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Mathematics for Electrical Engineering and Computing
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TLFeBOOK

Mathematics for Electrical Engineering and Computing
Mary Attenborough

AMSTERDAM BOSTON LONDON HEIDELBERG NEW YORK OXFORD PARIS SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO

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Newnes An imprint of Elsevier Linacre House, Jordan Hill, Oxford OX2 8DP 200 Wheeler Road, Burlington MA 01803 First published 2003 Copyright © 2003, Mary Attenborough. All rights reserved The right of Mary Attenborough to be identified as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988 No part of this publication may be reproduced in any material form (including photocopying or storing in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright holder except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London, England W1T 4LP. Applications for the copyright holder’s written permission to reproduce any part of this publication should be addressed to the publisher Permissions may be sought directly from Elsevier’s Science and Technology Rights Department in Oxford, UK: phone: (+44) (0) 1865 843830; fax: (+44) (0) 1865 853333; e-mail: permissions@elsevier.co.uk. You may also complete your request on-line via the Elsevier homepage (http://www.elsevier.com), by selecting ‘Customer Support’ and then ‘Obtaining Permissions’ British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloguing in Publication Data A catalogue record for this book is available from the Library of Congress ISBN 0 7506 5855 X
For information on all Newnes publications visit our website at www.newnespress.com
Typeset by Newgen Imaging Systems (P) Ltd, Chennai, India Printed and bound in Great Britain
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Contents

Preface

xi

Acknowledgements

xii

Part 1 Sets, functions, and calculus

1 Sets and functions

3

1.1 Introduction

3

1.2 Sets

4

1.3 Operations on sets

5

1.4 Relations and functions

7

1.5 Combining functions

17

1.6 Summary

23

1.7 Exercises

24

2 Functions and their graphs

26

2.1 Introduction

26

2.2 The straight line: y = mx + c

26

2.3 The quadratic function: y = ax2 + bx + c

32

2.4 The function y = 1/x

33

2.5 The functions y = ax

33

2.6 Graph sketching using simple

transformations

35

2.7 The modulus function, y = |x| or

y = abs(x)

41

2.8 Symmetry of functions and their graphs

42

2.9 Solving inequalities

43

2.10 Using graphs to find an expression for the function

from experimental data

50

2.11 Summary

54

2.12 Exercises

55

3 Problem solving and the art of the convincing

argument

57

3.1 Introduction

57

3.2 Describing a problem in mathematical

language

59

3.3 Propositions and predicates

61

3.4 Operations on propositions and predicates

62

3.5 Equivalence

64

3.6 Implication

67

3.7 Making sweeping statements

70

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vi Contents

3.8 Other applications of predicates 3.9 Summary 3.10 Exercises
4 Boolean algebra 4.1 Introduction 4.2 Algebra 4.3 Boolean algebras 4.4 Digital circuits 4.5 Summary 4.6 Exercises
5 Trigonometric functions and waves 5.1 Introduction 5.2 Trigonometric functions and radians 5.3 Graphs and important properties 5.4 Wave functions of time and distance 5.5 Trigonometric identities 5.6 Superposition 5.7 Inverse trigonometric functions 5.8 Solving the trigonometric equations sin x = a, cos x = a, tan x = a 5.9 Summary 5.10 Exercises
6 Differentiation 6.1 Introduction 6.2 The average rate of change and the gradient of a chord 6.3 The derivative function 6.4 Some common derivatives 6.5 Finding the derivative of combinations of functions 6.6 Applications of differentiation 6.7 Summary 6.9 Exercises
7 Integration 7.1 Introduction 7.2 Integration 7.3 Finding integrals 7.4 Applications of integration 7.5 The definite integral 7.6 The mean value and r.m.s. value 7.7 Numerical Methods of Integration 7.8 Summary 7.9 Exercises
8 The exponential function 8.1 Introduction 8.2 Exponential growth and decay 8.3 The exponential function y = et 8.4 The hyperbolic functions 8.5 More differentiation and integration 8.6 Summary 8.7 Exercises

72 73 74
76 76 76 77 81 86 86
88 88 88 91 97 103 107 109
110 111 113
116 116
117 118 120
122 128 130 131
132 132 132 133 145 147 155 156 159 160
162 162 162 166 173 180 186 187
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Contents vii

9 Vectors

188

9.1 Introduction

188

9.2 Vectors and vector quantities

189

9.3 Addition and subtraction of vectors

191

9.4 Magnitude and direction of a 2D vector – polar

co-ordinates

192

9.5 Application of vectors to represent waves

(phasors)

195

9.6 Multiplication of a vector by a scalar and unit

vectors

197

9.7 Basis vectors

198

9.8 Products of vectors

198

9.9 Vector equation of a line

202

9.10 Summary

203

9.12 Exercises

205

10 Complex numbers

206

10.1 Introduction

206

10.2 Phasor rotation by π/2

206

10.3 Complex numbers and operations

207

10.4 Solution of quadratic equations

212

10.5 Polar form of a complex number

215

10.6 Applications of complex numbers to AC linear

circuits

218

10.7 Circular motion

219

10.8 The importance of being exponential

226

10.9 Summary

232

10.10 Exercises

235

11 Maxima and minima and sketching functions

237

11.1 Introduction

237

11.2 Stationary points, local maxima and

minima

237

11.3 Graph sketching by analysing the function

behaviour

244

11.4 Summary

251

11.5 Exercises

252

12 Sequences and series

254

12.1 Introduction

254

12.2 Sequences and series definitions

254

12.3 Arithmetic progression

259

12.4 Geometric progression

262

12.5 Pascal’s triangle and the binomial series

267

12.6 Power series

272

12.7 Limits and convergence

282

12.8 Newton–Raphson method for solving

equations

283

12.9 Summary

287

12.10 Exercises

289

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viii Contents

Part 2 Systems

13 Systems of linear equations, matrices, and

determinants

295

13.1 Introduction

295

13.2 Matrices

295

13.3 Transformations

306

13.4 Systems of equations

314

13.5 Gauss elimination

324

13.6 The inverse and determinant of a 3 × 3

matrix

330

13.7 Eigenvectors and eigenvalues

335

13.8 Least squares data fitting

338

13.9 Summary

342

13.10 Exercises

343

14 Differential equations and difference equations

346

14.1 Introduction

346

14.2 Modelling simple systems

347

14.3 Ordinary differential equations

352

14.4 Solving first-order LTI systems

358

14.5 Solution of a second-order LTI systems

363

14.6 Solving systems of differential equations

372

14.7 Difference equations

376

14.8 Summary

378

14.9 Exercises

380

15 Laplace and z transforms

382

15.1 Introduction

382

15.2 The Laplace transform – definition

382

15.3 The unit step function and the (impulse) delta

function

384

15.4 Laplace transforms of simple functions and

properties of the transform

386

15.5 Solving linear differential equations with constant

coefficients

394

15.6 Laplace transforms and systems theory

397

15.7 z transforms

403

15.8 Solving linear difference equations with constant

coefficients using z transforms

408

15.9 z transforms and systems theory

411

15.10 Summary

414

15.11 Exercises

415

16 Fourier series

418

16.1 Introduction

418

16.2 Periodic Functions

418

16.3 Sine and cosine series

419

16.4 Fourier series of symmetric periodic

functions

424

16.5 Amplitude and phase representation of a Fourier

series

426

16.6 Fourier series in complex form

428

16.7 Summary

430

16.8 Exercises

431

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Contents ix

Part 3 Functions of more than one variable

17 Functions of more than one variable

435

17.1 Introduction

435

17.2 Functions of two variables – surfaces

435

17.3 Partial differentiation

436

17.4 Changing variables – the chain rule

438

17.5 The total derivative along a path

440

17.6 Higher-order partial derivatives

443

17.7 Summary

444

17.8 Exercises

445

18 Vector calculus

446

18.1 Introduction

446

18.2 The gradient of a scalar field

446

18.3 Differentiating vector fields

449

18.4 The scalar line integral

451

18.5 Surface integrals

454

18.6 Summary

456

18.7 Exercises

457

Part 4 Graph and language theory

19 Graph theory

461

19.1 Introduction

461

19.2 Definitions

461

19.3 Matrix representation of a graph

465

19.4 Trees

465

19.5 The shortest path problem

468

19.6 Networks and maximum flow

471

19.7 State transition diagrams

474

19.8 Summary

476

19.9 Exercises

477

20 Language theory

479

20.1 Introduction

479

20.2 Languages and grammars

480

20.3 Derivations and derivation trees

483

20.4 Extended Backus-Naur Form (EBNF)

485

20.5 Extensible markup language (XML)

487

20.6 Summary

489

20.7 Exercises

489

Part 5 Probability and statistics

21 Probability and statistics

493

21.1 Introduction

493

21.2 Population and sample, representation of data, mean,

variance and standard deviation

494

21.3 Random systems and probability

501

21.4 Addition law of probability

505

21.5 Repeated trials, outcomes, and

probabilities

508

21.6 Repeated trials and probability trees

508

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21.7 Conditional probability and probability

trees

511

21.8 Application of the probability laws to the probability

of failure of an electrical circuit

514

21.9 Statistical modelling

516

21.10 The normal distribution

517

21.11 The exponential distribution

521

21.12 The binomial distribution

524

21.13 The Poisson distribution

526

21.14 Summary

528

21.15 Exercises

531

Answers to exercises

533

Index

542

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Preface
This book is based on my notes from lectures to students of electrical, electronic, and computer engineering at South Bank University. It presents a first year degree/diploma course in engineering mathematics with an emphasis on important concepts, such as algebraic structure, symmetries, linearity, and inverse problems, clearly presented in an accessible style. It encompasses the requirements, not only of students with a good maths grounding, but also of those who, with enthusiasm and motivation, can make up the necessary knowledge. Engineering applications are integrated at each opportunity. Situations where a computer should be used to perform calculations are indicated and ‘hand’ calculations are encouraged only in order to illustrate methods and important special cases. Algorithmic procedures are discussed with reference to their efficiency and convergence, with a presentation appropriate to someone new to computational methods.
Developments in the fields of engineering, particularly the extensive use of computers and microprocessors, have changed the necessary subject emphasis within mathematics. This has meant incorporating areas such as Boolean algebra, graph and language theory, and logic into the content. A particular area of interest is digital signal processing, with applications as diverse as medical, control and structural engineering, non-destructive testing, and geophysics. An important consideration when writing this book was to give more prominence to the treatment of discrete functions (sequences), solutions of difference equations and z transforms, and also to contextualize the mathematics within a systems approach to engineering problems.
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Acknowledgements
I should like to thank my former colleagues in the School of Electrical, Electronic and Computer Engineering at South Bank University who supported and encouraged me with my attempts to re-think approaches to the teaching of engineering mathematics.
I should like to thank all the reviewers for their comments and the editorial and production staff at Elsevier Science.
Many friends have helped out along the way, by discussing ideas or reading chapters. Above all Gabrielle Sinnadurai who checked the original manuscript of Engineering Mathematics Exposed, wrote the major part of the solutions manual and came to the rescue again by reading some of the new material in this publication. My partner Michael has given unstinting support throughout and without him I would never have found the energy.
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Part 1 Sets, functions,
and calculus
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1

Sets and functions

1.1 Introduction

Finding relationships between quantities is of central importance in engineering. For instance, we know that given a simple circuit with a 1000 resistance then the relationship between current and voltage is given by Ohm’s law, I = V /1000. For any value of the voltage V we can give an associated value of I . This relationship means that I is a function of V . From this simple idea there are many other questions that need clarifying, some of which are:
1. Are all values of V permitted? For instance, a very high value of the voltage could change the nature of the material in the resistor and the expression would no longer hold.
2. Supposing the voltage V is the equivalent voltage found from considering a larger network. Then V is itself a function of other voltage values in the network (see Figure 1.1). How can we combine the functions to get the relationship between this current we are interested in and the actual voltages in the network?
3. Supposing we know the voltage in the circuit and would like to know the associated current. Given the function that defines how current depends on the voltage can we find a function that defines how the voltage depends on the current? In the case where I = V /1000, it is clear that V = 1000I . This is called the inverse function.
Another reason exists for better understanding of the nature of functions. In Chapters 5 and 6, we shall study differentiation and integration. This looks at the way that functions change. A good understanding of functions and how to combine them will help considerably in those chapters.
The values that are permitted as inputs to a function are grouped together. A collection of objects is called a set. The idea of a set is very simple, but studying sets can help not only in understanding functions but also help to understand the properties of logic circuits, as discussed in Chapter 10.

Figure 1.1 The voltage V is an equivalent voltage found by considering the combined effect of circuit elements in the rest of the network.

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4 Sets and functions
1.2 Sets
Figure 1.2 A Venn diagram of the sets E = {a, b, c, d, e, f, g}, A = {a, b, c}, and B = {d, e}.

A set is a collection of objects, called elements, in which the order is not important and an object cannot appear twice in the same set.
Example 1.1 Explicit definitions of sets, that is, where each element is listed, are:
A = {a, b, c} B = {3, 4, 6, 7, 8, 9} C = {Linda, Raka, Sue, Joe, Nigel, Mary}
a ∈ A means ‘a is an element of A’ or ‘a belongs to A’; therefore in the above examples:
3∈B Linda ∈ C
The universal set is the set of all objects we are interested in and will depend on the problem under consideration. It is represented by E .
The empty set (or null set) is the set with no elements. It is represented by ∅ or { }.
Sets can be represented diagrammatically – generally as circular shapes. The universal set is represented as a rectangle. These are called Venn diagrams.
Example 1.2 E = {a, b, c, d, e, f, g}, A = {a, b, c}, B = {d, e}
This can be shown as in Figure 1.2.
We shall mainly be concerned with sets of numbers as these are more often used as inputs to functions.
Some important sets of numbers are (where ‘. . .’ means continue in the same manner):
The set of natural numbers N = {1, 2, 3, 4, 5, . . .} The set of integers Z = {. . . −3, −2, −1, 0, 1, 2, 3 . . .} The set of rationals (which includes fractional numbers) Q The set of reals (all the numbers necessary to represent points on a line) R Sets can also be defined using some rule, instead of explicitly.
Example 1.3 Define the set A explicitly where E = N and A = {x | x < 3}. Solution The A = {x | x < 3} is read as ‘A is the set of elements x, such that x is less than 3’. Therefore, as the universal set is the set of natural numbers, A = {1, 2}
Example 1.4 E = days of the week and A = {x | x is after Thursday and before Sunday}. Then A = {Friday, Saturday}.
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Figure 1.3 A Venn diagram of a proper subset of B: A ⊂ B.

Sets and functions 5
Subsets
We may wish to refer to only a part of some set. This is said to be a subset of the original set.
A ⊆ B is read as ‘A is a subset of B’ and it means that every element of A is an element of B.
Example 1.5
E =N A = {1, 2, 3}, B = {1, 2, 3, 4, 5}
Then A ⊆ B Note the following points: All sets must be subsets of the universal set, that is, A ⊆ E and B⊆E A set is a subset of itself, that is, A ⊆ A If A ⊆ B and B ⊆ A, then A = B
Proper subsets
A ⊂ B is read as ‘A is a proper subset of B’ and means that A is a subset of B but A is not equal to B. Hence, A ⊂ B and simultaneously B ⊂ A are impossible.
A proper subset can be shown on a Venn diagram as in Figure 1.3.

1.3 Operations on sets
Figure 1.4 The shaded area is the complement, A , of the set A.

In Chapter 1 of the background Mathematics notes available on the companion website for this book, we study the rules obeyed by numbers when using operations like negation, multiplication, and addition. Sets can be combined in various ways using set operations. Sets and their operations form a Boolean Algebra which we look at in greater detail in Chapter 4, particularly its application to digital design. The most important set operations are as given in this section.
Complement
A¯ or A represents the complement of the set A. The complement of A is the set of everything in the universal set which is not in A, this is pictured in Figure 1.4.
Example 1.6 E =N A = {x | x > 5}
then A = {1, 2, 3, 4, 5}
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6 Sets and functions
Figure 1.5 A = {x |x < 5} and A = {x |x 5}.

Figure 1.6 The shaded area represents the intersection of A and B.

Figure 1.7 The intersection of two sets {1, 2, 4} ∩ {1, 5, 6} = {1}.

Figure 1.8 The
intersection of two sets: {a, b, c, d, e} ∩ {a, b, c, d, e, f, g, h, i, j} = {a, b, c, d, e}.

Figure 1.9 The intersection of the two sets: {−3, −2, −1} ∩ {1, 2} = ∅, the empty set, as they have no elements in common.

Figure 1.10 Disjoint sets A and B.

Example 1.7 The universal set is the set of real numbers represented by a real number line.
If A is the set of numbers less than 5, A = {x | x < 5} then A is the set of numbers greater than or equal to 5. A = {x | x 5}. These sets are shown in Figure 1.5.
Intersection
A ∩ B represents the intersection of the sets A and B. The intersection contains those elements that are in A and also in B, this can be represented as in Figure 1.6 and examples are given in Figures 1.7–1.10.
Note the following important points:
If A ⊆ B then A ∩ B = A. This is the situation in the example given in Figure 1.8.
If A and B have no elements in common then A ∩ B = ∅ and they are called disjoint. This is the situation given in the example in Figure 1.9. Two sets which are known to be disjoint can be shown on the Venn diagram as in Figure 1.10.
Union
A ∪ B represents the union of A and B, that is, the set containing elements which are in A or B or in both A and B. On a Venn diagram, the union can be shown as in Figure 1.11 and examples are given in Figures 1.12–1.15.
Note the following important points:
If A ⊆ B, then A ∪ B = B. This is the situation in the example given in Figure 1.13.
The union of any set with its complement gives the universal set, that is, A ∪ A = E , the universal set. This is pictured in Figure 1.15.
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Sets and functions 7

Figure 1.11 The shaded area represents to union of sets A and B.

Figure 1.12 The union of two sets: {1, 2, 4} ∪ {1, 5, 6} = {1, 2, 4, 5, 6}.

Figure 1.13 The
union of two sets: {a, b, c, d, e} ∪ {a, b, c, d, e, f, g, h, i, j} = {a, b, c, d, e, f, g, h, i, j}.

Figure 1.14 The
union of the two sets: {−3, −2, −1} ∪ {1, 2} = {−3, −2, −1, 1, 2}.

Figure 1.15 The shaded area represents the union of a set with its complement giving the universal set.
Figure 1.16 S is the set of students in a survey and V is the set of people who own a video. The numbers in the sets give the cardinality of the sets, n(S) = 100, n(S ∪ V) = 794, n(V) = 720, n(S ∩ V) = x .

Cardinality of a finite set
The number of elements in a set is called the cardinality of the set and is written as n(A) or |A|.
Example 1.8
n(∅) = 0, n({2}) = 1, n({a, b}) = 2
For finite sets, the cardinality must be a natural number.
Example 1.9 In a survey, 100 people were students and 720 owned a video recorder; 794 people owned a video recorder or were students. How many students owned a video recorder?
E = {x | x is a person included in the survey}
Setting S = {x | x is a student} and V = {x | x owns a video recorder}, we can solve this problem using a Venn diagram as in Figure 1.16.
x is the number of students who own a video recorder. From the diagram we get
100 − x + x + 720 − x = 794 ⇔ 820 − x = 794 ⇔ x = 26
Therefore, 26 students own a video recorder.

1.4 Relations and functions

Relations
A relation is a way of pairing up members of two sets. This is just like the idea of family relations. For instance, a child can be paired with its mother, brothers can be paired with sisters, etc. A relation is such that it may not always be possible to find a suitable partner for each element in the first set whereas sometimes there will be more than one. For instance, if we try to pair every boy with his sister there will be some boys who have no sisters and some boys who have several. This is pictured in Figure 1.17.
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8 Sets and functions

Figure 1.17 The relation boy → sister. Some boys have more than one sister and some have none at all.
Figure 1.18 An arrow diagram of the function y = 1/x .

