x, and we write y = log(x). Show that (a) log(10x) = log(x) + 1, (b) log(x1 x2 ) = 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 I studied Algebra. It is rather complicated to describe, and it makes more sense to use a calculator to do that calculation. However there is an interesting alternative method for calculating square roots in an iterative fashion which I will describe here. Given a positive number x, let y0 be a rough approximation to the square root of x. 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 y0 and possibly a different value of x. We will now demonstrate that the square root of 2 is not rational, that it can 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, we can divide by that factor, and so we may assume that they do not have a common factor. But then m2 = 2n2 which means that m must be even, say m = 2k where k is an integer. Then m2 = 4k 2 = 2n2 and n2 = 2k 2 from which it follows that n is also even, contradicting our assumption√that m and n have no common factor. This contradiction establishes that 2 is 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. This sense is quite special, since there are infinitely many rational numbers. While it is obvious that there is no real solution to the equation y 2 = −1, there is an interesting branch of Mathematics which deals with an extension of real numbers, called complex numbers, for which that equation has two solutions called imaginary. We will not discuss complex numbers in this text, even though they are especially useful in the study of Electricity and Aerodynamics. Pr 4.5.1 Find the square root of (x + y)2 − 4xy. √ Pr 4.5.2 Let en = yn2 − x be a measure of the error in estimating x using yn . Show that en+1 = e2n /4yn2 for the iteration rule of this section. Show that for yn > 0, en+1 ≥ 0 and for en ≥ 0, en+1 ≤ en /4. Chapter 5 Arithmetic and Geometric Progressions 5.1 Introduction There is a theory that the professions that people choose are much influenced by a combination of special talents and psychological needs. Personally, I have felt that one of the psychological factors that plays a role in the decision to become a mathematician is an underlying belief in magic which is 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 conjectures play an important role in the success of a mathematician’s work. For example, since 1 = 12 , 1+3 = 22 , 1+3+5 = 32 , 1+3+5+7 = 42 and 1 + 3 + 5 + 7 + 9 = 52 , it is natural for one who notices this and believes in magic to conjecture that this relation, indicating that the sum of the first n odd numbers is n2 is true for all integer values of n. In fact this conjecture is valid, but a practice of leaping to such conclusions inevitably leads to false ones. To compensate, mathematicians have had to develop a highly rigorous approach to checking or proving their conjectures. Thus, in what seems paradoxical, a belief in magic leads to a habit of very cold hard logical proof. Nevertheless, good mathematicians have always depended on flights of fancy to raise conjectures, some of which have subsequently been soundly established. Let us return to the example of the sum of the first n odd integers. For each positive integer i, the i-th even number is 2i and the i-th odd number is 2i − 1. Each odd number differs from the previous one by 2. Thus we are 31 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 the right hand side of the first equation, each term increases by 2, while for the second equation each term decreases by 2. Thus each of the n sums remains fixed 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, which is not as appealing to me, but very useful. We say that a sequence of real numbers, a1 , a2 , , · · · , an form an arithmetic progression if each number differs from the preceeding one by a fixed amount d. Then a2 = a1 + d, a3 = a2 + d, . . .. In general, an = an−1 + d = a1 + (n − 1)d. Examples are 1,3,5,7 and 1,2,3,4 and 2,0,-2,-4,-6. The sum of an arithmetic progression is s = a1 + a2 + · · · + an s = an + an−1 + · · · + 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 times half 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 have the conclusion that the sum satisfies s = n(n + 1)/2 which is often useful to know. Check it out for n = 5. Ex 5.2.1 For each integer n let an = an−1 + 2 and a1 = −5. Find the sum s = a1 + a2 + · + a100 . 5.3. GEOMETRIC PROGRESSIONS. 5.3 33 Geometric Progressions. A geometric progression is a sequence where each term is a multiple of the preceeding one by the same multiple or factor, r, which stands for the word ratio. Thus 1,2,4,8,16, and 3,1,1/3,1/9, and 2,-2,2,-2,2 are geometric progressions corresponding to the ratios 2, 1/3, and -1 respectively. We shall now derive an expression for the sum of a geometric progression. Notice that if a1 , a2 , a3 , · · · , an form a geometric progression, an = ran−1 = r · ran−2 = r2 an−2 . It is easy to see that an = rn−1 a1 for all positive integers n. Then the sum of a geometric progression can be written s = a1 + ra1 + r2 a1 + · · · + rn−2 a1 + rn−1 a1 and rs = ra1 + r2 a1 + r3 a1 + · · · + rn−1 a1 + rn a1 . These expressions for s and rs sum the same terms with the exception of the first term of s and the last term of rs. Subtracting we have (r − 1)s = (rn − 1)a1 and, if r 6= 1, s = a1 rn − 1 1 − rn = a1 . r−1 1−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 + · · · + 2n−1 = 2n − 1 and 1 + 1/2 + 1/4 + · · · + 2−(n−1) = (1 − 2−n )/(1 − 1/2) = 2 − ( 21 )n−1 . This last example raises an interesting issue on which we will rarely elaborate in this text, with the exception of the next section. That issue is one 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 = 2an−1 and a1 = 3. What is the sum s = a1 + a2 + · · · + a7 ? Pu 5.3.3 Suppose an = 2an−1 + 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% compounded monthly. That means that each month .5% is added to the principal or the principal is multiplied by 1.005. I borrow $100,000. (a) How much will I owe the bank at the end of 2 years? 