Mathematics

the Practical,

the Logical, and

the Beautiful,

by

Benjamin Baumslag and Frank Levin

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PREFACE

This book is intended for:

1. Readers who liked mathematics at school but never studied it further.

2. Young people with mathematical talent.

3. Teachers who are looking for inspirational material for school.

Each reader will study different parts of the book in different ways. Not many are likely to read every page, but then, not every visitor to an art museum looks at every painting. You do what you find enjoyable and find the time for; you may even think of coming several times. Chapters can be read in almost any order. Nor is it necessary to read the whole of a chapter. Many sections indicate optional material. In particular, the solved problems are optional. The more difficult calculations are often put into the solved problems, and even if one does not read these, one still gets an understandable account.

The aim of the book is expressed by its title, namely to give examples of mathematics which is practical and useful in everyday life, examples of beautiful mathematics, and to illustrate the logical arguments used in mathematics, i.e. proof. Practical topics include approximation and the use of the powers of 10 notation. Then there is percentages, and estimating various quantities with simple calculations (Chapter 4), some knowledge of graphs, some probability and statistics. There is also a chapter on units like Watts and horsepower.

Proof is demonstrated for instance by proving Pythagoras’s theorem, and using numbers to derive Euclidean Geometry.

There are some beautiful classical results1 such as examples of a finite Geometry and a Projective Geometry, the existence of an infinite number of primes, and the irrationality of the square root of 2. Fermat’s little and last theorems and the estimation of the number of primes less than a given number N, are discussed. Finally, we return to counting but this time counting infinite sets, and have the striking results of Cantor such as there are as many points on a line of length 1 as on a line of length 2.

Much could be added, but we have chosen to be brief in order to present a more easily comprehendible book. There is much of value and interest anyway.

We have tried as far as possible to provide mathematics, which the reader can verify for himself or herself, and not have to rely on our authority.

The book begins with topics that would normally be discussed in school, and ends with topics, which would normally appear in a university course on mathematics. The careful choice of material and presentation provides an account which is understandable by those who have studied secondary school mathematics. Because we give a self-contained account, the reader who has forgotten the school mathematics will be reminded of some of the details. What is

1 Until you have read the relevant chapters the following is only partly understandable.

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required of the reader is a flexible mind, curiosity, and the sort of patience and determination that is required to play bridge or chess or solve cross- word or sudoku puzzles.

The examples are from different countries, England or Sweden or the U.S.A. But these are only examples which are meant to illustrate various methods, and the reader will with their help apply these techniques to their own interests and needs.

The material in this book is not original to us.

There are many brilliant ideas in mathematics. If we can introduce you to some of them it will be an honor and a privilege.

AcknowledgementsWe gratefully acknowledge help from the following: David Baumslag, Pia Baumslag, Jeff Bourne, János Hegedüs, Geoffrey Howson, Lars-Göran Larsson, Marko Marin, Anatoliy Malyarenko, Julia Medin, Nils Mårtenson, Gunilla Sandberg, Abe Shenitzer, Naomi Yodaiken, Ralph Yodaiken.

Västerås and Swansea 2007

Benjamin Baumslag and Frank Levin

ISBN

© xxxxxxxxx

CONTENTS

PREFACE iii

Acknowledgements........................................................................................................iv

CONTENTS iv

1. PRELIMINARIES AND A LITTLE FUN 9

§1 How to read this book................................................................................................9

§2 Ask questions...........................................................................................................10

§3 Indispensable tools for reading this book................................................................10

§4 Sets, numbers and infinity........................................................................................11

§5 A litle fun.................................................................................................................12

2. SQUARING, CHECKING AND APPROXIMATION 14

§1 Introduction..............................................................................................................14

§2 Finding a square.......................................................................................................14

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§3 Checking..................................................................................................................15

§4 Powers of 10.............................................................................................................16

§5 Accuracy..................................................................................................................18

§6 Solved Problems......................................................................................................18

§7 Optional items..........................................................................................................20

3. MATHEMATICS AND COMPUTING 22

§1 Bytes and bits...........................................................................................................22

§2 Turing Machines......................................................................................................22

§3 Speed of operation...................................................................................................23

§4 Will the computer replace the mathematician?........................................................24

4. FERMI PROBLEMS 25

§1 Weight of a baby......................................................................................................26

§2 Number of people in the world................................................................................27

§3 Circumference of the earth.......................................................................................27

§4 Maximum number of inhabitants on the earth.........................................................27

§5 Viability of running a shop......................................................................................27

§6 General Comments on Fermi calculations...............................................................28

§7 Solved problems.......................................................................................................28

5. PERCENTAGES 33

§1 Definition and examples..........................................................................................33

§2 A fictitious company’s accounts..............................................................................35

§3 Calculating percentages...........................................................................................35

§4 Examples of judging figures in the news.................................................................36

§5 Keep an eye on the total figures...............................................................................37

§6 Abortions..................................................................................................................37

§7 Compound increases................................................................................................37

§8 Worked examples.....................................................................................................37

6. MEASUREMENT SENSE OR DIMENSIONAL ANALYSIS 40

§1 Length......................................................................................................................40

§2 Mass.........................................................................................................................41

§3 Time.........................................................................................................................42

§4 Speed:.......................................................................................................................42

§5 Acceleration.............................................................................................................44

§6 Force.........................................................................................................................45

§7 Work and Power, Joules and Watts..........................................................................45

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§8 Dimensional Analysis..............................................................................................46

§9 Solved problems.......................................................................................................48

§10 Additional Topics...................................................................................................49

7. MEASURING HEIGHTS OR TRIGONOMETRY 55

§1 Shadow heights........................................................................................................55

§2 The artist’s method...................................................................................................56

§4 Dropping a stone over a cliff....................................................................................59

§5 Solved problems and other optional topics..............................................................59

8. LOGARITHMS AND NATURAL LOGARITHMS 63

§1 Multiplying by adding. Logarithms.........................................................................63

§2 Doubling your money..............................................................................................65

§3 Euler’s e...................................................................................................................65

§4 The natural logarithm is roughly 2.3 times the ordinary logarithm.........................66

§5 Theory of logarithms................................................................................................67

§6 Optional worked examples.......................................................................................68

§7 Logarittihms and Planeetary Motion........................................................................68

§8 Earth quakes measured on the Richter scale............................................................69

§9 Some Historical remarks..........................................................................................70

9. COORDINATE GEOMETRY 72

§1 Coordinates..............................................................................................................73

§2 Graphs as a concise source of information..............................................................74

§3 Plotting a graph........................................................................................................75

§4 Equations and Geometry..........................................................................................78

§6 The Greeks and their curves.....................................................................................79

§7 Equations and conic sections...................................................................................81

§8 Applications of coordinate geometry.......................................................................81

§9 Solved Problems......................................................................................................82

§10 Solving problems in Geometry with algebra and vice - versa...............................82

10. SOLVING EQUATIONS AND GAUSS’S METHOD 84

§1 Solving an equation..................................................................................................84

§2 Method 1: Take a guess...........................................................................................84

§3 Method 2: Draw a graph..........................................................................................85

§4 Method 3: Do what your mathematics teacher told you..........................................85

§5 Two unknowns.........................................................................................................86

§6 Some general results................................................................................................87

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§7 Solved problems and other optional material...........................................................88

11. FUNCTIONS 91

§1 What do we mean by a set?......................................................................................91

§2 Functions defined geometrically..............................................................................92

§3 Awkward functions..................................................................................................93

§4 Functions described algebraically............................................................................94

§5 Contrast with the definition in the calculus books...................................................95

§6 What comes after 1, 2, 3? Could it be 34?...............................................................95

§7 Solved Problems and other Optional items..............................................................96

12. GEOMETRY 98

§1 Reducing the assumptions........................................................................................99

§2 Euclidean Geometry assuming only the real numbers...........................................100

§3 A finite Geometry..................................................................................................100

§4 A finite projective geometry..................................................................................101

§5 Hyperbolic geometry..............................................................................................102

§6 Sketch of the arguments for completing §2...........................................................103

13. PROBABILITY AND STATISTICS 105

§1 Probability..............................................................................................................105

§2 De Méré ’s Bet.......................................................................................................106

§3 The same birthday..................................................................................................107

§4 Two boys?..............................................................................................................107

§3 Three wise men......................................................................................................108

§5 Monte Carlo method of calculating approximately............................................108

§6 It is impossible to send a rocket to the moon.........................................................109

§7 Lotteries.................................................................................................................110

§8 Nuclear power plants exploding.............................................................................110

§9 Sampling................................................................................................................111

§10 General remarks about sampling..........................................................................112

§11 Our own beliefs....................................................................................................113

§12 Average................................................................................................................114

§13 Solved Problems..................................................................................................114

14. PYTHAGORAS’S THEOREM 116

§1 Proof of Pythagoras’s Theorem.............................................................................117

§2 The converse of Pythagoras’s theorem..................................................................119

§3 Pythagorean triples.................................................................................................119

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§4 Estimating the square root of a number.................................................................120

§5 The square root of 2 is irrational............................................................................121

§6 Distance to the Horizon..........................................................................................121

§7 Solved Problems....................................................................................................123

§8 Fermat’s Last Theorem..........................................................................................126

15. MODULAR ARITHMETIC. PRIME NUMBERS AND LOGARITHMS 128

§1 Modular arithmetic.................................................................................................128

§2 Congruence modulo n............................................................................................129

§3 Prime numbers.......................................................................................................131

§4 Gauss’s estimate of number of primes...................................................................131

§6 Fermat’s little theorem...........................................................................................132

§7 Proof of the casting out nines check......................................................................132

§8 ISBN check digit....................................................................................................133

§9 Make a cipher code for your credit cards...............................................................134

§10 Application to codes.............................................................................................134

§11 Solved Problems and other optional items...........................................................135

16. COUNTING 138

§1 What do we mean by a set?....................................................................................138

§2 The same cardinality..............................................................................................139

§3 Lines of lengths 1 and 2 have the same cardinality...............................................140

§4 A list.......................................................................................................................141

§5 The rational numbers have the same cardinality as the whole numbers................141

§6 An infinite set which does not have the same cardinality as the whole numbers. .142

§7 Is there a set of cardinality less than the reals but greater than the natural numbers?.....................................................................................................................................142

§8 Solved problems.....................................................................................................143

§9 Hilbert’s Infinite Hotel...........................................................................................144

§10 Russell’s Paradox.................................................................................................145

17. THE TRACHTENBERG METHOD OF MULTIPLICATION 146

§1 The Method............................................................................................................146

§2 Solved Problems and some optional ideas.............................................................149

18. BIBLIOGRAPHY 151

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Chapter 1

PRELIMINARIES AND A LITTLE FUN

Mathematics begins with a guess, just as naturally as love begins with a kiss.

§1 How to read this bookThis is a book for people who choose to read for fun and enlightenment. Doing mathematics voluntarily means you can pick and choose what you want to do.

Reading mathematics is slower than reading other subjects. You may expect to read a page in a few minutes if you read a novel: with mathematics you can be lucky to read a sentence at that speed. The subject is concentrated. So do not try to study too much at one sitting, it being better to learn a little well rather than a lot badly.

We have laid out this book in the best way for our minds. Since your mind is different, you may prefer to change the order. You may skip sections you find boring. Do so. But be prepared to return to them later, when maybe they make more sense.

The chapters are on the whole independent and so can be read in the order you prefer. . There are three main exceptions; Chapter 10 §1 to §4 inclusive on linear equations is needed for Chapter 12 §2 on Geometry. Chapter 9 §1 on coordinates is needed for Chapter 11 on functions.

Chapter 8 §1 to §4 inclusive on logarithms is needed for Chapter 15 §4 on the number of primes less than a given number.

Solved problems can be skipped at without loss of intelligibility. Among these problems will be more detailed or technical arguments and laving them out will make following the text easier. One can always return to them later if one feels like it. Many problems are of interest in themselves. Also often a solved problem can explain a difficulty-

Solved problems are also good for practice. It is more fun and instructive to do them oneself before reading the solution. Solving problems on one’s own is not so easy. You may find it useful to read the book “How to solve it” by Polya if you are interested. But in any case, the problems are optional.

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FIGURE 1. George Pólya (1887-1985), a talented mathematician of the 20th century. Towards end of his life he was much interested in the teaching of mathematics, see his book “How to solve it.”

§2 Ask questionsOne is sometimes admonished not to ask stupid questions. There are two ways of doing this, either ask no questions, i.e. give up learning, or else know the subject so well that you can avoid asking the stupid questions. In other words, ignore the advice to avoid asking stupid questions.

And don’t be afraid to estimate or to guess. Even a wild guess can help you. I would hesitate to give this advice in say Chemistry. You may run the danger of an explosion if you guess which chemicals to combine, but in mathematics nothing so drastic can occur.

§3 Indispensable tools for reading this bookPencil and paper are essential for reading this mathematics book and any other mathematics book. This is because the best and easiest way to follow an argument is to do the calculations yourself. Not only must you have pencil and paper to hand, you must use them all the time. The text in the book you use will give you a clue as to what you should be doing. Often it helps to write the definitions or the assertions in your own hand to absorb and understand them. We have assumed that enough of what you studied at school either remains or else you will be reminded of it as you read. If that is not the case, the book may be very hard to read, since it does not begin from the beginning. However we feel that most people will be able to cope.

In particular, we hope you remember the use of symbols in mathematics. To write a product like 34 we use in between, but if we have represented an unknown number by a symbol x say, and we take twice this quantity, we leave out the multiplication sign and simply write 2x. Thus if one is searching for an unknown quantity x and twice this quantity plus 4 is 10, then this is written briefly as 2x + 4 = 10.

Example of how symbols help understanding:

Here is a party trick. You ask somebody to do the following:

Think of a whole number between 1 and 9

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Multiply by 5. Add 3. Multiply by 2.

Then think of another whole number between 1 and 9 and add it to your total.

Give me your answer.

You can then at once tell the person the two numbers. All you have to do is to take away 6 from the total he gives you. Then the tens will be the first number thought of and the units the next number. For instance, suppose he thinks of the numbers 4 and 7. The first step he has to do is to multiply by 5 thus getting 20. To that he adds 3 getting 23. He then multiplies by two to get 46. Adding 7 gives a total of 53. So in telling him the numbers he thought of you subtract 6 to get 47. Thus 4 was the first number he thought of and 7 the second.

How it works is easy to see with the use of symbols

Thus let x and y denote the numbers thought of. Multiplying x by 5 gives 5x. Adding 3 gives 5x + 3. When we multiply by 2 we get 10x + 6. We then add y to get 10x + y + 6. When we subtract 6 we get 10x + y. Since x and y are numbers between 1 and 9, x becomes the tens and y becomes the units.

Another example of how symbols help understanding:

This is the hand method of learning the product of two numbers lying between 6 and 10

Baumslag’s father used this method for teaching young children who had mastered multiplying numbers lying between 1 and 5 how to deal with larger numbers. Place both hands in front of you with palms facing. The thumb in each hand represents 6, the next finger 7 and so on, till the little finger which represents 10. To find the product of two numbers, say 7 and 8, place the finger representing 7 on the left hand on the finger representing 8 on the right hand. Count the fingers touching and those up to and including the thumbs. In this case 5, and count 5 tens, i.e. 50. There are three fingers on the left hand, and 2 on the right hand which have not been counted. Multiply the two and three to get 6, and add to the 50 to get the product 56. This method helps children to multiply two numbers each lying between 6 and 10.

We can explain how this method works as follows: Let s be the finger on the left hand and t the finger on the right hand. Then this represents the product of 5 + s and 5 + t. This is 25 + 5s + 5t + st.

The calculation we do is to count the number of fingers from the touching fingers to the thumbs and this is (s + t), and count them as 10s, i.e. we get 10s + 10t. To this we add the product of the remaining fingers, i.e. 5 - s and 5 - t, getting 25 – 5s - 5t + st.

Adding this to the 10s + 10t gives us 25 +5s + 5t + st, i.e. the same as we got before. .

§4 Sets, numbers and infinityThe numbers 1,2,3, etc., are called whole numbers. A set is a synonym for a collection: for instance, the collection of all whole numbers. This is denoted by {1, 2, 3, …}, where the squiggly brackets are a convention for denoting a set. The objects or elements of the set are the whole numbers, 1, 2, 3, … and the dots are understood to mean that the list continues forever.

Modern Mathematics explains much in terms of sets, and we will do so as well in this book.

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The set of whole numbers is an infinite set. If you tried to list them in a finite number of steps you could not. An example of a finite set would be {1, 2, 3}.

The set of all atoms on the Earth, although huge, is not an infinite set. In theory one could list all the atoms, and after a while the list would come to a stop.

For centuries mathematics consisted of the study of numbers and geometry. This has long ago ceased to be the case, but we will stick to these parts of mathematics.

In addition to the whole numbers, we also have the fractions, numbers which are one whole number divided by another, like ¾. The fractions are also called rational numbers. The rational numbers also include the negative fractions.

The rational numbers can also be expressed as decimal expansions. These can be finite, or else infinite, like 1/3 = ,33333… But if a rational number’s decimal expansion is infinite, then it has a repeat pattern after a while. If we allow these, and all other possible decimal expansions, both positive and negative, then we obtain the set of all numbers These are called Real Numbers to distinguish them from the Imaginary Numbers, which involve the square root of – 1.

A useful word in mathematics is “theorem,” which means “important deduction or result.”

§5 A little fun.1. Multiplying any two numbers using only the two times table.

We can multiply any two numbers by multiplying solely by 2 and dividing solely by 2. We illustrate the method by working out 46 33.

We write the numbers in two columns. The next row is produced by multiplying the first number by 2 and dividing the second number by 2. We ignore any halves that appear. We continue in this way till we get to 1 in the right-hand column.

46 33

92 16

184 8

368 4

736 2

1472 1

We then add all the numbers in the left-hand column, which are opposite an odd number in the right-hand column. The result is the product of the two numbers. Thus in the example we have chosen, we get

1472 + 46 = 1518 which is the answer.

2. The game of Nim

The game of Nim is played with matches (or tooth-picks for non-smokers.)

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There are two players. One arranges the matches in as many piles as desired, with as many matches as desired in each pile. Each player at each move chooses a pile and takes as many matches as he or she wants. But each player must take at least one match on every move. At each move one must restrict oneself to one pile only. The LOSER is the person who takes the last match.

Now there is a strategy for winning. We take two special cases rather than the general case.

Two piles Nim

If there are only two piles and it is your turn, then if one pile has only one match, take the other pile away. If each pile has more than one match, then take matches away from the larger pile to leave two piles with the same number of matches. If on your turn both piles have the same number of matches, then you will lose if your opponent knows the strategy. Otherwise you can take away only one match from one pile, hoping that your opponent does not know the method.

One pile with one match and two other piles.

When there are three piles each with one match, the person to makes the first move loses. So if there are two piles with one match in each and a third pile with two or more matches, leave only one match in the larger pile.

If two piles have more than one match, check the number of matches in each.

Case a) The two piles have the same number of matches in them and this is more than one match. Take away the pile with one match to get the two pile situation described above.

Case b) The smaller of the two piles with more than one match has an odd number of matches. Take matches away from the larger pile so that it has one match less than the smaller.

Case c) The smaller of the two piles with more than one match has an even number of matches.

Take matches away from the larger pile so that it has one match more than the smaller.

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CHAPTER 2

SQUARING, CHECKING AND APPROXIMATION

Mathematicians like the rest of us make mistakes. But, being accustomed to checking, they usually detect their mistakes before it is too late.

§1 IntroductionAll of us have learnt the basic skills we need to do many arithmetic calculations.

We are suggesting that instead of letting our skills degenerate, we try to use them every day to think and make interesting conclusions. It is so seldom that we use these basic skills that we may now no longer be adept in adding, multiplying and dividing. Many think that does not matter, we can always use an electronic calculator, and so we can. But there is still a place for these basic skills, and most important, personal satisfaction in being able to do these calculations oneself. So in this chapter we will begin by practicing. In these initial sections, we will also find methods of checking the calculation.

§2 Finding a squareWhen we write (45)2 we mean 45 times 45. More generally, if x is any number, x2 means x times itself. We say “x-squared”, because it represents the area of a square of side x.

There is a quick way of working out the square of a number ending in 5. The general rule is: Remove the last digit, 5, multiply the remaining number, call it r, by r+1 and attach 25. For instance, to calculate (45)2, remove the 5, leaving 4, multiply 4 by 4 + 1 = 5 to get 20, attach 25 to get the answer 2025. (Why this method works is explained in problem 11 §7.)

Example: (995)2. (Some advice: Before reading the solution to an example, work through the problem yourself and read the solution only if you get stuck or wish to verify that you are correct.)

Solution: Take away the 5 to get 99 (r in this case), add 1 to get 100 (i.e. r + 1), multiply 99 and 100 to get 9900, and attach 25 to get the answer of 990025.

Of course you can calculate this result by multiplying 995 by 995, using the usual method of multiplying numbers, but this method is simpler and quicker.

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You may say, why should I bother with such a calculation. Is that not the reason for calculators? Yes, you are right. But using a calculator is tantamount to letting some authority tell you the result. Which is a pity, since mathematics is the one subject where you can rely on yourself, and not on authority.

§3 CheckingCasting out nines

In §2 we calculated the square of 995. Can we check the result?

It may seem over the top to bother about checking in this particular case, since it is not a very complicated calculation, but by discussing what happens in such a calculation we are preparing the ground for how to handle much more difficult problems. Mathematicians are keen to check their calculations, because often other things depend on them, and one wants to be absolutely certain. It is also surprisingly easy to make mistakes, even using electronic devices. Repeating a calculation is also a good method of checking, but often one tends to repeat the same mistakes when doing the second calculation.

If the following check fails, we will know that the result is false. If the check succeeds, the result may nevertheless be wrong. But this check is useful for all sorts of arithmetic calculations. It is called the method of casting out nines. But first, we must define the checksum of a number. We can illustrate by 867. We begin by adding the digits of 867: 8+6+7 = 21. Since the sum is greater than 8, we repeat the process with 21, that is, we add the digits of 21 to get 3. Since 3 is less than 8, the checksum of 867 is 3. One further point, if the number 9 occurs anywhere in our calculations, we replace it by 0. For example, the checksum of 9 is 0. Also, the checksum of 291 is 2+0+1, or 3,which we obtained by replacing the middle digit, 9, by 0.

The method is based on the following fact: If the result of a product is correct, then the product of the checksums of the factors must be the same as the checksum of the answer. As an example of the method, suppose we are to check that the product, 3523, is 805. We begin by replacing the numbers 35, 23 and 805 by their checksums, 8, 5 and 4, obtained by adding their digits. (For instance, 805 is replaced by 8+0+5 = 13, but since 13 is larger than 8, we further replace 13 by the sum of its digits, 1+3=4.) Multiply 8 and 5, the checksums of 35 and 23, to get 40. Add the digits of 40 to get 4. Since this matches the checksum of 805, the check is positive, and we have increased our confidence in the result, though the accuracy is not guaranteed.

The reason the method is called casting out nines is the rule that to obtain the checksums all 9s in the calculation are replaced by 0. For instance, applying the check to 99=81 gives 00 for the product, while the answer has checksum 8 + 1 = 9 which we replace by 0. Thus the check works.

We check the example in §2: 995995 is supposed to be 990025. We replace the 9’s with 0s to get for the sum of the digits 0 + 0 +2 + 5 = 7. The sum of the digits in 995 with 9s replaced by 0 is 5, and 5x5 is 25 which has checksum 7, the checksum of the answer. This increases our confidence in the result.

(Chapter 15 §7 gives an explanation of why casting out nines works.)

Example

Use the method of casting out nines to check whether 7438 =2,712.

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Solution: Sum of digits of 74 is 11, and the sum of digits in 11 is 2. Sum of digits in 38 is also 11, which becomes 2 when we sum its digits. The product of the two checksums is 4, and this should agree with the check on the given answer. The sum of the digits in the given answer is 2, namely 2 + 7 + 1 + 2 = 12, but the sum of the digits of 12 is 3, not 4. Thus, 7438 is not 2,712. Indeed, a more careful calculation gives the correct answer of 2,812.

Casting out 9s can also be used to check a sum of numbers against the total. For instance, to check that 75 + 236 = 305, replace 75 by its checksum, 3, and 236 by its checksum, 2. The sum of these two checksums is 5, which should be the checksum of 305. However, the checksum of 305 is 8, so there is a mistake in the addition.

Remark

The Welsh mathematician Jim Wiegold used to emphasize the importance of checking by his code of practice: Whenever he used a result he felt bound to check the proof of the result so as to ensure correctness of the previous result as well as his own. This is despite the fact that a referee has checked all articles printed. However, even Wiegold has not been able to carry out his code of practice always. The amount of checking is just too much in some cases.

Two Quick Checks

A very crude check of a product of two numbers is obtained by counting the number of digits in each factor and adding. Suppose the sum of these digits is S. Then the product should have either S or S - 1 digits. In §2 we claimed that 4545 was 2025. Thus the product we have calculated has 4 digits as it should. Although this is a very crude check, it does bring to light errors we might otherwise miss.

A similar but more accurate method is to use only single digits. We again take the example of the square of 45, which we calculated in §2.

(45)2 = 4545, and this is approximately 5040 =2000. We chose this approximation by increasing 45 to 50, i.e. a whole number of tens and then reducing the other factor 45 to 40, arguing that as we had increased one factor, we should compensate by reducing the other. Of course multiplying 50 by 40 is easy. The result 2000 is strikingly close to the result obtained in §2, that is, 2025. These two checks, casting out 9s and approximating, give further evidence that the method in §2 both works and probably there is no serious mistake in the calculation.

§4 Powers of 10

We have already explained in the meaning of 102 as the product of 10 and itself, i.e. 100. Similarly 103 is the product of three 10’s, and so on. Thus

102 = 10 ×10= 100

103= 10×10×10 = 1,000

104= 10×10×10 ×10 = 10,000

105= 10×10 ×10×10×10 = 100,000

106= 10×10×10×10×10 ×10 = 1,000,000

107= 10×10×10×10×0×10×10 = 10,000,000

and so on. These are called the powers of 10. A useful word is exponent: the exponent of 105

is 5; that of 107 is 7. Note that 105 is 1 followed by five 0s, 107 is 1 followed by seven 0s and

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so on. It seems reasonable to define 101 to be 1 followed by one 0, i.e. 10, and 100 as 1 followed by no 0s, i.e. 1. Thus101 = 10 and 100 = 1.

Multiplying these powers of 10 is easy: one simply adds the exponents. Thus 107×105

= 107+5 = 1012.

You will notice that an advantage of this way of writing is that it is much easier to comprehend, for instance, 109 rather than 1,000,000,000. Even more important, this notation gives one a way of expressing very large numbers. Problem 8 in §6 illustrates this.

FIGURE 1. Archimedes (287BC - 212BC). One of the greatest mathematicians of all times. He had an alternative of the power of tens notation to denote large numbers which enabled him to calculate the grains of sand in the entire universe (as known then).

A very useful method is to express a number as a product of a number lying strictly between 10 and 1 and powers of 10. Thus we can express 887 as 8.87102. The number 64789 is expressible as 6.4789104. The exponents make it very easy to compare these numbers. Obviously the one with exponent 4 is very much bigger than the one with exponent 2. Also, if the numbers were written out in detail, it could be rather awkward to perform calculations with them. For example, to square 65,000, if we rewrite this as 65×103, the answer, using our previous formula, is easily seen to be 4225×106, which can also be expressed as 4.225×109or, without exponents, 4,225,000,000.

The advantage of this calculation is that it is so simple to do; it also gives us a very good idea of the powers of 10 that are in the answer. As such it is a useful test. With it we will certainly discover large errors.

Example

(995)2. 995 is approximately 1103, and so the square is approximately

11031103 = 1106. If we look at the example in §2 we calculated (995)2 to be 990025, which is equal to 9.90025105. This is very close to 106.

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Negative powers of 10

Positive powers of 10 are useful for large numbers. Negative powers are used for small numbers. For instance, 10-5 means 1 divided by 105, i.e. 1/100,000. In general the same rule applies that multiplying by powers of 10, whether positive or negative or mixed, we simply add the exponents. Thus

10-5 10-1510-5= 10-25.

Rounding

The number 3.467 can be rounded to 2 decimal places by changing it to 3.47, i.e. we drop the last digit and if it is 5 or larger add 1 to the second decimal. If the last digit is less than 5, we simply drop it and leave the other digits unchanged.

Examples

To three decimal places we replace 18.8244 by 18.824, to 4 decimal places replace 5.67185 by 5.6719, to one decimal place replace 299.95 by 300.0. In this last example, we drop the 5 and add .1 to 299.9 thus getting 300.0.

In Science and Engineering most numbers are not exact, being the result of measurement, which is always subject to some inaccuracy. In the scientific notation that we have discussed above, the convention is that all digits given are correct, with the possible exception of the last digit, which could be 1 larger if the number has been rounded up. Thus 5.678108 means that the number lies between 5.6775108 and 5.6784108.

§5 AccuracyTaking 10 instead of 14 is an approximation. The error is 4/14, i.e. approximately 28%. So when we consider a product and approximate by taking the nearest single digit numbers we can incur large errors. When we take the product of two such numbers the errors compound. For instance, to calculate 1423, if we approximate by 1020 = 200, instead of the correct product of 322, the error is 122/322 i.e., an error of about 40%. The method of approximating by taking a single digit is subject to considerable errors. Often it is still useful to do so, but we recommend using two digit numbers to approximate, thus getting a much closer approximation. We recommend this because multiplying two digit numbers is relatively easy. In fact, if one uses the method advocated by Trachtenberg (described in Chapter 17) one can write down the answer in a few seconds.

In working out an approximation multiplying two numbers, note that the error is approximately the sum of the percentage errors in each of the factors as explained in 2 of §6. In the example above, approximate 1423 by 1020, the percentage errors of the factors are 28% and 13%, so the error in the product is approximately 41%. In the actual calculation we found an error of 40%.

§6 Solved Problems1. Practicing squaring a number that ends in 5. Find the squares of 85, 75, 125.

Solution: (85)2 : Drop the 5 to get 8, add 1 to what remains to get 9. Multiply 8 and 9 to get 72. Attach 25 to get the result of 7,225.

(75)2 : Drop the 5 to get 7, add 1 to what remains to get 8. Multiply 7 and 8 to get 56. Attach 25 to get the result of 5,625.

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(125)2 : Drop the 5 to get 12, add 1 to what remains to get 13. Multiply 12 and 13 to get 156. Attach 25 to get the result of 15,625.

2. Checking by casting out nines – Find 82×36. Check by the method of casting out nines.

Solution: 8236 = 2952 by direct calculation.

To check we sum the digits of the answer replacing 9 with a 0, getting 2 + 0 + 5 + 2 = 9 which we replace by 0.

The check for 82 is 8 + 2 = 10, 1 + 0 = 1 on adding the digits of 10. The check for 36 is 3 + 6 = 9, which we replace by 0.

We multiply the check digits for 82 and 36 to get 10 = 0, the same as the check digit for the answer. So our check does not indicate an error.

3. A quick check. Check problem 2 by checking the number of digits.

Solution: We add the number of digits in each factor: 82, has two digits and 36 has two digits. The answer should have 4 or maybe 4 – 1 = 3 digits. In fact the answer 2,952 has 4 digits.

5. Single digit check. Check problem 2 by replacing the product by one with single digit numbers.

Solution: To calculate 8236 we replace 82 by 80 and 36 by 40, the product is 3,200, which agrees reasonably well with 2,952.

6. The scientific notation. In the scientific notation, what is the meaning of 7.5104?

Solution: This means 7.5 times 104, which is 1 followed by four 0s, i.e. 10,000. Furthermore the result lies between 7.45104 and 7.54104.

7. Powers of 10. Use powers of 10 to multiply 2.5104 by 3.1103.2.53.1 = 7.75.

104 103 = 107 and so the product is

7.75107

8. Powers of 10 continued. Estimate the volume of a sphere with center the earth and extending to the moon. (This example illustrates how easily the powers of ten notation can handle very large numbers. Indeed, without the powers of ten notation we could not even give an answer.)

Solution: The volume of a sphere of radius r is 4 r3/3. The moon is approximately 1.6106 km. So the volume is 4(1.6)3 1018/3 = 17.157 1018 = 1.7157 1019 km3.

9. Negative powers of 10. What is the meaning of 10-3?

103 means 1 followed by 3 zeros. 10-3 means 1/103 = 1/1000 = 0.001.

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10. Errors. Calculate the percentage error in the calculation in example 5 above. Does it agree with the assertion that the error is approximately the sum of the percentage errors?

Solution: Replacing 82 by 80 gives a percentage error of (2/82)100, i.e. approximately 2%. Replacing 36 by 40 gives a percentage error of (4/36)100. i.e. approximately 11%, the sum gives an error of approximately 13%. The actual error in the answer is less than (3/32)100, i.e. approximately 10%. This agrees well with the calculated error.

§7 Optional itemsThe method of checking modulo 11

1. There is another useful method of checking which we can explain by means of problem 2 of §6. This is 8236.

For this check, we add the first digit to the third digit and then add the result to the fifth digit and so on. We then add the second digit to the fourth digit and then add the result to the sixth digit and so on. We then subtract the second sum from the first to get our check number.

For instance, 93,546 gets check number (6 + 5 + 9) – (4 + 3) = 20 – 7 = 13, which is then replaced by 3 – 1 = 2. Sometimes the check number will be negative, for instance the check number of 82 is 2 – 8 = - 6. In such a case, we add 11 to get a positive number. So 82 which had check number – 6 has check number 11 – 6 = 5.

Our problem is 8236. The first factor 82 has check number 5 as we have just explained. The other factor is 36, which has check number 6 –3 = 3. We then multiply the check number of the first factor by the check number of the second factor to get 15, which is further replaced by 5 – 1 = 4. In problem 2 the answer was 2,952, whose check number is (9 + 2) - (2+5) = 4, which agrees with our previous check digit. (See Chapter 15 §6 problem 2 for an explanation of why this method of checking works.)