Functions
Functions are relations where the pairing is always possible. Functions are like mathematical machines. For each input value there is always exactly one output value.
Calculators output function values. For instance, input 2 into a calculator, press 1/x and the calculator will display the number 0.5. The output value is called the image of the input value. The set of input values is called the domain and the set containing all the images is called the codomain.
The function y = 1/x is displayed in Figure 1.18 using arrows to link input values with output values.
Functions can be represented by letters. If the function of the above example is given the letter f to represent it then we can write

f

:

x

→

1 x

This can be read as ‘f is the function which when input a value for x gives the output value 1/x’ . Another way of giving the same information is:

f (x)

=

1 x

f (x) represents the image of x under the function f and is read as ‘f of x’. It does not mean the same as f times x.
f (x) = 1/x means ‘the image of x under the function f is given by 1/x’ but is usually read as ‘f of x equals 1/x’.
Even more simply, we usually use the letter y to represent the output value, the image, and x to represent the input value. The function is therefore summed up by y = 1/x.
x is a variable because it can take any value from the set of values in the domain. y is also a variable but its value is fixed once x is known. So x is called the independent variable and y is called the dependent
variable. The letters used to define a function are not important. y = 1/x is the
same as z = 1/t is the same as p = 1/q provided that the same input values (for x, t, or q) are allowed in each case.
More examples of functions are given in arrow diagrams in
Figures 1.19(a) and 1.20(a). Functions are more usually drawn using
a graph, rather than by using an arrow diagram. To get the graph the
codomain is moved to be at right angles to the domain and input and output values are marked by a point at the position (x, y). Graphs are
given in Figures 1.19(b) and 1.20(b).

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Sets and functions 9
Continuous functions and discrete functions applied to signals
Functions of particular interest to engineers are either functions of a real number or functions of an integer. The function given in Figure 1.19 is an example of a real function and the function given in Figure 1.20 is an example of a function of an integer, also called a discrete function.
Often, we are concerned with functions of time. A variable voltage source can be described by giving the voltage as it depends on time, as also can the current. Other examples are: the position of a moving robot arm, the extension or compression of car shock absorbers and the heat emission of a thermostatically controlled heating system. A voltage or current varying with time can be used to control instrumentation or to convey information. For this reason it is called a signal. Telecommunication signals may be radio waves or voltages along a transmission line or light signals along an optical fibre.
Time, t, can be represented by a real number, usually non-negative. Time is usually taken to be positive because it is measured from some reference instant, for example, when a circuit switch is closed. If time is used to describe relative events then it can make sense to refer to negative time. If lightning is seen 1 s before a thunderclap is heard then this can be described by saying the lightning happened at −1 s or alternatively that the thunderclap was heard at 1 s. In the two cases, the time origin has been chosen differently. If time is taken to be continuous and represented by a real variable then functions of time will be continuous or piecewise continuous. Examples of graphs of such functions are given in Figure 1.21.

Figure 1.19 The function y = 2x + 1 where x can take any real value (any number on the number line). (a) is the arrow diagram and (b) is the graph.
Figure 1.20 The function q = t − 3 where t can take any integer value (a) is the arrow diagram and (b) is the graph.

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10 Sets and functions
Figure 1.21 Continuous and piecewise functions where time is represented by a real number > 0. (a) A ramp function; (b) a wave (c) a square wave. (a) and (b) are continuous, while (c) is piecewise continuous.
A continuous function is one whose graph can be drawn without taking your pen off the paper. A piecewise continuous function has continuous bits with a limited number of jumps. In Figure 1.21, (a) and (b) are continuous functions and (c) is a piecewise continuous function. If we have a digital signal, then its values are only known at discrete moments of time. Digital signals can be obtained by using an analog to digital (A/D) convertor on an originally continuous signal. Digital signals are represented by discrete functions as in Figure 1.22(a)–(c)
A digital signal has a sampling interval, T , which is the length of time between successive values. A digital functions is represented by a discrete function. For example, in Figure 1.22(a) the digital ramp can be represented by the numbers 0, 1, 2, 3, 4, 5, . . . If the sample interval T is different from 1 then the values would be 0, T, 2T, 3T, 4T, 5T, . . . This is a discrete function also called a sequence. It can be represented by the expression f (t) = t, where t = 0, 1, 2, 3, 4, 5, 6, . . . or using the sampling interval, T , g(n) = nT , where n = 0, 1, 2, 3, 4, 5, 6, . . .
Yet another common way of representing a sequence is by using a subscript on the letter representing the image, giving fn = n, where n = 0, 1, 2, 3, 4, 5, . . . or, using the letter a for the image values, an = n, where n = 0, 1, 2, 3, 4, 5, . . . Substituting some values for n into the above gives a0 = 0, a1 = 1, a2 = 2, a3 = 3, . . .
As a sequence is a function of the natural numbers and zero (or if negative input values are allowed, the integers) there is no need to specify
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Sets and functions 11

Figure 1.22 Examples of discrete functions. (a) A digital ramp; (b) a digital wave; (c) a digital square wave.

the input values and it is possible merely to list the output values in order. Hence the ramp function can be expressed by 0, 1, 2, 3, 4, 5, 6, . . .
Time sequences are often referred to as ‘series’. This terminology is not usual in mathematics books, however, as the description ‘series’ is reserved for describing the sum of a sequence. Sequences and series are dealt with in more detail in Chapter 18.

Example 1.10 Plot the following analog signals over the values of t given (t real):

(a) x = t3 t 0

(b)



0

t3

y = t2− 3

3<t t >5

5

(c)

z

=

1 t2

t >0

Solution In each case, choose some values of t and calculate the function values at those points. Plot the points and join them.

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12 Sets and functions

(a) t

0 0.5 1 1.5 2 2.5 3 3.5

x = t3 0 0.125 1 3.375 8 15.625 27 42.875

These values are plotted in Figure 1.23(a). (b)
t 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 y 0 0 0 0 0 0.5 1 1.5 2 2 2 2 2

y=0

y =t−3

y=2

These values are plotted in Figure 1.23(b).
(c)
t 0.0001 0.001 0.01 0.1 1 10 100 1000 10000 z 108 106 104 100 1 0.01 10−4 10−6 10−8

These values are plotted in Figure 1.23(c).

Figure 1.23 The

analog signals described in

Example 1.10.

(a) x = t3 t 0

0

t3

(b) y = t2− 3

3<t t >5

5

(c) z = 1/t 2 t > 0

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Sets and functions 13

Example 1.11 Plot the following discrete signals over the values of t given (t an integer):

(a)

x

=

t

1 −

1

t >2

(b)



0

t4

y = −1/0t.1−50.25

4 < t < 10 t 10

(c) z = 4t − 2 t > 0

Solution In each case, choose successive values of t and calculate the function values at those points. Mark the points with a dot.
(a)

t 2 3 4 5 6 7 8 9 10 x 1 0.5 0.33 0.25 0.2 0.17 0.14 0.13 0.11
These values are plotted in Figure 1.24(a). (b)

t 34

5 6 7 8 9 10 11 12

y 0 0 −0.05 −0.08 −0.11 −0.12 −0.14 −0.15 −0.15 −0.15

y=0

y

=

1 t

−

0.25

These values are plotted in Figure 1.24(b).

(c)

y = −0.15

t 12 3 4 5 6 7 8 z 2 6 10 14 18 22 26 30
These values are plotted in Figure 1.24(c).

Undefined function values
Some functions have ‘undefined values’, that is, numbers that cannot be input into them successfully. For instance input 0 on a calculator and try getting the value of 1/x. The calculator complains (usually displaying ‘-E-’) indicating that an error has occurred. The reason that this is an error is that we are trying to find the value of 1/0 that is 1 divided by 0. Look at Chapter 1 of the Background Mathematics Notes, given on the accompanying website for this book, for a discussion about why division by 0 is not defined. The number 0 cannot be included in the domain of the function f (x) = 1/x. This can be expressed by saying
f (x) = 1/x, where x ∈ R and x = 0
which is read as ‘f of x equals 1/x, where x is a real number not equal to 0’.
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14 Sets and functions

Figure 1.24 The digital signals described in Example 1.11.

1

0

t4

(a) x = t − 1

t >2

(b) y = 1−/0t.1−50.25

4 < t < 10 t 10

(c) z = 4t − 2

t >0

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Sets and functions 15
Often, we assume that we are considering functions of a real variable and only need to indicate the values that are not allowed as inputs for the function. So we may write

f (x) = 1/x where x = 0

Things to look out for as values that are not allowed as function inputs are :
1. Numbers that would lead to an attempt to divide by zero 2. Numbers that would lead to negative square roots 3. Numbers that would lead to negative inputs to a logarithm.
Examples 1.12(a) and (b) require solutions to inequalities which we shall discuss in greater detail in Chapter 2. Here, we shall only look at simple examples and use the same rules as used for solving equations. We can find equivalent inequalities by doing the same thing to both sides, with the extra rule that, for the moment, we avoid multiplication or division by a negative number.

Example 1.12 Find the values that cannot be input to the following functions, where the independent variable (x or r) is real:
√ (a) y = 3 x − 2 + 5

(b) y = 3 log10(2 − 4x)

(c)

R

=

r + 1000 1000(r − 2)

Solution √
(a) y = 3 x − 2 + 5

Here x − 2 cannot be negative as we need to take the square root of it.

x−2 0⇔x 2

therefore, the function is √
y = 3 x − 2 + 5 where x 2

(b) y = 3 log10(2 − 4x). Here 2 − 4x cannot be 0 or negative else we could not take the logarithm.

2 − 4x > 0 ⇔ 2 > 4x ⇔ 2/4 > x

or

equivalently,

x

<

1 2

.

So

the

function

is

y = 3 log10(2 − 4x) where x < 0.5

(c)

R

=

r + 1000 1000(r − 2)

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16 Sets and functions

Here 1000(r − 2) cannot be 0, else we would be trying to divide by 0. Solve the equation for the values that r cannot take

1000(r − 2) = 0 r −2 = 0 r =2

The function is

R

=

r + 1000 1000(r − 2)

where r = 2

Example 1.13 Find the values that can be input to the following discrete functions where the independent variable is an integer

(a)

y

=

k

1 −4

where k ∈ Z

(b) f (k) =

1

where k ∈ Z

(k − 3)(k − 2.2)

(c) an = n2 where n ∈ Z

Solution

(a)

y

=

k

1 −4

Here k − 4 cannot be 0 else there would be an attempt to divide by 0. We get k − 4 = 0 when k = 4 so the function is:

y

=

k

1 −

4

where k = 4 and k ∈ Z

(b)

f (k)

=

(k

−

1 3)(k

−

2.2)

where k ∈ Z

Solve for (k − 3)(k − 2.2) = 0 giving k = 3 or k = 2.2. As 2.2 is not an integer then there is not need to specifically exclude it from the function input values, so the function is

f (k)

=

(k

−

1 3)(k

−

2.2)

(c) an = n2, n ∈ Z

where k = 3 and k ∈ Z

Here there are no problems with the function as any integer can be squared. There are no excluded values from the input of the function.

Using a recurrence relation to define a discrete function
Values in a discrete function can also be described in terms of its values for preceeding integers.
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Sets and functions 17

Example 1.14 Find a table of values for the function defined by the recurrence relation:

f (n) = f (n − 1) + 2

(1.1)

where f (0) = 0.
Solution Assuming that the function is defined for n = 0, 1, 2, . . . then we can take successive values of n and find the values taken by the function. n = 0 gives f (0) = 0 as given.
Substituting n = 1 into Equation (1.1) gives

f (1) = f (1 − 1) + 2 ⇔ f (1) = f (0) + 2 = 0 + 2 = 2 (using f (0) = 0)

hence, f (1) = 2. Substituting n = 2 into Equation (1.1) gives

f (2) = f (2 − 1) + 2 ⇔ f (2) = f (1) + 2 ⇔ f (2) = f (1) + 2 = 2 + 2 = 4 (using f (1) = 2)

hence, f (2) = 4. Substituting n = 3 into Equation (1.1) gives

f (3) = f (3 − 1) + 2 ⇔ f (3) = f (2) + 2 = 4 + 2 (using f (2) = 4)

hence, f (3) = 6. Continuing in the same manner gives the following table:

n 0 1 2 3 4 5 6 7 8 9 10 · · · n · · · f 0 2 4 6 8 10 12 14 16 18 20 . . . 2n · · ·

Notice we have filled in the general term f (n) = 2n. This was found in this case by simple guess work.

1.5 Combining functions

The sum, difference, product, and quotient of two functions, f and g

Two functions with R as their domain and codomain can be combined using arithmetic operations. We can define the sum of f and g by

(f + g) : x → f (x) + g(x)

The other operations are defined as follows:

(f − g) : x → f (x) − g(x) difference,

(f × g) : x → f (x) × g(x) product,

(f /g)

:

x

→

f (x) g(x)

quotient.

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18 Sets and functions
Figure 1.25 The function a : kilograms → money used in Example 1.16.

Example 1.15 Find the sum, difference, product, and quotient of the functions:
f : x → x2 and g : x → x6

Solution

(f + g) : x → x2 + x6

(f − g) : x → x2 − x6

(f × g) : x → x2 × x6 = x8

(f /g) :

x

→

x2 x6

= x−4

The specification of the domain of the quotient is not straightforward. This is because of the difficulty which occurs when g(x) = 0. When g(x) = 0 the quotient function is undefined and we must remove such elements from its domain. The domain of f /g is R with the values where g(x) = 0 omitted.

Composition of functions
This method of combining functions is fundamentally different from the arithmetical combinations of the previous section. The composition of two functions is the action of performing one function followed by the other, that is, a function of a function.
Example 1.16 A post office worker has a scale expressed in kilograms which gives the cost of a parcel depending on its weight. He also has an approximate formula for conversion from pounds (lbs) to kilograms. He wishes to find out the cost of a parcel which weighs 3 lb.
The two functions involved are:
a : kilograms → money and c : lbs → kilograms
a is defined by Figure 1.25 and the function c is given by
c : x → x/2.2
Solution The composition ‘a ◦ c’ will be a function from lbs to money. Hence, 3 lb after the function c gives 1.364 and 1.364 after the function
a gives e1.90 and therefore
(a ◦ c)(3) = e1.90.

Example 1.17 Supposing f (x) = 2x + 1 and g(x) = x2, then we can combine the functions in two ways.
1. A composite function can be formed by performing f first and then g, that is, g ◦ f . To describe this function, we want to find what happens
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Sets and functions 19 to x under the function g ◦ f . Another way of saying that is we need to find g(f (x)). To do this call f (x) a new letter, say y.
y = f (x) = 2x + 1
Rewrite g as a function of y g(y) = y2
Now substitute y = 2x + 1 giving g(2x + 1) = (2x + 1)2
Hence, g(f (x)) = (2x + 1)2 (g ◦ f )(x) = (2x + 1)2.
2. A composite function can be formed by performing g first and then f , that is, f ◦ g. To describe this function, we want to find what happens to x under the function f ◦ g. Another way of saying that is we need to find f (g(x)). To do this call g(x) a new letter, say y. y = g(x) = x2
Rewrite f as a function of y
f (y) = 2y + 1
Now substitute y = x2 giving f (x2) = 2x2 + 1
Hence, f (g(x)) = 2x2 + 1 (f ◦ g)(x) = 2x2 + 1.

Example 1.18 Supposing u(t) = 1/(t − 2) and v(t) = 3 − t then, again, we can combine the functions in two ways.
1. A composite function can be formed by performing u first and then v, that is, v ◦ u. To describe this function, we want to find what happens to t under the function v ◦ u. Another way of saying that is we need to find v(u(t)). To do this call u(t) a new letter, say y.

y

=

u(t )

=

t

1 −

2

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20 Sets and functions

Rewrite v as a function of y

v(y) = 3 − y

Now substitute y = 1/(t − 2) giving

v

1 t −2

=

3

−

t

1 −

2

=

3(t

− 2) − t −2

1

(rewriting the expression over a common denominator)

= 3t − 6 − 1 = 3t − 7

t −2

t −2

Hence,

v(u(t ))

=

3t − 7 t −2

(v

◦

u)(t )

=

3t − 7 t −2

2. A composite function can be formed by performing v first and then u, that is u ◦ v. To describe this function, we want to find what happens to t under the function u ◦ v. Another way of saying that is we need to find u(v(t)). To find this call v(t) a new letter, say y.

y = v(t) = 3 − t

Rewrite u as a function of y

u(y)

=

y

1 −2

Now substitute y = 3 − t giving

v(3

−

t)

=

(3

1 − t)

−

2

=

1

1 −

t

Hence,

u(v(t ))

=

1

1 −

t

(u

◦

v)(t )

=

1

1 −

t

Decomposing functions
In order to calculate the value of a function, either by hand or using a calculator, we need to understand how it decomposes. That is we need to understand to order of the operations in the function expression
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Figure 1.26 A function pictured as a machine represented by a box. x represents the input value, any value of the domain, y represents the output, the image of x under the function.
Figure 1.27 The function y = (2x + 1)3 decomposed into its composite operations.

Sets and functions 21
Example 1.19 Calculate y = (2x + 1)3 when x = 2
Solution Remember the order of operations discussed in Chapter 1 of the Background Mathematics booklet available on the companion website. The operations are performed in the following order:
Start with x = 2 then
2x = 4
2x + 1 = 5
(2x + 1)3 = 125
So, there are three operations involved
1. multiply by 2, 2. add on 1, 3. take the cube.
This way of breaking down functions can be pictured using boxes to represent each operation that makes up the function, as was used to represent equations in Chapter 3 of the Background Mathematics booklet available on the companion website. The whole function can be thought of as a machine, represented by a box. For each value x, from the domain of the function that enters the machine, there is a resulting image, y, which comes out of it. This is pictured in Figure 1.26.
Inside of the box, we can write the name of the functions or the expression which gives the function rule. A composite function box can be broken into different stages, each represented by its own box. The function y = (2x + 1)3 breaks down as in Figure 1.27.
y = (3x − 4)4 can be broken down as in Figure 1.28.

Figure 1.28 The function y = (3x − 4)4 decomposed
into its composite operations.

The inverse of a function
The inverse of a function is a function which will take the image under the function back to its original value. If f −1(x) is the inverse of f (x) then f −1(f (x)) = x (f −1 ◦ f ) : x → x
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22 Sets and functions
Figure 1.29 The top line represents the function f (x ) = 5x − 2 (read from left to right) and the bottom line the inverse function.

Example 1.20
f (x) = 2x + 1 f −1(x) = x − 1
2
To show this is true, look at the combined function f −1(f (x)) = (2x + 1 − 1)/2 = x.
Finding the inverse of a linear function
One simple way of finding the inverse of a linear function is to: 1. Decompose the operations of the function. 2. Combine the inverse operations (performed in the reverse order) to
give the inverse function. This is a method similar to that used to solve linear equations in Chapter 3 of the Background Mathematics Notes available on the companion website for this book.
Example 1.21 Find the inverse of the function f (x) = 5x − 2. The method of solution is given in Figure 1.29. The inverse operations give that x = (y + 2)/5. Here y is the input
value into the inverse function and x is the output value. To use x and y in the more usual way, where x is the input and y the output, swap the letters giving the inverse function as
y = x+2 5
This result can be achieved more quickly by rearranging the expression so that x is the subject of the formula and then swap x and y.

Example 1.22 Find the inverse of f (x) = 5x − 2.
y = 5x − 2 ⇔ y + 2 = 5x ⇔ y+2 =x 5 ⇔ x = y+2 5
Now swap x and y to give y = (x + 2)/5. Therefore,
f −1(x) = (x + 2)/5.

