SCHAUM’S OUTLlNE OF
THEORY AND PROBLEMS
OF
FINITE
MATHE RIATIC S
0
BY
SEYMOUR LIPSCHUTZ, Ph.D.
Associate Professor of Mathematics
Temple University
0
SC€IAUMfS OUTLINE SERIES
McGRAW-HILL BOOK COMPANY
New York, St. Louis. San Francisco, Toronto, Sydne)
Copyright @ 1966 b j McGraw-Hill, Inc. All Rights Reserved. Printed in the
United States of America. No p a r t of this publication may be reproduced.
stored in a rerrieval sy%tem, or transmitted, in any form or by any means.
electronic, mechanical, photocopying, recording. or otherwise, without the prior
written permissioii of the imblisher.
579hT
6 7 8 9 1 0 SHSH 7 5 4 3 2 1 0
Finite mathematics has in recent years become an integral part of the mathematical
background necessary for such diverse fields as biology, chemistry, economics, psychology,
sociology, education, political science, business and engineering. This book, in presenting
the more essential material, is designed for use as a supplement to all current standard
texts or as a textbook for a formal course in finite mathematics.
The material has been divided into twenty-five chapters, since the logical arrangement
is thereby not disturbed while the usefulness as a text and reference book on any of several
levels is greatly increased. The basic areas covered are: logic; set theory; vectors and
matrices: counting - permutations, combinations and partitions; probability and Markov
chains; linear programming and game theory. The area on vectors and matrices includes
a chapter on systems of linear equations; it is in this context that the important concept
of linear dependence and independence is introduced. The area on linear programming
and game theory includes a chapter on inequalities and one on points, lines and hyperplanes; this is done to make this section self-contained. Furthermore, the simplex method
is given for solving linear programming problems with more than two unknowns and fo r
solving relatively large games. In using the book it is possible to change the order of
many later chapters or even t o omit certain chapters without difficulty and without loss
of continuity.
Each chapter begins with a clear statement of pertinent definitions, principles and
theorems together with illustrative and other descriptive material. This is followed by
graded sets of solved and supplementary problems. The solved problems serve to illustrate
and amplify the theory, bring into sharp focus those fine points without which the student
continually feels himself on unsafe ground, and provide the repetition of basic principles
so vital t o effective learning. Proofs of theorems and derivations of basic results are
included among the solved problems. The supplementary problems serve as a complete
review of the material in each chapter.
More material has been included here than can be covered in most first courses. This
has been done t o make the book more flexible, to provide a more useful book of reference
and to stimulate further interest in the topics.
I wish to thank many of my friends and colleagues, especially P. Hagis, J. Landman,
B. Lide and T. Slook, for invaluable suggestions and critical review of the manuscript.
I also wish to express my gratitude t o the staff of the Schaum Publishing Company,
particularly to N. Monti, for their unfailing cooperation.
S. LIPSCHUTZ
Temple University
June, 1966
CONTENTS
Page
Chapter
1
PROPOSITIONS AND TRUTH T A B L E S . . ....................
Statements. Compound statements. Conjunction, p
Negation, -p. Propositions and truth tables.
Chapter
2
A
q.
1
Disjunction, p v q.
ALGEBRA OF PROPOSITIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
Tautologies and contradictions. Logical equivalence. Algebra of propositions.
Chapter
3
CONDITIONAL STATEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Conditional, p
Chapter
4
+ q.
18
Biconditional, p ++ q. Conditional statements and variations.
ARGUMENTS, LOGICAL IMPLICATION .....................
26
Arguments. Arguments and statements. Logical implication.
Chapter
5
S E T THEORY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
Sets and elements. Finite and infinite sets. Subsets. Universal and null sets.
Set operations. Arguments and Venn diagrams.
Chapter
6
PRODUCT SETS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
Ordered pairs. Product sets. Product sets in general. Truth sets of propositions.
Chapter
7
RELATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
Relations. Relations a s sets of ordered pairs. Inverse relation. Equivalence
relations. Partitions. Equivalence relations and partitions.
Chapter
8
FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66
Definition of a function. Graph of a function. Composition function. One-one
and onto functions. Inverse and identity functions.
Chapter
9
VECTORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
Column vectors. Vector addition. Scalar multiplication. Row vectors. Multiplication of a row vector and a column vector.
Chapter
10
MATRICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Matrices. Matrix addition. Scalar multiplication. Matrix multiplication.
Square matrices. Algebra of square matrices. Transpose.
90
CONTENTS
Page
Chapter
II
LINEAR EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
104
Linear equation in two unknowns. Two linear equations in two unknowns.
General linear equation. General system of linear equations. Homogeneous
systems of linear equations.
Chapter
12
DETERMINANTS OF ORDER TWO AND THREE. . . . . . . . . . . . 127
Introduction. Determinants of order one. Determinants of order two. Linear
equations in two unknowns and determinants. Determinants of order three.
Linear equations in three unknowns and determinants. Invertible matrices.
Invertible matrices and determinants.
Chapter
13
THE BINOMIAL COEFFICIENTS AND THEOREM . . . . . . . . . . 141
Factorial notation. Binomial coefficients.
angle. Multinomial coefficients.
Chapter
14
Binomial theorem.
Pascal's tri-
PERMUTATIONS, ORDERED SAMPLES ....................
152
Fundamental principle of counting. Permutations. Permutations with repetitions. Ordered samples.
Chapter
15
COMBINATIONS, ORDERED PARTITIONS . . . . . . . . . . . . . . . . . 161
Combinations.
Chapter
16
Partitions and cross-partitions. Ordered partitions.
TREE DIAGRAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
176
Tree diagrams.
Chapter
17
PROBABILITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
184
Introduction. Sample space and events. Finite probability spaces. Equiprobable spaces. Theorems on finite probability spaces. Classical birthday
problem.
Chapter
18
CONDITIONAL PROBABILITY. INDEPENDENCE . . . . . . . . . 199
Conditional probability. Multiplication theorem for conditional probability.
Finite stochastic processes and tree diagrams. Independence.
Chapter
19
INDEPENDENT TRIALS, RANDOM VARIABLES. . . . . . . . . . . 216
Independent or repeated trials. Repeated trials with two outcomes. Random
variables. Probability space and distribution of a random variable. Expected
value. Gambling games.
Chapter
20
MARKOV CHAINS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
233
Probability vectors. Stochastic and regular stochastic matrices. Fixed points
of square matrices. Fixed points and .regular stochastic matrices. Markov
chains. Higher transition probabilities. Stationary distribution of regular
Markov chains. Absorbing states.
Chapter
21
INEQUALITIES
..............................................
The real line. Positive numbers. Order. (Finite) Intervals. Infinite intervals.
Linear inequalities in one unknown. Absolute value.
257
CONTENTS
Page
Chapter
22
POINTS, LINES AND HYPERPLANES.. . . . . . . . . . . . . . . . . . . . . .
266
Cartesian plane. Distance between points. Inclination and slope of a line.
Lines and linear equations. Point-slope form. Slope-intercept form. Parallel
lines. Distance between a point and a line. Euclidean nz-space. Bounded
sets. Hyperplanes. Parallel hyperplanes. Distance between a point and a
hyperplane.
Chapter
23
CONVEX SETS AND LINEAR INEQUALITIES. . . . . . . . . . . . . . 281
Line segments and convex sets. Linear inequalities. Polyhedral convex sets
and extreme points. Polygonal convex sets and convex polygons. Linear
functions on polyhedral convex sets.
Chapter
24
LINEAR PROGRAMJIIKG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
300
Linear programming problems. Dual problems. Matrix notation. Introduction
t o the simplex method. Initial simplex tableau. Pivot entry of a simplex
tableau. Calculating the new simplex tableau. Interpreting the terminal
tableau. Algorithm of the simplex method.
Chapter
25
THEORY OF GAMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
321
Introduction to matrix games. Strategies. Optimum strategies and the value
of a game. Strictly determined games. 2 X 2 matrix games. Recessive rows
and columns. Solution of a matrix game by the simplex method. 2 X m and
i i i x 2 matrix games. Summary.
INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
337
Chapter
I
Propositions and Truth Tables
STATEMENTS
A statement (or verbal assertion) is any collection of symbols (or sounds) which is either
true or false, but not both. Statements will usually be denoted by the letters
P,
(1,
r,
...
The truth or falsity of a statement is called its truth value.
Example 1.1:
Consider t h e following expressions:
(i) P ar i s is in England.
(ii) 2
+2 = 4
(iii) Where a r e you going?
(iv) P u t the honiework on the blackboard.
6)
are statements; the first is false and the second is true.
The expressions (i) and
The expressions (iii) and [iv) a r e not statements since neither is either true o r false.
COMPOUND STATEMENTS
Some statements are composite, that is, composed of substatements and various logical
connectives which we discuss subsequently. Such composite statements are called compound statements.
Example 2.1:
“Roses a r e red and violets a r e blue” is a compound statement with substatements
“Roses a r e red” and “Violets a r e blue”.
Example 2.2:
“He is intelligent o r studies every night” is, implicitly, a compound statement with
substatements “He is intelligent” and “He studies every night”.
The fundamental property of a compound statement is that its truth value is completely
determined by the truth values of its substatements together with the way in which they
are connected to form the compound statement. We begin with a study of some of these
connectives.
CONJUNCTION, p A Q
Any two statements can be combined by the word “and” to form a compound statement
called the conjunction of the original statements. Symbolically,
P A 9
denotes the conjunction of the statements p and q, read “ p and q”.
Example 3.1:
Let p be “It is raining” and let q be “The sun is shining”.
Then p
A
q denotes the statement “It is raining and the sun is shining”.
The truth value of the compound statement p ~ satisfies
q
the following property:
[TI] If p is true and Q is true, then p A q is true; otherwise, 11 A q is false.
In other words, the conjunction of two statements is true only in the case when each substatement is true.
1
PROPOSITIONS AND TRUTH T A B L E S
2
Example 3.2:
[CHAP. 1
Consider the following four statements:
(i)
+2
2 +2
2 +2
2
Paris is in France and
(ii) P a r i s is in France and
(iii) P a r i s is in England and
(iv) Paris is in England and 2
= 4.
= 5.
= 4.
- 2 = 5.
By property [Ti], only the first statement is true. Each of the other statements
is false since at least one of its substatements is false.
A convenient way to state property [Ti] is by means of a table as follows:
Here, the first line is a short way of saying that if p is true and q is true then p A 4 is true.
The other lines have analogous meaning. We regard this table as defining precisely the
truth value of the compound statement p A q as a function of the truth values of p and of q.
DISJUNCTION, 11 q
Any two statements can be combined by the word “or” (in the sense of “and/or”) to
form a new statement which is called the disjunction of the original two statements.
Symbolically,
P”4
denotes the disjunction of the statements p and q and is read ‘ip or q”
Example 4.1:
Let p be “Marc studied French at the university”, and let q be “Marc lived in
France”. Then p v q is the statement “Marc studied French at t h e university o r
(Marc) lived in France”.
The truth value of the compound statement p
v
q satisfies the following property:
IT,] If p is true or q is true o r both p and q a r e true, then
q is true; otherwise p v q
13
is false.
Accordingly, the disjunction of two statements is false only when both substatements a r e
false. The property LT,] can also be written in the form of the table below, which we
regard as defining 11 v q:
Example 4.2:
Consider the following four statements:
+2
+2
2 +2
2+2
P a r i s is in France o r
2
= 4.
(ii) P a r i s is in France o r
2
= 5.
ii)
iiii) P a r i s is in England or
iiv) P a r i s is in England o r
= 4.
= 5.
By property [T2], only (iv) is false. Each of the other statements is t r u e since a t
least one of its substatements is true.
PROPOSITIONS A N D TRUTH T A B L E S
CHAP. 11
Remark:
3
The English word “or” is commonly used in two distinct ways. Sometimes
it is used in the sense of “ p o r (I or both”, i.e. a t least one of the two alternates
occurs, as above, and sometimes it is used in the sense of “ p o r 4 but not both”,
i.e. exactly one of the two alternatives occurs. For example, the sentence
“He ill go to Harvard o r to Yale” uses “or” in the latter sense, called the
e,cclusice disjunction. Unless otherwise stated, “01”’ shall be used in the
former sense. This discussion points out the precision we gain from o u r
symbolic language: p b (I is defined by its truth table and always means
“p andlor q”.
NEGATION, -11
Given any statement p , another statement, called the negatioii of 11, can be formed by
writing “It is false t h a t . . .” before p or, if possible, by inserting in p the word “not”.
Symbolically,
-1,
denotes the negation of p (read “not 71”).
Example 5.1 :
Consider the following three statements:
( i ) P a r i s is in France.
(ii, I t is false t h a t P a r i s is in France.
(iii) P a r i s is not in France.
Then (ii) and (iii) a r e each the negation of (i).
Example 5.2:
Consider the following statements:
(i) 2 - 2 = 5
(ii) It is false t h a t 2 - 2 = 5 .
+
(iii) 2 2 # 5
Then (ii) and (iii) a r e each the negation of (i).
The truth value of the negation of a statement satisfies the following property:
[T,] If p is true, then - p is false; if 11 is false, then -p is true.
Thus the truth value of the negation of any statement is always the opposite of the truth
value of the original statement. The defining property IT,] of the connective can also be
written in the form of a table:
P
F
I
-P
T
Example 5.3:
Consider t h e statements in Example 5.1. Obserre t h a t ti) is t r u e and (ii) and [iii),
each its negation, a r e false.
Example 5.4:
Consider the statements in Example 5.2. Observe t h a t ri) is false and (ii) and (iiii,
each its negation, a r e true.
PROPOSITIONS AND TRUTH TABLES
By repetitive use of the logical connectives (4,v,
and others discussed subsequently),
we can construct compound statements that are more involved. In the case where the
substatements p , Q, . . . of a compound statement P ( p , q, . . .) are variables, we call the
compound statement a proposition.
Now the truth value of a proposition depends exclusively upon the tru th values of its
variables, that is, the truth value of a proposition is known once the tru th values of its
variables are known. A simple concise way to show this relationship is through a truth
table. The truth table, for example, of the proposition - ( p A -q) is constructed as follows:
-
[CHAP.1
PROPOSITIONS AND T R U T H TABLES
4
P
Q
T
T
T
F
F I T
-
(P
A
-
4)
F / F
Step
F
1’
4
T
T
T
F
F
T
F
F
Step
-
-
4)
P
F
T
F
F
F
T
F
T
T
F
T
F
T
T
F
F
3
2
1
(P
A
T
T
F
F
1
F
F
F
T
F
T
F
Step
T
F
T
T
4
F
-
4
F
T
F
F
F
T
F
T
T
F
T
F
3
2
1
(11
A
T
T
F
F
1
5
PROPOSITIONS A N D T R U T H TABLES
CHAP. 11
Solved Problems
STATEMENTS
1.1. Let p be “It is cold” and let q be “It is raining.” Give a simple verbal sentence which
describes each of the following statements:
(1) -21, ( 2 ) P A Q, (3) P ” 4, (4)cl ” -21, ( 5 ) ”P A -4, (6)
Q.
--
-
In each case, translate A , v and
to read “and”, “or” and “ I t is false t h a t ” o r “not”,
respectively, and then simplify the English sentence.
(1) It is not cold.
(41 It is raining or i t is not cold.
(2) It is cold and raining.
( 5 ) I t is not cold and i t is not raining.
(3) It is cold or it is raining.
( G I I t is not true t h a t i t is not raining.
Let p be “He is tall” and let Q be “He is handsome.”
statements in symbolic form using 11 and q .
Write each of the following
(1) He is tall and handsome.
(2) He is tall but not handsome.
(3) It is false that he is short or handsome.
(4) He is neither tall nor handsome.
(5) He is tall, or he is short and handsome.
(6) It is not true that he is short or not handsome.
(Assume t h a t “He is short” means “He is not tall”, i.e. -11.)
(1) P
A
q
(3) -(-P
(2) P
A
-!I
(4) -P
A
v 4)
( 5 ) 1’
@q
(6) - ( - P
V
(“P
A
4)
” -q)
TRUTH VALUES OF COMPOUND STATEMENTS
1.3. Determine the truth value of each of the following statements.
(i) 3 + 2 = 7 and 4 f 4 = 8. (ii) 2 + 1 = 3 and 7 + 2 = 9. (iii) 6 + 4 = 10 and
1 + 1 = 3.
By property [TI], the compound statement “ p and q” is t r u e only when p and q a r e both true.
Hence: ti, False, iii) True, (iii) False.
1.4.
Determine the truth value of each of the following statements.
+
+
+
(i) Paris is in England or 3 4 = 7.
(ii) Paris is in France or 2 1 = 6.
(iii) London is in France or 5 2 = 3 .
By property [ T L ] ,the compound statement “ p o r q” is false only when p and q a r e both false.
Hence: li) True, (ii) True, (iii) False.
1.5.
Determine the truth value of each of the following statements.
(i) It is not true that London is in France.
(ii) It is not true that London is in England.
By property IT,], the t r u t h value of the negation of p is the opposite of the t r u t h value of p .
Hence: Ii) True, rii) False.
6
1.6.
[CHAP. 1
PROPOSITIONS AND T R U T H T A B L E S
Determine the truth value of each of the following statements.
(i) It is false that 2 + 2 = 4 and 1 + 1 = 5 .
(ii) Copenhagen is in Denmark, and 1 + 1 = 5 or 2 f 2 = 4.
(iii) It is false that 2 + 2 = 4 or London is in France.
(i)
+
+
+
+
The conjunctive statement “2 2 = 4 and 1 1 = 5” is false since one of i t s substatements
“ I 1 = 5” is false. Accordingly its negation, the given statement, is true.
+
(ii) The disjunctive statement “1 1 = 5 or 2 2 = 4” is t r u e since one of i t s substatements
“2 2 = 4” is true. Hence the given statement is t r u e since it is the conjunction of two
t r u e statements, “Copenhagen is in Denmark” and “1 1 = 5 o r 2 2 = 4”.
+
+
+
+
(iii) The disjunctive statement “ 2 2 = 4 o r London is in France” is t r u e since one of its substatements “2 2 = 4” is true. Accordingly its negation, the given statement, is false.
+
TRUTH TABLES OF PROPOSITIONS
1.7.
Find the truth table of
-13
,. q .
9
-
P
a
-
T
F
T
F
T
T
F
F
T
F
T
F
F
F
T
T
T
T
F
F
F
F
T
F
T
F
T
F
2
1
3
1
Step
P
A
P
Method 1
1.8.
Find the truth table of
-(]I
v
4).
T
F
T
T
F
F
F
T
F
F
T
F
Step
Method 2
Method 1
1.9.
Find the truth table of - ( p
P
T
T
F
F
I
a
I -9 I
T
F
T
F
F
T
F
T
PV
-a
\d)
1
-q).
-9)
-(?I”
T
T
F
F
F
T
T
p l q
I -
(P
v
-
a)
F
Method 1
1.10. Find the truth table of the following: (i) p
A
(4 v
r ) , (ii) ( p A 4) v ( p A T
Since there a r e three variables, we will need 23 = 8 rows in the t r u t h table.
) ~
C'HAP. I]
PROPOSITIONS AND TRUTH TABLES
P
q
T
T T
T I F
F T
F F
T T
T F
F T '
F Fl
q
r
T T
T T
T F
T F
F , T
F T
F ~
F F
T
F
T
F
T
F
F
T
F
P
PA(q"r)
7.
7
PA?.
i
T
T
T
F
F
F
F
T
T
T
T
F
F
F
F
F
T
T
F
T
T
T
F
MISCELLANEOUS PROBLEMS
1.11. Let A p q denote p / q and let Sp denote -p.
using A and N instead of A and -.
(9 P A -q
(iii) - p
(4 -(-P
(i)
p
(ii)
-(-p
(iii) - p
(iv) - ( p
= p
-q
A
4
A
A
A
-q)
A 1.)
A
~F
i F
Rewrite the following propositions
(-q
A
(iv) - ( p
Q)
T
F
T
F
F
F
A
A
-q)
A
1)
(-(I
A
-7,)
iYq = A p N q
q ) = -(.Vp
(-q
A
T
T
F
F
F
F
F
F
(-q
\
q) = -(ASpq) = NANpq
A
= Np
(Nq A r) = S p
A
-Y)
= -(ApXq)
A
A
( A N q r ) = ANpA'VqT
(ANqSr) = (NApNq) A ( A S q S r ) = ANApNqANqXr
Observe t h a t there a r e no parentheses in the final answer when A and .V a r e used instead of
A and -.
In fact, i t has been proved t h a t parentheses a r e never needed in any proposition using
A and N .
1.12. Rewrite the following propositions using
(i)
A
and
(i) N A m
(iii) ApNq
(ii) A N p q
(iv) ApAqr
NApq = S ( p A q) = - ( p
(ii) A N p q = A ( - p ) q = - p
A
A
q)
q
-
instead of A and
N.
(v) Iv'AANpql.
(vi) A S p A q N r
(iii) ApNq = A p ( - q ) = p
(iv) A p A q r = A p ( q
A 7,)
A
-q
= p
A
(q A r)
(v) N A A S p q r = S A A ( - p ) q r = N A ( - p fi q ) r = Nr(-p A q ) / r1 = -:(-p A q ) A r ]
(vi) A N p A q S r = A N p A q ( - r ) = A N p ( q A -7) = A ( - p ) ( q A - 1 . ) = -lJ A ( Q A - r )
Notice t h a t the propositions involving A and N a r e unraveled from right to left.
Supplementary Problems
STATEMENTS
1.13.
Let p be "Marc is rich" and let q be "Marc is happy". Write each of the following in symbolic form.
(i)
Marc is poor but happy.
(ii) Marc is neither rich nor happy.
(iii) Marc is either rich o r unhappy.
(iv) Marc is poor o r else he is both rich and unhappy.
8
1.14.
PROPOSITIONS AND TRUTH TABLES
[CHAP. 1
Let p be “E r i k reads Newsiccek”, let Q be “Erik reads Life” and le t r be “Erik reads Time”.
Write each of the following in symbolic form.
(I)
iii)
Erik reads Newsweek or L i f e , but not Tima.
Erik reads Newsweek and Lit’c, o r he does not read N e w s w e e k and T i m e .
( i i i ) I t is not t r u e t h a t Erik reads Newsweek but not Time.
(ivi I t is not t r u e t h a t Er i k reads Time or Life but not N e w s w e e k .
1.15.
1.16.
Let p be “Audrey speaks French” and let q be “Audrey speaks Danish”. Give a simple verbal
sentence which describes each of the following.
(i) P
v
4
(iii) p
(ii) 11
A
q
(iv) - p v -q
A
(v) - - p
-q
(vi) -(-P
A
-4
Determine the t r u t h value of each of the following statements.
(i,
3 f 3 = 6 and 1 + 2 = 5.
iiii
I t is not t r u e t h a t 3
iiii) It is true t h a t 2
+3 = 6
o r 1 f 2 = 3.
+ 2 # 4 and 1 + 2 = 3.
+ 3 # 6 o r 1+ 2 # 5.
iiv) It is not t r u e t h a t 3
TRUTH TABLES OF PROPOSITIONS
1.17.
Find the t r u t h table of each of the following.
ii) p
1.18.
iii) - p
-q,
v
A
-q,
iiii) - ( - p
A
q ) , (iv) -(-p
v -q)
Find the t r u t h table of each of the following.
(i) ( p
A
- q ) v r , (ii) - p v ( q A
(iii) ( p v -v)
-T),
A
(q v
-T),
(iv) - ( p v -9)
A
(-p v
7).
MISCELL-4NEOUS PROBLEMS
1.19.
Let A p q denote P A q and let S p denote -p.
using A and N instead of n and -.
(i) -p
1.20.
A
q,
(ii) - p
A
-q,
iiii) -(p
A
-q),
Rewrite the following propositions using
(i) N A p N q , (ii) A N A p q N y ,
(See Problem 1.11.) Rewrite the following propositions
(iv) ( - p
and
A
A
q) A
-T.
- instead of A
and N .
(iii) A A p N r A q N p , (iv) A S A N p A N q r N p .
Answers to Supplementary Problems
1.13.
[i) - p
1.14.
(i) (p v q )
1.15.
(i)
A
q,
(ii) - p
A -7,
A
-q,
(iii) p v - q ,
(ii) ( p A q ) v - ( p
A T),
(iv) - p v ( p A - q )
(iii) - ( p
A -T),
(iv) -[(Y v q) A -p]
Audrey speaks French o r Danish.
(ii) Audrey speaks French and Danish.
(iii) Audrey speaks French but not Danish.
(iv) Audrey does not speak French or she does not speak Danish.
( v ) I t is not t r u e t h a t Audrey does not speak French.
(vi\ I t is not t r u e t h a t Audrey speaks neither French nor Danish.
1.16.
(i) F, ( i i ) F, (iiii F, (iv) F
-p
1.18.
9
PROPOSITIONS AND TRUTH TABLES
CHAP. 11
F
F
F
T
F
p
q
)*
(i)
(ii)
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
F
T
T
T
F
T
F
F
T
F
F
T
T
T
T
F
T
A
-(-p
-q
F
T
A
4)
-(-pv
F
T
F
F
(iii) (ivi
T
T
F
T
F
T
F
T
F
F
F
F
T
T
F
F
1.19.
(i) Ah'pq, (iii A.YpNq, (iii) N A p N q , (ivi A A S p q N r
1.20.
(i) - ( p
A
-q),
(ii) - ( p
A
-q)
q)
A
-Y,
(iii) ( p A
-T)
A
(q A -p),
(iv) - [ - p n ( - p
A T)] A
-p
Chapter 2
Algebra of Propositions
TAUTOLOGIES AND CONTRADICTIONS
Some propositions P ( p , q , . . . ) ca,ntain only T in the last column of their truth tables,
i.e. a r e true for any truth values of their variables. Such propositions are called
tautologies. Similarly, a proposition P ( p ,(1,. . .) is called a cotitmdiction if it contains
only F in the last column of its truth table, i.e. is false for any truth values of its variables.
Example 1.1:
'
The proposition " p o r not p", i.e. p v -11, is a tautology and the proposition
" p and not p", i.e. p
-11, is a contradiction. This is verified by constructing
their t r u t h tables.
1' 1 -11
P " -p
I i l l P;-P
,"
1
T
F
F
T
~
TT
~
F
~
T
IF
Since a tautology is always true, the negation of a tautology is always false, i.e. is a
contradiction, and vice versa. That is,
Theorem 2.1:
If P ( p , q , . . .) is a tautology then -Pip,
conversely.
(I, ,
. . ) is a contradiction, and
Now let P ( p , (I, . . .) be a tautology, and let P , ( p , q, . . . ), P,(p, q , . . .), . . . be any propositions. Since P ( p ,q , . . . ) does not depend upon the particular truth values of its variables
p , 'I, . . ., we can substitute P , for p , P , for q, . . . in the tautology P ( p , q, . . . ) and still
have a tautology. In other words:
Theorem 2.2 (Principle of Substitution):
If P ( p , q , . . .) is a tautology, then P(P,,P,, . . . )
is a tautology f o r any propositions P,, P,~,, . . .
LOGICAL EQUIVALENCE
Two propositions P(p, q, . . . ) and Q ( p , q ,
simply equicnleiit o r equal, denoted by
P(l4 9 ,
. . .)
a r e said to be logically equivalent, or
. 1 = &(P, q, . . . )
* *
if they have identical truth tables.
Example 2.1:
The t r u t h tables of
-(]I
q i and - p v - q folio\\.
T
T
T
F
F
Accordingly, the propositions - ( p
"(p
10
T
T
A
A
q ) and -11 v - q
9)
-pV
-q
are logically equivalent:
ALGEBRA O F PROPOSITIONS
CHAP. 21
11
The statement
Example 2.2:
“It is false that roses a r e red and violets a r e blue”
can be written in the form - ( p A q ) where p is “Roses a r e red” and q is “Violets
a r e blue”. B y the preceding example, - ( p A q ) is logically equivalent to - p v - q ;
that is, the given statement is equivalent to the statement
“Either roses a r e not red or violets a r e not blue.”
ALGEBRA OF PROPOSITIONS
Propositions, under the relation of logical equivalence, satisfy various laws or identities
which are listed in Table 2.1 below. In fact, we formally state:
Propositions satisfy the laws of Table 2.1.
Theorem 2.3:
LAWS OF THE ALGEBRA OF PROPOSITIONS
Idempotent Laws
=
la. p v p
lb.
p
p
~
p
=
p
r
=
Associative Laws
2a.
ip
3a.
p 4 q
=
4U.
p
/
q)
J
=
d T
2b.
p v (q v r )
( p A q)
A
p
(q A r)
/\
Commutative Laws
3b.
q v p
p
=q q
~
~
p
Distributive Laws
(4
J
( p V 4)
T)
A
(p
4b.
u‘ T )
p
A
=
(q v r )
(p
A
q) v ( p A r )
Identity Laws
5a. p v f
=
p
6a. p
t
=
~
5b. p
6b.
t
~
t
e
p
= f
P A f
Complement Laws
7a.
p v -p
8a.
--p
5
=
f
7b. p ~ - p
f
8b.
p
--t
3
f, -f
t
De Morgan’s Laws
9a.
-(p
V
q)
3
-p
A
9b.
-q
-(p
A
q)
=
-P”
-g
Table 2.1
I n the above table, t and f denote variables which are restricted to the truth values
true and false, respectively.
12
[CHAP. 2
ALGEBRA O F PROPOSITIONS
Solved Problems
TAUTOLOGIES AND CONTRADICTIONS
2.1. Verify that the proposition p v - ( P A q ) is a tautology.
Construct the t r u t h table of p v - ( p
2.2.
4):
T
T
F
T
T
F
F
T
T
F
T
F
T
T
F
F
F
T
T
4
1
A
C
/\
4)
q ) is T f o r all values of p and q, i t is a tautology.
A
Construct the t r u t h table of
1 4
"(P
T
A
4)
P
q
Verify that the proposition ( p A q ) A -(p
p
"(P
p
Since the t r u t h value of p v - ( p
p
A
v
q ) is a contradiction.
( p A q ) A - ( p v q):
P A 4
I
P V Q
I
"(PVq)
1
(PAq)*-(pvQ)
T
T
T
F
F
T
F
F
T
F
T
F
F
F
F
F
T
F
F
F
Since the t r u t h value of
( p + q ) A - ( p v q ) is F for all values of p and q, i t is a contradiction.
LOGICAL EQUIVALENCE
2.3. Prove the Associative Law: (11 A q ) A r = p
A
(4A r).
Construct the required t r u t h tables:
P
-
1'
T
T
T
T
F
F
F
F
Since the t r u t h tables a r e identical, the propositions a re equivalent.
2.4.
Prove that disjunction distributes over conjunction; t h a t is, prove the Distributive
Law: 11 v ( q A r ) = ( p v q ) A ( p v Y).
Construct the required t r u t h tables:
ALGEBRA O F PROPOSITIONS
CHAP. 21
13
Since the t r u t h tables a r e identical, the propositions a r e equivalent.
2.5.
Prove that the operation of disjunction can be written in terms of the operations
of conjunction and negation. Specifically, p v (r = -(-?I A -(I).
Construct the required t r u t h tables:
Since the t r u t h tables a r e identical, the propositions a r e equivalent.
2.6.
There are exactly four non-equivalent propositions of one variable; the t r u t h tables
of such propositions follow:
P
1
Pi(P0)
1
P,(?J)
1
1
p,(P)
pi(?,)
T
F
1 i
Find four such propositions.
;;
Observe t h a t
~
F
Hence
2.7.
PI (11)
p v -P,
P2h)
=
T
PVT-P
P y P
T
F
p, P 3 ( p ) = - p ,
P4(p)
=
P A -11.
Determine the number of non-equivalent propositions of two variables p and 4.
The t r u t h table of a proposition P ( p ,q ) will contain 22 = 4 lines.
appear as follows:
In each line T or F can
In other words, there a r e 24 = 16 non-equivalent propositions of two variables p and
(I.
14
2.8.
ALGEBRA O F PROPOSITIONS
Determine the number of non-equivalent propositions of:
and 1’; (ii) 72 variables pl, p , , . . . ,p,].
(i) three variables 11, q
(i)
The truth table of a proposition P ( p , q , r ) will contain 2? = 8 lines. Since in each line T o r F
can appear, there a r e 2 8 = 236 non-equivalent propositions of three variables.
(iir
The truth table of a proposition P ( p l , . . . , p,) will contain
Z2” non-equivalent propositions of 71 variables.
l l
N E GAT1ON
2.9.
[CHAP. 2
Prove De Morgan’s Laws: (i) - ( p
A
(I) = -p
v
In each case construct the required truth tables.
(i’
~
F
F
-(?;Y)
F
F
T
F
T
t
ip
iq
T
T
F
T
2l’
(ii)
”(7:
lines; hence, as above, there are
-(TI
q)
= -p
A
-q.
-”-‘
t
F
F
T
t
t
- -p = p .
2.10. Verify:
1’
I
-P
I
--P
2.11. TJse the results of the preceding problems to simplify each of the following propositions:
(i) -(p
(1)
-[I1
V
(ii)
-(I),
v
-q)
(ii) -(-11
A
(iii) -(-11
v -9)
q)
-(-]I
r
* p A --q
a),
-1)
- -p
= p
= - -p =
=
v
-a).
*Y
v -y
v -y
A
(iii) -(-p
-q
p \ 4
2.12. Simplify each of the following statements.
It is not true that his mother is English or his father is French.
(ii) It is not true that he studies physics but not mathematics.
(iii) It is not true that sales are decreasing and prices are rising.
(i)
(iv) It is not true that it is not cold or it is raining.
(ii
Let p denote “His mother is English” and let q denote “His father is French”. Then the
given statement is - ( p v q ) . But - ( p v y) = - p A - q . Hence the given statement is logically
equivalent to the statement “His mother is not English and his father is not French”.
CHAP. 21
15
ALGEBRA O F PROPOSITIONS
(ii) Let p denote “He studies physics” and let y denote “He studies mathematics”. Then the
given statement is -(PA -4). But - ( p A - q ) = - p v - - y 5 - p v q. Hence the given statement is logically equivalent t o the statement “He does not study physics or he studies
mathematics”
(iii) Since -()I (11 5 - p v -q, the given statement is logically equivalent to the statement
“Sales a l e increasing or prices are falling”.
(iv) Since -(-p q )
“It is cold and it
=
IS
-q,
the given statement is logically equivalent to the statement
not raining”.
ALGEBRA OF PROPOSITIONS
2.13. Simplify the proposition ( p v q ) -11
listed on Page 11.
by using the laws of the algebra of propositions
Statement
(1) ( I ] \ 4)
=
-i
PA
-11
Reason
(1) Commutative law
(PV 4)
(2)
=
(-PAP)v(-P
(3)
=
=
fv(-p/\q)
(3) Complement law
-pAq
(4) Identity law
(4)
(2) Distributive law
(I)
2.14. Simplify the proposition p v ( p ~ q by
) using the laws of the algebra of propositions
listed on Page 11.
Statement
Reason
=
( P A t)v ( P A 4)
(2)
=
p A ( t v q)
(2) Distributive law
(3)
E
pAt
(3) Identity law
(4)
= p
(1)
(11
4)
(1) Identity law
(4) Identity law
2.15. Simplify the proposition + v q )
propositions listed on Page 11.
( - p ~ q ) by using the laws of the algebra of
Statement
(1) -(P \ 4 )
‘4
(‘11
Reason
(“PA -q)
4Q )
V
(”p
A
4)
(1) De Morgan’s law
-p
(3)
-pAt
(3) Complement law
-p
(4) Identity law
(4)
==
MISCELLANEOUS PROBLEMS
2.16. The propositional connective
“ p or q but not both”.
A
(-qv q)
y
(2)
is called the exclusive disjunction; p y q is read
(i) Construct a truth table for I-, y q.
.
(ii) Prove: p y q = ( p v q ) ~ - ( p ~ q ) Accordingly
the original three connectives A , v and -.
(i)
Distributive law
(2)
y
can be written in terms of
Now p 1 q is true if p is true or if q is true but not if both a r e true: hence the t r u t h table of
p y q is as follows:
ALGEBRA OF PROPOSITIONS
16
(iii
We construct the t r u t h table of
( pv q) A -(p
A
[CHAP. 2
q ) , by the second method, as follows:
P I Y / ( P
”
4)
A
-
(P
A
n)
T
F
F
T
F
F
T
T
F
F
T
F
T
T
F
T
T
T
T
F
F
F
F
F
F
T
F
1
2
1
4
3
1
2
1
F
T
F
Step
2.17. The propositional connective J, is called the joint denial; p J y is read “Neither p
nor q”.
(i) Construct a truth table f o r p J q .
(ii) Prove: The three connectives V,
connective J, as follows:
( a ) -13
(i)
A
and
- may
= P J P , ( b ) P A C ? = (PJP)J,(YLY),
be expressed in terms of the
(c) PJC?
= (PL4)J(PJq)*
Now p J q is t r u e only in the case tha t p is not true and y is not true; hence the truth table of
p y is the following:
1
F
ciij
; ; ”;‘
Construct the appropriate truth tables:
(a)
;;1 U
i ”%”
(b)
F
F
T
F
“ q
F
F
f
T
T
F
T
(pyqLq)
F
F
t
ALGEBRA OF PROPOSITIONS
CHAP. 21
17
Supplementary Problems
LOGICAL EQUIVALENCE
=
2.18.
Prove the associative law f o r disjunction:
2.19.
Prove t h a t conjunction distributes over disjunction: p
2.20.
Prove ( p v y)
2.21.
Prove p v ( p A q )
2.22.
Pkove - ( p v q ) v ( - p
2.23.
(i) Express
(ii) Express
2.25.
=
-p
A
=
-p
p
A
( p vy)v
T
A
p v ( q v r)
(qv r)
=
(p A q ) v ( q A r ) .
q by constructing the appropriate t r u t h tables fsee Problem 2.13).
by constructing the appropriate t r u t h tables (see Problem 2.14).
q)
=
-p
\i
in ternis of
A
and
-.
A
in terms of v and
-.
by constructing the appropriate t r u t h tables isee Problem 2.15).
Write the negation of each of the following statements a s simply as possible.
(i)
H e is tall but handsome.
(ii) H e has blond hair o r blue eyes.
(iii) He is neither rich nor happy.
(iv) He lost his job o r he did not go to mork today.
(v) Neither Marc nor Erik is unhappy.
(vi) Audrey speaks Spanish o r French, but not German.
ALGEBRA OF PROPOSITIOKS
2.26.
Prove t h e following equivalences by using the l a ~ of
s the algebra of propositions listed on P a g e 11:
(i) p ~ ( p v q =
) 11,
iii) ( p ~ q ) v - p = - p v q ,
f i i i ) p ~ ( - p v q )5 p A q .
Answers to supplementary Problems
=
-q),
(ii) P A q
-q,
(iii) p v q.
2.23.
(i) p v q
2.24.
(i) - p v q , (ii) p
2.25.
(iii) He is rich or happy.
2.26.
(i) p ~ ( p v q =
) (pvt)r\(pvq)
-(-p
P
=
- ( - p v -4).
(vi) Audrey speaks German but neither Spanish nor French.
=
p v ( f
(I)
=- p
f
E
p.
Chapter
3
Conditional Statements
CONDITIONAL, p -+ y
Many statements, particularly in mathematics, are of the form “If p then q”.
statements are called coriditioizal statements and are denoted by
Such
p+q
The conditional p +
Q
The truth value of p +
can also be read:
(i) p implies q
(iii)
(ii) p only if q
(iv) (I is necessary f o r p .
(I
23
is sufficient f o r
(I
satisfies:
[T,] The conditional p +
y is true except in the case that 1-I is true and q is false.
The truth table of the conditional statement follows:
Example 1.1 :
Consider the following statements:
If Paris is in France, then
2 1 - 2 = 4.
(iij If P a r i s is in France, then
2 + 2 = 5.
(iJ
(iii) If P a r i s is in England, then 2 + 2 = 4.
(iv) If P a r i s is in England, then 2 + 2 = 5.
By the property I T J ] , only (ii) is a false statement; the others a r e true. JVe emphasize t h a t , by definition, (iv) is a t r u e statement even though its substatements
“Paris is in England” and “ 2 f 2 = 5” a r e false. It is a statement of the type:
If monkeys a r e human, then t h e earth is flat.
Now consider the truth table of the proposition -11
J
q:
T
T
Observe that the above truth table is identical to the truth table of p - + y.
is logically equivalent t o the proposition -p v q:
p’q
Hence 1 1 3 4
= -pvq
I n other words, the conditional statement “If p then g” is logically equivalent t o the statement “Not 11 or q” which only involves the connectives \ and
and thus was already a
part of our language. We may regard p + q as an abbreviation f o r a n oft-recurring
statement.
-
18
CHAP. 31
19
CONDITIONAL S T A T E M E N T S
BICONDITIONAL, p * q
Another common statement is of the form ‘‘p if and only if q” o r , simply, “ p iff q“.
Such statements, denoted by
P
f-$
(I
are called bicoudifio/ial statements. The truth value of the biconditional statement p e q
satisfies the following property:
[T5J If p and Q have the same truth value, Then p - q
truth values, then p - q is false.
is true; if p and
(I
have opposite
The truth table of the biconditional follows:
1
P
T
T
F
F
Example 2.1 :
1
Q
P*9
T
F
T
F
T
F
F
T
Consider the following statements:
Paris is in France if and only if
2
2 = 4.
(ii) Paris is in France if and only if
2
2 = 5.
(i)
(iii) P a r i s is in England if and only if 2 - 2 = 4.