2. Demonstration of summation of percentage errors in a product.The percentage error in a product is approximately the sum of the percentage errors in each approximation.

Solution; Before giving the explanation we note that the product of two small numbers is very much smaller than each of the individual numbers. For instance, if we multiply .01 by .02, the result is .0002, which is considerably smaller than both .01 and .02. So the product of two small numbers can be neglected if we are looking for an approximate result.

Suppose now that we are multiplying two numbers, n and m, by approximating to n by

n + a and to m by m + b. Our approximate answer will then be (n + a) (m + b).

[As an example, say we are multiplying 2.9 by 4.8. Suppose we use n = 2.9 and a = .1, and m = 4.8 and b = .2. Thus instead of 2.94.8 we take 35.]

The difference between our approximate answer and the correct answer, nm, will be

(n + a) (m + b) - nm = nb + ma + ab, which is approximately nb + ma, if we assume that a and b are relatively small, and so ab can be neglected by the remark at the beginning of this solution. The percentages of error in the approximations are (a/n) 100 and (b/m) 100 and, for the product, (nb + ma)/(nm) 100. Finally, (nb + ma)/(nm) 100 = b/m100 + a/n100. That is, the sum of the percentage errors of each of the factors, which is what we were to prove.

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[If we repeat the above argument using the example we have chosen of 2.94.8, we have the difference between our approximate answer of 34 = 12 and the correct answer 2.94.8 will be (2.9 + .1) (4.8 + .2) – 2.94.8 = 2.9.2 + 4.8.1 + .1.2, which is approximately 2.9.2 + 4.8.1 since .1.2 can be neglected. The percentages of error in the approximations of the factors are (.1/2.9)100 and (.2/4.8)100 and, for the product, {2.9.2+4.8.1 2.94.8})100 = (.2/4.8)100 + (.1/2.9)100, that is the sum of the percentage errors in each of the factors]

3. Prove the quick method of squaring a number ending in 5 described in §2.

Note that 95 = 910 + 5. In general a number ending in 5 can be written 10n + 5. Thus

(10n + 5)2 = (10n + 5) (10n + 5) = 100n2 + 50n + 50n + 25 = 100(n2 + n) + 25.

Since (n2 + n) = n(n + 1), the result follows.

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Chapter 3

MATHEMATICS AND COMPUTING

This chapter is quite diffent from the others. Whereas most of what appears in the rest of the book will be correct in thousands of years, much of what appears here, and especially the specifications, will be out of date probably before the book is published.

§1 Bytes and bitsComputers’ memories are specified in bytes. A byte corresponds to a character, such as a number or a letter of the alphabet or a punctuation mark. The computer has a very limited vocabulary. It understands only 0 and 1. This is called a bit. A byte consists of 8 bits.

The computer has two types of memory. The first is called RAM memory, and stands for random access memory. It is the memory that the computer has for calculating and thinking, and corresponds to what we would normally use a sheet of paper for our calculations, which can then be thrown away. This, for instance, we would normally use for recording a telephone number when somebody phones. Later on we would transfer this to a telephone list, which is kept. This corresponds to the memory on the hard disc of the computer.

The byte is a small unit, and we have a number of other units to describe the memory capacity of a computer. A kilobyte is 1000 bytes, a megabyte is a million bytes and a gigabyte is one thousand million bytes. One thousand million is a billion and so a gigabyte is a billion bytes. It is not unusual for a computer to have 200 gigabytes of memory, i.e. it has more characters that it can remember than there are people in the world (6 billion.) With such a memory, it can remember 5 words describing each person in the world.

§2 Turing MachinesThe Turing Machine is a theoretical model of the computer. It is a very simple device, but then the computer is also a very simple device. A Turing Machine has a infinite strip of paper. It can make a markon the paper or delete a mark and then move one position to the

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left or one position to the right. It can also be in a number of states, and depending on its state, it will do other things.

Perhaps an example of how the Turing Machine adds will help. A Turing Machine to add has two states, State A and State B.

Initially it is given two numbers to add. The two numbers are indicated by a number of marks. There is a space in between them. For instance, 5 + 3 will appear on the infinite strip as follows:

.

The tape is read at the beginning by the machine in state A. If in this state the machine sees a mark, it moves one space to the right. It then reads the next item and if it is a mark it moves one step to the right again. If it sees a blank it moves one step to the right. It now changes to state B. If it sees a mark it moves one step to the left and makes a mark and then moves one step to the right and erases the mark. It then moves one step to the right and continues. If it sees a blank it simply stops. The result is of course that all the marks are now all together and there are now eight of them, and so the Turing Machine has added the two numbers to get the total of eight.

§3 Speed of operationFrom this description of the Turing Machine you get the impression that the computer is not very smart. Of course the Turing machine is only a theoretical model of the computr, a nd the computer works completely differently.

Perhaps a more usful way to think of the computr, is that it is a device that carries out th4e instructions of algorithms. An algorithm is a step by step procedure. At each step the algorithm tells one exactly what one has to do. As an example we will consider the algorithm of finding the highest common factor of two numbers. This is the largest number which divides both numbers exactly. For instance, the highest common factor of 54 and 30 is 6. This is easy to see by dividing the numbers mentally. A simple algorithm for doing this is as follows:

We prepare two columns. In the first row we write 54 in the first column and in the second we write 30. If the two numbers are the same the highest common factor is that common number. In this case of course 30 is the least number, and we write it in the same column in the new row. We subtract it from other number and place that in the same row under itself. We continue in this way till we get two equal numbers in a row. This the highest common factor.

Thus we have in this example

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54 30 (Begin with the two initial numbers.)

24 30 (30 unchanged and subtracted from 54.)

24 6 (24 unchanged and subtracted from 30.)

18 6 (6 unchanged and subtracted from 24.)

12 6 (6 unchanged and subtracted from 18.)

6 6 (6 unchanged and subtracted from 12.)

As the two numbers are now the same, that is the highest common factor.

As the computer is limited to carrying out such laborious processes this agaam suggests that it is rather stupid. How come then it is so effective.?

The answer lies in its ability to do each step incredibly fast. Typically performance is measured in giga cycles per second, i.e. a billion times per second. The word hertz after a famous physicist means cycles per second. Thus 2 gigahertz is a typical speed for computers.

The electricity in your house has a frequency of 50 or 60 hertz, radio waves are measured in kilocycles or at most megacycles. Per second. The more hertz the faster the computer can do calculations. So the computer is outstandingly clever because it does everything incredibly fast, even though it uses quite laborious methods. …..

§4 Will the computer replace the mathematician?This has already occurred in some respects.

Many people do not calculate the sum of or the product of two numbers, they use a pocket calculatior. At most shops nowadays the assistanct taking your money does not need to calculate how much change to give you. This is done automatically by the cash register. Nor does an accountant nowadays need to be quick and accurate at adding numbres, he or she simply uses a computing program which does all the adding automatically. Similarly the payment clerk, does not need to calculate your tax, it is all done automatically by the computer program.

At University the first two parts of mathematics that are usually studied are Calculus and Linaear Algebra. There are many computer programs that can do all that a clever student can do and more quickly and more accurately. Although students still study these two subjects, it is surely only a matter of time before the subjects will be modified and at the very least, much of the techniques and methods will prove to be redundant and will not be studied. How far this will go is hard to say. Att the moment, most mathematics lecturers have a built in tendency to teach the subject very much in the old way. But time will certainly change this and we can expect the computer to be used much more

Just how much we can leave to the computer is hard to say. We must avoid the danger that after a while there will be nobody who really understands the principles and we simply rely on the computer as an oracle. And of course there are going to be times when the computer is going to be wrong. Either because there are bugs in the program (problems and conflicts that arise which nobody thought of at the time) and also because conditions may have changed and so are no longer applicable.

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In this book we stick mainly to our own understanding, and do not rely on authority, whether it be computers or famous professors.

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Chapter 4

FERMI PROBLEMS

To scribble a few figures on the back of an envelope and yet get a reasonable approximation to something whose value you have no idea about, sounds a bit like cheating. It isn’t. It’s a Fermi calculation.

The Nobel Prize winner Enrico Fermi was a physicist who had great skill in estimating with little information. For instance, as a standard question he would ask his students, ”How many piano tuners are there in Chicago?” With ingenuity one could find an estimate. How good the estimate is depends on the skill of the estimator. At the first ever explosion of a nuclear bomb Fermi noted how a piece of paper had been blown away by the blast, which was many miles away, and produced very quickly an estimate of the yield. His result was remarkably accurate.

As another example of his methods, knowing the distance between Los Angeles and New York and the time difference, we will be able to estimate the circumference of the earth.

In this chapter we will solve some Fermi-type problems. The idea is to get numerical values with very little information, and, of course, with results that are only rough approximations, For instance, one of our calculations will be an estimate for the weight of the earth. After our calculation we will be able to replace a vague remark that “the earth weighs a great deal” with the remark that the earth weighs approximately so many kilograms. This is relatively easy to check against published figures, and it will turn out that our rough calculation is out by a factor of 2. But this is much more precise than the original estimate that “the earth weighs a great deal”.

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FIGURE 1 Enrico Fermi (1901-1954). Nobel Prize in Physics 1938

Most of the examples below can also be checked in reference books, which enables us to see the advantages and the limitations of our calculations. But there are many cases where there is no reliable answer, and then the Fermi calculations give us guidance.

To simplify the calculations, it will be useful to express numbers using powers of 10. We remind the reader of these powers, which were discussed in Chapter 2. By 103 we mean 10 multiplied by itself three times, or 1000. Generally, for any whole number x, 10x means ten multiplied by itself x times. For example, 107 means 1 followed by seven 0s, i.e. 10,000,000, and 3.8 107 means 3.8 multiplied by 10,000,000, i.e., 38,000,000.

The x in 10x is called an exponent. As we explained previously when we multiply powers of 10 we simply add the exponents, and if we divide by a power of 10, we subtract the exponents, e.g. 104105 = 109, and 106 divided by 102 is 10(6 - 2) = 104.

§1 Weight of a babyAssuming that a man of 2 meters height weighs 100 kilograms, how much should a baby 50 cm high weigh? Since the baby is a fourth of the height of the man, as a first guess one might divide 100 kilos by 4 to get 25 kilos. This is clearly too naïve, since weight depends on volume, and the baby is not only shorter, it is also not as wide. Moreover, among many other considerations, skin and bones are less dense. To get a better approximation think of two solid rectangular boxes, one of which is one-fourth the length, width and breadth of the other. The volume of the smaller one will be (1/4)3 = 1/64th of the larger. If the same reasoning applies to the baby, its weight should be 100/64, or approximately 1½ kilos.

This answer shows both the weaknesses and strengths of this type of calculation. Everybody knows that a 50 cm baby is likely to weigh about 3 kilograms: twice as much as our estimate of 1½ kilos. However, with very little effort we have obtained a rough value, which is somewhere near the right answer. Clearly we should not expect this simple approach to give us an accurate result. Nevertheless, it has provided us with a rough idea of the result.

There is another possible approximation, which is based on the body mass index. This is a way of checking on being underweight or overweight. The body mass index is obtained by dividing the weight in kilograms by the square of the height.

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For instance, consider the man who weighs 100 kilos and has a height of 2 meters. The square of his height is 4 and so the body mass index is his weight divided by 4, that is 25. And indeed, 25 is given empirically as the maximum body mass index for a person not to be overweight.

If we use this method for the baby, with its weight denoted by w, and its length 0.5 m, then, since (0.5)2 = 0.25, the body mass index for the baby is w/0.25. Assuming that w/0.25 = 25, the maximum for the ideal body mass index for a man, the baby’s weight must be 6 kilograms. Again, we are some distance from a reasonable result.

One can regard these rough calculations as, at least, giving us some quantitative information. This might be sufficient for our needs, but it will often be just the first step in trying to get a useful result.

§2 Number of people in the worldThere are something like 200 countries in the world, and Great Britain has some 60 million people, i.e. 6107 people. There are other countries much larger, like the USA, Russia, China, India, but many much smaller Assuming that all on average, 200 countries have half the population of Great Britain, this would make the total population of the world about

2003107 = 6109. This very crude calculation has given the correct result.

§3 Circumference of the earthEnrico Fermi came up with this clever way to deduce the circumference of the earth.

The distance from New York to Los Angeles is approximately 5,000 kilometers, and the difference in time is 3 hours. Since the earth is divided into 24 time zones, the distance from New York to Los Angeles corresponds to 1/8 the circumference of the earth. Hence, his estimate for the circumference of the earth is 85000 = 40,000 km. The equatorial circumference is in fact 40,074 km. Note that the circumference varies as the earth is not a perfect sphere.

Fermi’s argument is very much the same as that used by the Greek astronomer, Eratosthenes, about 240 B.C., who chose Aswan and Alexandria in Egypt instead of New York and Los Angeles, and came to a result of a little over 40,000 km. His method was based on the relative position of the sun at noon in the two places.

§4 Maximum number of inhabitants on the earthThe radius of the earth, r, is about 6103 km so the surface area of the earth is given approximately by the formula 4πr2 (the surface area of a sphere of radius ) i.e.

4 (36 106) = 452106 km = 4.52108 km2. However, since 70% of the earth’s surface is water, this leaves 30% of the area on dry land, that is, approximately 1.5108106 m2.

Then if we allow 100 m2 of space for each person, this comes to a maximum of 1.51012

people. It is interesting to compare this figure with the actual number of people inhabiting the earth today, something like 6109.

§5 Viability of running a shopSuppose you wish to earn a net income of £20,000 per year. Working a whole year with 40 hours a week, and 50 weeks gives 2,000 hours. This means that you need to make at least £10

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per hour. However, it would be foolish to regard that as sufficient, because there are always extra expenses. So we assume that what is needed is £20 per hour, just asking double since we do not have any better idea. Assuming that the profit margin is one third, we will need to sell some £60 per hour, or £5 per five minutes. This seems rather optimistic so we conclude that the profit margin must be higher, say 50%. With this profit margin you can reduce your average hourly sales to £40, or £3.50 per five minutes.

Of course, there’s a trade-off here, since increasing your prices may result in fewer customers.

§6 General Comments on Fermi calculationsIt would be foolish to use these rough calculations as being correct conclusions. But it is remarkable how often they give one quite a good idea.

It is a sobering thought that often the figures that are quoted by the authorities have been made by similar calculations. Your own calculations can give you some reason to agree or disagree.

Indeed, in general when you are given some official number, you should always add in your mind plus or minus 20%. We suggest this margin of error because we know that usually it is not possible to give a very exact number. There are always errors. It is also not uncommon for people to err in a direction that makes them look better.

It is important to regard a Fermi calculation as the first stage in a more detailed investigation. Also, of course, the more you know about a subject the more accurate you can make your Fermi calculation.

§7 Solved problems1. Our town has 120,000 inhabitants. What are the numbers of births and deaths?

Solution: Since the town does not seem to be growing or declining, to a first approximation the numbers of births and deaths should be about the same. If the average life span is 70, for our rough calculation we may assume that, on average, an inhabitant will die at age 70. Thus the number of deaths should be 120,000 divided by 70, i.e. about 1,700 per year.

2. How many hairdressers in our town of 120,000 people?

Solution: Most men need a haircut once a month. Most of the hairdressers take about quarter of an hour per haircut, presumably more for women, but then women have a haircut less frequently.

On average, in a town of 120,000 people each month there will be about 100,000 people needing a haircut. Divide by 20 to get the number per day, 5,000. Each hairdresser can do 20 haircuts a day. Divide by 20 to get 250 hairdressers. This can be checked by using the yellow pages to count the number of hairdressers. In our phone book the total is 120. So the calculation is wrong by a factor of 2. This is not bad for a rough calculation, but in any case, one thing we did not take into account is that many hairdresser salons employ more than one hairdresser. If we assume the average is 2, then the rough calculation should have been improved by dividing by 2: 250 divided by 2 is 125 hairdresser salons.

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3. Weight of a page of paper.

Solution: Paper is measured at weight per square meter. Suppose our paper weighs 80g per square meter. If the size of a page is 210297 mm, i.e. approximately

0.2m0.3m, then the area will be 0.06 square meters. Hence, the weight must be 800.06 or approximately 5 g. As a check we use a scale and found the weight to be 4 g.

4. Weight of the earth

Solution: The volume of a sphere of radius r is (4r3)/3. Since the radius of the earth is approximately 6,000km, and since 4 is approximately 4, the volume is roughly 4 (6103)3 = 4216109, or 864109 km3. The weight of one cubic centimeter of water is 1 gram, so the weight of one cubic meter of water is 106 grams, or 103 kg, which means that one cubic kilometer weighs 1012 kilograms.

If we guess that earth is four time as heavy as water, the weight of the earth must be approximately 86410941012 kilograms, or 34561021 kg or 3.4561024 kg. Checking on the internet we found a value of 5.976310 24 kg. for the weight of the earth. Our guess is out by a factor under 2, not too bad for a first approximation.

5. Quick calculation of the tax burden.

Solution: In a country with a tax rate of 20% and a value added or sales tax of 20%, for every income of 100 one has to pay 20 in tax. This leaves one with 80. A total purchase price of 80 breaks down to 67 plus 13 value added tax. So when one purchases an item for 80 one has to pay value added tax of about 13. The person who receives the 67 has to pay 20% tax on that, which gives a further 13 tax to the government. Thus, the total tax so far paid on the original 100 is

20 + 13 + 13 = 45.

6. Conservatives claim that reducing taxes will encourage sales, which, in turn, will result in more tax being collected despite the lower tax rate. How sound is the economics?

Solution: To verify this imagine that there is a 5% tax reduction, say, from 20% to 15%. Suppose this results in everybody earning 15% more, a rather extravagant estimate. Previously, if somebody earned 100 he paid tax of 20. Now the same person earns 115 and pays tax at 15%, i.e. 17.2. This is lower than the original 20, and results in a net loss for the government.

7. A person is selling a product at £100 with a profit of £50. He decides to hold a sale giving a 10% reduction. How many more of his product must he sell in order make the same profit as before?

Solution: Since the sale price will be £90, his new profit will be £40, instead of £50. Thus, one needs to sell 25% more to get the same result as before. Of course selling 25% more is a lot more work, so unless many sales result from this maneuver, in the long run it will not be a good idea

8 Three for the price of two.

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Some stores offer three items for the price of two. For instance, one is offered three books for the price of two.. In the following table, we calculate the profits. Take the example when the books are sold at a price of £10 each. The profit will of course depend on what the books cost the store. We will consider two possibilities, assuming first that each book costs £5. Then we consider the profit if each books costs £4.

Thus if each book costs the store £5, selling one at £10 gives a profit of £5. Selling three for the price of two costs the store £15, and they sell them for £20. This gives them a profit of £5. The rest of the following table has been calculated in a similar way.

Cost of book to the store Profit selling one book Profit 3 books for the price of 2

£5 £5 £5

£4 £6 £8

From the table we see that selling three for the price of two gives the same profit as selling just one book.. On the other hand, with a lower cost, of £4 for the store, three for the price of two gives a larger profit than selling simply one. Thus for this method of selling to be profitable for the store, there must be a substantial mark-up.

9. Pyramid selling.

There is a type of selling which sounds good for all participants. The first person sells franchises to 10 subagents, and gets a percentage of their profits. Each agent then sells franchises to 10 subagents and also gets a percentage of their sales. And so on. This system is untenable. Why?

Solution: Suppose we are dealing with a town of 100,000 people. Suppose there are 5 stages of agents and subagents. The very first in the chain has sold franchises to 10 subagents and each of these sells to10 more, making a total of 100 agents. Each of these 100 sells to 10 more, making a total of 1,000. Each of these again sells to 10 making a total of 10,000 participants. Each of these sells to 10 more, making a total of 100,000 at the fourth stage. At the fifth stage there would be 1,000,000 participants, and these are only the ones who have been appointed at the fifth stage, and do not even include all the others. In other words, after 5 steps the whole system collapses because there are not sufficiently many people to participate.

10. Estimate the lower of the two blood pressure readings for a person.

Solution: The blood pressure is measured by two readings in mm (millimeters) of mercury. When you are standing upright the lower pressure must be sufficient to keep the blood in the brain, otherwise you would faint. The heart is roughly 50 cm from the top of the head, and so the pressure must be sufficient to support a 50 cm column of blood. If we make a rough guess that blood and water weigh the same, we would need to support a 50 cm column of water. However, since we are not always at the minimum requirement, let us add 50% to get 75 cm of water.

Blood pressure is measured not in the lengths of water columns, but in the lengths of columns of mercury. To change from a column of water of 75 cm to a column of mercury divide by

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13.9 to get about 5.4 cm, which is 54 mm. Doctors would say that this is too low, and a value of 70 or 80 would be more reasonable.

11. Pension. How much must one save for a pension of T pounds per year?

Solution: Suppose you want to have a pension of T pounds per year. Once one reaches the retirement age of 65, one has probably not much more than 15 years left of life. So one will need 15T pounds in savings. Of course one will invest this sum, but the usual idea is to invest in very safe funds, which means that you can not expect a very high return, but hopefully enough to cover inflation and perhaps give you enough money for a few extra years if you live longer than 80.

As a check we note that the annuity rate is about 6% to 7% which is also about 1/15. This means that insurance companies that provide annuities want a payment of 15T to provide a payment of T pounds per year.

12. Pension. What percentage of his salary should a person save for a satisfactory pension?

Solution: If one receives a salary of S, in practice one hopes for a pension of S/2, which is regarded as satisfactory in England. Thus if we use the calculation in the preceding example, one should save a minimum of 15S/2. Here one can afford to take greater risks and thus get a greater return on the money one saves. One can at a guess expect to get double or even three times the money one has saved because of a reasonable return. (See the following problem for an explanation of this.) Let’s be cautious and say double. Thus one needs to save 15S/4. Assume that one works for 40 years. Thus each year one needs to save 15S/160, i.e., about 9% of salary.

13. If one saves one pound per year for 40 years at 5% interest, what is the final sum?

Solution: Note that 5% is a good average return, allowing for inflation and taxes.

After the first year one has (1 + .05) pounds. One then puts in another 1 in savings, making a total of (1 + (1 + .05)) pounds. Since this accumulates interest at 5%, after another year we will have (1 + .05) + (1 + .05)2 pounds. One then adds an extra saving of 1 to get a total, after the second year, of

1 + (1 + .05) + (1 + .05) 2 pounds.

Continuing in the same way, after 40 years we will accumulate a total of

1 + (1 + .05) + (1 + .05) 2 + …+ (1 + .05)40 pounds.

In order to calculate this sum, we replace (1 + .05) by the symbol r and thus the sum S we wish to calculate can be written as

S = 1 + r + r2 + … + r40.

Next, we multiply this by r we get

rS = r + r2 + … + r40 + r41.

The differences between rS and S are the 1 in the expression for S and r41 in the expression for rS. Hence, if we subtract S from rS, the common terms cancel out and we see that

rS – S = (r – 1)S = r41 – 1.

Hence, S = (r41 – 1)/(r – 1), and since r = 1.05, we see that

S = (7.392-1)/(1.05-1) = 6.392/.05 = 127.84.

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Without interest, we would of course have saved £1 each year for 40 years, i.e. £40. The interest has meant that we have something like 3 times the amount.

14 A twenty year old receives a prize of one million pounds which he puts in a bank. Should he give up his job and “live happily ever after”?

Solution: Assume his present job pays £25,000 per year. With a million pounds one would expect to have a more luxurious life, say £50,000 per year. This would last 20 years. So he would be 40 when he had used up all his winnings. Clearly our 20 year old must think this over more carefully. He should certainly consider interests. Suppose he invests in an interest bearing account at 6% and that inflation is 3% and tax is 20% per annum. So he is receiving 3% after allowing for inflation, and after tax, he is getting 2.4%. Thus he has an income of £24,000 per annum. It does not look as if he can afford to use £50,000 per annum. These calculations suggest that he seeks a financial adviser.

15. In September 2005 a hurricane threatened to destroy Houston Texas. It was essential to leave Houston. If you have a car, how urgent is it to leave?

Solution: To assess how urgent this was one could do the following rough calculation: Suppose that 1 million cars need to leave Houston. Not knowing Houston we have to guess.

It would be nice to have some basis for this guess, but as usual Fermi calculations are made on insufficient knowledge. Inhabitants of Houston would do better.

Suppose there are 5 ways of leaving the city and each road has 4 lanes. Suppose that with the emergency, traffic moves slowly, say 20 km per hour. Suppose we allow 20 m per car. Then in an hour each lane will take 20000/20 = 1000 cars, so five four lane highways will take 20000 cars per hour., A million cars will require about 50 hours, about 2 days. With only two day’s notice it would be sensible to leave as soon as possible. .

16. How long do you need in order to learn a foreign language?

Solution: To manage in a foreign language, one needs say 3,000 words. One can learn say 6 words in an hour. Hence one needs 500 hours. At ten hours a week, this is about 50 weeks, i.e. a year to get a useable knowledge of a language.

17. Estimate the proportion of teachers in the population.

Solution: Assuming that ages in the population vary from 0 to 80 and that they are equally distributed. In England one goes to school from the age of 5 to 17. That means the proportion of school children is 12/80 = .15. or 15%. Assuming that each pupil is in a class of 30 others, and each such class needs a teacher, then we have that the percentage of teachers must be ½%.

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Chapter 5

PERCENTAGES

½ could be a small or a large number. It depends what you compare it with. But ½% is small. That is the advantage of percentages. You know the importance of each item.

§1 Definition and examplesA simple but extremely valuable method of understanding numbers is to interpret them as percentages. This is especially so for figures one is not intimately concerned with. So, for instance, financial reports can be understood better by converting them to percentages. Similarly the budget for a country is more readily understood when expressed in terms of percentages.

As an example: In a prison population of 6,000, 12 prisoners escaped one year. The opposition called for the resignation of the Justice Minister who is also responsible for prisons. In percentage terms this means that 0.2% of the prison population have escaped, and even if this occurs every year, it is quite a small percentage, and calling for the resignation of the Minister of Justice seems a bit overboard.

Another example: The Rector of our university explained we were some ten thousands of pounds in the red. It was a tremendous figure. We were all shocked. So we asked the Rector what the deficit was as a percentage of the income. He had not thought of this but in the end said about 5%. This did not seem so serious a problem as it did at first. This is the value of percentages. They enable you to make sense of the figures. Nowadays with pocket calculators or spreadsheets they are easily calculated.

Examples

On holiday with a budget of £500 for a week a couple has estimated the following expenses.

Bus, tube and train travel £70

Food including restaurants £210

Museum and theatre charges £150

Miscellaneous £70

FIGURE 1. Holiday Money

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In the next table we have expressed the expenses in terms of percentages of the budget.

Bus, tube and train travel 14%

Food including restaurants 42%

Museum and theatre charges 30%

Miscellaneous 14%

FIGURE 2. Holiday money in percentages

The percentages are calculated by dividing each expense by 500, the total cost in pounds for the week, and multiplying the result by 100. Thus bus, tube and train travel percentage are calculated by (70/500) ×100 = 14. The sum of all percentages must be 100%. It is easy to understand the significance of the figures, for instance, 75% is three quarters, etc.

Percentages give a clearer way of seeing how the money is spent. For instance, these figures seem to indicate that it might be worthwhile spending less on food, by reducing the number of restaurant meals and eating more sandwiches prepared at home. This would leave more of the budget to spend on museums and theatre visits.

Also useful to keep in mind is the connection between expenditures and percentages of the weekly and daily budgets in the following table.

£ % per week % per day

10 2 14

50 10 70

100 20 140

500 100 700

FIGURE 3. Holiday expenditure as percentages of the total available to spend

Here again we have assumed a total weekly budget of £500. Spending £10 amounts to a percentage of the week’s budget of 10/500×100, i.e. 2%. But the daily budget is £500/7=£71.43. So £10 as a percentage of the daily budget is about 13%. This makes it easier to judge whether it is worth spending that £10 on a particular day.

In calculating a percentage one starts by deciding on the reference value. In this example we have chosen the total amount available for the week to be the reference. Choosing a different reference will result in different percentages and give different impressions, so it is important to consider carefully what one should use as a reference. In this case, we might have used the cost of the total holiday as a reference, i.e. the cost of travel, the cost of the hotel, and the cost of the food, museums, bus etc. In this case the idea was to help decide how £500 was to be used to enjoy the holiday, and so it seemed the right quantity to choose. In general, if the reference is denoted by R, the percentage for each item is calculated by the formula

(Item/R)100.

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Caution: Do not try to average percentages. For instance, if an investment goes up 100% one year, and then down by 50% the next year, the net result is not 25% (100 – 50)/2) the average of the two percentages, but 0%. For if you have £100, a 100% increase gives you £200, and a 50% decrease gives you £100, i.e. you are back where you started from. .

§2 A fictitious company’s accountsIn the following table we have listed a company’s accounts.

This year Last year

Sales £386,234 £320,234

Advertising £50,000 £25.000

Postage & telephone £35,874 £30,874

Wages £140,000 £130,000

Directors fees £70,000 £70,000

Consultant’s fees £568 £368

Accountant’s fees £1,865 £1,865

Profit £87,927 £62,127

Working capital £25,000 £25,000

FIGURE 4. Accounts of a fictitious company

We can re-express this in terms of percentages of the previous year’s results as in the following table:

This year/last year %Sales 120Advertising 200Postage &telephone 116Wages 108Directors fees 100Consultant’s fees 154Accountant’s fees 100Profit 141Working capital 100

FIGURE 5. Accounts re-expressed in percentages of previous year y

Thus we see at a glance that sales are up by 20% and profits are up by 40% and advertising has doubled. The other items have shown relatively no change.

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§3 Calculating percentagesWe can calculate the percentages quickly with the aid of a pocket calculator. Even quicker is to use a spreadsheet. If you are familiar with say Excel, then the following description may be sufficient, though usually handling computer programs is more easily done with help. To illustrate how this method works we will take the holiday budget described in Fig. 1.

Begin by opening Excel. In cell D2 type “Percentage with respect to R =”. In cell H2 put in 500, which is the reference figure we chose for the holiday budget. Type in the table beginning in cell C4 where you type in “Bus, tube and train travel.” In cell H4 type 70, and continue to type in the rest of Fig.1. The table for the budget is now in cells C4 to C7 with the numbers in cells H4 to H7,

Bus, tube and train travel 70

Food including restaurants 210

Museum and theatre charges 150

Miscellaneous 70

FIGURE 6. Example used to illustrate use of Excel for calculating percentage

In cell I3 type “%”. In cell I4 type = H4/$H$2*100 and press return. This calculates the percentage that 70 is of the reference you have placed in cell H2 (which in this case is 500). Go back to cell I4 and press ctrl and C at the same time. Then go down to the next cell I5 and press ctrl and v at the same time. You then go to the next row down to cell I6 and press ctrl and v at the same time. You go on doing this till you have covered the whole column. What Excel does is to copy your instruction of how to work out a percentage to each of the cells. It changes the H4 successively to H5, H6 and H7 as you go down the column, but the H2 which is the reference remains unchanged because Excel interprets the $ sign to mean leave this unchanged.

Thus, you will get the percentages in this way. If you should decide to change the reference R, go back to cell H2 and change it accordingly.

§4 Examples of judging figures in the newsIt is very difficult to comprehend large numbers, so it is particularly useful to use percentages to describe them. The problem is: what percentage of what? The following examples give ways of understanding the significance of the numbers.

Example: In the Swedish news two items were mentioned. The first was that there would be an extra amount of one million crowns to assist further employment. The other was that 2,300 million crowns was the estimate of how much money was spent on illegal drugs per year.

Since a reasonable salary in Sweden is a quarter of a million crowns per year, the million crowns correspond to the wages of 4 people in a year. We can see immediately that the million crowns is not worthwhile bothering about. It can’t make much difference to the overall job situation.

The second figure, of 2,300 million crowns should be considered in relationship to the population. Since the population of Sweden is 9 million or roughly 10,000,000, or 107, this

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means that on average 2.3×102 crowns are spent on illegal drugs by each person. Of course, only a small part of the total population can be taking drugs, so as a rough guess we can start by excluding half the people, those between the ages of 0 and 20 and those from 60 to 80. Of those remaining, at a rough guess, suppose a quarter take drugs. This means that these people are spending 2,000 crowns each per year. This is a considerable sum of money and so the drug consumption is significant.

§5 Keep an eye on the total figuresOccasionally percentages may deceive. Suppose for argument that we have a study of 2,000 people, half of whom drank water with meals, and half of whom did not. Suppose in the first group there were 3 cases of cancer, and in the other group there were two cases of cancer. Then we could claim:

“In a study of 2,000 people, those who drank water with meals had 50% more cancers than those who did not drink water with their meals.”

Strictly speaking the statement is correct, but totally misleading, in that the number of cases is not sufficient to make a sensible conclusion.

§6 AbortionsThe problem of allowing legal abortions is one of considerable importance. There are a number of different views, the most usual being

1. Abortion is killing and killing is not allowed and so abortion should be illegal.

2. While the foetus is inside the woman, it is really her right to decide what is to be done, and so abortion is acceptable.

3. Contraceptive and abortion advice encourages sexual intercourse and so should be banned.

4. In view of the need to avoid abortion; sexual and contraceptive advice should be freely available.

We do not wish to take sides but simply wish to point out the urgency of the problem. In Sweden the number of abortions per year is 30,000. Assuming that the ages of most people in the risk zone for needing an abortion are between 15 to 25, i.e. a ten-year span. If one argues that the population goes up to 80 that means that is roughly 1/8 of the population. Since Sweden has a population of 9 million, there are approximately one million people in that range and half of them are women. That means that one twentieth or about 5% of the women population per year are affected, a very large percentage.