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Sets and functions 23

Example 1.23 Find the inverse of

g(x)

=

1 2−x

where x = 2

Set

y

=

1 2−x

⇔

y(2 − x) =

1

⇔ 2y − xy = 1

⇔ 2y = 1 + xy

⇔ 2y − 1 = xy

⇔ xy = 2y − 1

⇔

x

=

2y − 1 y

where y = 0

⇔

x

=

2

−

1 y

Swap x and y to give y = 2 − (1/x) So
g−1(x) = 2 − 1 x = 0 x
To check, try a couple of values of x. Try x = 4,

g(x) = 1 = 1 = − 1 2−x 2−4 2
Perform g−1 on the output value −(1/2). Substitute g(4) = −(1/2) into g−1(x):

g−1

−1 2

=

2

−

1 −(1/2)

=

2

+

2

=

4.

The function followed by its inverse has given us the original value of x.

1.6 Summary

The range of a function
When combining functions, for example, f (g(x)), we have to ensure that g(x) will only output values that are allowed to be input to f . The set of images of g(x) becomes an important consideration. The set of images of a function is called its range. The range of a function is a subset of its codomain.
1. Functions are used to express relationships between physical quantities.
2. The allowed inputs to a function are grouped into a set, called the domain of the function. The set including all the outputs is called the codomain.
3. A set is a collection of objects called elements.
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24 Sets and functions

4. E is the universal set, the set of all objects we are interested in. 5. ∅ is the empty set, the set with no elements. 6. The three most important operations on sets are:
(a) intersection: A ∩ B is the set containing every element in both A and B;
(b) union: A ∪ B is the set of elements in A or in B or both; (c) complement: A is the set of everything, in the universal set,
not in A. 7. A relation is a way of pairing members of two sets. 8. Functions are a special type of relation which can be thought of as
mathematical machines. For each input value there is exactly one output value. 9. Many functions of interest are functions of time, used to represent signals. Analogue signals can be represented by functions of a real variable and digital signals by functions of an integer (discrete functions). Functions of an integer are also called sequences and can be defined using a recurrence relation. 10. To find the domain of a real or discrete function exclude values that could lead to a division by zero, negative square roots, or negative logarithms or other undefined values. 11. Functions can be combined in various ways including sum, difference, product, and quotient. A special operation of functions is composition. A composite function is found by performing a second function on the result of the first. 12. The inverse of a function is a function which will take the image under the function back to its original value.

1.7 Exercises

1.1. Given E = {a, b, c, d, e, f, g}, A = {a, b, e}, B = {b, c, d, f}, C = {c, d, e}.
Write down the following sets: (a) A ∩ B (b) A ∪ B (c) A ∩ C (d) (A ∪ B) ∩ C (e) (A ∩ C) ∪ (B ∩ C) (f) (A ∩ B) ∪ C (g) (A ∪ C) ∩ (B ∪ C) (h) (A ∩ C) (i) A ∪ C .
1.2. Use Venn diagrams to show that: (a) (A ∩ B) ∩ C = A ∩ (B ∩ C) (b) (A ∪ B) ∪ C = A ∪ (B ∪ C) (c) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) (d) (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) (e) (A ∩ B) = A ∪ B (f) (A ∪ B) = A ∩ B .
1.3. Let E = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and given P = {x|x < 5}, Q = {x|x 3} find explicitly: (a) P (b) Q (c) P ∪ Q (d) P (e) P ∩ Q.

1.4. Below are various assertions for any sets A and B. Write true or false for each statement and give a counter-example if you think the statement is false. (a) (A ∩ B) = A ∩ B (b) (A ∩ B) ⊆ A (c) A ∩ B = B ∩ A (d) A ∩ B = B ∩ A .
1.5. Using a Venn diagram simplify the following: (a) A ∩ (A ∪ B) (b) A ∪ (B ∩ A ) (c) A ∩ (B ∪ A ).
1.6. A computer screen has 80 columns and 25 rows: (a) Define the set of positions on the screen. (b) Taking the origin as the top left hand corner define: (i) the set of positions in the lower half of the screen as shown in Figure 1.30(a); (ii) the set of positions lying on or below the diagonal as shown in Figure 1.30(b).
1.7. A certain computer system breaks down in two main ways: faults on the network and power supply faults. Of the last 50 breakdowns, 42 involved network faults and 20 power failures. In 13 cases, both the power supply and the network were faulty. How many breakdowns were attributable to other kinds of failure?

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Sets and functions 25

(c) Confirm the following:

Figure 1.30 (a) Points lying in shaded area represent the set of positions on the lower half of the computer screen as in Exercise 1.6(a). (b) Points on the diagonal line and lying in the shaded area represent the set of positions for Exercise 1.6(b).

(i) (f −1 ◦ f ) : x → x (ii) (h−1 ◦ h) : x → x (iii) (f ◦ f −1) : x → x
(d) Using the results from sections (b) and (c), find the following:

1.8. Draw arrow diagrams and graphs of the following functions: (a) f (t) = (t − 1)2 t ∈ {0, 1, 2, 3, 4} (b) g(z) = 1/z z ∈ {−1, −0.5, 0.5, 1, 1.5, 2} (c) y = x x ∈ {−2, −1} 2x x ∈ {0, 1, 2, 3} (d) h : t → 3 − t t ∈ {5, 6, 7, 8, 9, 10}
1.9. Given that f : x → 2x − 1, g : x → (1/3)x2, h : x → 3/x
(a) Find the following:

(i) (h−1 ◦ h)(1) (ii) h(g(5)) (iii) g(f (4))
1.10. An analog signal is sampled using an A/D convertor and represented using only integer values. The original signal is represented by g(t) and the digital signal by h(t) sampled at t ∈ {2, 3, 4, 5, 6, 7, 8, 9, 10}. The definitions of g and h are as below
g(t) = t − 2.25 t < 5 6.8 − t t 5

(i) f (2)

(ii) g(3)

(iii) h(5)

(iv) h(2) + g(2) (v) h/g(5) (vi) (h × g)(2)

(vii) h(g(2))

(viii) h(h(3))

(b) Find the following functions:

h:2→0 h:5→2 h : 8 → −1

h:3→1 h:6→1 h : 9 → −2

h:4→2 h:7→0 h : 10 → −3

(i) f ◦ g (ii) g ◦ f (iii) h ◦ g (v) h−1

(iv) f −1

If e(t) is the error function (called quantization error), defined at the sample points, find e(t) and represent it on a graph.

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2

Functions and

their graphs

2.1 Introduction

The ability to produce a picture of a problem is an important step towards solving it. From the graph of a function, y = f (x), we are able to predict such things as the number of solutions to the equation f (x) = 0, regions over which it is increasing or decreasing, and the points where it is not defined.
Recognizing the shape of functions is an important and useful skill. Oscilloscopes give a graphical representation of voltage against time, from which we may be able to predict an expression for the voltage. The increasing use of signal processing means that many problems involve analysing how functions of time are effected by passing through some mechanical or electrical system.
In order to draw graphs of a large number of functions, we need only remember a few key graphs and appreciate simple ideas about transformations. A sketch of a graph is one which is not necessarily drawn strictly to scale but shows its important features. We shall start by looking at special properties of the straight line (linear function) and the quadratic. Then we look at the graphs of y = x, y = x2, y = 1/x, y = ax and how to transform these graphs to get graphs of functions like y = 4x − 2, y = (x − 2)2, y = 3/x, and y = a−x.

2.2 The straight line: y = mx + c

y = mx + c is called a linear function because its graph is a straight line. Notice that there are only two terms in the function; the x term, mx, where m is called the coefficient of x and c which is the constant term. m and c have special significance. m is the gradient, or the slope, of the line and c is the value of y when x = 0, that is, when the graph crosses the y-axis. This graph is shown in Figure 2.1(a) and two particular examples
shown in Figure 2.1(b) and (c).

Figure 2.1 (a) The graph of the function y = mx + c. m is the slope of the line, if m is positive then travelling from left to right along the line of the function is an uphill climb, if m is negative then the journey is downhill. The constant c is where the graph crosses the y-axis. (b) m = 2 and c = 3 (c) m = −1 and c = 2.
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Functions and their graphs 27

The gradient of a straight line

The gradient gives an idea of how steep the climb is as we travel along the

line of the graph. If the gradient is positive then we are travelling uphill as we move from left to right and if the gradient is negative then we are travelling downhill. If the gradient is zero then we are on flat ground. The gradient gives the amount that y increases when x increases by 1 unit. A straight line always has the same slope at whatever point it is measured. To show that in the expression y = mx + c, m is the gradient, we begin

with a couple of examples as in Figure 2.1(b) and (c) In Figure 2.1(b), we have the graph of y = 2x + 3. Take any two
values of x which differ by 1 unit, for example, x = 0 and x = 1. When x = 0, y = 2 × 0 + 3 = 3 and when x = 1, y = 2 × 1 + 3 = 5. The increase in y is 5 − 3 = 2, and this is the same as the coefficient of x in
the function expression. In Figure 2.1(c), we see the graph of y = −x + 2. Take any two
values of x which differ by 1 unit, for example, x = 1 and x = 2. When x = 1, y = −(1) + 2 = 1 and when x = 2, y = −(2) + 2 = 0. The increase in y is 0 − 1 = −1 and this is the same as the coefficient of x in
the function expression. In the general case, y = mx + c, take any two values of x which differ
by 1 unit, for example, x = x0 and x = x0+1. When x = x0, y = mx0+c and when x = x0 + 1, y = m(x + 1) + c = mx + m + c. The increase in y is mx + m + c − (mx + c) = m.
We know that every time x increases by 1 unit y increases by m.
However, we do not need to always consider an increase of exactly 1 unit in x. The gradient gives the ratio of the increase in y to the increase in x.

Therefore, if we only have a graph and we need to find the gradient then
we can use any two points that lie on the line. To find the gradient of the line take any two points on the line (x1, y1)
and (x2, y2).

The

gradient

=

change in change in

y x

=

y2 x2

− y1 − x1

Example 2.1 Find the gradient of the lines given in Figure 2.2(a)–(c) and the equation for the line in each case.

Solution

(a) We are given the coordinates of two points that lie on the straight line in Figure 2.2(a) as (0,3) and (2,5),

gradient

=

change change

in in

y x

=

5 2

− −

3 0

=

2 2

=

1.

To find the constant term in the expression y = mx + c, we find the value of y when the line crosses the y-axis. From the graph this is 3, so the equation is y = mx + c where m = 1 and c = 3, giving

y =x+3

(b) Two points that lie on the line in Figure 2.2(b) are (−1, −3) and (−2, −6). These are found by measuring the x and y values for some points on the line.

gradient

=

change change

in in

y x

=

−6 − (−3) −2 − (−1)

=

−3 −1

=

3.

To find the constant term in the expression y = mx + c, we find the value of y when the line crosses the y-axis. From the graph this

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28 Functions and their graphs

Figure 2.2 Graphs for Example 2.1.

is 0, so the equation is y = mx + c where m = 3 and c = 0 giving y = 3x

(c) Two points that lie on the line in Figure 2.2(c) are (0,2) and (3,3.5).

gradient

=

change change

in in

y x

=

3.5 − 2 3−0

=

1.5 3

=

0.5

To find the constant term in the expression y = mx + c, we find the value of y when the line crosses the y-axis. From the graph this is 2, so the equation is y = mx + c where m = 0.5 and c = 2 giving

y = 0.5x + 2

Finding the gradient from the equation for the line

To find the gradient from the equation of the line we look for the value of m, the number multiplying x in the equation. The constant term gives the value of y when the graph crosses the y-axis, that is, when x = 0.

Example 2.2 Find the gradient and the value of y when x = 0 for the following lines:

(a) y = 2x + 3, (b) 3x − 4y = 2,

(c)

x − 2y = 4,

(d)

x

−

1

=

1

−

y .

2

3

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Functions and their graphs 29

Solution
(a) In the equation y = 2x + 3, the value of m, the gradient, is 2 as this is the coefficient of x. c = 3 which is the value of y when the graph crosses the y-axis, that is, when x = 0.
(b) In the equation 3x − 4y = 2, we rewrite the equation with y as the subject of the formula in order to find the value of m and c.

3x − 4y = 2 ⇔ 3x = 2 + 4y
⇔ 3x − 2 = 4y
⇔ 3x − 2 = y 44
⇔ y = 3x − 1 42
We can see, by comparing the expression with y = mx + c, that m, the gradient, is 3/4 and c = −1/2. (c) Write y as the subject of the formula:

x − 2y = 4 ⇔ x = 4 + 2y
⇔ x − 4 = 2y
⇔ 2y = x − 4 ⇔ y = x −2
2
We can see, by comparing the expression with y = mx + c, that m, the gradient, is 1/2 and c = −2. (d) Write y as the subject of the formula

x−1 =1− y

2

3

⇔ x − 1 =1− y

22

3

⇔ 3x − 3 = 3 − y 22

⇔ y = 3 − 3x − 3 22

⇔ y = − 3x + 9 22

We can see, by comparing the expression with y = mx + c, that m, the gradient, is −3/2 and c = 9/2.

Finding the equation of a line which goes through two points

Supposing we have been given two points, (x1, y1) and (x2, y2), which lie on a line and we want to find the equation of that line. We already found that the gradient of the line is given by:

The

gradient

=

change in change in

y x

=

y2 x2

− y1 − x1

We know that the equation of a line is of the form y = mx + c, but we would like to express the equation just in terms of the two variables, x

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Functions and their graphs 31
Example 2.4
(a) Sketch the graph of y = 4x − 2. To find where the graph crosses the y-axis, substitute x = 0 into
the equation of the line:
y = 4(0) − 2 = −2.
This means that the graph passes through the point (0,−2). To find where the graph crosses the x-axis, substitute y = 0,
that is,
4x − 2 = 0
⇔ 4x = 2
⇔ x = 2 = 0.5. 4
Therefore, the graph passes through (0.5, 0). Mark the points (0,2) and (0.5,0), on the x- and y-axes and join the two points. This is done in Figure 2.3(a). (b) Sketch the graph of y = −4x When x = 0 we get y = 0, that is the graph goes through the point (0,0). In this case, as the graph passes through the origin, we need to choose a different value for x for the second point. Taking x = 2 gives y = −8, so another point is (2, −8). These points on marked on the graph and joined to give the graph as in Figure 2.3(b).

Figure 2.3 (a) The graph of y = 4x − 2. (b) The graph of y = −4x .

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32 Functions and their graphs

2.3 The quadratic function: y = ax 2 + bx + c

y = ax2 + bx + c is a general way of writing a function in which the highest power of x is a squared term. This is called the quadratic function and its graph is called a parabola as shown in Figure 2.4.
All the graphs, in this figure, cross the y-axis at (0, c). To find where they cross the x-axis can be more difficult. These values, where f (x) = 0, are called the roots of the equation. There is a quick way to discover whether the function crosses the x-axis, only touches the x-axis, or does not cross or touch it. In the latter case there are no solutions to the equation f (x) = 0. The three possibilities are given in Figure 2.4.

Crossing the x -axis
The function y = ax2 + bx + c crosses the x-axis when y = 0, that is, when ax2+bx +c = 0. The solutions to ax2+bx +c = 0 are examined in the Background Mathematics Notes available on the companion website for this book and are given by the formula
√ x = −b ± b2 − 4ac
2a

Figure 2.4 (a) The function y = ax 2 + bx + c. (a) Case 1 where there are 2 solutions to f (x ) = 0. (b) Case 2 where there is only one solution to f (x ) = 0. (c) Case 3, where there are no real solutions to f (x ) = 0.
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Functions and their graphs 33

Figure 2.5 Three quadratic
functions with two roots to the equation f (x ) = 0. Each satisfies b2 − 4ac > 0.(a) y = 2x 2 − 3, a = 2, b = 0, c = −3, b2 − 4ac = 0 − 4(2)(−3) = 24. (b) y = −x 2 + 5, a = −1, b = 0, c = 5, b2 − 4ac = 0 − 4(−1)(5) = 20.(c) y = −3(x − 2)2 + 1 ⇔ y = −3x 2 + 12x − 11, a = −3, b = 12, c = −11, b2 − 4ac = (12)2 − 4(−3)(−11) = 144 − 132 = 12.

From the graph, we can see there are three possibilities:
1. In Figure 2.4(a) where there are two solutions, that is, the graph crosses the x-axis for two values of x. For this to happen, the square root part of the formula above must be greater than zero:
b2 − 4ac > 0
Examples are given in Figure 2.5. 2. Only one unique solution, as in Figure 2.4(b). The graph touches the
x-axis in one place only. For this to happen, the square root part of the formula must be exactly 0. Examples of this are given in Figure 2.6. 3. No real solutions, that is, the graph does not cross the x-axis. Examples of these are given in Figure 2.7.

2.4 The function y = 1/x

The function y = 1/x has the graph as in Figure 2.8. This is called a hyperbola. Notice that the domain of f (x) = 1/x does not include x = 0. The graph does not cross the x-axis so there are no solutions to 1/x = 0.

2.5 The functions y = ax

Graphs of exponential functions, y = ax, are shown in Figure 2.9. The functions have the same shape for all a > 1. Notice that the function is always positive and the graph does not cross the x-axis so there are no solutions to the equation ax = 0.

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34 Functions and their graphs
Figure 2.6 Quadratic functions with only one unique root of the equation f (x ) = 0. Each satisfies b2 − 4ac = 0. (a) y = x 2 − 4x + 4, a = 1, b = −4, c = 4, b2 − 4ac = (−4)2 −4(1)(4) = 16−16 = 0. (b) y = −3x 2 − 12x − 12, a = −3, b = −12, c = −12, b2 − 4ac = (−12)2 − 4(−3)(−12) = 144 − 144 = 0. (c) y = x 2, a = 1, b = 0, c = 0, b2 − 4ac = (0)2 − 4(1)(0) = 0 − 0 = 0.
Figure 2.7 Quadratic functions with no real roots to the equation f (x ) = 0. In each case b2 − 4ac < 0. (a) y = x 2 + 2, a = 1, b = 0, c = 2, b2 − 4ac = (0)2 − 4(1)(2) = 0 − 8 = −8. (b) y = −x 2 + 2x − 2, a = −1, b = 2, c = −2, b2 − 4ac = (2)2 − 4(−1)(−2) = 4 − 8 = −4. (c) y = 3x 2 − 6x + 4, a = 3, b = −6, c = 4, b2 − 4ac = (−6)2 − 4(3)(4) = 36 − 48 = −12.

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Functions and their graphs 35

Figure 2.8 Graph of the hyperbolic function y = 1/x .

Figure 2.9 Graphs of functions y = ax : (a) y = 2x ; (b) y = 3x ; (c) y = (1.5)x .

2.6 Graph sketching using simple transformations

One way of sketching graphs is to remember the graphs of simple functions and to translate, reflect or scale those graphs to get graphs of other functions. We begin with the graphs below as given in Figure 2.10.
The translation x → x + a
If we have the graph of y = f (x), then the graph of y = f (x + a) is found by translating the graph of y = f (x) a units to the left. Examples are given in Figure 2.11.
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36 Functions and their graphs
Figure 2.10 To sketch graphs using transformations we begin with known graphs. In the rest of this section we use: (a) y = x ; (b) y = x 2; (c) y = x 3; (d) y = 1/x ; (e) y = ax .
The translation f (x ) → (x ) + A
Adding A on to the function value leads to a translation of A units upwards. Examples are given in Figure 2.12.
Reflection about the y -axis, x → −x
Replacing x by −x in the function has the effect of reflecting the graph in the y-axis – that is, as though a mirror has been placed along the axis and only the reflection can be seen. Examples are given in Figure 2.13.
Reflection about the x axis, f (x ) → −f (x )
To find the graph of y = −f (x), reflect the graph of y = f (x) about the x-axis. Examples are given in Figure 2.14.
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Functions and their graphs 37
Figure 2.11 Translations x → x + a. (a) (i) y = 1/x ; (ii) y = 1/(x + 2). Here x has been replaced by x + 2 translating the graph 2 units to the left. (b) (i) y = x 2; (ii) y = (x − 3)2, x has been replaced by x − 3 translating the graph 3 units to the right.