(iv) Paris is in England if and only if 2
+2
= 5.
By property [T5], the statements (i) and (iv) a r e true, and (ii) and (iii) a r e false.
Recall that propositions P ( p , (I, . . . ) and Q ( p , q, . . .) are logically equivalent if and
only if they have the same truth table; but then, by property [Ti],
the composite proposition P(p, (I,. . . ) cf Q ( p , Q, . . .) is always true, i.e. is a tautology. In other words,
Theorem 3.1:
P ( p , (I, . . , ) = Q ( p , q, . . .) if and only if the proposition
PiPt (I, . .)
f
is a tautology.
Conditional
Theorem 3.2:
P
q
P-’9
T
T
F
F
T
F
T
F
T
F
T
T
Converse
(I
A conditional statement p
equivalent.
+
T
T
F
T
-+
P
‘
3
&(P, q, . . . )
Inverse
-P
-9
Contrapositive
T
T
F
T
T
F
T
T
+
-q
+ ”11
q and its contrapositive -q
+
-p are logically
[CHAP. 3
CONDITIONAL STA TEMEN TS
20
Example 3.1:
Consider the following statements about a triangle A :
p
q
Note t h a t p
Example 3.2:
+
-
q:
If A is equilateral, then A is isosceles.
p:
If A is isosceles, then A is equilateral.
3
q is true, but q
+
p is false.
Prove: ( p + q ) If ~2 is odd then x is odd.
We show t h a t the contrapositive - q
-p, “If x is even then x2 is even”,
is true. Let J be even: then x = 2n where is a n integer. Hence x2 = (2n)(2n)=
2 ( 2 4 ) is also even. Since the contrapositive statement -q + - p is true, the
original conditional statement p q is also true.
-f
--f
Solved Problems
CONDITIONAL
3.1. Let p denote “It is cold” and let q denote “It rains”. Write the following statements
in symbolic form.
It rains only if it is cold.
A necessary condition for it to be cold is that it rain.
A sufficient condition f o r it to be cold is that it rain.
Whenever it rains it is cold.
(v) It never rains when it is cold.
(i)
(ii)
jiii)
(iv)
Recall t h at p + q can be read ‘‘p only if q”, “p is sufficient for q” or “ q is necessary f o r p”.
(i)
q+p
iii)p+q
(iii)q-p
(iv) Now the statement “Whenever it rains i t is cold” is equivalent to “If i t rains then i t is cold”.
Th at is, q + p.
( v ) The statement “It never rains when it is cold” is equivalent to “If i t is cold then it does not
rain”. Th at is, p + -q.
3.2.
Rewrite the following statements without using the conditional.
(i) If it is cold, he wears a hat.
(ii) If productivity increases, then wages rise.
Recall t h a t “If p then q” is equivalent to “Not p or q”.
(i)
It is not cold o r he wears a hat.
(ii) Productivity does not increase or wages rise.
3.3.
Determine the truth table of ( p
-$
q ) * ( p A 4).
F
T
C,HAP. 31
CONDITIONAL STATEMENTS
3.4.
Determine the truth table of - p
3.5.
Verify that (p q )
-+
p).
-+
( p v q ) is a tautology.
F
F
3.6.
(q
-+
21
T
F
T
F
F
F
T
T
Prove that the conditional operation distributes over conjunction:
P
P
Q
?“
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
-+
((I
A 79
= (P
-+
(I) A
q“r
P+(qfiTj
P-+9
T
F
F
F
T
F
F
F
T
F
F
F
T
T
T
T
T
T
F
F
T
T
T
T
(P r )
-+
T
t
4
BICONDITIONAL
3.7. Show that “ p implies (I and (I implies 11’’ is logically equivalent to the biconditional
“71 if and only if 4”; that is, (21 q ) A ( q .+ p ) = p cs q.
+
F
F I F l
3.8.
T
F
T
I
1
T
4
I
Determine the truth value of each statement.
2 + 2 = 4 iff 3 + 6 = 9
(ii) 2 2 = 7 if and only if 5
+1= 2
(iii) 1 + 1 = 2 iff 3 + 2 = 8
(iv) 1 2 = 5 if and only if 3
+1= 4
(i)
+
+
Now p e q is true whenever p and q have the same t r u t h value: hence (i) and (ii) a r e true
statements. but (iii) and (iv) a r e false, (Observe tha t (ii) is a true statement by definition of the
conditional, even though both substatements 2 2 = 7 and 5 1 = 2 a re false.)
+
+
22
3.9.
CONDITIONAL STATEMENTS
Show that the biconditional p * q
connectives v, A and -.
Now
=
q
TI+
P*q
can be written in terms of the original three
=
- p v q and q + p
-qv p ; hence by Problem 3.7,
(P+Q)A(q+P)
(“P‘-‘q)A(-qVP)
3.10. Determine the truth value of
(11
3.11. Determine the truth value of
( p t)-4) e ( 4
T
T
F
F
T
F
T
F
+
4 ) v - ( p H Wq).
F
T
T
F
F
T
F
T
[CHAP. 3
T
T
F
T
-+
p).
F
T
F
F
Method 1
I
Step
1
1
4)
F I T I T
T
F
T
F I T I F
F I F ~
T
T
F
TF
4 1 1 ( 2 / 1
3
Method 2
NEGATION
3.12. Verify by truth tables that the negation of the conditional and biconditional are as
follows: (i) - ( p
4) = 13 A -q,
(ii) - ( p t)4 ) = 11 H - 4 = -p t)q .
+
ii)
P
1
Q
T
T i :
F I T
Remark:
1
I
P+q
F
T
F
T
F
T
Since p + q
“(P’4)
E
1
-q
F
T
F
1
PA-4
F
T
F
- p v q , we could have used De Morgan’s law t o verify (ii as follows:
-(p + 4 )
-(-fJ
V 4) 5 - - T I
A -4
p A -q
C 0 N D IT I 0 N A L STATEMENTS
CHAP. 31
3.13. Simplify:
(i)
-01
(ii)
-(-]I
(iii)
-(-/I
(ii
-(/I
-i
-q)
=
=
(ii) - ( - ] I
-q),
- -q
p
-q)
q)
c)
- -p * q
-p
A
-
H
=
p
tf
q
=
IJ tf
(I
=
-q
-11
q ) , (iii) -(-p
23
+
-(I).
11
~
3.14. Write the negation of each statement as simply as possible.
(i) If he studies, he will pass the exam.
(ii) He sn-inis if and only if the water is warm.
(iii) If it snows, then he does not drive the car.
(i)
B y Problem 3.12, - ( p + q )
=
hence the negation of !ii is
-q:
11
H e studies and he will not pass the exam.
(ii) By Problein 5.12, - ( p
f 011 ow in g :
=
t)q )
p
=
-q
-p H q ; hence the negation of (ii) is either of the
H e swims if and only if the water is not warm.
H e does not swim if and only if the water is warm.
(iii) Note t h a t
-(]I+
-q)
=p
- -q
A
=
p
q.
Hence the negation of (iiii is
I t sno\vs and he drives the car.
3.15. Write the negation of each statement in as simple a sentence as possible.
(i) If it is cold, then he wears a coat but no sweater.
(ii) If he studies, then he will go t o college or to art school.
(i)
Let 11 he “ I t is cold”, q be “He weals a coat” and r be “He n e a r s a sweater”. Then the given
Now
statement can be written as p + ( q - 7 ) .
-[p
+
((I
-?)I
=_
11 A
-(q
=
A -?)
11
(-q
v
Y)
Hence the negation of (i) is
I t is cold and he wears a sweater
(ii) The g i ~ e nstatement is of the f o l m 1 1 - ( q v 1.).
- [ p + ((I
7)
=
p
A
01
no coat.
But
- ( q v Y)
=
11
-q
I
-r
Thus the negation of (ii) is
He studies and he does not go to college
01
t o a r t school.
CONDITIONAL STATEMENTS AKD VARIATIONS
3.16. Determine the contrapositive of each statement.
(i)
(ii)
(iii)
(iv)
If John is a poet, then he is poor.
Only if Marc studies will he pass the test.
It is necessary to have snow in order f o r Eric t o ski.
is not positive.
If ,I‘ is less than zero, then
(i)
The tontraposltive of 11- q is - q +
-11.
Hence the contrapositive of (i) is
If John is not poor, then he is not a poet.
(ii) The Riven statement is equivalent to “If Marc passes the test, then he studied”.
contrapositive of (ii) is
If Marc does not study, then he will not pass the test.
Hence the
24
CONDITIONAL S T A T E M E N T S
[CHAP. 3
(iiij The given statement i s equivalent to “If Eric skis, then it snowed”. Hence the contrapositive
of iiii) is
If it did not snow, then Eric will not ski.
(iv) The contrapositive of p - -q is
- -q
+
-p
=
q
-p.
+
Hence the contrapositive of fiv) is
If n: is positive, then x is not less than zero.
3.17. Find and simplify: (i) Contrapositive of the contrapositive of p + q. (ii) Contrapositive of the converse of p + q. (iii) Contrapositive of the inverse of p + q.
(i,
(ii,
-
-
-11 -t
-q
The contrapositive of p - t q is - q + -p. The contrapositive of - q + - p is
p + q , which is the original conditional proposition.
The converse of p + q is q + p . The contrapositive of q - t p is -p+ - q , which is the inverse
of p - t q.
(iii) The inverse of p + q is - p i -q.
which is the converse of p-+ q.
The contrapositive of - p +
-q
is
-
-q
+
-
-p
=
q + p,
In other words, the inverse and converse a r e contrapositives of each other, and the conditional
and contrapositive a r e contrapositives of each other!
Supplementary Problems
STATEMENTS
3.18.
Let p denote “He is rich” and let q denote “He is happy”. Write each statement in symbolic form
using p and q.
(i)
(ii)
If he is rich then he is unhappy.
H e is neither rich nor happy.
(iii,
It is necessary t o be poor in order to be happy.
(iv) To be poor i s to be unhappy.
(v)
Being rich is a sufficient condition to being happy.
(vi) Being rich is a necessary condition t o being happy.
(vii) One is never happy when one is rich.
(viii) He is poor only if he is happy.
(ix) To be rich means the same as to be happy.
(x) He is poor or else he is both rich and happy.
N o t e . Assume “He is poor” is equivalent t o -p.
3.19.
Determine the t r u t h value of each statement.
(i) If 5 < 3, then - 3 < -5.
(ii) I t is not true t h a t 1 1 = 2 iff 3
+
+ 4 = 5.
(iii) A necessary condition t h a t 1 + 2 = 3 i s t h a t 4 + 4 = 4.
+1= 5
(iv) I t is not t r u e t h a t 1
(vi
iff 3
+ 3 = 1.
If 3 < 5 , then - 3 < - 5 .
(vi) A sufficient condition t h a t 1 1.2 = 3 is t h a t 4
3.20.
+ 4 = 4.
Determine the t r u t h value of each Statement.
+3 = 5
2+1 = 3
+ 1 = 2.
5 + 5 = 10.
(i)
It is not t r u e t h a t if Z T 2 = 4, then 3
or 1
(ii,
If 2 1 . 2 = 4, then i t is not t r u e t h a t
and
(iii) If 2 + 2 = 4, then 3 + 3 = 7 iff 1’1
= 4.
CHAP. 31
3.21.
Write the negation of each statement in a s simple a sentence a s possible.
(i) If stock prices fall, then unemployment rises.
(ii) He has blond hair if and only if he has blue eyes.
(iii) If Marc is rich, then both Eric and Audrey a r e happy.
(iv) Betty smokes Kent o r Salem only if she doesn't smoke Camcls.
(v) M a r y speaks Spanish or French if and only if she speaks Italian.
(vi) If John reads Sewsweek then he reads neither Life nor Time.
TRUTH TABLES
3.22.
Find the t r u t h table of each proposition
3.23.
3.25.
ii) ( - p v q ) + p ,
Find the t r u t h table of each proposition
(4
-cl)
(-11
q ) , (ii) (-q v P )
(i) ( P
++
3.24.
25
CONDITIONAL S T A T E M E N T S
++
+
Find the t r u t h table of each proposition:
(p--q)+
(qv-p)],
(i) [ p r \ I - q + p ) ] A
-
Prove:
iii
(pnq)+v
=
+
(ii, q e ( - q ~ p ) .
-PI.
in) q e ( r - + - p ) ] v [ ( - q + p ) + + r ] .
( p + r ) v ( q + ~ ) . (iii z ~ + ( q + r =
) ( P A - ? ) + -q.
CONDITIONAL XVD VARIATIONS
3.26.
Determine the contrapositive of each statement.
(i) If he has courage he will win.
(ii) It is necessary to be strong in order to be a sailor.
(iii) Only if he does not tire will he win.
(iv) It is sufficient for i t to be a square in order to be a rectangle.
3.27.
Find:
ii) Contrapositive of p + -q.
(ii) Contrapositive of - p + q.
(iii) Contrapositive of the converse of p + -q.
iiv) Converse of the contrapositive of - p -+ -q.
Answers to Supplementary Problems
3.18.
(i) P
(ii) - p
-q
-+
A
(iii) q
iiv) - p
+
-q
-
iv)
-p
-q
q
p
(vii 4
+
3.19.
(i) T, (ii) T, (iii) F , (iv) F, (v) F, (vi) T
3.20.
(i) F, (ii, F , (iii) T
3.21.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
3.22.
(i) T T F F , (ii) F F F T
3.23.
(i) T F T T , iii) F T F T
3.24.
(i) F T F F , (ii) T T T F T T F T
3.25.
Hint. Construct the appropriate t r u t h tables.
3.26.
(i)
(ii)
(iii)
(iv)
3.27.
(i) q-f - p ,
p
Stock prices fall and unemployment does not rise.
He has blond hair but does not have blue eyes.
Marc is rich and E r i c or Audrey is unhappy.
Betty smokes K e r t or Salem, and Camels.
Mary speaks Spanish or French, but not Italian.
John reads Sczcsweek, and Life o r T i m e .
If
If
If
If
he does not win, then he does not have courage.
he is not strong, then he is not a sailor.
he tires, then he will not win.
it is not a rectangle, then i t is not a square.
iii) -q+ p, (iii) - p +
q, (iv) p + q
(vii) p 4 -q
(viii) -11
q
+
(ix) P
(x) -P"
4
(P A 4)
++
Chapter
4
Arguments, Logical Implication
ARGUMENTS
An aiyzrmerzt is a n assertion that a given set of propositions P I ,P,, . . ., P,I, called
premises, yields (has as a consequence) another proposition Q, called the conclz~sion. Such
a n argument is denoted by
The truth value of an argument is determined as follows:
[T,] An argument P I ,P,, . . . ,P,, - Q is true if Q is true whenever all the premises
P,, P2, . . . , P I Iare true; otherwise the argument is false.
Thus an argument is a statement, i.e. has a truth value. If an argument is true it is called
a valid argument; if a n argument is false it is called a f a l l a c y .
Example 1.1:
The following argument is valid:
11, 11
+
q
+
(Law of Detachment)
Q
The proof of this rule follows from the following t r u t h table.
P
T
T
F
F
26
I
Q j
T
F
T
F
P”4
T
F
T
T
ARGUMEKTS, LOGICAL IMPLICATION
CHAP. 41
I-,
4
T
[(P
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
T
T
F
F
F
F
Example 1.4:
T I T
T
T
F
F
F
F
T
T
T
T
T
F
T
F
ll
(n
+
r)]
T
T
F
T
F
F
F
F
T ’ T
F
T
T
F
T
F
T
F
T
T
T
F
T
T
T
F
T
4)
+
27
(P
+
T
T
T
T
T
T
T
T
+
T
T
T
T
F
F
F
F
T
F
T
F
T
T
T
T
:~
F
T
F
r)
T
F
T
F
T
F
T
F
The following argument is a fallacy:
For t h e proposition
table below.
F
F
F
l(p
2’;q
+
~
; ;
q)
T
T
1
P’q,
-P t. - q
- p ] .+ - q is not a tautology, as seen in the t r u t h
(p+q;A-l’l
T
T
~
[(p+q)i-pl+-q
T
F
T
Equivalently, the argument is a fallacy since, in Case (line) 3 of the t r u t h table,
p + q and - p a r e t r u e but - q is false.
An argument can also be shown t o be valid by using previous results as illustrated in
the next example.
Example 1.5:
We prove t h a t the argument
p
+
-q,
q t -p
Statement
(1) q is true.
( 2 ) p + - q is true.
(3) q - p is true.
(4) - p is true.
+
is valid:
Reason
(1)
(2)
(3)
(4)
Given
Given
Contrapositive of (2)
Law of Detachment (Example 1.11
using (1) and (3)
ARGUMENTS AND STATEMENTS
We now apply the theory of the preceding section to arguments involving specific
statements. We emphasize that the validity of a n argument does not depend upon the
truth values nor the content of the statements appearing in the argument, but upon the
particular form of the argument. This is illustrated in the following examples.
Example 2.1 :
Consider the following argument:
S,: If a man is a bachelor, he is unhappy.
S p : If a man is unhappy, he dies young.
.......................................
S : Bachelors die young.
Here t h e statement S below the line denotes the conclusion of t h e argument, and
the statements S1 and S, above the line denote the premises. We claim t h a t the
argument S1,S2 + S is valid. For t h e argument is of t h e form
p+q,q+r
t. P + T
where p is “He is a bachelor”, q is “He is unhappy” and T is “He dies young”;
and by Example 1.3 this argument (Law of Syllogism) is valid.
[CHAP. 4
ARGUMENTS, LOGICAL IMPLICATION
28
Example 2.2:
W e claim t h a t the following argument is not valid:
S1: If two sides of a triangle a r e equal, then the opposite angles a re equal.
SB: Two sides of a triangle a r e not equal.
......................................................................
S: The opposite angles a r e not equal.
F o r t h e argument is of the form p - t q , - p c -q, where p is “Two sides of a
triangle a r e equal” and q is “The opposite angles a r e equal”; and by Example 1.4
this argument is a fallacy.
Although the conclusion S does follow from S 2 and axioms of Euclidean geometry, the above argument does not constitute such a proof since the argument is
a fallacy.
Example 2.3:
We claim t h a t the following argument is valid:
S1:If 5 is a prime number, then 5 does not divide 15.
Sy: 5 divides 15.
.............................................
S: 5 is not a prime number.
F o r t h e argument is of the form p + -q, q t- - p where p is “ 5 is a prime
number” and q is “ 5 divides 15”; and we proved this argument is valid in Example 1.5.
We remark t ha t although the conclusion here is obviously a false statement,
the argument as given is still valid. It is because of the false premise Slthat
we can logically arrive a t the false conclusion.
Example 2.4:
Determine the validity of the following argument:
S1:If 7 is less tha n 4, then 7 is not a prime number.
Sy: 7 is not less tha n 4.
..................................................
S: 7 is a prime number.
W e translate the argument into symbolic
form. Let p be “7 is less tha n 4” and q be
“7 is a prime number”. Then the argument is
of the form
p-t-q,
-P
+ 4
T
T
T
The argument is a fallacy since in Case (line)
+ -q
and - p
a r e t r u e b u t q is false.
The f act t h a t the conclusion of the argument happens to be a true statement
i s irrelevant to the fa c t t h a t the argument is a fallacy.
4 of the adjacent truth table, p
LOGICAL IMPLICATION
A proposition P(p, q, . . .) is said to logically imply a proposition Q ( p , q ,
Q ( p , q, . . .) is true whenever P(p,q, . . . ) is true.
Example 3.1:
W e claim t h a t p logically implies p v q. For consider the
t r u t h tables of p and p v q in the adjacent table. Observe
t h a t p is t r u e in Cases (lines) 1 and 2, and in these Cases
p v q is also true. In other words, p logically implies
PVQ.
Now if Q ( p , 4,. . .) is true whenever P ( p ,q,
. . .)
. . .)
if
F
is true, then the argument
P ( P ,Q, . . .) I- Q@, 4, . . . )
is valid; and conversely. Furthermore, the argument P c Q is valid if and only if the
conditional statement P + Q is always true, i.e. a tautology. We state this result formally.
CHAP. 41
29
ARGUMENTS, LOGICAL IMPLICATION
Theorem 4.2: The proposition P(p, q , . . . ) logically implies the proposition & ( p , q, . . .)
if and only if
(i) the argument P(p,q ,
or, equivalently,
. ..)
(ii) the proposition P ( p ,q , . . . I
Remark:
Q ( p , q,
-
+
. . .)
is valid
Q ( p , q, . . . ) is a tautology.
The reader should be warned that logicians and many texts use the word
“implies” in the same sense as we use “logically implies”, and so they distinguish between “implies” and “if . . . then”. These two distinct concepts
are, of course, intimately related as seen in the above Theorem 4.2.
Solved Problems
ARGUMENTS
4.1. Show that the following argument is valid: p - q ,
4 t
t
11.
Method 1.
Construct the t r u t h table on the right. Now p e p is true in Cases
(lines) 1 and 4, and Q is t r u e in Cases 1 and 3: hence p w q and q a re true
simultaneously only in Case 1 where p is also true. Thus the argument
p* q , q F p is valid.
F
Method 2.
Construct the t r u t h table of
Since [ ( p w 4 ) A 91
--f
[ ( p t t q ) A 91 .+ p :
p is a tautology, the argument is valid.
4.2.
P
T
T
F
F
Q
[ ( P -
4)
A
T
F
T
F
T
T
F
F
T
F
T
T
T
F
T
F
F
F
F
T
1
2
1
3
Step
4.3.
Determine the validity of the argument
Construct the t r u t h table of
[(-]I+
q ) A py
-
-
P
T
F
T
T
F
T
F
T ’ F
T I F
T
T
T I T
T
T
F
F
2
1
4
1
F
41
+
-p+ q, p t -q.
-q:
2
F
F
T
ARGUMENTS, LOGICAL IMPLICATION
30
T
F
F
F
T
F
F
T
T
Since the proposition
is a fallacy.
[(-p
Observe t h at - p +
4.4.
T
F
F
F
q)
+
A ?I’
~i
-p
+
q, p
+ -q
q and p ar e both true in Case (line) 1 but in this Case -q is false.
Prove that the following argument is valid:
Method 1.
T
T
T
F
is not a tautology, the argument
-q
+
[CHAP. 4
p
+
-q,
T+
q, r
yw.
I-
Construct the following truth tables:
~
1 4
?,
I
r
1
1
p+-q
r+q
T
T
F
1
-p
‘
F
F
I F
;1;
T
F
T
T
Now p + -y, 7 + q and r a r e true siinultaneously only in Case (line) 5 , where - p is also true; hence
the given argument i s valid.
Method 2.
Construct the t r u t h table for the proposition
[(P+-Q)
A
(‘)“-’q)
A
r]
+
-P
I t will be a tautology, and so the given argument is valid.
Method 3.
Reason
Statement
(1) p
+
-y
(1) Given
( 2 ) Given
is true.
( 2 ) r + q is true.
(3) -q
+
-7-
(3) Contrapositive of (2)
(4) Law of Syllogism, using (1) and (3)
is true.
(4) p + - ~ is true.
(6) r is true.
( 5 ) Contrapositive of (4)
(6) Given
(7) Hence - p is true.
(7) Law of Detachment, using (5) and (6)
(5) r+-p
is true.
ARGUMENTS AND STATEMENTS
4.5. Test the validity of each argument:
(ii) If i t rains, Erik will be sick.
(i) If it rains, Erik will be sick.
It did not rain.
Erik was not sick.
. . . ......................
........ .,..............
*
It did not rain.
Erik was not sick.
F i r s t translate the arguments into symbolic form:
(i) P
+
Y,
-11
i-
-q
(ii) P
-+
q, - q
I-
-P
where p is “It rains” and q is “Erik is sick”. By Example 1.4, the argument (i) is a fallacy; by
Problem 4.2, the argument (ii) is valid.
CHAP. 41
4.6.
31
ARGUMENTS, LOGICAL IMPLICATION
Test the validity of the following argument:
If 6 is not even, then 5 is not prime.
But 6 is even.
................................
Therefore 5 is prime.
Translate the argument into symbolic form. Let p be
“6 is even” and let q be “5 is prime.” Then the argument is
of the form
-p-+-q,
P t-
4
F
Now in the adjacent t r u t h table, -11 + - q and p a r e both
t r u e in Case (line) 2; but in this Case q is false. Hence the
argument is a fallacy.
,
T
The argument can also be shown to be a fallacy by constructing the truth table of the
proposition [ ( - p + - 9 ) ~ p +/ q and observing t h a t the proposition is not a tautology.
The f act t h at the conclusion is a true statement does not affect the fa c t t h a t the argument
is a fallacy.
4.7.
Test the validity of the following argument:
If I like mathematics, then I will study.
Either I study or I fail.
.................................
If I fail, then I do not like mathematics.
F i r s t translate the argument into synibolic form. Let p be “ I like mathematics”, q be “I study”
and r be “I fail”. Then t h e given argument is of the form
p + q , q v r
F r+-p
To test the validity of the argument, construct the truth tables of the propositions
p + q , q v r and ~ + - p :
P
9
7,
P’4
9 “ r
T
T
T
T
T
F
T
T
T
T
T
F
T
T
F
T
F
T
F
F
F
F
T
T
T
F
F
T
T
T
T
F
-11
F
F
F
F
T
T
T
T
1
?’+
-p
F
T
F
T
T
T
T
T
Recall t h at an argument is valid if the conclusion is true whenever the premises a re true.
Now in Case (line) 1 of the above t r u t h table, the premises p + q and q v 7‘ a r e both true but the
conclusion 7.+ - p is false; hence the argument is a fallacy.
4.8.
Test the validity of the following argument:
If I study, then I ~7illnot fail mathematics.
If I do not play basketball, then I will study.
But I failed mathematics.
........................................
Therefore, I played basketball.
32
ARGCRIENTS, LOGICAL IMPLICATION
p
T
T
T
T
F
F
F
F
1
q
‘
I
I
I
I
’
r
l;
p+-q
-r
-r+ P
F
F
T
T T
T
T
~ T T
T
F
T
T
T
F
T
T
F
F
T
-q
T
T I F
T I F
F
F I T I T
F ~ F I
T
~
T
F
T
F
F
F
~
T
F
F
T
LOGICAL IMPLICATION
4.9. Show that p 4 (I logically implies p cs 4.
Construct the truth table for ( p A q ) + (p
Since
(IJ
q)
-
F
q):
Pi‘
p i q
F
T
F
( p ++ q ) is a tautology, p
-
[CHAP. 4
‘
(pAq)i(p-q)
q logically implies p
tf
q.
4.10. Show that p tf (r logically implies p + q.
Consider the truth tables of p
1’
:I:
F
and p + q :
g
I
4
T
P I F
I
P-q
T
F
F
T
I
P-)Q
T
F
T
T
4.11. Prove: Let P(p,q, . . .) logically imply Q ( p , q , . . .). Then for any propositions
P I ,P,, . . . , P(P,, P,, . . .) logically implies Q(P,,P,, . , ,).
By Theorem 4.2, if P ( p , q , . . . ) logically implies Q ( p , q , . . . ) then the proposition P ( p , q , . . . ) -t
Q ( p , q , , . .) is a tautology. By the Principle of Substitution (Theorem 2.2), the proposition
P ( P , ,P 2 , , . , I
Q(Pl,P,, . . .) is also a tautology. Accordingly, P(P,, P,, . . .) logically implies
QIP,,P,, . . . ).
+
CHAP. 41
33
ARGUMEKTS. L O G I C A L IMPLICATION
4.12. Determine the number of nonequivalent propositions P ( p ,q ) which logically imply
the proposition 11 ++ q.
TI-
Consdc,r the adjacent t r u t h tahle of p ++ q . Now P ( p , (I) logically imq IS t r u e
plies p t-t (J i f IJ H q is t r u e whenever P ( p , qi is true. But p
only in Cases linesi 1 and 4; hence Pi7i, q ) c a n n o t be t r u e in Cases 2 anti 3.
There a r e foul, iuch propositions which a l e listed below.
F
T
F
F
F
F
T
4.13. Show that pt-t-(1 does not logically imply
Method I .
Construct the t r u t h tables of 21
2'
1
T
T
F
A
I
r!
'
cf - q
-rl1
I
q.
TI+
and p
+
q:
I
p--q
l,+q
T
T
T
F
T
F
T
F
T
P
F
T
T I F
T
Supplementary Problems
ARGUMENTS
4.14.
Test the validity of each argument: l i )
4.15.
Test the validity of each argument,
4.16.
Test the validity of each argument: riI
q , p t- - q ;
-]I+
ti1 11- 9, r ' + -q
li+
-4,
Y--'
(ii) -11-
t- Y -
p , q t- - r :
ARGUMENTS AND STATERIENTS
1.17.
Test the validity of the argument:
If London is not in Denmark, then Paris is not in France.
But Paris is i n France.
.......................................................
Therefore, London is in Denmark.
4.18,
Test the validity of the argument:
If I study, then I will not fail mathematics.
I did not study.
..........................................
I failed mathematics.
-p;
- q , q I-
p.
(ii) p + - q , - r + - q
iii) p+ q ,
YV
- q , -r
t- p +
I-
-p.
-T.
34
4.19.
[CHAP. 4
ARGTJPVIENTS, LOGICAL IMPLICATION
Translate into symbolic form and test the validity of the argument:
icr) If G is even, then 2 does not divide 7.
Either 5 is not prime or 2 divides 7 .
But 5 is prime.
........................................
Therefore, 6 is odd (not e v e n ) .
( b ) On iiiy wife’s birthday, I bring her flowers.
Either it’s iny wife’s birthday or I work late.
I did not bring m y wife flowers today.
............................................
Therefore, today I worked late.
If I woi.k, I cannot study.
Either I work, o r I pass mathematics.
I passed mathematics.
(c)
........................................
Therefore, I studied.
i d ) If I work, I cannot study.
Either I study, o r I pass mathematics.
I worked.
........................................
‘Therefore, I passed mathematics.
LOGICAL IXIPLICATION
4.20.
Show t h a t rii p A q logically implies p , (ii) p v q does not logically imply p .
4.21.
Show t h a t iil q logically implies p - q, (ii) - p logically implies p + q .
4.22.
Show t h a t p
4.23.
Determine those propositions
4.24.
Determine the number of nonequivalent propositions P ( p , q ) which logically imply the proposition
p- q , and construct t r u t h tables for such propositions (see Problern 4.12).
( q v r ) logically implies ( p
15
?
q ) v r.
hich logically imply (i) a tautology, (ii) a contradiction.
Answers to Supplementary Problems
4.14.
(I)
fallacy,
(11) valid
4.15.
ti)
valid,
4.16.
(1)
valid, i i i i valid
4.17.
valid
4.18.
fallacy
4.19.
(a)
4.23.
I ) E\ e i y pioposition logically implies a tautology.
conti adiction.
4.24.
Theie a r e eight such propositions
(111 fallacy
-
p -q, - 7 v q, r + - p ; valid.
( b ) p + q , p v I * , - q + r ; valid.
( c ) p + - q , p v i , r F y, fallacy.
( d ) p + - q , q v T , p + 7 , valid.
(ii, Only a contradiction logically implies a
F
F
F
T
T
Chapter
5
Set Theory
SETS AND ELEMENTS
The concept of a set appears in all branches of mathematics. Intuitively, a set is any
well-defined list or collection of objects, and will be denoted by capital letters A , B, X , Y , . . , ,
The objects comprising the set are called its elements or menzbem and will be denoted by
lower case letters a,b , x , y , . . . . The statement ‘‘p is an element of A” or, equivalently,
“ p belongs to A” is written
p EA
The negation of p E A is written p 6!A.
There are essentially two ways to specify a particular set. One way, if it is possible,
is to list its members. For example,
A = {a,e, i, 0,u )
denotes the set A whose elements are the letters a,e,i,o,u. Note that the elements are
separated by commas and enclosed in braces { 1. The second way is t o state those properties which characterize the elements in the set. For example,
B = { x : x i s a n integer, x
>0}
which reads “ B is the set of x such that x is an integer and x is greater than zero,” denotes
the set B whose elements are the positive integers. A letter, usually x, is used to denote a
typical member of the set; the colon is read as “such that” and the comma as “and”.
Example 1.1:
The set B above can also be written as B = {1,2,3, . . .I.
Observe t h a t -6 4 B , 3 E B and
Example 1.2:
P
4 B.
The set A above can also be written as
A = {Z : x is a letter in the English alphabet,
E
is a vowel}
Observe t h a t b 6Z A , e E A and p fZ A .
Example 1.3:
Let E = {x : xz-33x f 2 = 0 ) . In other words, E consists of those numbers
which a r e solutions of the equation xz- 3% 2 = 0, sometimes called the solution
set of the given equation. Since the solutions of the equation a r e 1 and 2, we
could also write E = {1,2}.
+
Two sets A and B are equal, written A = B, if they consist of the same elements, i.e.
if each member of A belongs to B and each member of B belongs to A . The negation of
A = B is written A # B.
Example 1.4:
Let E = {x :
+
3x 2 = 0 } , F = ( 2 , l} and G = {l,2,2,1,6/3}.
Then E = F = G. Observe that a set does not depend on t h e way in which
its elements are displayed. A set remains the same if its elements a r e repeated or
rearranged.
x2-
35
36
[CHAP. 5
S E T THEORY
FINITE AND INFINITE SETS
Sets can be finite or infinite. X set is finite if it consists of exactly n different elements,
where 72 is some positive integer; otherwise i t is infinite.
Example 2.1:
Let M be the set of the days of the week. In other words,
M = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday]
Then M is finite.
Example 2.2:
Let Y = { 2 , 4, 6, 8 ,
Example 2.3:
Let P = {x : s is a river on the earth}. Although it may be difficult to count the
number of rivers on the earth, P is a finite set.
. , ,}.
Then Y is infinite.
SUBSETS
A set A is a subset of a set E or, equivalently, B is a superset of A , written
A C E or B > A
iff each element in A also belongs t o E ; that is, x E A implies x E B. We also say that
A is contailzed in B or B contains A . The negation of A c B is written A $! E or B $ A
and states that there is a n x E A such that x 6! E .
Example 3.1:
Consider the sets
A = (1,3, 5, 7,
C = {.c :
2
. . .},
is prime,
T
.
B = { 5 , 10, 15, 20, . . .I
> 2 } = { 3 , 5 , 7 , 1 1 , ...}
Then C c A since every prime number greater than 2 is odd. On the other hand,
B @ A since 10 E B but 10 @ A .
Example 3.2:
Let N denote the set of positive integers, 2 denote the set of integers, Q denote
the set of rational numbers and R denote the set of real numbers. Then
N c Z c Q c R
Example 3.3:
I
The set E = { 2 , 4 , 6 ] is a subset of the set F = {6,2,4}, since each number 2, 4
and 6 belonging to E also belongs to F . In fact, E = 8’. I n a similar manner
it can be shown t ha t every set is a subset of itself.
As noted in the preceding example, A C B does not exclude the possibility that A = B.
In fact, we may restate the definition of equality of sets as follows:
Definition:
I Two sets
A and E are equal if A
c B and E c A .
In the case that A c B but A = E , we say that A is a proper subset of B o r B contains
A properly. The reader should be warned that some authors use the symbol C for a subset
and the symbol c only for a proper subset.
The following theorem is a consequence of the preceding definitions:
Theorem 5.1:
Let A , B and C be sets. Then: (i) A c A ; (ii) if A C B and B
A = B ; and (iii) if A C E and B C C, then A C C.
C
A , then
UNIVERSAL AND NULL SETS
In any application of the theory of sets, all sets under investigation are regarded as
subsets of a fixed set. We call this set the universal set or universe of discourse and
denote it (in this chapter) by C .
Example 1.1:
I n plane geometry, the universal set consists of all the points in the plane.
Example 4.2:
I n human population studies, the universal set consists of all the people in the
world.
CHAP. 51
37
SET THEORY
It is also convenient to introduce the concept of the empty or null set, that is, a set
which contains no elements. This set, denoted by 8, is considered finite and a subset of
every other set. Thus, f o r any set A , $8 c A C U.
Then A is empty, i.e. A = @.
Example 4.3:
Let A = {x : x’ = 4, z is odd}.
Example 4.4:
Let B be the set of people in the world who a r e older than 200 years. According
to known statistics, B is the null set.
CLASS, COLLECTION, FAMILY
Frequently, the members of a set are sets themselves. For example, each line in a set
of lines is a set of points. To help clarify these situations, other words, such as “class”,
“collection” and “family” are used. Usually we use class o r collection for a set of sets,
and family f o r a set of classes. The words subclass, subcollection and subfamily have
meanings analogous to subset.
Example 5.1:
The members of the class {{2,3}, {2}, {5,6}}
Example 5.2:
Consider any set A . The power set of A , denoted by ‘?‘(A) or Z A , is the class of all
subsets of A . In particular, if A = {a,b , c}, then
a r e the sets {2,3}, {2} and { 5 , S}.
T(A) = { A , {a,b } , {a,c>, ( 6 , c>, { a } , Cb), {c>, @I
In general, if A is finite and has n elements, then “ ( A ) will have 2n elements.
SET OPERATIONS
The union of two sets A and B, denoted by A U B, is the set of all elements which
belong to A o r to B:
A U B = { x : x E A or x E B }
Here “or” is used in the sense of andlor.
The intersection of two sets A and B, denoted by A n B, is the set of elements which
belong to both A and B:
A n B = { x : x E A and x E B }
If A n B = 0,that is, if A and B do not have any elements in common, then A and B
are said to be disjoint or non-intersecting.
The relative co?nplernent of a set 5’ with respect to a set A or, simply, the difference of
A and B, denoted by A\B,
is the set of elements which belong to A but which do not
belong to B:
A\B
= { n : : x E A , x4B}
Observe that A \B
and B are disjoint, i.e. (A\B)
n B = $3.
The absolute complement or, simply, complement of a set A , denoted by Ac, is the set
of elements which do not belong t o A :
A“ = { x : x E U , x 4 A )
That is, Ac is the difference of the universal set U and A .
Example 6.1 :
The following diagrams, called Venn diagrams, illustrate the above set operations.
Here sets are represented by simple plane areas and U , the universal set, by the
area in the entire rectangle.
SET THEORY
38
A
U
A n B is shaded.
B is shaded.
A\B
Example 6.2:
[CHAP. 5
A c is shaded.
is shaded.
Let A = {1,2, 3 , 4 ) and B = {3,4,5, 6} where ’7, = {1,2,3, . . .}.
A
U
A\B
Then:
A n B = (3, 4)
B = (1, 2, 3, 4, 5, 6}
Ac = (5, 6 , 7, . ..}
= {1,2}
Sets under the above operations satisfy various laws o r identities which are listed in
Table 5.1 below. In fact we state:
Theorem 5.2:
Sets satisfy the laws in Table 5.1.
LAWS OF THE $LGEBRA OF SETS
la. A U A = A
2a.
(AuB)uC = Au(BUC)
3a. A U B = B u A
4a.
Idempotent Laws
lb.
AnA = A
Associative Laws
2b.
(AnB)nC = An(BnC)
Commutative Laws
3b. A n B = B n A
Distributive Laws
Au(BnC) = (AuB)n(AuC)
4b. A n ( B u C ) = ( A n B ) u ( A n C )
5a. A u @ = A
Identity Laws
5b. A n U = A
6a.
AUU = U
6b. A n @ = @
7a.
AuAc = U
8a.
(Ac)c = A
9a.
( A u B ) ~= AcnBc
Complement Laws
7b. A n A c = @
8b.
Uc =
13,
@c
=U
De Morgan’s Laws
9b. ( A nB)c = A c u B c
Table 5.1
Remark:
Each of the above laws follows from the analogous logical law in Table 2.1,
Page 11. For example,
A n B = { x : x E A and z E B } = { x : x E B a n d x E A } = B n A
CHAP. 51
39
SET THEORY
Here me use the fact that if 11 is I E A and q is x E B, then p
equivalent to q ~ p :P A C != q ~ p .
A
q is logically
Lastly we state the relationship between set inclusion and the above set operations:
Theorem 5.3:
Each of the following conditions is equivalent t o A c B:
(i) A n B = A
(iii) Bc c Ac
(v) B
(ii) A U B = B
u A‘
= Lr
(iv) A n Br = $3
ARGUMENTS AND VENN DIAGRAMS
Many verbal statements can be translated into equivalent statements about sets which
can be described by Venn diagrams. Hence Venn diagrams are very often used to determine
the validity of an argument.
Example 7.1 :
Consider t h e following argument:
S1: Babies a re illogical.
S p : Nobody is despised who can manage a crocodile.
S,: Illogical people a r e despised.
................................................
S : Babies cannot manage crocodiles.
(The above argument is adapted from Lewis Carroll, Sumbolic L o g i c ; he is also the
the set of babies is a subset of the
author of Alice in WoizclerLa?zcZ.) Now by S1,
set of illogical people:
babies
By S,, t h e set of illogical people is contained in the set of despised people:
Furthermore, by S 2 , the set of despised people and the set of people who can
manage a crocodile a r e disjoint:
despised people
people who can
manage
crocodiles
illogical people
But by t h e above Venn diagram, the set of babies is disjoint from the set of people
who can manage crocodiles, o r “Babies cannot manage crocodiles” is a consequence
of S,, S , and S3. Thus the above argument,
is valid.
s,, s,, s 3
+
s
[CHAP. 5
S E T THEORY
40
Solved Problems
SETS, ELEMENTS
5.1. Let A = ( x :3x = 6 ) . Does A = 2 ?
A is the set which consists of the single element 2, that is, A = {2}. The number 2 belongs
t o A ; it does not equal A . There is a basic difference between a n element p and the singleton
set { p } .