§7 Compound increasesIf you look back at prices and wages over say the last twenty years it is striking how much they have risen. This may be a consequence of the fact that we always think of increases in percentages per year. We expect our salaries to increase by a certain percentage each year. Similarly we accept with resignation but as being at least reasonable, an increase of say 5% each year in costs. However these costs are compounded, and occurring year after year they become large. For instance an increase of 5% per annum becomes a doubling in 14 or 15 years. Is it possible that the very concept of a percentage increase per year is the reason for the huge increases in costs and salaries?

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§8 Worked examples1. Your salary increases from £30,000 to £35,000. What is the percentage increase?

Solution:

The percentage increase is (5,000/30,000)×100 = 16.6%.

2. Your salary decreases from £30,000 to £25,000. What is the percentage decrease?

Solution

The percentage decrease is (5,000/30,000)×100 = 16.6%.

3. Your salary of £30,000 increases by 5%. What is your new salary?

Solution:

The increase is ((5/100)×30,000 = £1,500.

4. Your salary increases by 10% one year, only to fall by 10% the following year. Are you back to where you started?

Solution: For each £100 you received before the increase, you now receive £110. The decrease of 10% means that you now receive £99, so you are not back to where you started.

5. Your salary goes up 10% for two consecutive years. Is this the same as a 20% increase?

Solution: £100 increases to £110, which in turn increases to £121. Thus your salary after two increases of 10% is greater than after a single increase of 20%.

6. Using Fig. 4 of §2 calculate the percentages of expenditures with respect to sales for this year.

Solution:  Percentages with respect to SalesSales 100Advertising 13Postage & telephone 9Wages 36Director´s fees 18Consultant’s fees 0.1Accountant’s fees 0.48Profit 23Working capital 6

FIGURE 7. Calculating the percentages in Figure 4.

7. Estimating a percentage helps thinking.

In the Swedish election held in September 2006 an alliance of conservative parties went to the polls with the promise of reducing unemployment. They had a number of measures, which included making it cheaper for employers to employ a long time unemployed, but they were

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also going to reduce unemployment benefits. Was it smart for the unemployed to vote for this alliance?

Solution: Probably not. For one must always ask by how much could the new government reduce unemployment. It won’t be 100% and a reasonable guess would be 10% because on the whole it is difficult to make a big change in a society. This means that 10% of the unemployed would get a job but that the other 90% would have lower benefits.

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Chapter 6

MEASUREMENT SENSE OR DIMENSIONAL

ANALYSIS

If the units are right, the formula is often right.

This chapter is about length, time, force, work and power, fundamental physical concepts, which we will discuss informally to provide an intuitive feel for these concepts. Kilometres, kilograms, watts, horsepower, seconds: these are the substance of our discussion.

In the early 1900s, the advent of mass production required precise measurement. In the beginning of the motor industry each part was handcrafted to each car, but with the introduction of mass production methods it became important to build parts which were so standard that they would fit any unit on the production line. Precision measurements and, consequently, accurate units, became imperative.

The international system of units, normally denoted by SI (short for Système International), is most common, although the Imperial system is popular in America and Great Britain. We will concentrate almost exclusively on the SI system.

§1 LengthSince some measurements are very small and others very large, it is convenient to express these in terms of powers of 10, which we will now define. The convention is that 10, stands for 10 multiplied by itself n times if n is a positive number while 10-n is 1/10n. For example, 5X103 = 5000, while 5X10-3 = 0.005. This has the added advantage that calculations involving products become easier, since 10m×10n = 10m+n, for both positive or negative values of m and n. [These powers of 10 have also been discussed in Chapter 2, Section 4.]

The standard unit of length is the meter, abbreviated by m. Originally it was defined as one ten thousandth of the distance from the equator to the North Pole. Later a special rod was kept in Paris to be the standard for the meter. The present method involves using light to define the standard, but the technical details need not concern us here. All we need to know is that there is a system that ensures an accurate and uniform definition. Several prefixes are standard for SI units. We use kilo-, for “thousand”; centi-, for “hundredth”; milli- for “thousandth”. For example, a kilometer is a thousand meters.

A centimeter (cm) is one hundredth of a meter, that is, 1 cm = 10-2 m and 1 m = 102 cm, To have a reasonable understanding of the units of length, the width of a hand is about 10 cm, a

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meter is the measurement from the ground to your belly button, and the length of a small European car is about 4 meters. A three-story block of flats is about 10 meters high, which is also tree height. Also, while 100 meters is a short stroll, 100 meters vertically is extremely high, the height of a 30-story skyscraper.

Turning to measuring the smallest of distances, such as the atom, we need another unit called the angstrom. An angstrom is 10-8 cm., which is about the width of an atom. The smallest distance that can be seen in an electron microscope is three-fifths of an angstrom.

The following table lists a few other approximations.

Finger width 2 cmWidth of thumb 2.5 cmHeight of step in a flight of stairs 16 cmLength of a foot, or height of a head 25 cmFrom the foot till the knee 50 cm Width of kitchen units 60 cmFoot to navel 1 mTall man or height of door 2 mLength of small car 4 mOne story of a block of flats 3 mHeight of airplane flight 5 – 12 kmHeight of TV satellite 36,000 km

§2 MassThe SI system has a standard unit of mass, the kilogram, abbreviated kg, which is a fixed body kept under careful conditions

The standard mass is kept well protected and used only to make a very few secondary standards, which are themselves used to make further standards, and in this way the standard of mass, the kilogram, is uniform throughout the world.

The gram – written g - is the mass of a body one thousandth of the kilogram. To give some idea of masses we have the following examples:

MassA4 sheet of paper 4 gStandard letter 20 gSlice of bread 40gAn egg 60gBanana or apple 120 gMeal of two eggs, two slices of bread, and a potato 350 gA litre of milk 1 kgA packed suitcase 20 kgA man 70 kgA medium sized European car 800 kg

Substituting for the unit as shown in the following example is a useful technique for changing from one unit to another.

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Example: Changing units, milligrams to kilograms. Express 34 mg as kilograms.

The first step is to convert mg to g:

1000 mg = 1 g, so 1mg = 1/1000g = 10-3g.

We simply take the mg in 34mg and replace it by ×10-3g. Thus

34mg = 34 ×10-3g.

The next step is to convert g to kg:

1000g = 1 kg, so 1g = 1/1000 kg = 10-3 kg, or g = 10-3 kg.

To continue, we replace the g by ×10-3 kg. Hence,

34mg = 34 ×10-3g = 34 ×10-3 ×10-3kg = 34 ×10-6kg.

Note that we have simplified 10-3 ×10-3 to 10-6 .

§3 TimeThe SI unit of time is the second, and this is denoted by s. Originally this was defined to be the time for a pendulum of length one meter to swing from one extreme to the other. The modern measurement is based on the Cesium-133 atom, but we will not bother about the details in this book.

“One thousand and one”, “one thousand and two”, etc. spoken deliberately measures time in seconds. The number of heart beats is about 70 per minute, so that one heart beat is about one second. Counting fast from one to ten lasts about 2½ seconds.

We denote hours by the symbol h. The following table lists some time markers.

Hours worked in a year (40 hours/week) 2,000Hours in a year 8,760Time spanned by great grandfather, grandfather and father 100 years

§4 Speed:Speed is defined as distance traveled divided by the time traveled.

Speed = distance/time

In symbols, the speed of an object traveling a distance d in time t is

d/t.

Example: Calculating speed

A horse travels 56 kilometers in 3 hours. What is the speed of the horse?

Solution

Speed = distance/time = 56km/3 h = 18.67 km/h. Here is a table of speeds. Travel at 11 km/h 3 m/sTravel at 108 km/h 30 m/sMan running 100 meters in 10 seconds 36 km/hSpeed of sound 340 m/sSpeed of light 3 ×105 km/s

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For instance, if a man shouts out to you at a distance of 100 meters, the sound will take one third of a second to reach you. On the other hand, the light reflected from the man will take 102 times 1/3 ×10-8 seconds or 1/3 ×10-6 seconds to reach you. The sun is so far away that it takes 8 minutes for the light from the sun to reach the earth.

Estimating the speed of sound

Two people stand about 150 meters from one another. The first one blows a whistle, and starts his stopwatch. The other whistles back as soon as he hears the first one. When the first one hears the other whistle, he stops his stopwatch. Thus the sound has travelled 300 m, and if the time is about 1 second, the speed can be calculated as 300m/sec.

Another method of estimating the speed of sound is in the spirit of Chapter 4, using Fermi calculations, in which we guess and estimate a rough value.

We know that commercial aeroplanes fly at a speed of about 900km/h. This is less than the speed of sound, so this is a lower estimate for the speed of sound. Now 900km/h = 15km/ min = 250 m/s. So knowing the speed of commercial aircraft, we can estimate the speed of sound to be about 250m/s, which is roughly in agreement with the known value.

Estimating the speed of light by satellite

This is also a calculation in the spirit of Chapter 4. If you listen to a TV sending a report from a correspondent who is a considerable distance away so that the message comes via satellite, you will notice a pause between the question asked in the studio and the reply. This is partially due to the distance that the radio signal must travel. The radio signal travels at the speed of light. The satellite is normally in what is called a geostationary orbit, which is some 40,000 km above the earth. The signal must therefore travel from the sender in your country to the satellite, and then travel from the satellite to the correspondent. His reply must also travel up to the satellite and from the satellite back to the studio. That is a total journey of 4 ×40,000 km or 160,000 km. By dividing this distance by the delay in replying one can get a rough estimate of the speed of light. In one program we viewed, there was a ½ seconds delay, which gives an estimate for the speed of light to be 320,000 km/s.

Rohmer estimate of the speed of light

In 1676 the Danish astronomer Rohmer gave an estimate for the speed of light. He had measured when the Jovian moon would cross the face of Jupiter, and found that this time varied depending on how far away Jupiter was from the earth. He suggested that the time difference was due to the time it took for the light to travel to the earth and in this way gave the estimate, 227,000 km/s.

Some wind speeds

60 km/h is a strong breeze, large branches in motion, umbrellas handled with difficulty, telegraph wires can be heard whistling. Wind at 70 km/h is called a severe gale, and will cause structural damage, such as chimney pots being displaced and slates removed. A wind speed of 100 km/h is a violent storm; when it reaches 118 km/h it is classified as a hurricane, thankfully not very often experienced.

Examples of speed conversion

To convert m/s to km/h we note that 1km = 1000m, so that 1 m = 10-3 km and 1 s = 1/3600 h. Substituting these values for m and s, we get

1 m/s = 10-3 km/(1/3600) h = 10-3 X 3600 km/h = 3.6 km/h.

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Converting from km/h to m/sec is a bit easier. Since 1km = 1000m and 1hr=3600s, we conclude that 1 km/h = 1000m/3600s = 5/18 m/s.

§5 AccelerationAcceleration is defined as rate of change of speed. That is, if the speed at time t = t0 is s0, the acceleration necessary to achieve a speed of s1 at time t1 is defined to be

(s1 - s0)/(t1 - t0).

That is, the difference in speeds divided by the time.

Example 1: Acceleration.

A car running at 40km/h speeds up to 50 km/h in 2 minutes, i.e., 1/30 h.. The acceleration is given by (50 – 40) km/h/ (1/30) h = 300 km/h/h. This is usually written 300 km /h2.

Example 2: Acceleration.

Calculate the acceleration of a sports car if it moves from 0 to 100km/h in 4 seconds. The change in speed from 0 to 100 is 100km/h = 1/36 km/s. Since this occurs in 4 seconds, the acceleration is 1/144 km/s2. In this example we chose to work in seconds, whereas in the previous example we worked in hours. Either way is acceptable, but since the SI units include the second as a basic unit, it is probably better to use seconds.

FIGURE 1 Galileo Galilei (1564-1642). Introduced the modern scientific approach based on experiment or theory supported by experiment. Father of Mechanics the study of moving bodies, forces and gravitation. Also made magnificent discoveries in Astronomy.

Acceleration due to gravity

It is a remarkable fact that all bodies acted on by gravity fall to the earth at the same speed if air resistance is insignificant. Galileo (1564-1642) demonstrated this by dropping two grossly different sized cannon balls from the leaning tower of Pisa and observing when they struck the ground. The expert knowledge at that time was that the larger the mass the shorter the

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time of impact. When Galileo tried the experiment he proved this was wrong, and showed that the difference perceived earlier was due to air resistance.

One way in which to argue this logically is to consider a mass of 10 kg and regard it as being split into 10 masses of mass 1 kg each. Drop all ten one kg masses simultaneously and, of course, they will all fall to the ground in the same time. Two adjacent masses will not affect one another so if we glue all the masses together to form one 10 kg mass, they will still fall to the ground in the same time.

You may prefer to argue this case using Newton’s laws of motion, as given in §10 problem 3.

Experiment reveals that the acceleration of any mass when wind resistance is disregarded is 9.80 meters per second per second.

§6 ForceForce depends on two factors. You notice that a force is acting when a mass accelerates. The mass and the acceleration are both needed to define the force. It is defined by multiplying the mass by the acceleration. The SI unit of force is the Newton, which is the force needed to cause a mass of one kilogram to accelerate one meter per second per second. If you hold a mass of 100 grams in your hand, it is pulled to the earth with a force of approximately one Newton. This is because the acceleration, as we remarked above, is 9.80 meters per second per second, and so the force is 0.1 kg × 9.80 m/s2; that is 0.98 Newtons, or approximately one Newton. The symbol to denote a Newton is N.

§7 Work and Power, Joules and WattsIn this section we discuss the material in an intuitive way, and provide a more precise and detailed discussion in §10.

Work depends on two factors. There is a force you are struggling against and a distance that you move through, and indeed, work is defined to be the product of the force and the distance. For instance, if you lift a suitcase a meter high from the ground the work is less than if you lift two suitcases a meter high, or lift one suitcase 2 meters high. How long you take to do this does not change the work done; a second for the job entails the same amount of work as taking an hour, just as if you travel from one point to another, you still have travelled the same distance, whether it takes a day or an hour. It is the speed which changes if you take more or less time, not the work.

However, if you take the time into account, then instead of work you measure power. Power is the work divided by the time; so the shorter the time, the more the power.

Power is measured in watts – indicated by W. A kilowatt is a 1,000 W, and is denoted by kW.

Example: Lifting a suitcase weighing 25 kilogram one meter high (i.e. up to your waist) in one second is a power of roughly 250 W. If you do the same work in one half a second, the power is 500 W.

How exactly these ideas are defined and the values calculated appear in the optional material below in §10. Here the aim is to give an intuitive idea of these concepts.

Thus, a person at rest is working at a rate of something like 30 W. A normal size electric light bulb works at a rate of 60 W, and an electric kettle works at a rate of about 1,000 W, i.e. a kilowatt. The power of a human being, i.e. the rate at which he can work, is approximately 90 W.

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During the steam age it was the custom to visualize the power of the new steam engines in comparison with horses. This led to the unit of a horsepower. One horsepower is the equivalent of 746 W. Actually this corresponds to an idealized horse, and the value is a little optimistic, in that working at the rate of one horsepower is more the rate of work of one and a half horses.

The concept of horsepower gives one a graphic way of viewing power. When you boil a kettle of water in an electric kettle, the electricity is working at a rate that is more than that of a horse. A car driving fast at constant speed along a level road works at the rate of about 20 horsepower.

Cars can typically develop 100 horsepower. Super sports cars boast 350 horsepower and more.

The typical electricity consumption in a modern flat means that we have roughly the equivalent of a horse working for us for ten hours every day or 80 man-hours of labour. In other words, we have the equivalent of 8 slaves working for us every day for ten hours. No wonder modern man is like a king of only a century ago. [See §9 problem 3 for the calculation.]

Kilowatt hours: A kilowatt hour is a measurement of work. It is the work done at a rate of one kilowatt for one hour. For example, a typical electric kettle boiling water non-stop for an hour performs one kilowatt hour of work. It is denoted by kWh,

§8 Dimensional AnalysisA useful technique for checking a physics formula, or even getting a suggestion as to what the formula should look like, is obtained by using the fundamental dimensions of length [L], time [T], and mass [M]. The square brackets are used to indicate that we are talking about dimensions. It is a fundamental law in Physics that in any equation the dimensions on the right-hand side of the equation must be equal to the dimensions on the left-hand side of the equation.

Example 1

Find the dimensions of area and volume. What are the dimensions of speed and acceleration?

Solution: Area is calculated by multiplying length by breadth. Thus the dimension is [L]2. Volume is obtained by multiplying length by breadth by height with dimension [L]3. Speed is distance divided by time, giving us the dimension [L]/[T] or [L][T]-1. Acceleration is speed divided by time. Since the dimensions of speed, as we have just calculated, are [L][T] -1, for acceleration we must divide by time again, and the dimension must be [L][T]-2

Example 2

Find the units of force and work.

Solution: Force is defined to be the product of mass and acceleration. Acceleration has the units [L][T]-2, thus the units of force are [M][L][T]-2. Work is force, [M][L][T]-2, times distance, [L], so the units for work are [M][L]2[T]-2.

Example 3

A pendulum of length l is subject to gravity, which has acceleration g. Find the form of the formula for the period of the pendulum, that is, the time required for the pendulum to swing from its initial starting position back to its starting position.

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Solution: It seems reasonable to assume that the period will be a constant times a power say a of the length, l, of the pendulum and the b-th power of the acceleration due to gravity, g. That is, T should have the form

T = constant× lagb ,

where the powers a and b are to be determined. The dimensions of g are the dimensions of acceleration, i.e. [L][T]-2, and the dimensions of l are [L]. Hence, the dimensions of the right-hand side of the formula are [L]a[L]b[T]-2b = [L]a+b[T]-2b . These must match with the dimensions of the left-hand side, and since T has the units of [T], this means that [T] = [L]a+b[T]-2b. This will be possible only if -2b = 1, and a+b = 0, that is, b = - ½, and a = ½. Hence, the formula for the period must look like:

T = constant × l½g-½.

(Note that an exponent of ½ means the square root, and an exponent of - ½ means 1 divided by the square root.) In fact, the correct formula is

T= 2l½g-½ = 2(l/g).

Of course, this analysis does not give an exact result for the constant, but it is remarkable how frequently dimensional analysis points towards the right formula.

Example 4

A body in free fall (i.e., subject only to gravity and with no wind resistance) has an acceleration g due to gravity. Find the units in a formula relating distance fallen, x, and time of fall, t.

Solution: Assuming a formula of the form x = constant×gatb, where g is the acceleration due to gravity, in terms of units this becomes

[L] = [L]a[T]-2a[T]b.

Since the left-hand side of the equation has [L] to the power 1, a must also be 1. On the left-hand side [T] does not appear, but on the right-hand side we see that with a having the value 1, we have [T]b - 2 so this entails that b = 2. The actual formula is

x = ½ gt2.

We now come to what is at once the most wonderful and the most terrible of all formulae in Physics, namely the formula from Einstein’s paper on Relativity relating energy, E, mass, m, and the speed of light, c, in one formula:

E = mc2,

the product of the mass m and c2, which itself is the product of the speed of light by itself. Note that in Physics energy means the ability to do work, and the equation says that if the mass m were converted entirely into work, the amount of work would be given by the formula. Thus the units of energy are the same as the units of work.

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FIGURE 2. Albert Einstein (1879-1955), whose theoretical discoveriescontributed to the development of nuclear power and the nuclear bomb.

It is the most wonderful equation because it explains so much. For instance, the sun, which is the source of all energy and light on the earth, gives out an enormous amount of energy. For a long time it was thought that this was the result of burning: the same sort of burning we have on Earth, where matter combines with oxygen. That meant that it was possible to estimate how much energy the sun had, but when the calculation was done, it was obvious that the sun should have burnt out long ago. Ordinary burning could not explain the energy, which the sun emitted and had emitted for so many years. Einstein’s equation could. The energy of the sun came from the conversion of mass to energy.

Einstein’s equation also meant that we ourselves could convert matter into energy in our nuclear reactors, which has turned out to be a mixed blessing, with plenty of energy but with awful dangers. Einstein’s theory has also led to nuclear weapons with the ability to destroy our civilisation completely.

Example 5: Check that the dimensions on both sides of Einstein’s equation match.

Solution: The dimensions of E the left-hand side of the equation are those of work, which we found out in Example 2 immediately above, were [M][L]2[T]-2. The dimensions of the RHS are those of [M] multiplied by the dimensions of velocity squared, i.e.[L]2[T]-2, Hence the dimensions on both sides of the equation are the same.

§9 Solved problems1. What is the acceleration in going from 0 to 100 km/h in 10 seconds?

Solution. The acceleration is given by difference in speed divided by time. We need to use the same unit of time. 10 seconds is 1/360 of an hour. The difference in speed is 100. So the acceleration is

100/(1/360) = 3.6X104 km h-2.

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2. In a recent Winter Olympics the difference in time between the first and second places in the women’s skeleton slide was 0.65 seconds. Assuming that the final speed attained in the event was 120 km/h, what was the distance in meters between the two at the finish?

Solution. 120km/h = 120 ×1000 /3600 m /s = 33.33 m/sec. Hence, 0.65 seconds corresponds to

0.65 × 33.33 m = 21.66 m.

3. Assuming that your household uses 2,500 kWh per year, what is the average daily consumption? What is the equivalent in horses per day?

Solution. There are 365 days in a year, so the average daily consumption is 2500/365 = 6.8 kWh. One horsepower is 746 W. So it would require roughly 9 to 10 horses to work at the rate of 6.8 kW for an hour. So 9 or 10 horses working for an hour, or one horse working for 9 or 10 hours, will give about 6.8 kWh. In terms of manpower, or slave power, this means we would need 8 slaves working each day for each household for 10 hours, since a man can work at only about 90 W.

4. If a man drives at the rate of 50 km/h for the first 300 km of a trip, how fast must he drive over the next 300 km to have an average speed of 60 km/h for the 600 km journey?

Solution. At an average speed of 60 km/h for 600 km, it would take 10 hours to complete the journey. At 50 km/h, the first 300 km drive would have taken 6 hours, which leaves him 4 hours to complete the final 300 km. Hence, to make up the time he would have to drive at 75 km/h over the final 300 km.

5. One man can dig a hole in 2 days, another man can dig the hole in 3 days. How long will they take to dig the hole together?

Solution: There is an ingenious way of solving this problem. One calculates the rate of digging the hole. The first man has a rate of ½ a hole per day, the second man a rate of 1/3 a hole per day. Thus the combined rate is ½ + ⅓ = 5/6 of a hole per day. Thus it takes 6/5 of day to dig the hole.

§10 Additional Topics1. Definition of force

The force acting on a mass is defined by multiplying the mass M in kg by the acceleration a in m/s2, i.e.

Force = Ma.

The unit of force is the Newton. Thus a Newton is the force that accelerates a mass of 1 kg by 1 m/s2. For example, a mass of 15 kg which when acted on by a force accelerates 40 m/s2 has a force of 15 ×40 = 600 N acting on it.

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2. Newton’s laws of motion (published in 1687)

FIGURE 3. Newton (1643–1727) was co-discoverer of the Calculus and developeda far-reaching theory of moving bodies such as the motion of the planets.

Newton formulated three laws of motion. With these and the mathematical theory that Newton developed, which is called Calculus, Newton was able to explain all the laws of planetary motion. Newton’s methods are still fundamental in calculating, for instance, the movement of satellites as well as the laws of physics which regulate our everyday lives.

The first law states that a body continues moving in a straight line at the same speed unless acted on by a force. One result of this law was to dispel the commonly held belief that planets moved because angels or spirits were pushing them.

The second law defines force in terms of mass and acceleration as described in item 1 immediately above.

The third law is that to every action there is an equal and opposite reaction. That law governs the motion of rockets as well as collisions of billiard balls.

Newton also assumed that there is a force of attraction between two bodies with masses m1, m2, which is a constant G times the product of their masses divided by the square of the distance d between them, that is,

F = G m1m2/d2 .

The value of G has been found experimentally to be 6.67 ×10-11. The units of G are Nm2/kg2.where N stands for Newton, m for meter and kg for kilogram.

3. Explain why all bodies fall to the ground at the same time if air resistance is not a factor, using Newton’s Laws of Motion.

Solution; If M is the mass of the earth and r is the distance to the centre of the earth, then a mass m will experience a force of

GmM/ r2

according to point 2 above. Since the force equals ma where a is the acceleration, we have the equation

ma = GmM/r2.

If we cancel m from both sides, the result is

a = GM/r2

in which m doesn’t play a part. This means that all bodies accelerate at the same rate independent of their masses and, hence, will reach the ground at the same time if released simultaneously.

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4. Definition of work

Work is defined to be the product of the distance a force is moved through times the force. The unit of work is the joule, abbreviated to J.

The Joule is the work done in moving a force of 1 Newton a distance of 1 meter. Since the force due to gravity on a 100 gram mass is about 1 Newton, if you raise a 100 gram weight from the floor to you waist, you will have done one joule of work.

5. Definition of Watts

The rate of work in Joules per second is called the power. A Joule per second is also known as a Watt.

6. A man weighing 70 kilos walks up three flights of stairs in one minute. What is the work done and what is the rate of work, i.e. the power?

Solution: Every step is 16 cm high and there are 16 steps per level. That means 768 cm, i.e. roughly 8 m. He is lifting 70 kg. The force acting on the man is calculated by taking the mass and multiplying by the acceleration due to gravity, which is 9.81 m/s. Thus the force in Newtons is

709.81 = 686.7 N.

To calculate the work done we need to multiply the force by the distance, to get

8 686.7 = 5,493.6 J.

To find the rate or work we divide by the time in seconds, which is 60. Thus the result is

5,493.6 /60 = 91.56 J/s = 91.56 W.

7. Approximate the power in raising a shoe from the floor to a height of 2m. in three seconds.

Solution: A shoe of mass say 400 g =0 .4 kg has a force of 0.4× 9.81 N . i.e. the mass 0.4 kg multiplied by the acceleration due to gravity 9.81 m/s2. This is approximately 4 N. So the work done on lifting the shoe two meters is approximately 8J. Thus the power required is 8/3 J/s. or 2.67 W.

8. A car journey of 100 km takes an hour and uses 7 litres of fuel. What is the work done and what is the rate of work.

Solution: This is another of those Fermi type calculations. According to the label on a bottle of cooking oil it has an energy value of 3,700 kJ per decilitre. It would be better to know the value for a decilitre of petrol but this figure is not available, so we use the cooking oil as a rough estimate. Hence a litre of petrol has about 40,000 kJ and 7 litres about 300,000 kJ. Per second we have therefore been using 100 kJ per second, or. 100kw. Since a horsepower is roughly 740w or, even more roughly, a kilowatt , this corresponds to 100 horsepower. We know however that there are inefficiencies in engines, and so if we assume only one quarter of the energy is available at the wheels, the car is developing roughly 25 horsepower.

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9. Calories and conversions

Another unit of energy is the calorie. This is defined to be the amount of energy required to raise the temperature of one gram of water by one degree Centigrade. Since calories and Joules are measurements of energy one can convert the one to the other, and a calorie is the larger by approximately 4 times. More accurately,

1 calorie = 4.184 J.

Just as there is a kJ, there is also the concept of kilocalorie, a thousand times larger. This is also the unit used to measure human nutrition. It is often written as kcal or Calories, with a large C. A daily intake is roughly 2,000 kcal.

10. Calories of some foods per 100 gram.

Bread 230 kcal

Olive oil 884 kcal

Butter 700 kcal

Fried hamburger 280 kcal

Cheese 30% 400 kcal

Cottage cheese 4% 100 kcal

Sugar 400 kal

Baked soya beans 100 kcal

Thus, each gram of food gives between 1 and 9 kilocalories.

11. Estimate the weight of food eaten each day.

This is an exercise in the spirit of Chapter 4 Fermi Problems. Since one eats about 2,000 kcal per day, and food gives about 2 kcal per gram (rough guess based on the list of foods above), we need 2,000 divided by 2, i.e. 1,000 g or 1 kg of food per day

12 Estimate the rate of energy used by a person.

Solution: Since 2,000 kilocalories is consumed in 24 hours (and this corresponds to the minimum), the rate of energy is 2000/24 ×3600 kilo calories per second = about 6 calories per second. From point 10 immediately above, 1 calorie = 4.184 J. Multiply by 4.18 to get roughly 24 J per second or 24 W. This seems to corroborate the roughly 30 W given in the text above,

13. Given that 1W = 1J/s, that 1 calorie = 4.18 J, and that 1 horse-power = 745 W, …

a) Express J in terms of calories

b) Find the relationship between kJ and kW hours

c) Find kW in terms of horse-power.

Solution:

a) Since 1 calorie = 4.18 J, dividing by 4.18, 1J = 0.239 calories.

b) To express kW hours in terms of kJ. 1J/s = 1W, so 1 W hr = 3600J.

So 1kW hour = 3600 kJ. Hence 1 kJ = 1/3600 kW hours = 2.778 X 10-5 kW hours.

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c) 1 horsepower = 745 W. Thus 1000 horsepower = 745 kW. Thus dividing by 745, 1 kW = 1000/745 horsepower = 1.342 horsepower.

14. Temperature

The most commonly used temperature scales are the Centigrade (also called Celsius) and the Fahrenheit.

There is a third scale, called Kelvin. Zero degrees Celsius is approximately 273 degrees Kelvin, that is, 0 degrees Kelvin is 273 degrees below 0 Celsius. The Kelvin is of scientific interest because at a temperature of 0 degrees Kelvin molecules have no energy and have stopped vibrating.

The Celsius scale, which dates from 1743, is based on a temperature of 0 degrees for water freezing and 100 degrees for water boiling. On the Fahrenheit scale, which dates from 1724, water freezes at 32 degrees and boils at 212 degrees.

The founder of the scale, Fahrenheit, used 32 to avoid negative temperatures in winter. Also, Fahrenheit’s idea was to have a temperature of 100 degrees for the human body, which is close to the actual figure of 98.6 degrees Fahrenheit. Unfortunately, when he did his experiment, his assistant had a slight fever.

To convert from Centigrade to Fahrenheit multiply by 9/5 and add 32. To convert from Fahrenheit to Centigrade, subtract 32 and multiply by 5/9.

Example: Convert 20 degrees Celsius to Fahrenheit. We multiply 20 by 9 and divide by 5 to get 36. We then add 32 to get the Fahrenheit equivalent of 68 degrees Fahrenheit.

Convert 80 degrees Fahrenheit to Centigrade. We subtract 32 from 80 to get 48, multiply by 5 and divide by 9 to get approximately 27 degrees Celsius.

15. When does the Celsius reading equal the Fahrenheit reading?

Solution Let t denote the temperature on the Celsius scale which is to be the same reading on the Fahrenheit scale. Since t degrees Celsius is the same as (9/5)t + 32 degrees Fahrenheit, the problem is to find a t such that (9/5)t + 32 = t. Subtracting t from both sides gives (4/5)t = -32, so t = - 40. That is, when it’s -40 degrees Celsius it is also -40 degrees Fahrenheit.

16. Check the following verse which accentuates the bewildering variety of old English units:

A dozen, a gross and a score,

Plus three times the square root of four,

Divided by seven, plus five times eleven,

Is nine squared

And nothing more.

Solution: This is indeed an exercise in English units. A dozen is 12, a gross is 144 and a score is 20, and so their sum is 176, Three times the square root of 4 is 6, giving a total of 182. If we divide 182 by 7, we get 26. Add 5 times 11, i.e. 55, we get 81. 81 is 9 squared, as claimed.

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17. Derive the units of G.

Solution: We have the formula: force = GmM/r2. Force has the units [M][L][T]-2, and r has the units [L].

From the formula

[M][L][T]-2 = [G][M]2/[L]2, it follows that [G] = [M]-1[L]3[T]-2 .

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CHAPTER 7

MEASURING HEIGHTS OR TRIGONOMETRY

A knowledge of right-angled triangles with largest side 1

enables us to calculate the lengths of any other right-angled triangle

§1 Shadow heightsIt is easy to measure heights by means of shadows. If you want to decide on the height of a tree then compare the length of its shadow with the shadow of a vertical pole. If the shadow of the tree is four times longer than the shadow of the pole, then its height is 4 times the height of the pole.

Often it is easy to use your own shadow to make the comparison, since you know your own height. By calculating how many lengths of your shadows fit inside the shadow of the tree you can estimate the height of the tree. You can do the same for a building or any vertical structure.

This method for measuring heights is based on comparing similar triangles. Similar triangles are the same shape, but of different size; like two shirts of different sizes. More formally, two triangles ABC and ABC are similar if their angles are equal in pairs, as in Fig. 1, where angle A = angle A´, angle B = B´ and angle C = C´. If two triangles are similar, the one with sides of length x, y and z, then there is some number k so that the lengths of the other triangle’s sides are k×x, k×y, k×z..

Fig. 1 Similar triangles

How does this concern measuring heights from shadows? The reason comes about as follows. First of all, since the sun is so far away we may think of light from the sun as consisting of parallel rays.

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Suppose that the top of the tree we are concerned with is at the point T (see Fig. 2) and its bottom at B. The sun casts a shadow that extends from B to the point E, the end of the shadow. Thus we have a triangle BTE with a right angle at the point B since the tree is at right angles to the ground. By measuring the length BE of the shadow of the tree, we can determine the tree’s height BT if we have a reference.

The reference we obtain by measuring the shadow of a person standing upright, i.e. also at right angles to the ground. Suppose the top of the person is represented by the point P, the person is standing at A, and the person’s shadow is from A to Q as in Fig. 2.

Now the two triangles BTE and APQ are similar, because each has a right-angle and the angles TEB and PQA are equal, since the sun’s rays are parallel. This means that the third angle of each triangle is the same. We can therefore find the lengths of the triangle BTE by multiplying the lengths of the triangle APQ by a constant k. In particular, if we know that any side of BTE is k times the length of the corresponding side of APQ, this will mean that all sides of BTE are k times the sides of APQ.

For example, suppose the shadow cast by the person is 3 meters and that the tree has a shadow of 30 m, then the constant k must be 10. If the person has a height of 2 m, then the height of the tree must be 10 times the height of the man, i.e. 20 m.

