Mathematics
the Practical,
the Logical, and
the Beautiful,
by
Benjamin Baumslag and Frank Levin
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 results[1] 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 selfcontained account, the reader who has forgotten
the school mathematics will be reminded of some of the details. What is 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.
We gratefully
acknowledge help from the following: David Baumslag, Pia Baumslag, Jeff Bourne,
János Hegedüs, Geoffrey Howson, LarsGöran 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
1. PRELIMINARIES AND A LITTLE FUN
§3 Indispensable tools for reading this book
2. SQUARING, CHECKING AND APPROXIMATION
§4 Will the computer replace the mathematician?
§2 Number of people in the world
§4 Maximum number of inhabitants on the earth
§5 Viability of running a shop
§6 General Comments on Fermi calculations
5. PERCENTAGES
§2 A fictitious company’s accounts
§4 Examples of judging figures in the news
§5 Keep an eye on the total figures
6. MEASUREMENT SENSE OR DIMENSIONAL ANALYSIS
§7
Work and Power, Joules and Watts
7. MEASURING HEIGHTS OR TRIGONOMETRY
§4 Dropping a stone over a cliff
§5 Solved problems and other optional topics
8. LOGARITHMS AND NATURAL LOGARITHMS
§1 Multiplying by adding. Logarithms
§4 The natural logarithm is roughly 2.3 times the
ordinary logarithm
§7 Logarittihms and Planeetary Motion
§8 Earth quakes measured on the Richter scale
§2 Graphs as a concise source of information
§6 The Greeks and their curves
§7 Equations and conic sections
§8 Applications of coordinate geometry
§10 Solving problems in Geometry with algebra and
vice  versa
10. SOLVING EQUATIONS AND GAUSS’S METHOD
§4 Method 3: Do what your mathematics teacher
told you
§7 Solved problems and other optional material
11. FUNCTIONS
§2 Functions defined geometrically
§4 Functions described algebraically
§5 Contrast with the definition in the calculus
books
§6 What comes after 1, 2, 3? Could it be 34?
§7 Solved Problems and other Optional items
12. GEOMETRY
§2 Euclidean Geometry assuming only the real
numbers
§4 A finite projective geometry
§6 Sketch of the arguments for completing §2
13. PROBABILITY AND STATISTICS
§5 Monte Carlo method of calculating p
approximately
§6 It is impossible to send a rocket to the moon
§8 Nuclear power plants exploding
§10 General remarks about sampling
§1 Proof of Pythagoras’s Theorem
§2 The converse of Pythagoras’s theorem
§4 Estimating the square root of a number
§5 The square root of 2 is irrational
15. MODULAR ARITHMETIC. PRIME NUMBERS AND LOGARITHMS
§4 Gauss’s estimate of number of primes
§7 Proof of the casting out nines check
§9 Make a cipher code for your credit cards
§11 Solved Problems and other optional items
16. COUNTING
§3 Lines of lengths 1 and 2 have the same
cardinality
§5 The rational numbers have the same cardinality
as the whole numbers
§6 An infinite set which does not have the same
cardinality as the whole numbers
§7 Is there a set of cardinality less than the
reals but greater than the natural numbers?
17. THE TRACHTENBERG METHOD OF MULTIPLICATION
§2 Solved Problems and some optional ideas
18. BIBLIOGRAPHY
Chapter 1
Mathematics begins
with a guess, just as naturally as love begins with a kiss.
This 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.
figure 1. George Pólya
(18871985), 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.”
One 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.
Pencil 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 3´4 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.
Here is a party trick.
You ask somebody to do the following:
·
Think of a
whole number between 1 and 9
·
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.
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.
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. .
The 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.
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.”
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 righthand column.
46 
33 
92 
16 
184 
8 
368 
4 
736 
2 
1472 
1 
We then add all the
numbers in the lefthand column, which are opposite an odd number in the
righthand 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.
The game of Nim is
played with matches (or toothpicks for nonsmokers.)