34 CHAPTER 5. ARITHMETIC AND GEOMETRIC PROGRESSIONS 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 the present 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 2 years? 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 simply 2. 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 formerly pointed out that the decimal expansion of 1/3 does not terminate. It consists of an infinite sequence of threes. In mathematical terms, it can be regarded as 3/10 + 3/100 + 3/1000 + · · ·. By the rules of sums of infinite geometric progressions 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 that the infinite decimal consisting of all nines is equivalent to 1. In our problem sets we explore the fact that every rational number can be expressed in decimal notation as either having a finite number of nonzero digits after the decimal point, or having a repeating sequence of digits such as .25343434 . . . which is the sum .25 + .0034 · (1 + .01 + .0001 + · · ·) = .25 + .0034(1 − 1/100) which is obviously rational. Thus the fact that such a repeating sequence is rational is not surprising. What is a little less obvious is 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 mathematician and physicist. He was given a puzzle, and the method he used to solve it would determine whether he was primarily a physicist or mathematician. Two trains head toward each other at 50 miles per hour and 30 miles per 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 first train and continues to shuttle between them until it is crushed. How far has the fly flown? Von Neumann answered instantly. He was told that he was a physicist because the mathematician would have figured out how far the fly flew each time till he was crushed, and added the infinite series. A physicist would have used a much faster and simpler calculation. When this was explained to him, he admitted that he had added the infinite series. The simpler calculation is a puzzle for the reader. The infinite series is a problem which is complicated and difficult, but not very tricky. Try both approaches if you can think of the physicist’s. 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 what may seem very complicated otherwise. The down side is that it takes practice to become familiar with a great deal of notation, and the usefulness declines if there is little opportunity to use it. The standard notation for sums involves the greek letter capital sigma which follows, Σ. The sum of a sequence of terms such as a5 + a6 + · · · + a15 is represented by capital sigma with the lower and upper limits below and above the sigma. That is a5 + a6 + · · · + a15 = 15 X ai . i=5 In this notation the character i seems to come out of nowhere. What is it doing? It is called a dummy index. The right hand side is saying that we are adding all the numbers ai where the symbol i varies in steps of 1 from 5 to 15. It is a dummy because the right hand side would be the same if we changed i to j or any other symbol that had not previously been given a meaning. The sigma notation is useful, because the alternative we have been using is sometimes clumsy and misleading. I often like the alternative, but it is embarassing to write a1 + a2 + a3 + · · · + an−1 + an when n is one or two. One should keep in mind that the usual convention for the use of sigma has the upper index greater than or equal to the lower index. Otherwise there is some confusion. Using this notation, the result for arithmetic and 36 CHAPTER 5. ARITHMETIC AND GEOMETRIC PROGRESSIONS geometric progressions can be written in the following equations. n X a + (i − 1)d = n[2a + (n − 1)d]/2 i=1 n X i=1 ari−1 = a(1 − rn )/(1 − r) 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 is therefore 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 6= 0. Figure 6.1.1 15 10 y 5 0 x -5 -10 -1 0 1 2 3 4 More generally, we have an n-th degree polynomial which has the form y = Pn (x) = a0 xn + a1 xn−1 + · · · + an−1 x + an 37 38 CHAPTER 6. POLYNOMIALS when a0 6= 0, and n is a positive integer. If we plot y = x3 − 3x2 + x − 7, we see that y becomes large for large x and highly negative for large negative values of x. (See Figure 6.1.1.) As x increases y rises from highly negative values 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 can be represented by a third degree polynomial. We shall concentrate on the linear and quadratic cases. The form y = ax + b yields a straight line which intercepts the y-axis at 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 if a is positive. If a is negative the line slopes down and y decreases by |a| for each increase of 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 6= 0 is represented by a curve called a parabola where y rises as |x| gets large if a > 0 and y decreases as |x| gets large if a < 0. This parabola either never crosses the x-axis (y = 0), or it crosses it twice, or simply touches the 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. Since b 2 b2 a(x + 2a ) = a(x2 + ab x + 4a 2) = ax2 + bx + ax2 + bx + c = a(x + a(x + (x + b 2 2a ) b 2 2a ) b 2 2a ) b2 4a + (c − = b2 −4ac 4a = b2 −4ac . 