We can write this in the form of an equation in general. To find the constant k we simply divide the length of the shadow of the tree by the length of the shadow of the man, i.e.

length of shadow of tree/length of shadow of man.

We then multiply by the height of the man, so that

Height of tree = (shadow of tree/shadow of man) × (height of man),

FIGURE 2 Shadows

We checked the height of a building this way: By comparison, we climbed from floor one to floor two, measuring the number of steps. There were 16. Each step was 16 centimetres high, so the total height for one floor is 16×16 = 256 cm, i.e., roughly 2.60 meters. The three-storey flat was therefore 7.80 meters. But since the ground floor was about one meter from the ground, we made the height to be 8.80 meters, i.e. about 9 meters. This agreed with the height calculated from the shadows.

These are rough calculations, but the surveyor uses the same principle of similar triangles and measuring very accurately gets precise results.

§2 The artist’s methodA standard method used by artists is shown in Fig. 3. Here he holds his arm straight in front of him and compares the various heights on his pencil. For instance, if he is drawing a head, he gets the tip of the pencil at the top of the head he is drawing, and then places his thumb to be

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in line with the eyebrows. He then uses the pencil to mark this distance on his sketch. Then in a similar way, he measures the distance of the eyebrows and the tip of the nose, and then to the mouth and finally to the chin. In this way he gets accurate measurements. This is also based on similar triangles. The explanation is given in §5 problem 8.

FIGURE 3 The artist’s method

Right-angled triangles appear often in practice, for instance in surveying. Given a right-angled triangle ABC and knowing the angle at A and the length of the largest side h, we can with the aid of a table calculate the lengths of the other two sides. This is the idea of sine and cosine. The largest side of a right-angled triangle is also called the hypotenuse.

We would have to produce an infinite table if we listed all right-angled triangles. We reduce our list by considering right-angled triangles with hypotenuse 1. The triangle is then determined by one of its angles, since the sum of the angles of a triangle must be 180 degrees. The sine and cosine are simply the lengths of the two sides in a triangle with hypotenuse 1.

Thus in the right-angled triangle ABC in Fig. 4, AB has length 1.

FIGURE 4 Definition of sine and cosine

In Fig.4, the side BC, the side straight in front of the angle A, is called sine of the angle at A, and written sin(A). The side AC is called the cosine of the angle at A and written cos(A). (Note that it is customary to write the sine or cosine of a given angle without the final e.)

To make sure you remember which is which, you can think of sin as S in, i.e. short for straight in front. Also you can think of cos and that the C stands for the closer side.

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The sine and cosine are useful for surveying and navigating. It is interesting to note that the gps system, which from satellites determines the latitude and longitude, uses them as well.

Extensive tables of sine and cosine exist. Some of the values are listed below in Fig. 5.

Angle in degrees Sine Cosine

0º 0 1

15º 0.2588 0.9659

30º 0.5 0.8660

45º 0.7071 0.7071

60º 0.8660 0.5

75º 0.9659 0.2588

90º 1 0

FIGURE 5. Table of sines and cosines to 4 decimal places

There is also a useful approximation if we are dealing with very small angles. If d is the value of an angle in degrees, and d is small, then the value of sin(d) is approximately d/180. For instance, sin(0.180) is 0.001, or 0.00314, correct to 5 decimal places.

Now suppose we are given a right angled-triangle XYZ with Z a right angle, see Fig. 6. Then consider the right-angled triangle ABC with right angle at C and with angle A equal to angle X. Then triangle ABC is similar to triangle XYZ. But the sides of triangle ABC are already tabulated in the tables of sine and cosine. Since the hypotenuse of the triangle XYZ is h, which is h times the length of the hypotenuse of ABC, this tells us that the multiplication factor connecting the sides of XYZ to ABC is h. So to find the sides of triangle XYZ it is only necessary to multiply the sides of ABC by the length of the hypotenuse h of XYZ, that is the sides of XYZ are hsin(A) and hcos(A).

FIGURE 6. A right-angled triangle XYZ with hypotenuseh, and a similar triangle ABC with hypotenuse 1

Example 1: A right-angled triangle has angle 45 degrees. Calculate the length of the sides if the largest side, the hypotenuse, has length 14m.

Solution. The sides will be given by 14×sin(45) and 14×cos(45). From the above table the sine of 45 degrees is 0.7071, as is the cosine, so the length of either of the shorter sides is 14 ×0.7071 = 9.9.

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Example 2: A right-angled triangle has angle 30º. Calculate the length of the sides if the largest side has length 29m.

Solution. The sine of 30º is 0.5. Hence, the length of the side opposite the 30 degree angle is 29×sin(30) = 29 ×0.5 = 14.5 m. The length of the side adjacent to the 30 degree angle is 29 ×cos(30) = 29 ×0.8660 = 25.114 m.

Example 3: Determine the lengths of all the sides in the right-angled triangle in Fig. 7.

Solution: The sides are AC = 10×sin(15) = 2.588 and AB = 10×cos(15) = 9.658, from the table in Fig. 5.

FIGURE 7. A right-angled triangle with 15º angle at B and hypotenuse 10 m.

§4 Dropping a stone over a cliffOne way of estimating the height of a cliff, which is over a deserted stretch of water, is to drop a stone and with a stopwatch measure the time for the stone to reach the water. If this time is t, then the height is given by the formula

Height of cliff = ½gt2,

where g is the acceleration due to gravity, i.e. about 9.8 ms-2. (This formula was mentioned in Example 4 §8 of Chapter 6.) In our case we got a time of 1.5 s, and so the height of the cliff was

½ ×9.8 ×1.5 ×1.5 = 11.025 m.

Example: Using a ruler to calculate reaction time.

Use a centimetre ruler. One person holds the ruler at the top and the other holds his hand near the bottom, with forefinger and thumb almost clasping the ruler, but loosely, so that if the ruler is released at the top it will fall.

The first person suddenly says now, and releases the ruler. The second person must clasp his finger and thumb together so stopping the falling ruler. One then notes the distance the ruler has fallen before the second person reacts and grabs the ruler. The reaction time is then calculated from the formula

s = ½gt2

From this formula we see on multiplying by 2 and dividing by g that

2s/g = t2

We then calculate the time in seconds by measuring the distance s in cm. Then calculate 2s/g and take the square root. That is the reaction time. For g take 980 cm s-2.

In an actual test one of us stopped the ruler in 10 cm. Thus the reaction time was 0.14 s..

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§5 Solved problems and other optional topics1. Suppose you are measuring the height of a building with a shadow 20 meters long, and your shadow is 5 m. long. If you are 2 m tall, how high is the building?

Solution. We use the formula we found in §1 (with the slight modification that it is a building and not a tree that we are concerned with).

Height of building = (shadow of building/shadow of man) × (height of man)

If h is the height of the building, then h = 20/5 ×2 = 8 m.

2. Explain why sine of 0º is 0 and sine of 90º is 1.

Solution. For an angle of 0º it is not exactly clear how to define the sine. But if we take a right-angled triangle with hypotenuse 1 in which the angle is very small, we can see that the side, which defines the sine, becomes very small indeed, and the smaller the angle the smaller the side becomes. It seems reasonable to define the sine of 0º as 0.

Similarly we see what happens when the angle approaches 90º, where it is clear that the side opposite the angle becomes larger and larger and seems to be getting closer and closer to the hypotenuse 1. It is thus natural to define the sine of 90º as being 1.

3. Determine the sine of 30º and 60º without the use of tables.

Solution. The triangle in Fig. 8 has all three sides of length 1.

B

A C

DFIGURE 8. Equilateral triangle with sides of length 1.

The angles are also equal, so each measures 60º. A perpendicular dropped from the vertex B to the base bisects the base, that is, cuts the base in half, so AD = DC. and AD = 1/2. Also, BAD is a right-angled triangle with hypotenuse BA of length 1.

Since the angle BAC is 60º, the angle BAD is 30º, so sin(30º) = AD = 1/2. Also, since ADB is a right triangle, by Pythagoras’s theorem, AD2 + BD2 = AB2. Hence, (1/2)2 +BD2 = 1, so BD, the side opposite the 60º angle at A, is the square root of

1 - (1/2) 2 = 3/4,

which means cos(30º) = 0.866.

4. A surveyor measures the angle of a top of a building at 15 degrees at a distance of 20 meters. His instrument is at a height of 1.5m. What is the height of the building?

Solution: Refer to Fig. 9.

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We know that the triangle ABC is similar to a triangle XYZ with hypotenuse 1 and angle 15º, which has sides YZ = sin(15)= 0.2588 and XZ = cos(15) = 0.9659 as listed in Fig. 5. Thus, the sides of ABC are simply a multiple k of the sides of XYZ, so, in particular, AC = kXZ. Thus the factor k is calculated by taking

AC/XZ = 20/cos (15º) = 20/0.9659 = 20.706.

Thus, BC = 20.706× sin(15º)= 20.706 × 0.2588 = 5.359. To get the height of the building we must then add 1.5 meters, thus getting a final height of 6.859 meters.

FIGURE 9. A surveyor calculates the height of a building.

5. I stood on the top of a cliff and dropped a stone. It took 3 seconds before I heard the splash when the stone hit the water below. What is the height of the cliff?

Solution: The distance is calculated from the formula

distance fallen = ½gt2.

In this case, t = 3 and g, as we know, is 9.8 m/s2. Thus, the distance is ½×9.8×9 = 44.1 m.

6. Remark.

For the next two problems we will accept the following fact from physics: A force F at an angle a can be regarded as the result of two forces F1 and F2 acting at right angles to each other, as shown in Figure 10. In particular, the two forces are

F1 = F×sin(a) and F2 = F×cos(a).

FIGURE 10. The force F can be regarded as the sum of two forces at right-angles to one another

7. Explain why if the sides of a military battle tank are oblique rather than vertical, a shell is less able to penetrate it. For example, consider the shell hitting the tank at an angle of 90 and then, if the tank’s sides are slanted, at an angle of 30.

Solution; We use he remark in 6 above. If the shell hits a vertical side of a tank standing at an angle of 90 degrees with a force F, then the force acting at that point is going to be F. However, if the angle at which it is directed is 30 degrees, then the force can be regarded as the result of the two forces, F×cos(30) and F×sin (30). The force of F×cos(30) is parallel to the tanks armour, and so does not help to penetrate it. The other force of F×sin(30) is ½×F

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and this is the effective penetrating force. This explains why angling the tank’s armour helps reduce the shell’s impact.

8. Sunbathing at an angle rather than directly. Suppose you are lying in front of the sun with the rays coming in at 90 or at 30 degrees. The effect is considerably reduced according to sine of 30 degrees.

Solution: If the intensity of the sun’s radiation is I, then a similar result to 6 above holds for it, that is the intensity can be regarded as two intensities, at right-angles, the one of

I×sin(30) and I×cos(30), The intensity parallel to the skin does not cause any sunburn, and the only part that gives sunburn is I×sin(30) = ½×I.

The same type of argument helps to explain why, when the sun’s rays come in at lower angles, winter is colder than summer.

9. Explain how the artist’s method works,

FIGURE 10. The artist’s method explained

In Fig. 10 the bottom of the vertical object we are drawing is at the point D and the pencil is held vertically at the point A. The artist’s eye is at O.

The point C represents a point on the object which is at a height x above the bottom point. Seen by the artist this point is at a height of x on the pencil which is fixed at the point A. Since he triangles OAB and ODC are similar, we have that

x/OA = x/OC, or that x¨= (OA/OD) x.

Thus x¨ is proportional to x and with this method we get a direct scaling of the object to be drawn.

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O

B

A

x´

D

C

x

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Chapter 8

LOGARITHMS AND NATURAL LOGARITHMS

“The invention of logarithms saved astronomers a lot of trouble and doubled their lives” - Laplace

§1 Multiplying by adding. Logarithms

FIGURE 11. John Napier (1550-1617)

Multiplication is more time consuming than addition. To multiply two 8-digit numbers, we need to perform 64 multiplications and several additions. On the other hand, to add two 8-digit numbers we need only add 8 times. The difference is large and gets much larger with every increase in the number of digits, so after the invention of the telescope ushered in a revolution in astronomy, which, in turn, necessitated large and precise calculations, it became a matter of time before a method of simplifying calculations would be found. The answer was: logarithms.

John Napier in Edinburgh published the first table of logarithms in 1614. It is said that Henry Briggs, professor of geometry in Gresham College, London, was so impressed by Napier’s system of logarithms that he was speechless for fifteen minutes when they first met, gazing at Napier with admiration. Then he explained that he had travelled especially to see Napier and enquired, “… by what engine of wit or ingenuity you came first to think of this most excellent help in astronomy... “.

Ten years later, in partial collaboration with Napier, Briggs published a new table of logarithms, which he called “common logarithms”, based on an improved system still in use today.

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In Fig. 2 we have a short table of common logarithms. In each column the number on the right is the logarithm of the number on the left. Briggs produced a table of logarithms to 17 places, which meant it could be used to find products of up to 17-digit numbers, correct to 16 decimal places. This was a major effort and took almost the whole ten years. Since then, much larger tables to over 200 places have been constructed.

1.0 .000 2.0 .301 3.0 .477 4.0 . 602 5.0 .699 6.0 .778 7.0 .845 8.0. .903 9.0 .954

1.1 .041 2.1 .322 3.1 .491 4.1 .613 5.1 .708 6.1 .785 7.1 .851 8.1 .908 9.1 .959

1.2 .079 2.2 .342 3.2 .505 4.2 .623 5.2 .716 6.2 .792 7.2 .857 8.2 .914 9.2 .964

1.3 .114 2.3 .362 3.3 .519 4.3 .633 5.3 .724 6.3 .799 7.3 .863 8.3 .919 9.3 .968

1.4 .146 2.4 .380 3.4 .531 4.4 .643 5.4 .732 6.4 .806 7.4 .869 8.4 .925 9.4 .973

1.5 .176 2.5 .398 3.5 .544 4.5 .653 5.5 .740 6.5 .813 7.5 .875 8.5 .929 9.5 .978

1.6 .204 2.6 .415 3.6 .556 4.6 .663 5.6 .748 6.6 .820 7.6 .881 8.6 .934 9.6 .982

1.7 .230 2.7 .431 3.7 .568 4.7 .672 5.7 .756 6.7 .826 7.7 .886 8.7 .840 9.7 .987

1.8 .255 2.8 .497 3.8 .580 4.8 .681 5.8 .763 6.8 .833 7.8 .892 8.8 .944 9.8 .901

1.9 .279 2.9 .462 3.9 .591 4.9 .690 5.9 .771 6.9 .839 7.9 .898 8.9 .949 9.9 .996

10.0 1.000

FIGURE 2. Table of common logarithms

This is the way logarithms are used to multiply. If we want to find the product of two numbers, x and y, we look up their logarithms in a table of logarithms. The sum of these logarithms will be the logarithm of their product. The answer is the number with this logarithm, which then can be found from the table of logarithms, such as the one in the above table. For example, from the table,

logarithm of 2 = 0.301 and logarithm of 3 = 0.477.

The sum of these two logarithms, 0.778, will be the logarithm of the product of 2 and 3, which we find from the table to be the logarithm of 6. This is a trivial example, but, in general, we can use tables of logarithms to calculate more difficult products easily.

If we denote the common logarithms of x and y by Log(x) and Log(y), then what we said in the previous paragraph translates to the equation:

Log(x) + Log(y) = Log(xy).

Logarithms look like magic, but this is actually nothing but the law of exponents as explained in Chapter 2 §4. For the theory see the explanation in §5 of this chapter.

The initial impetus for logarithms as we said was the need to simplify multiplication. However, we can now do this extremely quickly with the aid of computers, so what is the point of studying logarithms? The answer is that logarithms are still extremely useful in theory. For example, as we shall see later (in Chapter 15) the idea of logarithms gives a formula for approximating the number of prime numbers less than a given number n. (A prime number is a number like 5 or 7 or 17 which is not itself the product of smaller numbers.

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An example of a non-prime number is 6, the product of 2 and 3.) It is difficult to see what the logarithm could possibly have to do with prime numbers, and the result is indeed a surprise.

Other uses of logarithms are in chemistry to define acidity of a substance and in acoustics, to define decibels, the unit of sound intensity. We will not discuss these but point out that they are important applications of the logarithm.

§2 Doubling your moneyAs an application of logarithms, suppose you invest one pound at x% interest per annum. A good approximation to the number of years required to double your money at this rate of interest is 69/x. For example, if the rate of interest is 7%, it will take just under 10 years to double your money. This rough approximation is more accurate with small values of x, but is still useful. We will explain how this works in §6, problem 1.

Example.

If you receive interest at 3%, how long will it be before you double your money?

Solution: Divide 69 by 3 to get 23 years.

As this is a long time, maybe in this case one would want to know how long it takes for your money to increase by 40%. The rough rule in this case is to divide 34 by the interest rate. Thus with a 3% interest rate, to increase your money by 40% will take approximately 34 divided by the interest rate, i.e. 34 divided by 3 or between 11 and 12 years. See §6 problem 2 for an explanation of why this works.

§3 Euler’s eCommon logarithms are not the only ones in use. There are other logarithms, called natural logarithms, which have many uses, mainly scientific. In Chapter 15 we will use this logarithm to explain a formula giving an estimate for the number of prime numbers less than a particular number. To understand the difference between the natural logarithms and common logarithms, we change direction. We shall define a new constant e, as famous in mathematics as the constant .

To see how e comes about imagine that you invest a unit of money, say, a pound or a dollar, at a rate of x per cent per annum. That means that your return at the end of the year will be

(1 + x/100).

If we replace x/100 by y, the return can be expressed as (1 + y). Next, suppose that the rate is x percent per year, but paid every six months. That means that every pound or dollar invested will return (1 + y/2) pounds or dollars after the first six months, and this money will itself get an interest of x/2 percent for the next six months. Hence, the total you will get for one year is will be

(1+y/2)×(1+y/2) = (1+y/2)2..

Similarly, if the rate is x% payable every 3 months, that is, 4 times a year, the return after 1 year will be

(1 + y/4) 4.

In general, if interest is paid at the end of n equal periods per year, the total return for one year will be

(1 + y/n) n.

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The table below shows the values of (1 + y/n) n for y = 1 and various values of n.

Value of n Value of (1 + 1/n) n

10 2.593742

100 2.704814

1000 2.716924

10000 2.718146

1,000,000 2.71828

FIGURE 3. Approximating (1+1/n)n

The question is, what happens as n increases further? Here we come face to face with fundamental and subtle concepts, which are the subjects in more advanced courses in mathematics. The best we can do is to calculate (1 + 1/n) n, with n as large as we can manage. We then hope that this is close to the correct result and that nothing unexpected will happen with larger n. Sometimes this method works reasonably well, sometimes it goes disastrously wrong. To be sure of what we are talking about requires a course after the first course in calculus. Neither Newton nor Leibniz, the discoverers of the calculus, had a complete understanding of these concepts, and indeed it has taken some 300 years after them to develop the proper ideas.

In this case, however, large n produces no surprises, and we will get a number approximately 2.718. This is the number the Swiss mathematician, Leonard Euler (1707-1783), called e. It has retained this name to this day.

§4 The natural logarithm is roughly 2.3 times the ordinary logarithmThe logarithms discovered by Briggs are called common logarithms. They are also called logarithms to the base 10, they are defined in terms of exponents of 10 and satisfy

Log(10) = 1.

Natural logarithms, usually denoted by ln(x), are defined in terms of exponents of e and have the same property as common logarithms, in that multiplication can be replaced by addition. They are also called logarithms to the base e because they are based on exponents of e rather than of 10. In particular, ln(e) = 1, and, as we already have said, the natural logarithm retains the important property that

ln(xy) = ln(x) + ln(y).

For ordinary calculations logarithms base 10 are more convenient, but for theoretical questions logarithms base e have important advantages. One disadvantage is that a table is much more difficult to construct than for common logarithms. Fortunately, there is a conversion factor: To obtain ln(x), multiply Log(x) by ln(10), which is approximately 2.3026.

An interesting approximation for ln(1 + x) when x is small is x. For instance, ln(1 + .001) = .000999950033, which is very close to .001 The smaller x is the better the approximation.

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If x is a very large number with n digits, a good approximation to Log(x) is (n-1). For instance, Log(5,613,678) = 6.749, to 3 decimal places. The error in approximating this as one less than the number of digits, i.e. by 6, is less than 1, so this is accurate to 20%. If the number has 100 digits, the error is at most 1, and so the approximation is correct to 1%. In general, the larger the number, the better the accuracy of this rough approximation.

§5 Theory of logarithmsTheory of logarithms. To explain Briggs’ common logarithms we need the concept of exponents, introduced in Chapter 2. For any number, a, and each whole number n, an is defined as the product of n copies of a and n is called the exponent or power of a. For example, 23 = 2×2×2 = 8. We define a1 = a, and a 0 = 1.

There are two laws of exponents we need:

a m×a n = a m + n, and (a m)n = a mn.

For example, 102×103 = 105, that is, the product of two 10s times the product of three 10s is the product of five 10s. Also, (102)3 = 106.

In Briggs’ system the logarithm of a number x, written as Log(x), is defined as the exponent of 10 which gives x, that is, Log(x) = y if

10y = x.

For example, Log(100) = 2, since 102 = 10×10 =100, Log(10) =1, since 101=10.

To define logarithms we will need to define 10y for exponents y other than whole numbers.

Fractions as exponents. Let n be a positive integer and let x = 1/n. Then ax is defined to be the nth root of a. Thus 4½ is that number whose square is 4, and, hence, 4½ = 2. If x = m/n, i.e., one positive integer divided by another, we define ax to be the m-th power of a(1/n). For instance,

23/2 = (2½)3 = (1.4142)3 = 2.828.

So far we have defined ax for all exponents which can be written in the form m/n, where m and n are whole numbers. Such numbers are called rational numbers. But there are many numbers which are not rational numbers (for example, the square root of 2, as explained in Chapter 14, §5. Nevertheless it is possible to define ax for any positive number x, but a precise definition uses the concept of limit, which we do not discuss here. Instead we will accept that this can be done, and we will also assume that the closer a rational number y is to x, the better ar is an approximation to ax. For example, the square root of 2 is approximately 1.414, which, is 1414/1000, so a√2 is approximately the 1414-th power of the 1000-th root of a.

Negative exponents. If the above law of exponents is to hold for both positive and negative numbers, then, ax × a-x should be ax-x = a0 = 1. Therefore, we define a-x = 1/ax. For example,

2-3 = 1/23 = 1/8.

Definition of logarithms to the base a. If ax = y, then x is defined to be the logarithm of y to the base a, and is written as loga(y). In particular, common logarithms can be defined as logarithms to the base 10.

Definition of the natural logarithm (also called logarithm to the base e). If ex = y then x is the natural logarithm of y, written as x = ln(y).

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Laws of logarithms

1. Log(x) + Log(y) = Log(xy)

To see this let Log(x) = x´and Log(y) = y´, Then, xy = 10x´10y´= 10x´+y´ so, by definition,

x´+ y´= Log(xy), i.e., Log(x) + Log(y) = Log(xy),

2. xLog(y) = Log(yx).

This follows from raising both sides of the defining equation, y = 10Log(y), to the power x. Thus, yx = (10Log(y))x = 10xLog(y), i.e., xLog(y) is the exponent of 10 which gives yx, so, by definition, Log(yx) = xLog(y).

§6 Optional worked examples1. Why does the rough rule for doubling your money work?

Solution: If the interest rate is x%, the return on M invested for n years is M(1 + x/100)n. If this is to double, (1 + x/100)n = 2. Taking the natural logarithm on both sides of the equation, the left hand side equals ln(1 + x/100)n = n ln(1 + x/100), and, since x/100 is small,

ln(1 + x/100) is approximately, x/100 – see §4.

On the other hand, the right-hand side must be ln(2) = .6931471.

Hence, we have the equation nx/100 = .69 to solve. Multiplying the right side of the equation by 100 and dividing by x we see that n is approximately 69 divided by x.

2. Why does the rough rule for the money increasing by 40% work?

Solution. An increase of 40%, means a multiplication factor of 1.4 instead of 2 in the above example, and since ln(1.4) is approximately .34, for 40% the above solution becomes nx/100 = .34, so n = 34/x.

3. Why is ln(x), the natural logarithm of x, equal to ln(10) times Log(x)?

Solution: By definition, x = eln(x). Also, x = 10Log(x), and since 10 = eln(10), x = (eln(10))Log(x). Hence, x = eln(10)Log(x), so ln(x) = ln(10)Log(x).

4. Given that Log(2) is approximately 0.3, find Log(5).

Solution. Since 2×5 = 10, Log(2×5) = Log(10) = 1. Hence, Log(2) + Log(5) = 1, so Log(5) is approximately 0.7.

5. Use 210 to find an approximation to Log(2).

Solution. Since 210 = 1024, 210 is approximately 103. Taking the 10th root of both sides gives, 100.3 = 2, so Log(2) = 0.3, approximately. Actually, to 4 decimal places, Log(2) = 0.3010.^

§7 Logarittihms and Planeetary MotionKepler’s LawsAstronomers by the 1600s had observed the planets, established their distance to the sun, and their period, i.e. the time it took for the planet to return to its orginal position. These figures are summarized in the table bleow. The period is measured in days, and the distance is measured with the distance of the earth to the sun as unit. Thus from the table, the

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distance of Jupiter from the sun is a little more than 5 times the distance of the earth to the sun.

Plantet Mean distance D to sun Period T of rotation in daysMercury 0.387 0.241Venus 0.723 0.615Earth 1.000 1.000Mars 1.524 1.881Jupiter 5.203 11.862Saturn 9.555 29.458

The relationship between T and D is not easy to see, but by condiering ln(T) and ln(D, it seems as if ln(T) is one and a half times ln(D). We have listed ln(D and ln(T) in the table below. We have calculated a fourth column by multiplying ln(D) by 3/2. This last column agrees well with the third column ln(T), with an error of at most 1 in a 1000..

Plantet ln(D) ln(T) 3/2ln(D)Mercury −0.949 −1.423 −1.424Venus −0.324 −0.486 −0.486Earth 0.000 0.000 0.000Mars 0.421 0.632 0.632Jupiter 1.649 2.473 2.474Saturn 2.257 3.383 3.386

Thus we have Kepler’s law

3/2ln(D) = ln(T)

This is the third of Kepler’s famous laws. We can simplify the left-hand side to ln(D3/2) and so if we calculate e to this power we get

D3/2 = T..

This is indeed a remarkable formula.

§8 Earth quakes measured on the Richter scale

Earthquakes are usually measured on the Richter scale. The details are involved but we will give a simplified version. The severity of an earthquake 100 kn away can be calculated as follows.To set up the scale, Richter decided on a certain reading S on the seismograph whould be taken as a standard. Then if a seismograph 100

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km away recorded I on the seismograph, the Richter reading was defined to be

Log(I/S).

That is, we divide the intensity I by the standard S and then take the logarithm, So if an earthquake gave a reading 10 times as large as the standard S, its Richter number would 1, if it was 100 times larger, its reading would be 2 and if it was a 1000 times larger, its reading would be 3.

The seismograph of course has to be specified and there has to be a way for dealing with earthquakes which are at other distances than 100 km. The purpose of this example is to illustrate the use of the logarithm for this scale. There is also a logarithmic scale used for sound and also one for the apparent magnitude of stars.

A Richter value of 4 corresponds to light eathrquakes, usually without significant damage, but easily noticed shaking a nd ratling.

A Richter value of 8 corresponds to a severe earthquake causing serious damage over an area of several hundreds o kilometres.,

A Richter ,value of 9 would be devastating over an area of many thousand kilometeres. A reading of 10 has fortunately not yet been recorded. ,

§9 Some Historical remarksAround 1600, Lippershay invented the telescope in Holland. At first this invention was classified as a military secret, but when Galileo got wind of it he fashioned one on his own. His observations led to great strides in astronomy, which, in turn, led to efforts by astronomers to simplify the many precise numerical calculations involved. The main problems came from spherical trigonometry with the need to multiply numbers with a large number of digits precisely.

In 1524, Stifel described the basic principles of logarithms, but did not carry his ideas through to constructing a table. Almost a hundred years later the Scot, John Napier, after working twenty years, published the first table. It was he who called his exponents logarithms. His table was an immediate success and made an impact similar to that made by the introduction of computers in our time.

Napier’s original system was not based on powers of 10. This had the big disadvantage that his table of logarithms was very long. It had to include logarithms of all numbers, not just from 1 to 10. Briggs realized the convenience of using powers of 10, and in 1624 published his table, which greatly simplified the use of logarithms.

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The advantage of using base 10 is in the construction of tables. A table of common logarithms of numbers from 1 to 10 is readily extendable to any number. The rule is to express the number as a number, say x, between 1 and 10, times 10 to some exponent. The common logarithm is then the exponent of 10 plus Log(x). For example, 5,613,678 = 5.613678 × 106, so

Log(5,613,678) = 6 + Log(5.613678).

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Chapter 9

COORDINATE GEOMETRY

Algebra was one thing, and Geometry another. Descartes made them one.

FIGURE 1. René Descartes (1596-1650)

The idea of describing the position of a point in a plane by giving its distances from each of two lines that are perpendicular to one another is deceptively simple. But it leads to two important consequences: The ability to visualise how quantities depend on one another, and also a link between algebra and geometry. A very difficult problem in algebra may in its geometrical translation prove to be solvable, and, vice-versa, a difficult geometrical problem may turn out to be easier to handle in its algebraic translation. We owe this ingenious idea to the French mathematician and philosopher, René Descartes (1596-1650). His methods have been expanded and improved by many other mathematicians, to make the impressive subject we have today.

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§1 Coordinates

FIGURE 2. Coordinates

In Fig. 2 we have drawn two straight lines, OX and OY, which intersect in a point O, and are at right-angles. The position of any point can be described by a pair of numbers, the first gives the horizontal distance of the point P to the vertical line OY, while the second gives the vertical distance of P from the horizontal line OX. Distances above OX are taken to be positive, while distances below OX are taken to be negative. Distances to the right of OY are taken to be positive, while distances to the left of OY are taken to be negative. The position of a point is indicated by bracketing these two numbers together. And this bracketed pair of numbers are called the coordinates of the point.. For instance, in Fig. 3, the coordinates of the point P are (2,1), those of the point Q are (1,2), those of R are (-2,-1.5), and the coordinates of the point S are (-3, 0.75). We call the line OX the x-axis and OY the y-axis, while the point O is called the origin. Any collection of points on the plane is called a graph.

FIGURE 3. Examples of coordinates

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S(-3,0.75)

Q(1,2)

R(-2,-1.5)

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§2 Graphs as a concise source of informationThe first important application is the listing of information. Figure 4 illustrates a graph of temperature scale conversions. For example, to find the equivalent on the Fahrenheit scale of 20 degrees Celsius we carry out the following:

FIGURE 4. Graph of degrees Celsius vs. degrees Fahrenheit

We locate 20 on the OC axis and draw a vertical line till it reaches the curve, which in this case is a straight line. The corresponding F value is 68. So 68ºF corresponds to 20º C. By using the same procedure in reverse, we can convert from Fahrenheit to Centigrade.

Another example is illustrated in Fig. 5. This enables us to convert from pounds, £, to euros, €. Using this graph the equivalent of 5 euros in pounds is at the intersection of the vertical line drawn from 5 on the €-axis to meet the curve. The corresponding value on the £-axis is 3.50, which is the corresponding value in pounds.

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FIGURE 5. Graph of pound versus euro

A graph like this is very useful when travelling abroad. It is easy to make one that is accurate at the time of travel. Simply check the number of euros corresponding to £10, €14, for example, and plot the point (10,14). Draw the straight line joining this point and the point O. This then gives you the required conversions. Of course, this scheme can be used for converting from other currencies, using the relevant correct rates.

As a third example of the conveying of information, consider Fig. 6, the 2005 U.K. postal rates for a 1st class letter, in which we can relate the weight of a letter to the postage charge to send it.

FIGURE 6. Cost versus weight of the package

§3 Plotting a graphExample 1 The number of cells in time t.

A biologist checks under a microscope the number of cells that he is growing in a culture. At a time t = 0 there is only one cell. The table below summarises his findings.

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Time t 0 1 2 3

Number of cells 1 2 4 8

We plot the points with coordinates (t, number of cells at time t). For instance, at time t = 2 we have 4 cells so we plot the point with coordinates (2,4).

After plotting all these points, we then draw a smooth curve joining them as shown in Fig. 7.

FIGURE 7. Graph of number of cells at a time t

Example 2

This is an example where a quantity y depends on another quantity x and we have the following table of values.

x 0 1 2 3 4

y 1 3 4 4.5 3

We again plot the points (x, y) and then draw a smooth curve joining these points as in Fig. 8.

FIGURE 8. Graph of Example 2

One important advantage of this visual representation is that although the data was given for 5 points only, this rough graph allows us to approximate a value of y for intermediate values of x. For instance, a reasonable guess for y for x = 3.5 would seem to be 4.6.

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Example 3: Stretched rubber band.

An elastic band has various weights attached to it and its length is measured. The graph is then sketched as in Fig. 9. We get a straight line. But we know if the weight is too great the elastic band will snap. So it is not possible to assume that the graph will continue as it seems to do from the beginning.

FIGURE 9. Graph of a stretched rubber band’s length as the load is increased.

Example 4: Carbon dating.

The graph in Fig.10 gives a way of finding out how old objects are by carbon-14 content.

All plants, animals and people absorb carbon-12, normal carbon, and to a much less extent, carbon-14, which is radioactive. While alive, both carbons exist in the same ratios in plants, animals and people. After death the level of normal carbon remains constant, but carbon-14 decays. In fact, carbon-14 has a half life of 5,700 years, that is, one-half will decay in 5,700 years, half of the rest will decay in another 5,700 years, etc. The amount of normal carbon in the specimen determines how much carbon-14 there was originally, so the remaining carbon-14 determines its age. The following graph illustrates the relationship between the percentage of the carbon-14 of the original remaining to the age. For instance, if just 5 percent of the original carbon-14 remains the specimen is 24,640 years old,

FIGURE 10. Carbon-14 dating.