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.
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.
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.
CHAPTER 2
Mathematicians like the rest of us make mistakes. But, being accustomed to checking, they usually detect their mistakes before it is too late.
All 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.
When we write (45)^{2}
we mean 45 times 45. More generally, if x
is any number, x^{2} means x times itself. We say “xsquared”,
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.)
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.
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.
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, 35´23, 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 9´9=81 gives 0´0 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: 995´995 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.)
Use the method of
casting out nines to check whether 74´38 =2,712.
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, 74´38 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.
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.
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 45´45 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} = 45´45, and this is approximately 50´40 =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.
We have already explained in the meaning of 10^{2} as the
product of 10 and itself, i.e. 100. Similarly 10^{3} is the product of
three 10’s, and so on. Thus
10^{2 }= 10 ×10= 100
10^{3}=
10×10×10 = 1,000
10^{4}=
10×10×10 ×10 = 10,000
10^{5}= 10×10
×10×10×10 = 100,000
10^{6}=
10×10×10×10×10 ×10 = 1,000,000
10^{7}=
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 10^{5
}is 5; that of^{ }10^{7} is 7. Note that 10^{5} is
1 followed by five 0s, 10^{7} is 1 followed by seven 0s and so on. It
seems reasonable to define 10^{1} to be 1 followed by one 0, i.e. 10,
and 10^{0 }as 1 followed by no 0s, i.e. 1. Thus10^{1 }= 10 and
10^{0 }= 1.
Multiplying these powers of 10 is easy: one simply adds the exponents. Thus 10^{7}×10^{5}
= 10^{7+5} =
10^{12}.
You will notice that
an advantage of this way of writing is that it is much easier to comprehend,
for instance, 10^{9} 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 8.87´10^{2}. The number 64789 is expressible
as 6.4789´10^{4}. 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 65×10^{3}, the answer, using our previous
formula, is easily seen to be 4225×10^{6}, which can also be expressed
as 4.225×10^{9}or, 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.
(995)^{2}. 995 is approximately 1´10^{3}, and so the square is
approximately
1´10^{3}´1´10^{3} = 1´10^{6}. If we look at the example in §2
we calculated (995)^{2} to be 990025, which is equal to 9.90025´10^{5. }This is very close to 10^{6}.
Positive powers of 10 are useful for large numbers.
Negative powers are used for small numbers. For instance, 10^{5} means
1 divided by 10^{5}, 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^{15}´10^{5}= 10^{25}.^{}
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.
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 5.678´10^{8} means that the number lies between 5.6775´10^{8 }and^{ }5.6784´10^{8}.
Taking 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 14´23, if we approximate by 10´20 = 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 14´23 by 10´20, 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%.
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.
(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.
Solution: 82´36 = 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 1´0 = 0, the same as the check digit for the answer. So our check does not indicate an error.
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.
Solution: To calculate 82´36 we replace 82 by 80 and 36 by 40, the
product is 3,200, which agrees reasonably well with 2,952.
Solution: This means 7.5 times 10^{4}, which is 1 followed by four 0s, i.e. 10,000. Furthermore the result lies between 7.45´10^{4} and 7.54´10^{4}.
2.5´3.1 = 7.75.
10^{4} ´10^{3} = 10^{7} and so the
product is
7.75´10^{7}
Solution: The volume of a sphere of radius r is 4p r^{3}/3^{. }The moon is approximately 1.6´10^{6} km. So the volume is 4p(1.6)^{3} ´10^{18}/3 = 17.157 ´ 10^{18 }= 1.7157´ 10^{19} km^{3}.
9. Negative
powers of 10. What is the meaning of 10^{3}?
10^{3} means 1 followed by 3 zeros. 10^{3}
means 1/10^{3} = 1/1000 = 0.001.
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.
The method
of checking modulo 11
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 82´36. 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.)
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 2.9´4.8 we take 3´5.]