4a2 b2 4a ) = 0 6.3. POLYNOMIALS AS PRODUCTS 39 Finally, if the discriminant d = b2 − 4ac ≥ 0, √ b b2 − 4ac x+ =± 2a 2a or √ b2 − 4ac . (6.1) 2a 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 no solution because there is no real square root of a negative number. Note that if 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 were two 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 don’t recommend forcing yourself to memorize this equation, but I would suggest recalling the idea of completing the 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 the square. x= 6.3 −b ± 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 = x1 x2 . Then y2 = 0 if and only x = x1 or x = x2 . Similarly, y3 = (x − x1 )(x − x2 )(x − x3 ) = x3 + a1 x2 + a2 x + a3 where a1 = −(x1 + x2 + x3 ) , a2 = x1 x2 + x1 x3 + x2 x3 , and a3 = −x1 x2 x3 . In English terms, −a1 is the sum of the xi , a2 is the sum of the products two at a time, and −a3 is the product of all three. This result generalizes for polynomials which are the product of n factors of the form (x − xi ). 40 CHAPTER 6. POLYNOMIALS A special case of interest is that of y2 = (x−x1 )2 = x2 −2xx1 +x21 . In that case there is only one solution of y2 = 0. That is x = x1 . But in this case x1 is considered a double solution, because (x − x1 ) can be factored out of the polynomial for y2 twice.. There is a sophisticated theorem in the analysis of complex variables that states that every polynomial of degree n can be factored into a0 times the product of n terms of the form (x − xi ), where the xi are complex numbers. Since we will concentrate on real numbers in this text, we shall not use that theorem directly, but it implies that every polynomial with real coefficients, a0 , a1 , · · · , an can be factored into a product of terms of the form (x − xi ) or ((x − ui )2 + vi2 ), where xi ,ui and vi are real. Since the latter term is always positive, the polynomial can be 0 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 rational number, the ratio of two polynomials y = P1 (x)/P2 (x) is called a rational function. If P1 and P2 have a common factor P0 (x), that can be factored out, so usually we assume that they have no common factor. Let us show, by example, how P1 can be divided by a linear polynomial P2 . The procedure is basically the same as the one used in ordinary long division. Let P1 (x) = 6x3 + 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 29/4 = 3x2 − x/2 − 1/4 + . 2x + 1 2x + 1 6.5. REDUCED FORM 41 Just as in ordinary long division, we indicate an algorithm which puts the above process into a regular format. 2x + 1 3x2 6x3 6x3 +x/2 +4x2 +3x2 x2 x2 −1/4 +0 · x +7 +0 · x +x/2 −x/2 −x/2 +7 −1/4 29/4 More 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). In that case, if P2 (x) = bx + c, and hence vanishes at x = x0 = −c/b, so will P1 . Otherwise we will have P1 (x) R = Q(x) + . P2 (x) bx + c If R 6= 0, the expression 1/(bx + c) behaves strangely for x near x0 . To appreciate this, simply plot y = 1/(2x + 1), and note that for x slightly larger than -1/2, y is very large, and for x slightly less than −1/2 , y is highly negative. For x far from -1/2, y becomes small. The plot consists of the two branches of a curve called a hyperbola. A similar process of division can also be carried out when the divisor is not linear. For example, if P2 (x) = 3x2 + 6x + 5, the procedures described above need be modified in only the most trivial fashion. In this case the remainder 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 and the remainder R. 6.5 Reduced Form In the arithmetic of fractions it is often desirable to represent the fraction in reduced form where the numerator and denominator are integers with no common factor. A similar reduction problem appears when dealing with square roots. For example, we can simplify √ 2+ 3 √ . y= 4−3 3 42 CHAPTER 6. POLYNOMIALS p √ If we note that (a + b (c))(a −√b c) = a2 − b2 c, we can multiply numerator and denominator of y by 4 + 3 3, to get √ √ √ (2 + 3)(4 + 3 3) = −17/11 − (10/11) 3, y= 16 − 27 which is a simpler looking form than that offered originally. Chapter 7 Analytic Geometry 7.1 Introduction Combining methods of Algebra and Geometry has often proved to be very insightful in both fields. We will not expand much in this area of Mathematics, but the theorem of Pythagoras is useful in calculating distances between two 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 is w part of the way from x1 toward x2 . In particular (x1 + x2 )/2 is half way between x1 and x2 , while (2x1 + x2 )/3 is only one third of the way from x1 to 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 1 + x1 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) 1 1 1 +w ≥ , x1 x2 (1 − w)x1 + wx2 with equality if and only if x1 = x2 . 43 44 7.2 CHAPTER 7. ANALYTIC GEOMETRY Theorem of Pythagoras The main point of this section is to announce the fact that the distance d between two points on a graph with coordinates (x1 , y1 ) and (x2 , y2 ) satisfies, d2 = (x2 − x1 )2 + (y2 − y1 )2 , p and hence the distance from (x, y) to the origin is x2 + y 2 . 