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§4 Equations and GeometryDescartes then had this brilliant idea: He set up a way of associating equations with curves.

Consider the equation y + x2 = 0. This is satisfied by a variety of values of x and y. Descartes idea was to consider all such values (x, y) that satisfied this equation and plot the corresponding points. For instance, (2,- 4) and (3,- 9). The collection of all such points form a curve as shown in Fig. 11. This also called the graph of the equation

xy

-4 -3 -2 -1 0 1 2 3 4

-15

-10

-5

0

FIGURE 11. Graph of y + x2 = 0.

In this way Descartes set up a relationship between equations and curves in the plane. Depending on the problem, it allows us to use geometrical techniques to assist in solving algebraic problems and algebraic techniques for solving geometrical problems. This study initiated by Descartes is known as analytic geometry or as coordinate geometry.

But first, let us consider some typical equations and the corresponding graphs.

Example

2x + 3y = 2. We consider the set of all points with coordinates (x, y), which satisfy this equation. To do this we draw up a table of values. We chose various values of x and then find the corresponding value of y to satisfy the equation. For instance, for x = 0, the equation2x + 3y = 2 simplifies to 3y = 2 and hence y = 2/3. We do the same for various values of x and in this way we obtain the table below:

x y0 2/31 02 -2/33 -4/3-1 4/3-2 2-3 8/3

Next, we plot these points and try to join these points with a smooth curve. The result is illustrated in Fig.12.

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FIGURE 12. The graph corresponding to 2x + 3y = 2.

In fact, it turns out that the graph of any equation of the form ax + by + c = 0, with a, b, c real numbers, is always a straight line (excluding the trivial case a = b = 0). Thus, the graph could have been more quickly drawn by joining any two points (x,y) which satisfy the equation.

The next example is the graph of x2 + y2 = 1. The corresponding curve is a circle, radius 1, centre at (0,0), as shown in Fig. 13. As in the above example, this can be verified directly by constructing a table of values and plotting the points.

FIGURE 13. The graph of the equation x2 + y2 = 1.

§6 The Greeks and their curvesTwo thousand years ago the Greeks studied the straight line, the circle and a group of curves, which are related to the circle, namely the conic sections. They studied these curves for their own sake and their work, certainly for about 2000 years, could only be described as entirely intellectual, without any application - useless knowledge. All this was to change. These curves became of fundamental importance in understanding our universe, and in Satellite TV.

What is a conic section? A conic section is the intersection of a double cone and a plane, see Fig. 14. When we talk about the cone we mean a hollow cone, something like a double dunce’s cap made out of paper. So the intersection with a plane produces a curve,

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FIGURE 14. The intersection of a plane and a conegives a curved line, which is called a conic section.

These curves were given the names parabola, ellipse, and hyperbola, depending on how they intersect the cone, but we shall not elaborate. We simply have drawn examples to illustrate the different types of curves in Fig. 14. In Fig.15 and Fig. 16 we have drawn the curves without showing how they arise from the intersection of a plane and the cone.

If you have a torch that has a well-formed cone of light coming from it, you can shine it in a darkened room on the ceiling, and by positioning it in different ways, you will get the circle, the ellipse and part of the hyperbola.

One type of these conic sections is easily drawn as follows. To draw a circle we can take a piece of string and tie its ends together, thus forming a loop. Put a nail in a board and holding a pencil in the loop at maximum extension, you can then draw a circle. If you take two nails, positioned at a and b, separated of course, and put the loop round the nails, again you can draw a closed figure, which is called an ellipse. Thus in this case there are two key points, the nails, and each of these is called a focus of the ellipse. (The plural of focus is foci).

FIGURE 15. Ellipse with foci at a and b.

A property of the ellipse is that if you shine a light from one focus to the circumference on the ellipse (for instance, if the circumference of the ellipse is in the form of a mirror), the light will be reflected in the other focus, as illustrated in Fig, 15. Here the straight line we have drawn from a is shown reflected in the circumference and then it passes through the other focus b:

An example of a parabola is the curve that a cannon ball traces out when shot out. See Fig. 16.

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FIGURE 16. A parabola, the path of a cannon ball.

§7 Equations and conic sectionsAs mentioned above, there is an association between equations and curves. Here is a short table with some examples.

Equation Curve

2x + 3y = 4 Straight line

x2 + y2 = 1 Circle with centre at O and radius 1

x2/4 + y2/9 = 1 Ellipse

xy = 1 Hyperbola

The table gives examples of what are general results. We state these results without proof.

The first is that any equation which involves only x and y and numbers, such as 2x + 3y = 4 (such equations are called linear) is a straight line.

Even more remarkable is the following theorem:

It concerns equations which involve only x, x2, y, y2 and xy and numbers, the so called quadratic equations, for instance, 7x + 4x2 +8y - 9y2 + 17 = 0. These differ from linear equations in that they must involve at lest one term which is a square or else the product of x and y.

Then the theorem states that the graph of such an equation will correspond to a conic section.

This is a very striking result. That an equation with x and y appearing only to the powers of 1 and 2 should have anything to do with taking a section of a cone is remarkable.

§8 Applications of coordinate geometryThe first major application of the conic sections occurred between 1609 and 1616 when the astronomer Kepler discovered three important rules of motion. The first was that the planets moved in ellipses with the sun at a focus. Moreover, the paths of comets like Halley’s comet also have the form of an ellipse.

Using his three rules of motion and his law of gravity, Newton was able in 1687 to show that Kepler's laws were a consequence. He used the connection between equations and curves that we have described above. The same method is used to determine the orbits of satellites. The reason why we can watch TV in so many different places is a consequence of these calculations.

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The reflecting property described in §6 is important for the antennas that are used to send and receive the TV signals. Newton had the idea of using the reflecting property of the parabola to design a telescope. The parabola has a focus, similar to the ellipse; but only one, not two. And the reflecting property means that parallel light will be reflected into the focus. The same property holds for radio waves. This is the basis of the reflecting antenna, the so-called parabolic antenna. The TV programs that are sent all round the earth use the same principle. A more down to earth application is in the reflectors for headlights in cars, in torches and in spotlights.

Another important application of this reflecting principle is medical. Lithotripsy eliminates the need for surgery to remove kidney stones. To pulverize the stones the lithotripter uses shock waves, which pass harmlessly through soft tissue. The patient is placed in an elliptical tank of water with the kidney stone at one focus. The shock waves are generated at the other focus. The procedure lasts about an hour during which time about 8,000 shock waves are administrated.

The property of reflection from one focus to the other also explains the workings of so-called whispering galleries, such as in St. Paul’s Cathedral and in the rotunda of the Capitol in Washington, where, if a person stands at one focus his voice is “reflected” to the other focus.

Finally to emphasise the importance of conic sections, even the paths of electrons rotating around the nucleus of an atom are ellipses.

§9 Solved Problems1. Draw the graph of y = x + 3.

Solution. The graph of any equation of the form y = ax + b is a straight line. Hence, the graph is determined by any two points on the graph. For example, the points (0,3) and (-3,0) are on the graph, so it we plot these points the graph is the straight line joining them

2. Draw the graph of y = 5x.

Solution. As in the previous example this is the graph of a straight line. Two points which satisfy y = 5x, are (0,0) and (1,5), so the graph is the line joining these points.

3. Draw the graph of y = 5x + 3.

Solution. A straight line joining, for example, the points (0,3) and (1,8).

§10 Solving problems in Geometry with algebra and vice - versaTo find a common solution to two equations, we draw the curves corresponding to each of the equations and observe intersections. In Figure 14 we have done this for the two equations,

y + x = 4 and y - x = 2.

From Fig. 17 we see that these two lines intersect in the point (1,3). Hence, x = 1, y = 3 is a common solution to the two equations.

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FIGURE 17. Solving two equations with a graph.

Algebraically, we can solve the two equations as follows: If we add the two equations, we have (y + x) + (y – x )= 2 + 4. i.e., 2y = 6 or y = 3. We then substitute y = 3 in the first equation to get 3 + x = 4, so x = 1, which agrees with our first solution.

A more difficult problem algebraically is to find a common solution of the two equations

x2 + y2 = 4 and y = x3 – x.

In Fig. 18, the graphs of the circle x2 + y2 = 4 and the curve y = x3-x have been plotted.

FIGURE 18. Common solutions to x2 + y2 = 4 and y = x3–x.

We can see from the graph that there are exactly two intersections, which are, approximately, (1.5,1.5) and (-1.5, -1.5).

The solutions suggest a bit more. Since for both points the x-coordinate and the y-coordinate are equal, we might investigate what happens if we set x = y in both equations.

If we set y = x in the first equation, the result is x2 + x2 = 4, that is, 2x2 = 4, and dividing both sides by 2 gives x2 = 2. Substituting y = x in the second equation gives x = x3 – x, or, 2x = x3, and dividing both sides by x gives 2 = x2, which agrees with what we found for the first equation. Thus, it is true that at the common solution, x = y, and x2 = 2. Therefore, more precisely, the common solutions are (+√2,+ √2) and (-√2, -√2 ). Note that √2 is approximately 1.414, not too far off our estimate from the graph.

Here we can see the use of geometry and algebra, each contributing to the solution.

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Chapter 10

SOLVING EQUATIONS AND GAUSS’S METHOD

Back substitution is the essence of this ingenious method.

§1 Solving an equationConsider the following problem: We are designing a budget hotel. An area, 25 meters long, is to contain 11 rooms; and each room is to be the same size, say, x-by-3.5 meters. The wall in each room is to be ¼ meter wide. What value of x will give the best possible size of each room? See Fig. 1.

FIGURE 1. Designing a hotel.

The number of walls is 10, and these add 10 ×¼ = 2.5 meters to the total. In addition there are eleven rooms each of width x, making a total of 11x. Thus the equation we need to solve for x is 11x + 2.5 = 25.

If you remember your school mathematics you will be able to solve this problem straight away. Otherwise, try guessing! Our mathematics teacher would have been horrified at this suggestion. “Don’t guess, boy,” he would say. We now know that he was wrong. Why not guess?

§2 Method 1: Take a guessAn easy first guess is: x = 1. When we substitute x = 1 in 11x + 2.5 = 25, the left hand side (abbreviated LHS) becomes 13.5, while the right hand side (RHS) is 25. So the first guess was a bit off, but it was not too bad. The next guess, so as to make the LHS bigger, is x = 2, which makes LHS = 24.5. This is still not quite correct since RHS = 25, but very good for a guess.

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The next guess, x = 2.5 makes the LHS = 30, which overshoots the mark of 25 by 5, and indicates that the actual answer is between 2 and 2.5, but very close to 2, since x =2 undershot the mark of 25 by only 0.5, while x = 2.5 overshot the mark by 5. We could continue in this way, trying say x = 2.1.

§3 Method 2: Draw a graphWe re-write the equation 11x + 2.5 = 25 by subtracting 25 from both sides of the equation, thus getting 11x –22.5 = 0. The problem can be formulated as finding the value of x to make y = 0 in the equation y = 11x - 22.5.

FIGURE 2. Solving the equation 11x + 2.5 = 25 by drawing the line y = 11x – 22.5

Fig. 2 is the graph of y = 11x - 22.5. The graphical solution to our problem can be obtained as follows:

First, we know from Chapter 9 that this graph is a straight line. The graph can be drawn by finding two distinct points on the line and joining them. For the first point we choose x = 1.5. Substituting x = 1.5 in the equation y = 11x – 22.5 we find y = 6. Thus the point (1.5, 6) lies on this straight line. For the second point we chose x = 2. Substituting x = 2 in the equation y = 11x – 22.5, shows that (2,0.5) is also a point on the line. We can then draw the graph by joining these two points with a straight line. (There is nothing special about the choices x = 1.5 and x = 2, any other choices would have done just as well.)

At the point where the straight line crosses the x-axis the value of y is 0, which means that the corresponding value of x is the value which solves the equation 0 = 11x – 22.5. From the graph we observe that y = 0 when x is approximately 2. This is a good approximation since when x = 2, y = 2222.5, that is, y = 0.5.

§4 Method 3: Do what your mathematics teacher told youThe method your mathematics teacher might have taught you for solving the above problem is based on the principle that equal mathematical operations done to both sides of an equality produce a new equality. Using this principle the solution to the problem proceeds in the following steps.

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Subtract 2.5 from both sides of 11x + 2.5 = 25 to get 11x = 25 2.5 = 22.5.

Finally, divide both sides by 11 to get x = 22.5/11, which is approximately 2.05. As a check we see that 11× 2.05 + 2.5 = 25.05, with an error of 0.05 in 25 or 0.22%.

§5 Two unknownsMathematicians are ambitious. Having solved a problem they immediately seek to generalise the problem. The problem we solved in the previous section involved one unknown, x. Why not solve problems with two unknowns? With two unknowns and one equation there can be an infinite number of solutions. However, this situation changes if we have to find a numbers xand y to satisfy two equations.

A typical example is to find values for x and y satisfying both equations

x + y = 2 (1)

2x+3y = 5 (2)

The problem here is to find values for x and y which will satisfy both equations.

FIGURE 3. C.F. Gauss (1777 – 1855), one of the greatest mathematicians.

The great German mathematician, C.F.Gauss, devised a systematic method for solving any number of equations in any number of unknowns. In this chapter we will apply it to two equations in two unknowns and to three equations in three unknowns in Section 7 Problem 5, but the method is easily adaptable to any number of equations and unknowns.

For two equations in two unknowns, x and y, Gauss’s method involves changing to an equivalent but simpler pair of equations having the same solution, the first equation of which still involves both unknowns but the second of which no longer involves one of the two unknowns.

The two operations we can use to do this are: (1) multiply any equation by a non-zero constant and (2) add or subtract one equation from another. Since both operations are reversible we can be sure that the method has neither eliminated a possible solution nor added

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a new solution to the system of equations. For our example, if equation (1) is multiplied by 2 this results in

2x + 2y =4

and subtracting this from equation (2), gives

2x – 2x + 3y – 2y = y = 5 – 4,

from which we see that y = 1 is the only possible value for y in the solution of the system. The value of x can now be found by substituting y = 1 for y in equation (1), so x + 1 = 2, or x = 1. Hence, the common solution is (1,1).

As with any method, it is advisable to check that x = 1, y = 1, solves both equations, which we do by substituting these values for x and y in the original equations and checking that the equations hold.

If we have three equations in three unknowns x, y and z, we can use the first equation to eliminate x from the second and third equations, giving a system in which the first equation is unchanged, but the second and third equations involve just the two unknowns y and z. We now use the above method to solve the system consisting of the new second and third equations for y and z. As above, the first of these two can be used to eliminate y from the third, leaving one equation in one unknown z. This can be easily solved for z, this value substituted back into the second equation to solve for y, and finally substitute both values obtained for y and z back in the first equation to solve for x.

That then is the general method. It means that the very last equation we obtain by this systematic elimination process has only one unknown. We then solve this equation, and with the value obtained for the last unknown, we substitute in the previous equations. This means that one of them has only one unknown. And so we can proceed. This method is called back-substitution, and the whole procedure is called Gaussian elimination.

Examples 5 and 6 in §7, below, give further examples of this method.

§6 Some general resultsThese general results will be useful in Chapter 12 on Geometry, so it is worthwhile noting them now. Observe that the way we derive them depemds only on the properties of the real numbers. This will be important in Chapter 12.

If we have two equations in two unknowns, then, the following possibilities arise:

1. There is no solution

As an example consider the system:,

2x + 3y = 1 (3)

4x + 6y = 6 (4)

We use our method of changing the equations by transforming the system into one in which x has been eliminated from one equation. In this case we can do this by subtracting twice the first equation from the second, which eliminates x from the second equation. Thus, our new system becomes

2x + 3y = 1 (3)

4x +6y – (4x + 6y) = 0x + 0y = 6-2 = 4 (4´)

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Our subtraction has successfully removed x, but at the same time removed y. The net result has been the equation 0 = 4, which is clearly impossible and means that our system can have no solution.

2. There is a unique solution

For example, consider the following system:

2x + 3y = 1  (5)

x + y = 1  (6)

If we subtract twice the second equation from the first, x will be eliminated from the first and our new system will be:

0x + y = 1 …(5)

x + y = 1 …..(6)

Solving the first equation for y we find y = 1.

We now use the method of back substitution putting in this value of y in equation (6) to get x – 1 = 1, so x = 2, y = 1 is the unique solution.

3. There are an infinite number of solutions

For instance

x y = 2. ….(7)

2x – 2y = 4 ….(8)

Again we change to a system with two unknowns in the first equation and one in the second. We do this by subtracting twice the first equation from the second. The result is the system

x – y = 2 …..(7)

0x + 0y = 0 …(8’)

In this case both x and y have disappeared in the second equation, but in contrast with the example in (1), the second equation (8) is true for all values of x and y. Thus the system has actually reduced to just the one equation, x – y = 2. In this case, we can choose any value, say, t for y, and then we find x = 2 + t is a solution to the system of equations, and this works for any value of t. Thus, we have an infinite number of solutions. What we have also proved is that the two equations are equivalent, so that any value of x and y which satisfies the one equation, will satisfy the other and vice-versa.

§7 Solved problems and other optional material1. Find a solution to the equation 2.5x - 6 = 5 by guessing.

Solution: We guess 10 because that way we get rid of the decimal. Then the left-hand side becomes 25 – 6 = 19, which is some way from 5. Guessing x = 4 gives a left-hand side which is 4. So perhaps x = 5, and this we see gives a left-hand side of 6.5. Next we try x = 4.5 and this gives a left-hand side of 5.25 which is quite close to the right-hand side.

2. Find a solution to the equation 2.5x - 6 = 5 by the school teacher’s method

Solution: Adding 6 to both sides gives 2.5x = 11. Divide by 2.5 to get x = 11/2.5 = 4.4.

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3. Find a solution to the equation 2.5x - 6 = 5 with a graph.

Solution. By subtracting 5 from both sides of the equation we get the equivalent equation of 2.5x 11 = 0. So we draw the graph of y = 2.5x – 11 and find out when it crosses the x-axis. We choose two points satisfying the equation and then we can draw the graph which we know must be a straight line. We choose convenient values for x, for instance x = 2 and x = 4. If x = 2 then y = 2.5x – 11 = 5 – 11 = 6 and so the point (2, 6) lies on the line. Similarly choosing x = 4 we see that the point (4, 1) lies on the line. Thus, we join these two points and get our graph . (Fig. 4)

FIGURE 4. C.F. Graph of y = 2.5x – 11.

We need to find where the curve touches the X-axis, and this is approximately at x = 4.4.

4. Why can we not divide by 0?

Solution: 6 divided by 3 is 2 is another way of saying that 23 = 6. In symbols a divides b means that there is a number c such that ac = b. Now 0 times any number is 0. For instance, 3 × 0 = 0 and 5 × 0 = 0. Thus 0 0 could be 3 or 5 or indeed any number. So dividing 0 by 0 we would not get a unique answer. If we try to divide a non-zero number by 0 we are in worse trouble. For instance, we cannot divide 1 by 0 since there is no number c such that 0 c = 1

5. Show that the following system of equations has no solution.

x + y + z = 4 (9)

2x + 3y + 2z = 5 (10)

x + 2y + z = 3 (11)

Solution: First we arrange that x is removed from equations (10) and (11) by subtracting twice equation (9) and then just equation (9) from (10) and (11) respectively, thus getting the three equations

x + y + z = 4 (9)

0x + y + 0z = 3 (10´)

0x + y + 0z = 1 (11´)

Next we subtract (10´) from (11´) to remove y from the last equation, getting the three equations

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x + y + z = 4 (9)

0x + y + 0z = 3 (10´´)

0x + 0y + 0z = 2 (11´´)

But this last equation is inconsistent, since the left-hand side is 0 and the right-hand side is 2. Thus these equations do not have a solution.

6. Solve the following system of equations for x, y and z:

x + y + 2z = 2 (12)

3x + 2y + 2z = 3 (13)

2x + 3y + 2z = 1 (14)

Solution: Subtract 3 times equation (12) from equation (13) to give the new equation (13’) and then subtract 2 times equation (12) from equation (14) to give the new equation (14’). This leaves equation (12) unchanged but replaces equations (13) and (14) by (13’) and (14’):

x + y + 2z = 2…… …..(12)

0×x - y - 4z = 3… …..(13´)

0×x + y - 2z = 3…… ..(14´)

Next solve the system consisting of the two equations (13’) and (14’), by adding equation (13´) to equation (14’) to get rid of the y. This replaces (14’) by the new (14”), which no longer involves y.

- y - 4z = - 3 (13´)

0×y - 6z = - 6 (14´´).

The solution to (14´) is z = 1. Substitute z = 1 in equation (13´) to get y = -1. Finally substitute z = 1 and y = -1 in equation (12) to get x = 1.

7. Why is –1 × 1 = +1? (Assume that 1 times any number leaves the number unchanged, that 0 times any number is 0 and that the distributive law a(b + c) = ab + ac holds for any choice of numbers a, b, and c, positive or negative.)

Solution.

(i) Since 1 times any number does not change the number, 1 × -1 = -1.

(ii) 1 + (- 1) = 0. Multiply both sides by –1. The right-hand side becomes 0. The left-hand side becomes using the distributive law –1 × 1 + (-1) × (-1) and this is equal to the right-hand side which is 0. Thus ,

–1 × 1 + (-1) × (-1) = 0,and since-1 × 1= -1, we see that -1 + (-1) × (-1) = 0 and so (-1) ×(-1) must be +1.

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Chapter 11

FUNCTIONS

The concept of function took centuries to evolve.Which is very surprising, for although it is very general,it is remarkably simple. It is fundamental to mathematics.

The concept of function is so central to mathematics, that we will devote this whole chapter to discussing it. Historically, it has not been easy to define this concept properly. Now that we have settled on a definition, it seems relatively straightforward, although abstract.

Knowledge of Chapter 9, Coordinate Geometry, is needed for this chapter. Indeed, we will use coordinates to define the concept of function. This is quite natural, as we shall see. We also need the concept of set.

From Chapter 9 we need to recall that we choose an origin O and draw two lines at right angles to one another, the x-axis OX and the y-axis OY. We specify the position of a point by means of its coordinates, i.e. the distances from the x-axis and the y-axis.

§1 What do we mean by a set?In the 1960s the New Mathematics was introduced into the schools. It caused much confusion since most parents did not know what it was about and so could not help their children. Also, it did not seem to be of much practical importance.

The New Mathematics was based on set theory. The idea of set theory is to consider collections of objects, for instance, the collection of all schoolteachers. Such a collection is called a set, and the objects in the set, in this case, schoolteachers, are called the elements of the set.

Almost any collection can be a set. For instance, the set consisting of the number 2 is a set consisting of a single element, 2. On the other hand, the set of all numbers is an example of an infinite set. The idea of set is indeed quite basic, and nobody has any difficulty with it. The difficulty comes with trying to express as much as possible of mathematics in terms of set theory.

Just as it is difficult for a foreign tourist to express himself in English with a limited vocabulary of only one thousand words, so is it a struggle to express complicated mathematical ideas in terms of set theory alone.

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§2 Functions defined geometricallyThe idea of a function is a crucial concept in mathematics, and so it is worthwhile defining it precisely. We begin with a geometrical definition and then provide an algebraic one. Obviously one can have only one definition, so we will have to decide on which one we will take as our primary definition. In fact, this will turn out to be the algebraic one, but the geometric one is a good start. The method was suggested by the Polish-Swedish mathematician, Richard Bonner.

We begin by choosing the usual coordinate system in the plane with x-axis OX and y-axis OY. We define a relation to be any set, R, of points in the plane. If these points have an extra property, namely, that any vertical line parallel to the y-axis OY meets R in at most one point, we call R a function. That is, if (x, y) is a point in the function R there cannot be a point (x, y’) in the function R with y’≠ y. The graphs in Fig. 1(a) and (b) illustrate sets of points which define functions, while in Fig.2 the set S consisting of the points on either the drawn circle or the drawn curve is not a function, since there are lines parallel to the y-axis which meet the points of the set in more than one point. For instance, the y-axis itself meets the set S in three points.

FIGURE 1.a. The points on the curve are a function.

FIGURE 1.b. The points on the curve are a function.

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FIGURE 2. The set consisting of the points on the circletogether with the points on the curve is not a function.

You may object that you have studied functions before, and this definition is not at all in accordance with the usual definition. That may be true, but be patient, and the connection will become apparent to you soon.

If f is a function, such as the one described by the points of the curve drawn in Fig. 1(a), we define f(x) to be the number y if the point (x, y) is an element in the relation, that is, (x, y) are the coordinates of the intersection of the line parallel to the y-axis which passes through x on the x-axis. According to the definition of function there is at most one such number y, and so f(x) is unique. However, it is also possible that there is no such point, in which case f(x) is not defined. The collection of all x for which f(x) is defined is called the domain of f.

§3 Awkward functionsThe advantage of the geometrical definition is that it is very easy to illustrate various kinds of functions. In Fig. 3 we have one awkward function. Its domain is the set of all real numbers between –3.5 and 7.2 with the exclusion of 0, because there is no point (0,y) in the function. We say, intuitively, that the function has a gap at x = 0.

FIGURE 3. A function with a gap.

The points of the function in Fig. 4 are (x,1), for all x less than or equal to 0, but for all x greater than 0 the points are (x,2). It does not have a gap at x = 0, but we say it has a jump at x = 0.

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FIGURE 4. A function with a jump.

In Fig. 5 we have a curve that abruptly changes direction at (0.2)

FIGURE 5. A function that changes direction abruptly.

Many of these functions, which we can draw so easily, are difficult to analyse algebraically. Curves with no gaps or jumps are said to be continuous. (This is not a proper definition but an intuitive one.) If, in addition, the curve does not change direction abruptly, it is called a smooth curve. (Again this is not an exact definition.) .

§4 Functions described algebraicallyHaving defined function geometrically, we will abandon this definition and instead give the algebraic definition. The geometrical definition of a function is as a collection or set of points with an extra condition. For the algebraic definition we take the coordinates of these points as our definition, so a relation R is now a set of coordinates. For instance, the set of all coordinates of the form (x, 2x) is a relation.

In the geometrical definition we had the additional requirement that every line parallel to the y-axis meets our function in at most one point. For our algebraic definition that is the same thing as saying that for any x, there is at most one y for which (x, y) belongs to the function.

We can now state our definition:

Definition. A function f is a set of coordinates (x, y), in which for any number x, if (x, y) and (x, y’) belong to f, then y = y’. Also, if (x, y) belongs to f, we write f(x) = y.

Note that the possibility that two or more different x’s can occur with the same y is not ruled out. But the essential requirement is that we cannot have the same element x occurring twice with two different second values y and y`. For example, the set {(1,2), (1,3)} is not a function.

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Example. The following are examples of functions.:

Function A: The set of all coordinates of the form (x, x+2) for all real numbers x.

Function B: The set of all coordinates (x, Log(x)) for all numbers x greater than 0.

Function C: The set of coordinates (x, 1) for all numbers x less than 5; (x, 3)for x greater than 5 but less than 6; and (x, 2) for all numbers x greater than 6.

Function D: The set {(1,1), (2,1)}

§5 Contrast with the definition in the calculus booksA function from a set A to a set B is defined to be a rule so that to each number in A is associated precisely one number in B.

This is a common definition. The disadvantages are that we do not know what a “rule” is, and what “associated to” means. Given something, which is supposed to be a function, we do not have a way of checking that it is. We would first have to check that we are given a “rule”. But what is a “rule”? What do we mean by “associated to”? Our algebraic definition is simpler. We check that we have a set, and we check that each element a occurs as the first entry in an element belonging to the set with only one b.

§6 What comes after 1, 2, 3? Could it be 34?In an IQ test if asked what comes after 1, 2, 3, one would be tempted to answer 4. But this is not the only logical answer. For instance, for the formula

f(x) = x + 5(x 1)(x 2)(x 3),

f(1) = 1 + 5×0×(1) ×(1) = 1, f(2) = 2+ 5×1×0×(1) = 2,f(3) = 3 + 5×2×1×0 = 3.

However, f(4) = 34. That is, for this function, the number that comes after 1, 2, 3, is 34.

Note that to check what happens when x takes on the values 1, 2 and 3 in the above formula, the calculations are easy to carry out, since each (x 1)(x 2)(x 3) is 0 for the values of 1, 2 and 3.

This approach was applied more generally by the French-Italian mathematician Lagrange. As an example of Lagrange’s method, suppose we have a function g(x) and we know that g(3) = 11, g(4) = 3 and g(6) = 10. What would be a reasonable estimate for g(5)?

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FIGURE 6. Lagrange (1736–1813).

Lagrange’s solution was to construct a formula of the form

f(x) = a(x 4)(x 6) + b(x 3)(x 6) + c(x 3)(x 4),

such that f(3) = 11, f(4) = 3, f(6) = 10. His solution was to make each term to consist of the product of two of the three factors, (x 3), (x 4) and (x 6). This clever trick makes it easy to determine a, b and c to produce the formula, since for each of 3, 4, 6 exactly one of the terms is not multiplied by zero.

To determine a we first set x = 3, which makes the second and third terms zero. What remains is a(3 4)(3 6) = 3a. Since we want f(x) to be 11 for x = 3, this means that 3a should be 11, that is, a = 11/3. Similarly, with x = 4, the first and third products in the formula are zero and what remains is the second term, b(4 3)(4 6) = -2b. Since f(x) = 3 when x = 4, this means -2b should be 3, so b = 3/2. Finally, with x = 6, the formula reduces to c(6 3)(6 4) = 6c. Since f(x) =10 when x = 6, f(x) = 6c = 10, so, c = 10/6 or 5/3. Hence, the desired formula is

f(x) = 11/3(x 4)(x 6) - 3/2(x 3)( x 6) + 5/3(x 3)(x 4).

Using Lagrange’s solution, an estimate for f(x) when x = 5 can now be obtained by substituting 5 for x in the formula. The answer is

f(5) = 11/3 + 3/2(2) + 5/3(2) = 11/3 + 3 + 10/3 = 8/3.

From this it is easy to see the pattern of the general solution, that is, a function f(x) which gives the values y1, y2,…, yn when x takes on the values x1, x2, …,xn,

§7 Solved Problems and other Optional items1. Find the domain of each function in each case given in the examples in §4.

Solution:

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For function A the domain is the set of all real numbers, since every real number appears as the first entry in the collection of coordinates.

For function B the domain is the set of all positive numbers.

For function C the domain is the set of all numbers except for 5 and 6.

For the function described in D, the domain is {1, 2}.

2. Show that the set {(1,1), (2,1), (3,1)} is a function.

Solution: This satisfies the first requirement of the definition of function in that we have a set of coordinates. The second requirement is that no first entry can occur with two different second entries. This is the case. Thus the given set is a function.

3. Show that the set {(1,1), (3,2), (3,1)} is not a function.

Solution: This is not a function because although we have a set of coordinates, which is the first requirement for a function, we have 3 appearing twice with two different second entries, namely 2 and 1, so the second requirement is violated.

4. Consider the functions A, B, C defined in §4. What are A(1), A(2), B(10), B(-10), C(4)?

Solution: A(1) = 1 + 2 = 3, A(2) = 2 + 2 = 4.

B(10) = Log(10) = 1.

On the other hand, B(10) is not defined.

C(4) is 1.

5. Let f be a function. What condition do you need for the set consisting of the reversals of f, that is, the set of all (y,x) for all (x,y) in f, to be a function?

Solution: If we take an example first it is easier to understand the argument. If f is the function consisting of the elements (1,1) and (2,1) then the reversals will be (1,1) and (1,2). This is not a function since 1 appears twice in the first position, once with 1 in the second position and once with 2 in the second position. This contradicts the second condition of a function.

The condition required to ensure that the reversals form a function is that no number appears in the second position in two or more distinct pairs.

6. Find a formula f such that f(10) = 1 and f(11) = 2 using Lagrange’s method as given in §6..

Solution: In this case we need just two terms, one with the factor (x – 10) and the other with the factor (x – 11), Thus, we put f(x) = a(x 10) + b(x 11), where a and b are two numbers to be determined. Since

f(10) = a(10 10)+b(10 11) = b,

to make f(10) = 1, we must have b = 1. Similarly,

f(11) = a(11 10) +b(11 11) = a,

so a must be 2. Hence, the formula is f(x) = 2(x 10) – (x 11 ) = x – 9.

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Chapter 12

GEOMETRY

It became apparent that Euclid’s remarkable geometry was not the only one.

FIGURE 7. Lobachevsky (1792 – 1856), oneof the founders of non-Euclidean Geometry.

In about 300 BC Euclid wrote The Elements, one of the most famous books ever written. Euclid collected and codified geometry and put forward a model for logical argument in mathematics. Euclid’s methods have essentially stood the test of time, and although there have been various improvements on some parts of his methods, his system is still very impressive.

Euclid’s system of geometry consists of the undefined objects called points and collections of points called lines. We say that a point P lies on a line if it is one of the line’s points. Euclid assumed certain extra statements, called axioms, without proof. He made various definitions, and thereafter always justified his arguments using only the undefined concepts and axioms that he had assumed. Euclid’s five axioms were essentially:

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1. There is one and only one line lying on two given distinct points.

2. Two lines are either identical, intersect in precisely one point or intersect in no point. (If two lines do not intersect in any point, we say they are parallel.)

3. Any two right angles are equal.

4. Given any point P and any positive real number R, there exists a circle with radius R and centre P. (A circle is defined as a collection of points each at the same distance R away from the centre.)

5. Given a line L and a point P outside L, there exists one and only one line passing through P and parallel to L.

(Strictly speaking this description of Euclid’s axioms is not correct, but it will be fine for our purposes.)