5.2.
Which of these sets are equal:
{T,
t , s } , {s,t, Y,S},
{ t ,s, t , r } ,
{s,r,s,
t}?
They are all equal. Order and repetition do not change a set.
5.3.
M'hich of the following sets are finite?
(iv) {x : x is an even number)
(i) The months of the year.
(ii) (1, 2, 3, . . ., 99, loo}
(v) (1,2, 3, * . . >
(iii) The number of people living on the earth.
The first three sets a r e finite; the last two sets a r e infinite.
5.4.
Determine which of the following sets are equal: 8, (01,{g}.
set
5.5.
Each is different from the other. The set ( 0 ) contains one element, the number zero. The
is the empty set. The set (0)also contains one element, the null set.
0 contains no elements; it
Determine whether or not each set is the null set:
(i) X = { x :x2 = 9, 2x = 4}, (ii) Y = {x:x # x}, (iii) 2 = { x :x + 8 = 8 ) .
(i)
There is no number which satisfies both
x2
= 9 and 2 2 = 4; hence X is empty, i.e. X = 0.
(ii) We assume t h a t any object is itself, so Y is also empty. In fact, some texts define the null
set as follows:
$3
E
{x:x#x)
+
(iii) The number zero satisfies z 8 = 8; hence 2 = (0).
since i t contains 0. That is, Z # @ .
Accordingly, 2 is not the empty set
SUBSETS
5.6. Prove that A = {2,3,4,5} is not a subset of B = ( x :x is even}.
It is necessary to show that at least one element in A does not belong to B . Now 3 E A and,
since B consists of even numbers, 3 6? B ; hence A is not a subset of B .
5.7.
Prove Theorem 5.l(iii): If A C B and B c C , then A c C.
x
5.8.
We must show t h a t each element in A also belongs to C. Let x E A . Now A c B implies
c C; hence x E C. We have shown t h a t x E A implies x E C, t h a t is, that A c C.
E B. But B
Find the power set "(8) of the set S = { 1,2,3} .
The power set T(S) of S is the class of all subsets of S; these a r e {1,2,3), {1,2}, {1,3), {2,3},
{l}, {2}, (3) and the empty set 0. Hence
T(S) = { S , C1,3}, {2,3}, {1,2}, C1), {2>, { 3 ) ,
Note t h a t there a r e 23 = 8 subsets of S.
CHAP. 51
5.9.
41
SET T H E O R Y
Let V = ( d ) , W = { c , d } , X = ( a , b , c ) , Y = { a , b } and 2 = ( a , b , d } . Determine
whether each statement is true or false:
(i) Y c X , (ii) W + Z , (iii) Z 3 17, (iv) V c X , (v) X = W , ( ~ 7 i ) W c Y .
(i)
Since each element in Y is a member of X , Y c X is true.
(ii) Now a E 2 but
(-1
4 W ; hence W
# 2 is true.
(iii) The only element in V i s d and it also belongs to 2 ; hence Z > V is true.
(iv) V is not a subset of X since d E V but d 4 X ; hence V C X is false.
(v) Now u E X but a 4 W ; hence X = W is false.
(vi) W is not a subset of Y since c E 68 but c 4 Y ; hence W C Y is false.
5.10. Prove:
If A is a subset of the empty set @, then A = @ .
8 is a subset of every set; in particular, @ C A . But, by hypothesis, A c
The null set
hence A = 0.
SET OPERATIONS
5.11. Let U = ( l , 2 , . . . , 8,9}, A = { l , 2 , 3 , 4 } , B = [2,4,6,8}
(i) Ac, (ii) A n C, (iii) ( A n C ) [ , (iv) A u B , (v) B\C.
(i)
and C={3,4,5,6}.
$3;
Find:
Ac consists of the elements in C t h at a re not in A ; hence Ac = {5,6, 7,8,9}.
(ii) A n C consists of the elements in both A and C; hence A n C = { 3 , 4 } .
(iii) ( A n C ) c consists of the elements in C tha t a r e not in A n C .
so (AnCC)c= {1,2,5,6,7,8,9}.
Now by iii), A n C = { 3 , 4 ) and
(iv) A U B consists of the elements in A or B ( o r both): hence A LB = { 1 , 2 , 3 , 4 , 6 ,8}.
(v) B\C
consists of the elements in B which a r e not in C ; hence B\C
5.12. In each Venn diagram below, shade:
(i)
= {S,S}.
[i) A U B, (ii) A n B.
A L E consists of those elements which belong to A or B (or both); hence shade the a re a in A
and in B as follows:
A U B is shaded.
(ii) A n B consists of the ar ea t h a t is common to both A and B . To compute A n B, first shade
A with strokes slanting upward t o the right (////) and then shade B with strokes slanting
downward t o the r i g h t (\\\\), as follows:
Then A n E consists of t h e cross-hatched area which is shaded below:
42
SET THEORY
[CHAP. 5
A n B is shaded.
Observe the following:
( a ) A n E is empty if A and E a r e disjoint.
( b ) A r E = B if B c A .
i c ) A , - I B = A if A C E .
5.13. In the Venn diagram below, shade: (i) B", (ii) ( A C B ) < ,(iii) (B\A)c,
(il
(iv) A' n B c .
Bc consists of the elements which do not belong to B : hence shade the a r e a outside B a s follows:
EC is shaded.
(ii) F i r s t shade A U B ; then iA1-Bj' is the a r e a outside A I J B :
A U B is shaded.
(iii, F i r s t shade B\A,
( AU E)' is shaded.
the area in B which does not lie in A ; then (B\A)c
is the area outside
E\A.
B \A
is shaded.
( E \ A)c is shaded.
(ivr F i r s t shade Ac, the area outside of A , with strokes slanting upward to the right ( / / I / ) , and
then Ac0iBc is the
then shade Bc with strokes slanting downward t o the right (\\\\I;
cross-hatched area:
S E T THEORY
CHAP. 51
A c and Bc a r e shaded.
43
AcnBc is shaded.
Observe t h at ( A LIB)"= ACnBc, a s expected by De Morgan's law.
5.14. I n the Venn diagram below, shade
(i)
(i) A n ( B U C), (ii) ( A n B ) U ( An C).
F i r st shade A with upward slanted strokes, and then shade B U C with downward slanted
strokes; now A n (BU C) is the cross-hatched area:
A and BUC a r e shaded.
A n ( B u C ) is shaded.
(ii) F i r st shade A nB with upward slanted strokes, and then shade A n C with downward slanted
strokes; now ( An B ) U ( A n C) is the total a re a shaded:
A n B and A 17 C are shaded.
( An B)U ( An C) is shaded.
Notice t h a t A n ( B uC ) = ( A n B ) u (A n C), a s expected by the distributive law.
B\A
= B n A". Thus the set operation of difference can be written in
terms of the operations of intersection and complementation.
5.15. Prove:
B\A
= {x: x E B , .,;@A) =
{x: x E B , x E A C } = B n A c
44
SET THEORY
[CHAP. 5
A n ( B U C) = ( A n B ) U ( An C).
5.16. Prove the Distributive Law:
An(BuC)
17
=
=x
=
=
{x : x € A ; x E B u C }
{x: x E A ; x € B or x E C }
{ x : % € A ,x E B ; o r x E A , x E C }
{ r e : x E A n B or a E A n C }
(AnB)u(AnC)
Observe t h a t in t h e third step above we used the analogous logical la w
P
5.17. Prove:
( A \B)
4 (Q
"4
=
(P A 4)
" (P
A
4
n B = @.
(A\B)nB
=
=
{ x : xEA\B,
xEB}
{ x : x E A , x 4 B ;z E B }
=!a
The last step follows from the fa c t t h a t there is no element
2
satisfying z E B and z 48.
5.18. Prove De Morgan's Law: ( A U B ) c= A c n B c .
( A u B ) ~=
{x: x 4 A u B }
= {x: x4A, x4B}
= {x: x E A c , x E B C }
= AcnBc
Observe t h at in the second step above we used the analogous logical law
"(p
V
4)
5.19. Prove: For any sets A and B, A n B C A
-p
C
A
-q
A UB.
Let z E A n B ; then x E A and x E B. I n particular, x E A . Since x E A n B implies x € A ,
A n B c A. Furthermore, if z E A , then x E A o r x E B , i.e. 3: E A u B . Hence A c AUB. In other
words, A n B c A c A u B .
5.20. Prove Theorem 5.3(i): A c B if and onlyif A n B = A .
Suppose A c B . Let x E A ; then by hypothesis, x E B. Hence x € A and x E B , i.e. x E A n B.
Accordingly, A c A nB. On the other hand, i t is always true (Problem 5.19) t h a t A n B C A .
Thus A n B = A .
Now suppose t h a t A n B = A . Then in particular, A C A nB.
A n B c B. Thus A c A n B c B and so, by Theorem 5.1, A c B.
But i t is always true t h a t
ARGUMENTS AND VENN DIAGRAMS
5.21. Show that the following argument is not valid by constructing a Venn diagram in
which the premises hold but the conclusion does not hold:
Some students are lazy.
All males are lazy.
.....................
Some students are males.
Consider the following Venn diagram:
43
SET THEORY
CHAP. 51
lazy people
students
males
Notice t h a t both premises hold, but the conclusion does not hold. For a n argument to be valid, the
conclusion must always be t r u e whenever the premises a r e true. Since the diagram represents a
case in which the conclusion is false, even though the premises a r e true, the argument is false.
It is possible to construct a Venn diagram in which the premises and conclusion hold, such as
lazy people
students
males
5.22. Show that the following argument is not valid:
All students are lazy.
Nobody who is wealthy is a student.
...............................
Lazy people are not wealthy.
Consider the following Venn diagram:
wealthy people
Now the premises hold in the above diagram, but the conclusion does not hold; hence the argument
is not valid.
5.23. For the following set of premises, find a conclusion such that the argument is valid:
S,: All lawyers are wealthy.
S,: Poets are temperamental.
S,: No temperamental person is wealthy.
....................................
S:
By S,, the set of lawyers is a subset of the set of wealthy people: and by S3, the set of wealthy
people and the set of temperamental people a r e disjoint. Thus
temperamental people
46
CHAP. 5
S E T THEORY
By S y , the set of poets is a subset of the set of temperamental people; hence
Thus the statement “No poet is a lawyer’’ o r equivalently “No lawyer is a poet” is a valid
conclusion.
The statements “NO poet is wealthy” and “No lawyer is temperamental”
conclusions which do not make use of all the premises.
are also valid
MISCELLANEOUS PROBLEMS
5.24. Let A = (2, (4,5}, 4). Which statements are incorrect and why?
(i) ( 4 , 5 ) C A , (ii) {4,5} E A , (iii) ((4,5}} C A .
The elements of A a r e 2, 1 and the set {4,5}. Therefore (iii is correct, but (i) is a n incorrect
statement. Furthermore, (iii) is also a correct statement since the set consisting of the single
element {4, 5 ) which belongs to A is a subset of A .
5.25. Let A = (2, {4,5}, 4}. Which statements are incorrect and why?
(il 5 E A , (ii) (5) E A , jiii) {5) c A.
Each statement is incorrect. The elements of A a re 2 , 4 and the set {4,5}; hence (ii and (ii)
a r e incorrect. There a r e eight subsets of A and (5) is not one of them; so (iii) is also incorrect.
5.26. Find the power set T(S) of the set
S = (3, { 1 , 4 ) } .
Note first t h a t S contains two elements, 3 and the set {1,4).
Therefore T(S) contains
22 = 3 elements: S itself, the empty set (3, and the two singleton sets which contain the elements
3 and { 1 , 4 } respectively, i.e. {3} and {{1,4}}. I n other words,
Tu(S) = {S,{3}, {{L4}}> c)}
Supplementary Problems
SETS, SUBSETS
5.27.
Let A = { 1 , 2,..., 8,9}, B = {2,4,6,8}, C = {1,3,5,7,9], D = {3,4,5} and E = { 3 , 5 } .
Which sets can equal X if we ar e given the following information?
C c but
# A.
(i) X and E a r e disjoint. (ii) X c D but X # B. (iii) X C A but x’ C. (iv)
x
x
5.28.
S t a te whether each statement is t rue or false:
(ii Every subset of a finite set is finite. rii) Every subset of an infinite set is infinite.
5.29.
Find the power set “ ( A ) of A = (1,2,3,4} and the power set T(i3) of B = {l,{ 2 , 3 ) , 4 ) .
CHAP. 51
5.30.
47
State whether each set is finite or infinite:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
5.31.
S E T THEORY
The
The
The
The
The
The
set
set
set
set
set
set
of
of
of
of
of
of
lines parallel to the 7 axis.
letters in the English alphabet.
numbers which a r e multiples of 5.
animals living on the earth.
numbers which are solutions of the equation x 2 i
circles through the origin (0,O).
+ 2 6 ~ -1 17x11
~
+ 7x3 - 10 = 0.
State whether each statement is true o r false:
(i) {1,4,3] = { 3 , 4 , 1 ]
(ii) {1,3,1,2, 3 , 2 ) c {I,2,3}
(iii) { 1 , 2 } = { 2 , 1 , 1 , 2 , 1 ]
(iv) C4) E CC4))
(v) {4} c CC4))
(vi) 8 c CC43)
SET OPERATIONS
5.32.
Let U = { u , b , c , d , e , f , g } , A = { a , b , c , d , e ) , B = { a , c , e , g } and C = { b , e , f , g } . Find:
( i ] AUC
(iii) C \ B
(v) C c n A
ivii) ( A \ B c ) c
iiir B n A
(vi) ( A \ C P
iviii) ( An A c ) c
(iv) B C ~ C
(i) W\V
(ii) V c U W
(iiii V n W c
(iv) Vc\Wc.
5.33.
I n the Venn diagrams below, shade
5.34.
Draw a Venn diagram of three non-empty sets A , B and C so t h a t A , B and C have the following
properties:
(iii) A c C, A # C, B n C = $3
( i ) A c B , C c B , AnC = 8
(iv) A c ( B n C ) , B c C , C I B , A Z C
(ii, A C E , CQB, AnC # 0
5.35.
The formula A \ B = A nBc defines the difference operation in terms of the operations of
intersection and complement. Find a formula that defines the union of two sets, A U B , in terms of
the operations of intersection and complement.
5.36.
Prove Theorem 5.3iiil: A c B if and only if A U B = B.
5.37.
Prove:
If A T E = (3, then A c B c .
5.38.
Prove:
Ac \ Bc = b' \A.
5.39.
Prove: A c B implies A u ( B \A)
5.40.
Prove: An(B\C)
= ( A n B )\(A?C).
Give
an
example
to
show
t h a t A ( B \ C) f ( AU B ) \ ( AJ C ) .
(ii)
= B.
(i)
-ARGUMENTS AND VENS DIAGRAMS
Determine the validity of each argument for each proposed conclusion.
5.41.
No college professor is wealthy.
Some poets a r e wealthy.
(i) Some poets a r e college professors.
(ii) Some poets a r e not college professors.
SET THEORY
48
5.42.
Determine the validity of each a r g ~ i m e n tf o r each proposed conclusion.
All poets a r e interesting people.
Audrey is a n interesting person.
5.43.
iiI
Audrey is a poet.
iiii
Audrey is not a poet.
Determine the validity of the argument f o r each proposed conclusion.
All poets a r e poor.
I n order to be a teacher, one must graduate from college.
No college graduate is poor.
(i)
Teachers a r e not poor.
(ii) Poets a r e not teachers.
(iii) If Marc is a college graduate, then he is not a poet.
5.44.
Determine the validity of the argument f o r each proposed conclusion.
A11 mathematicians a r e interesting people.
Some teachers sell insurance.
Only uninteresting people become insurance salesmen.
Insurance salesnien a r e not mathematicians.
Some interesting people a r e not teachers.
Some teachers a r e not interesting people.
Some mathematicians a r e teachers.
Some teachers a r e not mathematicians.
If Eric is a mathematician, then he does not sell insurance.
Answers to Supplementary Problems
5.27.
(i) C and E , (ii) D and E , [iii) A , E and D, (iv) None
5.28.
(i) T, (ii, F
5.30.
(i) infinite, (ii) finite, (iii) infinite, (iv) finite, (v) finite, ( v i ) infinite
5.31.
(i) T, (ii) T, (iii) T, (iv) T, (v) F, (vi) T
5.32.
AUC = U
(vj
(ii) B n A = { a , c , e }
(vi)
(i)
(iii) C \ B = { b , f}
(iv) BCUC = { b , d , e , f , g }
5.33.
(a)
CcnA = {a,c,d: = CL
( A \ C). = { b , e , f,d
(vii) ( A \Bc)c = { b , cI, t’, gl
(viii) (AnAc)c = U
[CHAP. 5
CHAP. 51
49
SET THEORY
W\V
Observe t h at V cu W = U and
5.34.
(i)
@
VnW
C
= 0 in case ( b ) where V C VV.
(iii)
@@
(ii)
5.35.
A u B = (AcnBc)C
5.40.
(ii)
A L ( E \ C) is shaded.
( A u B ) \ ( A u C) is shaded.
5.41.
(i) fallacy, (iii valid
5.42.
(i) fallacy, (iij fallacy
5.43.
(i) valid, (ii, valid, iiii) valid
5.44.
(i) valid, (ii) fallacy, riii) valid, (iv) fallacy, ivi valid, (vi) valid
The following Venn diagrams show why (ii) and (iv) a r e fallacies:
interesting
insurance
0
mathematicians
u
teachers
Chapter 6
Product Sets
ORDERED PAIRS
An ordeyed pair consists of two elements, say a and 0 , in which one of them, say a,
is designated as the first element and the other as the second element. Such an ordered
pair is written
(a, b )
Two ordered pairs (a, b ) and (c, d ) are equal if and only if a = c and b = d.
Example 1.1:
The ordered pairs ( 2 , 3 ) and ( 3 , 2 ) a r e different.
Example 1.2:
The points in the Cartesian plane in Fig. 6-1 below represent ordered pairs of
real numbers.
Example 1.3:
The set { 2 , 3 } is not a n ordered pair since the elements 2 and 3 a re not distinguished.
Example 1.4:
Ordered pairs can have the same first and second elements, such as (1,l), (4,4)and
(5,5).
Remark:
An ordered pair ( u , b ) can be defined rigorously as follows:
(a, b ) =
{@It
@,b)}
the key here being that {a} c { a , b J which we use to distinguish a as the
first element.
From this definition, the fundamental property of ordered pairs can be
proven:
(a,b ) = (c, d ) if and only if u = c and b = d
PRODUCT SETS
Let A and B be two sets. The iwoduct set (or Cartesian product) of A and B, written
A X B, consists of all ordered pairs (a,
b ) where a E A and b E B:
A x B = ( ( a , b ): a E A, b E B }
The product of a set with itself, say A X A , is sometimes denoted by A2.
Example 2.1:
The reader is familiar with the Cartesian plane R2 = R X R (Fig. 6-1 below).
Here each point P represents a n ordered pair ( u , b ) of real numbers and vice versa;
the vertical line through P meets the x axis a t a , and the horizontal line through P
meets t h e 7~ axis at b.
B
P
b
a
t -1
J::
1
2
Fig. 6-2
Fig. 6-1
50
8
A
CHAP. 61
PRODUCT SETS
51
Since A and B do not contain many elements, it is possible to represent A X B by
a coordinate diagram a s shown i n Fig. 6-2 above. Here the vertical lines through
the points of A and the horizontal lines through the points of B meet in 6 points
which represent A X B in the obvious way. The point P is the ordered pair (2, b ) .
In general, if a finite set A has s elements and a finite set B has t elements, then
A x B has s times t elements. If either A or B is empty, then A X B is empty. Lastly, if
either A or B is infinite, and the other is not empty, then A x B is also infinite.
PRODUCT SETS I N GENERAL
The concept of a product set can be extended to more than two sets in a natural way.
The Cartesian product of sets A , B, C, denoted by A X B X C, consists of all ordered triplets
(a,b, c ) where u E A, 2, E B and c E C:
A x B x C
=
((ct,b,c): u E A , b E B , c E C ]
Analogously, the Cartesian product of n sets A,, A,, . . . , A n , denoted by A, X A,
consists of all ordered n-tuples (a,,
a,, . . ., uil) where a, E A,, . . . , anE An:
A , x A, x
X
-
X
An,
. . ., an E A n }
Here an ordered ?i-tuple has the obvious intuitive meaning, that is, it consists of n elements,
*
x An =
{(cL,,. . .,an)
: a, E A,,
not necessarily distinct, in which one of them is designated as the first element, another
as the second element, etc. Furthermore,
(a,,
Example 3.1 :
. . .,an)
= ( b l , . . ., b J
a, = b,, . . ., ail= biz
iff
In three dimensional Euclidean geometry each point represents a n ordered triplet:
its x-component, its y-component and its z-component.
TRUTH SETS OF PROPOSITIONS
Recall that any proposition P containing, say, three variables p,
value to each of the eight cases below:
P I
q l
T I T
T
T
T
F
T
F
F
T
F
T
F
F
F
F
r
T
F
T
F
T
F
T
F
(r
and r , assigns a truth
52
[CHAP. 6
PRODUCT SETS
1-
The truth set of a proposition P, written S ( P ) , consists of those elements
of U for which the proposition P is true.
Example 1.1:
Following is the truth table of ( p + q )
A (q
p+q
T
T
T
T
T
(P+q)A\(Q+r)
T
F
T
T
T
F
T
T
T
F
F
F
T
F
T
T
T
Accordingly, the truth set of ( p -+ q )
T ( ( p+ q)
A
( q + Y))
A
T):
q+r
T
T
F
F
F
+
( q + T ) is
=
{TTT, FTT, FFT, FFF}
The next theorem shows the intimate relationship between the set operations and the
logical connectives.
Theorem 6.1:
Let P and Q be propositions.
Then:
(i) S(P + Q ) = 7 ( P )n S ( Q )
(ii) T ( P \ Q) = T ( P )U T ( Q )
(iii) T ( - P )
= (T(P))"
(iv) P logically implies Q if and only if S(P)c 7 ( Q )
The proof of this theorem follows directly from the definitions of the logical connectives
and the set operations.
Solved Problems
ORDERED PAIRS
6.1.
Let W = {John, Jim, Tom) and let
V = {Betty, Mary}. Find W X V .
TV x V consists of all ordered pairs (a,b ) where a E IT' and b E V . Hence,
W
6.2.
X
=
V
{(John, Betty), (John, Mary), (Jim, Betty), (Jim, ilIaryi, (Tom, Betty), (Tom, Rlary)}
(x+ y, 1) = (3, n.- y). Find r and y.
( x - u,1) = (3, x - y) then, by the fundamental
Suppose
If
.r+r/=3
and
property of ordered pairs,
l = x - y
The solution of these simultaneous equations is giverrby
.2:
= 2, y = 1.
CHAP. 61
6.3.
53
PRODUCT SETS
Let A = { a , b , c , d , e , f } and B = (a, e , i , o , u } . Determine
the ordered pairs corresponding to the points PI,P,, P ,
and P, which appear in the coordinate diagram of A X B
on the right.
The vertical line through P I crosses the A axis at b and the
horizontal line through PI crosses t h e B axis a t i; hence P I corresponds to the ordered pair ( b , i). Similarly, P , = ( a ,a), P , = (d, u)
and P, = ( e , e ) .
B
0
i
a
PRODUCT SETS
6.1. Let A = {1,2,3}, B = {2,4} and C = {3,4,5}. Find AX B
A convenient method of finding A
X
B
>;
X
A
C.
C is through the so-called “tree diagram” shown below:
The “tree” is constructed from the left to the right.
listed to the r i g h t of the “tree”.
6.5.
a b c d e f
A
X
B X C consists of the ordered triples
Let A = {a,b ) , B = {2,3) and C = {3,4}. Find:
(i) A x ( B u C), (ii) ( A x B ) u ( A x C ) , (iii) A x ( B n C), (iv) ( A X B ) fl ( A X C).
(i)
F i r st compute B u C = {2,3,4}.
A
(ii) F i r st find A
X
X (BUG‘)
B and A
X
Then
=
{ ( a , 21, (a,3), (a,4),( b , 2 ) , ( b , 3), ( b , 4):
C:
A xB
A xC
=
{ ( a , 2), (a,3), ( b , 2 ) , ( b , 3 ) )
=
{ ( a , 31, (a,4),( b , 3), ( b , 4))
Then compute the union of the two sets:
( A x B ) u ( A x C)
=
{ ( a , 21,
(a,3), ( b , 2), ( b , 31, ( a , 41, ( b , 4))
Observe, from (i) and (ii), t h a t
Ax(BuC) = (AXB)U(AXC)
(iii) Fi r st compute B r C = {3}.
Then
A x ( Br ,C ) = { ( a ,31, ( b , 3))
54
PRODUCT S E T S
[CHAP. 6
(iv) Now 4 X B and A X C were computed above. The intersection of A X B and A X C consists of
those ordered pairs which belong t o both sets:
( A x B ) n ( A x C ) = { ( a , 31, ( h3))
Observe from (iii) and (iv) that
Ax(BnC)= (AxB)n(AxC)
6.6.
Prove: A x ( B n C ) = ( A x B ) n ( A x C).
A x (BnC) =
{(x,y) : x € A , y E
= {(x,y) : x E A ,
=
=
6.7.
y
BnC}
E B , y E C}
{(x,Y) : (x,4 E A x B ,
( A x B ) n ( A x C)
(2,
Y) E A
Let S = { a , b } , W = { l , 2 , 3 , 4 , 5 , 6 } and V = { 3 , 5 , 7 , 9 } .
x c>
Find ( S x W ) n ( S X V ) .
The product set (SX W ) n ( S K V ) can be found by first computing S X W and S X V , and
then computing the intersection of these sets. On the other hand, by the preceding problem,
( S X W ) n ( S X V ) = S X ( W n V ) . NOW W n V = {3,5}, and so
(S x W ) n
6.8.
Prove: Let A
C
(Sx V ) = S x ( W n V ) = { ( a ,31, ( a , 5 ) , ( b , 3), ( b , 5))
B and C
C
D; then ( A X C) C ( B X D).
Let (x,y) be a n y arbitrary element in A X C; then x € A and y EC. By hypothesis, A c B and
C c D ; hence x E B and y E D. Accordingly, (x,y) belongs t o B X D. W e have shown t h a t
( x , y ) € A X C implies (z,y) E B X D ; hence ( A X C) C ( B X D).
TRUTH SETS O F PROPOSITIONS
6.9. Find the truth set of p A -q.
First construct the truth table of P A -q:
F
Note t h a t p
A
-q
is true only in the case t h a t p is true and q is false; hence
PA
-q)
= {TF)
6.10. Find the truth set of - p + q.
First construct the truth table of - p + q:
F
Now - p + q is true in the first three cases; hence
T(-p
+ q)
= {TT, T F , F T ]
PRODUCT SETS
CHAP. 61
P
rl
1 1
?’
P”4
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
6.12. Suppose the proposition P = P ( p ,q,
T ( P )of the proposition P.
, ,
55
I
( P V d A Y
T
T
F
T
T
F
T
T
F
F
T
F
F
F
. ) is a tautology.
Determine the t r u t h set
If P is a tautology, then P is true for any t r u t h values of its variables. Hence t h e t r u t h set
of P i s the universal set: T ( P ) = U .
6.13. Suppose the proposition P = P ( p , g , , . . ) is a contradiction.
set T ( P ) of the proposition P.
Determine the t r u t h
If P is a contradiction, then there is no case in which P is true, i.e. P is false f o r a n y t r u t h
values of its variables. Hence t h e t r u t h set of P is empty: T ( P ) = $3,
6.14. Let P = P ( p , q , . , .) and Q = Q ( p , q , . . , ) be propositions such t h a t P A Q is a
contradiction. Show t h a t t h e t r u t h sets T ( P ) and T ( Q )a r e disjoint.
If P A Q is a contradiction, then its t r u t h set is empty: T ( P A Q) =
0. Hence, by
Theorem 6.1(i),
T ( P ) n 7 ( Q ) = T ( P A Q ) = $3
6.15. Suppose t h a t the proposition P = P(p,g, . . .) logically implies the proposition
Q = Q ( p , g, . . .). Show t h a t the t r u t h sets T ( P ) and T ( - Q ) a r e disjoint.
Since P logically implies Q, 7 ( P ) is a subset of T ( Q ) . B u t by Theorem 5.3,
7 ( P )c T ( Q )
is equivalent to
T ( P )n ( 7 ( Q ) ) c = @
Furthermore, by Theorem 6.l(iii), ( T ( Q ) ) c = 7 ( - Q ) .Hence
7 ( P )n T ( - Q ) = @
56
‘CHAP. 6
PRODUCT S E T S
Supplementary Problems
ORDERED P-IIIRS, PRODUCT SETS
(u-2,
2 x t 1) =
:m
2 Find
) . x and y.
6.16.
Suppose
6.17.
Find the ordered pairs corresponding to the points P I , P,, P , and P , which appear below in the
coordinate diagram of {l, 2, 3 , 4 > Y ( 2 , 4 , 6 , S}.
(.I.-
1, ~
~
4
2
1
2
3
4
6.18.
Let W = {Mark,Eric, Paul: and let V = {Eric,David].
t i ) IT’ v T’, (ii) V X W , (iii) 1’1 = T’X V .
6.19.
Let A = {2,3}, B = { 1 , 3 , 5 ) and C = {3,4}.
in Problem 6.4, and then find A X B X C.
6.20.
Let S = { a , b , c}, T = { b , c, cl) and W = {a,d } . Construct the tree diagram of
then find S X T X W .
6.21.
Suppose t h a t the sets V , W and Z have 3, 4 and 5 elements respectively. Determine the number
of elenients in (i) V x W x 2, iii) Z x V X W , (iii) W X 2 X V .
6.22.
Let A = B r: C. Determine if either statement is true:
rii A A A = ( B x B ) n ( ( C x C ) , (ii) A x A = ( B x C ) ? ( C X B ) .
6.23.
Prove:
A
X
Find:
Construct the “tree diagram” of A
SX
X
C, as
?’ A TI.‘
and
X
E
(BUG‘) = ( A X B ) u ( A X C ) .
TRUTH SETS OF PROPOSITIONS
6.2-1.
Find the truth se t of p
6.25.
Find the t r u t h set of - p v -q.
6.26.
Find the t r u t h set of ( p v
6.27.
Find the truth set of ( p .+ q )
6.28.
Let P = P ( p ,q, . . .) and Q = Q ( p , q , . . .) be propositions such t h a t P v Q is a tautology.
t ha t the union of the t r u t h sets T ( P ) and T(Q)is the universal set: T ( P ) U T ( Q )= 7J.
6.29.
Let the proposition P = P ( p ,q , . . . ) logically imply the proposition Q = Q ( p , q, . . . ) . Show t h a t
the union of the t r u t h sets T ( - P ) and T ( Q )is the universal set: T ( - P ) u T ( Q ) = L‘.
t)- q ,
Q) i
A
-r.
( p ++ 9.).
Show
Answers to Supplementary Problems
= 2, u = 3.
6.16.
.I.
6.17.
P I = ( 1 , 3 ) , P, = (2,8), P; = ( 4 6 ) and P4 = (3,2).
6.18.
ii)
V A V = {(Mark, Eric), (Mark,David), (Eric, Eric), (Eric, David), (Paul, Eric), (Paul, David))
(ii) V
(iii) V
X
X
W = {(Eric, Ma rk), (David,Nark), (Eric, Eric), (David, Eric), (Eric, Paul), iDavid, Paul)}
V = ((Eric, Eric), (Eric, David), (David, Eric), (David, David)}
57
PRODUCT S E T S
CHAP. 61
6.19.
3, 3)
3, 4)
5 , 3)
5 . 4)
1, 3 )
1, 4)
3, 3)
3 , 4)
5, 3)
5 , 4)
The elements of A x E x C a r e the ordered triplets to the right
6.21.
Each has 60 elements.
6.22.
Both a r e true:
6.23.
A
X
(BuC) =
=
=
=
=
A
X
A = ( B X B ) n (C X C ) = ( B X C) n (C X 6’).
{ ( Y , y) : 2 E A , y EBUC}
{ ( x , ~ ) :% € A ;y E B o r y E C }
{ ( x , ~ :) x E A , ZJ E B : or .T E A , y E C}
{(x,y) : (x,y) E A X B ; o r (s,u)E A X C}
( A x B ) u ( A X C)
Observe t h a t the logical law p
A
(q \
T) E
( p 1, q ) v ( p A
= {TF. FT}
6.24.
T(p ts - q )
6.25.
T(--p v - q ) = { T F , F T , F F }
6.26.
T ( ( p v q)
-t
-T)
the tree diagram above.
= { T T F , T F F , FTF, F F T , FFF]
T)
was used in the third step above.
Chapter 7
Relations
RELATIONS
A binuisu i7elotion or, simply, relation R from a set A to a set B assigns to each pair
(a, b ) in A Y E exactly one of the following statements:
(i) “u is related to b”, written a R b
(ii) “u is not related to b”, written a$ Z,
A relation from a set A to the same set A is called a relation in A .
Example 1.1:
Set inclusion is a relation in any class of sets. For, given a n y pair of sets A and B ,
either A C B o r A d B .
Example 1.2:
Marriage is a relation from t h e set M of men t o the set W of women. For, given
a n y man m E -11 and any woman w E TV, either ?ii is married to ~ t ‘ or nz is not
married to w.
Example 1.3:
Order, symbolized by “<”, or, equivalently, the sentence “x is less than y”, is a
relation in any set of real numbers. For, given any ordered pair ( u , b ) of real
numbers, either
a<b
or
a Q b
Example 1.4:
Perpendicularity is a relation in the set of lines in the plane. For, given any pair
of lines a and b , either a is perpendicular t o b o r a is not perpendicular t o b.
RELATIOKS AS SETS OF ORDERED PAIRS
Now any relation R from a set A t o a set B uniquely defines a subset R“ of A
follows:
€2:: = { ( a ,b ) : a is related to b } = {(a,
b ) : aR b }
X
B as
On the other hand, any subset R,’of A x B defines a relation R from A to B as follows:
n I: 2, iff ( a , b ) E
In view of the correspondence between relations R from A to B and subsets of A X B , we
redefine a relation by
I Definition: 1
A relation R from A to B is a subset of A x B.
Example 2.1:
Let R be the following relation from A = {1,2, 3)
B = {a, b } :
R = CCL 4, (1, b ) , (370))
I l l 1
to
Then 1R a, 2 @ b , 3 R a, and 3 $ b. The relation R is displayed on the coordinate diagram of A X B on the right.
Example 2 . 2 :
Let R be the following relation in W = { a , b , c]:
R = { ( a , b ) , (a,c), (c,
Then a$a, u R b , c R c and c $ a .
58
4,(c, b ) )
a
1
2
3
CHAP. 71
Example 2.3:
59
RELATIONS
Let A be a n y set. The i d e x t i t y relation in A , denoted by 1 or P A , is the set of all
pairs in A X A with equal coordinates:
A.4 = { ( a , a ) : a E A }
The identity relation is also called the diagonal by virtue of its position in a coordinate diagram of A X A .
INVERSE RELATION
Let R be a relation from A to B. The inverse of €2, denoted by R-l, is the relation
from B to A which consists of those ordered pairs which when reversed belong to R:
R-’
Example 3.1 :
= { ( b ,a ) :
(a, b ) E R }
Consider the relation
in A = {1,2,3}.
R = {(1,2), (1,3), ( 2 , 3 ) )
Then
R-’ = ((2, I), (3,1), (3,2)$
a r e identical, respectively, to the relations
Observe t h a t R and R-’
in A , i.e.,
(a, b ) E R iff a
Example 3.2:
<b
and
<
and
>
( a , b ) E R-l iff a > b
The inverses of the relations defined by
“x is the husband of y”
‘‘x is taller than
and
y”
a r e respectively
“x is the wife of y”
and
“3:
is shorter than y”
EQUIVALENCE RELATIONS
A relation R in a set A is called .i.e.flexive if a R a , i.e. (a,a) E R, for every a E A .
Example 4.1:
(i)
Let R be the relation of similarity in the set of triangles in the plane. Then
R is reflexive since every triangle is similar to itself.
(ii) Let R be the relation < in any set of real numbers, i.e. (a,b ) E R iff a < b.
Then R is not reflexive since a Q a for any real number a.
A relation R in a set A is called symmetric if whenever a R b then b R a , i.e. if
(a, b ) E R implies ( b , a ) E R.
Example 4.2:
ii) The relation R of similarity of triangles is symmetric.
similar to triangle /3, then ,8 is similar t o 01.
For if triangle
01
is
(ii) The relation R = {(l,
3), (2,3), (2,2), (3, l ) } in A = {l,2,3} is not symmetric
since (2,3) E R but ( 3 , 2 ) 4 R .
A relation R in a set A is called t ~ a n s i t i v eif whenever a R b and b R c then a R c ,
i.e. if (a, 6) E R and ( b , c ) E R implies (a,c) E R.
Example 4.3:
(i)
The relation R of similarity of triangles is transitive since if triangle
similar to ,8 and /3 is similar to y, then a is similar to y.
01
is
(ii) The relation R of perpendicularity of lines in the plane is not transitive.
Since if line a is perpendicular to line b and line b is perpendicular to line c,
then a is parallel and not perpendicular to c.
A relation R is an equivalence r e l a t i m if R is (i) reflexive, (ii) symmetric, and
(iii) transitive.
Example 4.4:
By the three preceding examples, the relation R of similarity of triangles is a n
equivalence relation since it is reflexive, symmetric and transitive.
60
RELATIONS
[CHAP. 7
PARTITIONS
A partition of a set X is a subdivision of X into subsets which are disjoint and whose
union is X , i.e. such that each a € X belongs to one and only one of the subsets. The
subsets in a partition are called cells.
of subsets of X is a partition of X iff:
Thus the collection { A l ,A,, . . . ,
(ij X = A,UA,U
UA,,,; (ii) for any A i , A , , either A , = A j or A , n A j = (2
-
Example 5.1 :
a
+
Consider the following classes of subsets of X = { l , 2 , .
. . ,8,9}:
(i)
[{I,3,5}, { 2 , 6 ) , {4,8,931
!ii)
[{I, 3,5}, {2,4,6,83, {5,7,9}]
(iii) [{I,3,51, {2,4,6,8}, (7,911
Then (i) is not a partition of X since 7 E X but 7 does not belong to any of the cells.
Furthermore, (ii) is not a partition of X since 5 E X and 5 belongs t o both { l , 3,5}
and {5,7,9). On the other hand, ( 5 ) is a partition of X since each element of
belongs to exactly one cell.
x
EQUIVALENCE RELATIONS AND PARTITIONS
Let R be an equivalence relation in a set A and, for each a E A , let [a], called the
equivalence class of A , be the set of elements t o which a is related:
[a]
=
{X : ( a , ~E
) R}
The collection of equivalence classes of A , denoted by A I R, is called the quotient of A by R:
AIR
=
{[a]: a E A )
The fundamental property of a quotient set is contained in the following theorem.
Theorem 7.1:
Let R be an equivalence relation in a set A . Then the quotient set A I R
is a partition of A . Specifically,
(i) a E [ a ] , for every a E A ;
(iij [a] = [ b ] if and only if (a,b ) E R;
(iii) if [a]# [ b ] , then [a] and [ b ] are disjoint.
Example 6.1:
Let R , be the relation in Z, the set of integers, defined by
x = y (mod 5 )
which reads “z is congruent to y modulo 5” and which means t h a t the difference
x - y is divisible by 5 . Then R5 is an equivalence relation in Z. There a r e exactly
five distinct equivalence classes in Z I Rj :
A,
zz
{. . ., -10, -5, 0, 5 , 10,
. . .}
A , = { ..., -9, -4, 1, 6, 11, ...}
A , = { ..., -8, -3, 2 , 7 , 12, ...}
A , = { ..., -7, -2, 3, 8, 13, . . . }
A4
= { ..., -6, -1, 4, 9, 14, . . . }
+
Observe t h a t each integer x which is uniquely expressible in the form x = 5 q T
where 0 5 r < 5 is a member of the equivalence A , where r is the remainder. Note
t h a t the equivalence classes a r e pairwise disjoint and t h a t
2 = A, U A, U A, U A, U A,
61
RELATIONS
CHAP. 71
Solved Problems
RELATIONS
7.1. Let El be the relation < from A = (1 ,2 ,3 ,4 ) t o B = (1,3,5}, i.e. defined by
“x is less than y”.
(i) Write R as it set of ordered pairs.
(ii) Plot R on a coordinate diagram of ‘4x B.
(iii) Find the inverse relation l?*.
(i)
R consists of those ordered pairs
(a,b ) E A X B
for
< b ; hence
R = h l , 3 ) , ( 1 , 5 ) , (2,31, 12, 51, 1 3 ,51, (4 , 5 ))
which a
(ii) R is sketched on the coordinate diagram of A X E as
shown in the figure.
(iii) The inverse of R consists of the same pairs as a r e in R
but in the reverse order; hence
R-’
1
= ((3,1), ( 5 , 1 ) , (3,21, ( 5 , 2 ) , 1 5 , 3 ) ,(5,4))
Observe t h a t R-1 is the relation
greater than y”.
7.2.
IiEEE
1
>, i.e.
2
3
4
defined by ‘‘x is
Let €2 be the relation from E = [ 2 , 3 , 4 , 5 ) to F = {3,6,7,10) defined by “x divides y”.
(i) Write R as a set of ordered pairs.
(ii) Plot R on a coordinate diagram of E XF .
(iii) Find the inverse relation R-I.