The difference between our approximate answer and the correct answer, n´m, will be
(n + a)´ (m + b)  n´m = 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/m´100 + a/n´100. That is, the sum of the percentage errors
of each of the factors, which is what we were to prove.
[If we repeat the above argument using the example we have chosen of 2.9´4.8, we have the difference between our approximate answer of 3´4 = 12 and the correct answer 2.9´4.8 will be (2.9 + .1) ´(4.8 + .2) – 2.9´4.8 = 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 2.9´4.8})´100 = (.2/4.8)´100 + (.1/2.9)´100, that is the sum of the percentage errors in each of the factors]
Note
that 95 = 9´10 +
5. In general a number ending in 5 can be written 10n + 5. Thus
(10n + 5)^{2} = (10n + 5)
(10n + 5) = 100n^{2} + 50n + 50n + 25 = 100(n^{2} + n) + 25.
Since
(n^{2} + n) = n(n + 1), the result follows.
Chapter 3
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.
Computers’
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.
The 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 mark÷on the paper or delete a mark and then move one
position to the 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.
From 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
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. …..
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.
In this book we stick mainly to our own understanding, and do not rely
on authority, whether it be computers or famous professors.
Chapter 4
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 Fermitype 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”.
figure 1 Enrico Fermi (19011954).
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 10^{3} we mean
10 multiplied by itself three times, or 1000. Generally, for any whole number x,
10^{x} means ten multiplied by itself x times. For
example, 10^{7 }means 1 followed by seven 0s, i.e. 10,000,000, and 3.8´ 10^{7} means 3.8 multiplied by
10,000,000, i.e., 38,000,000.
The x in 10^{x}
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. 10^{4}´10^{5} = 10^{9}, and 10^{6}
divided by 10^{2} is 10^{(6  2)} = 10^{4}.
Assuming 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 onefourth 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.
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.
There are something
like 200 countries in the world, and Great Britain has some 60 million people,
i.e. 6´10^{7} 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
200´3´10^{7} = 6´10^{9}. This very crude calculation has given the correct result.
Enrico 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 8´5000 = 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.
The radius of the
earth, r, is about 6´10^{3} km so the surface area of the
earth is given approximately by the formula 4πr^{2} (the surface
area of a sphere of radius ) i.e.
4p ´(36 ´10^{6}) = 452´10^{6} km = 4.52´10^{8} km^{2}. However, since
70% of the earth’s surface is water, this leaves 30% of the area on dry land,
that is, approximately 1.5´10^{8}´10^{6} m^{2}.
Then if we allow 100 m^{2}
of space for each person, this comes to a maximum of 1.5´10^{12} people. It is interesting to compare this figure with
the actual number of people inhabiting the earth today, something like 6´10^{9}.
Suppose 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 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 tradeoff here, since increasing your prices may result in fewer customers.
It 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.
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.
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.
Solution: Paper is measured at weight per square meter. Suppose our paper weighs 80g per square meter. If the size of a page is 210´297 mm, i.e. approximately
0.2m´0.3m, then the area will be 0.06 square meters. Hence, the weight must be 80´0.06 or approximately 5 g. As a check we use a scale and found the weight to be 4 g.
Solution: The volume of a sphere of radius r is (4pr^{3)}/3.
Since the radius of the earth is approximately 6,000km, and since 4p/3
is approximately 4, the volume is roughly 4´ (6´10^{3})^{3 }= 4´216´10^{9}, or 864´10^{9} km^{3}. The weight of one cubic centimeter of water is 1 gram, so the weight of
one cubic meter of water is 10^{6} grams, or 10^{3} kg, which
means that one cubic kilometer weighs 10^{12} kilograms.
If we guess that earth
is four time as heavy as water, the weight of the earth must be approximately
864´10^{9}´4´10^{12} kilograms, or 3456´10^{21} kg or 3.456´10^{24} kg. Checking on the internet we
found a value of 5.9763´10 ^{24} kg. for the weight of the earth. Our guess is out
by a factor under 2, not too bad for a first approximation.