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 x p 2 and y, is x + y 2 . This statement uses several technical terms in geometry, 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 • • x y • z x y • Figure 7.2.1 is the basis of a proof of the theorem, which we outline without most of the details that justify the statement that the areas of the outer square, each of the inner triangles, and the inner square are (x + y)2 , xy/2, and z 2 respectively. Totalling the areas,we have (x + y)2 = z 2 + 4xy/2 but (x + y)2 = x2 + 2xy + y 2 , and it follows that x2 + y 2 = z 2 , where x and y are the sides of the right triangle with hypoteneuse z. 7.3. CIRCLES AND ELLIPSES 7.3 45 Circles and Ellipses The points (x, y) for which x2 + y 2 = r2 where r > 0, are r units away from the origin, and characterize a circle of radius r, and diameter 2r with center at the origin. The points (x, y) for which x2 y 2 + 2 =1 a2 b is called an ellipse with major and minor axes a and b. The major axis corresponds to the larger of a and b. One characteristic of an ellipse, is that it has two focal points and that the sum of the distances from the points on the ellipse to the focal points is constant. Figure 7.3.1 presents a circle with radius r = 1, and an ellipse with a = 3, and b = 2. The circle is a special or degenerate case of an ellipse where the two focal points coincide at the center. Figure 7.3.1 2 y 1 0 x -1 -2 -3 -2 -1 0 1 2 3 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, and how far it is from there. 46 CHAPTER 7. ANALYTIC GEOMETRY Pr 7.3.3 (a) Show that if a ≥ b and the p two focal points of the ellipse above are at (c, 0) and (−c, 0), then c = (a2 − b2 ), and that the sum of the distances to the focal points is 2a. (b) Establish that indeed for all points on the elipse, the sum of the distances is 2a. This problem is not easy. Use a pencil and two tacks, holding down the ends of a cord to draw an ellipse. Chapter 8 Simultaneous Equations 8.1 Two Linear Equations Example 8.1.1. Consider the two equations in which x and y appear in linear form (to the first power). 3x + 2y = 7 x + y = 9. Is there a solution which yields values of x and y for which both equations are satisfied? The answer is yes. We take y = 9−x from the second equation and substitute for that in the first equations yielding 3x + 2(9 − x) = x + 18 = 7 from which it follows that x = −11 and hence y = 20. To check we see that 3 · (−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 to the solution of the two simultaneous equations. Those equations are called linear equations because both x and y appear only in the first power. Is it always the case that two such equations have a unique soluton? No. For example, 3x + 2y = 7 and 6x + 4y = 9 represent two parallel lines which never intersect, and have no solution. On the other hand the equations 3x + 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 these equations in the form a1 x + a2 y = a0 b1 x + b2 y = b0 47 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 that condition is that a1 b2 6= a2 b1 . If a1 b2 = a2 b1 , then we have either no solution or infinitely many solutions depending on whether or not the two equations represent two parallel lines or the same line. Another approach to solving these two equations when a1 b2 6= a2 b1 is indicated 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 of y in the two equations are equal we can subtract the second equation from the first to obtain x = 10 from which it easily follows that 3y = 11 − 2x or y = −3, which checks with the original equations. We can deal with 3 linear equations in three variables x, y, and z by either of the two methods indicated above. We could substitute in one equation to find an expression for one of the variables in terms of the other two. Substituting that expression in the other two equations, leads to two equations in the two remaining variables. That is an extension of the substitution method, and we used that in Chapter 3. The analogue of the method of equating coefficients used above, is to multiply two equations by the coefficients which make those of one of the variables match. Subtracting the resulting two equations leads to a fourth equation with one of the three variables missing. Then apply the same technique to another pair of the original equations to equalize the coefficients of the variable that was cancelled before. Once more we get another equation in the other two variables. Either way we get two equations in two variables. We can use either method to solve for one of the variables. Use the remaining equations to solve for the other variables. This may sound a bit abstract, but trying the techniques on the problems will show how the method works. 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 parallel planes, they intersect in a straight line. If that straight line is not in the plane of the third equation, or parallel to that plane, the planes will intersect in one unique point. Otherwise we may have no solution or many possible solutions. The general theory of determinants and matrices was developed to solve n linear equations in n unknowns. We shall not elaborate on that theory here. 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 = 1 2x − 5y = −1 4x + 3y + 2z = 4. Pr 8.1.3 Find a solution of the two simultaneous equations below in terms of the coefficients assuming that a1 b2 6= a2 b1 . a1 x + a2 y = a0 b1 x + b2 y = b0 8.2 Nonlinear Equations We now consider a special example where one equation is linear, but the second is not. Example 8.2.1 Let 3x + 2y = 13 x2 + 2y 2 = 17 What are the values of x and y? Before solving these equations, it is well to note what they mean graphically. The first equation is a straight line with a y-intercept of 6.5, √ and the 17, 0) and second equation is an ellipse which goes through the points ±( √ ±(0, 8.5) and is centered at the origin. The solution or solutions correspond to those points, if any, where the line intersects the ellipse. 