(i)
Choose from the sixteen ordered pairs in E X F those in
which the first element divides the second; then
R = { ( 2 ,6), ( 2 , lo), ( 3 , 31, 13, 61, 1 5 , l O ) )
(ii) R is sketched on the coordinate diagram of E X F as
shown in the figure.
(iii) To find the inverse of R , write the elements of R but in
reverse order.
R-l = { ( 6,2) , (10,2), ( 3 , 3 ) ,f6,3i, 110,5)}
7.3.
Let d l = (a,b , c, d } and let R be the relation in M
consisting of those points which are displayed on the
coordinate diagram of M x M on the right.
d
Find all the elements in A l which are related to b,
that is, ( J . (x,b ) E R } .
b
(i)
(ii) Find all those elements in J l to which d is related, that is, {x : (d,x)€ R ) .
(iii) Find the inverse relation R-l.
(i)
C
a
a
b
c
d
The horizontal line through b contains all points of R in x h i c h b appears as the second
element (cr, b ) , ( b , b ) and (d,b ) . Hence the desired set is { a , b , d ] .
(ii) The vertical line through d contains all the points of R in Nhich ( I appears as the first element
(d,a)and ( t l , b ) . Hence { u , b \ i b the desired set.
(iii) F i i s t n r i t e R a s a set of ordered pails, and then write the pairs in reverse order
R
R-’
= ( ( a , b ) , ( b , a ) , (6, b), ( b , 4 , (c, c ) , ( d , (0, ( d , b ) )
= ( ( b , a),( a , b ) , ( b , b ) , (6b ) , (c, c), ( a , 4, ( b , 4 )
62
[CHAP. 7
RELATIONS
EQUIVALENCE RELATIONS
7.4. Let R be the relation 4 in N = {1,2,3, . . .}, i.e. (a, b ) E R iff a L b. Determine
whether R is (i) reflexive, (ii) symmetric, (iii) transitive, (iv) an equivalence
relation.
(i) R is reflexive since a f a for every a € N .
(ii) R is not symmetric since, for example, 3 f 5 but 5
(iii) R is transitive since a 5 b and b 6 c implies a 6 c.
a 3, i.e.
(3,5) E R b u t (5,3) 4 R.
(iv) R is not an equivalence relation since i t is not symmetric.
7.5.
Let R be the relation ( 1 (parallel) in the set of lines in the plane. Determine whether
R is (i) reflexive, (ii) symmetric, (iii) transitive, (iv) an equivalence relation.
(Assume that every line is parallel to itself.)
(i) R is reflexive since, by assumption, (Y / I a f o r every line a.
(ii) R is symmetric since if a ( 1 p then p 11 a, i.e. if the line a is parallel to the line /3 then /3 is
parallel to a.
iiii) R is transitive since if
Q
) j /3 and /3
1) y
then a I)y.
(iv) R is an equivalence relation since it is reflexive, symmetric and transitive.
7.6.
Let W = {1,2,3,4}. Consider the following relations in W :
R , = ((1,2 ) , (4,317(2,2),(2,1), (3,1))
R, = ((2,219 (2,3), (392))
R, = {(1,3))
Determine whether each relation is (i) symmetric, (ii) transitive.
(i)
Now a relation R is not symmetric if there exists an ordered pair (a, b ) E R
( b , a ) 4 R. Hence:
R , is not symmetric since (4,3) E Rl but (3,4) 4 R , ,
such that
R3 is not symmetric since (1,3)E R3 but ( 3 , l ) 4 R3.
On the other hand, R, is symmetric.
(ii) A relation R is not transitive if there exist elements a, b and c , not necessarily distinct, such
that
( a , b ) E R and ( b , c) E R but ( a , c ) 4 R
Hence R , is not transitive since
(4,3) E R1 and ( 3 , l ) E Rl
but
( 4 , l ) 4 Rl
but
(3,3) 4 R,
Furthermore, R, is not transitive since
(3,2) E R , and (2,3) E R,
On the other hand, R3 is transitive.
7.7.
Let R be a relation in A . Show that:
(i) R is reflexive iff
(i)
Recall t h a t
A
A C
R ; (ii) R is symmetric iff R = R-l.
A = {(a,a) : a E A } .
Thus R is reflexive iff
(a,a) E R for every a € A iff
c R.
(ii) Suppose R is symmetric. Let ( a , b ) E R , then ( b , a ) E R by symmetry. Hence ( a ,b ) E R-1,
and so R c R-1. On the other hand, let (a,b ) E R-1; then ( b , a) E R and, by symmetry,
( a , b ) E R. Thus R-1 c R , and so R = R-1.
Now suppose R = R-1.
symmetric.
Let
(a,b) E R; then
( b , a ) E R-1 = R.
Accordingly R is
7.8.
63
RELATIONS
CHAP. 71
Consider the relation R = ((1,l), ( 2 , 3 ) , (3,2)} in X = { 1 , 2 , 3 } . Determine whether
or not El is ( i ) reflexive, (ii) symmetric, (iii) transitive.
(ij R is not reflexive since 2 E X but 12,2j 4 R.
(ii) R is symmetric since R-1 = ((1,l), (3,2), (2,3)} = R.
(iii) R is not transitive since ( 3 , 2 ) E R and 12,3) E R b u t (3,3)4 R.
7.9.
Let N = [ l ,2 , 3 , . . . }, and let R be the relation = in N
(a,b ) (c, d ) iff ad = bc
Prove that R is an equivalence relation.
X
N defined by
--
F o r every (a, b ) E N X N , ( a , b )
( a , b ) since a6 = ba; hence R is reflexive.
Suppose (a,b ) = (c, d). Then ad = bc, which implies t h a t cb = da. Hence (c, d )
and so R is symmetric.
Now suppose (a,b )
( e , d ) and (c, d ) = ( e , f ) . Then ad = bc and cf = de. Thus
(a, b)
iatZ)(cf) = (be)(&)
and, by cancelling from both sides, at’ = b e . Accordingly, (a, b ) = (e, f ) and so R is transitive.
Since R is reflexive, symmetric and transitive, R is a n equivalence relation.
then the above relation R is,
Observe t h a t if the ordered pair (a,b ) is written as a fraction
b
a
c .
in fact, the usual definition of equality between fractions, i.e. 7 = - 1ff ad = bc.
j
d
7.10. Prove Theorem 7.1: Let R be an equivalence relation in a set A . Then the quotient
set A 1R is a partition of A . Specifically,
(i) a E [ a ] , for every a € A ;
(ii) [a] = rbl if and only if (a, b ) E R ;
(iii) if [ a ] + Ibl, then la] and [bl are disjoint.
Proot o f ( I ) , Since R is reflexive, (a, a ) E R f o r every a € A and therefore a € [ a ] .
Proof o t (11). Suppose ( a ,b ) E R. We want to show t h a t [ a ] = [ b ] . Let x E [b]; then ( b , x) E R.
But by hypothesis ( a , b ) E R and so, by transitivity, ( a ,a ) E R. Accordingly x € [ a ] . Thus [bl C [ a ] .
To prove t h a t a c , b ] , we observe t h a t ( a , b ) E R implies, by symmetry, t h a t ( b , a ) E R. Then by a
similar argument, we obtain [a]~ [ b l . Consequently, [a]= [ b ] .
On the other hand, if [a]= [ b ] ,then, by ( I ) , b E [b]= [a]; hence ( a , b ) E R.
Proof o t (1111. We prove the equivalent contrapositive statement
if ‘a1 bl # @ then [a] = [bl
If [a]r b] P $3, then there exists an element x E A with .c E [ a ] n [ b l . Hence (a,x)E R and
(b,x) E R. By symmetry, (x,b) E R and by transitivity, ( a , b ) E R. Consequently by (ii), [ a ] = lbl.
PARTITIONS
7.11. Let X = { a , b , c , d , e , f , g } , and let:
(i) A , = { a , c , e ) , A , = { b } , A , = { d , ~ } ;
(ii) B , = ( a , e , Y), B, = { c , d } , B, = { b , e, f } ;
(iii) C, = ( a , b , e , g } , C, = { c } , C, = { d , j ) ;
(iv) D,= { a , b , c, d, e , f, SI.
Which of
{ A , ,A 2 ,A 3 } , { B l ,B2,BJ), {C,, C,, C3}, {D,} are partitions of X ?
(i) { A , , A , , A , ] is not a partition of
(ii) { B , , E 2 , E , ] is not a partition of
(iii) {C,, C 2 ,C,] is a partition of X
X = C , d C 2 u C3 and the sets
(iv) {D,;is a partition of X .
X since t E X but f does not belong to either A l , A > , o r A , .
X since e E X belongs to both B , and B,.
since each element in X belongs t o exactly one cell, i.e.
a r e pairwise disjoint.
64
RELATIONS
[CHAP. 7
7.12.
Supplementary Problems
RELATIONS
7.13.
Let R be the relation in A = ( 2 , 3 , 4 , 5 ) defined by “x and ?/ a r e relatively prime”, i.e. “the only
comnion divisor of x and y is 1”.
(i) Write R as a set of ordered pairs. (ii) Plot R on a coordinate diagram of A X A . (iii) Find R-1.
7.14.
Let
N =
{l,2 , 3 ,
.. . ]
and let R be the relation in N defined by r
R = {(rr,y): x , y E N , r + 2 y = 8 }
(i) Write R as a set of ordered pairs.
7.15.
= 8, i.e.,
(ii) Find R - l .
Let C = { 1 , 2 , 3 , 4 , 5 } and let R be the relation in C consisting of the points displayed in the
following coordinate diagram of C X C:
(i)
State whether each is t r u e or false:
( a ) 1 R 4, ( b ) 2 R 5, (c) 3
(ii) Find the elements of each of the following subsets of C:
(a) {x : 3 R x}, ( b ) {x : (4,x ) E R } , ( c ) {x : ( x , 2 ) 62 R } ,
7.16.
+ 2y
Consider the relations
diagonal relation.
< and
5
in N = {l,2 , 3 , . . .}.
$ 1,
(d) 5
( d ) {Z : x R 5).
Show t h a t
<UA =
EQUIVALENCE RELAATIONS
7.17.
Let TY = {l,2 , 3 , 4 } . Consider the following relations in W :
Rl = {(1,1), ( 1 , 2 ) }
R , = {(1,1), ( 2 , 3 ) , (431))
R4 = ((1, I), (2,217 (3,311
R, = T.V X W
R , = {(1,3), ( 2 , 4 ) )
R, =
Determine whether or not each relation is reflexive.
8
$ 3.
f
where 1 is the
RELATIONS
CHAP. 71
65
7.18.
Determine whether o r not each relation in Problem 7.17 is symmetric.
7.19.
Determine whether o r not each relation in Problem 7.17 is transitive.
7.20.
Let R be the relation I of perpendicularity in the set of lines in the plane. Determine whether
R is Ii) reflexive, (ii) symmetric, (iii) transitive, (iv) a n equivalence relation.
7.21.
Let N = {l,2,3, . . . } and let
=
be the relation in N X N defined by
(a,b)
(i) Prore r is a n equivalence relation.
(c,d)
iff
a
+d
= b
+c
(ii) Find the equivalence class of (2,5), i.e. [ ( 2 , 5 ) ] .
7.22.
Prove: If R and S a r e equivalence relations in a set X , then R n S is also a n equivalence
relation in X .
7.23.
(i)
Show t h a t if relations R and S a r e each reflexive and symmetric, then R
reflexive and symmetric.
U
S is also
(ii) Give a n example of transitive relations R and S f o r which R U S is not transitive.
V = {l,Z,3 ) .
7.25.
Find all partitions of
7.26.
Let [ A , , A , , . . ., A,,,1 and [B,,
B,, . . ., B,] be partitions of a set X . Show t h a t the collection of sets
is also a partition (called t h e
[ A , n B , : i = l , ...,m , j = l , . . . ,n]
p ~ ~ , t i t i ~ofi l X
) .
CTOSS
7.13.
7.14.
7.15.
7.17.
7.18.
R,, R, and R , a r e the only symmetric relations.
7.19.
All the relations a r e transitive.
7.20.
(i) no, ( i i ) yes, iiii) no, (iv) no
7.21.
(ii) [(Z,5)l = {(a,b ) : a
= {(a,a
+ 5 = b + 2, a , b E N )
+ 3) :
a E NI = {(I,4), (2,5), (3,6), ( 4 , 7 ) , . . .I
1.23.
(ii) R = { \ l , 2 ) ) and S = {(2,3)}
7.24.
(i) no, (iii no, (iii) yes, (iv) yes
7.25.
[ { 1 , 2 , 3 ) ] , [{l:, {Z, 3)], [C2>, C1,3$], [{3), {1,2)1 and [{I},{z:, (311
Chapter 8
Functions
DEFINITION OF A FUNCTION
Suppose that to each element of a set A there is assigned a unique element of a set B ;
the collection, f , of such assignments is called a functioyz (or mapping) from (or on) A
into B and is written
f : ~ +B
or A & B
The unique element in B assigned to a E A by f is denoted by f’(a),and called the i m a g e of
a under f o r the value of f at a. The domain of f is A , the co-domain B. The m x g e of f,
denoted by f [ A ]is the set of images, i.e.,
f l A ] = { f ( a ): a € A }
Example 1.1:
Let f assign to each real number its square, t h a t is, for every real number
f ( x ) = x2. Then the image of -3 is 9 and so we may write f ( - 3 ) = 9 o r f : -3
2
:
let
9.
+
Example 1.2:
Let f assign to each country in the world its capital city. Here the domain of f is
t h e set of countries in the world; the co-domain is the list of cities of the world.
The image of France is Paris, t h a t is, f\Fra nc e ) = Paris.
Example 1.3:
Let A = { a , b , c , d ) and B = { a, b, c } .
a-b,
The assignments
b - t c , c + c and d + b
define a function f from A into B. Here f(a) = b, f(b) = c, f(c) = c and f(c1) = b.
The range of f is { b , c ) , tha t is, f [ A ]= { b , c } .
Example 1.4:
Let R be t h e set of real numbers, and let f : R + R assign to each rational number
the number 1, and to each irrational number the number -1. Thus
f(x)
=
1 if 3: is rational
-1 if x is irrational
The range of f consists of 1 and -1: f [ R ]= {I,-1).
Example 1.5:
Let A = {a, b , c , d ) and B = {x,y,z}.
f : A+B.
The following diagram defines a function
Here f ( a ) = y, f(b) = N , f(c) = x and f ( d ) = y.
and t h e co-domain a r e identical.
66
Also f [ A ]= B , t h a t is, the range
CHAP. S]
F CN CTIO N S
67
GRAPH OF A FUNCTION
To each function J : A B there corresponds the relation in A X B given by
((a,/'(a)) : a E A )
We call this set the g r a p h of f . Two functions f : A B and g : A B are defined to be
equal, written f = y, if ?(a)= g(a) for every a E A , that is, if they have the same graph.
Accordingly, we do not distinguish between a function and its graph. The negation of
J = g is written i = y and is the statement:
there exists an ci E A f o r which f ( a ) g ( a )
-+
+
-+
-
A subset f of A x B , i.e. a relation from A to B , is a function if it possesses the folloving
property:
[F] Each a E A appears as the first coordinate in exactly one ordered pair (a, b ) in f .
Accordingly, if ( a , b ) E f, then t ( a ) = b ,
Example 2.1:
Let f : A -+ B be the function defined by the diagram in Example 1.5. Then the
graph of f is the relation
{ ( a ,Y), ( b , x), (c, x ) , ( d . L/)S
Example 2.2:
Consider the following relations in A = {1,2,3]:
f is a function from A into A since each member of A appears as the first coordinate in exactly one ordered pair in f ; here f(1)= 3, f ( 2 ) = 3 and f(3) = 1. g is not
a function from A into A since 2 € A is not the first coordinate of a n y pair in g
and so g does not assign any image to 2. Also h is not a function from A into A
since 1 € A appears as the first coordinate of two distinct ordered pairs in h ,
(1,3) and (1,2). If h is t o be a function i t cannot assign both 3 and 2 to the
element 1 E A .
Example 2.3:
-
A real-valued function ,f : R R of the form
f ( x ) = ax t b
(or: defined by y = a.c 4- b )
is called a linear f u x c t i o x : its graph is a line in the Cartesian plane RZ,
The g r ap h of a linear function can be obtained by plotting ( a t least)
two of its points. F o r example, to obtain the graph of f j x ) = 2%- 1,
set up a table with a t least t w o values of x and the corresponding
values of f ( x ) as in the adjoining table. The line through these
points, (-2, -5), (0, -1) and (2,3), is the graph of f as shown in the
diagram.
Graph of f(x) = 2 x
-1
68
[CHAP. 8
FUNCTIONS
Esaiiiple 2.4:
A real-valued function f’ : R + R of the form
. . . a,,-,:c
f ( l ) = q)T’* alX,--l
+
+
+
:c
+ a,
is called a p o l y z o m i u l f m c t i o n . The g r a p h of such a function is
sketched by plotting various points and drawing a smooth continuous
curve through them.
Consider, for example, the function f i x ) = .c’ - 2.1; - 3. Set up
a table of values f o r s and then find the corresponding values f o r
f ( x ) as in the adjoining table. The diagram shows these points and
the sketch of the graph.
-2
-1
0
1
2
3
4
I f(x)
5
0
-3
--1
-3
0
5
t
Graph of f(x) =
COMPOSITION FUNCTION
Consider now functions f : A
+
E and g : B
+
- 2.1: - 3
C illustrated below:
Let a E A ; then its image f ( a ) is in E , the domain of g . Hence we can find the image of
f ( a ) under the function g, i.e. g ( f ( a ) ) . The function from A into C which assigns t o each
a E A the element g(f(a)) E C is called the composition o r product of f and g and is
denoted by g 0 f . Hence, by definition,
( g o f ) ( a ) = !ma))
Example 3.1:
Let f : A
.+ E
and g : B
C be defined by the following diagrams:
f
A
W e compute ( g 0 t’) : A
3
+
B
C
g
C by its definition.
(gof)(4 =
s(f(a))= d u )
= t
( 9 0 f N b ) = g ( f ( b ) ) = g(z) =
?“
iu O f ) ( C ) = g ( f ( c ) ) = u ( u ) = t
Notice t h a t the composition function g 0 t‘ is equivalent to “following the arrows”
from A to C in the diagrams of the functions j and g.
CHAP. 81
Example 3.2:
69
FUNCTIONS
Let R be the set of real numbers, and let f : R
follows:
f(2)
=
R and
g(x) = z
and
22
+
g :R
-+
R be defined a s
+3
( f o g ) ( 2 ) = f ( d 2 ) ) = f(5) = 25
Then
= d f ( 2 ) ) = g(4) = 7
( g Of)@)
Observe t h a t the product functions f o g and g o f a r e not the same function.
We compute a general formula f o r these functions:
( f o g ) ( x ) = f(g(z)) = f ( x t 3 ) = ( x
( g o f ) ( x ) = g ( f ( x ) ) = g(22) =
+ 3)2 =
+3
x2
+ 62 + 9
ONE-ONE AND ONTO FUNCTIONS
A function f : A + B is said to be one-to-one (or: one-one or 1-1)if different elements
in the domain have distinct images. Equivalently, f :A + B is one-one if
f ( a ) = f(a') implies
Example 4.1:
Consider the functions f : A
ing diagram:
A
f
-
a = a'
B , g :B
B
+
C and h : C
C
B
+
D defined by the followh
D
a
W
Now f is not one-one since the two elements a and c in its domain have the same
image 1. Also, g is not one-one since 1 and 3 have the same image y. On the other
hand, h is one-one since the elements in the domain, x, y and x, have distinct images.
A function f : A + B is said t o be onto (or: f is a function from A onto B or f maps
A o n t o B ) if every Z, E B is the image of some a E A . Hence f : A B is onto iff the range
of f is the entire co-domain, i.e. f [ A ]= B.
+
Example 4.2:
Consider the functions f, g and h in the preceding example. Then f is not onto
since 2 € B is not the image of a n y element in the domain A . On the other hand,
both g and h a r e onto functions.
Example 4.3:
Let R be the set of real numbers and let f : R R, g : R + R and h : R
defined as follows:
f ( x ) = 22, g(z) = z3 - x and h ( z ) = x2
+
The graphs of these functions follow:
f ( x ) = 2"
g(2) = 23 - x
h(x) = 22
+
R be
[CHAP. 8
FUNCTIONS
70
The function f is one-one; geometrically, this means t h a t each horizontal line does
not contain more than one point of f. The function g is onto; geometrically, this
means t h a t each horizontal line contains a t least one point of g. The function h
is neither one-one or onto; f o r h(2) = h ( - Z ) = 4, i.e. the two elements 2 and -2
have t h e same image 4, and h[R]is a proper subset of R ; f o r example, -16 4 h [ R ] .
INVERSE AND IDENTITY FUNCTIONS
In general, the inverse relation f - I of a function f C A X B need not be a function.
However, if f is both one-one and onto, then f - ' is a function from B onto A and is called
t,he inverse function.
Example 5.1:
Let A = { a , b , c ) and B = { r ,s, t}.
Then
f = { ( a ,s), ( b , t ) , ( c , 41
is a function from A into B which is both one-one and onto.
seen by the following diagram of f :
This can easily be
Hence t h e inverse relation
f-1
=
((5,
a ) , ( 4 JJ), ( r ,4)
is a function from B into A . The diagram of
Observe t h a t the diagram of
ing the arrows.
f-'follows:
f-'can be obtained
from the diagram of f by revers-
For any set A, the function f :A + A defined by f ( x ) = x, i.e. which assigns to each
element in A itself, is called the i d e n t i t y function on A and is usually denoted by 1, or
simply 1. Note that the identity function 1, is the same as the diagonal relation: 1, = A,.
The identity function satisfies the following properties:
Theorem 8.1:
For any function f : A
+
B,
lBOf = f =
Theorem 8.2:
f o l ,
If f : A + B is both one-one and onto, and so has a n inverse function
f-' : B + A , then
f-lof
= 1, and f o f - ' = 1,
The converse of the previous theorem is also true:
Theorem 8.3:
Let f : A-+ B and y : B
+
A satisfy
g o f = 1,
and
f o g = 1,
Then f is both one-one and onto, and y = f - ' .
FU I\:C TIONS
CHAP. 81
71
Solved Problems
FUNCTIONS
8.1. State whether or not each diagram defines a function from A = ( a , b , c } into
B = {r,y,z],
m@gJm
z
C
z
C
(i)
C
(ii)
2
(iii)
(i) No. There is nothing assigned t o the element b € A .
(ii) No. Two elements, x and x , a r e assigned t o c € A .
(iii) Yes.
8.2.
Rewrite each of the following functions using a formula:
(i) To each number let f assign its cube.
(ii) To each number let g assign the number 5.
(iii) To each positive number let li assign its square, and to each non-positive number
let 11 assign the number 4.
(i) Since f assigns l o a n y number ?. its cube .v3, f can be defined by f(.c) = x3,
(ii) Since g assigns 5 to a n y number s, we can define g by g(..) = 5.
(iii) Two different rules a r e used to define k as follows:
li (?.)
8.3.
=
$2
4
if x > 0
ifx"0
Let f , g and 11 be the functions of the preceding problem. Find:
(i) f(4),f'(-2), f(@); (ii) ~ ~ ( 4 g1(,- 2 ) , ~(0);(iii) h(4), h(-2), 4 0 ) .
(i)
Now ,t'(:cs)
=
x3
(ii) Since y(x) = 5
f o r every number
2;:
hence f(4) = 43 = 64. J ( - 2 ) = (-2)s = -8,
5, g(-2) = 5 and g(0) = 5.
h(4) = 42 = 16. On the other hand, if
(iii) If 3: > 0, then h(x) = x2; hence
h(a) = 4: thus h(-2) = 4 and h i O ) = 4.
8.4.
Let A = ( 1 , 2 , 3 , 4 , 5 } and let J:,4
f(0)
O3 = 0.
y(4) =
f o r every number x ,
+
x
f
0,
then
A be the function defined in the diagram:
(i) Find the range of f.
(ii) Find the graph of f , i.e. write f as a set of ordered pairs.
The range f[AI of the function f consists of all t h e image points. Kow only 2, 3 and 5 appear
as the image of a n y elements of A ; hence f[A] = { 2 , 3 , 5 ) .
(ii) The ordered pairs ( a , f ( a ) ) ,where a € A , form the graph of f. Now f(1) = 3, f ( 2 ) = 5,
f(3) = 5, f(4) = 2 and f(5) = 3; hence f = {(1,3), ( 2 , 5 ) , (3,5), (4,2), (5,3)}.
(i)
72
8.5.
FUNCTIONS
[CHAP. 8
Determine whether o r not each relation is a function from
Let X = (1,2,3,4>.
X into X.
ti) f = ii2,3), (1,4), (2, I), ( 3 , 2 ) ,14,4)}
iii) g = “3, l),(4,2), (1,1))
iiii) h = j(2, l ) , (3,4), (1,4),
j2,11, (4,4)}
Recall t h a t a subset f of X X X is a function f : X
X if and only if each a E X appears
as the first coordinate in exactly one ordered pair in f .
iir
No. Two different ordered pairs ( 2 , 3 ) and ( 2 , l ) in f’ have the same number 2 as their first
coordinate.
i i i l No. The element 2 E X does not nppear as the first coordinate in a n y ordered pair in 8 .
iiiii Yes. Although 2 E X appears a s the first coordinate in two ordered pairs in h, these two
ordered pairs a r e equal.
--f
8.6.
Find the geometric conditions under which a set f of points on the coordinate diagram
of A x B defines a function f : A B.
+
The requirement t h a t each u E A appear a s the first coordinate in exactly one ordered pair
in f is equivalent to the geometric condition t h a t each vertical line contains exactly one point in f .
8.7.
:m-
Let TV = { a , b, c, d } . Determine whether the set of points in each coordinate diagram
of T I ’ x IT’ is a function froin IT’ into W .
tM
b
a
a
b
c
d
a
a
(i)
b
c
f f!Ff
a
d
a
(ii)
b
c
d
(iii)
No. The vertical line through b contains two points of the set, i.e. two different ordered pairs
( b , b ) and ( b , d ) contain the same first element b.
( i i ) KO. The vertical line through c contains no point of the set, i.e. c E W does not appear a s
the first element in any ordered pair.
(iiir Yes. Each vertical line contains exactly one point of the set.
(i)
GRAPHS OF REAL-VALUED FUNCTIONS
8.8. Sketch the graph of f ( x ) = 3 x - 2 .
This is a linear function: only two points (three a s a check) a r e needed t o sketch its graph.
Set up a table with three values of s, say, x = -2, 0, 2 and find the corresponding values of f(r):
t’(-2) = 3(-2)
-
2 = -8,
((0) = 3(0) - 2 = -2,
f(2) = 3(2) - 2 = 4
D r a u the line through these points as in the diagram.
X
10, - 2 )
CHAP. 81
8.9.
FUNCTIONS
Sketch the graph of (i) f ( x ) = 2 t .L‘
-
73
6 , (ii) g(x) = x? - 3x2 - x
+ 3.
In each case, set up a table of values f o r x and then find the corresponding values of t ’ ( ~ ) .
Plot the points In a coordinate diagram, and then d r a w a smooth continuous curve through the
points
(ii)
x
I gis)
-2
-15
0
3
0
-3
0
15
-1
0
1
2
3
4
Graph of f .
Graph of g.
COMPOSITION OF FUNCTIONS
8.10. Let the functions f : A
-+
B and y : B
-+
C be defined by the diagram
(i) Find the composition function y o I: : A
(i)
.+
C. (ii) Find the ranges of f , g and y o f .
We use the definition of the composition function to compute.
(Y {)(a) = M a ) ) = g(u) = t
(gof)(b) = g(f(b)) = g(r) = s
(Y
t’)(C)
= g ( f ( c ) ) = s(l/)= t
Note t h a t we arrive at the same answer if we “follow the arrows” in the diagram:
c+y +t
a --t u + t , b + x + s ,
(ii) By the diagram, the images under the function f a r e x and y, and the images under g a r e
T , s and t : hence
and
range of g = { Y , s, t }
range of f = {.c, g }
By
the images under the composition function a r e t and s. hence
range of g o f = {s, t )
i ~ ,
Note t h a t the ranges of g and y
8.11. Consider the functions
f
c
t’ a r e different.
= { ( 1 , 3 ) ,( 2 , 5 ) , (3,317 (4,1), (592))
g = { ( I ,4),(2, I), (3911, (4,
a, 15,311
from X = j l , 2 , 3 , 4 , 5 } into X. (i) Determine the range of f and of g.
the composition functions g o f and f o g .
(ii) Find
74
FUNCTIONS
(ij
[CHAP. 8
The range of a function is the set of image values, i.e. the set of second coordinates: hence
r an g e o f f = ( 3 , 5 , 1 , 2 }
and
range of g = { 4 , 1 , 2 , 3 }
8.12. Let the functions f and g be defined by f(x) = 2 x f l and g(x) = x 2 - 2 .
formulas defining the composition functions (i) g 0 t' and (ii) f o g .
(i)
Compute g 0 f as follows:
(g 0 f i ( r j = g ( f ( z ) ) = g(2.2.
+ 1) = ( 2 +~ 1)2 - 2
= 422
Find
+ 42 - 1.
Observe t h a t the same answer can be found by writing
y
= f ( ~=
) 2.t
+1
and
z = g(y) = y2 - 2
and then eliminating y from both equations:
x =
(iil
Compute
fog
as follows:
Note t h a t f o g # g
u2
-2 =
(2%
+ 1)2 - 2
(fog)(x) = f(g(x))=
f(x'-
= 4x2
+ 4.1 - 1
2 ) = 2 ( n 2 - 2 ) + 1 = 2%'- 3.
of.
8.13. Prove the associative law for composition of functions: if
A A B % C % D
then ( J z o g ) o f = h o ( g o f ) .
( ( h 0s) 0
For every a E A ,
({l
= (hog)(f(a)) = h(u(f(aj))
(Y f ) j ( a ) = h((sO f)(a)) = h ( g ( f ( a ) ) )
Hence ( h 0 9 )0 f = h o ( g o f ) , since they each assign the same image to every a E A . Accordingly,
we may simply write the composition function without parentheses: h 0 g 0 f .
ONE-ONE AND ONTO FUNCTIONS
8.14. Let A = ( a , b , c , d , e J , and let B be the set of letters in the alphabet.
functions f , g and h from A into B be defined as follows:
f
9
a+ r
b+a
a+ x
b+Y
C+S
C + X
b+c
c+e
d-+r
d+Y
d-+r
e-+e
e+x
e+s
(iii)
(9
Are any of these functions one-one?
(ii)
h
a+
a
Let the
CHAP. 81
75
FUNCTIONS
Recall t h a t a function is one-one if it assigns distinct image values t o distinct elements in
the domain.
(i) No. For f assigns T to both a and cl.
(ii) No. F o r g assigns x to both a and e .
(iii) Yes. F o r h assigns distinct images to different elements in the domain.
8.15. Determine if each function is one-one.
(i) To each person on the earth assign the number which corresponds to his age.
(ii) To each country in the world assign the latitude and longitude of its capital.
(iii) To each book written by only one author assign the author.
(iv) To each country in the world which has a prime minister assign its prime
minister.
(i)
(ii)
(iii)
(iv)
No. Many people in the world have the same age.
Yes.
No. There a r e different books with the same author.
Yes. Different countries in the world have different prime ministers.
8.16. Prove: If f : A+ B and g : B + C are one-one functions, then the composition
function g 0 f : A C is also one-one.
+
Let (g o f ) ( a ) = ( g o f ) ( a ’ ) ; i.e. &“(a)) = g(j(a’)). Then f(a)= f(a’) since g is one-one.
Furthermore, a = a‘ since f is one-one. Accordingly, g o f is also one-one.
8.17. Let the functions f : A
i
A
+
B, g : B + C and h : C + D be defined by the diagram.
C
B
(i) Determine if each function is onto.
(i)
The function f : A
The function g : B
+
--f
h
D
(ii) Find the composition function h o g o f .
B is not onto since 3 E B is not the image of any element in A.
C is not onto since z E C is not the image of any element in B.
The function h : C + D is onto since each element in D is the image of some element of C.
(ii) Now
a+Z+x+4,
b+l+y+6,
c+Z+x+4.
Hence
h o g o f = {(a,4),(b,6),(~,4)}.
8.18. Prove: If f : A + B and g : B -+ C are onto functions, then the composition function
g o f : A + C is also onto.
Let e be any arbitrary element of C. Since g is onto, there exists a b E B such t h a t g ( b ) = c.
Since f is onto, there exists a n a E A such t h a t f ( a ) = b . But then
( g o f ) ( 4 = g(f(a\) = g ( b ) = c
Hence each c E C is the image of some element a € A . Accordingly, g 0 f is a n onto function.
;CHAP. 8
FUNCTIONS
76
INVERSE FUNCTIONS
8.19. Let W = { 1 , 2 , 3 , 4 , 5 } a n d l e t f : W - + W ,g : W + W and h : W - + W b ed efi n ed b y
the following diagrams:
Determine whether each function has a n inverse function.
In order for a function to have a n inverse, the function must be both one-one and onto.
Only h is one-one and onto; hence h , and only h, h a s a n inverse function.
8.20. Let f : R -+ R be defined by f ( x ) = 2x - 3. Now f is one-one and onto; hence f has
an inverse function f - l . Find a formula f o r
f F 1 .
Let y be the image of x under the function f :
y = f ( x ) = 2x - 3
Consequently, 2; will be the image of y under the inverse function
in the above equation:
x = (y 3)/2
f-1.
Solve for x in terms of y
+
J-W
Then
= (~+3)/2
is a formula defining t h e inverse function.
8.21. Let A = (a,b, c, d } . Then f = ((a,
b), ( b , d ) , (c,a), ( d , c)} is a one-one, onto function
from A into A . Find the inverse function f - ' .
To find the inverse function
in reverse order:
which is the inverse relation, simply write each ordered pair
f-1,
= { ( b , 4, (4b), ( a ,c ) , ( c , 4)
f-'
The function assigns to any element the square of the element minus 3 times the element plus 2.
(a)
=
t(-3)
(-3)2
-
3(-3)
( b ) f(2) = (2)2 - 3(2)
+2
+ +
+ 2 = 9 9 2 = 20
= 0, f(-4) = (-4)2 - 3(-4)
f(2) - f(-4)
1(y)
=
(L/)2 -
3(y)
(a2)2 -
(t)
(X2)Z -
-
3(a*)
-
-
-
/I)
= 30.
= 0 - 30 = -30
+ 2 = y2 3y + 2
+ 2 = a1 3a* + 2
t(.')
=
3(x2) + 2 = a4 32' + 2
+ 2 = y2 - 2yz +
(t ) f ( y - x ) = (y z)2 - 3(y
= (x + h)2 3(a + h ) + 2 = x2 + 2xh + h'
( 9 ) f(.c +
(c)
( d ) f(a2) =
+2
- 2)
-
22 -
3y
+ 32 + 2
+2
- 3%- 3h
Then
FUNCTIONS
CHAP. 81
(h) f ( x
+ 3)
=
(X
+ 3)'
- 3(x
+ 3) + 2 =
+2
(N?
(i) f ( 2 x - 3 ) = ( 2 2 - 3)2 - 3(2x - 3)
77
+ 62 + 9) - 3% - 9 + 2 = x2 + 3%+ 2
+ 9 - 6 x + 9 + 2 = 4x2 - 18% + 20
= 4r2 - 122
( j ) Using ( h ) and (i), we have
+ f ( +~ 3) = (429 - 18s + 20) + (6+ 3%+ 2 ) = 5 ~ ' - 15%+ 22
- 6x3 + 1 0 ~ ' 3%
(k) f ( 6 -3~ + 2 ) = ( 9 -3%+ 2)2 - 3 ( z * - 3%+ 2) + 2 =
(1) f(f(x)) = f ( d - 3~ + 2) = x4 + 10x2 - 3%
( m ) f(f(x + 1)) = f([(x + 1)2 - 3 ( x + 1) + 21) = f([x2+ 22 + 1 - 3 x - 3 + 21)
f ( 2 -~ 3)
~4
-
623
=
( n ) By (g), f ( z
f(z' - x)
+h)
= x2
=
Using ( n ) ,we have
- x)'
-
3(s'-
x)
+2
=
x4
- 2x3 - 2x2
+ 2xh + h2 3%- 3h + 2. Hence
+ 2 ~ +h h' 3.2: - 3 h + 2) (
~
2
3~ i- 2 ) = 2xh
($2-
-
+
+
[ f ( x h) - f ( z ) ] / h = (2rh h2 - 3 h ) h = 2r
8.23. Prove Theorem 8.1:
+ 32 + 2
-
f ( 2 Ih ) - t ( x ) =
(0)
(22
For any function f :A
+
B,
+ h2
-
3h
+h-3
= f = f~ 1,.
lnof
Let a be an y ar b i t r ar y element in A ; then
( 1 0!)(4
~
= 1~(f(a)=
) f(a)
Hence 1,o f = f = f
0
1,
and
(folA)(a)
= f(la(a)) = f ( a )
since they each assign f ( a ) to every element a E A .
Supplementary Problems
FUNCTIONS
8.24.
8.25.
S t a t e whether each diagram defines a function from { l , 2 , 3 } into { 4 , 5 , 6 ) .
Define each function by a formula:
(i)
To each number let f assign its square plus 3.
(ii) To each number let g assign its cube plus twice the number.
(iii) To each number greater t h an o r equal to 3 let h assign the number squared, and to each
number less than 3 let h assign the number -2.
8.26.
Determine the number of different functions from { a , b } into {l, 2 , 3 ) .
8.27.
Let f : R + R
(iv) f ( x - 2 ) .
8.28.
Let g : R
+
be defined by
R be defined by
f(x) =
g(x) =
z2 - 4 2
+ 3.
Find
(i) f ( 4 ) ,
(ii) f(-3),
(iii) f ( y - 2 x ) ,
78
FUNCTIONS
[CHAP. 8
8.30.
Let the function g assign t o each name in the set {Betty, Martin, David, Alan, Rebecca} the number
of different letters needed t o spell the name. Find the g r a p h of g , i.e. write g a s a set of ordered
pairs.
8.31.
Let it’ = [ 1 , 2 , 3 , 4 } and let g : W
+
be defined by the diagram
(iii) Find
iii) Plot g on the coordinate diagram of W X TT’.
( i i Write g a s a set of ordered pairs.
the range of g .
8.32.
TIT
:mfffF ##
Let V = { 1 . 2 , 3 , 4 } . Determine whether the set of points in each coordinate diagram of V X V
is a function from V into 1’.
4
3
2
2
1
1
1
2
3
4
1
2
(i)
3
4
1
2
(ii)
3
4
1
(iii)
2
3
4
(iv)
GRAPHS OF REhL-VALUED FUNCTIONS
8.33.
Sketch the graph of each function:
(i) f(z)= 2, (ii) g(x) = 4%- 1, (iii) k ( z ) 1 2%’- 4%- 3.
8.34.
Sketch the g r a p h of each function:
(i) f ( x ) =
:c3
- 3%-I- 2,
(ii) g(z) = :ci - 10x2
+ 9,
0 if x = O
(iii) h ( x ) =
-if
x#O
COMPOSITION OF FUNCTIONS
8.35,
The functions f :A + B, g : B + A , h : C + B, F : B + C, G : A + C a r e pictured in the diagram
below.
Determine whether each of the folloLving defines a product function and if i t does, find its domain
and co-domain: (i) g o f , iii) h o f , (iii) F o f , (iv) G o f , (v) g c h , (vi) F o h , (vii) h o G o g ,
rviiil li 0 G.
8.36.
@Q@-
The following diagrams define functions .f, g and h which map the set {1,2,3,4} into itself.
4
[it
(ii)
4
Find the ranges of f, y and h .
Find the composition functions
4
4
4
(1) f o g , (2) h o j , (3)
g2,
i.e. gog.
4
8.37.
79
FUKCTIONS
CHAP. 8 )
+ +
Let f : R 3 R and g R + R be defined by Ji.r) = z2 3% 1 and g ( r ) = 2s - 3. Find fomiulas
defining the product functions (i) f c g , iii) g 3 f , (iii) g o g , (iv) f o f .
ONE-ONE, ONTO .iKD INVERSE FUNCTIONS
8.38
Let f : X
- 17.
Which conditions define a one-one function:
J ( n l = f ( b ) implies a = b
(iii) f ( a ) # J ( b ) iniplies a # b
iiir n = b implies f(n) = f(b)
(iv) a f b implies J ( a ) # f ( b )
li)
8.39.
(i) State whether o r not each function i n Problem 8.36 is one-one
(ii) State whether or not each function in Problem 8.36 is onto.
8.10.
Prove Theorem 8.2: If f : A B is one-one and onto, and so has an inverse function
(i) f - 1 o . f = 1, and (ii) f o f - 1 = 1,.
8.41.
Prove: If t : A
8.42.
Let f : R
8.43.
Let
8.14.
Let R be an equivalence relation in a non-empty set A . The function from A into the quotient
set AIR is defined by ~ ( a=) [ a ] , the equivalence class of a. Show t h a t 17 is a n onto function.
8.45.
Prove Theorem 8.3: Let f : A + B
f is one-one and onto, and g = f - 1 .
-f
+
g :R
+
4
E and g : B
+
A satisfy y
of
Find a formula f o r the inverse function
and y : E
.+
A
(i) No,
iii) Yes,
iiii) No
8.25.
8.26.
Nine.
8.27.
(i) 3, (iil 24, ( i i i ) y2 - 4xy
8.28.
(i) g ( 5 ) = 10. (ii) y(0) = 2, (iii) g(-2) = 0
8.29.