Figure 2.12 Translations f (x ) → f (x ) + A. (a) y = 1/x ; (ii) y = 1/x + 2. Here the function value has been
increased by 2 translating the
graph 2 units upwards. (b) (i) y = x 2; (ii) y = x 2 − 2. The function value has had 2
subtracted from it, translating
the graph 2 units downwards.

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38 Functions and their graphs
Figure 2.13 Reflections x → −x . (a) (i) y = ax , a > 1; (ii) y = a−x , x has been replaced by −x to get the second function. This has the effect of reflecting the graph in the y-axis. (b) (i) y = (x /2) + (1/2); (ii) y = −(x /2) + (1/2), x has been replaced by −x , reflecting the graph in the y-axis.
Figure 2.14 Reflections f (x ) → −f (x ). (a) (i) y = x 2; (ii) y = −x 2. The function value has been multiplied by −1 turning the graph upside down (reflection in the x-axis). (b) (i) y = 2x ; (ii) y = −2x . The second function has been multiplied by −1 turning the graph upside down.
Scaling along the x -axis, x → ax
Multiplying the values of x by a number, a, has the effect of: squashing the graph horizontally if a > 1 or stretching the graph horizontally if 0 < a < 1. Examples are given in Figure 2.15.
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Functions and their graphs 39
Figure 2.15 Scalings x → ax . (a) (i) y = x 2; (ii) y = [(1/2)x ]2; (iii) y = (2x )2. The second function has x replaced by (1/2)x which has stretched the graph horizontally (the multiplication factor is between 0 and 1). The third function has replaced x by 2x, which has squashed the graph horizontally (the multiplication factor is greater than 1). (b) (i) y = 2x ; (ii) y = 2(1/2x); (iii) y = 22x . The second function has replaced x by (1/2)x which has stretched the graph horizontally. The third function has x replaced by 2x which has squashed the graph horizontally.
Figure 2.16 Scalings f (x ) → Af (x ). (a) y = 1/(x + 2); (b) y = 2/(x + 2) (c) y = 1/[3(x + 2)]. The second graph has the function values multiplied by 2 stretching the graph vertically. The third graph has function values multiplied by 1/3 squashing the graph vertically.
Scaling along the y -axis, f (x ) → Af (x )
Multiplying the function value by a number A has the effect of stretching the graph vertically if A > 1, or squashing the graph vertically if 0 < A < 1. Examples are given in Figure 2.16.
Reflecting in the line y = x
If the graph of a function y = f (x) is reflected in the line y = x, then it will give the graph of the inverse relation. Examples are given in Figure 2.17.
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40 Functions and their graphs

Figure 2.17 Reflections in the line y = x produce the inverse relation. (a) (i) y = 2x ; (ii) y = log2(x ). The

second graph is obtained from the first is only one value of y for each value of

bx.y(rbe)fl(ei)cytin=√g ixn2t;h(eii)dyot=ted±l√inxe.yT=hexs.eTchoendingvrearpseh

is is

a function as there found by reflecting

the first graph in the line y = x . Notice that y = ± x is not a function as there is more that one possible

value of y for each value of x > 0.

In Chapter 1, we defined the inverse function as taking any image back to its original value. Check this with the graph of y = 2x in

Figure 2.17(a): x = 1 gives y = 2. In the inverse function, y = log2(x), substitute 2, which gives the result of 1, which is back to the original

value. However,

the

inverse

of

y

=

x2,

y

±

√ x,

shown

in

Figure

2.17(b),

is

not a function as there is more than one y value for a single value of x.

To understand this problem more fully, perform the following experiment. On a calculator enter −2 and square it (x2) giving 4. Now take

the square root. This gives the answer 2, which is not the number we first

started with, and hence we can see that the square root is not a true inverse

of squaring. However, we get away with calling it the inverse because it

works if only positive values of x are considered. To test if the inverse

of any function exists, draw a line along any value of y = constant.

If, wherever the line is drawn, there is ever more than one x value

which gives the same value of y then the function has no inverse func-

tion. In this situation, the function is called a ‘many-to-one’ function.

Only ‘one-to-one’ functions have inverses. Figure 2.18 has examples

of functions with an explanation of whether they are ‘one-to-one’ or

‘many-to-one’.

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Functions and their graphs 41

Figure 2.18 (a) (i) y = x 3. This function has only one x value for each value of y as any line y = cons√tant only cuts the graph once. In this case, the function is one-to-one and it has an inverse function (ii) y = 3 x is the inverse function of y = x 3. (b) y = 1/x , x = 0, has only one x value for each value of y as any line y = constant only cuts the graph once. It therefore is one-to-one and has an inverse function – in fact, it is its own inverse! (to see this reflect it in the line y = x and we get the same graph after the reflection). (c) (i) y = x 4. This function has two values of x for each value of y when y is positive (e.g. the line y = 16 cuts the graph twice at x = 2 and at x = −2). Th√is shows that there is no inverse function as the function is many-to-one. (ii) The inverse relation y = ± 4 x .

2.7 The modulus function, y = |x | or y = abs(x )

The modulus function y = |x|, often written as y = abs(x) (short for the absolute value of x) is defined by
y = x x 0 (x positive or zero) −x x < 0 (x negative)
The output from the modulus function is always a positive number or zero.
Example 2.5 Find | − 3|. Here x = −3, which is negative, therefore y = −x = −(−3) = +3.
An alternative way of thinking of it is to remember that the modulus is always positive, or zero, so simply replacing any negative sign by a positive one will give a number’s modulus or absolute value. | − 5| = 5, | − 4| = 4, |5| = 5, |4| = 4. The graph of the modulus function can be found from the graph of y = x by reflecting the negative x part of the graph to make the function values positive. This is shown in Figure 2.19.
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42 Functions and their graphs

Figure 2.19 The graph of the modulus function y = |x | obtained from the graph of y = x . (a) The graph of y = x with the negative part of the graph displayed as a dotted line. This is reflected about the x-axis to give y = −x for x < 0. (b) The graph of y = |x |.

2.8 Symmetry of functions and their graphs

Functions can be classified as even, odd, or neither of these.
Even functions
Even functions are those that can be reflected in the y-axis and then result in the same graph. Examples of even functions are (see Figure 2.20):
y = x2, y = |x|, y = x4.
As previously discussed, reflecting in the y-axis results from replacing x by −x in the function expression and hence the condition for a function to be even is that substituting −x for x does not change the function expression, that is, f (x) = f (−x).
Example 2.6 Show that 3x2 − x4 is an even function. Substitute −x for x in the expression f (x) = 3x2 − x4 and we get
f (−x) = 3(−x)2 − (−x)4 = 3(−1)2(x)2 − (−1)4(x)4 = 3x2 − x4.
So, we have found that f (−x) = f (x) and therefore the function is even.

Odd functions
Odd functions are those that when reflected in the y-axis result in an upside down version of the same graph. Examples of odd functions are (see Figure 2.21):
y = x, y = x3, y = 1 x
Reflecting in the y-axis results from replacing x by −x in the function expression and the upside down version of the function f (x) is found by multiplying the function by −1. Hence, the condition for a function to be odd is that substituting −x for x gives −f (x), that is,
f (−x) = −f (x).
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Functions and their graphs 43
Figure 2.20 y = x 2, y = |x |, and y = x 4 are even functions.

Figure 2.21 (a) y = x , (b) y = x 3, (c) y = 1/x are odd functions. If they are reflected in the y-axis they result in an upside down version of the original graph.

Example 2.7 Show that 4x − (1/x) is an odd function. Substitute x for −x in the expression f (x) = 4x − (1/x) and we get

f (−x)

=

4(−x) −

1 −x

=

−4x

+

1 x

=

−

4x

−

1 x

.

We have found that f (−x) = −f (x), so the function is odd.

2.9 Solving inequalities

For linear and quadratic functions, y = f (x), we have discussed how to find the values where the graph of the functions crosses the x-axis, that is how to solve the equation f (x) = 0. It is often of interest to find ranges of values of x where f (x) is negative or where f (x) is positive. This means solving inequalities like f (x) < 0 or f (x) > 0, respectively.
Like equations, inequalities can be solved by looking for equivalent
inequalities. One way of finding these is by doing the same thing to both
sides of the expression. There is an important exception for inequalities

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44 Functions and their graphs
that if both sides are multiplied or divided by a negative number then the direction of the inequality must be reversed.
To demonstrate these equivalences begin with a true proposition 3 < 5 or ‘3 is less than 5’. Add 2 on to both sides and it is still true 3 + 2 < 5 + 2, i.e. 5 < 7.
Subtract 10 from both sides and we get 5 − 10 < 7 − 10, i.e. − 5 < −3 which is also true.
Multiply both sides by −1 and if we do not reverse the inequality we get (−1)(−5) < (−1)(−3), i.e. 5 < 3 which is false. However, if we use the correct rule that when multiplying by a negative number we must reverse the inequality sign then we get: (−1)(−5) > (−1)(−3), i.e. 5 > 3 which is true. This process is pictured in Figure 2.22.
Note that inequalities can be read from right to left as well as from left to right: 3 < 5 can be read as ‘3 is less than 5’ or as ‘5 is greater than 3’ and so it can also be written the other way round as 5 > 3.
Using a number line to represent inequalities
An inequality can be expressed using a number line as in Figure 2.23. In Figure 2.23(a), the open circle indicates that 3 is not included in the set of values, t < 3. In Figure 2.23(b), the closed circle indicates that −2 is included in the set of values, x −2. In Figure 2.23(c), the closed circle indicates that the value 4.5 is included in the set y 4.5.
Figure 2.22 On the number line, numbers to the left are less than numbers to their right: −5 < −3 . If the inequality is multiplied by −1 we need to reverse the sign to get 5 > 3.

Figure 2.23 Representing inequalities on a number line.

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Functions and their graphs 45
Figure 2.24 The solution to 2t + 3 < t − 6 is given by t < −9.
Figure 2.25 The solution to x + 5 4x − 10 is found to be x 5, here represented on a number line.
Figure 2.26 The solution to 16 − y > −5y is found to be y > −4, here pictured on a number line.
Example 2.8 Find a range of values for t, x, and y such that the following inequalities hold (a) 2t + 3 < t − 6 (b) x + 5 4x − 10 (c) 16 − y > −5y Solution (a) 2t + 3 < t − 6 ⇔ 2t − t + 3 < −6 (subtract t from both
sides) ⇔ t < −6 − 3 (subtract 3 from both sides) ⇔ t < −9. This solution can be represented on a number line as in Figure 2.24. (b) x + 5 4x − 10 ⇔ +5 3x − 10 (subtract x from both sides) ⇔ 15 3x (add 10 to both sides) ⇔ 5 x (divide both sides by 3) ⇔ x 5. This solution in represented in Figure 2.25. (c) 16 − y > −5y ⇔ 16 > −4y (add y to both sides) ⇔ −4 < y (divide by − 4and reverse the sign) ⇔ y > −4. This solution is represented in Figure 2.26.
Representing compound inequalities on a number line
We sometimes need a picture of the range of values given if two inequalities hold simultaneously, for instance x 3 and x < 5. This is analysed
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46 Functions and their graphs
Figure 2.27 (a) x 3 and x < 5. (b) x > 6 and x > 2 combines to give x > 6. (c) x < 5 or x 7. (d) x < 2 or x 0.
in Figure 2.27(a) and we can see that for both inequalities to hold simultaneously x must lie in the overlapping region where 3 x < 5. 3 x < 5 is a way of expressing that x lies between 3 and 5 or is equal to 3. In the example in Figure 2.27(b), x > 6 and x > 2, and for them both to hold then x > 6.
Another possible way of combining inequalities is to say that one or another inequality holds. Examples of this are given in Figure 2.27(c) where x < 5 or x 7 and this gives the set of values less than 5 or greater than or equal to 7. Figure 2.27(d) gives the example where x < 2 or x 0 and in this case it results in all numbers lying on the number line, that is, x ∈ R.
Example 2.9 Find solutions to the following combinations of inequalities and represent them on a number line. (a) x + 3 > 4 and x − 1 < 5, (b) 1 − u < 3u + 2 or u + 2 6, (c) t + 5 > 12 and −t > 24.
Solution (a) x + 3 > 4 and x − 1 < 5 We solve both inequalities separately
and then combine their solution sets x + 3 > 4 ⇔ x > 1 (subtracting 3 from both sides) x − 1 < 5 ⇔ x < 6 (adding 1 to both sides) So the combined inequality giving the solution is x > 1 and x < 6, which from Figure 2.28(a) we can see is the same as 1 < x < 6. (b) 1 − u < 3u + 2 or u + 2 6 We solve both inequalities separately and then combine their
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Functions and their graphs 47

Figure 2.28 Solutions to compound inequalities as given in Example 2.8 represented on a number line.

solution sets.

1 − u < 3u + 2

⇔ −1 < 4u (subtracting 2 from both sides)

⇔ −1 < u 4
⇔ u > −1 4
u+2 6

(dividing both sides by 4)

⇔ u 4 (subtracting 2 from both sides)

Combining the two solutions gives u > −1/4 or u 4 and this
is represented on the number line in Figure 2.28(b) where we can see that it is the same as u > −1/4. (c) t + 5 > 12 and − t > 24

t + 5 > 12
⇔ t > 7 (subtracting 5 from both sides)
− t > 24
⇔ t < −24 (multiply both sides by −1 and reverse the inequality sign)

Combining the two solutions sets gives t > 7 and t < −24 and we can see from Figure 2.28(c) that this is impossible and hence there are no solutions.

Solving more difficult inequalities
To solve more difficult inequalities, our ideas about equivalence are not enough on their own, we also use our knowledge about continuous functions. In the previous chapter, we defined a continuous function as one that could be drawn without taking the pen off the paper. If we wish to solve the inequality f (x) > 0 and we know that f (x) is continuous then we can picture the problem graphically as in Figure 2.29. From the graph
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48 Functions and their graphs

Figure 2.29 The graph of a continuous function. To solve for f (x ) > 0, we first find the values where f (x ) = 0. On the graph these are marked as a, b, c, and d. If the function is above the x-axis then the function values are positive, if the function lies below the x-axis then the function values are negative. The solution to f (x ) > 0 is given by those values of x for which the function lies above the x-axis, that is, y positive. For the function represented in the graph the solution to f (x ) > 0 is x < a or b < x < c or x > d
.

Figure 2.30 Solving t 2 − 3t + 2 < 0 (Example 2.10).

we can see that to solve the inequality we need only find the values where f (x) = 0 (the roots of f (x) = 0) and determine whether f (x) is positive or negative between the values of x where f (x) = 0. To do this, we can use any value of x between the roots. We are using the fact that as f (x) is continuous then it can only change from positive to negative by going
through zero.

Example 2.10 Find the values of t such that t2 − 3t < −2. Write the inequality with 0 on one side of the inequality sign

t2 − 3t < −2 ⇔ t2 − 3t + 2 < 0 (adding 2 to both sides)

Find the solutions to f (t) = t2 − 3t + 2 = 0 and mark them on a number line as in Figure 2.30.
Using the formula

√

t = −b ±

b2 − 4ac 2a

where a = 1, b = −3, and c = 2 gives

√ t = −3 ± 9 − 8
2

⇔t = 3±1 2

⇔ t = 3 + 1 or t = 3 − 1

2

2

⇔ t = 2 or t = 1

Substitute values for t which lie on either side of the roots of f (t) in order to find the sign of the function between the roots. Here we choose 0, 1.5, and 3.

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Functions and their graphs 49
When t = 0
t2 − 3t + 2 = (0)2 − 3(0) + 2 = 0 + 0 + 2 = 2
which is positive, giving f (t) > 0. When t = 1.5
t2 − 3t + 2 = (1.5)2 − 3(1.5) + 2 = 2.25 − 4.5 + 2 = −0.25
which is negative giving f (t) < 0. When t = 3
t2 − 3t + 2 = (3)2 − 3(3) + 2 = 9 − 9 + 2 = 2
which is positive, giving f (t) > 0. By marking the regions on the number line, given in Figure 2.30, with
f (t) > 0, f (t) < 0, or f (t) = 0 as appropriate we can now find the solution to our inequality f (t) < 0 which is given by the region where 1 < t < 2.
Example 2.11 Find the values of x such that (x2 − 4)(x + 1) > 0. Solution The inequality already has 0 on one side of the inequality sign so we begin by finding the roots to f (x) = 0, that is,
(x2 − 4)(x + 1) > 0
Factorization gives
(x2 − 4)(x + 1) > 0 ⇔ (x − 2)(x + 2)(x + 1) = 0
⇔ x = 2, x = −2, or x = −1. So the roots are −2, −1, and 2. These roots are pictured on the number line in Figure 2.31.
Substitute values for x which lie on either side of the roots of f (x) in order to find the sign of the function between the roots. Here we choose −3, −1.5, 0, and 2.5.
When x = −3
(x − 2)(x + 2)(x + 1) = 0 gives (−3 − 2)(−3 + 2)(−3 + 1) = (−5)(−1)(−2) = −10, giving f (x) < 0.
When x = −1.5
(x − 2)(x + 2)(x + 1) gives (−1.5 − 2)(−1.5 + 2)(−1.5 + 1) = (−3.5)(0.5)(−0.5) = 0.875, giving f (x) > 0.
When x = 0
(x − 2)(x + 2)(x + 1) gives (0 − 2)(0 + 2)(0 + 1) = (−2)(2)(1) = −4, giving f (x) < 0.

Figure 2.31 Solving (x 2 − 4)(x + 1) > 0 (Example 2.11).

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50 Functions and their graphs
When x = 2.5
(x − 2)(x + 2)(x + 1) gives (2.5 − 2)(2.5 + 2)(2.5 + 1) = (0.5)(4.5)(3.5) = 7.875, giving f (x) > 0.
These regions are marked on the number line as in Figure 2.31 and the solution is given by those regions where f (x) > 0. Looking for the regions where f (x) > 0 gives the solution as −2 < x < −1 or x > 2.

2.10 Using graphs to find an expression for the function from experimental data

Linear relationships
Linear relationships are the easiest ones to determine from experimental data. The points are plotted on a graph and if they appear to follow a straight line then a line can be drawn by hand and the equation can be found using the method given in Section 2.2.

Example 2.12 A spring is stretched by hanging various weights on it and in each case the length of the spring is measured.

Mass (kg) 0.125 0.25 0.5 1 2 3 Length (m) 0.4 0.41 0.435 0.5 0.62 0.74

Approximate the length of the spring when no weight is hung from it and find the expression for the length in terms of the mass.
Solution First, draw a graph of the given experimental data. This is done in Figure 2.32.
A line is fitted by eye to the data. The data does not lie exactly on a line due to experimental error and due to slight distortion of the spring with heavier weights. From the line we have drawn we can find the gradient by choosing any two points on the line and calculating

change in y change in x .

Taking the two points as (0,0.28) and (2,0.58), we get the gradient as

0.58 2

− −

0.38 0

=

0.2 2

=

0.1.

The point where it crosses the y-axis, that is, where the mass hung on the spring is 0 can be found by extending the line until it crosses the y-axis. This gives 0.38 m.
Finally, the expression for the length in terms of the mass of the attached weight is given by y = mx + c, where y is the length and x is the mass,

Figure 2.32 The data for length of spring against mass of the weight as given in the Example 2.11. The line is a fitted by eye to the experimental data and the equation of the line can be found using the method of Section 2.2.

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Figure 2.33 (a) Graph of sound intensity against time as given in Example 2.13. (b) Graph of log10(intensity) against time and a line fitted by eye to the data. The line goes through the points (0,0) and (0.2, −2.2).