50 CHAPTER 8. SIMULTANEOUS EQUATIONS We proceed algebraically. From the first equation, we have y = 6.5−1.5x. Substituting that in the second equation, we have x2 + 2(6.5 − 1.5x)2 = x2 + 84.5 − 39x + 4.5x2 = 17. Collecting terms with the same powers of x, we have 5.5x2 − 39x + 67.5 = 0 √ which 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 of y are y = 2 and y = 4/11. Pr 8.2.1 Solve for x and y . 2x − y = 6 xy = 3. Pr 8.2.2 Find the value of b for which the line y = 2x + b is tangent to the parabola y = x2 . That means that the line touches the parabola, but does not cross it at the point of contact. What is the point of contact? Chapter 9 Combinations and Permutations 9.1 Introduction Combinations and permutations are useful tools for the study of Probability involving equally likely events. While we won’t pursue the field of Probability, we will find the elementary results very useful to derive the Binomial Theorem which provides a representation of an expression for (a + b)n for integer values of n. 9.2 A Fundamental Rule A fundamental rule is that if one thing can be done in m ways, and a second thing can be done in n ways for each of the first m ways, then the two things can 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 possibilities systematically, we get 1R,1B,2R,2B,3R,3B,4R,4B, where R stands for 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 be the first choice and then a second card from among those remaining. How many 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 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 we could select the second card from the same four cards, we would have had 16 possible pairs. We now consider a third example, a variation of Example 9.2.2. Example 9.2.3. Suppose that we wish to select a committee of 2 individuals from a group of 4 people. How many possible such committees are there where 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 which they were selected, we would have the previous list of 12 pairs. But each committee of 2 individuals is counted twice in that list. There are only 12/2=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 these integers will be called a permutation of those integers. In the same way, an ordered arrangement of any n distinct items is a permutation of these items. How many permutations or different arrangements can be made of the n integers or of any n distinct items? There are n ways of selecting the first item. For each of these there are (n − 1) ways of selecting the second item. Thus there are n(n − 1) ways of selecting the first pair. For each of these there are n−2 ways of selecting a third item. Proceeding in this way, it is clear that altogether there are 1·2·3 · · · (n−1)·n different permutations of the n items. For convenience we introduce the term n factorial to represent the product of the first n integers. The notation for n factorial is given by the 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 positive integers 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 out of n items where r is a positive integer less than n. But our derivation of n factorial incidentally answered this question. It is clear that the number of 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! 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 group of distinct individuals. In Example 9.2.3 above, a committee represented a subgroup of 2 of the 4 individuals, where the order was not important, and hence a combination of 2 out of 4. How many combinations of r out of n distinct members of a group are there? The amount Pn,r represented the number of arrangements of r items of the group of n where order was important. But for any choice of r 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 n to, and including, n − r + 1, and the denominator is the product of the first r integers. The equation above highlights one important property of the number of combinations. That is, for 0 ≤ r ≤ n Cn,r = Cn,n−r . From a certain point of view, this property is not surprising. For each committee of r members selected, there is a corresponding committee of n − r members who were not selected. The Pascal Triangle below is another interesting representation of combinations. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Each horizontal line is formed by starting with a 1 in the middle for the apex. Then each following line is obtained by adding the two numbers immediately to the right and left on the preceeding line. The n-th line below the 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(n−1),(r−1) + C(n−1),r 54 CHAPTER 9. COMBINATIONS AND PERMUTATIONS Note that we have defined C0,0 = 1. Two properties of the Pascal Triangle that are worth noting are that the sum of the elements of the n-th row 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 because of the symmetry in those rows. But it is not immediately obvious that we should expect 1 − 4 + 6 − 4 + 1 = 0 These properties are a consequence of the binomial theorem. described in the next section. Combinations and permutations play an important role in probability theory, on which we shall not expend much text. But