(i) Yes, iii) No, iiii) Yes, (iv) No
8.30.
g = {(Betty, 4 , (RIartin, 6), (David, 4 ) , (Alan, 3 ) , (Rebecca, 5))
8.31.
(i) g = C(1,2), (2,3), (3, I), (4,3)),
8.32.
(i) No, (ii\ No, (iii) Yes, (iv) No
8.33.
+ 4x2 - 4y + 8x + 3,
+
-2
-3
Graph of f
:R
.+
R.
g-1 : R
3
R.
f-1
satisfy g 0 t' = 1,4 and f o g = lg. Then
Answers to Supplementary Problems
8.24.
then
= l A ,then f is one-one and g is onto.
R be defined by f ( x ) = 3 x - 7 . Find a formula f o r the inverse function
R be defined by g(z) = x? -t 2.
f-1,
(iv) x2 - 8% 4- 15
(iii) : 2 , 3 , 1 )
-- 4
-- 2
# C H A P .8
FC N C TI 0N S
80
(iii)
-3
-5
-3
#Graph of h
8.34.
- 46
Graph o f f
-15
Graph of
g
'---4
Graph of h
8.35.
(i) g
of
:A
+
A , (ii) No, (iii) F
;1 to;^ ) I
,I
of
:A
+
C,
8.37.
(i)
iii)
4
(I/(Z)
1
{1,3}
+ 6r - 1
(iii) (y 0 g ) ( x ) = 4x - 9
(ivl (f o f ) ( x ) = x4 6x3
+
range of f = {1,2,4},
range of g = { 1 , 2 , 3 , 4 } , range of h
( , f o g ) ( x )= 4 2 2 - 6 ~ + 1
(ii) (g 0 f)(x) = 2x2
(iv) No, (v) g 0 h : C + A , (vi) F 0 h : C -t C,
(vii) h 0 G 0 g : B + B , (viii) h 0 G : A B
8.36.
lii
+
+ 14x2 + 152 + 5
8.38.
( i ) Yes, Iii) No, (iii) No, (iv) Yes
8.39.
ti) Only g is one-one.
8.42.
I-1 ( x ) = (x+ 7)/3
8.43.
g-l
9;)
(3:)
= 2-
(ii, Only g is onto.
Chapter 9
Vectors
COLUMN VECTOKS
A column rector ~r is a set of numbers u l , u,,. . . ,zcn written in a column:
ZG
The numbers
ti,
=
are called the components of the vector u.
Example 1.1:
The following a r e column vectors:
The first two vectors have two components; whereas the last two vectors have three
components.
Two column vectors 2 i and v a r e equal, written u = v, if they have the same number of
components and if corresponding components are equal. The vectors
\3i'
are not equal since corresponding elements are not equal.
Example 1.2:
Let
Then, by definition of equality of column vectors,
= 4
= 2
x - l = 3
x-y
z+y
Solving the above system of equations gives x = 3, y = -1, and x = 4.
Remark:
In this chapter we shall frequently refer t o numbers as scalars.
VECTOR ADDITION
Let u and 2' be column vectors with the same number of components. The sum of ZL
and v, denoted by ti -t v , is the column vector obtained by adding corresponding components:
81
[CHAP. 9
VECTORS
82
Note that ii + 1: has the same number of components as
with different numbers of components is not defined.
11
and
2'.
The sum of two vectors
Example 2.1:
Example 2.2:
(-a;
The sum
+ f,-54)l//
is not defined since the vectors have different numbers of components.
A column vector whose components are all zero is called a zero vector and is also denoted
by 0. The next Example shows that the zero vector is similar to the number zero in that,
f o r any vector 11, u f O = u.
r) p
SCALAR RIULTIPLICATION
The product of a scalar lc and a column vector u,denoted by k - u or simply kx, is the
column vector obtained by multiplying each component of 21 by It:
k
* 11
=
k
=
4/
un
Observe that
ZJ
and 1c.u have the same number of components. We also define:
-u
= -1.21
and
u-
2j
= z!
t
(-v)
Example 3.1:
The main properties of the column vectors under the operations of vector addition and
scalar multiplication are contained in the following theorem:
CHAP. 9
83
VECTORS
Theorem 9.1:
Let V T be
L the set of all 1%-componentcolumn vectors, and let
and let k , k r be scalars. Then:
(i)
(ii)
(iii)
(ivj
(v)
(vi)
(vii)
(viii)
(21 t
-L
19)
I' t 2%
ZI,I - , 20
E V,
w = u + (c t 2c), i.e. addition is associative;
v + u,i.e. addition is commutative:
+
0 = 0 u = u;
LI T ( - 1 1 )
= (-u) z/ = 0:
l;(u t 2') = ku kv;
ZI
-t
(k+~
+
' ) I I=
+
ku + lc'u;
(kk'jrc = k(lc'u);
l u = 21.
The properties listed in the above theorem are those which are used to define an
abstract niathematical system called a linear spuce 3r rector space. Accordingly, the
theorem can be restated as follon-s:
Theorem 9.1:
The set of all ,i-component column vectors under the operations of vector
addition and scalar multiplication is a vector space.
ROM' VECTORS
Analogously, a row vector
2)
is a set of numbers
=
ZL
(ul, u,,. . . ,
u,,, . .,u,
written in a row:
ifI,
21,)
The numbers ut are called the coiiiyoilents of the vector 21. Two row vectors are equal if
they have the same number of components and if corresponding components are equal.
Observe that a row vector is simply an ordered ri-tuple of numbers.
The sun1 of two row vectors ?( a n d 21 with the same number of components, denoted by
u + r , is the row vector obtained by adding corresponding components from 11 and u:
= (ul, u,,. . . , u,,) (vl, v2, . . ., w,,) = (211 + w l , u2+v2,. . . , 20, + 2yJ
1'
2'
T
+
The product of a scalar lc and a row vector u,denoted by l;*u or simply kri, is the row
vector obtained by multiplying each component of 21 by k :
k-u =
k(/,ll,u , ,
We also define
- R = -1.u
as v e did f o r column vectors.
Example 4.1:
. . . ) un) = (lW1, kLI,, . . rCuJ
.)
and
u-zt = u + ( - w )
+
( 0 , 5, -7, 11) = (I, 3, -4, 7 )
= (5, -10, 15, -20)
5 * (1, -2, 3, -4)
(1,-2, 3, -4)
We also have a theorem f o r row vectors which corresponds to Theorem 9.1.
Theorem 9.2:
The set of all 11-component row vectors under the operations of vector
addition and scalar multipIication satisfies the properties listed in Theorem
9.1, that is, is a vector space.
MULTIPLICATION O F A ROW VECTOR AND A COLUMN VECTOR
If a row vector u and a column vector v have the same number of components, then
their product, denoted by 21 * 1' or simply uv, is the scalar obtained by multiplying corresponding elements and adding the resulting products:
86
VECTORS
ROW VECTORS
9.6. Let 11. = (2,-7,1),
iiii) -3u, (iv) -w.
1i1
ii)
= (-3,0,4) and
= (0,5,-8).
IL'
(i) u + v , (ii) ZI fw,
Find:
Add corresponding components:
+
u
I
2:
[CHAP. 9
2,
= 12, - 7 , 1)
+ (-3,
0, 4) = ( 2 - 3, - 7 + 0 , 1 + 4 ) = (-1, - 7 , 5 )
Add corresponding components:
2,
+
= ( - 3 , 0 , 4)
20
+ ( 0 , 5, - 8 )
= (-3 f 0, 0 + 5 , 4 - 8 )
iiiii Multiply each component of u by the scalar -3.
-3u
= (-3, 5, -4)
= -3(2,-7,l)
= (-6,21,-3).
iiv) Xultiply each component of w by -1, i.e. change the sign of each component:
-W
9.7.
= -(O, 5, - 8 )
= ( 0 , -5, 8)
Let u,v and w be the row vectors of the preceding problem.
(ii) 22~ 3v - 5w.
+
Find (i) 3u-4v,
F i r s t perform the scalar multiplication and then the vector addition.
9.8.
(i)
3%- 42, = 3(2, - 7 , l ) - 4(-3,0,4)
iii)
2x
+ 3v - 5w
+
+
= (6, -21.3)
+ (12,0, -16)
= (18, -21, -13)
3(-3, 0, 4) - 5(0, 5, - 8 )
= (4, -14, 2 ) (-9, 0, 12) (0,-25, 40)
= (4 - 9 i0,-14
0 - 25, 2 12 440) = (-5, -39, 54)
= 2(2, -7, 1)
+
+
+
(2x,3, v) = (4, x + x , 22).
Find x, y and x if
Set corresponding components equal to each other to obtain the system of equations
2x = 4
3 = x + x
y = 22
Then x = 2, x = 1 and y = 2 .
9.9.
Find x and
?-J
if
s(1,l)-C y(2,-1)
= (1,4).
First multiply by the scalars x and y and then add:
x ( l , 1) f y(2, -1)
= ( z , x)
+ (2y, -y)
= (x
+ 2y, x - y)
= (1, 4)
Now set corresponding components equal to each other t o obtain
x t 2 y = 1
x--l/
= 4
Solve the system of etpations to find x = 3 and y = -1.
RilULTIPLICATION OF A ROW VECTOR AND A COLUMN VECTOR
(i)
Multiply corresponding coinponents and add:
(2, -3,6)(-!)
=
2.8
+ ( - 3 ) * 2 + 6*(-3)
=
16 - 6 - 18
=
-8
CHAP. 9J
\' E CT 0R S
87
(ii) The product is not defined since the vectors have different numbers of components.
(iii) Multiply corresponding components and add:
/ 4
=
(3,-5,2,1)
9.11. Let u = (1, li., -3)
+ (-51.1 + 2 * ( - 2 )
3.4
\-!I
f
and v =
2'
.
=
4- 1 . 5
12 - 5
Determine k so that
ZL*U
-
4
+5
=
8
= 0.
Compute z( * r and set i t equal to 0:
21.2,
= (l,k,-3)
:i
= 1-2
-5
T
(-5l.k
= 2
- (-3).4
-
5 k - 12 = -10 - 5k = 0
= 0 is k = -2.
The solution t o the equation -10-5k
RIISCELLANEOUS PROBLEMS
9.12. Prove Theorem 9,1(i): F o r any vectors
v and w, (uA u ) + zv = u
11,
+ (v +w).
Let i t , , uZ and zc, be the ith components of I ! , 21 and w respectively. Then u,+ v, is the 7th
component of z( - 2: and so (24% 71,) - ti', is the zth component of ( I ( - 2:) - z ~ . On the other hand,
v,+ w, is the zth component of v ZL' and so z(,
( v , f w,)is the ith component of u ( t , + zc).
But u,, T~ and u',a r e numbers for which the associative law holds. t h a t is,
+
+
+ vt)
= u + (v +
(u,
Accordingly, (z(
- 1 % )- w
+
A
w,=
i(,
+
+ (v,+ w J ,
f o r each i
since all their corresponding components a r e equal.
11)
9.13. Let 0 be the number zero and 0 the zero vector.
Show that, for a n y vector u,OIL = 8.
Method I :
Let ? I , be the ith component of u. Then 0 i t i = O is the ith component of Ozi.
component of 0(( is zero, Ozt is the zero vector, i.e. Ozt = e.
Method 2:
We use Theorem 9.1:
Now adding -0"
0 u = (0
+ 0)zc
= ou
Since every
+ ou
to both sides gives
e =
Ozt
+ (-010
9.14. Prove Theorem 9.l(vi): ( k
+ k')u
= 021
+ [Oa+ (-Ou)]
= kzi
= OH
e =
OI(
+ k'u.
Observe t h a t the first plus sign refers to the addition of the two scalars k and k' whereas
the second plus sign refers to t h e vector addition of the two vectors k u and k'u.
Let u, be the ith component of u . Then (k +k ')zi1 is the ith component of ( k + k')u. On the
other hand, kii, and k'u, a r e the ith components of the vectors kzi and k'", respectively, and so
ku, k'u, is the ith component of the vector kit - k'u. B u t k , k' and i!, a r e numbers; hence
+
(k
Thus (k - k')zi =
/;I(
+ k'zi,
+ k')zc, =
+
k z ~ , k'ul,
f o r each z
a s corresponding components a r e equal.
9.15. The norm 1 u 1 of a vector u = ( u l , I [ , , . . . , u,J is defined by
IIull =
du; +u;+ . * .
+v;
VECTORS
88
[CHAP. 9
Supplementary Problems
COLUMN VECTORS
9.16.
Compute:
(i)
(-!)+
-1
9.17.
Compute:
(i) 3
f
(ii)
-7
I-:)
+ 5 (1;)
- 2
(ii) 3
(iii) -
(1;)
-4
9.19.
Determine x and y if:
9.20.
Determine
9.21.
2, y
(i)
and x if
Determine r and y if
(.
z
x
+
(t)
=
+
=
2~
y
("6
(ii) x
(i)
= 2
(-t]i + .(8) (-:) .
=
(22).
ROW VECTORS
9.22.
Let u = (2,-1,0,-3),
(iI 32(, (iij
IL
+ 71,
9.23.
Find
2
and y if
9.24.
Find
2,
u and
x if
w = (1,-1, -1,3)
(iii) 2u - 3c, (ivj
and w = (1,3,- 2 , Z ) .
524 - 3V
- 4W, (v)
--2l
+ 2 V - 22c.
x(1,Z) = - 4 ( y , 3 ) .
x(l,l,O)
+ y(2,0, -1) + z ( O , l , l )
Find:
= (-1,3,3)
MULTIPLIC.-ITION OF A ROW VECTOR AND A COLUMN VECTOR
- 4
VECTORS
CHAP. 91
9.26.
Determine lc so t h a t
9.27.
Prove: If u = ( u I , . . , u J i s a row vector such t h a t
n components, then u is the zero vector.
89
(1, k , -2, -3)
.
u.2)
= 0 f o r every column vector v with
MISCELLANEOUS PROBLEMS
9.28.
Prove Theorem 9 . l ( v ) :
F o r a n y vectors u and v and scalar
9.29.
Prove Theorem 9.1tvii):
For a n y vector
zi
9.30.
Find
I / (2, -3,
1
ii)
11
( 3 , 5 , -4)
11,
(ii)
6, 4)
and scalars k,k',
li,
k(ic T
IS)
= k u + kv.
( k k ' ) ~= k ( k ' u ) .
Answers to Supplementary Problems
9.16.
9.17.
9.18.
9.19.
(i) x = 2, y = 4
9.20.
z = 2 , y=3, x=-2
9.21.
x = O , y = 0 or x = - 2 ,
9.22.
(ii) x = -1, y = - 3 / 2
y=-4
+
3% = (6, -3,0, -9), u ZI = (3, -2, -1,O),
2v - 2w = ( - 2 , - 7 , 2 , 5 )
-u
+
9.23.
9.24.
x = l , y=-l,
9.25.
(i) 7
9.26.
9.30.
2 = 2
(ii) Not defined
(iii) 0
22t - 3~ = (1,1,3,-15),
52t - 3~ - 4~ = (3, -14,11, - 3 2 ) ,
Chapter 10
Matrices
MATRI C E S
A rnatyiy is a rectangular array of numbers; the general form of a matrix with Tn rows
and n columns is
,\ . . . . . . . . . . . . . . . . . .%:::
'\
...
all
a13
* * *
a21
a22
a23
* * *
UnLl
am2
am3
*
.'
ainn
1
We denote such a matrix by
(a,,),,,,
o r simply
(a,])
and call it an n z x "rz matrix (read ('nz by n"). Note that the row and column of the element
atj is indicated by its first and second subscript respectively.
Example 1.1:
Consider the 2
X
3 matrix
-;3
(:
Its rows are (1, -3,4) and (0,5, -2) and its columns a r e
In this chapter, capital letters A , B , . . . denote matrices whereas lower case letters
a, b, . . . denote numbers, which we call scalars. Two matrices A and E are equal, written
A = E , if they have the same shape, i.e. the same number of rows and the same number of
columns, and if corresponding elements are equal. Hence the equality of two m
matrices is equivalent to a system of mn equalities, one for each pair of elements.
Example 1.2:
The statement
22
( x+ yy
+w
x-l.0
'I
=
X
n
(; :)
is equivalent to the system of equations
-I
x+u
=
=
2i+w =
2-20
=
x-y
3
1
5
4
The solution of the system of equations is x = 2,
Remark:
u = 1,
2
= 3 , w = -1.
A matrix with one row is simply a row vector, and a matrix with one column
is simply a column vector. Hence vectors are a special case of matrices.
MATRIX ADDITION
Let A and B be two matrices with the same shape, i.e. the same number of rows and
of columns. The sum of A and B , written A E , is the matrix obtained by adding corresponding elements from A and B:
+
90
all
a12
91
i\IA T R I CE S
CHAP. 101
.
+
bll
be1
b12
bml
bmz
"
+
+
+
\
. . . aan +
azl + bzl a22 + b.2
................................. I
+ b,i,l an,?+
amn+ b,,,,/
all
'
bz2
...
.................
amn
a12
bll
2112
* *
aln
blll
212,
bi7,2
anti
*
Note that A + B has the same shape as A and B. The sum of two matrices with different
shapes is not defined.
3 0 -6)
2 -3
11
Example 2.1:
Example 2.2:
=
1+3 -2+0
3+f-6)
0 + 2 4 + ( - 3 ) 1'5
The sum
is not defined since the matrices have different shapes.
A matrix whose elements are all zero is called a zero matrix and is also denoted by 0.
Part (iii) of the following theorem shows the similarity between the zero matrix and the
scalar zero.
Theorem 10.1:
For matrices A , B and C (with the same shape),
+ + C = A + ( B+ C), i.e. addition is associative;
(ii) A + B = B + A , i.e. addition is commutative;
(iii) A + 0 = 0 + A = A.
(i)
( A B)
SCALAR MULTIPLICATION
The product of a scalar lc and a matrix A , written k A or Ak, is the matrix obtained
by multiplying each element of A by lz:
all
kall
kazl
a12
.................
Gn1
am2
-
*
kaln
ka22
1
.
9
+
. kaI n
.
kU2,
....................
ka,l
a
kam2
* '
ka,,,,
Note that A and /<Ahave the same shape.
Example 3.1:
/1
3(4
-2
3
0)
-5
=
(:::
3.(--2)
3.3
We also introduce the following notation:
A
- A = (-l)A
and
-
3.(-5)
B = A
+ (-B)
The next theorem follows directly from the above definition of scalar multiplication.
Theorem 10.2:
For any scalars k , and lz., and any matrices A and B (with the same shape),
(i) (klk2)A = kl(kpA)
(iv) 1 . A = A , and OA = 0
(v) A + (-A) = ( - A ) + A = 0.
(ii) k l( A + B ) = k1A k1B
(iii) (kl k2)A = klA kzA
+
+
+
92
[CHAP. 10
RI AT RICE S
Using (iii) and (iv) above, we also have t h at
A + A = 2A, A + A + A = 3A, , . .
MATRIX PIULTIPLICATION
Let A and B be matrices such that the number of columns of A is equal to the number
of rows of B . Then the product of A and B , written A B , is the matrix with the same
number of rows as A and of columns as B and whose element in the ith row and jth column
is obtained by multiplying the ith row of A by the j t h column of B:
+
+
+
where ciJ = aiibij a22b23 - * * atpbpJ.
I n other words, if A = (atj)is an m x p matrix and B = ( b t l ) is a p
A B = ( c i J ) is the nz X n matrix f o r which
czj = atlblj
&2b& f * ' .
azpbpj
=
azkbkj
+
+
X
n matrix, then
$
k=l
If the number of columns of A is not equal to the number of rows of B , say A is m x p
and B is q X 71 where p # q, then the product A 6 is not defined.
val
tal
Example 4.1:
+ sb,
+ xbl
m2
ta2
+ sb,
+ ub,
ra3
ta,
+ sb,
+
ttb,
Example 4.2:
(t :)(: :)
=
1.1+1.3 1*2+1°4)
(0*1+2*3 0.2+2*4
=
(' ;)
We see by the preceding example that matrices under the operation of matrix multiplication do not satisfy the commutative law, i.e. the products A B and B A of matrices need
not be equal.
Matrix multiplication does, however, satisfy the following properties:
Theorem 10.3:
(i)
(AB)C = A ( B C )
(ii) A ( B + C ) = A B
(iii) ( B + C ) A = B A
+ AC
+ CA
(iv) k ( A B ) = ( k A ) B = A ( k B ) , where k is a scalar.
We assume that the sums and products in the above theorem are defined.
Remark:
In the special case where one of the factors of A B is a vector, then the product
is also a vector:
CHAP. 101
h I 4 TRICE S
Example 4.3:
(2, - 3 , 4 )
-;I
=
(2 * 1 T (-3)
*
5
93
+ 4 . (-2),
2 . (-3) t ( - 3 )
+
1* (-1)
(-3) * 2
5*(-1)+0*2
1-2) (-1)
4*2
1 -3
+
)
=
*
0
+ 4.4)
=
(-21,lO)
“I)
Example 4.4:
A system of linear equations, such a s
is equivalent t o the matrix equation
T h a t is, any solution to the system of equations is also a solution to the matrix equation, and
vice versa.
SQUARE MATRICES
A matrix with the same number of rows as columns is called a square matrix. A
square matrix with 71 rows and n columns is said to be of order 11, and is called a n n-square
matrix. The main diagonal, or simply diagonal, of a square matrix A = (at,) is the
numbers a,,, aZ2, . . ., ~ l , ~ ~ .
Example 5.1:
The matrix
is a square matrix of order 3.
The numbers along the main diagonal a r e 1, -4 and 2.
The n-square matrix with 1’s along the main diagonal and 0’s elsewhere, e.g.,
I
10
\,o
0
0
1
0
0
1
is called the unit matrix and will be denoted by I . The unit matrix I plays the same role
in matrix multiplication as the number 1 does in the usual multiplication of numbers.
Specifically,
Theorem 10.4:
For any square matrix A ,
AI = I A = A
94
RIA TRI CE S
[CHAP. 10
ALGEBRA OF SQUARE MATRICES
Recall that not every two matrices can be added or multiplied. However, if we only
consider square matrices of some given order n, then this inconvenience disappears.
Specifically, any 12 x n matrix can be added to o r multiplied by another n x n matrix, or
multiplied by a scalar, and the result is again an n X n matrix.
In particular, if A is any n-square matrix, we can form powers of A :
A 2 = A A , A 3 = A 2 A , . . . and A n = Z
We can also form polynomials in A . That is, for any polynomial
f(x) =
+
Uo
we define f ( A ) to be the matrix
f ( A ) zz aoZ
Uln:
+
u2x2
+
+
*..
+ alA + uzA2 +
*
a
UnXn
+ anAn
*
In the case that J ( A ) is the zero matrix, then A is said to be a zero o r r o o t of the polynomial J(.T).
Example 6.1:
Let A = /1
(3
If
f ( x ) = 2x9
');
-4
-
3~
+ 5,
=
A2 =
then
g(x) = x2
On the other hand, if
g(A)
then
+ 3x - 10
(-:ii) +
3(' 3
then
')
-4
-
lo(:
;)
=
('0
'1
0
Thus A is a zero of the polynomial g(z).
TRANSPOSE
The transpose of a matrix A , written At, is the matrix obtained by writing the rows
of A , in order, as columns:
a1
u2
a1
bl
b2
* * *
Note that if A is an m x n matrix, then .4t is an n X nz matrix.
Example 7.1:
The transpose operation on matrices satisfies the following properties.
Theorem 10.5:
+
(i) ( A B)t = A t
(ii) (At)t= A
+ Bt
(iii) ( ~ c A=)kAt,
~ for k a scalar
(iv) (AB)t= F A t .
CHAP. 101
MATRICES
95
Solved Problems
MATRIX ADDITION AND SCALAR MULTIPLICATION
10.1. Compute:
11 -1
1
-3
2
1 -2
(i)
Add corresponding elements:
4
5
1+1
4+0
6
2+(-1)
6+(-5)
5+3
(ii) The sum is not defined since the matrices have different shapes.
(iii) Add corresponding elements:
1
2
0 -5
3
-3
1 -1
-5
2
-
-1)
6
0
-2
-3)
1
0
+ 3 2 + (-5) (-3) + 6
+ 2 (-5) + 0 1 + (-2)
4 -3
(i)
+ (-3)
3
-1 -4
-5
2
(-1)
Multiply each element of the matrix by the scalar 3:
;)
3(-:
3.2
(3.(-3)
=
3
3:;)
6 12‘
(-9
3)
=
(ii) Multiply each element of the matrix by the scalar -2:
(-2) 1
(-2) 7
-Z(i 0 -1
= ( ((-2)
- 2 ) **20 (-2)*(-3))
(-2) ’ (-1)
-”)
1-2
;)
-14
(,,-;
=
(iii) Multiply each element of t h e matrix by -1, or equivalently change the sign of each element
in the matrix:
2 -3
10.3. Compute:
-:-:)
3%;
-
1 -2
2 \’o -1
-3)
5
+
qo
1 -2)
1 -1
-1
Fi r st perform the scalar multiplication, and then the matrix addition:
3(23
-: -:>
1 -2
- 2 ( 0 -1
-15
6
9
6
0
3)
-12
+ (-2) + 0
9
+0 +4
-3)
5/
+
2
(-2
0
+ +
{1
’
4
2
-1
6)
-10
+
(0
4
4
-4
+ 6 + (-8)
+ (-10) + (-4)
3
-15
4 4
0 2
(-4) -12
+ +
-1
-2)
-4
-8)
4
-7
-CHAP. 10
$1A TRI CE S
96
10.4. Find x, y, x and w if:
x
6
2,)
\-1
+
4
jz+w
First write each side as a single matrix:
($ :;,)
(
=
.?:
X Sx W
+ 4- 1
X+ZJ
3
+
y + 6
2x+3
Set corresponding elements equal t o each other to obtain the four linear equations
22 = 4
32 = 2 + 4
2y = 6 f x
3y = z + u + 6
or
22 = Zt' - 1
3X = z t ? C - l
zc: = 3
3w = 220 3
+
The solution of this system of equations is
z
= 2,
?J
= 4, z = I , w = 3.
MATRIX NIULTIPLICATION
10.5.
Let (Y x s) denote a matrix with shape r X s. When will the product of t w o matrices
( r x s ) ( t X ZL) be defined and what will be its shape?
The product ( r X s ) ( t X u)of an Y X s matrix and a t X u matrix is defined if the inner numbers
s and t a r e equal, i.e. s = t. The shape of the product will then be the outer numbers in the given
order. i.e. T X ZL.
10.6. Let (Y x s) denote a matrix with shape r x s. Find the shape of the following products
if the product is defined:
(2 x 3)(3 x 4)
(iv) ( 5 x 2)(2X 3)
(ii) (4x 1)(1x 2)
(iii) (1X 2)(3x 1)
(v) (4 x 4)(3 x 3)
(vi) (2 x 2 ) ( 2 X 4)
(i)
In each case the product is defined if the inner numbers a re equal, and then the product will
have the shape of the outer numbers in the given order.
(i)
The product is a 2 X 4 matrix.
(ii) The product is a 4 X 2 matrix.
(iii) The product is not defined since the inner numbers 2 and 3 a r e not equal.
(iv) The product is a 5
X
3 matrix.
(v) The product is not defined since the innar numbers 4 and 3 a r e not equal.
(vi) The product is a 2 X 4 matrix.
I\' ")
10.7. Let A = 2
(i)
-1
"
and B = ( 3
-2
-')
6
'
Find (i) A B , (ii) BA.
Now A is 2 X 2 and B is 2 x 3, so the product matrix A B is defined and is a 2 x 3 matrix.
To obtain the elements in the first row of the product matrix A B , multiply the first row
( l , 3 ) of A by the columns
,
(-:)
-*6 :I
= (
1.2 - 3.3
1.0
and
+ 3.(-2)
(-$
of B , respectively:
1*(-4)
,
+ 3.6)
=
(11 -6
14)
To obtain the elements in the second row of the product matrix A B , multiply the second row
(2,-1) of A by the columns of 6,respectively:
CHAP. 101
97
MATRICES
1
3 \ /
=
11
( 2 * 2 4-(-1).3
Thus
-C
(-1)*(-2)
=
AB
(ii) Now B is 2 X 3 and A i s 2
BA is not defined.
X
14
-6
2.0
(111 -6
2
14)
-14)
2. Since the inner numbers 3 and 2 a re not equal, the product
10.8. Let
(i)
(-1)*6
2.(-4)
Find (i) A B , (ii) BA.
Now A is 1 X 2 and B is 2 X 3, so the product A B is defined and is a 1 X 3 matrix, i.e. a row
vector with 3 components.
To obtain the elements of the product A B , multiply the row of A by each column of B :
AB
= (2,1)
(i
-:-:)
=
(2.1
+ 1.4, 2 . ( - 2 ) + 1 . 5 ,
2.0
+ 1.(-3))
=
(6,1,-3)
(ii) Now B is 2 x 3 and A is 1 X 2. Since the inner numbers 3 and 1 a r e not equal, the product
BA is not defined.
1
2 -1
1
0
10.9. Let A = i
\-3
(i)
4
)
andB=
(i
-f
-:I.
Find (i) A B , (ii) BA.
Now A is 3 X 2 and B is 2 X 3, so the product A B is defined and is a 3 X 3 matrix. To obtain
the first row of the product matrix AB, multiply the first row of A by each column of B,
respectively:
\
=I:
2.1
+ (-1)*3
2.(-2)
+ (-1)*4
+ (-1)*O
2.(-5)
=
r
-I0?
To obtain the second row of the product matrix A B , multiply the second row of A by each
column of B , respectively:
-8
1.lt0.3
-10
=
1*(-2)+0*4 1*(-5)&0-0)
-1 -8
1 -2
-8
-10
-5)
To obtain the third row of the product matrix A B , multiply the third row of A by each
column of B, respectively:
-3
4/\
=I:
(-3)
-1
1
1 4 * 3 (-3)
+
*
-8
-2
(-2)
+4
i/
4
-1
Thus
AB
=
(-3)
*
-10
-5
(-5)
- -I
-1
+4
*
0
9
-8
-10
22
--:
-8
-10
(ii) Now B is 2 x 3 and A is 3 X 2, so the product BA is defined and is a 2 X 2 matrix. To obtain
the first row of the product matrix BA, multiply the first row of B by each column of A ,
respectively:
[CHAP. 10
AT AT RICE S
98
=(
1 '2
+ (-2)
'1
+ (-5)
*
1 * (-1) + (-2)
(-3)
*
0
+ (-5)
4
=
;"
To obtain the second row of the product matrix B A , multiply the second row of B by each
column of A :
15
=
(3.2
-21
+ 4 . 1 + 0*(-3)
Thus
3*(-1)
=
15
(10
+ 4.0 + 0.4 )
=
15
(10
-21
-3)
-21
-3)
Remark: Observe t h a t in this case both A B and B A a re defined, but A B and B A a re not equal;
in fact, they do not even have the same shapes.
(i) Determine the shape of AB. (ii) Let cij denote the element in the ith row and
j t h column of the product matrix AB, that is, A B = (el!). Find: c23, c14, C P I and CIZ.
(ii
Since A is 2 X 3 and B is 3 X 4, the product A B is a 2
X
4 matrix.
(ii) Now c I j is defined as the product of the ith row of A by the j t h column of B .
Hence:
/ o\
/ 1\
C14
= (2,-1,O)
= 2 ~ 1 + ( - 1 ) ~ ( - 1 ) + 0 ~=
0 2+1+0
-1
= 3
0:'
c21
=
~ 1 21
(ii)
(i)
=
(2, -1,
(
-3
, (Tb)
1.1
+ 0 . 2 + (-3)*4
= 2*(-4)
=
+ (-l)*(-1) + 0 . 0
1
+ 0 - 12
= -8
= -11
+1+0
= -7
'\
5 / -7)
The first factor is 2 X 2 and the second is 2 X 2, so the product is defined and is a 2 X 2 matrix:
1*4+6*2
1*0+6*(-l)
(-3) 4 5 2 (-3) 0
5 ' (-1)
+
+
-2
-5
99
RIA TRICES
CHAP. 101
(ii) The first factor is 2 X 2 and the second is 2 X 1, so the product is defined and is a 2 X 1 matrix
/ 1.2+6.(-7)
)
+ 5*(-7) /
\(-3)*2
-
(iii) Now the first factor is 2 X 1 and the second is 2 X 2. Since the inner numbers 1 and 2 a r e
distinct, the product is not defined..
(iv) Here the first factor is 2 X 1 and the second is 1 X 2, so the product is defined and is a 2
matrix:
(v) The first factor is 1 X 2 and the second is 2
which we frequently write as a scalar.
(2, -1)
(-:)
j: 1,
(2. 1
=
Y,
2
so the product is defined and is a 1 X 1 matrix
+ (-1). (-6))
=
181
=
8
/
SQUARE MATRICES
+
. Find (i) A', [ii) As, (iii) f ( A ) , where J ( x ) = 2x3 - 42, 5 .
\ 4 -3
(iv) Show t h a t A is a zero of t h e polynomial g(x) = M ? 2s - 11.
10.12. L e t A =
(i)
A2
I
= AA
=
(: -;)(: -$
+
+
1.2
4.2
1.1
2.4
(-3)*4
4.1
9
\-8
(-3).(-3)
1.(-4) f 2.17
4 * (-4)
(-3) * 17
1 . 9 $- 2.(-8)
(-3) * (-8)
4 9
+ 2.(-3)
+
+
-41
IT,
1 G O -67
(iii) To find J ( A ) ,first substitute A for x and 51 f o r the constant 5 in the given polynomial
= 2x3 - 4 x - 5 :
m
Then multiply each matrix by its respective scalar:
Lastly, add the corresponding elements in the matrices:
-
1
/-14-4+5
\ l 2 0 - 16
0
+
60-8+0
)
-133 + 1 2 - 5,
-13
104
52')
-117
(iv) Now A is a zero of g(x) if the matrix g ( A ) is the zero matrix. Compute g ( A ) as was done
f o r J ( A ) ,i.e. first substitute A for .z and 111 f o r the constant 11 in g ( x ) = x2+ 22 - 11:
Then multiply each matrix by the scalar preceding it:
>IAT RI CE S
100
;CHAP. 10
Lastly, add the corresponding elements in the matrices.
s(A)
=
/
+ 2 - 11
+0
9
(-8 t 8
+
-4
4 i- o:,
17 - 6 - 111
-
Since g i A ) = 0, A is a zero of the polynomial g(z).
10.13. Let B =
[' -4"
3
0
and let j ( x ) = x2 - 4x
- 3.
Findf(B).
-1
Fi r s t coinpute B2:
2
=
BL
BB
2
=
+
+
+
+
+
+
1.1 2 . 3
0.0
3.1
(-4).3
5.0
0 * 1 (-1) * 3 2 * 0
-3
2
+
+
+
2.(-4) T 0*(-1)
1.2
3.2
(-4).(-4) - 5*(-1)
0-2
(-1) * (-4) - 2 * (-1)
+
+
+
+
+
+
1.0
2.5
0.2
3.0
(-4)*5
5.2
0.0
(-1)*5
2.2
-1/
Then
f ( B ) = 6' - 4B
+ 31
2
=
0
-
(-:
-6
1;
-3
=
10.14. Let -4 = /1
j 4 -3
.
(-9
-1
0
0
1
-;:)+
-1
7-4+3
- 12 - 0
-3-0+0
-6-8+0
17+ 16+3
2+4+0
10-0+0
-10 - 20 + O )
-1-8+3
Find a non-zero column vector
2~
= (-21
6
-3
-14
3;
-;:I
-6
such that AZL= 3u.
=
F i r s t set up t h e matrix equation Au = 3u:
Write each side as a single matrix (column vector):
Set corresponding elements equal to each other t o obtain the system of equations
+
x
321 = 3 2
4x - 3y = 3 y
or
-2.1.
+ 321
= 0
4~ - 621 = 0
The two linear equations a r e the same (see Chapter 11) and there exist a n infinite number of
solutions. One such solution is N = 3,
has the property t h a t A u = 3 2 ~ .
= 2.
In other words, the vector u =
("2>
is non-zero and
X A TR I C E S
CHAP. 101
TRANSPOSE
101
1 0 1 0 \
10.15. Find the transpose At of the matrix
A =
Rewrite the rows of A a s the columns of At:
A t = [l
"\ .
0
3
4
0
5
4
10.16.Let A be an arbitrary matrix. Under what conditions is the product AAt defined?
Suppose A is an m X n matrix: then At is n x m . Thus the product AAt is always defined.
Observe t h a t AtA is also defined. Here AAt is a n m X m matrix, whereas AtA is a n n X n matrix.
10.17. Let A =
/j 3
")
-1
4
Find (i) A A t , (ii) AtA.
*
(i -:).
1
At =
To obtain At, rewrite the rows of A as columns:
-
=
+ 2 * 2 + 0.0
+ (-1)*2 + 4 . 0
+
+
i
+
+
+
1.3
2*(-1)
0.4
3.3
(-1)*(-1) -I-4 . 4 )
1.1
(3-1
1.1
3.3
2.1
(-1).3
0.1 + 4 - 3
3\
+ 3*(-1)
++(-l)*(-l)
4*(-1)
1.2
2.2
0.2
=
Then
(n
++( -41 .) .44
2.0
0.0
2;)
10 -1
12
12 -4
16
PROOFS
10.18. Prove Theorem 10.3(i): (AB)C = A ( B C ) .
Let
A = ( a , ] ) , B = ( b j k ) and
C = (chi).
A B = S = (s,k)
Furthermore, let
T = (t,J. Then
and
BC =
m
s,k
tji
=
=
allblk
bjicii
a12brk
f aambm, =
-k
+ b@c,i +
"
'
+
x a,,bJh
1=1
n
bjncnL
=
2
b,hChl
h=l
Now multiplying S by C, i.e. ( A B ) by C, the element in the ith row and lth column of the matrix
(AB)G is
n
n
m
S,1C11
S,$21
..' f s l , I C n ~ = 2 S l k c k [ =
2 (aLIb&?d
+
+
k=l
k = l )=1
On the other hand, multiplying A by T , i.e. A by BC, the element in the ath row and Zth column
of t h e matrix A(BC) is
Since the above sums a r e equal, the theorem is proven.
102
MATRICES
10.19. Prove Theorem 10.3(ii): A ( B
+ C)
= AB
[CHAP. 10
+ AC.
Let A = ( a i j ) , B = (bik) and C = (cjk). Furthermore, let D = B
and F = AC = ( f i k ) . Then
+
djk
=
eik
= ailbl,
bjk
fiirc
+
+
aiibzk
+
alacak
allClk
m
+
*. *
+
. '*
=
aimbn,k
CLijbjk
j=1
ai~rlcink
x
=
j=1
Hence the element in the ith row and kth column of the matrix A B
eik
+
??1
=
fik
+ C = (djk),E = A B = (eiJ
aijbJi,
j=1
+
m
2
j=1
m
x
=
CLijCjk
aijcjk
+ AC
+
aij(bjk
j=1
is
Cjk)
On the other hand, the element in the ith row and kth column of the matrix A D = A ( B
aildlk
Thus A ( B
f
CLi2dzk
+ C) = A B + AC
f
' ' '
+
m
ailndmk
=
in
2
=
a,,dj,
i=1
3
U,j(bjk
j=1
+
+ C)
is
Cjk)
since the corresponding elements are equal.
10.20. Prove Theorem 10.5(iv): ( A B ) t= P A t .
Let A =
matrix A B is
(a,])
and
B = (bJ.
Then the element in the ith row and j t h column of the
allb,,
+
a22b23
+ .
* *
+
(11
annbm,
Thus ( 1 ) is the element which appears in the j t h row and ith column of the transpose matrix ( A B ) t .
On the other hand, the j t h row of B f consists of the elements from the j t h column of B:
(blJ
bzt
".
(21
bWIJ)
Furthermore, the ith column of A f consists of the elements from the ith row of A :
i'i
(3)
aim
Consequently, the element appearing in the j t h row and ith column of the matrix BtAt is the
product of ( 2 ) by ( 3 ) which gives ( 1 ) . Thus (AB)t = BtAt.
Supplementary Problems
MATRIX OPERATIONS
For Problems 10.21-23, let
-:),
+ C,
C =
(
-
-3
2
0
5 -1
1
-4
0
10.21.
Find:
(i) A
10.22.
Find:
ii) A B , (ii) A C , (iii) A D , (iv) BC. (v) BD, (vi) CD.
10.23.
Find:
(i) A t , (ii) AtC, (iii) DtAt, (iv) B t A , (v) DtD, (vi) DDt.
+
B , (ii) A
(iii) 3A - 4B.
0
")
3
and
D =
(,-!I
103
$1AT R I CE S
CHAP. 101
SQUARE MATRICES
10.24.
g ( ~z
)
10.25.
l2
Let A =
')
(ii) If f ( x ) =
(i) Find A2 and A3,
-1 .
8, find g ( A ) .
j3
.(.? - .C -
">
23
- 3 9 - 2r
+
+ 4,
(i) If f ( x ) = 2.c2 - 4.c 3, find f ( B ) . (ii) If g ( r ) =
fj 15 3 .
(iii) Find a non-zero column vector ti = f " such t h a t Bu = 6 2 4 .
Let B =
r2
\ Y)
10.26. Matrices A and E a r e said t o commute if A B = B A . Find all matrices
10.27.
with
(t t).
Let
A =
2\
1)
b
,
(E
find f ( A ) . (iii) If
- 42 - 12, find g ( B ) .
s>
which commute
Find An.