Functions and their graphs 51
m is the gradient and c is the value of y when x = 0, that is, where there is no weight on the string. This gives
length = 0.1 × mass + 0.38
The initial length of the spring is 0.38 m.
Exponential relationships
Many practical relationships behave exponentially particularly those involving growth or decay. Here it is slightly less easy to find the relationship from the experimental data, however, it is simplified by using a log–linear plot. Instead of plotting the values of the dependent variable, y, we plot the values of log10(y). If the relationship between y and time, t, is exponential as we suspected then the log10(y) against t plot will be a straight line.
The reason this works can be explained as follows. Supposing y = y0 10kt where y0 is the value of y when t = 0 and k is some constant; then, taking the log base 10 of both sides, we get log10(y) = log10(y010kt )
= log10(y0) + log10(10kt ).
As the logarithm base 10 and raising to the power of 10 are inverse operations, we get
log10(y) = log10(y0) + kt As y0 is a constant, the initial value of y, and k is a constant then we can
see that this expression shows that we shall get a straight line if log10(y) is plotted against t. The constant k is given by the gradient of the line and log10(y0) is the value of log10 y where it crosses the vertical axis. Setting Y = log10(y), c = log10(y0) Y = c + kt
which is the equation of the straight line.
Example 2.13 A room was tested for its acoustical absorption properties by playing a single note on a trombone. Once the sound had reached its maximum intensity, the player stopped and the sound intensity was measured for the next 0.2 s at regular intervals of 0.02 s. The initial maximum intensity at time 0 is 1.0. The readings were as follows:
Time(s) 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Intensity 1.0 0.63 0.35 0.22 0.13 0.08 0.05 0.03 0.02 0.01 0.005
Draw a graph of intensity against time and log(intensity) against time and use the latter plot to approximate the relationship between the intensity and time. Solution The graphs are plotted in Figure 2.33 where, for the second graph (b), we take the log10 (intensity) and use the table below:
Time 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 log10 0 −0.22 −0.46 −0.66 −0.89 −1.1 −1.3 −1.5 −1.7 −2 −2.3 (intensity)
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52 Functions and their graphs

We can see that the second graph is approximately a straight line and
therefore we can assume that the relationship between the intensity and time is exponential and could be expressed as I = I010kt . The log10 of this gives

log10(I ) = log10(I0) + kt.
From the graph in Figure 2.33(b), we can measure the gradient, k. To do this we calculate
change in log10(intensity) change in time

giving

−2.2 − 0 0.2 − 0

=

−11

=

k.

The point at which it crosses the vertical axis gives

log10(I0) = 0 ⇔ I0 = 100 = 1. Therefore, the expression I = I010kt becomes I = 10−11t .

Power relationships
Another common type of relationship between quantities is when there is a power of the independent variable involved. In this case, if y = axn where n could be positive or negative then the value of a and n can be found by drawing a log–log plot.
This is because taking log10 of both sides of y = axn gives
log10(y) = log10(axn) = log10(a) + n log10(x)
Replacing Y = log10(y) and X = log10(x) we get:
Y = log10(a) + nX,
showing that the log–log plot will give a straight line, where the slope of the line will give the power of x and the position where the line crosses the vertical axis will give the log10(a). Having found a and n, they can be substituted back into the expression
y = axn.

Example 2.14 The power received from a beacon antenna is though to depend on the inverse square of the distance from the antenna and the receiver. Various measurements, given below, were taken of the power received against distance r from the antenna. Could these be used to justify the inverse square law? If so, what is the constant, A, in the expression:

p

=

A r2

Power received (W)

0.39 0.1 0.05 0.025 0.015 0.01

Distance from antenna (km) 1 2 3 4

5

6

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Functions and their graphs 53

Figure 2.34 (a) Plot of power against distance (Example 2.14). (b) log10(power) against log10(distance). In (b), the line is fitted by eye to the data, from which the slope of the graph indicates n in the relationship P = Ar n. Two points lying on the line are (0, −0.38) and (0.7, −1.8).
Solution To test whether the relationship is indeed a power relationship, we draw a log–log plot. The table of values is found below:

log10(power) −0.41 −1 −1.3 −1.6 −1.8 −2

log10(distance) 0

0.3 0.5 0.6 0.7 0.78

Graphs of power against distance and log10(power) against log10(distance) are given in Figure 2.34(a) and (b).
As the second graph is a straight line, we can assume that the relationship is of the form P = Arn where P is the power and r is the distance.
In which case, the log–log graph is

log10(P ) = log10(A) + n log10(r). We can measure the slope by calculating

change in log10(P ) change in log10(r)
and, using the two points that have been found to lie on the line, this gives

−1.8 − 0.7

(−0.38) −0

=

−2.03.

As this is very near to −2, the inverse square law would appear to be
justified. The value of log10(A) is given from where the graph crosses the vertical
axis and this gives

log10(A) = −0.38 ⇔ A = 10−0.38 ⇔ A = 0.42.

So the relationship between power received and distance is approximately

P

=

0.42r −2

=

0.42 r2 .

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54 Functions and their graphs

2.11 Summary

1. The linear function y = mx + c has gradient (slope) m and crosses the y-axis at y = c.
2. The gradient, m, of a straight line y = mx + c is given by:

m

=

change change

in in

y x

and this is the same along the length of the line.
3. The equation of a line which goes through two points, (x1, y1) and (x2, y2) is:

y − y1 y2 − y1

=

x − x1 x2 − x1

where y2 = y1.

4. The graph of the quadratic function y = ax2 + bx + c is called a parabola. The graph crosses the y-axis (when x = 0) at y = c.
5. There are three possibilities for the roots of the quadratic equation a x2 + bx + c = 0 Case I: two real roots when b2 − 4ac > 0, Case II: only one unique root when b2 − 4ac = 0, Case III: no real roots when b2 − 4ac < 0.
6. By considering the graphs of known functions y = f (x), for
instance, those given in Figure 2.10, and the following transfor-
mations, many other graphs can be drawn. (a) Replacing x by x + a in the function y = f (x) results in
shifting the graph a units to the left. (b) Replacing f (x) by f (x) + A results in shifting the graph A
units upwards. (c) Replacing x by −x reflects the graph in the y-axis. (d) Replacing f (x) by −f (x) reflects the graph in the x-axis
(turning it upside down). (e) Replacing x by ax squashes the graph horizontally if a > 1
or stretches it horizontally if 0 < a < 1. (f) Replacing f (x) by Af (x) stretches the graph vertically if A >
1 or squashes it vertically if 0 < A < 1. (g) Reflecting the graph of y = f (x) in the line y = x results in
the graph of the inverse relation.
7. A function may be even, or odd, or neither of these.
(a) An even function is one whose graph remains the same if reflected in the y-axis, that is, when x → −x. This can also
be expressed by the condition

f (−x) = f (x)

Examples of even functions are y = x2, y = |x|, and y = x4. (b) An odd function is one which when reflected in the y-axis,
that is, when x → −x, gives an upside down version of the original graph (i.e. −f (x)). This can also be expressed as the
condition:

f (−x) = −f (x)

Examples of odd functions are y = x and y = x3.
8. Not all functions have true inverses. Only one-to-one functions have inverse functions. A function is one-to-one if any line y = constant drawn on the graph y = f (x) crosses the function only once. This means there is exactly one value of x that gives each value of y.

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Functions and their graphs 55
9. Simple inequalities can be solved by finding equivalent inequalities. Inequalities remain equivalent if both sides of the inequality have the same expression added or subtracted. They may also be multiplied or divided by a positive number but if they are multiplied or divided by a negative number then the direction of the inequality sign must be reversed.
10. To solve the inequalities f (x) > 0, f (x) < 0, f (x) 0, or f (x) 0, where f (x) is a continuous function, solve f (x) = 0 and choose any value for x around the roots to find the sign of f (x) for each region of values for x.
11. Graphs can be used to find relationships in experimental data. First, plot the data then: (a) If the data lies on an approximate straight line then draw a straight line through the data and find the equation of the line. (b) If it looks exponential, then take the log of the values of the dependent variable and draw a log–linear graph. If this looks approximately like a straight line then assume there is an exponential relationship y = y010kt , where k is given by the gradient of the line and log10(y0) is the value where the graph crosses the vertical axis. (c) If the relationship looks something like a power relationship, y = Axn, then take the log of both sets of data and draw a log–log graph. If this is approximately like a straight line, then assume there is a power relationship and n is given by the gradient of the line and log10 A is the value where the graph crosses the vertical axis.

2.12 Exercises

2.1. Sketch the graphs of the following:

(a) y = 3x − 1, (b) y = 2x + 1,

(c) y = −5x,

(d)

y

=

1 2

x

− 3.

In each case state the gradient of the line.

2.2. A straight line passes through the pair of points given. Find the gradient of the line in each case.

(a) (0, 1), (1, 4)

(b) (1, 1), (2, −4)

(c) (−1, −1), (6, 3) (d) (1, 4), (3, 4)

2.3. A straight line graph has gradient −5 and passes through (1,6). Find the equation of the line.
2.4. In Figure 2.35 are various graphs drawn to the scale 1 unit = 1 cm. By finding the gradients of the lines and where they cross the y-axis, find the equation of the line.

Figure 2.35 Straight line graphs for Exercise 2.4.

2.5. A straight line passes through the pair of points given. Find the equation of the line in each case.
(a) (0, 1), (−1, 4) (b) (1, 1), (−2, −4) (c) (1, 1), (6, 3) (d) (−1, −4), (−3, −4)

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56 Functions and their graphs

2.6. Find the values of x such that f (x) = 0 for the
following functions (a) f (x) = x2 − 4, (b) f (x) = (2x − 1)(x + 1), (c) f (x) = (x − 3)2, (d) f (x) = (x − 4)(x + 4), (e) f (x) = x2 + x − 6, (f) f (x) = x2 + 7x + 12, (g) f (x) = 12x2 − 12x − 144.
Using the fact that the peak or trough in the parabola,
y = f (x), occurs at a value of x half-way between the values where f (x) = 0 then sketch graphs of the
above quadratic functions.

2.7. By considering transformations of simple functions sketch graphs of the following:

(a) y

=

1 x−(1/2)

,

(c)

y

=

1 2

x3

,

(e) y = (2x − 1)2

(g) y = log2(x + 2), (i) y = 4x − x2.

(b) y = 3.2−x , (d) y = −3(1/2)x , (f) y = (2x − 1)2 − 2, (h) y = 6 − 2x ,

2.9. By substituting x → −x in the following functions

determine whether they are odd, even, or neither of

these:

(a)

y

= −x2 +

1 x2

where x = 0,

(b) y = |x3| − x2,

(c)

y

=

−1 x

+ log2(x)

where x

> 0,

(d)

y

=

−1 x

+ x + x5,

(e) y = 6 + x2,

(f) y = 1 − |x|.

2.10. Draw graphs of the following functions and draw the

graph of the inverse relation in each case. Is the inverse

a function?

(a) f (t) = −t + 2, (b) g(x) = (x − 2)2,

(c)

h(w)

=

w

4 +

. 2

2.11. Find the range of values for which the following inequalities hold and represent them on a number line.

(a) 10t − 2 31,

(b) 10x − 3x > −2,

(c) 3 − 4y 11 + y, (d) t + 15 < 6 − 2t.

2.8. Consider reflections of the graphs given in Figure 2.36 to determine whether they are even, odd, or neither of these.

2.12. Find the range of values for which the following hold and represent them on a number line: (a) x − 2 > 4 or 1 − x < 12, (b) 4t + 2 10 and 3 − 2t < 1, (c) 3u + 10 > 16 or 3 − 2u > 13.
2.13. Solve the following inequalities and represent the solutions on a number line:
(a) x2 − 4 < 5, (b) (2x − 3)(x + 1)(x − 5) > 0, (c) t2 + 4t 21, (d) 4w2 + 4w − 35 0.

2.14. For the following sets of data, y is thought to depend exponentially on t. Draw log–linear graphs in each case and find constants A and k such that y = A10kt . (a)
y 75 48 30 19 12 7 t1 2 3 4 5 6

Figure 2.36 Graphs of functions for Exercise 2.7.

(b)
y 2 4.2 8.5 18 35 73 t 0.1 0.2 0.3 0.4 0.5 0.6

2.15 An experiment measuring the change in volume of a gas as the pressure is decreased gave the following measurements:

P (105 N m−2) 1.5 1.4 1.3 1.2 1.1 1

V (m3)

0.95 1 1.05 1.1 1.16 1.24

If the gas is assumed to be ideal and the expansion is adiabatic then the relationship between pressure and volume should be:
pV γ = C
where γ and C are constants and p is the pressure and V is the volume. Find reasonable values of γ and C to fit the data and from this expression find the predicted volume at atmospheric pressure, p = 1.013 × 105 N m−2.

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3

Problem solving

and the art of the

convincing

argument

3.1 Introduction
Figure 3.1 A simple circuit.

Mathematics is used by engineers to solve problems. This usually involves developing a mathematical model. Just as when building a working model aeroplane we would hope to include all the important features, the same thing applies when building a mathematical model. We would also like to indicate the things we have had to leave out because they were too fiddly to deal with, and also those details that we think are irrelevant to the model. In the case of a mathematical model the things that have been left out are listed under assumptions of the model. To build a mathematical model, we usually need to use scientific rules about the way things in the world behave (e.g. Newton’s laws of motion, conservation of momentum and energy, Ohm’s law, Kirchoff’s laws for circuits, etc.) and use numbers, variables, equations, and inequalities to express the problem in a mathematical language.
Some problems are very easy to describe mathematically. For instance: ‘Three people sitting in a room were joined by two others, how many people are there in the room in total?’ This can be described by the sum 3 + 2 =? and can be solved easily as 3 + 2 = 5.
The final stage of solving the problem is to translate it back into the original setting – the answer is: ‘there are 5 people in the room in total’.
Assumptions were used to solve this problem. We assumed that no one else came in or left the room in the meantime and we made general assumptions about the stability of the room, for example, the building containing it did not fall down. However, these assumptions are so obvious that they do not need to be listed. In more complex problems it is necessary to list important assumptions as they may have relevance as to the validity of the solution.
Another example is as follows: ‘There are three resistors in series in a circuit, two of the resistors are known to have resistance of 3 and 4 ohm, respectively. The voltage source is a battery of 12 V and the current is measured as 1 A. What is the resistance of the third resistor?’
To help express the problem in a mathematical form we may draw a circuit diagram as in Figure 3.1.
The problem can be expressed mathematically by using Ohm’s law and the fact that an equivalent resistance to resistances in series is given by the sum of the individual resistances. If x is the unknown value of the
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58 Problem solving and the art of the convincing argument
third resistance and V = RI where R = R1 + R2 + R3, we obtain:
12 = (3 + 4 + x)1
The expression of the mathematical problem has taken the form of an equation where we now need to find x, the value of the third resistance.
The main assumptions that have been used to build this mathematical model are:
(1) There are such things as pure resistors that have no capacitance or inductance.
(2) Resistances remain constant and are not affected by any possible temperature changes or other environmental effects.
(3) The battery gives a constant voltage that does not deteriorate with time.
(4) The battery introduces no resistance to the circuit.
These assumptions are simplifications that are acceptable because although the real world cannot behave with the simplicity of the mathematical model, the amount of error introduced by making these assumptions is small.
Once we have the solution of a mathematical model then it should be tested against a real-life situation to see whether the model behaves reasonably closely to reality. Once the model has been accepted then it can be used to predict the behaviour of the system for input values other than those that it has been tested for.
The stages in solving a problem are as follows:
(1) Take to real problem and express it as a mathematical one using any necessary scientific rules and assumptions about the behaviour of the system and using letters to represent any unknown quantities. Include an account of any important assumptions and simplifications made.
(2) Solve the mathematical problem using your knowledge of mathematics.
(3) Translate the mathematical solution back into the setting of your original problem.
(4) Test the model solutions for some values to check that it behaves like the real-life problem.
Most mathematical problems are expressed by using equations, or inequalities, differential or difference equations, or by expressing a problem geometrically or a combination of all of these. We might need to incorporate a random element which results in the need to use a probabilistic model. In many of the following chapters we will look at the modelling process in more detail as we come across new mathematical tools and the situations in which they are used. To perform the entire modelling cycle properly, we need to be able to test our results in a reallife situation in order to reconsider assumptions used in the model. This would require access to engineering situations and tools. For this reason, engineering mathematics books tend to concentrate on those models that are commonly used by engineers. Many of the applied problems presented in the following chapters however do present an opportunity to move from an English language description of a problem to a mathematical language description of a problem, which is an important step in the modelling process.
In this chapter, we will look at translating a problem into mathematical language and, for the main part of the chapter, we concentrate on solving a mathematical problem and the reasoning that is involved in so doing.
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Problem solving and the art of the convincing argument 59
To solve the problem using your knowledge of mathematics, we need to use the ideas of mathematical statements and how to decide whether, and express the fact that, one statement leads logically on to the next. We shall mainly use examples of solving equations and inequalities although the same ideas apply to the solving of all problems.

3.2 Describing a problem in mathematical language

The stages in expressing a problem in mathematical language can be summarized as:
(1) Assign letters to represent the unknown quantities. (2) Write down the known facts using equations and inequalities, and
using drawings and diagrams where necessary. (3) Express the problem to be solved mathematically.
This is not a simple process because it involves a great deal of interpretation of the original problem. It is useful to try to limit the number of unknowns used as much as possible, or the problem may appear more difficult than necessary.

Example 3.1 Express the following problem mathematically: A web development company employs a freelance web designer and a freelance graphic designer to put up listings for new businesses on to their virtual business park website. Business customers are charged e200 per year for a listing. The fixed costs of the web development company amount to e2000 per week over 52 weeks in the year. The web designer charges e80 per listing and the graphic designer e100 per listing and both can prepare these at the rate of 2.4 listings in a day. The freelancers work for up to 200 days per year. How many listings does the company need in the first year to break even?
Solution The mathematical problem can be expressed by firstly assigning letters to some of the unknown quantities and then write down all the known facts as equations or inequalities.
First assign letters: Total number of listings of businesses on the park in the first year is L. LW is the number of listings prepared by the web designer and LG is the number of listings prepared by the graphic designer. The costs are K per year and the profit is P . The known facts can be expressed as follows:
L = LW + LG
This expresses the fact that the total number of listings L is made up of those prepared by the web designer and those prepared by the graphic designer. As there are up to 200 working days in a year and they both do a maximum of 2.4 listings per day.
0 LW 480, 0 LG 480
The costs, K, are; fixed costs of 2000 × 52, plus the cost of the freelance web designer at 80LW, plus the cost of the freelance graphic designer at
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60 Problem solving and the art of the convincing argument
100LG. This can be expressed as:
K = 104 000 + 80LW + 100LG.
We need to relate the profit to the other variables. As the profit is 200 multiplied by the number of jobs minus the total costs, we get:
P = 200L − K
Finally, we must express the mathematical problem that we would like to solve. For the web development business to make a profit in the first year then the profit must be positive, hence we get the problem expressed as: Find the minimum L such that P > 0.
Example 3.2 Express the following in mathematical language: A car brake pedal, as represented in Figure 3.2(a) is pivoted at point A. What is the force on the brake cable if a constant force of 900 N is applied by the driver’s foot and the pedal is stationary.
Solution First, we assign letters to the unknowns. Let F = the force on the brake cable.
In order to write down the known facts we need to consider what scientific laws can be used. As the force applied on the pedal initially provides a turning motion then we know to use the ideas of moments. The moment of a force about an axis is the product of the force F and its perpendicular distance, x, to the line of action of the force. Furthermore, as the pedal is now stationary, then the moments must be balanced so the clockwise moment must equal the anti-clockwise moment.
To use this fact, we need to use two further measurements, currently unknown, the perpendicular distance from the line of action of the force provided by the driver to the axis, A. This is marked as x1 m on the diagram in Figure 3.2(b). The other distance is the perpendicular distance from the line of action of the force on the cable to the axis A. This is marked as x2 m in Figure 3.2(b).
We can now write down the known facts, involving the unknowns x1, x2, and F . From the right angle triangle containing x1, we have

Figure 3.2 (a) A representation of a car brake pedal. (b) The same diagram as (a) with some unknown quantities marked and triangles used to formulate the problem.