There are a few ideas implicit in Euclid’s axioms. In particular, there is the assumption that all is known about the real numbers. So we could legitimately say that Euclid assumed not only the five axioms above but also axioms for the real numbers.

It was long thought that Euclid’s axioms were the only possible ones because they seemed to mirror reality, but from the first there were questions about the fifth axiom, the so-called parallel axiom. In fact, Euclid himself seemed to have some doubts about it and avoided its use as much as possible in his proofs.

Various attempts were made to determine if it was a consequence of the other four. Finally, about 2100 years after Euclid’s death, Gauss, Bolyai and Lobachevsky established the validity and properties of other system of geometries resulting from various modifications of the parallel axiom, with Lobachevsky publishing first in 1829.

FIGURE 7. Euclid (325 BC – 265 BC).

§1 Reducing the assumptionsWe will present an argument for Euclidean geometry, which will enable us to dispense with the axioms 1,2 and 5 above. (It is also possible to dispense with Axioms 3 and 4; but as it is more technical we shall not discuss them.) All we assume is the real numbers.

What is the advantage of this?

In mathematics we are always concerned about whether our system is consistent - the fewer assumptions the better. If a contradiction appears, then at least we know that it depends on the axioms of the real numbers and not on the additional assumptions 1 to 5 above.

In any logical system, including mathematics, it is desirable to reduce the number of assumptions as far as possible. This principle is often described as Occam’s razor. The idea is that one cuts off successively each assumption which one can do without, deleting the unnecessary with the precision of cutting off something with a sharp razor.

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§2 Euclidean Geometry assuming only the real numbersWe have already seen in Chapter 9 that there is a close connection between geometry and algebra, which was pioneered by Descartes. We will go one step further.

We define the Euclidean Geometry to be the set of all pairs (x, y) with x and y real numbers. By real numbers we mean all numbers, including whole numbers, fractions, numbers with an infinite decimal expansion as well as a finite one, positive and negative numbers. The points of the plane are these pairs. Previously the points were dot-like objects we could see on a page. Now we abandon that idea, and simply define the points to be the pairs.

In Chapter 9, we noted that the lines were the set of points that satisfied equations like

2x + 3y = 4. More generally, for fixed numbers a, b, and c, the set of all points (x, y) that satisfy ax + by = c. An additional requirement was that a and b were not both 0. This is now our definition of a line: for each choice of a, b and c, the set of all (x, y) that satisfy

ax + by = c with not both of a and b zero.

Two lines are said to be parallel if they have no points in common. With this definition, the axioms 1, 2, and 5 above are consequences of results on linear equations that we established in Chapter 10. (These results were derived solely using the properties of numbers.) The intersection of two lines is the set of points that satisfy the equations of both lines. As we know from Chapter 10, there are three possibilities:

(1) There is an infinite number of points that satisfy both equations, in which case the two lines are exactly the same.

(2) The equations have only one point in common.

(3) The two lines have no point in common. Thus, the first axiom is a consequence of the solution of linear equations. The other axioms are also consequences of manipulating linear equations, as is indicated in §6 at the end of this chapter.

§3 A finite GeometryBy a finite geometry we mean a finite collection of objects called points and a collection of objects called lines that satisfy axioms 1, 2 and 5. (Axioms 3 and 4 fall by the wayside here.) A finite geometry with 6 lines and 4 points is sketched in Fig. 3.

FIGURE 8. A finite geometry.

There are precisely 4 points, A, B, C, D, and there are six ‘lines’:

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L1, consisting of the two points, A and C.

L2, consisting of the two points, B and D.

L3, consisting of the two points, C and D.

L4, consisting of the two points, A and B.

L5, consisting of the two points, B and C.

L6, consisting of the two points, A and D.

In Fig. 8 we have drawn the sides and diagonals of the square ABCD, to indicate the lines. But this representation is misleading, since each line consists only of the end points and not the other points we have drawn along the line joining them. Thus in particular, AC and BD are parallel, since they have no points in common.

It is now easy to check that with this definition of the geometry, any two points determine one and only one line, and that two lines either intersect in one point or else have no point in common, i.e. are parallel. Verifying the fifth axiom is again a matter of checking each case separately. For instance, D is a point outside the line BC and the line DA is the only line parallel to BC that passes through D. In Fig. 8 the two do not look parallel, but they are, since the definition of parallel is that there are no points in common.

§4 A finite projective geometryAxiom 5 given above is called the parallel axiom. It was felt to be different from the others, and there were many attempts to prove that it was a consequence of the other axioms. Finally it was realised that it was not, and an alternative geometry, called projective geometry, was constructed. The parallel axiom is spectacularly false in this geometry because in this geometry there are no parallel lines. That is, every two lines are either identical or else intersect in precisely one point. Specifically, we will describe a geometry of points and lines satisfying the following conditions.

1. Any two distinct points determine one and only one line.

2. Any two distinct lines meet in one and only one point.

As you can see, in this geometry there will be a symmetry between lines and points, in that we can interchange these two words in any true sentence and automatically obtain another true sentence.

The smallest example of a finite projective geometry with these conditions is shown in Fig, 4. This again is a geometry with only a finite number of points, 7 in this case namely A, B, C, …, G, although there are also examples of infinite projective geometries.

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FIGURE 9. A projective geometry.

There are also 7 “lines”, each of which consists of just 3 points. Six of these lines have been depicted as ordinary lines, but in actual fact each line consists of only 3 points. As with the above example, the connecting lines we have drawn are only a device for enabling us to recognise what the lines are. The seventh line also consists of 3 points, and to indicate which three points they are we have drawn a circle.

It is a laborious task but rather straightforward to check that all the axioms are satisfied. Note that any two lines meet in only one point, and that any two points determine a unique line. Note also that every pair of lines has a point in common. Thus, there are no parallel lines.

§5 Hyperbolic geometryIn the above section we have seen that we have a system of points and lines in which the laws of geometry hold with the exception of the parallel axiom. This means, in particular, that the parallel axiom is not a consequence of the other axioms. In this section we will see another example of a ‘non-euclidean’ geometry. However, in hyperbolic geometry the parallel axiom is replaced by:

Through any point outside a given line there is an infinite number of lines passing through that point which are parallel to the given line.

For this example we define our geometry to consist of all the points on a given circle, including the circumference.

The lines are now ordinary line segments in the circle including the end points on the circumference, but no points outside the circle, as in Fig. 5. Any two points can be joined by one and only one line. Any two lines either meet in precisely one point, or else do not meet at all, i.e. are parallel. However, the parallel axiom does not apply. Instead we see that given any point P outside a line L, there are an infinite number of lines passing through P and parallel to L. This is illustrated in Fig. 5. AB is a given line, and P is a point outside AB. We have sketched two of an infinite number of lines that are parallel to AB and pass through the point P.

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FIGURE 10. A hyperbolic geometry.

We may summarise in the following table.

GEOMETRY NO. OF PARALLEL LINES THROUGH A POINT OUTSIDE A GIVEN LINE

Euclidean One and only one

Hyperbolic An infinite number

Projective No two distinct lines are parallel

FIGURE 11. The parallel axiom and various geometries.

The hyperbolic geometry in this section obviously does not satisfy Euclid’s fourth axiom, namely that there is a circle of radius R with centre at any given point P. But such a hyperbolic geometry remarkably does exist, although we do not discuss it here.

§6 Sketch of the arguments for completing §2In this section we give a few extra details of what is required to complete the arguments of §2. We need to show for instance that for a point outside a given line there is a parallel line passing through that point. We take as example the line

2x + 5y = 2 (1)

and the point (1, 2). We put x = 1 and y = 2 in the left-hand side of equation (1) to get 12. We then consider the line whose equation is

2x + 5y = 12 (2)

Now the line with equation (2) has no points in common with the line of equation (1) and iso is parallel to it. Furthermore the point (1, 2) lies on the line of equation (2), and so we have found a line parallel to line (1) that passes through the point (1, 2).

The general case proceeds in the same way. Thus given the line with equation

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B

A

PP

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ax + by = c (3)

and the point (u, v) outside the given line. Then au +bv differs from c. So the line with equation

ax + by = au +bv

is a line which contains the point (u, v) but has no point in common with the line (3). So we have found a line parallel to (3) which contains the point (u, v).

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Chapter 13

PROBABILITY AND STATISTICS

“To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of.” Sir R A Fisher

§1 ProbabilityWhen we leave Pure Mathematics and go to real life, everything becomes less certain. The mathematical way of treating uncertainty is to use probability1. We assign a real number, p, between 0 and 1, called the probability, to each event. The closer p is to 1, the more we think the event is likely to occur, while the closer it is to 0, the less likely. A probability of 0.5 means the event is as likely to occur as not to occur. However, probability is not just a mathematical pastime, there are many important scientific applications, for instance, in medicine, genetics, as well as for the insurance industry.

If we consider tossing a coin, there are two possible outcomes to the experiment: either a head appears or a tail appears. If the coin is fair these events are equally likely, and the probability of a head appearing is 0.5. If we toss the coin a large number of times, we would expect the proportion of times the head appears to be roughly a half, and the larger the number of tosses, the closer the proportion of heads occurring should be to 0.5.

More generally, if an experiment is repeated a large number of times, say r times, and of these r times a particular event occurs s times, then the quotient s/r should approximate to p, the probability of the event occurring. The more frequently the experiment is repeated, i.e. the larger r is, the better the approximation.

A good example to have in mind is a pointer mounted on an axle as shown in Figure 1. The pointer is free to spin, and the circular face is divided into 4 equal sectors. The experiment is to spin the pointer, and the result is the sector where the pointer stops. (We assume that there is no problem when the pointer stops on a border.) Since there are 4 sectors there is a compelling reason to assume the probability of the pointer stopping on any particular letter is 1/4. If we spin the pointer 100 times and the pointer stops on b 80 times, we might suspect, for instance, that the face might be warped.

1 We present only a very restricted and simplified account of probability.

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FIGURE 1. A pointer spinning freely about O.

There are two crucial ideas we need. The first is that of independence. Two results of an experiment are independent if the occurrence of the one has no influence on the occurrence of the other. For example, the pointer stopping on a in one spin and the pointer stopping on b in the second spin. These ideas are illustrated by the examples below.

Two events are mutually exclusive, if the occurrence of either automatically excludes the occurrence of the other. For example, in the spinning of the wheel in Figure 1, the events of b appearing or of c appearing in one spin are mutually exclusive..

1. Mutually exclusive events

For example, in a throw of a die (the singular form of dice) the probability of a 3 appearing is 1/6, and the probability of a 4 appearing is also 1/6. In a single throw these two results are mutually exclusive, since if 3 appears then 4 cannot have appeared. Hence, in a game you are playing, if you win if either a 3 or a 4 comes up in a single throw, the probability of your winning is 1/6 + 1/6 =1/3.

To put it more generally, if A and B are mutually exclusive, the probability that either A or B occurs in a single trial is p + q where p is the probability of A occurring and q is the probability of B occurring.

That an event should occur and that an event should not occur are two mutually exclusive events. If the probability of the event occurring is p and the probability of it not occurring is q, then the probability of either occurring is p + q. But of course the event must either occur or not occur, so the probability of this is 1. Hence

p + q = 1, or q = 1 – p.

Example: We throw a die. The probability that a 3 will appear is 1/6, since the die has 6 sides and any one of them is just as likely to occur as any other. The probability that 3 will not appear is 1 1/6 = 5/6.

2. Independent events

For example, the probability that in two throws of a die a 3 appears in the first throw and 4 in the second is 1/6 times 1/6, or 1/36.

More generally, If A and B are independent, then the probability that in two trials A occurs in the first trial and B occurs in the second trial is p times q, where p is the probability of A occurring and q is the probability of B occurring. We can use these rules to solve a very early problem in gambling:

§2 De Méré ’s BetA famous 17th century problem in probability arose from gambling. De Méré wagered that in a throw of 4 dice at least one 6 would appear. To work out the probability we begin by working out the probability that no 6 will appear. The probability of this for a single throw of

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a die is 1 – 1/6 = 5/6. For four throws the probability of no 6 appearing is 5/6×5/6 ×5/6 × 5/6, that is, approximately 0.48. Hence, the probability of at least one 6 coming up becomes 1 0.48 = 0.52. Thus de Méré had a slight advantage with his bet.

The solution to this problem was given by two French mathematicians, Blaise Pascal (1623-1664) and Pierre de Fermat (1601 – 1665), and was an early contribution to the theory of probability. Sometimes the arguments on probability give unexpected results as the following examples show:

§3 The same birthdayIn the above problem we used the strategy that to solve a probability problem involving “at least one”, it is usually much easier to work out the probability for “not any”, which is the opposite of “at least one”. This trick is especially useful for the next problem, the probability that in a given group of people there are at least two people with the same birthday.

We begin with the numerically easier problem of the probability of finding two people whose birthday lies in the same month. Suppose we have a group of 5 people, say, A, B, C, D, E, in the same room. What is the probability that at least two were born in the same month?

The above strategy suggests that we should first work out the probability that there is no pair born in the same month. We start with the first person, A. The probability that B was not born in the same month as A will be 11/12. The probability that C was not born in either of the first two birthday months is 10/12. The probability that D was not born in any of the birthday months of the first three is 9/12. Similarly the probability for E not being born in the same month as A, B, C or D is 8/12. Thus the probability that no two of the five were born in the same month is

11/12 × 10/12 × 9/12 × 8 /12 = 0.38.

It follows that the probability that at least two were born in the same month will be 1 – 0.38 = 0.62. With this example in mind, we turn our attention to the birthday problem.

We can work out the probability that in a group of 30 people there are two with the same birthday in the same way, the only difference being that the calculations are longer. The probability that no two of them have the same birthday is 364/365 × 363/365 ×… × 336/365 = 0.293684, so the probability that least two have the same birthday is 1 0.293684 = 0.706316.

§4 Two boys?Here are three questions about a family with two children.

a) What is the probability that the family has two boys?

b) What is the probability that the family has two boys if at least one is known to be a boy?

c) What is the probability that the family has two boys if we know the first-born was a boy?

One would guess that the answers for b) and c) would be both ½, but this is not so.

Answers

a) We assume that the probability of having a boy is ½. Thus, the probability that both are boys is ½ × ½ = ¼, since it reasonable to assume that the two events are independent. We can also solve this problem by listing the possibilities. We indicate

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(boy. boy) briefly as bb, and (boy, girl) as bg etc. There are 4 possibilities: bb, bg, gb, and gg. Each of these is equally likely and (boy, boy) is one of the 4. Hence the probability is ¼, just as we calculated previously.

b) The possibilities are now gb, bb and bg since we know there is at least one boy, so the probability of bb is 1/3.

c) The possibilities are bg and bb since we know the first child is a boy. Thus we have a probability of ½ for bb.

The next calculation concerns the probability of three wise men coming to the correct conclusion.

§3 Three wise menSuppose we have three wise men, and that each has a probability of 2/3 of making a correct judgement. Suppose that we ask all three to judge a particular problem. What is the probability of the majority judgement being correct? We assume that the wise men act totally independently of one another.

Solution: There are 8 possibilities as follows:

All of the judges are right, which we indicate by the symbol (R,R,R) meaning that the first judge is right, the second is right and the third is right.

Then all but one of the judges is right, e.g. (R,R,W) i.e. the first judge is right, the second is right and the third is wrong. The other cases when only one judge is wrong are (R,W,R), (W,R,R).

The cases where two are wrong and one is right are (R,W,W), (W,R,W) and (W,W,R).

Finally the situation when all are wrong is denoted by (W,W,W).

We have a correct majority decision in the following cases:

(R,R,R), with a probability of 2/3 ×2/3 ×2/3 = 8/27

(R,R,W), with a probability of 2/3 ×2/3×1/3 = 4/27

(R,W,R), with a probability of 2/3 ×1/3 ×2/3 = 4/27

(W,R,R), with a probability of 1/3 ×2/3 ×2/3 = 4/27.

Therefore the probability of a correct majority decision is the sum of these probabilities, i.e. 20/27. This is better than the probability that one judge will get the right result, which is 2/3 = 18/27, which one would think would be the case, since a decision by a panel of three judges should be fairer than by one judge alone.

§5 Monte Carlo method of calculating approximatelyDraw a circle inside a square as in Fig. 2. The area of the circle is r2 whereas the area of the square is 4r2 where r is the radius of the circle. Thus the ratio of the area of the circle to the area of the square is /4. If you place dots at random in the square the ratio of the number of dots inside the circle divided by the total number of dots, call it x, is close to /4. To get an approximation for π simply multiply x by 4. .

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FIGURE 2. A circle of radius 1 inside a squarewhich can be used to give an estimate for p.

Of course, the difficulty in doing this is to ensure the dots are placed at random. One possibility for this is to position a pencil above the square, varying the position from the top side, middle side, right side, etc. Then with closed eyes and an unsteady hand guide the pencil down to the paper. We did this 200 times and obtained a result of 3.12 for . This is in error by less than 1%, not bad for such a simple method.

[To remember the value of to 7 decimal places, remember the sentence “How I need a drink, alcoholic of course, …” Then count the number of letters in each word: How (3), I (1), need (4), a (1), drink (5), alcoholic (9), of (2), course (6) to get a value of of 3.1415926.]

If one repeats the experiment and uses a good method for choosing points at random, a better value for can be obtained.

This is an example of a method used to calculate quantities.

These methods are called Monte Carlo methods. The Monte Carlo method for calculating to a large number of decimal places is not very practical, but for some problems Monte Carlo methods are very useful.

In Monte Carlo approximations, one should have a good method of choosing ‘at random’. One method for choosing random numbers was suggested by the mathematician von Neumann. For example, to get a series of 4 digit numbers, choose a 4-digit number, say 3678, and then square it to get 13527684. Take a four-digit number from the middle 5276 in this case and square it in turn to get 27836176. Take the four digit number 8361 from the middle and so on. In this way we get a series of 4 digit numbers which are approximately random.

These numbers are not entirely random choices. They are called pseudo-random numbers, but they have properties which make them suited to Monte Carlo methods.

§6 It is impossible to send a rocket to the moonIn the 1950s it was pointed out that it would not be possible to send a rocket to the moon. The argument was that a successful launch depended on a large number of items functioning perfectly. But every item had an admittedly small chance of failing. For instance, if the probability of success is 0.999 for each of 100 components every one of which must function correctly if the rocket will be a success, then the probability of success is the product of the 100 probabilities, i.e. 0.99100 = 0.905. Thus the probability of failure would be about 0.1. In actual practice, there were even more than a 100 essential components, which makes the probability of failure even greater.

And yet we do have successful rocket launches. The reason for this is that we have found methods of improving the reliability of components, having double safety where necessary. The calculations were right, and since they were taken into account, the necessary safety features were put in place.

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§7 LotteriesIn a national lottery your chances of winning are very small. Still people take a chance, since the cost is low and somebody must win, and it could be you. If you do win, your rewards are enormous. So many feel it is worth the chance. Strictly speaking though, you know that you will lose money in the long run.

The converse reasoning is used for events that have very serious consequences. Atomic power stations are according to the calculations very unlikely to explode. If they do though, the consequences are so terrible that many argue it is not worth the risk, small though it may be. [Note that in a lottery we are apt to treat a very low probability of winning as still giving us a chance. In atomic power calculations, we tend to interpret a low probability as meaning the event will never happen.]

The difficulty is the very nature of probability. If you conclude that your power station will explode once in a thousand years, what this means is that if you repeat the experiment again and again, in the long run the number of explosions will be one in a thousand years. There is therefore no contradiction to the fact that the explosion might occur today. On the other hand, we will only know if our calculations have been right in 100,000 years. Not very practical.

Another problem is calculating probabilities. In general, this is mainly a complex guessing method, obviously more sophisticated that the methods of Chapter 4, Fermi Calculations, but in theory much the same thing. There are assumptions, guesses based on similar situations, but which are not exactly the same. Also conditions change, the staff may be replaced over the years by worse staff, the money to run the plant may decrease and maintenance may not be up to the same standard.

§8 Nuclear power plants explodingIn March 1979 the power plant in Harrisburg Pennsylvania U.S.A. had a major accident. This led to the following from the Swedish humorist, Tage Danielsson:

“Those probability calculations are not really reliable. They are completely different before and after.

I mean for example, before Harrisburg … then it was extremely unlikely that that which happened at Harrisburg could happen, but as soon as it happened, then the probability shot up to not less than 100%, so that it became nearly absolutely certain that it had happened.

Yes, but only NEARLY certain. That is what is so peculiar. Because it is still as if they believe that that which happened in Harrisburg, it was so unbelievably unlikely, that in reality it probably has not happened.

So if that which happened at Harrisburg really happened, so presumably, now the risk that it should happen again is now so ridiculously low, that it is nearly certain that it can not happen again.”

Despite Tage Danielson’s comments, there is no contradiction to the calculations for Harrisburg being absolutely correct. Even if the time required for an accident is a 1000 years, this does not exclude an accident happening today. All that we can say is that if all these power plants are run for many hundreds of thousands of years, then on average the number of accidents will occur once in a thousand years. We would require millions of years to verify that this is correct Nor is there anything which guarantees that once an unlikely event has happened, that it will not be repeated tomorrow.

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Indeed, only seven years (not a thousand years as one had hoped) after the accident at Harrisburg, on April 25th -26th, 1986, reactor number four exploded at Chernobyl in the Ukraine. Many died either immediately or else very soon thereafter as a result of the intense radiation they received. Many more died later of cancer, and a large number of people are still suffering from the effects of the radiation, including also those who suffer from the remaining radioactivity and were not even alive when Chernobyl exploded.

In both Harrisburg and Chernobyl, there can be some suspicion that the calculations of probabilities were influenced by politics. The discrepancy between the calculations of how safe the plants were and reality has been explained by the authorities as due to the human factor. Safety precautions were disregarded and the reactors were being subject to conditions not previously envisaged. However this points to a weakness in the probability calculations, since they should have estimated for the human factor as well.

§9 SamplingThe idea of sampling is that we take a population that we want to investigate. The best would be to investigate every person. But this is too time consuming and expensive. So we investigate a proportion instead. We arrange the investigation so that everybody in the population has an equal chance of being selected, hoping in this way that the result will then be representative of the whole. (This simplified theoretical approach is in practice never used. Instead there are more sophisticated and practical methods which we will not discuss here.)

The sampling method for elections has been fairly successful. The number of people needed in the sample has been calculated to be at least 1,111. In fact 1,500 has been used in the Gallup polls to predict who will win the American election and with what proportion of the votes cast. This number is chosen so that the difference between the actual proportion of votes and the predicted proportion can be accurate to 0.03 with 95% confidence. The interesting point is that the number 1,111 does not depend on the size of the population being sampled, provided it is big. As always this is a probabilistic argument, which means that if this type of surveying is used many times, in the long run we will be inside the margin of 0.03 for 95% of our predictions. So there is always uncertainty in this process.

How well this has worked is shown by the figures below.

Year Winner Final prediction

Proportion

Result

Proportion

Prediction - result

Within range

0.031936 Roosevelt .557 .625 -.068 No

1940 Roosevelt .520 .550 -.030 Yes

1944 Roosevelt .515 .533 -.018 Yes

1948 Truman .445 .499 -.054 No

1952 Eisenhower .510 .554 -.044 No

1956 Eisenhower .595 .578 .017 Yes

1960 Kennedy .510 .501 .009 Yes

1964 Johnson .640 .613 .027 Yes

1968 Nixon .430 .435 -.005 Yes

1972 Nixon .620 .618 .002 Yes

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1976 Carter .480 .500 -.02 Yes

1980 Reagan .470 .508 -.038 No

1984 Reagan .590 .591 -.001 Yes

1988 Bush .560 .539 .021 Yes

1992 Clinton .490 .432 .058 No

1996 Clinton .520 .501 .019 Yes

Note that the results have been inside the correct limits of 0.03 in 11 predictions out of a total of 16 predictions, i.e. only in about 69% of the elections; not as good as the predicted 95%.

Remark: According to the theory, the sample needs to be at least 1,111 to achieve an accuracy of 0.03. If you want to halve that error, you will need 4 times the number in your sample, i.e. a sample of 4,444, to get an error within 0.015.

We do not wish to go into the theory, but point out the interesting fact that the concepts we have introduced before play an important part in the argument. In particular, graphs and even the numbers e and play significant roles in the theory of sampling.

§10 General remarks about samplingWorking with statistics is tricky and it is sensible to have a professional statistician in right from the beginning. It makes a great difference who is carrying out the survey.

One cannot have much confidence in a survey carried out by, say, a food manufacturer who finds that his product is liked by most of the people he asks. If you do not have an independent investigator you will not get a reliable result. You have to remember that when the result is of importance, there will be some people who are prepared to lie, and a larger number of people who are prepared to bend the truth. Even if the person in charge of the survey is independent, he may often want another job from the same organisation and will therefore be mindful not to offend.

One of the most important things to check is that the sample selected for the investigation is representative. In the case of the simple method we described in §9, this means that each member of the population has an equal chance of being selected. Theoretically it may seem easy but in practice it is extremely difficult. For instance, if you watch sampling on the street, the investigator is supposed to stop passers-by at random. It is not difficult to see that a young man who is undertaking the sampling seems to favour asking young and attractive women. Also, not everybody is prepared to stop and answer the questions. Some people simply rush off. There are of course careful and responsible investigators who take considerable pains to try to get a representative sample.

On the 11th September 2001 two hijacked planes flew into the twin towers of the World Trade Center in New York, destroying the buildings and killing some 3,000 people. The television that night showed a jubilant crowd filmed in one country exulting in the destruction. A friend then said that he thought the people in that country were dreadful. But he failed to check the main points of sampling: Did he know exactly why they were jubilant? Had they really considered the event carefully? And then, how many were in that cheering group? It was after all a sample, and so one would like the number to be reasonable in comparison with the size of the country. One would also want it to be a representative sample. But this seems most unlikely, given the status of independent reporting in many countries. In any event, the people who appeared would be those who were the most radical.

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Sampling based on questionnaires has further possible difficulties. There is a psychological aspect to choosing the questions to ask. For example, “Are you in favor of building the new town hall?” will get a different response to “Are you in favor of destroying the beautiful woods and building the town hall in its place?”

A final point

It is always more satisfying when we have other evidence rather than only statistical evidence. For instance, if a certain drug is shown statistically to help against cancer, it is encouraging if experiments with test tubes show that the drug reduces a cultured cancer. However, often we do not have such extra information. Many times then the statistical argument is the best we have, we have to make the best of an unsatisfactory situation.

§11 Our own beliefsOur own beliefs are based partly on what we have learned or been told and also on our own experiences. We generally feel that our own experiences are the best and most reliable, they are things we have seen for ourselves.

Yet these beliefs are also based on a sample, the people we have seen and the conclusions we have come from this sample. For instance, people are happy to give opinions about Jews, Muslims, Roman Catholics. When you ask, you realise their opinions are based on a very small sample of what they are trying to characterise.

Think about it. Even including all the people you know, how many do you know well enough to be able to answer questions of age, occupation, salary, political beliefs, religious beliefs, attitude to the aged and the young, education, taxation, state medicine, personal relations, etc? Is it as many as 200? However many people you know reasonably well it is but a fraction of the population. Generalising on such flimsy information is suspect. So do not be too dogmatic about things that you yourself know.

For instance, in estimating the number or murderers in a population it is unlikely that you have a sufficiently large base. Since we know no murderers, we would have to conclude that there are no murders at all. Then too you will find that often assertions made are often meaningless. Thus people will say that Jews are stingy, or Scotsmen are stingy.

The first thing is to know just what it is we are trying to decide. For a start, it is very difficult to know how we decide whether somebody is stingy. What does it mean? Does it mean that one never treats somebody to lunch? Does it mean that one does not give to charity? Does it mean that one is very stingy in dealing with one’s own family? How stingy one is can presumably depend on how rich one is. When it is a struggle to make ends meet it is possibly not sensible to be free with one’s money. It is almost impossible to say what stingy means. Many things one thinks one knows the meaning of become quite vague when one tries to track them down. The mathematicians’ insistence on good definitions becomes helpful.

[However, this insistence on definition is not appropriate in every situation of life. For instance, there is a moving and romantic song, which Nat King Cole used to sing. He sang “I love you, for sentimental reasons.” When you analyse this, you realise that it is nonsense. To love somebody is to have a powerful feeling for them. And then the reason given, for sentimental reasons: why that is simply saying that you have a strong feeling for somebody because you have as reason feelings. In other words, that is no reason at all. But do not tell that to your loved one while dancing in your arms.]

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There are other reasons to suspect our own generalizations. There is the problem of documentation. Normally one does not write down and record details with care, one simply has a vague feeling. One would not normally accept a survey that worked on this basis.

There are also things one just does not know about one’s friends. For instance, Dr. Kinsey’s report on sexual behavior gave unexpected results, because nobody had gone about asking their friends and acquaintances questions about their sex life.

Finally, we do not know whether the people we have seen in our lives are a representative sample.

§12 AverageWhen one has a large number of measurements, for instance, the I.Q. of a sample of adults, then by giving a single number, an average, we try to summarize all the measurements.

There are two very commn averages. The first is called the mean, and is the sum of all the measurements divided by the number of measurements.

Example: We will take a very small list of I.Q. readings, to illustrate the method.

Suppose the sample of IQ’s we are given is the five numbers 90, 92, 95,100, 115. The mean is obtained by summing these numbers and dividing by the number of readings. In this case the answer is 492/5 = 99.4.

Another average is the median. This is obtained by listing all the measurements in increasing order. If there are an odd number of measurements we take the one in the middle. That is half the measurements are smaller than the median, and half are larger. If there are an even numbr of measurementss, we take the mean of the two middle ones. Thus in both cases, the median is larger than half the measurements and smaller than the othr half. In the above example where there are an odd number of measurements, the median is the middle one, 95. So in this case, and indeed very often, the mean and the median are different.

Use of the mean in shopping: It is helpful to have a quick way of approximating your shopping bill at the supermarket in advance. This you can do by finding the mean of your shopping for a few shopping trips by simply dividing the total cost by the number of items. Then you can roughly work out the cost at subsequent shopping trips by multiplying the mean by the number of items. For instance, if the mean is £2, and you buy 20 items, you would expect to pay about £40. This rough check helps you decide whether the total bill was approximately correct, or, if you are short of cash, whether you have enough to pay at the check-out or had better buy less to avoid embarrassment.

§13 Solved Problems1. What is the probability that in a roll of three dice all three will turn up 6?

Solution. The three results are independent of each other. Therefore, the answer is the product of each individually, 1/6 × 1/6 × 1/6 = 1/216.

2. In a throw of a single die twice what is the probability that the number on the second throw will be less than that on the first throw?

Solution. The total number of possible results is 6 × 6 = 36. The number of desired outcomes is 5 if the first throw is a 6, 4 if the first throw is a 5, etc. Hence, the total number of desired outcomes is 5 + 4 + 3 + 2 + 1 = 15, so the answer is 15/36.

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3. There are five socks in a drawer, 3 white and 2 black. You pick 2 socks at random. What is the probability that both are white?

Solution: Let a, b, c be the three white socks, and x, y the two black ones. If you choose two, the possibilities are (a, b), (a, c) and also (b, c) for picking a pair of white socks. Other possibilities are (a, x), (b, x), (c, x), (a, y), (b, y), (c, y) and finally (x, y). Thus in all there are ten possible outcomes and of these 3 are a pair of white socks. So the probability of picking a pair of white socks is 0.3.

4. Inside a bag there is one disc. You know that the color of the disc is either white or black, but you do not know which. Put another disc, this time white, into the bag. You shake the bag, reach in and pull out a disc that turns out to be white. What is the probability that the remaining disc is white? (This is a problem due to Lewis Carol author of Alice in Wonderland.)

Solution. Since there are only two colors you might expect the probability to be ½, but this is not the case.

Call the original disc D. and the added disc A. There are three possible outcomes. If D is removed, A is left and a white disc remains in the bag. If A is removed then D is left. It can be white or black with equal probability.

Therefore, in all, there are two possibilities in which white is left and one possibility in which black is left, that is, a total of three equally likely outcomes. Thus, the probability of white being left in the bag is 2/3.

5. Investigate the following by noting the weather each day for a long period.

The weather prediction that the weather tomorrow is the same as it is today is correct in 70% of the time. Remarks: You will have to decide how long to keep on observing and recording, and what is meant by saying that the weather is the same as the day before.

6. The other box.

There is a contest in which there are three closed boxes. Inside one is a large sum of money. You are asked by the game-master to choose a box. He then opens one of the the boxes (not the one you have chosen) and reveals that there is no money in that box. He then asks you whether you want to stick to your original choice or choose the other box. Show that it is best to change to the other box.

Solution. If you stick to your original choice then the chances of having chosen the cor-rect box is 1/3. If you change to the other box and your orginal choice was correct, you will now score a miss. If it was incorrect, you will now score a win. In other words, changing boxes means that a win is converted to a loss and a loss is converted to a win. So the chances of scoring a win if you change boxes is the same as the chances of scoring a miss on your original choice. The chance of this is of course 2/3. So if you change your mind, your chances of a win are now 2/3. So it is better to change boxes

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Chapter 14

PYTHAGORAS’S THEOREM

Five centuries before the birth of Christ, Pythagoras and his mystic philosophic sect grappled with problems of pure mathematics

FIGURE 1. Portrait of Pythagoras (585 BC–500 BC).

One of the first theorems one meets in school is Pythagoras´s theorem. (We remind the reader that a theorem means an important result in mathematics.) This concerns triangles with one angle a right angle, i.e., 90 degrees, as at C in Fig. 2. It is attributed to the school established by Pythagoras. The theorem states that if the sides are x, y, and z, with z the largest side and the triangle is right-angled we have the famous formula x2 + y2 = z2.

FIGURE 2. A right-angled triangle with sides x, y and z.