PROOFS
(k,
+ k,)A
= klA
+ k,A.
10.28.
Prove Theorem 10.2iiii):
10.29.
Prove Theorem 10.4[iii): k(AB) = ( k A ) B = A ( k B ) .
10.30.
Prove Theorem 10.5(i): ( A
+ B)t = A + + B t .
Answers to Supplementary Problems
(-! -;
10.21.
(i)
10.22.
(i) Not defined.
5
(ii) (T1
10.23.
10.24.
10.26.
4
(i) A2 = (Io
3
(')
-5)
18
(ii) Not defined.
,
17
11 -12
4
-12
2)
(
iiiii
(iii)
-2
-3
1 0
(i) (-1
3)
2
(ii) Not defined.
A3
= (26
1 )
Only matrices of the form
27
(iii) ( 9 , 9 ) ( i d
Is\
-1)
(
4
0
-3
(ii, f ( A ) =
(8 i)commute with
-7
-6
12
(y:
-1)
-1$
(vi) Not defined.
(-: -!)
4 -2
(v) 14
(vi)
(iii) g ( A ) =
-31
Chapter 11
linear Equations
LINEAR EQUATION IN TWO UNKNOWNS
A linear equation in two unknowns x and y is of the form
ax + by = c
where a,b , c are real numbers. We seek a pair of numbers
the equation, that is, for which
ak, bk, = c
N
= (k,, k2) which satisfies
+
is a true statement. Solutions of the equation can be found by assigning arbitrary values
to x and solving f o r y (or vice versa).
Example 1.1:
Consider the equation
2r+y = 4
If we substitute :c = -2
2*(-2)+y
= 4
or
in t h e equation, we obtain
= 4
-4+y
Hence (-2,8) is a solution. If we substitute
equation, we obtain
2*3+y = 4
or
6 + y = 4
y = 8
or
or
N
= 3 in the
y
= -2
Hence (3, -2) is a solution. The table on the right lists six
possible values for z and the corresponding values for y ,
i.e. six solutions of the equation.
T
2
3
Now any solution a = (k,,k,) of the linear equation ax b y = c determines a point in the Cartesian
plane R'. If a,b are not both zero, the solutions of
the linear equation correspond precisely to the points
on a straight line (whence the name linear equation).
+
Example 1.2:
+
Consider the equation 2 x
y = 4
of the preceding example. On the
r i g h t we have plotted the six solutions of t h e equation which appear
in t h e table above. Note t h a t they
all lie on the same line. We call this
line t h e g r a p h of the equation since
i t corresponds precisely to the solution set of the equation.
Graph of 2 x + y = 4
TWO LINEAR EQUATIONS I N w o UNKNOWNS
We now consider a system of two linear equations in two unknowns x and p:
+ b,y =
uPx+ b2y =
U,X
104
c,
c2
2
0
-2
CHAP. 111
105
L I N E A R EQUATIONS
A pair of numbers which satisfies both equations is called a simultaneous solution of the
given equations o r a solution of the system of equations, There are three cases which
can be described geometrically. (Here we assume that the coefficients of x and y in each
equation are not both zero.)
t
= -3
x+2y = 3
2 2 f 2y = 6
x+y = 1
3x 3y = 3
(a)
(b)
(4
z-cy
x-y
= 1
+
Fig. 10-1
1.
T h e system has exactly one solution. Here the lines corresponding to the linear equations intersect in one point as shown in Fig. 1 0 - l ( ~ above.
)
2.
T h e system has n o solutions. Here the lines corresponding to the linear equations are
parallel as shown in Fig. lO-l(b) above.
3.
T h e systenz has an infinite number o f solictions. Here the lines corresponding t o the
linear equations coincide as shown in Fig. 10-l(c) above.
Now the special cases 2 and 3 can occur if and only if the coefficients of
coefficients of y are proportional:
bl
- - -
x and the
b2
a2
There are then two possibilities:
(i) If the constant terms of the equations are in the same proportion, i.e.
-
_ -b 1 -
a,
-
Cl
c2
b2
then the lines are coincident. Hence the system has an infinite number of solutions
which correspond to the solutions of either equation.
(ii) If the constant terms of the equations are not in the same proportion, i.e.
then the lines are parallel.
Example 2.1:
a, _
-b1
f -
a,
b2
CZ
C1
Hence the system has no solution.
Consider t h e system
+ 6y
3%
2%
+ 4y
= 9
= 6
Note t h a t 3/2 = 6/4 = 9/6. Hence t h e lines are coincident and the system has an
infinite number of solutions which correspond t o t h e solutions of either equation.
L I N E A R EQUATIONS
106
[CHAP. 11
Particular solutions of t h e first equation and hence of the system can be
obtained as shown in Example 1.1. For example, substitute x = 1 in the first
equation t o obtain
3*1+62/ = 9
or
3f6y = 9
6y = 6
or
or
y = 1
Thus x = 1 and y = 1 or, in other words, t h e pair of numbers a = (1,l) is a
solution of the system. By substituting other values f o r x (or y) in either equation, we obtain other solutions of t h e system.
Example 2.2:
Consider the equations
3x
+ 62/
= 9
2x -t. 4y = 5
Note t h a t 312 = 614 but 312 = 614 # 915. Hence the lines are parallel and the
system has no solution.
On the other hand, if the coefficients of the unknowns are not proportional, i.e.
a,
bl
f a2
b2
then case 1 occurs; that is, the system has a unique solution. This solution can be obtained
by a process known as elimination, i.e. by reducing the equations in two unknowns to an
equation in only one unknown. This is accomplished by:
(i) multiplying each of the given equations by numbers such that the coefficients of
one of the unknowns in the resulting equations are negatives of each other;
(ii) adding the resulting equations.
We illustrate this method in the following example.
Example 2.3 :
Consider the equations
(1)
32
+ 2y
(2)
22
-
= 8
5y = -1
Note t h a t 3 / 2 i
.21-5,
so the system h a s a unique solution.
and ( 2 ) by -3 and then add in order to “eliminate” 2:
2 X (1):
-3 X ( 2 ) :
+ 4y =
- 6 ~+ 1571 =
6%
16
3
19y = 19
Addition:
Multiply ( 1 ) by 2
or
y =1
Substitute y = 1 in ( 1 ) to obtain
3x+2*1 = 8
or
3x+2 = 8
or
3% = 6
or
2 = 2
Thus x = 2 and y = 1 or, in other words, the pair 01 = ( 2 , l ) is the unique solution to t h e given system.
GENERAL LINEAR EQUATION
We now consider a linear equation in any arbitrary number of unknowns, say
xl,x,, . . ., x I 1 . Such an equation is of the form
a,x, a p , * . . a?txll= b
+
+
+
where a,,a,, . . .,a,,, b are real numbers. The numbers at are called the coefficients of xi,
and b is called the constant of the equation. An n-tuple of real numbers (Y = (kl, k,, , , . ,k,),
i.e. a n n-component row vector, is a solution of the above equation if, on substituting
ki for xl,the statement
a,k, a2k2 . - . a,kI1 = b
+
is true. We then say that
N
+
+
satisfies the equation.
LINEAR EQUATIONS
CHAP. 111
Example 3.1 :
Consider t h e equation
Now
107
= 3
x+2y-4x+w
01 = (3,2,1,0) is a solution of the equation since
3+2.2-4*1+0=3
or
3 + 4 - 4 + 0 = 3
or
3 = 3
is a t r u e statement. On the other hand, /3 = ( 1 , 1 , 2 , 2 ) is not a solution of the
equation since
1+2*1-4*2+2 = 3
or
1+2-8+2 = 3
or
-3 = 3
is not a t r u e statement.
A linear equation is said to be d e g e n e m t e if the coefficients of the unknowns are all
zero. There are two cases:
(i) The constant is not zero, i.e. the equation is of the form
0x1 0x2 . . . + Oxn = b, with b i
0
Then there is no solution to the linear equation.
+
+
(ii) The constant is also zero, i.e. the equation is of the form
0x1 0x2 * * * O X , = 0
Then every n-tuple of real numbers is a solution of the equation.
+
+
+
GENERAL SYSTEM O F LINEAR EQUATIONS
A system of nz equations in n unknowns n.,, x2, . . .,xn is of the form
a21 x1
+
+
XI
+
all
x1
am1
U22XZ
+
+
am2 ~2
+
U1zX2
* '
'
* ' '
* '
+
+
+ amnxn
alnxn
=
Z;I
~ 2 n x n
=
bz
=
bm
= (kl, k,, . . . , k J which
where the ail, b, are real numbers. 4 n n-tuple of numbers
satisfies all the equations is called a solution of the system.
We first consider the special case where all the above equations are degenerate, i.e.
where every a,, = 0. There are two cases:
O(
(i) If one of the constants b, # 0, i.e. the system has an equation of the form
0x1 0x2 .
Ox, = b,, with b, # 0
then this equation, and hence the system, has no solution.
+
+ -- +
(ii) If every constant bt= 0, i.e. every equation in the system is of the form
0x1 0x2
. * Oxn = 0
then each equation, and hence the system, has every n-tuple of real numbers as a
solution.
+
Example 4.1:
The system
+
+
ox
ox
+ o y + o x + ow
+ oy + o x + ow
Ox+Oy+Ox+Ow
= 0
= 4
= -1
has no solution since one of the constants on the right is not zero.
On t h e other hand, the system
ox+ o y + o x + o w =
ox + oy o x ow =
ox oy o x ow =
has every 4-tuple a = (kl, k,, k,,k,) as a solution
+ +
+ + +
right a r e zero.
0
0
0
since all the constants on the
108
LINEAR EQUATIONS
[CHAP. 11
In the usual case where the equations a r e not all degenerate, i.e. where one of the
a,,#O, we reduce the system ( 1 ) to a simpler system which is equivalent to the original
system, i.e. has the same solutions. This process of reduction, known as (Gauss) elimination, is as follows:
(i)
Interchange equations and the position of the unknowns so that all#O.
(ii) Multiply the first equation by the appropriate non-zero constant so that a,,= 1.
(iiij For each i > 1, multiply the first equation by -ai, and add i t to the ith equation
so that the first unknown is eliminated.
Then the system ( 1 ) is replaced by an equivalent system of the form
XI
. . . + al,x,
bl
aZ2x2 -+ a2353 + - . . + a&x,, = b;
....................................
uLx.2
Example 4.2:
(g3X3
+
am353
+
+ a:;"& -+
+
+
* . *
b:
=
am,5,,
Consider the system
2 ~ + 6 y - x
=
4
3x-22y-
=
1
5%
+ 9y
x
- 22
= 12
4 so t h a t the system is replaced by
Multiply the first equation by
x+3y-*x
3%-22y-
x
+ 9y -
5x
2x
=
2
=
1
= 12
Multiply t h e new first equation by -3 and add it to the second equation so t h a t
t h e system is replaced by
x + 3 y - * x
=
2
-1ly
+ 9y
52
+
32
= -5
-
2x
=
12
Lastly multiply the first equation by -5 and add it to the third equation so t h a t
the system is replaced by the following which is in the desired form:
=
x + 3 y - + z
-1ly
-6y
Example 4.3:
+
+
42
=
$2
=
2
-5
2
Consider the system of the preceding example. Here we reduce the system to a
simpler form without first changing the leading coefficient to 1. This method has
the advantage of minimizing the number of fractions appearing.
Multiply the first equation by 3 and the second by -2 and then add t o elimin a t e x from the second equation:
3 X first:
GX
1 8~ 3x = 12
-2 X second:
-6.2:
+
+
4x1
+ 22
= -2
22r/ - x = 10
Addition:
Now multiply the first equation by 5 and the third equation by -2 to eliminate
from t h e third equation:
5
-2
X
first:
X third:
Addition:
lox
T
30y - 52 =
-lox - 18y
12y
20
+ 4x
= -24
-
=
2
-4
3
CHAP. 111
L I K E A R EQUATIONS
109
Thus the system is equivalent t o
2x
+
61/ - x =
22y
-
122)
-
2
r+ 3u-3z=
4
10
z = -4
22y - z = 10
1 2 y - z = -4
or
z =
Continuing the above process, we obtain the following fundamental result:
Theorem 11.1:
Every system of m equations in n unknowns can be reduced t o an
equivalent system of the following form:
x1 f c12x2
C13s3
* ' ' f C ~ T f
~ T' ' '
ClnT,, = dl,
?"
??%
x2
c.323
. . . czrx7 * . . c2,zxn = dz
+
+
s j
+
+
+
'
+
.+
c3rrr
Xr
+
+
+
+
* * *
t C?,,Xn
= dS
+ . . . + c77,xn= a,
0 =
d T + l
o = d ,
Remark:
A system of linear equations in the above form, where the leading coefficient
is not zero, is said to be in echelon form.
The solutions of a system of equations in the echelon form ( 2 ) can easily be described
and found.
. . . , d m are not all zero, i.e. there is a n
(i) Inconsisteizt Equutiotis. If the numbers
equation of the form
Oxr Oxz . . . Ox,, = dk, with d h = 0
+
+
+
then the system is said to be inconsistent, and there a r e no solutions.
(ii) Consistent Equations. If the numbers d,,,, . . . , d m a r e all zero, then the system is
said to be consistent and there does exist a solution. There are two cases:
( a ) r = n , that is, there are as many non-zero equations as unknowns. Then we can
successively solve uniquely for the unknowns xr, x r p l ,. . ., x, and so there exists
a unique solution for the system.
( b ) r < n , that is, there are more unknowns then there are non-zero equations. Then
we can arbitrarily assign values to the unknowns x,,,, , , .,xn and then solve
uniquely f o r the unknowns xr, , . . , x,. L4ccordingly there exists a n infinite number
of solutions.
The following diagram shows the various cases:
System of linear equations
I
Inconsistent
I
I
I
solution
solution
I
L I N E A R EQUATIONS
110
Example 4.4:
[CHAP. 11
Consider the system
x + 2 y - 3x = 4
O x + y + 2x = 5
ox
+ Oy + ox
x+2y-332 = 4
y+2x = 5
or
= 3
0 = 3
Since the system has a n equation of the form 0 = c with c # 0, the system is
inconsistent and has no solution.
Example 4.5:
ox+
y + 2 z = 5
y+2x = 5
z = 1
or
Ox+Oy+ x = 1
o x oy 02 = 0
+ +
yt2x = 5
or
x = l
o = o
Since there is no equation of the form 0 = c, with c f 0, t h e system is consistent.
Furthermore, since there a r e three unknowns and three non-zero equations, the
system h as a unique solution. Substituting x = 1 in the second equation we obtain
y + 2 . 1 = 5
or y + 2 = 5
or y = 3
Substituting z = 1 and y = 3 in the first equation, we obtain
= 4
xf2.3-3.1
or
x + 6 - 3 = 4
or
x + 3 = 4
or
s = l
Thus x = 1, y = 3 and x = 1 or, in other words, the ordered triple ( 1 , 3 , 1 ) is
the unique solution of the system.
Example 4.6:
Consider the system
z+2y-3x+
w = 4
02+ y + 2 2 + 3 w = 5
02
o y + O x + Ow = 0
2+2y-3x+
w = 4
y+2x+3w = 5
or
+
o = o
The system is consistent, and since there a r e more unknowns tha n non-zero equations t h e system has a n infinite number of solutions. I n fact, we can arbitrarily
give values to z and zu and solve for x and y. To obtain a particular solution substitute, say, x = 1 and w = 2 in the second equation to obtain
or
y+ 2*1+ 3*2 =5
y+2+6
= 5
or
y + 8 = 5
or
y = - 3
z = 1 and w = 2 in the first equation to obtain
Substitute y = -3,
x+2*(-3)-3*1+2
= 4
or
= 4 or x - 7 = 4
x-6-3+2
o r x = 11
Thus x = 11, y = -3, x = 1 and w = 2 or, in other words, the row vector
(11,-3,1,2) is a specific solution to the system.
To obtain the general solution to the system substitute, say, z = a and zu = b
in t h e second equation t o obtain
y+2u+3b
Substitute y = 5 - 2a
-
= 5-2a-3b
or
3b, x = u and t*! = b into the first equation to obtain
x+2(5-2a-3b)-3a+b
or
= 5
= 4
z+10-7u-5b
or
= 4
~+10-4~-6b-33a+b = 4
or
s
= 7a+5b-6
Thus t h e general solution of the system is
(7a
+ 5b
-
6, 5 - 2a - 3b, a, b),
where a and b are real numbers
Frequently, the general solution is left in terms of z and w (instead of a and b )
as follows:
or,
( 7 ~ + 5 ~ - 6 5, - 2 ~ - 3 ~ 2,
, W)
x = 7 2 f 5 %-~ 6
y = 5 - 2x - 3w
111
L I N E A R EQUATIONS
CHAP. 111
The previous remarks give us the next theorem.
Theorem 11.2:
A system of linear equations belongs to exactly one of the following cases:
1. The system has no solution.
2 . The system has a unique solution.
3. The system has an infinite number of solutions.
Recall that, in the special case of two equations in two unknowns, the
above three cases correspond to the following cases described geometrically:
1. The two lines are parallel.
2. The two lines intersect in exactly one point.
3. The two lines are coincident.
HOMOGENEOUS SYSTEMS OF EQUATIONS
A system of linear equations is called homogeneous if it is of the form
+
+
allxl
a21
x1
u12x2
+
' *
u22x2
-t
' ' '
.
+ u1nxn
+ uznx,
=
=
0
0
(3)
. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .
+
+
.
+
am, x n = 0
that is, if all the constants on the right are zero. By Theorem 11.1, the above system is
equivalent to a system of the form
am1
X1
+
ClZX.
X2
x1
+
+
* * *
- . *
urn222
*
+ C1,Zr +
+
+
* . *
C2,Xr
* * *
+
+
ClnXn
=
CznXn
= 0
0,
7' L
m
o = o
Clearly, the above system is consistent since the constants are all zero; in fact, the system
trivial solution. It has a
non-trivial solution only if r < n. Hence if we originally begin with fewer equations than
unknowns in ( 3 ) , then T < n and so the system always has a non-trivial solution. That is,
( 3 ) always has the zero solution (0, 0 , , , . 0) which is called the
Theorem 11.3:
A homogeneous system of linear equations with more unknowns than
equations has a non-zero solution.
Example 5.1 :
The homogeneous system
x+zy-3x+
2x+
w = 0
x-zw
= 0
y-3x+5w
= 0
x-sy+
has a non-trivial solution since there a r e more unknowns than equations.
Example 5.2:
We reduce the following system to echelon form:
z+ yx = 0
x+ y - z = 0
2%-3y-t x = 0
x-4y+z2 = 0
to
-5y
-5y
+ 3x = 0
+ 32 = 0
to
x+yx = 0
y - " x5 = 0
o = o
L I N E A R EQUATIONS
112
‘CHAP. 11
The system has a non-trivial solution since we finally have only two non-zero
equations in three unknowns. F o r example, choose r = 5 ; then y = 3 and TC = 2.
In other words ( 2 , 3 , 5 1 is a particular solution.
Example 5.3:
We reduce the following system to echelon f o r m :
x + u - r = 0
x+ y- r = 0
to
2y+r = 0
2x+4y2 = 0
-y-z
= 0
3x
2y
22 = 0
+ +
to
x + y 2 = 0
y++z = 0
-32
= 0
z = 0
UA*% = 0
r = O
Since in echelon form there a r e three non-zero equations in t h e three unknowns,
the system has a unique solution, the zero solution ( 0 , 0 , 0).
z + y -
to
MATRICES AND LINEAR EQUATIONS
The system ( 1 ) of linear equations is equivalent t o the matrix equation
. . . . . .. . . . . . . . . . .
am1
dm2
* * .
amn
in that every solution to (1) is a solution to the above matrix equation and vice versa. In
other words, if A = ( a , , ) is the matrix of coefficients, X = (x,)is the column vector of
unknowns, and B = ( b J is the column vector of constants, then the system ( I ) is equivalent
to the simple matrix equation
AX = B
In particular, the homogeneous system of linear equations ( 3 ) is equivalent to the matrix
equation
AX = 0
Example 6.1:
The following system of linear equations and the matrix equation a r e equivalent:
2 x2-+2 3y -y 5- 54 xx
= 37 ’.
(; -; I;)(:)
=
(;I
In studying linear equations, it is usually simpler t o use the language and theory of
matrices.
Theorem 11.4: The general solution of a non-homogeneous system of linear equations
A X = B is obtained by adding the general solution of the homogeneous
system A X = 0 to a particular solution of the non-homogeneous system
A X = B.
Proof. Let a , be a fixed solution of A X = B. If p is any solution of the homogeneous
system A X = 0, then
A(a,+/3) = A a , + A P = B + O = B
That is, the sum nl + p is a solution to the non-homogeneous system A X = B.
On the other hand, suppose a, is a solution of A X = B distinct from a,. Then
A(a2- a , ) = Aa, - Aa, = B - B = 0
That is, the difference N., - a1 is a solution of the homogeneous system A X = 0. But
n2
=
a,
+ ( a 2 - a,)
hence every solution of A X = B can be obtained by adding a solution of AX=O to the
particular solution a1 of A X = B.
CHAP. 111
113
LINEAR EQUATIONS
Theorem 11.5:
If n l , a?, , . . , all a r e solutions to the homogeneous system of linear equations A X = 0, then every linear combination of the
of the form
k,a,
+ k p , + . . . + li,,a,,
where the k Lare scalars,
is also a solution of the homogeneous system AX = 0.
Proof. We a r e given t h a t
A(k,n,- . . .
Accordingly, k,a,
+ ...
Aa, = 0, . . ,,
+ k,{a,,) =
T
klAnl + .
ila,,
= 0.
Hence
. + k,Aa,
= k,O
+ . . . + k,,O
= 0
k p r l is also a solution of the homogeneous system A X = 0.
In particular, every multiple ke of any solution a of A X = 0 is also a solution of AX = 0.
Consider the homogeneous system in Example 5.2:
Example 6.2:
x = 0
y-
.r+
2.r - 3y
+
x = 0
= 0
.r-4y+2x
= 1 2 , 3 , 5 ) is a solution of the above. Hence every multiple of a ,
As was noted,
such a s ( 4 , 6 , 1 0 ) and i -6, -9, -15), is also a solution.
(Y
We now restate Theorem 11.2 using the language of matrix theory, and give an independent proof of the theorem.
If A X = B has more than one solution, then it has an infinite number of
solutions.
Proof. Let a and 13 be distinct solutions to A X = B. Then the difference N - ,5 is a
non-trivial solution to the homogeneous system A X = 0:
A ( a - P ) =z An -- AP = B - B = 0
Theorem 11.2:
Hence every multiple k ( ~ - p )of a - p
scalar k,
is a solution of A X = O .
n
+
Accordingly, f o r each
- p)
k ( N
Thus AX = B has a n infinite number of solutions as
is a distinct solution to A X = B.
claimed.
r)+t)
VECTORS AND LINEAR EQUATIONS
The system ( I ) of linear equations is also equivalent to the vector equation
+
X1
x2
an,1
...
+
=
xs
am2
bln
&n
In other words, if ( Y ~ , . . . , an and p denote the above column vectors respectively, then
the system (1) is equivalent to the vector equation
Xp,
+
Sp2
If the above equation has a solution, then
o r is said to depend upon the vectors
+.
p
’
’
A XnNn
=
p
is said to be a linear combination of the vectors
We restate this concept formally:
The vector p is a linear combination of the vectors
scalars k,, . . . , kll such that
p = k l N l 4- k2a2+
* * .
+ k,a,
al,
. . . , N~
if there exist
[CHAP. 11
L I N E A R EQUATIONS
114
i.e. if there exists a solution to
p =
where the
Remark:
X,
X,a,
+ X2a2 + . . . + 5"an
are unknown scalars.
The above definition applies to both column vectors and row vectors, although
our illustration was in terms of column vectors.
ExampIe 7.1:
Let
I;
/3=
i),
(ul
=
:-4
(i),
a2
=
Then /3 does depend upon the vectors a,, ue and
+ 7a2
p = -4q
(a)
a3
and
a3 =
($
since
- a3
t h a t is, the equation ( o r system)
2 = x+y+x
3 = x+y
-4 = x
has a solution ( - 4 , 7 , -1).
:"u [I+. .
VECTORS AND HOMOGENEOUS LINEAR EQUATIONS
I[
The system ( 3 ) of homogeneous linear equations is equivalent to the vector equation
XI
+
x2
+ Xn
f)
=
anlz
ami
Clmn
That is, if a,, a2, . . . ,an denote the above column vectors respectively, then the homogeneous
system ( 3 ) is equivaIent to the vector equation
X,a1
+ x2a2 +
*
. . + Xnafl = 0
If the above equation has a non-zero solution, then the vectors a l , . . .,an are said to be
dependent; on the other hand, if the above equation has only the trivial (zero) solution,
then the vectors are said to be independent. We restate this concept formally:
1 Definition: I The vectors
al, a*, .
. . , a l l are
not all zero, such that
kp,
dependent if there exists scalars k,, k,,
+ kznz +
* * '
+kp,
. . . , kn,
= 0
i.e. if there exists a non-trivial solution to
Z,a,
+ X2a2 +
where the xi are unknown scalars.
in dependent.
Remark:
' *
+ X,Ian =
0
Otherwise, the vectors are said to be
The above definition applies to both column vectors and row vectors although
our illustration was in terms of column vectors,
LINEAR EQUATIONS
CHAP. 111
Example 8.1:
115
The only solution to
or
:c+y+z
.c+y
x
is the trivial solution s = 0, ?/ = 0 and x = 0.
pendent.
Example 8.2:
= 0
= o
= o
Hence the three vectors a r e inde-
The vector equation lor' system of linear equations)
x+2y+
x = 0
z f 3 y t 3 x = 0
has the non-trivial solution ( 3 , - 2 , l ) .
Accordingly, the three vectors a r e dependent.
Solved Problems
LINEAR EQUATIONS IN TWO UNKNOWNS
11.1. Determine three distinct solutions of 2n. - 3y = 14 and plot its graph.
Choose any value for either unknown, say
2.(-2)
-
3y = 14
Thus x = -2 and y = -6
or
-4
-
2:
= -2.
3u = 14
Substitute z = -2 into the equation t o obtain
or
-3y
= 18
or
y
= -6
or, in other words, the pair (-2, -6) is a solution.
Now substitute, say, z = 0 into the equation t o obtain
2.0 - 3y = 14
Thus (0, -14/3)
= 14
-3u
or
or
y = -14/3
is another solution.
Lastly, substitute. say, y=O into the equation to obtain
2x-3.0
= 14
or
2x = 14
or
z
= 7
Hence (7,O) is still another solution.
Now plot the three solutions on the Cartesian plane R2 as shown below; the line passing through
these three points is the graph of the equation.
t5
Y
116
LINEAR EQUATIONS
[CHAP. 11
3x-2y = 7
x+2y = 1*
11.2. Solve the system:
Since 311 # -212, the system has a unique solution. Now the coefficients of y a re already
the negatives of each other: hence a d d both equations:
32 - 2y = 7
x+2y = 1
Addition:
Substitute
.r
or
= 8
4x
.1:=2
= 2 into the second equation to obtain
2+2y = 1
2y = -1
or
u
or
-4
= -$
-3)
Thus :2 = 2 and y =
or, in other words, the pair (Y = ( 2 ,
is the solution of the system.
Check pour answer by substituting the solution back into both original equations:
= 7
3*2-2*(-$)
and
2+2*(-3) = 1
2-1
or
7 = 7
or
= 1
1 = 1
or
(I) 2 x + 5 y = 8
11.3. Solve the system:
Note t h at 213
6+1 = 7
or
( 2 ) 3.2: - 2y = -7
hence the system ha s a unique solution.
and then add:
# 51-2:
( I ) by 3 and (2) by -2
3
-2
62
X (1):
X (2):
-6%
+ 15y
+
= 24
4~ = 14
19y = 38
Addition:
To eliminate x, multiply
y = 2
or
Substitute y = 2 into one of the original equations, say ( I ) , t o obtain
2xl-5.2 = 8
23:+10 = 8
or
22 = -2
or
Hence .c = --1 and y = 2 or, in other words, the pair ( - 1 , Z j
x = -1
or
is the unique solution to the system.
Check your answer by substituting the solution back into both original equations:
(1):
2*(--1) 4- 5 . 2 = 8
or
-2 f 10 = 8
(2):
3.(-1)-2*2
= -7
or
-3-4
= -7
8 = 8
or
or
-7
= -7
Multiply ( I ) by 2 and
We could also solve the system by first eliminating y as follows.
( 2 ) by 5 and then add:
2 i< ( 1 ) :
4% 1Oy = 16
+
Substitute
2
5 X (2):
15%- 1Oy = -35
Addition:
19x
= -19
or
x
= -1
= -1 in ( 1 ) to obtain
2*(-1)
Again we get (-1,Z)
11.4. Solve the system:
+ 5y
= 8
or
-2+5y
= 8
or
6~ = 10
or
y = 2
as a solution.
(I)
x -2y = 5
( 2 ) -32
+ 6y
= -10'
hence the system does not have a unique solution. But
-2
5
-1 _ - -3
6
-10
Hence the lines a r e parallel and the system ha s no solution.
Note t h a t 1/-3 = -216:
+---
LINEAR EQUATIONS
CHAP. 111
11.5. Solve the system:
117
(I) 5 x - 2 ~= 8
(2) 3 x + 4 y = 10
hence the sybtem has a unique solution. To eliminate y, multiply
Note t h a t 5/3 f -2/4;
( 1 ) by 2 and add it t o ( 2 ) :
2
X (1):
10s - 4y = 16
32
(2):
Addition:
+ 4y
13x
= 10
or
=26
x = 2
Substitute x = 2 in either original equation, say ( 2 ) , to obtain
3 . 2 + 4 y = 10
or
6
+ 4y
= 10
4y = 4
or
or
y =
1
Thus the pair 1 2 , l ) is t h e unique solution t o the system.
Check the answer by substituting the solution back into both original equations:
(1):
5.2-2-1
(2):
3.2+4.1
11.6. Solve the system:
= 8
= 10
10-2
or
or
x-2v = 5
( 2 ) - 3 ~ + 6 2 j = -15
=
8
or
6 + 4 = 10
or
8 =
8
10 = 10
(I)
*
Note t h at
Accordingly, the two lines a r e coincident. Hence the system has an infinite number of solutions
which correspond to the solutions of either equation.
Particular solutions can be found as follows: Let y = 1 and substitute in ( 1 ) to obtain
x - 2 * 1 = 5
or
or
x - 2 = 5
x = 7
Hence ( 7 , l ) is a particular solution. Let 11 = 2 and substitute in ( 1 ) t o obtain
x - 2 * 2 = 5
or
x - 4 = 5
or
x = 9
Then ( 9 , 2 1 is another specific solution of the system. And so forth.
The general solution to the system is obtained as follows. Let y = a and substitute in ( 1 ) t o
obtain
x-2y= 5
or
z = 5 + 2 a
Accordingly, the general solution to the system is (5
GENERAL SPSTERIS OF LINEAR EQUATIONS
11.7. Solve the following system in echelon form:
x+2y-32+4w = 5
O x + y f 5 x I - 2 ~= 1
or
+ 2a, a ) ,
x
where a is any real number.
+ zy - 32 + 4w
?/ + 5x +22u
= 5
= 1
0 = 2
02+Oy+O2+Ow = 2
The system is inconsistent since i t has a n equation of the form 0 = c with c # 0. Hence the
system has no solution.
11.8. Solve the following system in echelon form:
x+3y-22X=
4
on:+ y - & = 2
or
o x + o y + x = -1
ox o y + ox = 0
+
4
y-52=
2
x = -1
x+32j-22=
o =
0
[CHAP. 11
LINEAR EQUATIONS
118
The system i s consistent since it has no equation of the form O = c with c f 0 ; hence the
system has a solution. Furthermore, since there a r e three unknowns and three non-zero equations,
the system has a unique solution.
Substitute x = -1 into the second equation to obtain
= 2
y-5*(-1)
y+5 = 2
or
y =
or
-3
Then substitute y = -3 and x = -1 into the first equation to obtain
x+3.(-3)-2*(-1)
= 4 or z - 9 + 2 = 4 o r 2 - 7 = 4 or x = 11
Thus x = 11, y = -3 and z = -1 or, in other words, the ordered triple (11,-3, -1) is the unique
solution t o the system.
11.9. Solve the following system in echelon form:
x+2y-32+4w
ox
x+2y-%+4w
+
+
= 5
y +5 5 x +2 w = 1
= 5
or
y +5x
2242 = 1
ox+oy+oz+ow = 0
o = o
The system is consistent and so has a solution. Furthermore, since there a r e four unknowns
and only two non-zero equations, the system h a s a n infinite number of solutions which can be
obtained by assigning ar b i t r ar y values to the unknowns x and tu.
To find the general solution to the system let, say, x = a and w = b, and substitute into the
second equation to obtain
y + 5 a + 2 b = 1 or y = 1 - 5 a - 2 b
Now substitute x = a, w = b and y = 1- 5a - 2b into the first equation to obtain
= 5
x+2(1-5a-2b)-3a+4b
or
z+2-10a-4b-3~+4b
or
x+2-13a=5
or
= 5
x=3+13a
Thus the general solution is
(3
+ 13a, 1 - 5a - 2b, a, b)
where a and b a r e real numbers
To find a particular solution t o the system let, say, a = 1 and b = 2 to obtain
(3
+ 13
*
-
1, 1 - 5 1 - 2 * 2, 1, 2)
or
(16, -8, 1, 2)
N o t e . Some texts leave the general solution in terms of x and w as follows:
x = 3
+ 13x
= 1-5x-2w
where z and w a r e a ny real numbers
132, 1- 5x - 2w,x , w)
y
(3
+
z + ~ Y - ~= x-4
(2) 5x - 3y - 72 = 6
(I)
11.10. Solve the following system:
(3) 3x-2y+32
.
= 11
Reduce the system t o echelon form by first eliminating x from the second and third equations.
Multiply (I) by -5 and add to ( 2 ) to eliminate z from the second equation:
-5 x (I):
-5x - 1oy
+ 202
5%- 3y -
(2):
Addition:
-13y
'72
+ 13x
= 20
= 6
= 26
or
y -x
= -2
Now multiply ( 1 ) by -3 and add to (3) to eliminate x from the third equation:
-3
X
(I):
(3):
Addition:
- 3 ~- 6y t 1 2 2 = 12
3x - 2y
+
32 = 11
-8y 4- 152 = 23
CHAP. 111
L I N E A R EQUATIONS
119
Thus t h e original system is equivalent to the system
+ 2y -
x
4x = -4
x = - 2
y -
-8y i152 = 23
Next multiply the second equation by 8 and add to the third equation to eliminate y from the third
equation:
8 X second:
8y - 82 = -16
third:
+ 15x
-8y
Addition:
=
23
or
7
7 x =
x = l
Thus the original system is equivalent t o the following system in echelon form:
= -4
x t2y-4x
z = -2
y--
x =
1
The system is consistent since i t has no equation of the form 0 = c with c # 0; hence the system
has a solution. Furthermore, since there a r e three unknowns and also three non-zero equations,
t h e system has a unique solution.
Substitute z = 1 into t h e second equation to obtain
= -2
y-1
y = -1
or
Now substitute z = 1 and y = -1 into the first equation to obtain
x+2.(-1)-4.1
= -4
or
= -4
x-2-4
or
2-6
= -4
or
z
= 2
Thus x = 2 , y 2 -1 and z = 1 or, in other words, the ordered triple ( 2 , -1, 1) is the unique solution
to the system.
11.11. Solve the following system:
(1)
+2y-%
( 2 ) -3x
+
= -1
y - 2x = -7
5 x 4-3y-42 = 2
(3)
.
Reduce to echelon form by first eliminating x from the second and third equations. Multiply
( 1 ) by 3 and add t o ( 2 ) to eliminate z from the second equation:
3
x
3s
(1):
-3x
(2):
+ 6y -
+
Addition:
y -
7y
-
9~ =
-3
2x =
-7
112 = -10
Now multiply ( 1 ) by -5 and add to (3) to eliminate x from the third equation:
-5
- 5 ~- 1Oy
X (1):
52
(3):
+
Addition:
+ 1 5 ~=
5
3y - 4x = 2
+ llx
-7y
= 7
Thus the system is equivalent to t h e system
x + 2y
-
7y
-
-7y
32
=
-1
11z = -10
+ 11z
=
7
However, if we add the second equation t o the third equation, we obtain
O x + Oy
+ Ox
= -3
or
0 = -3
Thus the system is inconsistent since it gives rise to an equation 0 = c with c f 0; accordingly, the
system has no solution.
120
'CHAP. 11
LINEAR EQUATIONS
~ + 2 y - 3 ~ =
6
(I)
11.12. Solve the following system:
( 2 ) 2x - y
+ 4x
=
2
.
(3) 4 x + 3 1 ~ - 2 =
~ 14
Reduce the system to echelon form by first eliminating :c from the second and third equations.
Multiply ( 1 ) by -2 and add t o ( 2 ) '
-2
-2a - 4y f 6.2. = -12
X (1):
22 - y f 42 =
(2):
Addition:
-5y
Multiply ( 1 ) by -4 and add i t t o (3):
-4
X (1):
-4.2.
(3):
42
-
8y
2
+ 102
= -10
+ 122
= -24
+ 3y -
22 =
y
or
y - 22
-
14
-5y f 102 = -10
Addition:
22 = 2
or
= 2
Thus the system i s equivalent to
x + 2 y - 3 ~ = 6
y-22
= 2
y-22
= 2
~ f 2 u - 3 ~6
or simply
y-22
= 2
( S o t c . Since the second and third equations a r e identical, we can disregard one of them. I n fact,
if we multiply the second equation by -1 and add to the third equation we obtain 0 = 0.)
The system is now in echelon form. The system i s consistent and since there a r e three unknowns
but only two non-zero equations, the system ha s a n infinite number of solutions which can be
obtained by assigning ar b i t r ar y values to 2.
To obtain the general solution t o the system let, say, z = a .
second equation to obtain
y-22a=2
or y = 2 + 2 a
Then substitute z = a into the
Now substitute z = a and y = 2 + 2a into the first equation t o obtain
n+2(2+2a)-3a
= 6
.?+4-44a-3~ = 6
or
r + 4 + a = 6
or
or
x = 2-a
Thus the general solution t o the system is
(2 - a , 2
+ 2a, a )
where a is any real number.
To obtain a particular solution of the system let, say, a = l and substitute into the general
solution t o obtain ( 2 - 1 , 2 + 2 . 1 , 1) o r (1, 4, 1).
11.13. Solve the system:
+
= 1
(2) 32 - 9 - 42 = 7
( 3 ) 5~ + 2 1 ~- 62 1 5
(1) 2x
2/ - 3x
.
Method 1.
Reduce to echelon form by first eliminating x from the second and third equations. Multiply
( I ) by -3 and (2) by 2 and then add to eliminate from the second equation:
-3 X ( 1 ) :
2
X (2):
-62 - 3y
6x
+ 92
- 22/-
Addition:
-5y
+
= -3
82
= 14
2
= 11
Multiply ( 1 ) by -5 and (5) by 2 and then add to eliminate z from the third equation:
-5 X ( 1 ) :
2 x (3):
Addition:
-10% - 51/
+ 152
= -5
10% i 4y - 122 = 10
-1/+
32 =
5
or
y - 32
= -5
CHAP. 111
LINEAR EQUATIONS
121
Thus the system is equivalent to the system
2.1-
+ y - 3x
+ x
-5u
y-3x
=
1
1
22+2/-&=
= 11
o r , interchanging equations,
= -5
y-32
= -5
+
x = 11
-5y
Now niultiply the second equation by 5 and add to the third equation t o eliminate y from the third
equation:
5 X second:
5y - 152 = -25
third:
+
-5y
Addition:
=
x
11
= -14
-14x
or
= 1
x
Thus the original system is equivalent t o the system
+
23:
- 3x =
1
y - 3 % = -5
1
x =
Substitute z = 1 into t h e second equation to obtain
= -5
y-3.1
Substitute
2
= 1 and y = - 2
22+(-2)-331
Thus
1: = 3, y
= 1
or
= -5
y-3
or
y=--2
into the first equation to obtain
or
= 1
2x-2-3
or
= 1
22-5
or
22 = 6
or
z=3
= -2 and x = 1 is the unique solution to the system.
Method 2.
Consider y as the first unknown and eliminate y from the second and third equation.
( 1 ) to ( 2 ) to obtain
5 ~ - 7 2= 8
Add
Multiply ( 1 ) by -2 and add to (3):
-2
-4%
X (I):
-
5x
(3):
Addition:
2y
+ 62
+ 2y - 62
x
= -2
=
5
=
3
Thus the system is equivalent to
22+y-3x
Substitute
1: =
y - 3 2 + 2 ~ = 1
= 1
5%
- 7x = 8
X
= 3
or
-7%
+ 5~
= 8
r = 3
3 into the second equation t o obtain
-72+15
= 8
or
-72
= -7
or
x
= 1
Substitute x = 3 and z = 1 into the first equation to obtain
y - 3 + 6 = 1
Thus
N
or
y + 3 = 1
or
y = - 2
= 3, y = -2 and z = 1 is the unique solution of the system.
VECTORS AND LINEAR EQUATIONS
11.14. Reduce the following vector equation to a n equivalent system of linear equations
and solve:
LINEAR EQUATIONS
122
[CHAP. 11
Multiply the vectors on the right by the unknown scalars and then add:
Set the corresponding components of the vectors equal to each other to form the system
x + y + x =
2
x + y
=-4
X
= 7
2
z T y + x =
y + x = -4
or
7
x =
Substitute r = 7 into the second equation to obtain
y f 7
Substitute
-c
= -4
y = -11
or
= 7 and y = -11 into the first equation to obtain
x - l 1 + 7 = 2
or
or
x - 4 = 2
z = 6
Thus x = 7, y = -11 and x = 6 is the solution of the vector equation.