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.
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.
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
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 markup.
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.
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 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.
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 15´T 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 15´T to provide a payment of T pounds per year.
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 15´S/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 15´S/4. Assume that one works for 40 years. Thus each year one needs to save 15´S/160, i.e., about 9% of salary.
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 + r^{2}
+ … + r^{40}.
Next, we multiply this by r we get
rS = r + r^{2} + … + r^{40
}+ r^{41}.
The differences between rS and S are the 1 in
the expression for S and r^{41} 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 = r^{41} – 1.
Hence, S = (r^{41} – 1)/(r –
1), and since r = 1.05, we see that
S = (7.3921)/(1.051) = 6.392/.05
= 127.84.
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.
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.
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. .
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%.
Chapter 5
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.
A 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.
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
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.
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. .
In 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 reexpress this in terms of
percentages of the previous year’s results as in the following table:

This year/last year % 
Sales 
120 
Advertising 
200 
Postage &telephone 
116 
Wages 
108 
Directors fees 
100 
Consultant’s fees 
154 
Accountant’s fees 
100 
Profit 
141 
Working capital 
100 
figure 5. Accounts reexpressed 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.
We 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.
It 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 10^{7}, this means that on average 2.3×10^{2} 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.
Occasionally 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.
The problem of allowing legal abortions is one of considerable importance. There are a number of different views, the most usual being
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 tenyear 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.
If 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?
Solution:
The percentage increase is (5,000/30,000)×100 = 16.6%.
Solution
The percentage decrease is (5,000/30,000)×100 = 16.6%.
Solution:
The increase is ((5/100)×30,000 = £1,500.
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.
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%.
Solution:

Percentages with respect to Sales 
Sales 
100 
Advertising 
13 
Postage & telephone 
9 
Wages 
36 
Director´s fees 
18 
Consultant’s fees 
0.1 
Accountant’s fees 
0.48 
Profit 
23 
Working capital 
6 
figure 7. Calculating the percentages in Figure
4.
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 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.
Chapter 6
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.
Since 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 1/10^{n}. For example, 5X10^{3} = 5000, while 5X10^{3}
= 0.005. This has the added advantage that calculations involving products
become easier, since 10^{m}×10^{n} = 10^{m+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 = 10^{2}
cm, To have a reasonable understanding of the units of length, the width of a
hand is about 10 cm, a meter is the measurement from the ground to your belly
button, and the length of a small European car is about 4 meters. A threestory
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 30story 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
threefifths of an angstrom.
The following table
lists a few other approximations.
Finger width 
2 cm 
Width of thumb 
2.5 cm 
Height of step in a flight of stairs 
16 cm 
Length of a foot, or height of a head 
25 cm 
From the foot till the knee 
50 cm 
Width of kitchen units 
60 cm 
Foot to navel 
1 m 
Tall man or height of door 
2 m 
Length of small car 
4 m 
One story of a block of flats 
3 m 
Height of airplane flight 
5 – 12 km 
Height of TV satellite 
36,000 km 
The 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:

Mass 
A4 sheet of paper 
4 g 
Standard letter 
20 g 
Slice of bread 
40g 
An egg 
60g 
Banana or apple 
120 g 
Meal of two eggs, two slices of bread, and a
potato 
350 g 
A litre of milk 
1 kg 
A packed suitcase 
20 kg 
A man 
70 kg 
A 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.
The first step is to
convert mg to g:
1000 mg = 1 g, so 1mg = 1/1000g = 10^{3}g.
We simply take the mg
in 34mg and replace it by ×10^{3}g. Thus
34mg = 34 ×10^{3}g.
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^{3}g = 34 ×10^{3}
×10^{3}kg = 34 ×10^{6}kg.
Note that we have
simplified 10^{3} ×10^{3 }to 10^{6 }.
The 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 Cesium133 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) 
2,000 
Hours in a year 
8,760 
Time spanned by great grandfather,
grandfather and father 
100 years 
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.