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Example

Suppose we have a right-angled triangle with two sides of length 1. Then the length of the third side which we can call h is according to Pythagoras’s theorem, given by h2 = 12 + 12 = 2. Thus, h is 2.

§1 Proof of Pythagoras’s TheoremWhen you compare two shirts that are identical except for their sizes, they have the same shape, the only difference being that the one is larger than the other. This idea is even simpler for triangles, We say that two triangles as in Fig. 3, are similar if their angles are equal in pairs that is, A=A, B=B, C=C. The only difference in these triangles is in the scale, one is larger than the other.

The important property of similar triangles necessary for the proof of Pythagoras’s theorem is that given two similar triangles, the one is a scaled version of the other. That is, if the sides of the one triangle are of length x, y, z then the sides of the other are kx, ky, kz where k is a fixed number2. For example, in Fig. 3, the number k is approximately ¾, that is, A'B' = ¾ × AB, A'C' = ¾ × AC, B'C' = ¾ × BC. (Similar triangles were also discussed in Chapter 7)

FIGURE 3. These triangles are similar since their angles are equal in pairs.

We start our proof of Pythagoras’s theorem by considering the right-angled triangle ABC in Fig. 4. The longest side is always the side opposite the right angle and is called the hypotenuse. The length of AC is x, the length of BC is y and the length of the hypotenuse is z. The angle at A is a, the angle at B is b and the angle at C the right angle.

We begin by proving the theorem for triangles with hypotenuse having length 1. Later we will show how to extend this result to any right-angled triangle.

Case 1: z = 1

FIGURE 4. Beginning the proof of Pythagoras’s theorem bydrawing CD perpendicular to AB. Note that a = a´and b = b´.

2 It is possible to prove this property from more basic axioms, but we have chosen to assume it without proof.

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We start by drawing CD so that CD makes an angle of 90º with AB. We now have three triangles with right angles, namely, the original triangle ABC, the triangle ACD and the triangle BDC. It turns out that triangle ACD and triangle ABC are similar, that is, their angles are equal in pairs. Triangle ACD has one angle a, the angle at A, and shares this angle with triangle ABC. It also has an angle of 90 degrees at D. Since triangle ABC also has a right angle, we now have shown that these two triangles have two pairs of equal angles. Because the angles of a triangle add up to 180 degrees, the angle b must be b. Since the triangles ACD and ABC have the same angles, they are similar.

If the triangles ACD and ABC do not look similar, this is because the triangles are facing in different directions. In Fig. 5 we’ve repositioned the triangles, to help convince the reader that the triangles are indeed similar.

FIGURE 5. The triangles repositioned so that it clearer to see they are similar.

Similarly triangles ABC and DBC have the angle b in common and each has one angle a right angle and hence the angle a' must equal a, so the triangles are similar. Since the three triangles are similar, in each case the corresponding sides of the two smaller triangles are multiples of the sides of the original triangle ABC. In triangle ADC the side x is the hypotenuse, which must correspond to the hypotenuse of ABC. Since the hypotenuse of ABC has length 1 then the length of the AC, the hypotenuse of ADC, is x × 1, it follows that the other two sides of ADC are x times the corresponding sides of ABC. In particular, since AD corresponds to AC, AD is x times the length of AC, that is, the length of AD is x2.

If we make the similar comparison for ABC and BDC, we see that since the hypotenuse of BDC is y, the sides of BDC are y times the corresponding sides of ABC. In particular, the length of BD is y times y (the length of CB), that is, the length of BD is y2. Finally, since AD + DB = AB, x2 + y2 = 1 = 12. This shows that the theorem works for right-angled triangles with hypotenuse 1.

Case 2

Now we lift the restriction that the hypotenuse is 1 and consider a right-angled triangle with sides x. y and hypotenuse z, not necessarily 1. The triangle with sides 1/z times this triangle is similar to this triangle, so it is right-angled, also. The hypotenuse of this new triangle has length z/z = 1, and, as we have just shown, Pythagoras’ theorem is true in this triangle. Since the other two sides have length x/z and y/z, this means that

(x/z)2 + (y/z)2 = 1.

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Multiplying this equation by z2, we get x2 + y2 = z2, which proves Pythagoras’s theorem for an arbitrary right-angled triangle.

There are many other proofs of Pythagoras’s Theorem, for instance see §7 problem 11 below.

§2 The converse of Pythagoras’s theoremFirst, what do we mean by a converse? “If P then Q” is the general form of the statement “if P is true then so is Q”, the converse is: “If Q then P”. For example, if a man is a millionaire, he can afford to buy some socks. Here P is ‘a man is a millionaire’, and Q is ‘he can afford to buy some socks’. The converse would be: If he can afford to buy some socks, then he is a millionaire. We can see from this example that a converse need not be true, but for Pythagoras’ theorem the converse is true. That is, if the sides of a triangle satisfy the relation x2 + y2 = z2, the triangle is a right- angled triangle. The proof runs as follows.

FIGURE 6. Proof of the converse of Pythagorus´ Theorem.

Proof: Suppose we have a triangle ABC with sides x, y and z with the property that

x2 + y2 = z2. We next construct a triangle A´B´C´ as follows: side A´C´ is of length x, side B´C´ of length y, and angle C´ is a right angle. By Pythagoras’ theorem, the square of the side A´B´ must be x2 + y2, which, in our case, is z2, the square of the side AB. Hence, the third side of the right-angled triangle A´B´C´ has length z, which, by another of Euclid’s theorems tells us that the two triangles are identical. In particular, angle C is a right angle. This completes the proof of the converse of Pythagoras’ theorem.

§3 Pythagorean triplesWhole numbers, (x, y, z), which satisfy the equation x2 + y2 = z2 are called Pythagorean triples. Ancient people a thousand years before Pythagoras knew that (3,4,5) is such a triple. Then and even today, using ropes of length 3, 4 and 5 enables builders to ensure that walls are constructed at right angles. However, there are infinitely many Pythagorean triples. In fact, for any choice of m and n, Euclid showed that if

x = m2 n2, y = 2mn and z = m2 + n2 (1),

then (x, y, z) is a Pythagorean triple. Moreover, he showed that every Pythagorean triple can be obtained this way, that is, by a suitable choice of m and n. For example, with m = 2 and n = 1, x = 4 1, y = 4, z = 4 + 1, which gives the triple (3, 4, 5). With m = 3, n = 2, the triple is (5, 12, 13). The verification that the formula (1) always gives a Pythagorean triple will be demonstrated in Problem 6 in §7.

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BC

A´

BC´

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§4 Estimating the square root of a numberPythagoras’s Theorem is often used to calculate the length of a side of a right-angled triangle. For instance, if we have a right-angled triangle with unknown hypotenuse and sides of 3 and 1, then we know the hypotenuse must be of length the square root of 12 + 32, i.e. the square root of 10. (We remind the reader that the square root of a number x is a number whose square equals x. For example, the square root of 4 is 2 and the square root of 49 is 7.) Thus it is useful to have a way of calculating square roots. Often, we can only approximate to a square root, since it is often a number with a large number or even an infinite number of decimal places.

There is a useful and efficient method for approximating to square roots. The process is to start with a guess, call it g1, then by using a suitable formula improve g1 to obtain a better approximation, g2. The new approximation, g2, is further improved, and so on, until the desired degree of accuracy is obtained.

More precisely, for any number n, if g is any approximation to the square root of n, then

½(g + n/g)

will give a closer approximation for the square root of n.

For example, to approximate to the square root of 3, if our first guess is g1 = 1, our improved approximation for the square root of 3 will be g2 = ½(1 + 3/1) = 2. This approximation will be further improved to g3 by applying the formula to g2.

g3 = ½(g2 + 3/g2) = ½(2 + 3/2) = 7/4 = 1.75.

The square of 1.75 is 3.0625, which is pretty good after only three approximations. The next approximation is g4 = ½(1.75 + 3/1.75) = 1.73214, which is even better since (1.73214)2 = 3.0003. The next approximation is

g5 = ½(g4 + 3/g4) = ½(1.73214 + 3/1.73214) = 1.7320508

and (1.7320508)2 = 2.9999999.

Example: Estimate the square root of 10.

As a first guess we take g1 = 3. (We could have chosen g1 = 1, but the closer our first guess is, the more time saved.) The next approximation, g2, is given by g2 = ½(3 + 10/3) = 3.17. The next approximation, g3, comes from g3 = ½(3.17 + 10/3.17) = 3.1623. The next approximation will be g4 = 3.1622776.

As a final check, (g1)2 = 9, (g2)2 = 10.05, (g3)2 = 10.0001, (g4)2 = 9.9999996, which illustrates how efficient the method is.

§5 The square root of 2 is irrationalThe ancient Greeks, and, in particular, the Pythagoreans, thought of numbers as being either whole numbers, like 476, or fractions, that is one whole number divided by another, such as 5/8 or 2/3. But they also thought of numbers as corresponding to the lengths of lines. To their disgust, the use of Pythagoras’s theorem introduced lines whose lengths could not possibly be whole numbers or fractions. For instance, as we showed at the beginning of this chapter, the hypotenuse of a right-angled triangle with the two smaller sides both 1, is 2. The following proof given by Euclid shows that the square root of 2 cannot be a fraction.

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Proof

Suppose that the square root of 2 could be the fraction p/q, say, where we may further assume we have chosen q as small as possible. Since p2/q2 = 2, p2 = 2q2. This means that p2 is even. Since the square of an odd number is always odd, and p2 is an even number, p must be even, that is, p is a multiple of 2, for instance, p = 2r. Replacing p by 2r in p2 = 2q2, gives 4r2 = 2q2. Dividing both sides by 2 we see that 2r2 = q2. This means that q2 is also an even number, and since the squares of odd numbers are odd, q must be even, say q = 2s, and p/q = 2r/2s. Cancelling 2 from 2r and 2s, we see that p/q = r/s, so the square root of 2 is r/s. However, s is the half of q, this contradicts the choice of q as being the least possible. Hence our original assumption that the square root of 2 can be expressible as one whole number divided by another is false.

We say that a fraction or a whole number is a rational number, while a number which is neither a fraction nor a whole number is called an irrational number. We have therefore shown that the square root of 2 is an irrational number. So Pythagoras’s theorem meant that the Greeks had to accept that there were numbers that could not be expressed as fractions of whole numbers.

The fact that irrational numbers could be handled like rational numbers, that is, they could be combined like rational numbers and even used in calculations with rationals, was discovered by Hippasus of Metapontum. The story is that when he announced his discovery to a group of Pythagoreans while on a boat, they threw him overboard

§6 Distance to the HorizonBecause of the curvature of the earth, we can only see a limited distance ahead. Imagine you were standing at a height of h above the ocean and looking out to the horizon. Using Pythagoras’s theorem the distance to the horizon can be calculated from the radius R of the earth. We are making the assumption that the earth is a perfect sphere, which is not true but good enough for our calculation.

In Fig. 7 the earth is represented by a circle with its centre O, and the observer is at the point A, a distance of h above the ground. The line AB is the greatest distance that can be seen. It touches the circle at the point B.

By a well-known theorem of geometry a line which touches the circle at one point makes an angle of 90 degrees with the radius drawn to the point of contact (in this case B) from the centre of the circle (O in this case).

FIGURE 7. Using Pythagoras’s theorem to calculate the distance to thehorizon. R is the radius of the earth, h the height above sea.-level.

Since the angle ABO is 90 degrees, we can apply Pythagoras’ theorem to the triangle ABO:

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(OA)2 = (AB)2 + (OB)2, that is,

(R + h)2 = s2 + R2

where s = AB is the distance to the horizon. By an algebraic identity (derived in §7 Problem 8 below) the left-hand side equals R2 + 2Rh+ h2, so the equation can be rewritten as

R2 + 2Rh+ h2 = s2 + R2.

After subtracting R2 from both sides, the equation simplifies to

2Rh + h2 = s2

Since R is over 6300 km and h is relatively small, as a very good first approximation we can simplify this equation to

2Rh = s2 .

This means that

s = (2Rh).

We know that the value of R is 6371 km, so 2R is 12742.

For example, standing on a beach with your eyes 2 meters above sea level, h = 0.002 km, so the distance to the horizon is (12742 × 0.002) = 5.05 km.

If we stand on a mountain 1 km above sea-level, then the distance to the horizon is 112.88 km, which is the square root of 12742.

As another example, if we stand on one of the world’s highest mountains at 8 km above sea-level, then the distance to the horizon is 319 km.

Since radio waves have similar properties to light waves, what happens to them? They must also be affected by the curvature of the earth and so even if they are transmitted from a mountain 8 km high, they can only travel 319 km. And yet, Marconi in 1901 sent a signal across the Atlantic, from Poldhu in Cornwall to St. John’s in Newfoundland, a distance of approximately 3,000 km.

Fortunately for Marconi, the radio waves were reflected in a layer of electrically charged particles (later called the Heaviside layer), above the earth, and in this way short wave radio signals can be sent long distances. Since this discovery was not made until after Marconi’s experiment, he could not have known of this phenomenon, which means that he was extremely lucky. Nowadays radio, telephone and television often use satellites to bounce the signals long distances, instead of the more erratic Heaviside layer. .

§7 Solved Problems1. A right-angled triangle has its two smaller sides 2 and 3. Calculate the length of the hypotenuse.

Solution: We use Pythagoras’s theorem to find the square of the hypotenuse is

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22 + 32 = 4 + 9 = 13.

Thus the hypotenuse is 13.

2. Find the altitude (i.e. vertical height) of a triangle with all three sides equal to 1, as in Fig. 7.

FIGURE 8. Finding the altitude of a triangle with three sides equal to 1.

Solution: In Fig. 8, the three sides AB, BC and AC are all 1. The altitude, BD, bisects AC so each half is 1/2. Since the triangle BCD is a right-angled triangle, the sides satisfy the equation

BD2 + 1/4 = 1.

Hence, BD2 = 3/4, and BD = 0.866.

3. Is a triangle with sides 10, 15, 19 a right-angled triangle?

Solution. The squares of the sides are 100, 225, 381. Since 381 is not equal to 225 + 100 = 325, the triangle is not right-angled.

4. Find the square root of 12.

Solution. Take 3 as a first approximation. The next approximation is g2 = 1/2(3 + 12/3) = 3.5. The next approximation is g3 = 1/2(3.5 + 12/3.5) = 3.464; the next is g3 = 1/2(3.464 + 12/3.464) = 3.46410. The next is g4 = 1/2(3.4641 + 12/3.4641) = 3.46410. As a final check, (3.46410)2 = 11.999988

5. For any two numbers a, b, show that (a + b)2 = a2 + 2ab + b2, and (a - b)2 = a2 - 2ab + b2.

Solution. For any three numbers x, y, z,

x(y + z) = xy + xz.

If we apply this to

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

with x = a, y = b and z = (a + b), we have

(a + b)2 = (a + b) ×(a + b) = a(a + b) + b(a + b) =

a2 + ab + ba + b2 =a2 + 2ab + b2.

For (a b)2 we use the fact that the product of two negatives is a positive and the product of a negative and a positive is a negative. This gives

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

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6. For any two whole numbers m and n, show that x = m2 n2, y = 2mn, z = m2+n2 is a Pythagorean triple. That is, verify that equation (1) in §3 holds. (Use problem 5 above.)

Solution. With a = m2 and b = n2, applying the result from Problem 5 gives:

x2 = (m2 n2)2 = m4 2m2n2 + n4,y2 = (2mn)2. = 4m2n2.

We use Problem 5 to calculate

z2 = (m2 + n2)2 = m4 + 2m2n2 + n4,

so that x2 + y2 = z2.

7. Use equation (1) in §3 to find two Pythagorean triples both of which include the number 24.

Solution. With y = 2mn, x = m2 - n2, and z = m2 + n2, then (x, ,y, z) is a Pythagorean triple. For the problem we start by choosing numbers m and n such that 2mn = 24, that is, mn = 12. So, with m = 4 and n = 3, y = 24. This gives the triple, x =m2 n2 = 16 9 = 7, z = m2 + n2 = 16 + 9 = 25, so one triple is (24, 7, 25).

A second one comes from another choice of m, n with 2mn = 24. For instance, withm = 12, n = 1, y = 2mn = 24, x = 143, z = 145, and the triple is (143, 24, 145).

8. Show that (R + h)2 = R2 + 2Rh+ h2 with the aid of problem 5 above. (This identity is needed in §6 above.)

Solution. We substitute R for a and h for b in Problem 5 and the result then follows.

9. Give a geometrical interpretation of Pythagoras’ theorem.

Solution: Since x2 is the area of a square constructed on the side of length x, y2 is the area of the square constructed on the side of length y, and z2 is the area of the square constructed on the side of length z, we have the relationship between the areas of these squares, namely the sum of the areas of the two smaller squares is equal to the area of the larger square. See Figure 9.

FIGURE 9. Pythagoras’s Theorem in terms of squares.

This is indeed the form in which the theorem was originally discussed.

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10. Prove Pythagoras’s theorem using the fact (see problem 11 below) that the areas of similar right-angled triangles are proportional to the squares of the hypotenuses, i.e. there exists a constant k such that the area = k(hypotenuse)2 for any two similar right-angled triangles, where k depends on the triangles.

Solution: Let ABC be a right-angled triangle with right angle at C and drop a perpendicular CD to AB. Then exactly as we argued in §1 above, triangles ABC, CAD and CBD are similar. (The equal angles are indicated in Figure 10.) So

area of ABC = kz2, area of ACD = kx2 and area of CBD = ky².

So kz2 = area of ABC = area of ACD + area of CBD = kx2 + ky².

Thus kz2 = kx2 + ky2 and dividing by k we get Pythagoras’s theorem,

z2 = x2 + y².

Figure 10 Alternative proof of Pythagoras’s Theorem

11. Prove using the ideas in §3 Chapter 7 that the area of similar right-angled triangles is proportional to the square of the hypotenuse,

Solution: From Chapter 7 §3 we have the following diagram.

Fjigure 11 Proving the area is proportional to the square of thee hypotenuse

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So if we have a similar triangle with hypotenuse of length h, the other two sides have length hcos(A) and hsin(A).

The area is half the base times the height, and so the area is

½hcos(A) hsin(A) = ½ cos(A) sin(A)h2. Put ½ cos(A) sin(A) = k and we have the required relationship for any right-angled triangle similar to .triangle ABC. This concludes the proof.

12. One reason why one cannot see far on the Earth is as we have seen in is that light travels in straight lines. Another is that the eye needs an object sufficiently large to see.

At 6 meters one should be able to read a letter of size 88 mm, as one can find out by checking an eye chart. How high would an object need to be if one is to read it from 10 km?

Solution: At one meter one can read a letter of size 88/6 mm. So at 10 km. The letter would be

10100088/6 mm = 880/6 m = 146.7 m approximately. For a rough approximation, if we say we need 4 m per floor of a skyscraper, this is a skyscraper of about 36 floors.

§8 Fermat’s Last TheoremThe existence of Pythagorean triples led people to investigate the possibility of finding non-zero whole numbers x, y and z such that

xn + yn = z2

for whole numbers n larger than 2.

FIGURE 12. Andrew Wiles, 1953 – , who finally proved Fermat’s Last Theorem.

Pierre de Fermat (1601-1665) believed he had a proof that no such whole numbers could exist for any n other than 2. Fermat had a habit of using any book at hand to jot down a proof. Shortly before his death he wrote in the margin of a book that he had a proof of this, but that there was not enough room in the margin to show the details. This is all that is known about his “proof”. Since that time the statement that there is no triple of whole numbers for n greater than 2 has been known as “Fermat’s Last Theorem”. In the intervening 300 years many specific examples of values for n have been found for which no triples exist and very many prominent mathematicians have tried to get a proof, without success. Finally, in 1995 Andrew Wiles,, after a labour of many years, showed that Fermat’s Last Theorem is, in fact, true. That is, there is no whole number n larger than 2 for which the equation xn + yn = zn holds for whole numbers x , y and z.

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Chapter 15

MODULAR ARITHMETIC.

PRIME NUMBERS AND LOGARITHMS

Look at the clock, close your eyes and fall asleep 12 or 24 hours.You will not notice a difference when you look at the clock again.

§1 Modular arithmeticIf you divide any whole number by 2 there are two possible remainders, 0 or 1. In general if you divide a whole number by a given whole number n there are n possible remainders, 0, 1, …, n - 1. For instance, for n = 3, the remainders can be 0, 1, 2: Thus we get the remainder 0 on dividing 6 by 3, 1 on dividing 7 by 3, 2 on dividing 8 by 3, and so on. C.F. Gauss observed that for any whole number n the remainders could be used to form a finite system in which addition and multiplication could be easily defined. For n = 3, the addition and multiplication tables are given in Fig. 1.

+ 0 1 2

0 0 1 2

1 1 2 0

2 2 0 1

X 0 1 2

0 0 0 0

1 0 1 2

2 0 2 1

FIGURE 1. Addition and multiplication tables modulo 3.

The entries in Fig. 1 are obtained by ordinary addition and multiplication, but the answers have been replaced by remainders after dividing by 3. For example, 2 times 2 is 4, but the

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entry in the table is 1, the remainder on dividing 4 by 3. Likewise, 1 plus 2 is 3, but the entry in the table is 0, since there is remainder 0 after dividing 3 by 3.

Similar tables work for remainders for any n. For example, for n = 2, the tables are

X 0 1

0 0 0

1 0 1

+ 0 1

0 0 1

1 1 0

The notation Gauss used, and which is still used today, was to call the system arithmetic modulo n. To say for instance that 17 divided by 3 has a quotient of 5 with a remainder of 2 is the equivalent of the equation

17 = 5×3 + 2.

In other words, division and remainders are defined in terms of multiplication and addition. In general, if a divided by n has a remainder of r and a quotient of q, then

a = qn + r.

We also require that r should be less than n. For instance, if a = 149 and n = 6, then q = 24 and r = 5. The equation becomes

149 = 24×6 + 5.

Most of the usual rules of arithmetic remain valid for the system of remainders for any n. In “normal” arithmetic though, the product of two non-zero numbers is not zero. In modular arithmetic this can fail. For example, in arithmetic modulo 6, 2 × 3= 0. (This is because 2 × 3 = 6 which has remainder 0 when divided by 6).

Another difference is in establishing an order, i.e., less than or greater than. For ordinary numbers if a < b, then a + c < b + c. For modulo 2, for instance, if we try to define 0 < 1, then we should find 0 + 1 < 1 + 1 = 0 so 1 would be less than 0, a contradiction. A similar contradiction comes about if we try to define 0 > 1. That means that it is not possible to define an order on the integers modulo 2.

§2 Congruence modulo nIf the clock says 7 and we look at it 12 hours later it again says 7. We say that 7 + 12 is congruent to 7. We can generalise this idea as follows:

two whole numbers a and b are said to be congruent modulo n if their difference a – b is divisible by n. In symbols we write a ≡ b (mod n). For example, 4 ≡ 1(mod 3) and 8 ≡ 3(mod 5).

The symbol behaves as one would expect. For example, if a ≡ b (mod n), then a + c ≡ b + c (mod n) and a × c ≡ b × c (mod n). Also, if a b and c d (mod n), then a + c b + d (mod

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n) and ac bd (mod n). As you can see, in many respects ‘’ acts like ‘=’. (Proofs are given in Problem 1 of §11.)

The idea of congruence can be used to make calculations involving days of the week. We can number the days of the week, with Sunday 0, Monday 1, Tuesday 2, Wednesday 3, Thursday 4, Friday 5 and Saturday 6. Adding 7 days leaves the day number unchanged. Thus 7 days after Friday, which is day 5, we get back to Friday, again day 5. In the symbol for congruence, this is 5 + 7 5 (modulo 7).

Problem 1: If today is a Tuesday, what day will it be 139 days from now?

Solution: Tuesday is day 2, and 2 + 139 = 141. Since 141 divided by 7 leaves the remainder 1, 141 ≡ 1 (mod 7). Hence, 139 days after day 2, is day 1, i.e. Monday.

Problem 2: If the 1st of January 2005 is a Saturday, what day is the 9 th of September, 2005?

Solution: We need to count the days in the months. A useful way of remembering the number of days in any one month is as follows:

If the month is an even number less than 8, then it has 30 days. If it is an odd number less than 8 it has 31 days. If the month is 8 or more, then the even numbers have 31 days while the odd numbers have 30 days. The exception to this rule is February, which has 28 days usually, and 29 days in a leap year, that is, a year, which is divisible exactly by 4.

Since 31 3(mod 7), 30 2(mod 7) and 28 0(mod 7), a month of 31 days contributes 3 to the total of days counting from 1st January, 2005, a month of 30 days contributes 2 and February contributes 0 to the total of days.

Thus, modulo 7, for the day on which the 1st of September falls we sum:

3 for January,

0 for February,

3 for March,

2 for April,

3 for May,

2 for June,

3 for July,

3 for August.

The total is 19 5 (mod 7). Thus, September the 1st is calculated as follows:

The 1st of January is a Saturday, i.e. day 6. To that we must add the 5 we have just calculated modulo 7. Now 6 + 5 = 11 4 (mod 7), so the 1st of September is day 4, a Thursday. To get to the 9th September, we have to add 8 more days (to take us from the 1st to the 9th). So we calculate 8 + 4 = 12 5 (mod 7). We conclude the 9th September is day 5, a Friday.

Problem 3: I was born on 1st January 1945. On what day of the week was I born?

(Assume 1st January 2005 was a Saturday.)

From 1945 to 2005 is 60 years. Of these 60 years, 15 are leap years and 45 are ordinary years. An ordinary year has 365 days, and 365 1(mod 7). Each leap year has 366 days, and 366 2(mod 7).

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To determine the day number of 1st January, 1945, we work backwards from 1st January, 2005, day 6. For example, since 2004 was a leap year, 1st January, 2004, fell on day 6 2, Thursday. Since 2003 was a normal year, 1st January, 2003, fell on day 6 2 1, Wednesday. Thus, working back through 45 normal years and 15 leap years, 1st January, 1945, fell on day 6 – 45×1 – 15×2 = 6 – 75 6 5 1(mod 7), since 75 5(mod 7). Hence, 1st January, 1945, was day 1, a Monday.

Note: If one has a birthday on the 29th of February, first work from the 1st of May and then take the day before. Note also that when part of a leap year is included, either at the beginning or end of your reckoning, it counts as 1 modulo 7 only if the 29th of February is included. Otherwise it counts as 0 modulo 7. For instance, from May 1st 2000 to May 1st 2001 February counts as 0 modulo 7, while from February 1st 2000 to February 1st 2001 it counts as 1.

§3 Prime numbers The number 2 cannot be expressed as the product of two numbers, both of which are different from 2. We say that 2 is a prime number. Other examples of prime numbers are 3, 5, 7, 11, 19, 23, 29, 31, etc. On the other hand, 4 = 2×2 is not a prime. Nor is 6 = 2×3 and 8 = 4×2 etc. We say that a whole number greater than or equal to 2 is a prime if it cannot be expressed as the product of two smaller whole numbers.

Since prime numbers occur less and less frequently the higher we count, the obvious question is: do the primes go on forever, or is there a last prime? The ancient Greeks first asked this question and Euclid gave an elegant answer – see §11 Problem 5. Euclid showed that, in fact, there are an infinite number of primes.

Another interesting question is that of prime pairs. Numbers like 3 and 5, primes separated by 2, are called prime pairs. Other examples are 29 and 31, 57 and 59.

The obvious question about prime pairs is: are there an infinite number of prime pairs? The answer is: no one knows. A very large number of prime pairs have been found, but after hundreds of years of working on the problem, no one has come up with a proof one way or the other.

In the 17th century a German mathematician, C. Goldbach, noted that very many even numbers are the sum of two odd primes; indeed he could not find any exceptions. For instance, 40 = 29 + 11, and 32 = 19 + 13. Using modern computers millions of even numbers have been tested and all have been shown to be the sum of two odd primes. However, once again no one knows for sure if this is true for all even numbers. Here we see a contrast between the reasoning in real life.

In real life we know that everybody will age and die. How do we know this? We have records for centuries which show that this is what has happened. So if mathematicians used the same sort of argument as we use in real life, they would accept Goldbach’s conjecture as being true. But they do not.

§4 Gauss’s estimate of number of primesWe remind the reader that in Chapter 8 on Logarithms we described the natural logarithm, ln(n), and we pointed out that this is approximately 2.3 times the number of digits in n. We need this to understand Gauss´s theorem on the number of prime numbers less than n: Gauss showed that the number of primes less than a number n, which he called π(n), is given approximately by n/ln(n). This approximation becomes better the larger n is. We must leave this result unproven, but in the following table you can see how the result compares in some

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cases. The first column lists the number n under consideration, the second gives the number π(n) of primes less than n, and the third is the percentage error between the difference of π(n) and n/ln(n).

n π(n) % error using n/ln(n)

102 25 12

103 168 13.6

104 1,229 12

108 5,761,455 6

1012 37,607,912,018 3.8

1016 279,238,341,033,925 1.4

§6 Fermat’s little theoremIn the sixteen hundreds Fermat noticed the following result about prime numbers, which we will exemplify modulo 7. Take the 7th power of any number not divisible by 7, say 3. Then 37 = 2187 = 7 × 312 + 3 3 (modulo 7). Thus 3 raised to the seventh power reduces to 3 modulo 7. In general, Fermat’s little theorem says that if p is prime, then the p-th power of any number is congruent to the number modulo p.

This is also the basis of a test to determine if a number is not a prime, that is, if the p-th power of any number is not congruent to the number modulo p, then p is not a prime.

FIGURE 2. Pierre de Fermat (1601-1665), an amateur mathe-matician who came up with some astonishing discoveries.

§7 Proof of the casting out nines checkIn Chapter 2 we described a method of checking multiplication. In this we calculated a check digit for each of the two numbers we were multiplying and a check digit for the answer. If the

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product of the two check digits gave a different result than the check digit for the answer, then something was wrong.

The check digit was calculated by adding the digits repeatedly. For instance, 34×26, the check digit of 34 is 3 + 4 = 7, and the check digit for 26 is 2 + 6 = 8. Their product is 7×8 = 56, which has check digit 5 + 6 = 11 which can in turn be reduced by adding its digits 1 + 1 = 2. Thus, our check digit is 2. Suppose that we want to check the result of an original calculation, 34×26 = 884. The check digit of 884 is 8 + 8 + 4 = 20, which in turn has check digit 2 + 0 = 2. Thus this check digit agrees with the check digit we obtained from the two factors.

We would like to explain how this check works. It actually depends on working modulo 9. We know that modulo 9, if

a b (modulo 9) and c d (modulo 9) then

ab cd (modulo 9).

All we need to note is that in adding the digits of a number, we are calculating modulo 9, that is, any number is congruent to the sum of its digits modulo 9. For instance, if a number has two digits e and f, so that the number is written as ef, what we really mean is the number e×10 +f , i,e, e represents the tens digit and f the units digit. (For example, if e = 4 and f = 2, the number 42 is 4×10 + 2.) Thus,

ef = e × 10 + f = e × (9 + 1) + f = 9e + e + f 0 + e + f e + f ,

since 9×e ≡ 0(mod 9).

If we have a three-digit number, say efg, then e represents the hundreds, f the tens and g the units digit. Thus efg = (99 + 1)e + (9 + 1)f + g = 99e + 9f + e + f + g e + f + g (modulo 9) since 99e 9f 0 (modulo 9).

§8 ISBN check digit.All books have a distinct number, called the ISBN number which is used to refer uniquely to the book. Since the number is large and mistakes can occur, the ISBN has a built in check digit. This check digit is obtained by calculating modulo 11. This means that the symbols we need are 0, 1, 2, …, 10. We use X to denote 10.

For instance, suppose a book has ISBN number 0-306-40615- 2. The final 2 is a check number. With such a long number mistakes can occur, and a particularly common error is the interchange of two adjacent numbers. For example, we may have interchanged the 6 and 1 and thought the number was 0-306-40165-2. Without the check number we would not notice this, but with the check number we can.

The way the check number is caclulated is as follows: We calculate modulo 11 and in order to detect the interchanging of two numbers, we multiply the first number with 1, the second with 2,the third with 3 etc. Thus for the correct number 0-306-40615 we calculate

1×0 + 2×3 + 3×0 + 4×6 + 5×4 + 6×0 + 7×6 + 8×1 + 9×5

= 0 + 6 + 0 + 24 + 20 + 0 + 42 + 8 + 45

= 145

= 13×11 + 2

Since 145 ≡ 2(mod 11), the check digit is 2, and the ISBN number is 0-306-40615-2, If we interchange 6 and 1, and we do the whole calculation again, we get

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1×0 + 2×3 + 3×0 + 4×6 + 5×4 + 6×0 + 7×1 + 8×6 + 9×5 = 150.

Here we have underlined the numbers which are different from the previous calculation. Thus interchanging the 6 and the 1 produces 150 which is congruent to 7, modulo 11, and not the check digit.

§9 Make a cipher code for your credit cardsDecide on your own 4 digit number; one you can remember easily. Avoid your birthday and other numbers which other people might know. For example, suppose you decide to use 3186 as your fixed number. Find the numbers which are the negatives modulo 10 of each of the digits in your fixed number. Thus …

the negative of 3 is 7 since 3 + 7 0 (modulo 10)

the negative of 1 is 9 since 1 + 9 0 (modulo 10)

the negative of 8 is 2 since 8 + 2 0 (modulo 10)

the negative of 6 is 4 since 6 + 4 0 (modulo 10).

So the list of negatives is 7924.

Suppose that the four-digit code of your credit card is 1234. Add the negatives to each of these digits in turn modulo 10. Thus we get 7 + 1 8, 9 + 2 1, 2 + 3 5, 4 + 4 8. So we get 8158. This is the number you can write down on a piece of paper or in a notebook, When you need to enter your code word, take 8158 and add your fixed code 3186 digit by digit modulo 10, 8 + 3 1, 1 + 1 2, 5 + 8 3 and 8 + 6 4, so the correct code 1234 is recovered.

With this system you do not have to worry if your notebook is lost.