11.15. Write the vector v = (1,-2,5) as a linear combination of the vectors e l = (1,1,l),
e, = ( 1 , 2 , 3 ) and e3 = (2, -1,l).
We wish t o express
ZI
as v = x e l
(1, -2, 5)
+ ye, + xe3,
with x, y and z a s yet unknown. Thus we require:
+ y ( l , 2 , 3) + 4 2 , -1,
x, x) + (Y, 2Y, 3Y) + (22,
=
x(1, 1, 1)
=
(2,
=
(x+y+2x, x+2y-x,
-2,
1)
4
xf3yiz)
Hence
s+
x+y+2x=
1
y-22=
x+2y-
x = -2
.L:+31/-
x =
y-32
or
2y-
5
1
= -3
z =
4
x f y + 2 x =
or
or
1
= -3
y-32
2
x =
Substitute z = 2 into t h e second equation t o obtain
= -3
y-3.2
or
y-6
= -3
or
y = 3
Substitute z = 2 and y = 3 into the first equation to obtain
s+ 3+ 2. 2=
1
or
u = -6e,
Consequently,
11.16. Write
1
x + 3 + 4 =
or
\-5
Set
I'
(-%)
or
[-:I,
(::)
e2 =
(1;) and e3 =
as a linear combination of the e, using unknowns x, y and z: w
{-:)
= x f 2
+ y
(1;) +
(-:) (-:)
=
x =-6
+ 3e2 + 2e3
as a linear combination of el =
2' =
x +7 = 1
+
f
%el+ y e 2
(-5) (
=
[-:I.
+~ e 3 .
z+2y+z
2 x - Y f 72
-32 - 4y - 52
)
CHAP. 111
123
LINEAR EQUATIONS
Hence
=
2
y+7%=
3
x t 2 y f
22-
-3X
-
41/ -
2
52
x+2y+
= -1
= 1
-5y+5x
= -5
2
x =
z+2y+
or
2 y - 22
or
2
x =
g
y - z =
-5y+5x
= -1
z t 2 y + x = 2
or
= 1
y--x
o = "
The system is inconsistent and so has no solution. Accordingly, u cannot be written a s a linear
combination of the vectors e l , e2 and e 3 .
DEPENDENCE AND INDEPENDENCE
11.17. Determine whether the vectors il,1,-l), (2, -3,l) and (8, - 7 , l ) are dependent or
independent.
F i r st set a linear combination of the vectors equal to the zero vector:
(0, 0, 0)
= ~ ( 11,
, -1)
+
~ ( 2-3,
, 1)
+ ~ ( 8-7,
, 1)
Then reduce the vector equation to an equivalent system of homogeneous linear equations:
(0, 0, 0)
=
( r ,2, - x )
=
(X
+
( 2 ~-,3 ~ ,II)
+
(82, - 7 2 ,
+ 2u 4- 8x, x - 3y - 72, + y +
-2
Z)
X)
r+2y+8x = 0
2-3y-7x
= 0
-x+
y+ 2 = 0
01,
Lastly, reduce the system to echelon form. Eliminate
z+2y+
3y
+
from the second and third equation to obtain
r+2y+8x = 0
y+32 = 0
8x = 0
-5y - 152 = 0
5
or
o = o
y+3x = 0
92 = 0
= 0
y+3x = 0
z+2y+8z
or
Since there a r e three unknowns but only two non-zero equations, the system ha s a non-trivial
solution. Thus the original vectors ar e dependent.
Remark: We do not need to solve the system to determine dependence or independence; we
only need to know if a non-zero solution exists.
11.18. Determine whether the vectors
(-:I, [:)
and
\-3
are dependent or independent.
\-1
F i r st set a linear combination of the vectors equal to the zero vector; and then obtain a n
equivalent system of homogeneous linear equations:
or
-u/
-32
-1
xi2y+3x = 0
I
-22
-32-
+ 3y + 2x
y +
2
= 0
= 0
Eliminate z f r o m the second and third equations to obtain
z + 2 y + 32 = 0
x+2y+32 = 0
+ 82 =
5y + 10% =
7y
0
0
-3% -
+Z
124
[CHAP. 11
LINEAR EQUATIONS
Eliminate y from the third equation to obtain
z+2y+3x
= 0
y+22
= 0
-62
= 0
sf2yi-32
= 0
= 0
z = o
or
y+2x
In echelon form the system has three unknowns and three non-zero equations; hence the system
has the unique solution x = O , y = 0, z = 0, i.e. the zero solution. Hence the original vectors a re
independent.
Reduce the problem to a homogeneous system of equations by setting a linear combination of
the vectors equal t o the zero vector:
+ +
y
32 f 2v - 8~
-2% 4- 2y - 72 f UJ
32
y - 6v - 7~
-42
5y
22 - 5 v
4w
x
+
+ +
X+
+
y + 3 2 + 2 ~ - 8 ~= 0
-22+2y-72
3 ~ +y
-42f
$. W
= 0
- 6 ~ - ' 7 ~= 0
52/+22-55v+4w
= 0
Now the system of homogeneous equations ha s five unknowns but only four equations. Hence,
by Theorem 11.3, the system has a non-trivial solution and so the vectors a r e dependent.
11.20. Let vl,v2, . . .,vm be n-component vectors. Show that if the number of vectors is
greater than the number of components, i.e. m > 12, then the vectors are dependent.
Note, as in the preceding problem, t h a t the vector equation
x1v1
+
22v2
+ ... + X m W m
= 0
gives rise t o a homogeneous system of linear equations in which the number of unknowns equals
the number of vectors and the number of equations equals the number of components. Hence if
the number of vectors is greater than the number of components, then the system ha s a non-zero
solution by Theorem 11.3 and so the vectors a r e dependent.
CHAP. 111
LINEAR EQUATIONS
125
Supplementary Problems
LINEAR EQUATIONS IN TWO UNKNOWNS
11.21.
Solve:
22 - 5y = 1
32 2y = 11
+
11.22.
Solve:
22
32
11.23.
+ 4y
+ 6y
2s
5s
(ii)
= 10
(ii)
= 15
{
+ 3y
+ 7y
= 1
= 3
(iii)
4s - 2y = 5
3y = 1
-62
+
(iii)
32
2
i
-
2y = 7
= -16
+ 31/
22 - 3y = 5t
32
y = 2t
+
Solve:
GENERAL SYSTEMS OF LINEAR EQUATIONS
11.24.
Solve:
22+ y-3% = 5
32 - 2y
22 = 5
5 2 - 3y - z = 16
+
11.25.
Solve:
+ 3y - 2x
= 5
2y
3% = 2
42- y+4x = 1
22
z
+
-
x + 2 y + 3x = 3
2x
3y
8x = 4
3x
2y
172 = 1
+ +
+ +
+
+
22
3y = 3
2-2y = 5
32
2y = 7
11.26.
Solve:
z+2y+2x =
32-2y2 =
(i) 22 - 5y 3% =
x+4y+62 =
+
2
5
-4
0
Z + 5 y + 4 ~ -1 3 =
~ 3
(ii) 32 - y 22
5w = 2
4w = 1
22+2y+3x-
+ +
VECTORS AND EQUATIONS
11.27. Write v = (9, 5, 7) as a linear combination of the vectors el = (1,1,1), e2 = (1,-1,0) and
e3 = (2,0,1).
11.28.
Determine whether the vectors (1, -2, 3, l), (3, 2, 1, -2)
independent.
11.29.
Determine whether the vectors
11.30.
Prove: If the vectors el, e2 and e3 are independent, then any vector v can be written in a t most
one way as a linear combination of the vectors e l , e2 and e3.
(-i), (i) (-!)
and
and
(1, 6, -5, -4)
are dependent o r
are dependent or independent.
[CHAP. 11
LINEAR EQUATIONS
126
Answers to Supplementary Problems
(ii) ( 2 , -l), o r x = 2 , y = -1
=3, y =1
11.21.
(i) 13, 11, o r
11.22.
(ii (5 - 2 a , a), or x = 5 - 2 a , y = a
11.23.
(ii 16, 8)
11.24.
(i) (1, - 3 , -2)
11.25.
(ii (3, -1)
5
[ii) (5, 2 )
(ii) (-a+ 2b, 1
11.27. w = 2e, - 3e2
11.28.
Dependent
11.29.
Independent
21.30.
Proof:
(iii) (-1 - 7% 2
(ii) no solution
(ii (2, 1, -1)
(iii) ( t , -t),
iii) no solution
or z = -1, y = -5
or z = t , y = --t
(iii) ( 5 , 2 )
+
+ 2a, a ) , o r
x = -1-77x
y=2+2x
z = -x+2w
2u - 2 b , a , b ) , o r
(iii) (6 - 2a - 5b, a, b, 2b - l), or
11.26.
(iii) (-1, -5),
y=1+2x-2w
6-2y-52
<(w5 =
=22-1
(ii) no solution
+ 5e3
Suppose w = alel
+ a2e2+ u3e3
0 = w
-
and w = b,e,
v = (al - bl)el
+ b,ez + b3e3.
Subtracting, we obtain
+ (a2- b,)e, + (a3- ba)ea
Since the vectors a r e independent, the coefficients must be zero, i.e.
al - b , = 0 ,
a,
Accordingly, a, = b,, u2= b , and u3 = b,.
- b2
= 0
and
u3 - b3 = 0
Chapter 12
Determinants of Order Two and Three
INTRODUCTION
To every square matrix there is assigned a specific number called the determinant of
the matrix. This determinant function, first discovered in the investigation of systems
of linear equations, has many interesting properties in its own right.
In this chapter we define and investigate the determinant of square matrices of order
two and three, including its application to linear equations. Determinants of matrices
of higher order are beyond the scope of this text.
DETERMINANTS OF ORDER ONE
Write det(A) or ' A (for the determinant of the matrix A ; it is a number assigned to
square matrices only.
The determinant of a 1 x 1 matrix ( a ) is the number a itself: det(a) = a.
Observe that in the linear equation in one unknown x,
ax = b
we can view a as a 1 x 1 matrix (a) applied to the one-component vector x. If det(a) # 0,
1
However, if det(a) = 0, i.e.
i.e. if a # 0, then the equation has the unique solution x = ;b.
if a = 0, then the equation has no solution if b f 0, and every number is a solution if b = 0;
hence the solution is not unique.
As we shall subsequently see, the preceding result holds true in general, i.e. det(A) # 0
is a necessary and sufficient condition f o r the linear equation A x = b to have a unique
solution.
w
DETERMINANTS OF ORDER TWO
The determinant of the 2 x 2 matrix
(: :)
is denoted and defined as follows:
Example 1.1:
We emphasize that a square array of numbers enclosed by straight lines is not a
matrix but denotes the number that the determinant function assigns to the enclosed
array of numbers, i.e. t o the enclosed square matrix.
127
DETERMINANTS O F ORDER TWO AND T H R E E
128
[CHAP. 12
A very important property of the determinant function is that i t is multiplicative;
that is,
Theorem 12.1:
Remark:
The determinant of a product of matrices is the product of the determinants of the matrices: det(AB) = det(A) det(B).
The diagram on the right may help the reader obtain the
determinant of a 2 x 2 matrix. Here the arrow slanting
downward connects the two numbers in the plus term of
the determinant ad - be, and the arrow slanting upward
connects the two numbers in the minus term of the
determinant ad - be.
(X
-
+
LINEAR EQUATIONS IN TWO UNKNOWNS AND DETERMINANTS
Consider two linear equations in two unknowns:
a1x
a2,r
+
+ bey
b1y
=
c1
= cz
Let us solve the system by eliminating y. Multiply the first equation by b, and the second
equation by -bl, and then add:
bp
-bl
X
first:
x second:
albzx
-arblx
Addition:
+ b1bey
=
- blb2y
= -blcz
(a1bzra2b1)x =
b2C1
b2C1
- blcz
Now the system has a unique solution if and only if the coefficient of x in this last equation
is not zero. i.e.
Observe that D is the determinant of the matrix of coefficients of the system of equations.
In this case, where D # 0, we can uniquely solve for x and y as quotients of determinants
as follows:
Here D, the determinant of the matrix of coefficients, appears in the denominator of both
quotients. The numerators N Z and N y of the quotients for x and y, respectively, can be
obtained by substituting the column of constant terms in place of the column of coefficients
of the given unknown in the matrix of coefficients.
Example 2.1:
Solve using determinants:
2x - 3y = 7
3x 5y = 1
+
The determinant D of the matrix of coefficients is
51
-3
D = ) 3
= 2.5 - 3.(-3)
= 10 f 9 = 19
CHAP. 12)
129
D E T E R M I N A N T S O F ORDER TWO AND T H R E E
Since D f 0, t h e system has a unique solution. To obtain t h e numerator LV,
replace, in t h e matrix of coefficients, t h e coefficients of x by t h e constant terms:
To obtain t h e numerator N, replace, in the matrix of coefficients, t h e coefficients
of y by the constant terms:
l3
2
=
= 2.1
7
-
3.7
= 2
-
21
= -19
Thus the unique solution of the system is
Example 2.2:
Solve using determinants:
,2x = 5 + y
\I
3+2y+3% = 0
F i r s t a r r a n g e the equations in standard form:
22- y = 5
3 x f 2 y = -3
The determinant D of the matrix of coefficients is
Since D # 0, the system has a unique solution.
Now,
= 5 . 2 - (-3)(-1)
NZ =
=
10 - 3
= 7
Thus the unique solution of the system is
Example 2.3:
Solve using determinants:
{::I:; 1 ;
The determinant D of the matrix of coefficients is
= 13
-'
-61
= 2-(-6) - 3-(-4)
= -12
+ 12
= 0
Since D = 0, t h e system does not have a unique solution, and we cannot solve the
system by determinants. (The discussion of the previous chapter shows t h a t the
system h a s no solution since 2/3 = -41-6 f 7/5.)
a1
bl
c1
a2
b2
CP
a3
bs
c3
=
alb2c3
+ a2bscl + a3blcZ
-
alb3c2
-
aZblc3
-- a3bZcl
[CHAP. 1 2
DETERMINANTS O F ORDER TWO AND T H R E E
130
or
which is a linear combination of three determinants of order two whose coefficients (with
alternating signs) are the first row of the given matrix. Note that each 2 x 2 matrix can
be obtained by deleting, in the original matrix, the row and column containing its coefficient:
a1
+
=
2(-20+2)
-
3(0-2)
- 4(0+4)
=
0
(-4)
-36
-4
+
6 - 16
=
-46
-24
- 14
72
=
Example 3.2:
'
Remark:
=
2(-12-00)
- (10+4)
+
=
3(0-24)
-
-110
Problems 12.8 and 12.18 show that the determinant of a 3 x 3 matrix can be
expressed as a linear combination of three determinants of order two with
coefficients from any row or from any column, and not just from the first row
as exhibited here.
D
=
a1
bl
c1
a2
b2
cz
a3
b3
c3
#
0
131
D E T E R M I N A N T S O F ORDER TWO AND T H R E E
CHAP. 121
In this case, the unique solution of the system can be expressed as quotients of determinants,
bl
d:! bz
ds b3
dl
N,
1
c1
dl
dz
d3
a1
,
cz
NU
=
a2
c3
a3
c1
CZ
dt
dz
bi
,
N,
=
c3
~
uz bz
a3
u1 b3
d3
Example 4.1:
Solve the following system by determinants:
2r+
x = 3
y-
u f x = 1
2 +
= 4
2-22y-3x
The determinant D of t h e matrix of coefficients is obtained as follows:
=
2(-3+2)
- 1(-3-1)
Since D # 0, the system h as a unique solution.
f o r x, y and x respectively:
N,
=
1;
3
1
-1
11
-3
-
N?, =
= 2(-3-4)
- 3(-3-1)
N,
= 2(4+2)-1(4-1)+3(-2-1)
-2
=
+
-2
4
+
=
3
5
We evaluate S,, S, and N,, the numerators
= 3(-3+2)
1
=
- l(-2-1)
1(-3-4)
-
= -3
1(-2-4)
= -14
- l(4-1)
=
+7+6
= 10
+ 12 - 3
= -5
1 2 - 3 - 9
= 0
Thus the unique solution is
w,
2 = - - -
D
-
10 = 2 ,
5
= -N =
, - =
-5
D
5
-1,
x = ? N= - = O
o
D
5
INVERTIBLE MATRICES
A square matrix A is said to be invertible if there exists a matrix B with the property
that
A B = B A = I , the identitymatrix
Such a matrix B is unique; f o r
AB1= B1A = I and ABz = BZA = Z
implies
B I = BIZ = B I ( A B P=) (B1A)BZ= IBa
B2.
132
DETERMINANTS O F ORDER TWO AND T H R E E
[CHAP. 12
We call such a matrix B the iwaerse of A and denote it by A-l. Observe that the above
relation is symmetric; that is, if B is the inverse of A , then A is also the inverse of B.
Example 5.1:
3
+
-5
2)
=
2.3
S*(-l)
(1*3+3.(-1)
3 - 5
2 5
(-1
2)(1
3)
=
3.2
(-5)al
((-1)*2+2.1
(f
:)(-1
+
2*(-5)
5.2
1*(-5)+3*2)
+
=
1
(0
=
(i
+
3.5
(-5)*3
(-1)*5+2.3)
0
1)
It is known that AB = Z if and only if BA = Z (Problem 12.17); hence it is necessary to
test only one product to determine whether two given matrices are inverses, as in the
next example.
Example 5.2:
0
6
-1
1*(-11)+0*(-4)+2.6
2.(-11)
(-1)*(-4)
3.6
4*(-11)f 1.(-4)+8.6
+
+
1.2+0*0+2.(-1)
(-l).O
3*(-1)
2.2
4*2+1*0+8.(-1)
+
+
1*2+0.1+2*(-1)
2.2
(-1)'l
3*(-1)
4*2+1*1+8*(-1)
+
+
Thus the two matrices a r e invertible and a r e inverses of each other.
Example 5.3:
Find the inverse of
(i
.
We seek scalars x,y, z and w f o r which
y+2w = 0
3y+4w = 1
o r which satisfy
The solutions of the systems a r e
given matrix is
(pi-:)
.
2: = -2,
z
= 312 and y = 1, u)= -112.
Thus the inverse of the
We test our answer:
INVERTIBLE MATRICES AND DETERMINANTS
We now calculate the inverse of a general 2 X 2 matrix A =
x , y , x and w such that
ax + bx
ex + dx
(: i ).
We seek scalars
bw
+ dw
a y f
cy
whjch reduces t o solving the following two systems of linear equations in two unknowns:
ax + bx = 1
cx + dx = 0
ay+bw = 0
cy+dw = 1
CHAP. 121
DETERMINANTS OF ORDER TWO AND T H R E E
133
Note first that one equation in each system is homogeneous and has (0,O)as a solution,
whereas the other equation does not have (0,O) as a solution. In other words, the lines
in each system are not coincident and so each system has either a unique solution or no
solution. Furthermore, the matrix of coefficients in each system is identical to the original
matrix A . Accordingly, the systems have a solution (and so A has an inverse) if and only if
the determinant of A is not zero. We formally state this result (which holds for matrices
of all orders):
Theorem 12.2: A matrix is invertible if and only if its determinant is not zero.
Usually a matrix is said to be singular if its determinant is zero, and nonsingular
otherwise. Hence the preceding theorem can be restated as follows:
Theorem 12.3: A matrix is invertible if and only if it is nonsingular.
Now if the determinant of the above matrix A is not zero, then we can uniquely solve
for the unknowns 2 , y, x and w as follows:
In other words, we can obtain the inverse of a 2 x 2 matrix, with determinant not zero, by
(i) interchanging the elements on the main diagonal, (ii) taking the negative of the other
elements, and (iii) dividing each element by the determinant of the original matrix.
Example 6.1:
Find the inverse of the matrix A =
(i i ) .
Now det(A) = IAI = 2 * 5 - 4 3 = 10 - 12 = -2. Hence the matrix does
have a n inverse. To obtain A-1, interchange the elements 2 and 5 on the main
diagonal and take the negative of the other elements to obtain the matrix
(-:
-;I
Now divide each element by -2, the determinant of A , t o obtain A-I:
Solved Problems
DETERMINANTS OF ORDER TWO
12.1. Compute the determinant of each matrix:
134
[CHAP. 12
D E T E R M I N A N T S O F O R D E R TWO A N D T H R E E
12.2. Solve for
R'
and y, using determinants:
2x+ y = 7
(i,
3x-5y
= 4
12.3. Determine those values of k for which
Hence 2k=O
is zero.
ax-2by
(ii)
o r k = 0 , and k - 2
3ax-5by
1:
= c
= 2c
,
where a b # 0
,",=o*
= 0 or k = 2 .
T h a t is, if k = O or k = 2 , the determinant
12.4. Prove Theorem 12.1 (in the case of 2 x 2 matrices): The determinant of a product
of matrices is the product of the determinants of the matrices, i.e. det(AB) =
det(A) det(B).
Let
A = \
and
B =
det(AB) = (ar
\t
s\
Zt/
' andso AB =
+ b t ) ( c s + d u ) - (as -t bu)(cr+ d t )
+ adrzc. + bcst + bdtu - acrs adst - bcru
= acrs
= adm
and
Thus
-
+ bcst
-
bdtu
a d s t - bcru
det(A) det(B) = ( a d - b c ) ( ~ u - s t ) =
d N A B ) = det(A) det(B).
-
adrzc
-
adst
-
bcru
+ best
DETERMINANTS O F ORDER TWO AND TH R EE
CHAP. 121
135
DETERMINANTS OF ORDER THREE
12.5. Find the determinant of each matrix:
1
(i)
4
0
2
-2
5
3
3
-1
11
=
$ 3
= l(2-15) - 2(-4+0)
= 4(2+6)
+
= 2(10-6)
12.6. Solve, using determinants:
1
4
0
l(0
+
+ 3(20+O)
+ 15) - 2(0-10)
3(5-3)
+ 4(-2+2)
-2
5
+ 8 + 60
= -13
+
= 32
2
= 8
15
+
+6+
20
0
= 55
1
67
= 14
3y+2x = x + l
3% 2 2 = 8 - 5y
32-1 x x - 2 ~
+
F i r s t arrange the equation in standard f o r m with the unknowns appearing in columns:
2.1:+3y- 2 = 1
3% 5y 22 = 8
:c - 2y - 32 = -1
+ +
Compute the determinant D of the matrix of coefficients:
D
-1
2
-3
3
5
=
1 -2
1
= 2(-15+4)
- 3(-9-2)
- (-6-5)
= -22
+ 33 +
11 = 22
Since D # 0, the system h a s a unique solution, To compute N,, N, and N,, replace the coefficients
of the unknown in the matrix of coefficients by t h e constant terms:
N,
=
3
5
=
~
= 1 ( - 1 5 r 4 ) - 3(-24+2)
- 1(-16*5)
= -11
~
-1
N,
-1
2
-2
-3
12
1 -1
3
8
2
1 -1 -3
~
1
= 2(-24+2)
- 1(-9 -2)
- 1(-3-8)
= -44
+ 66 + 11
+ 11 + 11
= 66
= -22
136
DETERMINANTS O F ORDER TWO AND T H R E E
N, =
I
3
5
1 -2
-1
1
-
= 2(-5+16)
12.7. Consider the 3 x 3 matrix
::).
(Et !:
a3
3(-3-8)
b3
+ 1(-6-5)
= 22
[CHAP. 12
+ 33 - 11
= 44
Show that the diagram appearing below,
c3
where the first two columns are rewritten to the right of the matrix, can be used to
obtain the determinant of the given matrix:
-
-
+ + +
-
Form the product of each of t,he three numbers joined by an arrow slanting downward, and
precede each product by a plus sign as follows:
+ albZc3 + blcza3 + Clazb,
Now form the product of each of the three numbers joined by an arrow slanting upward, and
precede each product by a minus sign as follows:
-
b l c1
a2 b2 c2
a3 b3 c3
a1
'
= alh,c3
a,b2cl
+
-
b3c2al - c3a2b,
+ cla2b3 - a3b2cl -
h1~2a3
b3c2al - c3a2bl
12.8. We represented the determinant of a 3 x 3 matrix A as a linear combination of
determinants of order two with coefficients from the first row of A . Show that
det(A) can also be represented as a linear combination of determinants of order two
with coefficients from (i) the second row of A, (ii) the third row of A .
(i)
(ii)
bl c1
a2 b2 c2
a1
a3
b3
c3
i
= alb2c3
+ u2h3cl +
a3blc2 - alb3c, - a 2 h l c 3
= - ~ Z ( b , c 3 - b 3 ~ 1 )+ b Z ( ~ 1 ~ 3~
3 ~ 1-)
-
a3b2c,
cz(alb3 - aab,)
CHAP. 12j
DETERMINANTS O F ORDER TWO AND THREE
137
Observe t h a t the 2 X 2 matrix accompanying any coefficient can be obtained by deleting, in the
original matrix, the row and column containing the coefficient. Furthermore, the signs accompanying the coefficients form a checkerboard pattern in t h e original matrix:
Remayk: We can also expand the above determinant in terms of the elements of any column
in a n analogous way (Problem 12.18).
INVERTIBLE MATRICES
12.9. Find the inverse of
(i 35) .
Method 1.
We seek scalars
2, y, x
and w f o r which
(; :)(: :)
(; ;)
=
(322 x ++ 5232
Or
3y
2y
+ 5w
+ 3w
1
=
(: ;)
o r which satisfy
+ 52
+ 32
32
22
= 1
= 0
{
and
3 y + 5w = 0
2y+3w = 1
To solve the first system, multiply the first equation by 2 and the second equation by -3 and then add:
2 X first:
-3
X
62
+ 102
= 2
-62 - 9x = 0
second:
Addition:
x = 2
Substitute z = 2 into t h e first equation t o obtain
3x+5*2 = 1
32+10 = 1
or
or
32 = -9
z = -3
or
To solve the second system, multiply the first equation by 2 and the second equation by -3
then add:
2 X first:
6y low = 0
and
+
-3
X
second:
-6y - 9w = -3
Addition:
Substitute
?L'
w = -3
= -3 in the first equation to obtain
3 y f 5*(-3) = 0
or
3 y - 15 = 0
or
31,' = 15
or
y = 5
Thus the inverse of the given matrix is
Method 2.
We use the general formula f o r the inverse of a 2 X 2 matrix. F i r s t find t h e determinant of
the given matrix:
[CHAP. 12
DETERMINANTS O F ORDER TWO AND T H R E E
138
Kow interchange the elements on the main diagonal of the given matrix and take the negative of
the other elements to obtain
(-2” -35)
Lastly. divide each element of this matrix by the determinant of the given matrix, t h a t is, by -1:
(-; -35)
The above is the required inverse.
o r which satisfy
I
2x1 f 3y1 - X i = 1
21
2y,
2, = 0
-71 - y1
32, = 0
+
2x2
+
+
22
-22
+ 3y2 + 2y2 + 22
- yg + 3x2
Xg
= 0
= 1
= 0
2x3
x3
-23
+ 3y3 + 2y3 +
-
763
f
23
= 0
23
=Z
323
0
= 1
We solve each system by determinants. Observe t h a t the matrix of coefficients in each system is
the original matrix. Its determinant D is
D =
1
2
1
-1
3
2
-1
f ( = 2(6+1) - 3(3+1)
-
1(-lf2)
= 14 - 12 - 1 = 1
-1
Since D # 0, each system has a solution and so the given matrix has a n inverse.
each of the three systems follows:
x1 = 7, y1 = -4,
x1 =
1;
x2 = -8,
Thus the inverse of t h e given matrix is
2/2
= 5, zg = -1;
xg = 5, y3 = -3,
The solution of
23
1
1
CHAP. 121
DETERMINANTS OF ORDER TWO AND T H R E E
139
Supplementary Problems
12.11. Find the determinant of each matrix:
12.12.
Solve each system using determinants:
(32
(i) 1 4 x
+ 5y
-
= 8
2y = 1
2 2 - 3 y = -1
42
7y = -1
+
( ii)
5x = 2y - 7
3y = 4 4z
+
(iii)
12.13. Find t h e determinant of each matrix:
1
3
-2
-4
-2
-1
1 -3
12.14.
Solve each system using determinants:
+
2 2 f 3 = yJ-322
2%- 5y
22 = 7
z
2y - 42 = 3
3x - 4y - 6 2 = 5
+
(ii)
3 y + 2 = 2-222
12.15.
Find the inverse of each matrix:
12.16.
Find the inverse of
[-:
4
(i)
,
(; i) (ii) (
fSote:
2
-3
3)
There will be 9 unknowns.)
-2
12.17.
Prove: A B = Z if and only if B A = I , where A and B a r e square matrices and I is the identity
matrix.
12.18.
Express the determinant of a 3 X 3 matrix as a linear combination of determinants of order two
with coefficients from ti) t h e first column, (ii) the second column, (iii) the third column.
Answers to Supplementary Problems
12.11.
(i) -18,
iii) -15,
12.12.
21
29
(i) x = -,
y =26
26
(iii) 8,
(iv) 1, (v) 44,
5 y =1
(ii) x = - 13'
13
(vi) -57
(iii)
13 Y = -8
= -7 '
7
140
D E T E R M I N A N T S O F ORDER TWO AND T H R E E
12.13.
(i) 21, (ii) -11, (iii) 100, (iv) 0
12.14.
ii) x = 5 , y = 1, x = 1
(ii) z = z
+ y,
y = -2.
The determinant of this system = 0.
[CHAP. 12
13
Chapter
The Binomial Coefficients and Theorem
FACTORIAL NOTATION
The product of the positive integers from 1 to n inclusive is denoted by n ! (read
‘‘R factorial”):
n! = 1.2.3.....(n-2)(n-l)n
We can also define n ! recursively as follows:
and
l! = 1
n! = n - ( n - l ) !
It is also convenient to define O! = 1.
Example 1.1:
2! = 1.2
= 2,
3 ! = 1 0 2 . 3 = 6,
5 ! = 5 * 4 ! = 5.24
B! _
-
Example 1.2:
8.7*6!
6!
6!
= 120,
1) =
-T+
no?- 1)’‘ ‘(7%- r
6 ! = 6 * 5 ! = 6.120
= 8 . 7 = 56
12.11.10
+
T
Example 1.3:
?Z(n - 1). * *(?Z
4! = 1 . 2 . 3 . 4
n(12 -
1). . .(n- T l ) ( n - r ) ( n ( n- ~ ) (-n r - q . a . 3
+ 1) = n ( n - - l ) . . . ( n - r + l )
1* 2 * 3 ’ . . ( r - l ) r
1
- =
r!
=
= 24,
= 720
12*11.10*9!
9!
- 12!
9!
n!
- 1). * a 3 ’ 2 1 (n- r ) !
2*1
n!
-._
(n-T)!
1 T !
n!
r!(n-v)!
BINOMIAL COEFFICIENTS
The symbol
(:)
(read “nC r ” ) , where r and n are positive integers with r
defined as follows:
(:)
6
n, is
n(n- l ) ( n - 2) * - - ( n- r + 1)
1 . 2 . 3 . * * ( T - l)r
=
These numbers are called the binomial coe,tficients in view of the theorem in the following
section.
Example 2.1:
Note that
(:)
has exactly r factors in both the numerator and the denominator.
Observe, by Example 1.3, that
n(n- 1)* - . (n- T 1)
1.2’3.*.(r-l)r
+
141
-
n!
r ! (n - ?.) !
142
T H E BINOMIAL C O E F F I C I E N T S A N D T H E O R E M
But n - (R- r ) =
1';
Theorem 13.1: ( n
[CHAP. 13
hence we have the following important relation:
") (
=
or, in other words, if a t Z, = ?a then
:I
1
Example 2.2:
Compute ( l o ) . By definition,
\ 7
On the other hand, 10 - 7 = 8 and so we can also compute (lo' as follows:
\7/
Observe t h a t the second method saves space and time.
Example 2.3:
Compute 'I4)
!\ 12
and
Now 1 4 - 1 2 = 2 and 2 0 - 1 7 = 3; hence
(20'l.
17 //
2 0 . 1 9 - 1 8 = 1140
1.2.3
Remark:
and the fact that 0 ! = 1, we can extend the
Using the above formula for
definition of
(a)
(:)
to the following cases:
n!
=
O!n!
= 1
and, in particular,
O!
= 1
O!O!
BINOMIAL THEOREM
The Binomial Theorem, which can be proved by mathematical induction, gives the
general expression for the expansion of (a b)qt.
+
Theorem 13.2 (Binomial Theorem) :
(a
+ b)"
=
+ na"-'l, +
n(n- 1)
+ 1.2a'Z,"-'
?2(72 - 1)
,2
+
+
anP2b2
?2abn-l
+
b"
n(n - l)(n- 2) a n - 3 b 3 + . . ,
1e2,3
CHAP. 131
143
T H E BINOMIAL C O E F F I C I E N T S A N D THEOREM
+
The following properties of the expansion of (a b)" should be observed:
(i) There are
11
+ 1 terms.
(ii) The sum of the exponents of a and b in every term is n .
(iii) The exponents of n decrease term by term from n to 0; the exponents of b increase
term by term from 0 to n.
(iv) The coefficient of any term is
(y,) where
k is the exponent of either a or b.
(This
property follows from Theorem 13.1.)
(v) The coefficients of terms equidistant from the ends are equal.
Example 3.2:
Expand (2.c
+ 3 y 3 5 and simplify.
Observe t h a t a and b in the Blnom1al Theorem correspond in this case to 22 and 3y2 respectively. Thus
5.4
5.4
(2x+ 3y2)5 = ( 2 . r ) ;
5 ( 2 ~ ) ~ ( 3 y2 ) ~ *( 22 r ) ? ( 3 y ~ 1). 2~ ( 2 . ~ ) ' ( 3 y ~ ) ~ 5 ( 2 ~ ) ( 3 y ~ ) (~3 ~ / ~ ) ;
=
=
+
+
+
+
2 j . 2 + 5 . 2 4 ~ 4 . 3 ! / ~ + 10 * 2 3 r s * 3 2 y 4 + 1 0 . 2 2 x L . 3 3 y 6 + 5 * 2 ~ * 3 ~ +
y 8351/l0
3 2 ~ :+ 2 4 0 x 4 ~ 2+ 7 2 0 . ~ ' ? ( +
~ 1 0 8 0 +
~ ~810.c$
~ ~ - 2432/'O
PASCAL'S TRIANGLE
The coefficients of the successive powers of a + b can be arranged in a triangular a r r a y
of numbers, called Pascal's triangle, as follows:
+ b)O
=
1
a
+
a2
+
2ab
a3
+
3a2b
+
3ab'
+
b3
a4
t
4a3b
+
6a2b2
+
4ab3
-C
+
5a4b
(a
(a-L b)' =
(a
(a
T
b)3 =
( a + b)4 =
(a
+ b)5
( a + b)G =
=
a6
+ b)'
a5
- 6a5b
=
1_
1
1
b
1
1
b2
1
b4
+ lOa3b2 + 10aQb3+ 5ub4 + b5
15a4b2 + 20a3b3 + 15a2b4 + 6ab5 +
1
1
b6
5
6
1
2
3
1
3
1
'g
lo4 5
8
8
1
5
1
6
1
............................
................................................................
Pascal's triangle has the following interesting properties.
(i)
The first number and the last number in each row is 1.
(ii) Every other number in the array can be obtained by adding the two numbers appearing directly above it. For example: 10 = 4 f 6 , 15 = 5 10, 20 = 1 0 f 10.
+
Now since the numbers appearing in Pascal's triangle are the binomial coefficients, we
can rewrite the triangular a r r a y of numbers as follows:
[CHAP. 13
T H E BINOMIAL C O E F F I C I E N T S A N D THEOREM
144
(30)
3 j,
!,3
(:)
(;)
'
(
0
"
)
i')
(;)
(i'
...................................................
(fi)
(:)
(
;
)
.
$
1
)
(:)
' . . (nill\l
:":
...
();:
...
l)
............................................................................
(nil)
(%;I)
In order t o establish the above property (ii) of Pascal's triangle, it is necessary to prove
the following theorem.
Theorem 13.3:
( ) (I/L1) +
n+l
=
In the next example we prove a special case of this theorem; the general proof of
Theorem 13.3 is given as a solved problem.
Example 4.1:
We prove
(7)
=
'11
+ (I;)
/,
which is Theorem 13.3 when n = l l and
T =
7.
Now
7
5
Multiply the first fraction by - and the second fraction by 5 to obtain the same denominator,
7
and then add:
7-11!
7*6!5!
-
+ -7 5! -*151. !4 !
+
7-11! 5-11!
7 ! 5!
=
-
5'11!
7-11!
m +7-! 5!
(7+5)*11!
7! 5!
-
1-2 - 1-1 !
7 !5 !
-
12!
7! 5 !
The general proof of Theorem 13.3 is very similar t o the proof given above.
MULTINOMIAL COEFFICIENTS
Let
n1,112,
. . . . nr be non-negative integers such that
n
. . . . n,
?il
+ n2 + . . . + n, = n.
is defined as follows:
n
nl, n2, . . . . n,
i=
n!
nl!na! * . . n,!
Then the
CHAP. 131
145
THE BINOMIAL COEFFICIENTS A N D THEOREM
These numbers are called the ?nultiiioniial coefficients in view of the following theorem
which generalizes the binomial theorem.
Theorem 13.4:
n
( u ~ t a z f ~ ~ ~ t=a , ) ~
Example 5.1:
(
2
3
-
7!
-
2 ! 3! 2!
-
= 210
7*6.5.4.3*2*1
2-1.3.2.1.2.1
-
al"1aF.. .
Example 5.2:
Expand (a
nl
+ n2 + n3 =
-b
A
c ) ~ . Consider all triplets of non-negative integers (nl,nz,n,) for which
3:
=
a3
+ 3a'b + 3a2c + Gnbc
b3
+ 3bzc + 3ub2 +
c3
3ac2
+ 3bc2
Solved Problems
FACTORIAL
13.1. Compute 4 ! , 5!, 6!, 7! and 8!.
4! = 1 ' 2 ' 3 ' 4
5! = 1 . 2 . 3 . 4 - 5
= 5 - 2 4 = 120
= 5.4!
6! = 1 - 2 . 3 . 4 . 5 . 6
13.2. Compute:
7! = 7 * 6 ! = 7 . 7 2 0 = 5040
= 2.1
13 !
(i) -,
11!
8! = 8 * 7 ! = 8.5040 = 40,320
= 6 * 5 ! = 6.120 = 720
(ii)
7!
m,
1
13.3. Write in terms of factorials: (i) 27.26, (ii) 14. 13. 1 2 , (iii) 5 6 .
(i) 27.26
(ii)
= 27*26*25!
25!
-
3
25!
ll!
11!
1
- 14!
14 * 1 3 * 12
1 4 * 1 3 * 12 * 11!
6 -_
55!
(iii) 56 = 5
_
55!
- 56!
- -
55!
146
[CHAP. 13
THE BINORIIAL C O E F F I C I E N T S A N D THEOREM
13.4. Simplify: (i)
72 !
__ -
(i)
- I)!
(J1
+ I)!
n!
m,
iiii
11
+
(n 1)!
( n- 1)!
(iv) ____
(n-I)!'
(I2
2) ! *
.
n(n- l ) ( n - 2 ) . ' . 3 * 2 ' 1 - n
(72 - 1)(n- 2 ) ' ' ' 3 * 2 * 1
+
___
(n-l)!
(n+ l ) n ( n - 1 ) ( ) ~2 -) . * . 3 * 2 *1 - ( n + l ) * n = ?"+n
(n-l)(?z-2).'~3'2*1
( n T l ) ! - ( n + l )( n. n- . (l n) !- l ) !
or, simply,
- ( n + 1) n =
(71
(iii) -__
-
_ - n(n-
-
?L!
or, simply,
l)!
(n-l)!
-
%
~~
(n-l)! -
-
m-
iivi
(n - 1) !
(n-l)! ( n + 2) !
(n
+ 2 ) ( n+ 1) ' IZ
~~
* (11
-
- 1)!
BINOMIAL COEFFICIENTS
13.5. Compute:
(i)
( y),
' 121' ,
(ii)
'1
4
+
1
(iii)
/'
+n
1
-
-
1) * n
(n 2)(n
n2
123
+ 3n2 + 2%
(y).
Recall t h a t there a r e as m a n y factors in the numerator a s in the denominator.
13.6. Compute:
ti)
18)
\5/
(i)
(:),
(ii)
8.7.6.5.4
1.2.3.4-5
(;),
(iii)
(y).
= 56
Note t h at 8 - 5 = 3; hence we could also compute
a s follows:
(8)1
\5/
8.7.6
(ii)
Now 9 - 7 = 2; hence
(16p)
(iii) Now 1 0 - 6 = 4; hence
(140)
=
10.9.8'7
= 1.2.3.4
= 210.
13.7. Prove:
16!
16!
11
3
to
(?)
Multiply the first fraction by
6
3
and the second by
obtain the same denominator in both fractions; and then add:
):I(
+
6-16!
11*16!
=
6*5!.11!
6!.11.10!
-
6-16! 11*16!
6!*11!
+
-
-
6-16!
G!*11!
(6+11)*1G! 6 ! *11!
+ -1G 1! - 1 6 !
e l l !
17-* 16!
6! 11!
-
17!
6!*11!