A horse travels 56
kilometers in 3 hours. What is the speed of the horse?
Speed = distance/time
= 56km/3 h = 18.67 km/h. Here is a table of speeds.
Travel at 11 km/h 
3 m/s 
Travel at 108 km/h 
30 m/s 
Man running 100 meters in 10 seconds 
36 km/h 
Speed of sound 
340 m/s 
Speed of light 
3 ×10^{5} km/s 
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 10^{2} 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.
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.
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.
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.
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.
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.
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.
Acceleration is
defined as rate of change of speed. That is, if the speed at time t = t_{0}
is s_{0}, the acceleration necessary to achieve a speed of s_{1}
at time t_{1} is defined to be
(s_{1 }_{ }s_{0})/(t_{1}
 t_{0}).
That is, the
difference in speeds divided by the time.
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 /h^{2}.
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/s^{2}. 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
(15641642). 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.
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 (15641642) 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 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.
Force 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/s^{2}; that is 0.98
Newtons, or approximately one Newton. The symbol to denote a Newton is N.
In 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.
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
manhours 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 nonstop for an hour performs one kilowatt hour
of work. It is denoted by kWh,
A 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 righthand side of the equation must be equal to
the dimensions on the lefthand side of the equation.
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}
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}.
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.
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 bth power of the acceleration due to gravity, g.
That is, T should have the form
T = constant× l^{a}g^{b}^{ },^{}
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
righthand side of the formula are [L]^{a}[L]^{b}[T]^{2b}
= [L]^{a+b}[T]^{2b} . These must match with the dimensions of
the lefthand 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 × l^{½}g^{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= 2pl^{½}g^{½} = 2pÖ(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.
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×g^{a}t^{b}, where g is the acceleration due to gravity, in terms of units this becomes
[L] = [L]^{a}[T]^{2a}[T]^{b}.
Since the lefthand side of the equation has [L] to the power 1, a must also be 1. On the lefthand side [T] does not appear, but on the righthand 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 gt^{2}.
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 = mc^{2},
the product of the mass m and c^{2}, 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.
figure 2. Albert Einstein
(18791955), whose theoretical discoveries
contributed 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 lefthand 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.
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) = 3.6X10^{4} km h^{2}.
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.
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.
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.
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.
The force acting on a mass is defined by multiplying the mass M in kg by the acceleration a in m/s^{2}, 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/s^{2. }For example, a mass of 15 kg which when acted on by a force accelerates 40 m/s^{2} has a force of 15 ×40 = 600 N acting on it.
figure 3. Newton (1643–1727) was codiscoverer of the Calculus and
developed
a farreaching 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 m_{1}, m_{2}, 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 m_{1}m_{2}/d^{2}_{ }.
The value of G has been found experimentally to be 6.67 ×10^{11}. The units of G are Nm^{2}/kg^{2}.where N stands for Newton, m for meter and kg for kilogram.
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/ r^{2}
according to point 2 above. Since the force equals ma where a is the acceleration, we have the equation
ma = GmM/r^{2}.
If we cancel m from both sides, the result is
a = GM/r^{2}
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.
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.
The rate of work in Joules per second is called the power. A Joule per second is also known as a Watt.
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
70´9.81 = 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.
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/s^{2}. 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.
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.
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.
·
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.
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
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,
a)
Express J in terms of
calories
b)
Find the relationship
between kJ and kW hours
c)
Find kW in terms of horsepower.
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.
c) 1 horsepower = 745 W. Thus 1000 horsepower = 745 kW. Thus dividing by 745, 1 kW = 1000/745 horsepower = 1.342 horsepower.
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.
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.
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.
Solution: We have the formula: force =
GmM/r^{2}. 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
}.
CHAPTER 7
A knowledge of rightangled triangles with largest side 1
enables us to calculate the lengths of any other rightangled triangle
It 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 A¢B¢C¢ 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.
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 rightangle 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 threestorey 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.