§10 Application to codesIt may seem far-fetched, yet it is a fact, that this modulo arithmetic has an extremely important practical application. This is for the sending of messages which we want to be confidential. Banks for instance do this millions of times daily. The idea is to send messages that cannot be decoded by anyone but the person for whom they are intended. The RSA code, which is the most important in every day use, works modulo pq where p and q are two large prime numbers. To decode the messages we need to use Fermat’s Little Theorem as described in §6. Fermat would certainly have been surprised that his theoretical result of 400 years ago is now used practically.

Another type of code is an error correcting code. When messages are sent errors can arise. For instance, if you send a message from a satellite circling Mars some of the message might be garbled. But by choosing a suitable code, we can correct the message. As a simple (but totally unrealistic) example of such a code, we could send each letter three times: So the message “ttjheheef” can be corrected to “ttthhheee”, i.e., “the”. This method of correcting the message will only work if there are relatively few errors, for instance if we receive thjhhheee we could not decide that the first letter of the decoded word was t or h or j for instance. The modulo arithmetics we discussed in §1 provide the building stones of suitable codes.

Such codes also make sure that digital information such as on CD discs is correctly decoded.

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§11 Solved Problems and other optional items1. Proof that if a ≡ b and c ≡ d (mod n), then (i) a + c ≡ b + d (mod n) and (ii) ac ≡ bd (mod n).

Solution. Since a ≡ b (mod n), a b is divisible by n. Similarly, c d is divisible by n. Therefore, (a b) + (c d) is divisible by n. However, (a b)+(c d) can be rewritten as

(a + c) (b + d). Hence, (a + c) (b + d) is divisible by n, which means that a + c ≡ b + d (mod n). This proves (i).

For (ii), we start again with the fact that (a b) and (c d) are divisible by n. Hence, multiplying (a - b) by c and (c - d) by b still produces terms divisible by n, so

(a b)c + (c d)b = ac bc + cb db = ac bd is divisible by n, that is, ac ≡ bd (mod n).

2. Explanation of the modulo 11 multiplication check as described in Chapter 2 §7 item 1.

Solution. In the proof of the casting out nines check we used the fact that 10 is congruent to 1 modulo 9. Modulo 11, however, 10 is congruent to –1. This means that a number like 43, which is 4×10 + 3, is congruent to 4×(1) + 3, that is, to find the number congruent to 43 modulo 11, subtract the second digit from the first. Since 10 ≡ 1 (mod 11), 100 ≡ +1 (mod 11). Hence, 742, which is 7×100 + 4×10+2, is congruent to 7×1 + 4× (1)+2, or 7 + 2 4 = 5. Likewise, 1000 ≡ 1, 10,000 ≡ +1, etc. So to find the number congruent to a given number modulo 11, starting from the left, add the digits in the odd positions and subtract those in the even positions.

3. Construct the addition and multiplication tables for numbers modulo 4.

Solution.

4. The first of January, 2005 was a Saturday. Use this fact to calculate the day of birth of a man who was born on 1st of December, 1987.

Solution: We denote the day of the birth by the letter x. We begin by calculating the day on the first of December 2004.

In total there are 17 years and of these, 5 are leap years. For each leap year we add 2(modulo 7) and 1 for each ordinary year. So we need to add 10 for the leap years, and 12 for the ordinary years. That is a total of 22 or 1 modulo 7. We must add 31 days to bring us to the 1st January 2005. This with the extra 1 makes a total of 32 days, and 32 is congruent to 4 modulo 7. Thus since Saturday is day 6, x + 4 = 6 or x = 2. Thus the day of birth is a Tuesday.

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+ 0 1 2 3

0 0 1 2 3

1 1 2 3 0

2 2 3 0 1

3 3 0 1 2

X 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

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5. Prove Euclid’s theorem: There are an infinite number of prime numbers.

To understand this proof remember that every whole number is either a prime itself or else is divisible by a prime number.

Euclid’s argument is explained by the following example. Take all the primes up to say 7. Multiply them together to get P = 2×3×5×7 = 210. To this add 1 getting P + 1 = 211. Now if you divide this number by any of the primes from 2 to 7 we get a remainder of 1. For instance, 211 divided by 3 is 70 with a remainder of 1. If P + 1 is a prime, we have a number larger than 7 which is a prime. Otherwise, there must be a prime which divides P + 1. It cannot be any of the primes from 2 to 7, as we know there is always a remainder of 1 when we divide by any such prime. Hence there must be a prime larger than 7 that divides P + 1.

The above example gives the clue to complete the proof in general for any prime p. We will prove that for every such prime p, there is a larger prime.

We multiply all the primes up to and including p to get the number P. Then add 1 to get P + 1. Just as in the example above, when we divide P + 1 by any prime less than or equal to p, we get a remainder of 1. This means that either P + 1 is itself a prime or else there is a prime larger than p which divides P + 1. Either way, this means that for the given prime p, there is always a larger prime. In other words, we have shown that there are an infinite number of primes.

6. A check for prime numbers.

Any number is either prime or divisible by a prime. Thus, to check if a number, k, is prime we could test by dividing k by all the primes less than k. This test can be considerably shortened by noting that if k is not prime it is divisible by at least two primes. If k would be divisible by two primes, p and q, each of which were larger than the square root of k, then pq would be larger than k, which is not possible. Thus, if k is not prime there must be a prime dividing k that is less than the square root of k, so we can confine ourselves to testing primes that are less than the square root of k.

For instance, to test if a number less than 10,000 is prime, it is only necessary to try dividing by the primes less than 100.

7. Sieve of Eratosthenes.

Eratosthenes, who was made director of the library at Alexandria in 235 BC, described a method of listing all odd primes up to any desired number, N. The method, called the sieve of Eratosthenes, proceeds as follows.

1. Write down all odd numbers from 3 to N.

2. Underline 3 and cross every third number after 3 (9, 15, etc.).

3. Underline the next uncrossed number, 5, and cross every 5th number after 5.

4. Continue by underlining the next uncrossed number, 7, crossing every 7th number, then underlining the next uncrossed number, 11, crossing every 11th number, etc.

5. Continue applying the above for 13, 17, 19, etc., but only up to the square root of N.

6. The remaining uncrossed numbers are all the odd primes less than N.

For example, for N = 50, we need only continue the crossing up to 7, since √50 is under 8. The method proceeds as follows.

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3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49.

The uncrossed numbers are the primes less than 50.

8. When Gauss was a pupil his teacher posed the problem of summing the numbers from 1 to 100.

The teacher expected this task would occupy the class for the full period, but Gauss finished in a few minutes. His trick was to write down the sum 1+2+3+4+5+ ... +98+99+100, and immediately under, the sum in reverse order. The end result was

1 +2 +3 +4 +5 + ... +96+97+98+99+100

100+99+98+97+96+.... +5 + 4 +3 +2 +1

He then added each of the 100 vertical pairs to get 101+101+101+101+ ... +101+101+101+101+101, which is 100×101.Since each sum is the same, Gauss had found that 100×101 is twice the sum of the numbers from 1 to 100, so the answer was 100×101/2 = 5050.

9. Justify the formula: 1 + 2 + 3+ ... + N = N(N + 1)/2.

Solution: Write down the numbers 1+2+3+4+5+ ... +N, and immediately under, the numbers in reverse order. It looks like

1 + 2 + 3 +... + N

N + (N 1) + (N 2) +....+ 1.

Add each consecutive vertical pair to get (N + 1) + (N + 1) + ... + (N + 1). In total we are adding N numbers each of which is (N + 1). The total sum is N(N + 1). This is double the sum we want to calculate, so the sum is N(N + 1)/2.

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Chapter 16

COUNTING

When Noah loaded the Ark, the animals went on in pairs, one male and one female. So we know that there were as many males as females, even though we do not know how many animals were on board. This is the essence of saying two collections have the same number of elements.

Counting is the first mathematics children learn. When we count, we count the objects in a collection, for instance the number of cars in a parking lot. In this, our last chapter, we make the idea of counting precise. We begin first with the concept of set, which is another name for collection. We have mentioned set several times before in this book and here we repeat the essentials.

§1 What do we mean by a set?1

In the 1960s the New Mathematics was introduced in the schools. It caused much confusion since the parents did not know what it was about and so could not help their children. Also it did not seem to have a practical value.

The New Mathematics was based on set theory. The idea of set theory is to consider collections of objects, for instance the collection of all schoolteachers. This we call a set and the objects in the set, in this case schoolteachers, we call elements of the set.

Almost anything can be a set. For instance, there is the set of all vegetables. The idea of set is indeed quite basic, and nobody has any difficulty with it. The difficulty comes with trying to express as much as possible of mathematics purely in terms of set theory. Mathematicians want to base their subject on as few assumptions as possible, and being able to derive results relying only on set theory would put the subject on a more rigorous and logical basis. This is where the difficulty arises. Just as it is difficult for a learner to express himself in English with a limited vocabulary of only a few hundred words, so is it a struggle to express complicated mathematical ideas in terms of set theory alone.

We remind the reader that finite sets have elements that can be listed, and this list will come to an end, such as {1, 2, 3, 4}. The set of all whole numbers, {1, 2, 3, …} never comes to an end, since if x is a whole number, so is x + 1.

1 This is a repeat of §1 of Chapter 11.

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In addition to the whole numbers, we also have the fractions, numbers which are one whole number divided by another, like ¾. The fractions are also called rational numbers. The rational numbers also include the negative fractions. Rational numbers can be expressed as finite or recurring decimals. The real numbers include not only these, but also those numbers with infinite non-recurring decimal expansions, again including both positive and negative numbers. These are also examples of infinite sets.

In this chapter we will be looking at sets and explaining what we mean by saying two sets have the same number of elements. But the term “same number of elements” is usually reserved for finite sets. For infinite sets we shall say that the sets have the same cardinality.

The man who made set theory very interesting for mathematics was George Cantor. Most of the ideas in this chapter are due to him.

FIGURE 1. George Cantor (1845-1896).

§2 The same cardinalityIn this section we will define what is meant by saying two sets have the same cardinality. We will take for granted the idea of an ordered pair (a, b) of two elements. By ordered we mean that the order we write the elements down matters, that is, (a, b) differs from (b, a) if a and b differ.

Take two sets A = {a, b, c} and B = {apple, banana, cherry}. These squiggly brackets are the standard notation for indicating the elements of a set. How do we test that A and B have the same number of elements? This is so obvious that it is difficult to know how to explain it properly. But if mathematics is to make sense we must be able to explain what we mean.

This is what we do. We match each element of A with precisely one element of B. For example, we pair the a in A with the apple in B, b with banana and c with cherry. Since everything must be expressed in the language of set theory, we form a new set, C, with elements made up of pairs as follows:

C = {(a, apple), (b, banana), (c, cherry)},

and observe that each of the elements of A and B occurs in one and only one of the elements of C. This is the essential idea of our definition of saying two finite sets have the same number of elements.

It is analogous to claiming that the number of people in a theatre is equal to the number of seats. This will be true if every person is sitting, if every chair is occupied, and only one person is sitting on each chair.

So we are ready for our definition by applying the same idea to arbitrary sets, both finite and infinite.

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Two sets A and B have the same cardinality if there is a set C whose elements are ordered pairs with the first entry in each ordered pair coming from A and the second from B. Every element of A must appear as the first element in exactly one of the elements of C. Every element of B must appear exactly once as the second element of an element in C.If two sets A and B have the same cardinality, we write A = B and say that the cardinality of A is equal to the cardinality of B. If A and B are also finite, we also say they have the same number of elements.

§3 Lines of lengths 1 and 2 have the same cardinalityThe elements of these lines are their points, and each is of course infinite.

To prove that they have same cardinality, it is essential in what follows to use the definition exactly and precisely. We can either consider the points on the lines or the corresponding numbers. L1 is the set of all real numbers lying between 0 and 1, with both 0 and 1 included. L2 is the set of all real numbers between 0 and 2, with both 0 and 2 included. According to our definition to show that L1 and L2 have the same cardinality we need to form a new set C consisting of ordered pairs with every element of L1 appearing once and only once as the first element in one of the ordered pairs and every element of L2 appearing once and only once as the second element of one of the ordered pairs. This we can do both algebraically and geometrically.

Algebraically we define the set C as the set of all ordered pairs (x, 2x), for every number x lying between 0 and 1.

We need to check

1. Each element of L1 appears as the first element in one of the ordered pairs in C. It does, since according to our definition of the set C for each x in L1 there is the ordered pair (x, 2x) in C.

2. Every element, say y, in L2 appears as the second element in an ordered pair in C. Since y lies between 0 and 2, ½y lies between 0 and 1, and, hence, by our definition of C the element (½y, y) lies in C.

3. Each x in L1 and each y in L2 appears exactly once in C. By our definition of C, each x in L1 appears only once in C, in (x, 2x), and each element y of L2 appears only once in C, in (½y, y).

Hence by our definition, L1 and L2 have the same cardinality. We can also see this geometrically (see Figure 2).

FIGURE 2. A line of length 1 has the same cardinality as a line of length 2.

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We draw two parallel lines, AB, of length 1, to represent L1, and the line CD of length 2, to represent L2, and extend AC and BD to meet at the point O. We construct a set C of ordered pairs as follows:

For each point x on L1 we draw the line through O to x and extend it till it meets L2 in the point y. We then construct C as the set of all of ordered pairs (x, y). It is easy to see that every point x of L1 appears in one of the ordered pairs in C and that each element y in L2

appears as the second element of a pair in C, and furthermore it is clear that each x in L1

appears only once and that each point y in L2 appears only once. We have in fact showed that the cardinality of the two lines is the same.

Intuitively this conclusion seems strange. Since L1is only half the length of L2, we should not expect it to have the same cardinality. However, the argument is perfectly correct based on the definitions of lines and points. The only thing you can say in opposition is that you do not like these definitions, but this is a matter of personal taste. With our definitions, L1 and L2 do have the same cardinality.

§4 A listA useful way of proving that a set S has the same cardinality as the set of whole numbers, N = {1, 2, 3, …} is to list the elements of S. By this we mean that there is a first element, a second element and for every further element there is an immediate successor element. Also, every element of S occurs somewhere in the list. By analogy, imagine an infinite stairway ascending to heaven, which has a first step and for every step a following step, and all of the steps eventually get included in this way. Then each step will correspond to a number, first, second, etc., which makes the cardinality of the set of steps the same as the whole numbers. The elements of S are matched up with the whole numbers in the same way.

§5 The rational numbers have the same cardinality as the whole numbersWe can apply the method of the previous section to prove that Q+, the set of positive rational numbers (i.e., the set of all fractions p/q, formed by dividing one positive whole number by another), and N, the set of whole numbers, have the same cardinality.

We begin our list with all the rational numbers which can be written in the form p/q where p and q are positive whole numbers and p + q = 2. Next, we add to the list all the rational numbers which can be written in the form p/q, where p and q are positive whole numbers, and p + q = 3, next p + q = 4, etc. At each step of adding to the list we check back to delete any repeated numbers (e.g. 2/3 and 4/6).

Thus the only number of the form p/q with p + q = 2 is1/1. Next we add to the list all positive fractions, p/q, with p + q = 3. These are 1/2 and 2/1. Our list now consists of, 1, ½, 2. The next step is to add to the list all fractions, p/q with p + q = 4. These are 1/3, 2/2 and 3/1 (we list them in the order p = 1, 2, 3, …). Since 2/2 already appears in the list in the form 1/1 or 1, we delete 2/2, and our list now consists of 1, ½, 2, 1/3, 3. If we continue in this way, always increasing the list, eventually every positive rational number will appear in this list. Since our method lists all the positive rational numbers, this means that the set of whole numbers and the set of positive rational numbers have the same cardinality.

What we have effectively done is construct a set C consisting of the ordered pairs

(1, 1), (2, ½), (3, 2), (4, ⅓), (5, 3), … .

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§6 An infinite set which does not have the same cardinality as the whole numbersThe results in §3 and §5 above might tempt us to try to show that all infinite sets have the same cardinality as the whole numbers. This is not the case.

Let S be the set of all finite and infinite decimals with only 1s and 0s as entries, such as

0.11100011… and 0.0001110101….

Since there are an infinite number of possible decimals, there are clearly an infinite number of elements in S.

We begin by assuming the opposite to what we want to prove, we assume that S has the same cardinality as the whole numbers. By §4 above, this means that the elements of S can be listed. Thus, assume that the elements in S have been listed. This means we have a first ele-ment s1 of S, a second element s2, a third element s3, a fourth element s4, and so on. Based on our list we now construct an element, s, as follows:

The first decimal place of s will be 1 if the first decimal in s1 is 0, or 0 if the first decimal in s1 is 1. Hence, s is not equal to s1. The second decimal place in s is chosen to be 1 if the second decimal place in s2 is 0 and 0 if the second decimal place in s2 is 1. So far, s is not equal to s1

or to s2. We continue by a similar mismatching of the third decimal of s and the third decimal of s3, so s is also not equal to s3, and so on. In general, the n-th decimal place of s will be 1 if the n-th decimal place of sn is 0 and 0 if the n-th decimal place is 1. This insures that s will not be equal to any number in S, But since s is an infinite decimal with entries 0 and 1, it must belong to S. This is a contradiction which proves that our original assumption, that the elements of S could be listed, is false.

§7 Is there a set of cardinality less than the reals but greater than the natural numbers?In this section we shall sketch some ideas, without being precise and without any detail. Firstly it is possible to introduce a definition of inequality and say that the cardinality of A is less than the cardinality of B. This we write in symbols as A < B. With this notation, what we have shown in §6 above is that N < S, where N is the set of whole numbers and S is the set of infinite decimals with 0 and 1 as entries as described in the last section. From this with some quite involved argument, we can deduce that N < R where R is the set of all real numbers.

The natural question is: ”Is there a set X such thatN < X < R”? The answer to this problem is that without adding an extra axiom one cannot say. It is even possible to add an axiom saying that such an X exists or alternately that such an X does not exist, and both are equally consistent. It is a bit like Geometry, where you can have a Euclidean Geometry or else a Projective or even a Hyperbolic Geometry. The mathematician who came to this conclusion was Paul Cohen.

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FIGURE 3. Paul Cohen, 1934-, solved the problem ofis there a set “in between” the real numbers and the natural numbers.

§8 Solved problems1. Show that the set of even whole numbers has the same cardinality as the whole numbers, N = {1, 2, 3, …}.

Solution: Let S be the set of even numbers, {2, 4, 6, ...}. The set C of ordered pairs, {(x, 2x)} satisfies the definition for N and C to have the same cardinality. Alternatively, it is easy to list the elements of S: 2, 4, 6, ... .

2. The cardinality of points on an infinite line is the same as that of points on a semicircle.

Solution: In Fig. 4, the line through P and Q extends infinitely far in both directions. The line LN is parallel to the line through P and Q. The semicircle, center at 0, ends at L and N but does not contain the end points L and N. Unlike the line, its length is finite.

For every point x on the semicircle we draw a line from O through x and extend it till it meets the line PQ in the point y. The set of pairs (x, y) constructed this way shows that PQ and the semicircle LN have the same cardinality. This is because every x on the semicircle appears in this collection, and every y on the infinite line appears. Each x on the semicircle and each y on the line will appear precisely once.

Fig 4. The semicircle which is of finite lengthand the infinite line have the same cardinality

3. The set of squares of whole numbers has the same cardinality as the set of whole numbers.

Solution. Let C be the set of all ordered pairs. C = {(x, x2)}. Again it is clear that this set C satisfies the conditions required. Every whole number appears as the first entry in one of the pairs of C and every square appears as the second entry in one of the pairs in C. Furthermore, no element appears in the second entry with two different first entries.

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4. Let N be the set of whole numbers. Add another element, say the letter a to form a new set M. Show that M and N have the same cardinality.

Solution: We can list the elements of M starting with a: a, 1, 2, 3, 4, …etc. This shows that the set M has the same cardinality of elements as N.

5. (Very hard) Let S be the set of points in a square with side 1 and T be the set of points on one side. Show that S and T have the same cardinality.

Solution. The points in S can be written as ordered pairs of numbers between 0 and 1, as we did in the coordinate system in Chapter 9, and the points in T as decimals between 0 and 1. The plan is to match a pair of decimals (0.abc… , 0.efg…) in S with a single decimal 0.aebfcg… in T.

For example, (0.23053… , 0.42540…) will be matched with 0.2432055430… , formed by alternating the digits of .23053… and .42540…, starting with the first digit of the left element of the pair and the first digit of the right element of the pair. This process not only matches up every ordered pair with a single number in T, but it is easy to reverse the process to reconstruct the unique pair in S from a decimal of T by unraveling the decimal.

Note. The above argument requires a slight addition, due to a possible confusion between finite decimals and infinite decimals ending in infinitely many 9s. The problem is that there is no difference between 0.3, for example, and 0.299999…, with an infinite repetition of 9.

To see this let x = 0.2999… , then 100x = 29.999… , and 10x = 2.999… . Hence, 100x – 10x = 29 – 2 , so 90x = 27, and x = 27/90 = 0.3. This means that if we allow all possible decimal expansions, finite decimals and infinite decimals, like 0.123 and 0.12299999…, with 9 repeated infinitely all the time, the same number is repeated. The solution is to rewrite all finite decimals as decimals ending with infinitely many 9s and not allow any finite decimals.

§9 Hilbert’s Infinite Hotel

FIGURE 5. David Hilbert. Born: 23 Jan 1862 in Königsberg, Prussia(now Kaliningrad, Russia). Died: 14 Feb 1943 in Göttingen, Germany.

Example 1 in §8 emphasizes the difference between an infinite set and a finite set. There a part of a set was shown to have the same cardinality as the whole set. This could not occur with a finite set.

One of the most outstanding mathematicians of his time, David Hilbert, had a way of illustrating this. In Hilbert’s Infinite Hotel his argument goes as follows: Suppose you come to his hotel with rooms numbered 1, 2, 3, …, i.e., numbered by the whole numbers. If all rooms are full and another guest comes what can he do? He simply moves the person in room 1 to room 2, moves the person in room 2 to room 3, moves the person in room 3 to room 4, etc. He goes on doing this and, since he can continue this process forever, everyone will be accommodated by giving (the now empty) room 1 to the new guest. This is essentially the same as Problem 4 of §8 above

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§10 Russell’s ParadoxIt turned out tha set theory was not as hoped a flawless, logical establishment of mathematics. There was a serious problemwjich is explained below.

Note that the elements of a set can be sets, e.g. the set S whose elements are the sets {1}, {1, 2} i.e. the set S = {{1}, {1, 2}}. Note that S is not an element of itself, i.e. S does not belong to itself.

It is possible for a set to belong to itself,. for instance, if T is the set whose elements are sets with at least two elements, then T has for example elements {1, 2} and {1, 3} and hence itself has at least two elements. So T is an element of itself.

And of course, a set either belongs to itself or not. This apparently innocent remark gives rise to a problem.

Consider R the set of all sets which do not belong to themselves. Then if R does not belong to itself, then by its very definition it belongs to itself. If R does belong to itself, then by the definition of R we conclude that R does not belongs to itself. A contradiction. This contradiction was pointed out by Bertrand Russell, and led to the development of a more sophisticated set theory. There has even been a suggeston that another theory, Category Theory, should be used instead of set theory as a basis for mathematics. But that is another story

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CHAPTER 17

THE TRACHTENBERG METHOD OF

MULTIPLICATION

Multiplying two digit numbers quickly without much writing

§1 The MethodAs we have seen in Chapter 2, the method of replacing numbers with single digit numbers in a calculation gives a quick and rough check of the result, but many times it is a little too rough. Obviously, replacing with two-digit numbers would improve the accuracy. The main reason for not replacing by two-digit numbers is the complexity it adds to the computations. However, multiplying two 2-digit numbers is quick and easy if you use the method advocated by Jakow Trachtenberg, which we will explain in this chapter.

Not everybody will want to spare the time to learn this method, arguing that one can always use a calculator, or else multiply the numbers by the long method, but you, as we, may enjoy the method.

The explanation may seem long, but with practice you’ll be able to write down the two factors and the answer directly underneath. For instance, you would write

35

28 980

and would not need any further calculation. In fact, you should be able to do it as fast, or even faster, than you could with an electronic calculator.

With more effort you could be able to apply the method to products of numbers with more digits and simply write down the answer in a short time. For the present, however, we will concentrate on the product of two digit numbers.

We will always consider two-digit numbers in terms of tens and units. For example, the tens digit of 72 is 7 and the units digit of 72 is 2.

For our first example of Trachtenberg´s method consider the following product.

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59

The product of the tens digits of the two numbers, in this case. 57= 35, is called the first opposite product, and the product of the units digits of the two numbers, in this case, 89= 72, is called the second opposite product. The product of the tens digit of the first number and the units digit of the second number, in this case, 79= 63, is called the first cross product, while the product of the units digit of the first number and the tens digit of the second, in this case, 85 = 40, is called the second cross product. The product is calculated in 4 steps:

STEP 1

The first number we write down is the tens digit of the first opposite product, 75 = 35, in this case, 3.

78 59

3

STEP 2

Sum the units digit of the first opposite product, 35, to the tens digits of the two cross products, 63 and 40. In this case this is 5 + 6 + 4 = 15. The method is to attach the units digit, 5, of this sum to the digit of the last step, 3. However, as this sum, 15, has one tens digit we will have to carry 1 to change the 3 to a 4, and then attach the 5. The result, at the end of this step, is

78

59 45

STEP 3

Sum the units digit of the first cross product, 63, the units digit of the second cross product, 40, to the tens digit of the second opposite product, 72. In this case, the sum is

3 + 0 + 7 = 10.

As in Step 2, we attach the units digit of 10, which is 0, to 45, the result from step 2, and add the units digit, 1, of 10 to the last digit of 45.

So we need to carry one and thus change the 45 to 46 and attach 0 to obtain

78

59 460

STEP 4

Finally, we attach the units digit of the second opposite product, 72, which is 2, to get

78

59 4602

146 (152) 146

146

which is the answer.

With some practice this becomes quick and easy. Most of the above steps are done in our head and all we do is write down the answer in the four steps as follows:

Step 1

78

59 3 the tens digit from the first opposite product 5X7 = 35, i.e. 3

Step 2

78

59 45 the units digit of the sum of the first opposite product 7X5 = 35, i.e. 5 and

the tens digits of the two cross products hich are 7X9 = 63 and 5X8 = 40, i.e. 5 + 4 + 6 = 15. Since this more than 9 we need to carry the tens to the 3 which we calculated in the first step. Thus we get 45.

Step 3

78

60 460 the sum of the tens digit of the second opposite product and the units digits of

the two cross products, adding the tens digit, if there is one, of the sum to the result of the 2nd step

Step 4

78

59 attach the units digit of the second opposite product.

4602

In practice we do not rewrite the product, but simply write the answer in one step, since each successive step consists of attaching a digit or else if we get a number larger than 9, we carry the tens digit over to the previous digit and attach the units digit. Thus the above calculation is done as follows:

78

59 4602

EXAMPLE: 3942

STEP 1

The tens digit of the first opposite product, 34 = 12 is 1. Write down 1.

STEP 2

The units digit of the first opposite product, 34 = 12, 2, plus

the tens digit of the first cross product, 32, 0, plus

147 (152) 147

147

the tens digit of the second cross product, 49 = 36, 3,

The sum is: 2 + 0 + 3 = 5. Since there is no tens digit, we simply attach the units digit, 5, to get 15.

STEP 3

The units digit of the first cross product, 32 = 6, 6, plus

the units digit of the second cross product, 49 = 36, 6, plus

the tens digit of the second opposite product, 92 = 18, 1

The sum is: 6 + 6 + 1 = 13

As we have a tens digit, 1,we have to change the 15 from step 2, to 16 and then attach 3 to get 163.

STEP 4

The final digit is obtained from the units digit of the second opposite product, 92 = 18, i.e. 8. So attach the 8 to get the final answer, 1638.

§2 Solved Problems and some optional ideas1. Use the method of Trachtenberg to calculate 5645.

Solution:

STEP 1

The tens digit of the first opposite product, 5×4, i.e. 2

Write 2.

STEP 2

The units digit of the first opposite product, 54 = 20, i.e. 0

The tens digit of the first cross product 55 = 25, i.e. 2

The tens digit of the second cross product 64 = 24, i.e. 2

The sum is 0 + 2 + 2 = 4

Attach 4 to 2 to get 24.

STEP 3

The units digit of the first opposite product, 56 = 30 i.e. 0.

The tens digit of the first cross product, 6×4, i.e. 2

The tens digit of the second opposite product 5×6, i.e. 3

The sum is 0 + 2 + 3 = 5

Attach 5 to 24 to get 245

STEP 4

The units of the second opposite product, 56 = 30, i.e. 0.

Attach 0 to 245 to get 2520, as the answer.

148 (152) 148

148

2. Use the method of Trachtenberg to calculate 7723.

Solution:

STEP 1

The tens digit of the first opposite product, 72 = 14, i.e. 1

Write 1.

STEP 2

The units digit of the first opposite product, 72,i.e. 4

The tens digit of the first cross product 7×3, i.e. 2

The tens digit of the second cross product 2×7, i.e. 1

The sum is 4 + 2 + 1 = 7

Attach 7 to 1 to get 17.

STEP 3

The units digit of the first cross product, 7×3, i.e. 1

The units digit of the second cross product, 2×7, 4

The tens digit of the second opposite product 7×3, 2

The sum is 1 + 4 + 2 = 7

Attach 7 to 17 to get 177.

STEP 4

The final digit: attach the units digit of the second opposite product, 7×3, i.e. 1, to 177 to get the answer 1771.

3. Approximation using two digits. Multiply 346 785.

Solution: We approximate to 346 by 350 and to 785 by 790. Next, we multiply 3579 using the Trachtenberg method and then add two 0’s. The result is 276,500. The correct result is 271,610. That is, our approximation has an error of about 2%.

By comparison, using single digits we would approximate to the answer by 300 800 = 240000, with an error of approximately 12%.

4. How good is approximation with two digits

Solution: In the worst case, we will replace say 105 by 100, with an error of 5%. And we potentially have both factors are off by 5%. Since in the product the error is approximately the sum of the percentage errors, with this method the error could be as much as 10%.

For example, if we approximate to 105×105 by 100×100, we would be approximating to 105×105 = 11,025 by 100×100 = 10,000, which is an error of 1000 out 10,000, which is 10%.

5. Explain why the Trachtenberg method works.

In order to understand why Trachtenberg’s method works, we introduce a useful notation. We write a two digit number a in the form a1a0, so that a1 is the number in the 10 column and a0 is the number in the unit column. This means that with this notation,

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149

a = 10a1 + a0 and´b = 10b1 + b0 so that

ab = (10a1 + a0)(10b1 + b0) = 100a1b1 + 10a1b0 +10a0b1 + a0b0

Thus the tens of a1b1 give a the number in the 1000 column, the 100 column is given by the units of a1b1 and the tens of a1b0 and the tens of a0b1. The ten column comes from the units of a1b0 and the units of a0b1 and the tens of a0b0. Finally the units are contributed by the units of a0b0. This is exactly the Trachtenberg method.

10. Historical note:

Trachtenberg, a Russian, developed several similar fast methods for mathematics, of which this is just one, to occupy his time while a political prisoner of the Nazis in a concentration camp during the Second World War. Since he was denied writing implements, his methods had to be based on mental arithmetic. His was a dramatic story and he was narrowly saved from execution by his wife who managed to bribe him a transfer to another concentration camp.

BIBLIOGRAPHY

There are many interesting popular books on mathematics. We list a few below, but we have found almost all books to be very good. The ones below include some very good classics.

Ball, Rouse & Coxeter, H.S.M. (1987). Mathematical recreations and essays. 13 ed. New York: Dover Publications.

Bell, Eric Temple (1986). Men of mathematics : the lives and achievements of the great mathematicians from Zeno to Poincaré. 1. Touchstone ed. New York: Simon & Schuster.

Courant, Richard (1996). What is mathematics? : an elementary approach to ideas and methods. 2 ed. rev. by Ian Stewart. New York: Oxford Univ. Press. ISBN: 0-19-510519-2 (pbk)

Davis, Philip J.& Hersh, Reuben (1998). The mathematical experience ; with an introduction by Gian-Carlo Rota. New ed. Boston: Mariner Book.

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Dunham, William (1990). Journey through genius : the great theorems of mathematics. New York: Wiley.

Gowers, Timothy (2002). Mathematics: a very short introduction. Oxford: Oxford University Press. ISBN: 0-19-285361-9.

Gullberg, Jan (1997). Mathematics : from the birth of numbers. Technical illustrations: Pär Gullberg. New York: W.W. Norton.

Hardy, Godfrey Harold (1992). A mathematician's apology. Canto ed. Cambridge: Cambridge Univ. Press.

Hogben, Lancelot (1967). Mathematics for the million. Revised ed. London: Pan.

Littlewood, John Edensor (1986). Littlewood's miscellany. Rev. ed. by Béla Bollobás, Cambridge : Cambridge Univ. Press. [Tidigare utgiven 1953 som A mathematician's miscellany. London: Methuen]

Kordemskij, Boris A (1992). The Moscow puzzles : 359 mathematical recreations. Edited and with an introduction by Martin Gardner. New York: Dover Publications.

Körner, T. W. (1996). The pleasures of counting. Cambridge: Cambridge University Press.

Stewart, Ian (1998). The magical maze : seeing the world through mathematical eyes. London: Phoenix.

Stewart, Ian (1992). The problems of mathematics. 2. ed. Oxford: Oxford University Press.

Pólya, George (2004). How to solve it : a new aspect of mathematical method. [With a new foreword by John H. Conway]. Princeton : Princeton University Press.

Stillwell, John (2001). Mathematics and its history. 2. ed. New York: Springer.

Wells, David Graham (1991). The Penguin dictionary of curious and interesting geometry. [Illustrated by John Sharp]. Harmondsworth: Penguin.

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