combinations are also fundamental 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 note that (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 + a2 b + 2a2 b + 2ab2 + ab2 + b3 = a3 + 3a2 b + 3ab2 + b3 . Note that the coefficients of the terms in (a + b)2 and (a + b)3 are the combinations that appear in the second and third rows below the apex of the Pascal Triangle. This is no accident. We shall derive (a + b)n = Cn,0 an + Cn,1 an−1 b + · · · + Cn,i an−i bi + · · · + Cn,n bn . This equation can also be represented using the summation symbol, as (a + b)n = n X Cn,i an−i bi . i=0 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 9.5. BINOMIAL THEOREM 55 is 2n . By taking a = 1, b = −1 , we obtain the fact that the sum of the items with alternating signs is zero. There are several ways of deriving the theorem. I choose the following, which is easier to follow if one writes (a+b) in a vertical column n times. How do we multiply these n terms to get (a + b)n ? We sum all possible products where one of the items a and b is taken from each row. A typical product is an−i bi where i of the n terms in the product were b’s and the rest were a’s. How many such terms are there? As many as the number of ways we can select i out of n rows to carry the b. In other words there are Cn,i products of the given form √ and the binomial expansion is validated. Pu 9.5.1 The quantity ( 2 + 1)500 is irrational, and has an infinite decimal representation. Find the digit in the 95-th location after the decimal point. Pr 9.5.2 The fact that (x + y)m (x + y)n = (x + y)m+n implies a formula involving Cn,r when coefficients of xm+n−k y k are matched. Find that formula, and check it out for a simple case. 56 CHAPTER 9. COMBINATIONS AND PERMUTATIONS Chapter 10 Summary An essential element in the application of Algebra to solving relevant problems is the ability to translate those problems into symbolic statements, using a convenient notation which makes potentially complex relationships relatively simple and transparent. A major tool is that of manipulating symbols that represent quantities that are unknown or unspecified. 10.1 Notation In ordinary conversation, it is possible to speak imprecisely with a certain amount of ambiguity, and still have oneself clearly understood. The listener often uses the context to deduce the actual intended message, sometimes even when the conversation has errors. The algebraist who writes equations is expected to be quite precise and, except for some conventions, very little ambiguity is tolerated. The product of two numbers x and y is designated by xy or x · y. If we have x multiplying y + z, we can not write x · y + z, because that would be interpreted as x multiplying y and then adding z to that product. Instead, we use parentheses, as in x(y + z), to separate y + z from x. For more complicated expressions we sometimes find parentheses of different sizes, brackets and braces to be useful to make visually transparent what is going on. In algebraic notation we tend to avoid using symbols which have several characters. Instead we often find it useful to use subscripts to indicated several variables of a similar kind. Superscripts are usually used to represent exponents or powers. Thus xn+1 = x · xn , and x1/n is the n-th root of x, i.e. a number y for which the n-th power, y n = x. It is customary to use the symbols i, j, k, l, m, n to represent integers, 57 58 CHAPTER 10. SUMMARY but this is not always the case. In more advanced mathematics, many symbols tend to be used in conventional ways. For example the greek lower case pi, π = 3.14159... relates the circumference and area of a circle to its diameter. The letter e = 2.71828 . . . has a special interpretation. There are so many conventional symbols that mathematicians make use of much of the Greek alphabet. Even the Hebrew letter aleph has a special conventional interpretation. The product of the first n integers is called factorial n and designated by the symbol n!. Here n is a positive integer, but it is convenient to define 0! = 1!. As in conversation, the context will often tend to eliminate ambiguity. For example, in the expression x1 + x2 + · · · + xi + · · · + xn both i and n are understood to be positive integers. The expression represents the sum of the first n variables xi where i is a dummy index running from the integers 1 to n. This is not a very precise notation because it would be misleading for n = 1 or 2. A more precise notation for that expression that is used often in more advanced Mathematics is n X xi i=1 where use is made of the capital greek sigma. We have seldom used this notation in order to minimize notational innovation. Inequalities are represented by >, ≥, <, ≤, 6=, which stand for greater than, greater than or equal, less than, less than or equal, and not equal respectively. The absolute value of x which represents the distance from x to 0 is given by |x|. One of my major problems in communication as a statistician dealing with scientists is that we often have different conventional notations for the same thing. In Algebra we often use letters in the last part of the alphabet for unknown quantities, and in the first part of the alphabet for as yet unspecified quantities which are often known by the user of the algebra. Thus y = ax + b is often used in cases where a and b are known, to indicate a relation between x and y. In Physics the symbols for v often stands for volts or velocity, while i stands for current and m for mass. The symbol c stands for the velocity of light, while in Algebra it is often used to represent a constant. In the problem sets, we will occasionally come across these different uses of conventional symbols, but this should not be so frequent as to cause difficulty. 