A 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 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
Rightangled triangles appear often in practice, for instance in surveying. Given a rightangled 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 rightangled triangle is also called the hypotenuse.
We would have to produce an infinite table if we listed all rightangled triangles. We reduce our list by considering rightangled 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 rightangled 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.
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 dp/180. For instance, sin(0.180°) is 0.001p, or 0.00314, correct to 5 decimal places.
Now suppose we are given a right angledtriangle XYZ with Z a right angle, see Fig. 6. Then consider the rightangled 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 rightangled triangle XYZ with
hypotenuse
h, and a similar triangle ABC with hypotenuse 1
Example 1: A rightangled 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.
Example 2: A rightangled 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
rightangled 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 rightangled triangle with 15º angle
at B and hypotenuse 10 m.
One 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 = ½gt^{2},
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 = ½gt^{2}
From this formula we see on multiplying by 2 and dividing by g that
2s/g = t^{2}
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..
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.
Solution. For an angle of 0º it is not exactly clear how to define the sine. But if we take a rightangled 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.
Solution. The triangle in Fig. 8 has all three sides of length 1.
B
A C
D
figure 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 rightangled 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, AD^{2} + BD^{2} = AB^{2}. Hence, (1/2)^{2} +BD^{2} = 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.
Solution: Refer to Fig. 9.
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.
Solution: The distance is calculated from the formula
distance fallen = ½gt^{2}.
In this case, t = 3 and g, as we know, is 9.8 m/s^{2}. Thus, the distance is ½×9.8×9 = 44.1 m.
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 F_{1} and F_{2} acting at right angles to each other, as shown in Figure 10. In particular, the two forces are
F_{1} = F×sin(a) and F_{2} = F×cos(a).
figure 10. The force F
can be regarded as the sum of two forces at rightangles to one another
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 and this is the effective penetrating force. This explains why angling the tank’s armour helps reduce the shell’s impact.
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 rightangles, 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.
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.
DDddddddddddd
B
A
x´
D
x
Thus x¨ is proportional to x
and with this method we get a direct scaling of the object to be drawn.
O
Chapter 8
“The invention of logarithms saved
astronomers a lot of trouble and doubled their lives”  Laplace
figure 11. John Napier (15501617)
Multiplication is more time consuming than addition. To multiply two 8digit numbers, we need to perform 64 multiplications and several additions. On the other hand, to add two 8digit 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.
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 17digit 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. An example of a nonprime 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.
As 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.
Common 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 p.
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}.
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 (17071783), called e. It has retained this name to this day.
The 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.
If x is a very large number with n digits, a good approximation to Log(x) is (n1). 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.
Theory 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, a^{n} is
defined as the product of n copies of a and n is called
the exponent or power of a. For example, 2^{3} = 2×2×2 =
8. We define a^{1 }= 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, 10^{2}×10^{3}
= 10^{5}, that is, the product of two 10s times the product of three
10s is the product of five 10s. Also, (10^{2})^{3} = 10^{6}.
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
10^{y }= x.
For example, Log(100) = 2, since 10^{2} = 10×10 =100, Log(10) =1, since 10^{1}=10.
To define logarithms we will need to define 10^{y }for exponents y other than whole numbers.
Fractions as exponents. Let n be a positive integer and let x = 1/n. Then a^{x} is defined to be the nth root of a. Thus 4^{½} is that number whose square is 4, and, hence, 4^{½} = 2. If x = m/n, i.e., one positive integer divided by another, we define a^{x} to be the mth power of a^{(1/n)}. For instance,
2^{3/2} = (2^{½})^{3} = (1.4142)^{3} = 2.828.
So far we have defined a^{x} 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 a^{x} 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 a^{r} is an approximation to a^{x}. For example, the square root of 2 is approximately 1.414, which, is 1414/1000, so a^{√2} is approximately the 1414th power of the 1000th root of a.