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Problem solving and the art of the convincing argument 61
(converting 210 mm = 0.21 m), cos(40◦) = x1
0.21
From the right angle triangle containing x2, we have (converting 75 mm = 0.075 m), cos(15◦) = x2
0.075
The moments of the forces can now be calculated and equated. The clockwise moment is 800x1 and the anti-clockwise moment is given by F x2 and hence we have:
800x1 = F x2
Finally, we need to express the problem we are trying to solve. In this case it is simply ‘what is F ?’.
Note that in both Examples 3.1 and 3.2, certain modelling assumptions had been used in order to formulating the ‘natural language’ description of the problem that we were given. For instance, it is probable that the business park listings for the business park in Example 3.1 are not all identical and therefore average figures for times and costings have been used. Similarly, in Example 3.2 no mention has been made of friction would provide an extra force to consider. Here we have only considered the transition from natural language and accompanying diagrams to the mathematical problem. We have implicitly assumed that the modelling process can be performed in two stages. From real-life problem to a natural language description which incorporates some simplifying assumptions, and then from there to a mathematical description. In reality modelling a system is much more involved. We would probably repeat stages in this process if we decided that the mathematical description was too complex and return to the real-life situation in order to make new assumptions.
We are now in a position to discuss mathematical statements and how to move from the statement of the problem to finding the desired solution.

3.3 Propositions and predicates

When we first set up a problem to be solved, we write down mathematical expressions like:

2 + 3 =?

(3.1)

and

12 = (3 + 4 + x) · 1

(3.2)

These are mathematical statements with an unknown value. Statements containing unknowns (or variables) are called predicates. A predicate can be either true or false depending on the value(s) substituted into it. When values are substituted into a predicate it becomes a simple proposition. If in Equation (3.1) we substitute 5 for the question mark we get:

2 + 3 = 5 ⇔ 5 = 5 which is true.

If, however, we substitute 6 we get:

2 + 3 = 6 ⇔ 5 = 6 which is false.

2 + 3 = 5 and 2 + 3 = 6 are examples of propositions. These are simple statements that can be assigned as either true or false. They contain no

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62 Problem solving and the art of the convincing argument
unknown quantities. Notice that if we simply rewrite a proposition or predicate we use ‘≡’ or ‘⇔’ to mean ‘is equivalent to’ or ‘is the same as’.
In Equation (3.2) if we substitute 4 for x we get:
12 = 11 which is false
but, if we substitute 5 for x we get
12 = 12 which is true.
Example 3.3 Assign true or false to the following: (a) (3x − 2)(x + 5) = 10 where x = 1 (b) 5x2 − 2x + 1 = 25 where x = −2 (c) y > 5t + 3 where y = 2 and t = −3
Solution (a) Substitute x = 1 in the expression and we get:
(3 · 1 − 2)(1 + 5) = 10 ⇔ 1(6) = 10 ⇔ 6 = 10 which is false.
(b) Substituting x = −2 into 5x2 − 2x + 1 = 25 gives
5(−2)2 − 2(−2) + 1 = 25 ⇔ 20 + 4 + 1 = 25 ⇔ 25 = 25 which is true.
(c) Substituting y = 2 and t = −3 into y > 5t + 3 gives
2 > 5(−3) + 3 ⇔ 2 > −15 + 3 ⇔ 2 > −12 which is true.
Like functions, predicates have a domain which is the set of all allowed inputs to the predicate. For instance, the predicate 1/(x − 1) = 1, where x ∈ R, has the restriction that x = 1, as letting x equal 1 would lead to an√attempt to divide by 0, which is not defined.
x − 2 = 25 where x ∈ R has the restriction that x 2, as values of x less than 2 would lead to an attempt to take the square root of a negative number, which is not defined.

3.4 Operations on propositions and predicates

Consider the problem given in Example 3.1. Notice that the conditions that we discovered when writing down the known facts must all be true in any solution that we come up with. If any one of these conditions is not true then we cannot accept the solution. The first condition must be true and the second and the third, etc.
Here we have an example of an operation on predicates. In Chapter 1, we defined an operation on numbers is a way of combining two numbers to give a single number. ‘And’, written as ∧ is an operation on two predicates or propositions which results a single predicate or proposition.
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Table 3.1 Truth table for the operation ‘and’. This table can also be expressed by T ∧T ⇔ T, T ∧F ⇔ F, F ∧T ⇔ F, F ∧ F ⇔ F . T stands for ‘true’ and F stands for ‘false’

p

q

p∧q

T

T

T

T

F

F

F

T

F

F

F

F

Table 3.2 Truth table
for ‘or’, This table can
also be expressed by T ∨T ⇔ T, T ∨F ⇔ T, F ∨T ⇔ T, F ∨F ⇔F

p

q

p∨q

T

T

T

T

F

T

F

T

T

F

F

F

Table 3.3 The truth table for ‘not’, ¬. This table can also be expressed by ¬T ⇔ F , ¬F ⇔ T

p

¬p

T

F

F

T

Problem solving and the art of the convincing argument 63
Therefore, to express the fact that both L 0 and L = LW + LG we can write
L 0 ∧ L = LW + LG
and the compound statement is true if each part is also true. As propositions can only be either true (T) or false (F), all possible
outcomes of the operation can easily be listed in a small table called a truth table. The truth table for the operation of ‘and’ is given in Table 3.1. p and q represent any two propositions, for instance, for some given values of L, LW, and LG, p and q could be defined by:
p: ‘L 0’ q: ‘L = LW + LG’
Another important operation is that of ‘or’. One example of the use of this operation comes about by solving a quadratic equation. One way of solving quadratic equations is to factorize an expression which is equal to 0.
To solve x2 − x − 6 = 0, the left-hand side of the equation can be factorized to give (x − 3)(x + 2) = 0.
Now we use the fact that for two numbers multiplied together to equal 0 then one of them, at least, must be 0, to give:
(x − 3)(x + 2) = 0 ⇔ (x − 3) = 0 or (x + 2) = 0
‘or’ can be written using the symbol ∨. The compound statement is true if either x − 3 = 0 is true or if x + 2 = 0 is true. Therefore, to express the statement that either x − 3 = 0 or x + 2 = 0 we can write:
(x − 3) = 0 ∨ (x + 2) = 0
∨ is also called ‘non-exclusive or’ because it is also true if both parts of the compound statement are true. This usage is unlike the frequent use of ‘or’ in the English language, where it is often used to mean a choice, for example, ‘you may have either an apple or a banana’ implies either one or the other but not both. This everyday usage of the word ‘or’ is called ‘exclusive or’.
The truth table for ‘or’ is given in Table 3.2. A further operation is that of ‘not’ which is represented by the symbol ¬. For instance, we could express the sentence ‘x is not bigger than 4’ as ¬(x > 4). The truth table for ‘not’ is given in Table 3.3.
Example 3.4 Assign truth values to the following: (a) x − 2 = 3 ∧ x2 = 4 when x = 2 (b) x − 2 = 3 ∨ x2 = 4 when x = 2 (c) ¬(x − 4 = 0) when x = 4 (d) ¬((a − b) = 4) ∨ (a + b) = 2 when a = 5, b = 3
Solution
(a) x − 2 = 3 ∧ x2 = 4 when x = 2
Substitute x = 2 into the predicate, x − 2 = 3 ∧ x2 = 4 and we get 2 − 3 = 3 ∧ 22 = 4.
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64 Problem solving and the art of the convincing argument
The first part of the compound statement is false and the second part is true. Overall, as F ∧ T ⇔ F , the proposition is false.
(b) x − 2 = 3 ∨ x2 = 4 when x = 2
Substitute x = 2 into the predicate and we get
2 − 2 = 3 ∨ 22 = 4
The first part of the compound statement is false and second part is true. As F ∨ T ⇔ T , the proposition is true.
(c) ¬(x − 4 = 0) when x = 4
When x = 4 the expression becomes:
¬(4 − 4 = 0) ⇔ ¬T ⇔ F
(d) ¬((a − b) = 4) ∨ (a + b) = 2 when a = 5, b = 3 ¬((a − b) = 4) when a = 5, b = 3 gives ¬(5 − 3 = 4) ⇔ ¬F ⇔ T (a + b) = 2 when a = 5, b = 3 gives (5 + 3) = 2 ⇔ 8 = 2 ⇔ F
Overall, T ∨ F ⇔ T so ¬((a − b) = 4) ∨ (a + b) = 2 when a = 5, b = 3 is true.
Example 3.5 Represent the following inequalities on a number line: (a) x > 2 ∧ x 4 (b) x < 2 ∨ x 4 (c) ¬(x < 2)
Solution (a) x > 2 ∧ x 4. To represent the operation of ‘and’, find where
the two regions overlap (Figure 3.3a). x > 2 ∧ x 4 can also be represented by 2 < x 4. (b) x < 2 ∨ x 4. To represent the operation of ∨, ‘or’, take all points on the first highlighted region as well as all points in the second highlighted region and any end points (Figure 3.3b). (c) ¬(x < 2). To represent the operation of ‘not’ take all the points on the number line not in the original region (Figure 3.3c). This can also be expressed by x 2.

3.5 Equivalence

We can now express an initial problem in terms of a predicate, probably an equation, a number of equations, or a number of inequalities. However, to solve the problem we need to be able to move from the original expression of the problem toward the solution. In Chapter 3 of the Background Mathematics Notes, available on the companion website for this book, we discussed how to solve various types of equations and introduced the idea of equivalent equations. In Chapter 2 we also looked at equivalent inequalities. In both cases we used the idea that in moving from one expression to an equivalent expression the set of solutions remained the
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Problem solving and the art of the convincing argument 65
Figure 3.3 The operations (a)∧ (and); (b) ∨ (or); and (c) ¬ (not).
same. In general, two predicates are equivalent if they are true for exactly the same set of values. We use our knowledge of mathematics to determine what operations can be performed that will maintain that equivalence. The rule that can be used to move from one equation to another was given as: ‘Equations remain equivalent if the same operation is performed to both sides of the equation’. In the case of quadratic equations we can also use a formula for the solution or use a factorization and the fact that: ab = 0 ⇔ a = 0 or b = 0. In passing from one equation to an equivalent equation we should use the equivalence symbol. This then makes a mathematical sentence: x+5=3
⇔ x =3−5 can be read as ‘The equation x + 5 = 3 is equivalent to x = 3 − 5’.
In all but the most obvious cases, it is a good practice to list a short justification for the equivalence by the side of the expression. x+5=3
⇔ x = 3 − 5 (subtracting 5 from both sides) ⇔ x = −2. Because of the possibility of making a mistake, the solution(s) should be checked by substituting the values into the original expression of the problem. To check, substitute x = −2 into the original equation giving −2 + 5 = 3 which is true, indicating that the solution is correct.
Example 3.6 Solve the following equation: x − 3 = 5 − 2x.
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66 Problem solving and the art of the convincing argument

Solution

x − 3 = 5 − 2x

⇔ x + 2x − 3 = 5 (adding 2x to both sides)

⇔ 3x = 8

(adding 3 to both sides)

⇔

x

=

8 3

(dividing both sides by 3).

Check by substituting x = 8/3 into the original equation:

8 3

−

3

=

5

−

2

×

8 3

⇔

−

1 3

=

−

1 3

,

which is true.

We looked at methods of solving inequalities in Chapter 2. The rules for finding equivalent inequalities were: ‘Perform the same operation to both sides’; but in the case of a negative number when multiplying or dividing the direction of the inequality sign must be reversed. To solve more complex inequalities, such as f (x) > 0, f (x) < 0, where f (x) is a continuous but non-linear function, then we solve f (x) = 0 and then use a number line to mark regions where f (x) is positive, negative or zero. The important thing in the process is to present a short justification of the equivalence. Finally, when the set of solutions has been found, some of the solutions can be substituted into the original expression of the problem in order to check that no mistakes have been made.

Example 3.7 Solve the following inequalities:
(a) 3x − 1 < 6x + 2 (b) x2 − 5x > −6

Solution

(a) 3x − 1 < 6x + 2

⇔ −1 < 6x + 2 − 3x (subtracting 3x from both sides)

⇔ −1 − 2 < 3x

(subtracting 2 from both sides)

⇔ − 3 < 3x 33
⇔ −1 < x

(dividing both sides by 3)

⇔ x > −1

Check: Test a few values from the set x > −1 and substitute into 3x − 1 < 6x + 2 Try x = 0: this gives −1 < 2 ⇔ T Try x = 2: this gives 3(2) − 1 < 6(2) + 2 ⇔ 5 < 14 ⇔ T
(b) x2 − 5x > −6
Write the inequality with 0 on one side of the inequality sign
x2 − 5x > −6 ⇔ x2 − 5x + 6 > 0 (adding 6 to both sides)
Find the solutions to f (x) = 0 where f (x) = x2 − 5x + 6 and mark them on a number line as in Figure 3.4.
√ x2 − 5x + 6 = 0 ⇔ x = 5 ± 25 − 24
2

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Problem solving and the art of the convincing argument 67

Figure 3.4 Solving x 2 − 5x + 6 > 0.

(using the formula for solution of quadratic equations)

⇔ x = 5+1 ∨x = 5−1

2

2

⇔ x =3∨x =2

Using the fact that the function is continuous, we can substitute values for x which lie on either side of the roots of f (x) = 0 in order to find the sign of the function in that region. Here, we choose 0, 2.5, and 4 and find that
when x = 0: f (x) = x2 − 5x + 6 = 6, so f (x) > 0
when x = 2.5: f (x) = x2 − 5x + 6 = 6.25 − 12.5 + 6 = −0.25, so f (x) < 0
when x = 4: f (x) = x2 − 5x + 6 = 16 − 20 + 6 = 2, so f (x) > 0

These regions are marked on the number line as in Figure 3.4 and this gives the solution to f (x) > 0 as x < 2 ∨ x > 3.
Check: A check is to substitute some values from the solution set x < 2 ∨ x > 3 into the original predicate x2 − 5x > −6
Substitute x = 1, this gives 1 − 5 > −6 ⇔ −4 > −6 ⇔ T Substitute x = 5, this gives 25 − 25 > −6 ⇔ 0 < −6 ⇔ T
It therefore appears that this solution is correct.

3.6 Implication

We previously described one method of finding equivalent equations as that of ‘doing the same thing to both sides’. This was rather simplistic but a useful way of seeing it at the time. There are only certain things that can be ‘done to both sides’ like adding, subtracting, multiplying by a non-zero expression, or dividing by a non-zero expression that always maintain equivalence. There are also many operations that can be performed to both sides of an equation which do not give an equivalent equation but give an equation with the same solutions and yet more besides. In this situation we say that the first equation implies the second equation. The symbol for implies is ⇒.
An example of implication is given by squaring both sides of the equation
x − 2 = 2 ⇒ (x − 2)2 = 4
The first predicate x − 2 = 2 has only one solution, x = 4, the second predicate has two solutions x = 4 and x = 0. By squaring the equation we have found a new equation which includes all the solutions of the first
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68 Problem solving and the art of the convincing argument
equation, and has one more beside. Implication is expressed in English by using phrases like ‘If . . . then . . .’.
An expression involving an implication cannot always be turned the other way around in the same way as those involving equivalence can. An example of this is given by the following statement. It is true that: ‘If I am going to work then I take the car’ which can be written using the implication symbol as:

‘I am going to work’ ⇒ ‘I take the car’

However, it is not true that:

‘If I take the car then I am going to work’

This is because there are more occasions when I take the car than simply going to work.
More examples are:

‘I only clean the windows if it is sunny’ ‘I am cleaning the windows’ ⇒ ‘it is sunny’

This does not mean that ‘If it is sunny then I clean the windows’, as there are some sunny days when I have to go to work or just laze in the garden, or I am on holiday.
An implication sign can be written, and read, from left to right
‘It is sunny’ ⇐ ‘I am cleaning the windows’
which I can still read as ‘I am cleaning the windows therefore it is sunny’ or I could try rearranging the sentence as ‘Only if it is sunny will I clean the windows’.
The various ways of expressing these sentiments can get quite involved. The important point to remember is that p ⇒ q means that q must be true for all the occasions that p is true, but q could be true on more occasions besides. Going back to equations or inequalities:

Figure 3.5 P is the solution
set of p, Q is the solution set of q. p ⇒ q means that P ⊆ Q. D is the domain of p and q.

p⇒q
means that the solution set, P, of p is a subset of the solution set, Q, of q. This is pictured in Figure 3.5.
We can now see that for two equations or inequalities to be equivalent then p ⇒ q and q ⇒ p. This means that their solution sets are exactly the same (Figure 3.6).

Figure 3.6 P is the solution set of p and Q is the solution set of q. Then p ⇔ q means that P = Q.

Example 3.8 Fill in the correct symbol in each case either ⇒, ⇐ or ⇔

(a) x2 − 9 = 0 . . . x = −3

(b)

x

=

−

1 2

.

.

.

(1/(2x

− 5)(x

+ 1))

=

−

1 3

where

x

∈

R,

x

=

5,

x = −1

(c) (x − 3)(x − 1) > 0 · · · x > 3 ∨ x < 1

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Problem solving and the art of the convincing argument 69

Solution (a) x2 − 9 = 0 · · · x = −3 Solving x2 − 9 = 0 gives x2 − 9 = 0 ⇔ (x − 3)(x + 3) = 0
⇔ x = 3 ∨ x = −3.
Hence, −3 is only one of the solutions of the first equation so the correct expression is x2 − 9 = 0 ⇐ x = −3.

(b)

x

=

−

1 2

.

.

.

(1/(2x

− 5)(x

+ 1))

=

−

1 3

where x

∈

R,

x

=

5,

x = −1

Solving

(2x

−

1 5)(x

+

1)

=

−1 3

gives

(2x

−

1 5)(x

+

1)

=

−1 3

⇔

−3 = (2x − 5)(x + 1)

(multiplying both sides by (2x−5)(x + 1) and as x = 5, x = 1)

⇔ −3 = 2x2 − 3x − 5 (multiplying out the brackets)

⇔ 2x2 − 3x − 2 = 0

(adding 3 on to both sides of the equation) √
⇔ x = 3 ± 9 + 16 4

(using the quadratic formula to solve the equation)

⇔ x = 3±5 4

⇔

x

=

2

∨

x

=

−

1 2

The second predicate

(2x

−

1 5)(x

+

1)

=

−1 3

has

more

solutions

than

the

first

predicate

x

=

−

1 2

.

Thus,

the

correct

expression is:

x

=

−1 2

⇒

(2x

−

1 5)(x

+

1)

=

−1 3

where x ∈ R, x = 5, x = −1.

(c) (x − 3)(x − 1) > 0 · · · x > 3 ∨ x < 1 Solve the inequality on the left by firstly solving f (x) = 0:

(x − 3)(x − 1) = 0 ⇔ x = 3 ∨ x = 1

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70 Problem solving and the art of the convincing argument Choosing values on either side of the roots, for example 0,2,4 gives

f (0) = (−3)(−1) = 3, i.e. f (x) > 0

f (2) = (−1)(1) = −1, i.e. f (x) < 0

f (4) = (1)(3),

i.e. f (x) > 0

Figure 3.7 Solving (x − 3)(x − 1) > 0.

This is then marked on a number line as in Figure 3.7. As the solution to (x − 3)(x − 1) > 0 is x < 1 ∨ x > 3; we have
therefore shown that (x − 3)(x − 1) > 0 ⇔ x > 3 ∨ x < 1.