10.2. BASIC LAWS 10.2 59 Basic Laws Fundamental properties of real numbers that lead to useful methods are given below. x + y = y + x commutative law of addition xy = yx commutative law of multiplication x + (y + z) = (x + y) + z associative law of addition x(yz) = (xy)z associative law of multiplication x(y + z) = xy + xz distributive law x + (−x) = 0 inverse for addition x · x−1 = 1 inverse for multiplication for x 6= 0 If x is positive and y is real, there is a unique positive z for which z = xy , with the following properties. xy · xu = xy+u for x > 0 xy · uy = (xu)y for x > 0 and u > 0 (xy )u = xyu for x > 0 x0 = 1 for x 6= 0 x−1 = 1/x for x 6= 0 For positive and negative integer exponents x need not be positive for xn to satisfy the above rules, but 00 is not defined. For inequalities, we have x + y > 0 if x > 0 and y ≥ 0 xy > 0 if x > 0 and y > 0 xy < 0 if x > 0 and y < 0 60 CHAPTER 10. SUMMARY 10.3 Major Results We have introduced the Binomial Theorem which states that for positive integers n, (a + b)n = an + Cn,1 an−1 b + · · · + Cn,i an−i bi + · · · + bn where n! i!(n − i)! and 0! = 1, and for positive integers n, n! or n factorial is the product of the first n integers. The number of combinations of n items taken r at a time is Cn,r . The number of permutations of n items taken r at a time is Cn,i = Pn,r = Cn,r r! = n!/(n − r)! The sum of an arithmetic progression is given by s = (a + l)/n where a, l, and n are the first and last terms of the sequence of n terms. The sum of a geometric progression of n terms with initial term a and ratio r 6= 1 is given by s = a(rn − 1)/(r − 1) = a(1 − rn )/(1 − r). The linear equation ax + b = 0 has the solution x = −b/a if a 6= 0. The quadratic equation ax2 + bx + c = 0 has the two solutions p (b2 − 4ac) 2a 2 if a 6= 0 and d = b − 4ac > 0. By completing the square we can write x= −b ± ax2 + bx + c = a(x − b/2a)2 − (b2 − 4ac)/4a Most of the material in this summary should be so obvious to the reader of this text that very few feats of memory are required. A few of the items worth memorizing are the formula for the binomial theorem, the solution of the quadratic equation, and the sums of the arithmetic and geometric progressions. More important than memorizing these formulae, which can be looked up in references, are the ideas and methods used in deriving them. Chapter 11 Solutions and Hints for Puzzles 11.1 Solutions for Exercises and Problems Pr 2.3.1 2/3 + 4/15 − 3/10 = 20/30 + 8/30 − 9/30 = 19/30 7/3 + 5/12 + 1/18 = 84/36 + 15/36 + 2/36 = 101/36 19/30 19·6 101/36 = 5·101 = 114/505. Ex 2.5.1 (a) 1.886; (b) 1.024; (c) -97.202; 6.5604 (d) 29.230 Pr 3.1.1 Let x be the original value in dollars. Simon paid 1.01x. He received .99(1.2).8x. His loss is 1.01x − .99(1.2).8x = 200, .0596x = 200, and x = 3, 356. Pr 3.1.2 Let B = rate for Bill, and J = rate for John. Then 3B = 5J = 1. The combined rate is B + J, and the time is t where t(B + J) = 2 and t = 2/(1/3 + 1/5) = 30/8 = 15/4. Ex 3.1.3 B + J = 100 and 2B = J/2. Then B = J/4; 5J/4 = 100; and J = 80 and B = 20. Ex 3.1.4 J = B + 10; B + 50 = 2(J + 10); B + 50 = 2(B + 20); B = 10; Bill has 10 + 5 · 10 = 60 then and John has 20 + 5 · 2 = 30 then. Pr 3.3.1 (x + y) + z = (y + x) + z = y + (x + z) = y + (z + x) using the commutative, associative, and commutative laws in order. Pr 3.3.3 (a) no. 3 ⊕ 2 = 9 while 2 ⊕ 3 = 8. (b) yes. x y = |x/y − y/x| = |y/x − x/y| = y x. Pr 3.3.4 Let m be the number of additional months needed to break even. 30 + 5m = 10 + 7m; 2m = 20; m = 10. The cumulated losses, in thousands of dollars, is 20 + 18 + 16 + · · · + 2 = 110. Win.com will go bankrupt unless it can raise $10,000 more. 61 62 CHAPTER 11. SOLUTIONS AND HINTS FOR PUZZLES Pr 3.3.5 e = mv 2 /2. Pr 3.3.6 px + .25y = .1(x + y); (.1 − p)x = (.25 − .1)y; y = [(.1 − p)/.15]x. Pr 3.3.7 (a) Let i1 = 9k1 + r1 , i2 = 9k2 + r2 where 0 ≤ r1 ≤ 8, 0 ≤ r2 ≤ 8. Then i1 − i2 = 9(k1 − k2 ) + (r1 − r2 ), and r1 − r2 is divisible by 9. This is possible only if r1 = r2 (b) If the digits of i are m0 , m1 , m2 , · · · , mn , then i = m0 + 10m1 + 100m2 + · · · + 10n mn and s = m0 + m1 + m2 + · · · + mn . Then i − s = 9m1 + 99m2 + · · · + 99 . . . 9mn is divisible by 9 and i and s have the same remainder. (c) If i1 = 9k1 + r1 and i2 = 9k2 + r2 , then (i1 + i2 ) − (r1 + r2 ) is divisible by 9, and i1 + i2 and r1 + r2 have the same remainder. (d) If i1 = 9k1 + r1 and i2 = 9k2 + r2 , then i1 i2 = 81k1 k2 + 9(r1 + r2 ) + r1 r2 and i1 i2 − r1 r2 is divisible by 9, and i1 i2 and r1 r2 have the same remainder. Pr 3.4.1 (a) (x1 +x2 )−(y1 +y2 ) = (x1 −y1 )+(x2 −y2 ) > 0 since (x1 −y1 ) > 0 and (x2 − y2 ) ≥ 0. (b) xz − yz = (x − y)z > 0. (c) xz − yz = (x − y)z < 0 . Pr 3.4.2 (a) z/xy > 0. Then z/y − z/x = (x − y)(z/xy) > 0. (b) similar proof except that −z/xy > 0. Ex 3.4.3 4 Pr 4.2.1 (1 + 1/4)4 is larger than (1 + 1/2)2 . Pr 4.4.1 The googol is much larger. Pr 4.4.2 (b) If y1 = log(x1 ) and y2 = log(x2 ), x1 = 10y1 , x2 = 10y2 and x1 x2 = 10y1 +y2 and y1 + y2 = log(x1 x2 ). Note that we have used exponents that are not necessariy rational. (a) follows from (b) since log(10) = 1. (c) If y = log(x), x = 10y , and xz = (10y )z = 10yz . Pr 4.5.1 ±(x − y). Pr 4.5.2 yn+1 = 1 x yn + 2 