Negative exponents. If the above law of exponents is to hold for both positive and negative numbers, then, a^{x}^{ }× a^{x} should be a^{x}^{x} = a^{0 }= 1. Therefore, we define a^{x} = 1/a^{x}. For example,
2^{3 }= 1/2^{3} = 1/8.
Definition of logarithms to the base a. If a^{x} = y, then x is defined to be the logarithm of y to the base a, and is written as log_{a}(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 e^{x} = y then x is the natural logarithm of y, written as x = ln(y).
Laws of logarithms
To see this let Log(x) = x´and Log(y) = y´, Then, xy = 10^{x´}10^{y´}= 10^{x´+y´} so, by definition,
x´+ y´= Log(xy), i.e., Log(x) + Log(y)
= Log(xy),
This follows from raising both sides of the
defining equation, y = 10^{Log(y)},
to the power x. Thus, y^{x}
= (10^{Log(y)})^{x} = 10^{xLog(y)}, i.e., xLog(y) is the exponent of 10 which gives y^{x}, so, by definition,
Log(y^{x}) = xLog(y).
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 righthand 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.
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.
Solution: By definition, x = e^{ln(x)}. Also, x
= 10^{Log(x)}, and since 10 = e^{ln(10)}, x
= (e^{ln(10)})^{Log(x)}. Hence, x = e^{ln(10)Log(x)},
so ln(x) = ln(10)Log(x).
Solution. Since 2×5 = 10, Log(2×5) = Log(10) = 1. Hence, Log(2) + Log(5) = 1, so Log(5) is approximately 0.7.
Solution. Since 2^{10} = 1024, 2^{10 }is approximately 10^{3}. Taking the 10^{th} root of both sides gives, 10^{0.3} = 2, so Log(2) = 0.3, approximately. Actually, to 4 decimal places, Log(2) = 0.3010.^
Kepler’s Laws
Astronomers 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 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 days 
Mercury 
0.387 
0.241 
Venus 
0.723 
0.615 
Earth 
1.000 
1.000 
Mars 
1.524 
1.881 
Jupiter 
5.203 
11.862 
Saturn 
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) 
3/2´ln(D) 
Mercury 
−0.949 
−1.423 
−1.424 
Venus 
−0.324 
−0.486 
−0.486 
Earth 
0.000 
0.000 
0.000 
Mars 
0.421 
0.632 
0.632 
Jupiter 
1.649 
2.473 
2.474 
Saturn 
2.257 
3.383 
3.386 
Thus we have Kepler’s law
3/2´ln(D) = ln(T)
This is the third of Kepler’s famous laws.
We can
simplify the lefthand side to ln(D^{3/2}) and so if we calculate
e to this power we get
D^{3/2} = T..
This is indeed a remarkable formula.
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 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. ,



Around 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.
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 × 10^{6}, so
Log(5,613,678)
= 6 + Log(5.613678).
Chapter 9
Algebra was one thing, and Geometry another. Descartes made them one.
figure 1. René Descartes (15961650)
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, viceversa, 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 (15961650). His methods have been expanded and improved by many other mathematicians, to make the impressive subject we have today.
figure 2. Coordinates
In Fig. 2 we have drawn two straight lines, OX and OY, which intersect in a point O, and are at rightangles. 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 xaxis and OY the yaxis, while the point O is called the origin. Any collection of points on the plane is called a graph.
S(3,0.75) R(2,1.5) P(2,1) Q(1,2)
figure 3. Examples of coordinates
The 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.
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 1^{st} 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
Example 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.
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.
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 carbon14 content.
All plants, animals and people absorb carbon12, normal carbon, and to a much less extent, carbon14, 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 carbon14 decays. In fact, carbon14 has a half life of 5,700 years, that is, onehalf 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 carbon14 there was originally, so the remaining carbon14 determines its age. The following graph illustrates the relationship between the percentage of the carbon14 of the original remaining to the age. For instance, if just 5 percent of the original carbon14 remains the specimen is 24,640 years old,
figure 10. Carbon14 dating.
Descartes then had this brilliant idea: He set up a way of associating equations with curves.