3.7 Making sweeping statements

In Chapter 1 of the Background Mathematics Notes, available on the companion website for this book, we made some statements about numbers which we stated were true for all real numbers. Some of these were the commutative laws:
a + b = b + a and ab = ba
and the distributive law:
a(b + c) = ab + bc.
There is a symbol which stands for ‘for all’ or ‘for every’ which allows these laws to be expressed in a mathematical shorthand
∀a, b ∈ R a + b = b + a ∀a, b ∈ R ab = ba ∀a, b, c ∈ R a(b + c) = ab + bc.
Rules, such as the commutative law, are axioms for numbers and need not be proved true. However, more involved expressions, such as ∀a, b ∈ R (a − b)(a + b) = a2 − b2 need to be justified.
If the symbol ‘for all’ is used with a predicate about its free variable then it becomes a simple proposition which is either true or false. To show that an expression is true we use our knowledge of mathematics to write equivalent expressions until we come across an expression which is obviously true (like a = a). To prove it is false is much easier. As we have made a sweeping statement about the expression and said it is true for all a, b then we only need to come across one example of numbers which make the expression false.
Example 3.9 Are the following true or false? Justify your answer. (a) ∀a, b ∈ R, a3 − b3 = (a − b)(a2 + ab + b2) (b) ∀t ∈ R, where t = 1, t = −1 1/(t + 1) = (t − 1)/(t2 − 1) (c) ∀x ∈ R, where x = 0 (x2 − 1)/x = x − 1
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Problem solving and the art of the convincing argument 71

Solution (a) ∀a, b ∈ R a3 − b3 = (a − b)(a2 + ab + b2)

Looking at the right-hand side of the equality we have
(a − b)(a2 + ab + b2) = a(a2 + ab + b2) − b(a2 + ab + b2) (taking out the brackets) = a3 + a2b + ab2 − ba2 − ab2 − b3 (taking out the remaining brackets) = a3 − b3 (simplifying)

We have shown that the right-hand side is equal to the left-hand side ∀a, b ∈ R a3 − b3 = (a − b)(a2 + ab + b2) ⇔ a3 − b3 = a3 − b3

which is true. Therefore ∀a, b ∈ R a3 − b3 = (a − b)(a2 + ab + b2)

is true. (b) ∀t ∈ R where t = 1, t = −1 1/(t + 1) = (t − 1)/(t2 − 1)

Take the right-hand side of the equality

t −1 t2 − 1

=

(t

t −1 − 1)(t +

1)

=1 (t + 1)

(factorizing the bottom line)
(dividing the top and bottom line by t − 1 which is allowed as t = 1)

Hence

1

t −1

1

1

t + 1 = t2 − 1 ⇔ t + 1 = t + 1

which is true. Thus,

1

t −1

∀t ∈ R where t = 1, t = −1 t + 1 = t2 − 1

is true.

(c)

∀x ∈ R where x = 0

x2 − 1 x =x−1

To show this is false, substitute a value for x, for example, x = 2. When x=2

x2 − x

1

=

x

−

1

becomes

4 − 1 = 2 − 1 ⇔ 3 = 1 ⇔ F.

2

2

As the predicate fails for one value of x then

∀x ∈ R where x = 0 (x2 − 1)/x = x − 1

is false.

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72 Problem solving and the art of the convincing argument
Another useful symbol is ∃, which means, ‘there exists’. This can be used to express the fact that every real number has an inverse under addition. Hence, we get
∀a ∈ R, ∃b, a + b = 0.
If the symbol ∃ is used with a predicate about its free variable, it becomes a simple proposition which is either true or false. In the case of the example given concerning the inverse, this is an axiom of the real numbers and we can just state it is true. Other statements involving existence will need some justification. Proving existence is simpler than disproving it. If I were to state ‘There exists a blue moon in the universe’, to prove this to be true I only need to find one blue moon but to disprove it I must find all the moons in the universe and show that not one of them is blue.
In other words, to show that some value exists which makes a certain predicate into a true proposition then we only need to find that value and demonstrate that the resulting proposition is true. To show that no value exists, however, is more difficult and if the domain of interest is a set of numbers we need to present an argument about any member of the set.
Example 3.10 Are the following true or false? Justify your answer.
(a) ∃x ∈ R, (x + 2)(x − 1) = 0 (b) ∃x ∈ R, x2 + 4 < 0
Solution
(a) ∃x ∈ R, (x + 2)(x − 1) = 0
To show this is true, we only need find one value of x which makes the equality correct. For instance, take x = −2: when x = −2, (x + 2)(x − 1) = 0 becomes (−2 + 2)(−2 − 1) = 0 ⇔ 0 = 0, which is true.
Therefore, ∃x ∈ R, (x + 2)(x − 1) = 0 is true.
(b) ∃x ∈ R, x2 + 4 < 0
Trying a few values for x (e.g. −1, 0, 20, −2) we might suspect that this statement is false. We need to present a general argument in order to convince ourselves of this.
x2 is always positive or zero, that is, x2 0 for all x. If we then add on 4, then for all x, x2 + 4 4 and as 4 is bigger than 0.
x2 + 4 > 0 for all x; hence, ∃x ∈ R, x2 + 4 < 0 is false.

3.8 Other applications of predicates

Predicates are often used in software engineering. Some simpler applications are: (a) To express the condition under which a program block will be carried
out (or a loop will continue execution). (b) To express a program specification in terms of its pre- and post-
conditions.
Example 3.11 Express the following in pseudo-code: print x and y if y is a multiple of x and x is an integer between 1 and 100 inclusive.
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3.9 Summary

Problem solving and the art of the convincing argument 73

Solution Pseudo-code is a system of writing algorithms which is similar to some computer languages but not in any particular computer language. We can use any symbols we like as long as the meaning is clear.
y is a multiple of x means that if y is divided by x then the result is an integer. This can be expressed as

y x

∈Z

The condition that x must lie between 1 and 100 can be expressed as x 1 and x 100. Combining these conditions gives the following interpretation for the algorithm:

y if x ∈ Z ∧ x 1 ∧ x 100 then
print x, y endif

Example 3.12 A program is designed to take a given whole positive number, x, greater than 1, and find two factors of x, a and b, which multiplied together give x. a and b should be whole positive numbers different from 1, unless x is prime. Express the pre- and post-conditions for the program.
Solution Pre-condition is x ∈ N ∧ x > 1. The post-condition is slightly more difficult to express. Clearly ab = x
is a statement of the fact that a and b must multiply together to give x. Also a and b must be elements of N. a and b cannot be 1 unless x is prime, this can be expressed by
(a = 1 ∧ b = 1) ∨ (x is prime).
Finally, we have the post-condition as
a · b = x ∧ (a ∈ N) ∧ (b ∈ N) ∧ ((a = 1 ∧ b = 1) ∨ (x is prime)).

(1) The stages in solving a real-life problem using mathematics are: (a) Express the problem as a mathematical one, using any necessary scientific rules and assumptions about the behaviour of the system and using letters to represent any unknown quantities. This is called a mathematical model. (b) Solve the mathematical problem by moving from one statement to an equivalent statement justifying each stage by using relevant mathematical knowledge. (c) Check the mathematical solution(s) by substituting them into the original formulation of the mathematical problem. (d) Translate the mathematical solution back into the setting of the original problem. (e) Test the model solutions for some realistic values to see how well the model correctly predicts the behaviour of the system. If it is acceptable, then the model can be used to predict more results.
(2) A predicate is a mathematical statement containing a variable. Examples of predicates are equations and inequalities.
(3) If values are substituted into a predicate it becomes a simple proposition which is either true or false.
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74 Problem solving and the art of the convincing argument
(4) The three main operations on predicates and propositions are ∧, ∨, ¬, and these can be defined using truth tables as in Tables 3.1–3.3.
(5) Two predicates, p, q, are equivalent (p ⇔ q) if they are true for exactly the same set of values.
(6) p ⇒ q means ‘p implies q’, that is, q is true whenever p is true. If p, q are equations or inequalities and p ⇒ q then all solutions of p are also solutions of q and q may have more solutions besides.
(7) The symbol ∀ stands for ‘for all’ or ‘for every’ and can be used with a predicate to make it into a simple proposition, for example, ∀a, b ∈ R, a2 − b2 = (a + b)(a − b), which is true.
(8) The symbol ∃ stands for ‘there exists’ and can also be used with a predicate to make it into a simple proposition, for example, ∃x ∈ R, 3x = 45, which is true.

3.10 Exercises

3.1. Assign T or F to the following

(a) 2x + 2 = 10 when x = 1

(b) 2x + 2 = 10 when x = 2

(c) 3x2 + 3x − 6 = 0 when x = 1

(d) 1 − t2 = −3 when t = −2

(e) t − 5 = 6.5 ∧ t + 4 = 2.5 when t = 1.5

(f) u + 3 = 6 ∧ 2u − 1 = 4 when u = 3

(g) 3y + 2 = −2.5 ∨ 1 − y = 1 when y = −1.5

(h) ¬(x2 − x + 2 = 0) when x = −1

(i) ¬(t − 2 = 4 ∧ t = 3)

(j) ¬(t − 2 = 4) ∧ (t = 3)

(k)

¬

3t

−

4

=

6

∨

1

−

t

=

−2

1 3

when

t

=

3

1 3

.

3.2. Solve the following, justifying each stage of the solution
and checking the result. (a) 3 − 2x = −1, (b) 1 − 2t2 = 1 − 10t (c) 50t − 11 = −25t2, (d) 30y − 13 = 8y2 (e) 10t − 4 −3, (f) 10 − 4x > 12.

3.3. Find the range of values for which the following hold
and represent them on a number line. (a) x +3 5∨1−2x > 3, (b) 2−4t 3∧2−t < 1 (c) ¬(2x + 3 9).

3.4. Fill the correct sign ⇒, ⇐ or ⇔ or indicate none of these. Assume the domain is R unless indicated

otherwise.

(a) 3x2 − 1 = 0 · · · x = √1

√

3

(b) x − 1 = 5 · · · x = 26 (where x 1)

(c) t2 − 5t = 36 · · · (t − 4)(t − 9) = 0

(d) (2x − 2)/(x − 3) = 1 · · · 3x + 4 = −x (where x = 3)
(e) 3x = 4 · · · (3x)2 = (4)2 (f) t + 1 = 5 · · · (t + 1)3 = 53
(g) (x +1)(x −3) =√(x −3)(x +2) · · · (x +1) = (x +2) (h) x − 1 = 25 · · · x − 1 = 5 where x 1

(i) w/(w2 − 1) = 1 · · · 1/(w − 1) = 1 where w = 1 and w = −1
(j) (x − 1)(x − 3) < 0 · · · (x − 3) < 0 ∨ (x − 1) < 0 (k) x > 2 ∨ x < −2 · · · x2 > 4.

3.5. Determine whether the following statements are true or
false and justify your answer. (a) ∀a, b ∈ R, a4 − b4 = (a − b)(a3 − a2b + ab2 − b3) (b) ∀a, b ∈ R, a3 + b3 = (a + b)3
(c) ∀x ∈ R, x = 0 1/(1/x) = x (d) ∃t, t ∈ R, t2 − 3 = 4 (e) ∃t, t ∈ R, t2 + 3 = 0.

3.6. Write the following conditions using mathematical
symbols: (a) x is not divisible by 3, (b) y is a number between 3 and 60 inclusive, (c) w is an even number greater than 20, (d) t differs from tn−1 by less than 0.001.

3.7. Express the following problems mathematically and

solve them:

(a) A set of screwdrivers cost e10 and hammers

e6.50. Find the possible combinations of maximum

numbers of screwdriver and hammer sets that can

be bought for e40.

(b) An object is thrown vertically upwards from the ground with an initial velocity of 10 m s−1. The

mass of the object is 1 kg. Find the maximum height

that the object can reach using

(i)

Kinetic energy (KE) is given by

1 2

mv2

,

where

m is its mass and v its velocity.

(ii) The potential energy (PE) is mgh, where m is

the mass, g is the acceleration due to gravity (which can be taken as 10 m s−2), and h is the

height.

(iii) Assuming that no energy is lost as heat due to

friction, then the conservation of energy law

gives KE + PE = constant.

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Problem solving and the art of the convincing argument 75

3.8. A road has a bend with radius of curvature 100 m. The road is banked at an angle of 10◦. At what speed should
a car take the bend in order not to experience any side
thrust on the tyres? Use the following assumptions:
(a) The sideways force needed on the vehicle in order to
maintain it in circular motion (called the centripetal force) = mv2/r where r is the radius of curvature of the bend, v the velocity, and m the mass of the vehicle.
(b) The only force with a component acting sideways
on the vehicle, is the reactive force of the ground.
This acts in a direction normal to the ground (i.e. we
assume no frictional force in a sideways direction). (c) The force due to gravity of the vehicle is mg, where
m is the mass of the vehicle and g is the acceleration due to gravity (≈ 9.8 m s−2). This acts vertically downwards. The forces operating on the vehicle and
ground, in a lateral or vertical direction, are pictured
in Figure 3.8.

Figure 3.8 A vehicle rounding a banked bend in the road. R is the reactive force of the ground on the vehicle. The vehicle provides a force of mg, the weight of the vehicle, operating vertically downwards. The vehicle needs a sideways force of mv 2/r in order to maintain the locally circular motion.

TLFeBOOK

4

Boolean algebra

4.1 Introduction

Boolean algebra can be thought of as the study of the set {0, 1} with the operations + (or), . (and), and − (not). It is particularly important because of its use in design of logic circuits. Usually, a high voltage represents TRUE (or 1), and a low voltage represents FALSE (or 0). The operation of OR (+) is then performed on two voltage inputs, using an OR gate, AND(.) using an AND gate and NOT is performed using a NOT gate. This very simple algebra is very powerful as it forms the basis of computer hardware.
You will probably have noticed that the operations of ∧ (AND), ∨ (OR), and ¬ (NOT) used in Chapter 3 for propositions are very similar to the operations ∩ (AND), ∪ OR, and (NOT) (complement) used for sets. This connection is not surprising as membership of a set, A, could be defined using a statements like ‘3 is a member of A’ which is either TRUE or FALSE. In simplifying logic circuits, use is made of the different interpretations that can be put upon the operations and variables. We can use truth tables, borrowed from the theory of propositions, as given in Chapter 3, or we can use Venn diagrams, borrowed from set theory, as given in Chapter 1.
The first thing we shall examine in this chapter is what do we mean by an algebra and why are we able to skip between these various interpretations. Then we look at implementing and minimizing logic circuits.

4.2 Algebra

Before we look at Boolean algebra, we will have a look at some ideas about algebra:
(a) What is an algebra? (b) What is an operation? (c) What do we mean by properties (or laws or axioms) of an algebra?
An algebra is a set with operations defined on it. In Chapter 1 of the Background Mathematics Notes, available on the companion website for this book, we looked at the algebra of real numbers and defined an operation is a way of combining two numbers to give a single number. We could therefore define an operation as a way of combining two elements of the set to result in another element of the set.
Example 4.1 The set of real numbers, R, has the operations + and ., for example,
3 + 5 = 8 and 3 · 4 = 12
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4.3 Boolean algebras

Boolean algebra 77
and we could combine any two numbers in this way and we would always get another real number.
Example 4.2 Consider the set of sets in some universal set E , for example,
E = {a, b, c, d, e}
A = {a, d}, B = {a, b, c}
then
A ∩ B = {a} and A ∪ B = {a, b, c, d}.
The operations of ∩ and ∪ also result in another set contained in E . In both of these examples, the operations are binary operations because they use two inputs to give one output. There is another sort of operation which is important, called a unary operation, because it only has one input to give one output. Consider Example 4.2: A = {b, c, e} gives the complement of A. This is a unary operation as only one input, A, was needed to define the output A . If we can find a rule which is always true for an algebra then that is called a property, (law or axiom) of that algebra. For example, (3 + 5) + 4 = 3 + (5 + 4) is an application of the associative law of addition which can be expressed in general in the following way for the set of real numbers:
∀a, b, c ∈ R, (a + b) + c = a + (b + c)
If we can list all the properties of a particular algebra then we can give that algebra a name. For instance, the real numbers with the operations of + and . form a field.
Sets as a Boolean algebra
The sets contained in some universal set display a number of properties which can be shown using Venn diagrams.
Example 4.3 Show, using Venn diagrams, that, for any 3 sets A, B, C in some universal set E ,
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
Solution This can be shown to be true by drawing a Venn diagram of the left-hand side of the expression and another of the right-hand side of the expression. Operations are performed in the order indicated by the brackets and the result of each operation is given a different shading. This is done in Figure 4.1(a) and (b). The region shaded in Figure 4.1(a) representing A ∩ (B ∪ C) is the same as that representing (A ∩ B) ∪ (A ∩ C) in Figure 4.1(b), hence, showing that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
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78 Boolean algebra
Figure 4.1 (a) A Venn diagram of A ∩ (B ∪ C). (b) A Venn diagram of (A ∩ B) ∪ (A ∩ C).

Table 4.1 The truth tables defining the logical operators

p q p ∧ q p q p ∨ q p ¬p

TT

T

TT

T

TF

TF

F

TF

T

FT

FT

F

FT

T

FF

F

FF

F

In the same way, other properties can be shown to be true. A full list of the properties gives:
For every A, B, C ⊆ E

(B1) A ∪ A = A

A∩A = A

Idempotent

(B2) A ∪ (B ∪ C)

A ∩ (B ∩ C)

Associative

= (A ∪ B) ∪ C

= (A ∩ B) ∩ C

(B3) A ∪ B = B ∪ A

A∩B=B∩A

Commutative

(B4) A ∪ (A ∩ B) = A

A ∩ (A ∪ B) = A

Absorption

(B5) A ∪ (B ∩ C)

A ∩ (B ∪ C)

Distributive

= (A ∪ B) ∩ (A ∪ C) = (A ∩ B) ∪ (A ∩ C) laws

(B6) A ∪ E = E

A∩∅=∅

Bound laws

(B7) A ∪ ∅ = A

A∩E =A

Identity law

(B8) A ∪ A = E

A∩A =∅

Complement

laws

(B9) ∅ = E

E =∅

0 and 1 laws

(B10) (A ∪ B) = A ∩ B (A ∩ B) = A ∪ B De Morgan’s

laws

Notice that all the laws come in pairs (called duals). A dual of a rule is given by replacing ∪ by ∩ and ∅ by E and vice versa.

Propositions
We looked at propositions in Chapter 3. Propositions can either be given a value of TRUE (T) or FALSE (F). Examples of propositions are 3 = 5 which is false and 2 < 3 which is true. The logical operators of AND, OR, and NOT are defined using truth tables, which we repeats in Table 4.1.
Properties of propositions and their operations can be shown using truth tables.
Example 4.4 Show, using truth tables, that for any propositions p, q, r
(p ∧ q) ∧ r = p ∧ (q ∧ r)
Solution The truth tables are given in Table 4.2. Note that there are eight lines in the truth table in order to represent all the possible states (T, F) for the three variables p, q, and r. As each can be either TRUE or FALSE, in total there are 23 = 8 possibilities. To find (p ∧ q) ∧ r, p ∧ q is performed first and the result of that is ANDed with r. To find p ∧(q ∧r) then q ∧r is performed first and p is ANDed with the result. As the resulting columns are equal we can conclude that
(p ∧ q) ∧ r ⇔ p ∧ (q ∧ r)

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Boolean algebra 79
Table 4.2 A truth table to show (p ∧ q) ∧ r ⇔ p ∧ (q ∧ r ). The fifth column gives the truth values of (p ∧ q) ∧ r and the seventh column gives the truth value of p ∧ (q ∧ r ). As the two columns are the same we can conclude that (p ∧ q) ∧ r ⇔ p ∧ (q ∧ r )

p q r p ∧ q (p ∧ q) ∧ r q ∧ r p ∧ (q ∧ r )

TTT T

T

T

T

TTF T

F

F

F

TFT F

F

F

F

TFF F

F

F

F

FTT F

F

T

F

FTF F

F

F

F

FFT F

F

F

F

FFF F

F

F

F

Table 4.3 A truth table to show ¬(p ∧ q) ⇔ (¬p) ∨ (¬q). The fourth column gives the truth values of ¬(p ∧ q) and the seventh column gives the truth value of (¬p) ∨ (¬q). As the two columns are the same we can conclude that ¬(p ∧ q) ⇔ (¬p) ∨ (¬q)

p q p ∧ q ¬(p ∧ q) ¬p ¬q ¬p ∨ ¬q

TT

T

TF

F

FT

F

FF

F

F

F

F

F

T

F

T

T

T

T

F

T

T

T

T

T

Example 4.5 Show that for any two propositions p, q: ¬(p ∧ q) ⇔ (¬p) ∨ (¬q)

Solution The truth table is given in Table 4.3. It turns out that all the properties we listed for sets are also true for
propositions. We list them again, for any three propositions p, q, r

(B1) p ∨ p ⇔ p (B2) p ∨ (q ∨ r)
⇔ (p ∨ q) ∨ r (B3) p ∧ q ⇔ q ∨ p (B4) p ∨ (p ∧ q) ⇔ p (B5) p ∨ (q ∧ r)
⇔ (p ∨ q) ∧ (p ∨ r) (B6) p ∨ T ⇔ T (B7) p ∨ F ⇔ p (B8) p ∨ ¬p ⇔ T (B9) ¬F ⇔ T (B10) ¬(p ∨ q) ⇔ ¬p ∧ ¬q

p∧p⇔p p ∧ (q ∧ r) ⇔ (p ∧ q) ∧ r p∧q ⇔ q∧p p ∧ (p ∨ q) ⇔ p p ∧ (q ∨ r) ⇔ (p ∧ q) ∨ (p ∧ r) p∧F ⇔ F p∧T⇔p p ∧ ¬p ⇔ F ¬T ⇔ F ¬(p ∧ q) ⇔ ¬p ∨ ¬q

Idempotent Associative
Commutative Absorption Distributive laws
Bound laws Identity laws Complement laws 0 and 1 laws De Morgan’s laws

Notice again that all the laws are duals of each other. A dual of a rule is given by replacing ∨ by ∧ and F by T, and vice versa.