yn 2 yn+1 = x2 1 2 yn + 2x + 2 4 yn −x = 1 2 x2 yn − 2x + 2 4 yn 2 yn+1 " # " # 11.1. SOLUTIONS FOR EXERCISES AND PROBLEMS 63 en+1 = (yn − x/yn )2 /4 = e2n /4yn2 . Since e2n and yn2 are nonnegative, en+1 ≥ 0. If en ≥ 0, en = yn2 − x < yn2 (assuming x > 0). In that case, en /yn2 < 1, and en+1 = e2n /4yn2 ≤ en /4. In fact the rate at which en approaches 0 accelerates rapidly, for en+1 is approximately e2n /4x, and as en becomes small, e2n becomes considerably smaller. Ex 5.2.1 s = 100(−10 + 99 · 2)/2 = 9400. Ex 5.3.1 (a) s = a1 ; (b) s = 0. Ex 5.3.2 s = 3(27 − 1) = 381. Pr 5.3.4 (a) 100, 000(1.005)24 = 112, 716; (b)v = 4, 000(1.005)−1 [1 − (1.005)−24 ]/[1 − (1.005)−1 ] = 90, 251 ; (c) (1.005)24 (100, 000 − v) = 10, 989. Pr 5.4.2 For some problems it is easier to keep track of important issues by using notation, even for quantities that are known. We will let v0 , v1 and v2 represent the velocities of the fly and the two trains. Suppose that they are d miles apart and the fly goes from the first train to the second in time t1 . It goes a distance d − t1 v2 at a velocity v0 , while the remaining distance apart becomes d1 = d − t1 (v1 + v2 ). Thus d − t1 v2 = t1 v0 and t1 = d/(v0 + v2 ). On the return to the first train, it takes time t2 = [d − t1 (v1 + v2 )]/(v0 + v1 ) by the same reasoning. It follows that the fly has traveled v0 (t1 + t2 ) = 2dv02 /[(v0 + v1 )(v0 + v2 )] = a and the distance apart has been reduced from d to d∗ = d−(t1 +t2 )(v1 +v2 ) = d(v0 −v1 )(v0 −v2 )/[(v0 +v1 )(v0 +v2 )] = rd. The total distance traveled by the fly shuttling back and forth is the sum of the infinite geometric series, a+ar +ar2 +· · · = a/(1−r) = dv0 /(v1 +v2 ) = 100. Pr 6.2.1 (a) x = 3, −2; (b) x = −4; (c) no real solution Pr 6.2.2 3x2 + 5x − 2 = 3(x + 5/6)2 − 2 − 3 · 25/36 = 3(x + 5/6)2 − 49/12 which is obviously minimized when x = −5/6 with the minimum value of −49/12. Pr 6.4.1 Q = 2x2 + 3x + 1; R = 3x + 4. Pr 7.1.1 [(1−w)x21 +wx22 ]−[(1−w)x1 +wx2 ]2 = w(1−w)(x1 −x2 )2 ≥ 0 with equality if and only if x1 = x2 . This means that the chord connecting two points on the parabola y = x2 lies above the curve since (1 − w)x1 + wx2 = x1 + w(x2 − x1 ) is the point w part of the way from x1 toward x2 . Pr 7.1.2 [(1 − w)/x1 + w/x2 ] − 1/[(1 − w)x1 + wx2 ] = w(1 − w)(x1 − x2 )2 /x1 x2 [(1 − w)x1 + wx2 ] ≥ 0. Pr 7.3.1 The squared distance from a point on the line to the origin is 2 = x2 + (2x + 7)2 = 5x2 + 28x + 49 which attains its minimum value x2 + y√ of 1.4 5 = 3.13 at (x, y) = (−2.8, 1.4). 64 CHAPTER 11. SOLUTIONS AND HINTS FOR PUZZLES Pr 7.3.2 The squared distance from a point on the parabola to the √ origin is 2 2 2 4 2 x + (x − 4) = x − √ 7x + 16 which attains its minimum value of 3.75 = 1.936 at (x, y) = (± 3.5, −.5). Pr 7.3.3 (a) The sum of the distances 2a. The √ to (a, 0) is (a − c) + (a 2+ c) = 2 2 2 sum of the distances to (0, b) is 2 b + c = 2a. Therefore a = b + c2 . (b) Let the two distances from (x, y) be d1 and d2 where d21 = (x − c)2 + y 2 = x2 +y 2 +c2 −2cx and d22 = (x+c)2 +y 2 = x2 +y 2 +c2 +2cx. We wish to show that d1 = 2a − d2 or d21 = 4a2 − 4ad2 + d22 or 4ad2 = 4a2 + d22 − d21 = 4a2 + 4cx or d2 = a + cx/a or d22 = x2 + y 2 + c2 + 2cx = a2 + 2cx + c2 x2 /a2 . But y 2 = b2 (1 − x2 /a2 ). It suffices to show that x2 + b2 − b2 x2 /a2 + c2 = a2 + c2 x2 /a2 or x2 (1 − b2 /a2 − c2 /a2 ) = a2 − b2 − c2 . Since b2 + c2 = a2 , both sides of the last equation are equal to 0. Pr 8.1.1 (x, y) = (2, 3) Pr 8.1.2 (x, y, z) = (3/4, 1/2, −1/4). Pr 8.1.3 Multiplying the first equation by b2 and the second by a2 and then subtracting we find x = (a0 b2 − a2 b0 )/(a1 b2 − a2 b1 ). Similarly, we get y = (a1 b0 − a0 b1 )/(a1 b2 − a2 b1 ). √ √ Pr 8.2.1 There √ are two√solutions for (x, y). These are [(3 + 3)/2, (3 − 3)] and [(3 − 3)/2, (3 + 3)]. Pr 8.2.2 x2 = 2x + b must have one solution. The discriminant,4 + 4b = 0. Thus b = −1 and (x,y)=(1, 1). Pr 9.4.2 (a)C13,6 C10,2 = 77, 220; (b) C21,8 = 203, 490. Pr 9.5.2 Σki=0 Cm,i Cn,k−i = Cm+n,k where Cn,j = 0 for j > n. 11.2 Hints for Puzzles Pu 2.3.2 If you think that a piece is too big you should cut it down. Pu 5.3.3 Let an = 2n bn or cn = an + 1 = 2cn−1 . Pu 5.4.1 Reduce the case to m < n for the rational m/n. Apply long division. Wait till the remainder coincides with a previous remainder after the decimal point. Pu 5.4.2 It√takes one hour for √ the trains to meet. 500 = m + n 2 where m and n are integers. Pu 9.5.1 ( 2 + 1) √ √ ( 2 − 1)500 = m − n 2 < 10−100 is very small. 11.3 Solutions for Puzzles Pu 2.3.2 If there are 3 boys, the first is asked to cut out what he regards as a fair share. The next boy can cut it down if he thinks the piece is too large. 11.3. SOLUTIONS FOR PUZZLES 65 He then passes it to the third who can take it if he thinks it is fair or better. If he thinks it is too small, it goes to the last person who cut into the pie. The remaining boys divide what is left of the pie. This solution also works for more than three boys. Pu 5.3.3 an = 2n − 1 and sn = 2(2n − 1) − n. Pu 5.4.1 In the long division process the remainder must be an integer less than n, and so within n steps one of the remainders must coincide with a former one after the decimal point. Then the decimal representation begins to repeat. Pu 5.4.2 The fly√flies one hour √ at 100 miles per hour. 100 miles. Pu 9.5.1 (m + n 2) + (m − n 2) = 2m is an integer with just zeros √ after the decimal point. Subtracting the very small second term, m − n 2 , leaves well over 100 nines after the decimal point.