Consider the equation y + x^{2} = 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
figure 11. Graph of y + x^{2} = 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
equation
2x + 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 
y 
0 
2/3 
1 
0 
2 
2/3 
3 
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.
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 x^{2} + y^{2} = 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 x^{2}
+ y^{2} = 1.
Two 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,
figure 14. The intersection of a plane and a
cone
gives 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 wellformed 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.
figure 16. A parabola, the path of a cannon
ball.
As 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 
x^{2} + y^{2} = 1^{} 
Circle with centre at O and radius 1 
x^{2}/4
+ y^{2}/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, x^{2}, y, y^{2} and xy and numbers, the so called quadratic equations, for instance, 7x + 4x^{2} +8y  9y^{2} + 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.
The 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.
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 socalled 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 socalled 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.
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
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.
Solution. A straight line joining, for example, the points (0,3) and (1,8).
To 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.
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
x^{2 }+ y^{2} = 4 and y = x^{3 }– x.
In Fig. 18, the graphs of the circle x^{2 }+ y^{2} = 4 and the curve y = x^{3}x have been plotted.
figure 18. Common solutions to x^{2 }+ y^{2}
= 4 and y = x^{3}–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 xcoordinate and the ycoordinate 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 x^{2 }+ x^{2} = 4, that is, 2x^{2} = 4, and dividing both sides by 2 gives x^{2} = 2. Substituting y = x in the second equation gives x = x^{3} – x, or, 2x = x^{3}, and dividing both sides by x gives 2 = x^{2}, which agrees with what we found for the first equation. Thus, it is true that at the common solution, x = y, and x^{2} = 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.
Chapter 10
Back substitution is the essence of this ingenious method.
Consider 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, xby3.5 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?
An 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.
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.
We rewrite 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 xaxis 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.
The 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.
· 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%.
Mathematicians 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 nonzero 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 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 backsubstitution, and the whole procedure is called Gaussian elimination.
Examples 5 and 6 in §7, below, give further examples of this method.
These 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:
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 = 62 = 4 (4´)
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.
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.
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 viceversa.
Solution: We guess 10 because that way we get rid of the decimal. Then the lefthand side becomes 25 – 6 = 19, which is some way from 5. Guessing x = 4 gives a lefthand side which is 4. So perhaps x = 5, and this we see gives a lefthand side of 6.5. Next we try x = 4.5 and this gives a lefthand side of 5.25 which is quite close to the righthand side.
Solution: Adding 6 to both sides gives 2.5x = 11. Divide by 2.5 to get x = 11/2.5 = 4.4.
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 xaxis. 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 Xaxis, and this is approximately at x = 4.4.
Solution: 6 divided by 3 is 2 is another way of saying that 2´3 = 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 nonzero 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
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
x + y + z = 4 (9)
0x + y + 0z = 3 (10´´)
0x + 0y + 0z = 2 (11´´)
But this last equation is inconsistent, since the lefthand side is 0 and the righthand side is 2. Thus these equations do not have a solution.
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.
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 righthand side becomes 0. The lefthand side becomes using the
distributive law –1 × 1 + (1) × (1) and this is equal to the righthand 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.
Chapter
11
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 xaxis OX and the yaxis OY. We specify the position of a point by means of its coordinates, i.e. the distances from the xaxis and the yaxis.
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.
The 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 PolishSwedish mathematician, Richard Bonner.
We begin by choosing the usual coordinate system in the plane with xaxis OX and yaxis 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 yaxis 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 yaxis which meet the points of the set in more than one point. For instance, the yaxis 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.
figure 2. The set consisting of the points on
the circle
together 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 yaxis which passes through x on the xaxis. 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.
The 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.
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.) .
Having 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 yaxis 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.
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)}
A 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.
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 FrenchItalian 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)?
figure 6. Lagrange (1736–1813).
Lagrange’s solution was to construct a
formula of the form
f(x) = a(x  4)(x  6) +