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80 Boolean algebra

Table 4.4 The operations of AND (.), OR (+) and NOT (−) defined for any variables a, b taken from the Boolean set {0, 1}

a b a.b a b a + b a a¯

00 0 00

0

01

01 0 01

1

10

10 0 10

1

11 1 11

1

The Boolean set {0, 1}
The simplest Boolean algebra is that defined on the set {0,1}. The operations on this set are AND (.), OR (+), and NOT (−). The operations can be defined using truth tables as in Table 4.1, shown again in Table 4.4. This time notice that the first two are usually ordered in order to mimic binary counting, starting with 0 0, then 0 1, then 1 0, then 1 1. This is merely a convention and the rows may be ordered any way you like. a and b are two variables which may take the values 0 or 1.
This now looks far more like arithmetic. However, beware because although the operation AND behaves like multiplication, 0.0 = 0, 0.1 = 0, 1.0 = 0 and 1.1 = 1 as in ‘ordinary’ arithmetic, the operation OR behaves differently as 1 + 1 = 1.
All the laws as given for sets and for propositions hold again and they can be listed as follows:
For any three variables a, b, c ∈ {0, 1}

(B1) a + a = a (B2) a + (b + c)
= (a + b) + c (B3) a + b = b + a (B4) a + (a.b) = a (B5) a + (b.c)
= (a + b).(a + c) (B6) a + 1 = 1 (B7) a.1 = a (B8) a + a¯ = 1 (B9) 0¯ = 1 (B10) (a + b) = a¯.b¯

a.a = a a.(b.c) = (a.b).c
a +b = b+a a.(a + b) = a a.(b + c) = (a.b) + (a.c) a.0 = 0 a+0=a a.a¯ = 0 1¯ = 0 (a.b) = a¯ + b¯

Idempotent Associative
Commutative Absorption Distributive laws
Bound laws Identity laws Complement laws 0 and 1 laws De Morgan’s laws

We often leave out the ‘.’ so that ‘ab’ means ‘a.b’. We also adopt the convention that . takes priority over + hence miss out some of the brackets.
Example 4.6 Evaluate the following where +, ., and − are Boolean operators. (a) 1.1.0 + 0¯.1 (b) (1.1¯) + 1 (c) (1¯ + 1).0 + (1 + 1).0

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Boolean algebra 81
Solution (a) We use the convention that . is performed first:
1.1.0 + 0¯.1 = 0 + 1.1 = 0 + 1 = 1 (b) (1.1¯) + 1 = (1.0) + 1 = 0 + 1 = 1 (c) (1¯ + 1).0 + (1 + 1).0 = 1.0 + 1.0 = 0 + 0 = 0
The algebraic laws can be used to simplify a Boolean expression.

Example 4.7 Simplify abc + a¯bc + bc¯

Solution
abc + a¯b + bc¯ = (a + a¯)bc + bc¯ = 1.bc + bc¯ = bc + bc¯ = b(c + c¯) =b

(using a distributive law) (using a complement law) (using an identity law) (by one of the distributive laws) (using a complement and identity law)

Although it is possible to simplify in this way, it can be quite difficult to spot the best way to perform the simplification; hence, there are special techniques used in the design of digital circuits which are more efficient.

4.4 Digital circuits
Figure 4.2 The three basic gates; NOT (−), AND (.), and OR (+).

Switching circuits form the basis of computer hardware. Usually, a high voltage represents TRUE (or 1) while a low voltage represents a FALSE (or 0). Digital circuits can be represented using letters for each input.
There are three basic gates which combine inputs and represent the operators NOT(−), AND (.), and OR (+). These are shown in Figure 4.2.
Other gates
Other common gates used in the design of digital circuits are the NAND gate, (ab), that is, not(ab), the NOR gate, (a + b), that is, not(a + b) and the EXOR gate, a ⊕ b, (exclusive or) a ⊕ b = ab¯ + ab
These gates are shown in Figure 4.3.
Implementing a logic circuit
First, we need to simplify the expression. Each letter represents an input that can be on or off (1 or 0). The operations between inputs are represented by the gates. The output from the circuit represents the entire Boolean expression.

Figure 4.3 Three other
common gates; NAND (ab), NOR (a + b), and EXOR (a ⊕ b).

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82 Boolean algebra

Example 4.8 Implement abc¯ + ab¯ + ac¯. Solution We can use absorption to write this as ab¯ + ac¯ and this can be implemented as in Figure 4.4 using AND, OR, and NOT gates. Alternatively, we can use the distributive and De Morgan’s laws to write the expression as:
ab¯ + ac¯ = a(b¯ + c¯) = abc
which can be implemented using an AND and a NAND gate.
Minimization and Karnaugh maps
It is clear that there are numbers of possible implementations of the same logic circuit. However, in order to use less components in building the circuit it is important to be able to minimize the Boolean expression. There are several methods for doing this. A popular method is using a Karnaugh map. Before using a Karnaugh map, the Boolean expression must be written in the form of a ‘sum of products’. To do this, we may either use some of the algebraic rules or it may be simpler to produce a truth table and then copy the 0s and 1s into the Karnaugh map. Example 4.9 is initially in the sum of product form and Example 4.10 uses a truth table to find the Karnaugh map.
Example 4.9 Minimize the following using a Karnaugh map:
ab + a¯b + ab¯
and draw the implementation of the resulting expression as a logic circuit.
Solution Draw a Karnaugh map as in Figure 4.5(a). If there are two variables in the expression then there are 22 = 4 squares in the Karnaugh map. Figure 4.5(b) shows a Karnaugh map with the squares labelled term

Figure 4.4 (a) An
implementation of abc¯ + ab¯ + ac¯ = ab¯ + ac¯ . (b) An alternative
implementation using ab¯ + ac¯ = abc.

Figure 4.5 (a) A two-variable Karnaugh map representing ab + a¯ b + ab¯ (b) A two-variable Karnaugh map with all the boxes labelled. (c) A Karnaugh map is like a Venn diagram. The second row represents the set a and the second column represents the set b.
TLFeBOOK

Figure 4.6 A two-variable
Karnaugh map representing ab + a¯ b + ab¯ .

Boolean algebra 83
by term. Figure 4.5(c) shows that the map is like a Venn diagram of the sets a and b. In Figure 4.5(a) we put a 0 or 1 in the square depending on whether that term is present in our expression. Adjacent 1s indicate that we can simplify the expression. Figure 4.6 indicates how we go about the minimization. We draw a line around any two adjacent 1s and write down the term representing that section of the map. We are able to encircle the second row, representing a, and the second column, representing b. As all the 1s have now been included we know that a + b is a minimization of the expression. Notice that it does not matter if one of the squares with a 1 in it has been included twice but we must not leave any out. The implementation of a + b is drawn in Figure 4.7.

Figure 4.7 An implementation of a + b.

Example 4.10 Minimize c(b + (ab)) + c¯ab and draw the implementation of the resulting expression as a logic circuit.
Solution First, we need to find the expression as a sum of products. This can be done by finding the truth table and then copying the result into the Karnaugh map. The truth table is found in Table 4.5. Notice that we calculate various parts of the expression and build up to the final expression. With practice, the expression can be calculated directly for instance when a = 0, b = 0, and c = 0 then c(b + (ab)) + c¯ab = 0(0 + (0.0)) + 0¯.0.0
= 0(0 + 1) + 1.0 = 0.
Draw a Karnaugh map as in Figure 4.8(a) and copy in the expression values as found in Table 4.5. There are three variables in the expression, therefore, there are 23 = 8 squares in the Karnaugh map.

Table 4.5 A truth table to find ab + a¯ b + ab¯

a b c ab c¯ (ab) c¯ ab b + (ab) c(b + (ab)) c(b + (ab)) + c¯ ab

000 0 1 1 0

1

0

0

001 0 0 1 0

1

1

1

010 0 1 1 0

1

0

0

011 0 0 1 0

1

1

1

100 0 1 1 0

1

0

0

101 0 0 1 0

1

1

1

110 1 1 0 1

1

0

1

111 1 0 0 0

1

1

1

Figure 4.8 (a) A three-variable Karnaugh map representing c(b + (ab)) + c¯ ab. (b) A three-variable Karnaugh map with all the boxes labelled. (c) A Karnaugh map is like a Venn diagram. The third and fourth rows represent the set a and the second and third rows represent the set b. c is represented by the second column.
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84 Boolean algebra
Figure 4.9 A three-variable Karnaugh map representing c(b + (ab)) + c¯ ab.
Figure 4.10 An implementation of c + ab.

Figure 4.8(b) shows a Karnaugh map with the squares labelled term by term. Figure 4.8(c) shows the Venn diagram equivalence with sets a, b, and c. In Figure 4.8(a) we put a 0 or 1 in the square depending on whether that term is present, as given in the truth table in Table 4.5.
Adjacent 1s indicate that we can simplify the expression. Figure 4.9 indicates how we go about the minimization. We draw a line around any four adjacent 1s and write down the term representing that section of the map. The second column represents c and has been encircled. Then we look for any two adjacent 1s. We are able to encircle the third row, representing ab. As all the 1s have now been included we know that c + ab is a minimization of the expression. An implementation of c + ab is drawn in Figure 4.10.
Example 4.11 Minimize abc¯ + a¯bd + abcd¯ + ab¯c¯d + abc using a Karnaugh map and draw the implementation of the resulting expression as a logic circuit.
Solution Draw a Karnaugh map as in Figure 4.11(a). There are four variables in the expression therefore there are 24 = 16 squares in the Karnaugh map. Figure 4.11(b) shows a Karnaugh map with the squares labelled term by term. Figure 4.11(c) shows the Venn diagram equivalence with sets a, b, c, and d. In Figure 4.11(a), we put a 0 or 1 in the square depending on whether that term is present in our expression. However, the term abc¯ involves only three out of the four variables. In this case, it must occupy two squares. As d could be either 0 or 1 for ‘abc¯’ to be true, we fill in the squares for abc¯d and abc¯ d¯. The number of squares to be filled with a 1 to represent a certain product is 2m where m is the number of missing variables in the expression. In this case, abc¯ has no d term in it so the number of squares representing it is 21.
Adjacent 1s indicate that we can simplify the expression. Figure 4.12 indicates how we go about the minimisation. We draw a line around any eight adjacent 1s of which there are none. Next we look for any four adjacent 1s and write down the term representing that section of the map. The third row represents ab and has been encircled. The middle four squares represent bd and have been encircled. Then we look for any two adjacent 1s. The bottom two squares of the second column represent ac¯d. As all the 1s have now been included we know that ab+bd +ac¯d is a minimization of the expression. This is implemented in Figure 4.13.

Figure 4.11 (a) A four variable Karnaugh map representing abc¯ + a¯ bd + abcd¯ + ab¯ c¯ d + abc. (b) A four-variable Karnaugh map with all the squares labelled. (c) A Karnaugh map is like a Venn diagram. The third and fourth rows represent the set a and the second and third rows represent the set b. c is represented by the third and fourth columns and d by the second and third columns.
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Boolean algebra 85

Figure 4.12 A four-variable
Karnaugh map representing abc¯ + a¯ bd + abcd¯ + ab¯ c¯ d + abc .

Figure 4.13 An implementation of ab + bd + ac¯ d .

Figure 4.14 To display the digits 0–9 a seven-segment LED display may be used. For instance, the number 1 requires the segments labelled q and r to light up and the other segments to be off.
Table 4.6 A truth table giving the logic control signals for the lamp drivers for the LED segments pictured in Figure 4.14

Digit displayed

Circuit inputs

Segments

abcdpqr s t

uv

0

0000 1 1 1 1 1 1 0

1

0001 0 1 1 0 0 0 0

2

0010 1 1 0 1 1 0 1

3

0011 1 1 1 1 0 0 1

4

0100 0 1 1 0 0 1 1

5

0101 1 0 1 1 0 1 1

6

0110 0 0 1 1 1 1 1

7

0111 1 1 1 0 0 0 0

8

1000 1 1 1 1 1 1 1

9

1001 1 1 1 1 0 1 1

–

1010 XXXXXXX

–

1011 XXXXXXX

–

1100 XXXXXXX

–

1101 XXXXXXX

–

1110 XXXXXXX

–

1111 XXXXXXX

Example 4.12 To display the digits 0–9, a seven-segment light emitting diode (LED) display may be used as shown in Figure 4.14. The various states may be represented using a four-variable digital circuit. The logic control signals for the lamp drivers are given by the truth table given in Table 4.6. The X indicates a ‘don’t care’ condition in the truth table. The column for the segment labelled p can be copied into a Karnaugh map as given in Figure 4.15. Wherever a 1 appears in the truth table representation for p there is a 1 copied to the Karnaugh map. Similarly, the 0s and the ‘don’t care’ crosses are copied. Minimize the Boolean expression for p using the Karnaugh map.
Solution The minimization is represented in Figure 4.15(b). We first look for any eight adjacent squares with a 1 or a X in them. The bottom
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86 Boolean algebra
Figure 4.15 (a) A Karnaugh map for the segment labelled p in Figure 4.14. This has been copied from the truth table given in Table 4.5 (b) A minimization of the Karnaugh map. The ‘don’t care’ Xs may be treated as 1s if it is convenient but they can also be treated as 0s.

two rows are encircled giving the term a. Now we look for groups of four. The central four squares represent bd and the third column represents cd. Finally, we can count the four corner squares as adjacent. This is because two squares may be considered as adjacent if they are located symmetrically with respect to any of the lines which divide the Karnaugh map into equal halves, quarters, or eighths. This means that squares that could be curled round to meet each other, as if the Karnaugh map where drawn on a cylinder, are considered adjacent and also the four corner squares. Here, the four corner squares represent the term b¯ d¯. Hence, the minimization for p gives
p = a + cd + bd + b¯ d¯.

4.5 Summary

(1) An algebra is a set with operations defined on it. A binary operation as a way of combining two elements of the set to result in another element of the set. A unary operation has only one input element producing one output.
(2) A Boolean algebra has the operations of AND, OR, and NOT defined on it and obeys the set of laws given in Section 4.3 as (B1)–(B10). Examples of a Boolean algebra are: the set of sets in some universal set E , with the operations of ∩, ∪ and ; the set of propositions with the operations of ∧, ∨, and ¬; the set {0, 1} with the operations ‘.’ +, and −.
(3) Logic circuits can be represented as Boolean expressions. Usually, a high voltage is represented by 1 or TRUE and a low voltage by 0 or FALSE. There are three basic gates to represent the operators AND (.), OR (+), and NOT (−).
(4) A Boolean expression may be minimized by first expressing it as a sum of products and then using a Karnaugh map to combine terms.

4.6 Exercises

4.1 Show the following properties of sets using Venn
diagrams: (a) A ∪ (A ∩ B) = A (b) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(i) p ∧ q (ii) p ∨ q (iii) ¬p ∨ q (iv) p ∧ ¬q (v) ¬(p ∧ q) (vi) ¬p ∨ ¬q. (b) Given that p is true and q is false, what is the truth
value of each part of section (a)?

4.2 p = ‘It rained yesterday’ q = ‘I used an umbrella yesterday’ (a) Construct English sentences to express, (∧ ⇔ ‘and’, ∨ ⇔ ‘or’, ¬ ⇔ ‘not’)

4.3 Show the following properties of propositions using truth tables (a) p ∨ (p ∧ q) ⇔ p (b) ¬(p ∨ q) ⇔ ¬p ∧ ¬q.

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Boolean algebra 87

4.4 Using Venn diagrams or truth tables find simpler expressions for the following: (a) ab + ab¯ (b) (ab)(ac) (c) a + abc (d) (ab)a
4.5 (a) Draw implementations of the following as logic circuits: (i) a¯ b¯ + ab¯ (ii) a + bc (iii) a¯ + ab (iv) a¯ b¯c
(b) If a = 1, b = 0 and c = 1, what is the value of each of the expressions in section (a)?
4.6 Minimize the following expressions and draw their logic circuits: (a) ab¯ c¯ + ab¯ + abc (b) abc(a + b + c) + a (c) (a + c)(a + b) + ab (d) (a + b)(a + d) + abc¯ + abd¯ + abcd
4.7 Obtain a Boolean expression for the logic networks shown in Figure 4.16.
4.8 Consider the LED segment labelled r in Figure 4.14 given in the text. Follow the method given in

Example 4.12 to find a minimized expression for r and draw its logic network.
Figure 4.16 (a,b) Logic networks for Exercise 4.7.

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5

Trigonometric

functions and

waves

5.1 Introduction

Waves occur naturally in a number of situations: the movement of disturbed water, the passage of sound through the air, vibrations of a plucked string. If the movement of a particular particle is plotted against time, then we get the distinctive wave shape, called a sinusoid. The mathematical expression of a wave is found by using the trigonometric functions, sine and cosine. In Chapter 6 of the Background Mathematics Notes on the companion website for this book we looked at right angled triangles and defined the trigonometric ratios. The maximum angle in a right angled triangle is 90◦ so to find the trigonometric functions, sin(t), cos(t), and tan(t) where t can be extended over the real numbers, we need a new way of defining them. This we do by using a rotating rod. Usually the function will be used to relate, for instance, the height of the rod to time. Therefore, it does not always make sense to think of the input to the cosine and sine functions as being an angle. This problem is overcome by using a new measure for the angle called the radian, which easily relates the angle to the distance travelled by the tip of the rotating rod.
Waves may interfere with each other, as for instance on a plucked string, where the disturbance bounces off the ends producing a standing wave. Amplitude modulation of, for instance, radio waves, works by the superposition of a message on a higher signal frequency. These situations require an understanding of what happens when two, or more cosine or sine functions are added, subtracted, or multiplied and hence we also study trigonometric identities.

5.2 Trigonometric functions and radians

Consider a rotating rod of length 1. Imagine, for instance, that it is a position marked on a bicycle tyre at the tip of one of the spokes, as the bicycle travels along. The distance travelled by the tip of the rod in 1 complete revolution is 2π (the circumference of the circle of radius 1). The height of the rod (measured from the centre of the wheel), y, can be plotted against the distance travelled by the tip as in Figure 5.1.
Similarly, the position to the right or left of the origin, x, can be plotted against the distance travelled by the tip of the rod as in Figure 5.2. Figure 5.1 defines the function y = sin(t) and Figure 5.2 defines the function x = cos(t).
This definition of the trigonometric function is very similar to that used for the ratios in the triangle, if the hypotenuse is of length 1 unit. The definitions become the same for angles up to a right angle if radians
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