CHAP. 131
T H E BINOMIAL C O EFFICIEN TS AND THEOREM
147
(The technique in this proof is similar t o tha t of the preceding problem and Example 4.1.)
n!
11 !
denomi1):
r + I)!
- r ) ! To obtain nt h-ersame
)+(:)
=
+l
nator in both fractions, multiply the first fraction by? and the second fraction by
Hence
Now
I
(r-
I/\ T - 1)
-
* (ti
-
+
r ! (n-
n--r+l'
r
-
1) n !
?-!(n-y+l)!
(n-T+
T. * 1 2 !
~~-
r!(n-r+l)!
+
r - n ! t (n-r+l)*n!
Y! (n-r+l)!
__
Y!
(iLtl)?i!
-
oz-r+m
-
T
-
Ir -- (12 - r - l)]* n !
r ! (n-rTfl)!
( n+ 1)!
! ( n- ?' 1 7
+
BINOMIAL THEOREM
13.9. Expand and simplify: (z + 3 7 ~ ) ~ .
-
+
+
+
13.13. Given (a b)6 = a6 6a5b+ 15a4bbL
+ 20a3b3 15a2b4-L 6ab5 b6.
and ( a b)'.
-
Expand
(a+ b)'
We could expand (u+ b)7 and ( a T b ) k using the binomial theorem. However, given the expansion of (a- b ) F , it is simpler to use Pascal's triangle to obtain the coefficients of (a+ b)' and
( a b ) h . F i r s t n r i t e down the coefficients of (a b)6 and then compute the next two rows of the
triangle
+
+
THE BINOMIAL COEFFICIENTS AND THEOREM
148
1
6
20
15
[CHAP. 13
1
6
Recall t h at each number is obtained by adding the two numbers above, e.g.
28 = 7 21, 56 = 35 21. From the above diagram we obtain
+
+
+ b)7
( a + b)8
(a
=
=
35 = 1 5 + 2 0 ,
+ 7a6b + 21ajbz + 35a4b3 + 35a3b4 + Zla'b5 + 7ab6 + b7
a6 + 8a7b + 28a6bz + 56a5b3 + 70a4b* + 56a3b5 + 28a2b6 + 8ab7 + b8
a7
+
13.14. (i) How does (a - b)" differ from (a b)"? (ii) Use Problem 13.11 t o obtain (2%- y2)5.
ii)
The expansion of ( a - b)n will be identical to the expansion of ( a + b)" except t h a t the signs
in ( a - b)" will alternate beginning with plus, i.e. terms with even powers of b will be plus
and terms with odd powers of b will be minus. This follows from the fa c t tha t ( - b ) r = br if
T is even and ( - b ) r = -br
if T is odd.
( i i ) By Problem 13.11,
+ 80x4~2+ 80x3~4+ 40x2y6 + 10xy6 +
j u s t alternate the signs in (2s + y*)?
= 32x5 - 80&42/2 + 8 0 ~ -~ 4Ox2y6
2 -~
9
+ 1 0 % -~ y'~O
+
(2% Y ' ) ~ = 3225
y10
To obtain (2%- y2)5,
(
13.15. Obtain and simplify the term in the expansion of (239 - y3)8 which contains do.
Nethod 1.
The general term in the expansion of (2&-y3)8
(;)
Hence the term containing
210
(222)s-r
has
(:)
(-y3)'
~16-2"=
is
26-r %16-2r
(- Y3)7
d o or 16 - 27 = 10 or
T
= 3, t h a t is, is the term
Method 2.
Each term i n the expansion contains a binomial coefficient, a power of 2x2 and a power of -y3;
hence first write down the following:
To obtain
x10,
the power of 2x2 must be 5; hence insert the exponent 5 to obtain
()
(2x215 (
3 3 )
Now the sum of the exponents must add up to 8, the exponent of
(-us) must be 3. Insert the 3 to obtain
( ) (2x95
(222-y3)8;
hence the exponent of
(-1/3)3
Last, the binomial coefficient must contain 8 above and either exponent below.
term is
(:)
(2x2)s (-y3)3
= - 1792x10~9
Thus the required
CHAP. 13!
THE BINOMIAL COEFFICIENTS A N D THEOREM
13.16. Prove:
24 =
Expand (1
16
149
=
+ 1)4 using the binomial
theorern:
71
In\
13.17. Prove the Binomial Theorem 13.2: (a+b)" =
\a"-'@.
r = ~\y)
The theorem is true f o r n = 1, since
+ b)n and prove it is true f o r
( a + b)(a + b).
( a + b ) [ a n + (Y)an-lb + ... +
We assume the theorem holds f o r ( a
(a
+ b)nrl
=
=
+ (:)an-rb~
+
...
+
( a + b)n+'.
(Tnl)
an-rtlbr-l
(7) a bn-' + bil 3
N o w the term in the product which contains br is obtained from
[(TnJ +
But, by Theorem 13.3,
Note that ( a
(91
(?'
+ b ) ( a + b)n is a polynomial of
degree n
+ 1 in b.
Consequently,
which was to be proved.
MULTINOMIAL COEFFICIENTS
13.18. Compute:
iii) ( 4 , 2 8 2 , O )
8!
- 8 * 7 * 6 * 6 * 4 * 3 * 2 * 1= 420
= 4!2!2!0! - 4*3*2*1*2*1*2.1*1
(iii) The expression
( ;\, 2 > h as no meaning since 5 + 3 + 2 + 2
5,
# 10.
br
150
[CHAP. 13
THE BINOMIAL C O E F F I C I E N T S AND T H E O R E M
13.19. Show that
,
Observe t h a t the expression
Hence
implicitly implies t h a t n, in2 = n or n2 = n - n,.
n!
n, ! (n- n,)!
n!
n l ! n,!
-
13.20. Find the term in the expansion of ( 2 ~ ~ - 3 x y ~ +which
z ~ ) ~contains
xll
and y4.
The general term of the expansion is
2a3;:3a
1'
(-3)b~byZb 2%
(a, b , c /
-
2 a ( - 3 ) b % 3 a C b @ b z2c
(a, b, c /
Thus the term containing x l l and ~4 h a s 3a+ b = 11 and 2 b = 4 or, b = 2 and a = 3.
since a - b c = 6, we have c = 1. Substituting in the above gives
+
Supplementary Problems
FACTORIAL
13.21.
Compute:
(I)
9!,
13.22.
Compute
Ii),
:14!
16'
(ii) l o ! ,
iiii) 11!
14!
(ii) -
8'
ll!' iiii), :l o !
13.23. Write in terms of factorials:
13.24.
Simplify,
lo!
(iv) - .
13!
ii) 2 1 * 2 3 - 2 2 - 2 1 ,
( n tl ) !
72
!
(i) 7
(ii,
,
__
n.
(Ti - 2 ) ! '
(iii)
(ii)
(n- 1) !
1
(iii) 42.
10.11-12 '
(iv)
(n-r
(?z -
+ lI )) !!.
-
BINOMIAL AND MULTINOMIAL COEFFICIENTS
13.26.
(7) (7)
=
+
iii) (z2- 3y)4,
(iii) (+a
Show iwithout using Theorem 13.3) t h a t
a
);(
BINOMIAL 1 S D MULTINOMIAL THEOREMS
13.28.
Expand and simplify:
(i) 12x
y"3,
+ 2b)5,
(iv) (2az- b)6.
13.29. The eighth row of Pascal's triangle is a s follows:
1
8
28
56
70
Compute the ninth and tenth rows of the triangle.
56
28
8
1
Also,
13.30.
[n)
0
Show t h a t
+
(7) + (;)
+
-
(;)
-
!\
13.31.
13.32.
151
T H E BINOMIAL C O E F F I C I E N T S A N D THEOREM
CHAP. 131
(c;
Show t h a t
(;)
+
...
+
/71\
,3/
(;I
T
3
.
+
(E)
=
2n.
t
(;)
=
0.
.
Find the term in the expansion of (2x2 - +2/3)6 which contains x*.
+
13.33. Find the term in the expansion of ( 3 ~ ~ 2 2) : which contains yG.
13.34.
Find the term in the expansion of (2.j.'
13.35.
Find the term in the expression of (ru - ~2
y?
-
+ 32)' which contains
T~
and z4.
- 2 z ) 6 which
~3
and us.
contains
Answers to Supplementary Problems
13.21. (i) 362,880
13.22.
(i) 240
13.23.
(i) 24!/20!
13.24. (i) n
13.25.
iiii 2184
13.27. (i) 504
13.28. (i)
-n
n2
+
iiii) l / [ ? t ( n l)(n
(iv) 15
iiiii 91
IV)
(iv)
(12
- r ) ( n- r 4-1)
(vi) 816
1140
+ 1 2 ~ ~ ; 1 / b~ y 4 + y6
7
(iii) a"/32
+ 81yq
+ 5 d b / 8 + 5u3b2 + 20a2b3 + 40ab4 + 32b5
(iv) 64a"
-
(ii) x8
+ 2)]
iiii) 180
iiii 210
8 d
(iii) 42!/41! or 7 ! / 5 !
iiii x(x - 1) =
( i i ) 35
(iv) 111716
(iii) 1/90
rii) 9 ! / 1 2 !
+1
(ij 10
(iii) 39,916,800
iii 3,628,800
-
1 2 . ~ 6-~5 4 x 4 ~-~108x2y3
192a"Jb
A
240a8b2 - 160a6b3 - 60a4b4 - l W b 5
8
1
13.29.
1
1
+ l)'].
13.30. Hint.
Expand (1
13.31. Hint.
Expand (1 - l ) n .
13.32. 7OxS.y"
13.33. 945x3yGz8
13.34.
-(105/4)~4~'~4
13.35.
-240~3~5~'
36
9
10
45
56
28
126
84
120
70
210
252
+
56
126
210
b6
28
84
9
36
120
1
8
45
1
10
1
[CHAP. 14
PE R 31 U T A T I 0 N S, 0RD E R E D SA 11P L E S
154
Theorem 14.4:
The number of permutations of n objects of which n, a r e alike, n2 a r e
alike, . . ., n,.are alike is
n!
nl! nz! * . * n,!
We indicate the proof of the above theorem by a particular example. Suppose we
want to form all possible 5 letter words using the letters from the word “DADDY”. Now
there are 5 ! = 120 permutations of the objects DI, A, D2, Di, Y, where the three D’s are
distinguished. Observe that the following six permutations
D1DrD3AY,D2DD1D8AY,
D3D1D2AY,D1D3DzAY, D2D3D1AY, and D3D2D1AY
produce the same word when the subscripts a r e removed. The 6 comes from the fact that
there are 3 ! = 3 . 2 . 1 = 6 different ways of placing the three D’s in the first three
positions in the permutation. This is true for each of the three possible positions in which
the D’s can appear. Accordingly there are
5 ! - 120
= 20
3! - 6
different 5 letter words that can be formed using the letters from the word “DADDY”.
~
Example 3.1:
How many 7 letter words can be formed using the letters of t h e word “ B E N Z E N E ” ?
We seek t h e number of permutations of 7 objects of which 3 a r e alike (the three E’s),
and 2 a r e alike ( t h e t w o N’s). By Theorem 14.4, there a r e
7!
3 ! 2!
~
such words.
Example 3.2:
-
7.6.5.4-3.2.1
3.2.1-2.1
= 420
How many different signals, each consisting of 8 flags hung in a vertical line, can
be formed from a set of 4 indistinguishable red flags, 3 indistinguishable white
flags, and a blue flag? We seek t h e number of permutations of 8 objects of which
4 a r e alike and 3 a r e alike. There a r e
8.7-6.5.4.3.2.1
L -different signals.
1!3 !
= 280
4.3*2.1*3.2*1
ORDERED SAMPLES
Many problems in combinatorial analysie and, in particular, probability, are concerned
with choosing a ball from a n urn containing n balls (or a card from a deck, or a person
from a population). When we choose one ball after the other from the urn, say T times,
we call each choice a n ordered sample of size r. We consider two cases:
(i) Sampling with replacewzeizt. Here the ball is replaced in the urn before the next
ball is chosen. Now since there are n different ways to choose each ball, by the
fundamental principle of counting there a r e
r times
&
n - n - n. . . n = rzl
different ordered samples with replacement of size
T.
(ii) Sampliizg without replucemcizt. Here the ball is not replaced in the urn after it
is chosen. Thus there are no repetitions in the ordered sample. I n other words,
a n ordered sample of size 7- without replacement is simply an r-permutation of the
objects in the urn. Accordingly, there a r e
:HAP. 141
P E R M U T A T I O S S . ORDERED SAMPLES
P(n, r ) = njn - 1 ) ( 1 2 - 2 ) . ’ .(n- r + 1) =
155
12 !
___
(12-r)!
different ordered samples of size r without replacement from a population of
n objects.
Example 4.1:
In how many ways can one choose three cards in succession from a deck of 52 cards
iii with replacement, lii’l without replacement? If each card is replaced in the
deck before the next card is chosen, then each card can be chosen in 52 different
ways. Hence there a r e
5 2 * . 5 2 . 5 2 = 523 = 140,608
different ordered samples of size 3 with replacement.
In case there is no replacement, then the first card can be chosen in 52 different
ways; the second card can be chosen in 51 different ways; and, lastly, the third
card can be chosen in 50 different ways. Thus there a r e
5 2 - 5 1 - 5 0 = 132,600
different ordered samples of size 3 without replacement.
Solved Problems
14.1. There are 4 bus lines between A and B ; and 3 bus lines between B and C.
(i) In how many ways can a man travel by bus from A to C by way of B ?
(ii) In how many ways can a man travel roundtrip by bus from A to C by way of B ?
(iii) In how many ways can a man travel roundtrip by bus from A to C by way of B,
if he doesn’t want to use a bus line more than once?
(i)
There a r e 4 ways to go from A to B and 3 ways to go from E to C; hence there a r e
4 * 3 = 1 2 ways to go from A t o C by way of B.
(ii)
Method 1.
There a r e 12 ways to go from A to C by way of B , and 1 2 ways to return. Hence there
a r e 12 * 12 = 144 ways t o travel roundtrip.
Method 2.
The man d l travel from A to B t o C to B to A . E n t e r these letters with connecting
arrows as follows:
A+E+C+B+A
Above each arrow write the nuinher of v a y s t h a t p a r t of the t r i p can be traveled; 4 ways
from A t o E . 3 ways from B to C , 3 ways from C to B , and 4 ways from B to A :
A%B$C-%’B%A
Thus there a r e 4 * 3 * 3 . 4 = 144 ways to travel roundtrip.
(iii) The man will travel from A t o E to C to E to A . E n t e r these letters with connecting arrows
as follows:
A+E+C+B+A
The man can travel 4 ways from A t o E and 3 ways from E t o C; b u t he can only travel
2 ways from C to B and 3 ways from R t o A since he doesn’t want to use a bus line more
than once. E n t e r these numbers above the corresponding arrows a s follows:
A h E % C % B & A
Thus there a r e 4 - 3 . 2 . 3 = 72 ways t o travel roundtrip without using t h e same bus line
more than once.
PERJIUTATIONS, ORDERED S A N P L E S
156
[CHAP. 14
14.2. If repetitions are not permitted, (i) how many 3 digit numbers can be formed from
the six digits 2, 3, 5 , 6, 7 and 9 ? (ii) How many of these are less than 400? (iii) How
many are even? (iv) How many are odd? (v) How many are multiples of 5 ?
0to represent a n a rbitra ry number, and then
In each case d r aw three boxes
write in each box the number of digits t h a t can be placed there.
(ij
The box on the left can be filled in 6 ways; following this, the middle box can be filled in
---
5 ways: and, lastly, t h e box on the right can be filled in 4 ways:
a r e 6 . 5 . 4 = 120 numbers.
I 6I I5 I I4 I.
Thus there
(ii) The box on the left can be filled in only 2 ways, by 2 or 3, since each number must be less
than 400: the middle box can be filled in 5 ways; and, lastly, the box on the right can be filled
I?]
T&i .
in 4 ways:
Thus there a r e 2 5 . 4 = 40 numbers.
(iii, The box on t h e right can be filled in only 2 ways, by 2 or 6, since the numbers must be even;
the box on the left can then be filled in 5 ways; and, lastly, the middle box can be filled in
4 ways:
Fl
PI. Thus there a r e
5 4 * 2 = 40 numbers.
(ivj The box on t h e right can be filled in only 4 ways, by 3, 5 , 7 o r 9, since the numbers must be
odd: the box on the left can then be filled in 5 ways; and, lastly, the box in the middle can
FI
be filled in 4 ways:
(v)
.
Thus there a r e 5 * 4 * 4 = 80 numbers.
The box on the r i g h t can be filled in only 1 way, by 5 , since the numbers must be multiples
of 5 : the box on the left can then be filled in 5 ways; and, lastly, the box in the middle can
be fiIled in 4 ways:
,
Thus there a r e 5 * 4 1 = 20 numbers.
14.3. Solve the preceding problem if repetitions are permitted.
(ij
Each box can be filled in 6 ways:
[q
.
Thus there a r e 6 6 6 = 216 numbers.
(ii) The box on the left can still be filled in only two ways, by 2 o r 3, and each of the others in
Thus there a r e 2 * 6 6 = 72 numbers.
6 ways:
Lf._l .
(iii) The box on the r i g h t can still be filled in only 2 ways, by 2 or 6, and each of the others in
6 ways:
. Thus there a r e 6 * 6 2 = 72 numbers.
-
F1
(iv) The box on the right can still be filled in only 4 ways, by 3, 5, 7 or 9, and each of the other
Thus there a r e 6 * 6 * 4 = 144 numbers.
boxes in 6 ways:
El (41.
(vj
The box on t h e left can still be filled in only 1 way, by 5, and each of the other boxes in
6 ways:
Fl
.
Thus there a r e 6 * 6 1 = 36 numbers.
14.4. In how many ways can a party of 7 persons arrange themselves
(i) in a row of
7 chairs? (ii) around a circular table?
(ij
The seven persons can ar r ange themselves in a row in 7 6 . 5 4 * 3 2 1 = 7 ! ways.
(ii) One person can s i t at any place in the circular table. The other six persons can then a rra nge
themselves in 6 * 5 4 3 * 2 1 = 6! ways around the table.
This is a n example of a circular permutation. I n general, n objects can be arranged in
a circle in (n - l)(n- 2) . . 3 2 1 = (n- l)!ways.
-
9
157
PERMUTATIOKS, ORDERED S A M P L E S
CHAP. 141
14.5. (i) In how many ways can 3 boys and 2 girls sit in a row? (ii) In how many ways
can they sit in a row if the boys and girls a r e each to sit together? (iii) I n how
many ways can they sit in a row if just the girls are to sit together?
(9
The five persons can sit in a row in 5 ~ 4 . 3 . 2 . 1 = 5! = 120 ways.
(ii) There a r e 2 ways to distribute them according to sex: BBBGG o r GGBBB. I n each case the
boys can sit in 3 - 2 . 1 = 3! = 6 ways, and t h e girls can sit in 2 * 1 = 2! = 2 ways. Thus,
altogether, there a r e 2 * 3 ! 2 ! = 2 6 * 2 = 24 ways.
-
-
(iii) There a r e 4 ways to distribute them according to sex: GGBBB, BGGBB, BBGGB, BBBGG.
Note t h a t each way corresponds t o the number, 0, 1, 2 o r 3, of boys sitting to the left of the
girls. In each case, the boys can sit in 3 ! ways, and the girls in 2! ways. Thus, altogether,
there a r e 4 * 3 ! * 2 != 4 . 6 - 2 = 48 ways.
14.6. Solve the preceding problem in the case of r boys and s girls.
left in factorials.)
(i)
The
P
+s
persons can s i t in a row in
(T
(Answers are to be
+ s) ! ways.
(ii) There a r e still 2 ways to distribute them according to sex, the boys on the left or the girls
on the left. In each case the boys can sit in r ! ways and the girls in s! ways. Thus,
altogether, there a r e 2 * r ! s! ways.
-
+
1 ways t o distribute them according t o sex, each way corresponding t o the
(iii) There a r e r
number, 0,1,. . . , r , of boys sitting to the left of the girls. In each case the boys can sit in
r ! ways and the girls in s ! ways. Thus, altogether, there a r e ( r 1)* r ! * s! ways.
+
14.7. How many distinct permutations can be formed from all the letters of each word:
(i) them, (ii) that, (iii) radar, (iv) unusual, (v) sociological.
(i)
4 ! = 24, since there a r e 4 letters and no repetition.
(ii)
4r
= 12,
2!
(iii)
5!
= 30,
2! 2!
since there a r e 4 letters of which 2 a r e t .
since there a r e 5 letters of which 2 a r e r and 2 a r e a ,
7'
(iv) 3! = 840, since there a r e 7 letters of which 3 a r e u.
12!
(") 3 ! 2 ! 2 ! 2 ! ' since there a r e 12 letters of which 3 a r e 0 , 2 a r e c, 2 a r e i, and 2 a r e 1.
14.8. How many different signals, each consisting of 6 flags hung in a vertical line, can
be formed from 4 identical red flags and 2 identical blue flags?
This problem concerns permutations with repetitions.
there a r e 6 flags of which 4 a r e red and 2 a r e blue.
6!
There a r e - = 15 signals since
4! 2 !
14.9. In how many ways can 4 mathematics books, 3 history books, 3 chemistry books and
2 sociology books be arranged on a shelf so that all books of the same subject are
together?
F i r s t the books must be arranged on the shelf in 4 units according t o subject matter:
0
a.
The box on the left can be filled by a n y of the four subjects; the next by
a n y 3 subjects remaining; the next by any 2 subjects remaining; and the box on the right by the
l a s t subject:
a
.
Thus there a r e 4 . 3 - 2 - 1 = 4! ways to a r r a n g e the books
on t h e shelf according to subject matter.
In each of the above cases, the mathematics books can be arranged in 4 ! ways, the history
books in 3 ! ways, the chemistry books in 3 ! ways, and the sociology books in 2 ! ways. Thus,
altogether, there a r e 4 ! 4 ! 3 ! 3 ! 2 ! = 41,472 arrangements.
[CHAP. 1 4
PERM U TAT I 0 N S, ORDERED SAM P L E S
158
14.10. Solve the preceding problem if there are r subjects, A,,A,,
s,, s2,. . . ,s, books, respectively.
. . . , A 7 , containing
Since there a r e r subjects, the books can be arranged on the shelf in r ! ways according to
subject matter. In each case, the books in subject A , can be arranged in sl! ways, the books in
subject A ? in s,! ways, . . ., the books in subject A , in s,! ways. Thus, altogether, there a r e
r ! s1 ! s,! . . . s, ! arrangements.
14.11. (i) In how many ways can 3 -4mericans, 4 Frenchmen, 4 Danes and 2 Italians be
seated in a row so that those of the same nationality sit together?
(ii) Solve the same problem if they sit a t a round table.
(i)
The 4 nationalities can be arranged in a row in 4! ways. In each case, the 3 Americans can
be seated in 3 ! ways, the 4 Frenchmen in 4! ways, the 4 Danes in 4! ways, and the 2 Italians
in 2 ! ways. Thus, altogether, there a r e 4! 3 ! 4! 4! 2 ! = 165,888 arrangements.
(ii)
The 4 nationalities can be arranged in a circle in 3! ways (see Problem 14.4 on circular
permutations). In each case, the 3 Americans can be seated in 3 ! ways, t h e 4 Frenchmen in
4! ways, the 4 Danes in 4! wags, and the 2 Italians in 2! ways. Thus, altogether, there a r e
3 ! 3 ! 4 ! 4 ! 2 ! = 41,472 arrangements.
14.12.Find the total number of positive integers that can be formed from the digits
1, 2, 3 and 4 if no digit is repeated in any one integer.
Note t h a t no integer can contain more than 4 digits. Let sl, sq, ss and s4 denote the number
of integers containing 1 , 2, 3 and 4 digits respectively. We compute each si separately.
Since there a r e 4 digits, there a r e 4 integers containing exactly one digit, i.e. s1= 4 . Also,
since there a r e 4 digits, there a r e 4 * 3 = 1 2 integers containing two digits, i.e. s2 = 12. Similarly,
there a r e 4 * 3 2 = 24 integers containing three digits and 4 * 3 * 2 * 1 = 2 4 integers containing four
digits, i.e. s:]= 24 and s,$= 24. Thus, altogether, there a r e s1 s2 sg sq = 4 1 2 24 24 = 64
integers.
-
+ + +
+ + +
14.13. Suppose an urn contains 8 balls. Find the number of ordered samples of size 3
(i) with replacement, (ii) without replacement.
(i)
Each ball in the ordered sample can be chosen in 8 ways; hence there are 8 8 8 = 83 = 512
samples with replacement.
(ii,
The first ball in the ordered sample can be chosen in 8 ways, the next in 7 ways, and the last
in 6 ways. Thus there a r e 8 * 7 * 6 = 496 samples without replacement.
14.14. Find n if
(i)
(i) P(n, 2) = 72, (ii) P(n, 4) = 42P(?z, 2), (iii) 2P(n, 2)
P ( n , 2 ) = n(n-1) = n2--n; hence n 2 - n = 72 or n * - n - 7 2
+ 50 = P(2n, 2).
= 0 or ( n - 9 ) ( n + 8 )
= 0.
Since n must be positive, the only answer is n = 9.
(ii) P ( n , 4) = n ( n - l ) ( n - 2)(n - 3) and P(n, 2) = n(n- 1). Hence
n ( n - l)(n- 2)(n - 3) = 42n(n - 1 )
or
+6
n2 - 5%
= 42
or
n2
or, if n # 0, # 1, (n - 2)(n - 3) = 42
- 5% - 36 = 0
or
( n - 9 ) ( n + 4) = 0
Since n must be positive, the only answer is n = 9.
(iii) P ( J z2, ) = ?z(n- 1) = n2- ?z and P ( h , 2) = 2742% - 1) = 4n2 - 2%. Hence
2 ( n Z - n) t 50 = 4nz - 212
or
2x2
- 2%
+ 50
= 4n* - 2%
Since n must be positive, the only answer i s n = 5.
or
50 = 2n2
or
n2
= 25
PERMUTATIONS, ORDERED SAMPLES
CHAP. 141
14.15. Find the relationship between f 12
159
and p(n, r ) .
?'
Recall t h a t
(I)
P(n, T)
n!
and
= ,).! ( n - r ) !
=
n!
________
(n- r ) !
~
P(?i,T ) =
r!
n!
-.(n-r)!
T !
nl
(n-r)!.
-
r!
Multiply P ( n , r ) by
12
!
r ! (?Z--T)!
$ to
obtain
= r!(:)
Supplementary Problems
14.16.
(i) How many automobile license plates can be made if each plate contains 2 different letters
followed by 3 different digits? (ii) Solve the problem if the first digit cannot be 0.
14.17.
There a r e 6 roads between A and B and 4 roads between B and C.
(i)
In how many ways can one drive from A to C by way of B ?
(ii) In how many ways can one drive roundtrip from A to C by way of B ?
(iii) In how many ways can one drive roundtrip from A to C without using the same road more
than once?
14.18.
Find the number of ways in which 6 people can ride a toboggan if one of three must drive.
14.19.
(i)
Find the number of ways in which five persons can sit in a row.
(ii) How many ways a r e there if two of the persons insist on sitting next t o one another?
14.20.
Solve the preceding problem if they sit around a circular table.
14.21.
Find the number of ways in which a judge can award first, second and third places in a contest
with ten contestants.
14.22.
(i) Find the number of four letter words t h a t can be formed from the letters of the word HISTORY.
(ii) How many of them contain only consonants? (iii) How many of them begin and end in a
consonant? (ivj How many of them begin with a vowel? (vj How many contain t h e letter Y ?
(vi) How many begin with T and end in a vowel? (vii) How many begin with T and also contain S ?
(viii) How many contain both vowels?
14.23.
How many different signals, each consisting of 8 flags hung in a vertical line, can be formed from
4 red flags, 2 blue flags and 2 green flags?
14.24.
Find the number of permutations t h a t can be formed from all the letters of each word: (i) queue,
(ii) committee, (iii) proposition, (iv) baseball.
14.25.
(i)
Find the number of ways in which 4 boys and 4 girls can be seated in a row if t h e boys and
girls a r e to have alternate seats.
(ii) Find the number of ways if they sit alternately and if one boy and one girl a r e t o sit in
adjacent seats.
(iii) Find the number of ways if they sit alternately and if one boy and one girl must not sit in
adjacent seats.
14.26.
Solve the preceding problem if they sit around a circular table.
PERM U T A T I 0 N S, ORDERED SARI PLE S
160
[CHAP. 14
14.27.
An urn contains 10 balls. Find the number of ordered samples ( i ) of size 3 with replacement,
(ii, of size 3 without replacement, fiii) of size 4 with replacement, (iv) of size 5 without replacement.
14.28.
Find the number of ways in which 5 large books, 4 medium-size books and 3 small books can be
placed on a shelf so t h a t all books of the same size a r e together.
14.29.
Consider all positive integers with 3 different digits. (Note t h a t 0 cannot be the first digit.)
(i, How many a r e greater than 7001 iii) How many a r e odd? (iii) How many a r e even? (iv) How
many a r e divisible by 5 ?
14.30.
(i) Find the number of distinct permutations t h a t can be formed from all of the letters of the word
E L E V E N . fii) How many of them begin and end with E l f i i i ) How many of them have the 3 E's
together? (iv) How many begin with E and end with N ?
Answers to Supplementary Problems
14.16.
= 468,000
(i) 2 6 . 2 5 . 1 0 . 9 - 8
iii) 2 6 . 2 5 . 9 . 9 . 8
= 421,200
14.17.
14.18.
3 . 5 * 4 . 3 . 2 * 1 = 360
14.19.
(i) 5 ! = 120
(ii) 4 . 2 ! * 3 ! = 48
14.20.
(i) 4! = 24
(ii) 2 ! 3 ! = 1 2
14.21.
1 0 . 9 . 8 = 720
14.22.
14.23.
9'
-
14.24.
(ii)
14.25.
(i) 2 * 4 ! * 4 ! = 1152
14.26.
(i) 3 ! * 4 ! = 144
14.27.
(i) 10 * 10 10 = 1000
(iii) 10 * 10 1 0 10 = 10,000
(ii, 1 0 ' 9 . 8 = 720
(iv) 10 * 9 * 8 7 * 6 = 30,240
2 ! 2! 2!
3 ! 5 ! 4 ! 3 ! = 103,680
14.29.
(i)
(iii)
~
2! 3! 2!
(ii) 2 . 7 * 3 ! . 3 ! = 504
(ii) 2 * 3 ! * 3 ! = 72
14.28.
3 ' 9 . 8 = 216
45,360
= 1,663,200
(iii) 1152-504
fiii) 144-72
8!
(iv) _ _ = 5040
2! 2! 2!
= 648
= 72
(ii) 8 ' 8 . 5 = 320
(iii, 9 ' 8 . 1 = 72 end in 0, and 8 ' 8 . 4 = 256 end in the other even digits; hence, altogether,
7 2 + 2 5 6 = 328 a r e even.
(iv) 9.8.1 = 72 end in 0, and
divisible by 5.
14.30.
G!
(i) 7 = 120
3.
(ii) 4! = 24
8.8.1 = 64 end in 5; hence, altogether, 7 2 + 6 4 = 136 a r e
iiii) 4 * 3! = 24
(iv)
4!
= 12
2!
-
Chapter
Combinations, Ordered Partitions
Theorem 15.1:
C(n,T ) =
P(n, r ) =
~
r!
abc
a b c , a c b , bac, bca, c u b , c b a
abd
a b d , a d b , b a d , bda, d a b , dba
acd
a c d , a d c , c a d , c d a , d u e , dca
bcd
bcd, bdc, c b d , c d b , d b c , c k b
n!
T ! (n- T ) !
161
15
162
COMBINATIOKS, ORDERED PARTITIONS
n!
Recall that the binomial coefficient
We shall use C(n,r) and
Example 1.3:
(:)
[CHAP. 15
I
. hence
I
interchangeably.
How many committees of 3 can be formed from 8 people?
essentially, a combination of the 8 people taken 3 a t a time.
Each committee is,
Thus
different committees can be formed.
Example 1.4:
A f ar mer buys 3 cows, 2 pigs and 4 hens from a man who h a s 6 cows, 5 pigs and
8 hens. How many choices does the fa rm e r have?
(i
The f ar mer can choose the cows in /6' ways, the pigs in
ways, and
\3
the hens in
ways. Hence altogether he can choose the animals in
1
PARTITIONS AND CROSS-PARTITIONS
Recall (Page 60) that a partition of a set X is a subdivision of X into subsets which
are disjoint and whose union is X, i.e. such that each a E X belongs to one and only one
of the subsets. In other words, the collection {A,,A,, . , . , A m }of subsets of X is a partition of X iff
(ii) foranyA,,A,, either A l = A 3 or A t n A j = @
(i) X = A , u A 2 u
UA,;
The subsets of a partition are called cells.
9
Example 2.1:
.
.
Consider the following classes of subsets of X = { 1 , 2 ,
6)
. . ., 8,9}:
[{I, 3,51, { 2 , 6 } , {4,8,931
(ii) [{I,3,5), { 2 , 4 , 6 , 8 ) , { 5 , 7 , 9 ) ]
(iii) [{I,3,51, { 2 , 4 , 6 , 8 } , (7,911
Then (i) is not a partition of X , since 7 6 X but 7 does not belong to a ny cell in (i).
Also (ii) is not a partition of X since 5 € X and 5 belongs t o both { 1 , 3 , 5 } and
{5,7,9}. On the other hand, (iii) is a partition of X since each element of X
belongs t o exactly one cell.
Suppose {A,,A,, ..., A,.} and {B,,B,, , . .,Bs}
are both partitions of the same set X.
Then the class of intersections {At n B , } form a new partition of X called the cross-partition.
Example 2.2:
Let { A , B, C, D} be the partition of the undergraduate students at a university
into freshmen, sophomores, juniors and seniors, respectively; and let {M, F } be
t h e partition of the students into maies and females, respectively. The crosspartition consists of the following:
A n M : male freshmen
A n F : female freshmen
B n M : male sophomores
B n F : female sophomores
C n M : male juniors
D n M : male seniors
C n F : female juniors
D n F: female seniors
CHAP. 151
163
COMBINATIONS, ORDERED P A R T I T I O N S
ORDERED PARTITIONS
Suppose an urn A contains seven balls numbered 1 through 7. We compute the
number of ways we can draw, first, 2 balls from the urn, then 3 balls from the urn, and
lastly 2 balls from the urn. In other words. we want to compute the number of ordered
partitions
(AI, A2, A3)
of the set of 7 balls into cells A , containing 2 balls, A, containing 3 balls and A , containing
2 balls. We call these ordered partitions since we distinguish between
({I, 21, {3,4,5), ( 6 7 ) ) and
(C6,7), {3,4,51, {1,21)
each of which determines the same partition of A .
Now we begin with 7 balls in the urn, so there are
(3
ways of drawing the first 2 balls,
i.e. of determining the first cell A , ; follon.ing this, there are 5 balls left in the urn and so
mays of drawing the 3 balls, i.e. of determining the second cell A , ; finally,
/c,
\
ways of determining the last cell A , .
there are 2 balls left in the urn and so there are
Hence there are
5 . 4 . 3 2.1
7.6
210
-
different ordered partitions of A into cells A, containing 2 balls, A, containing 3 balls,
and A, containing 2 balls.
We state the above result in its general form as a theorem.
Theorem 15.2:
Let A contain n elements and let n,, n2,. . . , nr be positive integers with
12,1?1,4 ..
n? = 12. Then there exist
-+
(;$n
i 2 n 1
)( n
nz
). . .
( n - nl - n2 -
* *
- -nr-l
nr
-
1
different ordered partitions of A of the form (A,,A,, . . . , A r ) where A ,
contains n, elements, A , contains n, elements, . . ., and A , contains nT
elements.
Now observe that
7!
-.-.2!5!
5!
2!
3!2! 2 ! 0 !
-
7!
2!3!2!
since each numerator after the first is cancelled by the second term in the denominator
of the previous factor. Similarly,
(nnl)(i n - nl ) ( n
- nl - nz
n2
-
-
n!
n3
-,
n,!(n- n l ) !* nz!( n- n1- n2)!. n3!( n - n, - n2- n3)!
n!
nl ! n,! ns! . - n,!
...
~
n,!
The above computation together with the preceding theorem gives us our next theorem.
CO R I B I N A T I 0 N S, 0 RD E R E D P A R T IT I 0 K S
164
Theorem 15.3:
[CHAP. 15
Let A contain 71 elements and let nl, 712, . . . , n , be positive integers with
111
n2 . . . n, = n. Then there exist
+ +
+
n!
n l !na! ns! . . . n,!
different ordered partitions of A of the form (A1,Az, . . .,A,) where A1 contains nl elements, A>contains n2 elements, . . . , and Ar contains nr elements.
Example 3.1:
In how many ways can 9 toys be divided between 4 children if t h e youngest child
is to receive 3 toys and each of t h e others 2 toys?
We wish to find the number of ordered partitions of t h e 9 toys into 4 cells
containing 3, 2, 2 and 2 toys respectively. By Theorem 15.3, there a r e
3 ! 2 ! 2 ! 2!
such ordered partitions.
= 2520
Solved Problems
COMBINATIONS
15.1. In how many ways can a committee consisting of 3 men and 2 women be chosen
from 7 men and 5 women?
ways, and the 2 women can be chosen from the
7.6.5 5-4
350 ways.
ways. Hence the committee can be chosen in
The 3 men can be chosen from the 7 men in
5 women in(')
2)
Find the number of ways 4 balls can
be drawn from the bag if (i) they can be of any color, (ii) 2 must be white and
2 black, (iii) they must all be of the same color.
3.5.2. A bag contains 6 white balls and 5 black balls.
(ii
The 4 balls (of a n y color) can be chosen from the 11 balls in the urn in
330 ways.
(ii)
The 2 white balls can be chosen in
ways, and the 2 black balls can be chosen in
'
150 ways of drawing 2 white balls and 2 black balls.
Thus there a r e
(iii) There a r e
(:)
= 15 ways of drawing 4 white balls, and
balls. Thus there a r e 15
)"
= 5 ways of drawing 4 black
+ 5 = 20 ways of drawing 4 balls(4,of the same color.
15.3. A delegation of 4 students is selected each year from a college to attend the National
Student Association annual meeting. (i) In how many ways can the delegation be
chosen if there are 12 eligible students? (ii) In how many ways if two of the
eligible students will not attend the meeting together? (iii) In how many ways if
two of the eligible students are married and will only attend the meeting together?
(y)
12 11 ' 10 * 9 = 495 ways.
1.2.3-4
(i,
The 4 students can be chosen from the 12 students in
(ii,
Let A and B denote the students who will not attend the meeting together.
=
165
COMBINATIONS, O R D E R E D PARTITIONS
CHAP. 151
Method 1.
If neither A nor B is included, then the delegation can be chosen in
210 ways.
+ 240
10-9*8.7
= 1.2.3.4
-
If either A o r B , but not both, is included, then the delegation can be chosen in
2 * ( y ) = 2 . 11.02' 9. .38
210
(140)
1
240 ways.
Thus, altogether, the delegation can be chosen in
= 450 ways.
Method 2.
If A and B a r e both included, then the other 2 members of the delegation can be chosen in
1'"
= 45 wags.
Thus there a r e 495 - 45 = 450 ways t h e delegation can be chosen if
(2,
A and B a r e not both included.
(iii) Let C and ll denote the married students. If C and D do not go, then t h e delegation can be
chosen in
(120\'
/""
\4:
210 ways.
~
If both C and D go, then the delegation can be chosen in
~
= 45 wags.
Altogether, the delegation can be chosen in 210 t 45 = 255 wags.
15.4. There are 12 points A , B, . . . in a given plane, no three on the same line. (i) How
many lines are determined by the points? (ii) How many of these lines pass through
the point A ? (iii) How many triangles are determined by the points? (iv) How
many of these triangles contain the point A as a vertex?
Since two points determine a line, there a r e
('2")
=
1 2 * 11
1
.
2=
66 lines.
To determine a line through A , one other point must be chosen: hence there a r e 11 lines
through A .
12 - 1 2 * 1 1 * 1 0
Since three points determine a triangle, there a r e ( 3 ) 1.2.3
= 220 triangles.
Method 1.
To determine a triangle with vertex A , two other points must be chosen; hence there a r e
('2' ) =
= 55 triangles with A a s a vertex.
Method 2.
There a r e
(131 )
,
= 165 triangles without A a s a vertex.
=
Thus 220 - 165 = 55
I
of the triangles do have A a s a vertex.
15.5. A student is to answer 8 out of 1 0 questions on an exam. (i) How many choices
has he? (ii) How many if he must answer the first 3 questions? (iii) How many
if he must answer at least 4 of the first 5 questions?
(i)
The 8 questions can be selected in
lo - lo
( 8 ) - (2)
-
= 45 ways.
(ii) If he answers the first 3 questions, then he can choose the other 5 questions from the last
7 questions in '7
(5)
-
7
-
(2)
-
7.6
=
1.2
21 ways.
(iii) If he answers all the first 5 questions, then he can choose the other 3 questions from the last
5 in
(i)
= 10 ways.
On the other hand, if he answers only 4 of the first 5 questions, then
he can choose these 4 in
from the last 5 in
25 ways.
5(i 4)
= (5)
= 5 ways, and he can choose the other 4 questions
(i) = (f)
= 5 ways; hence he can choose the 8 questions in 5 . 5 =
Thus he h a s a total of 35 choices.
166
[CHAP. 15
COMBINATIOXS, ORDERED PARTITIONS
15.6. How many diagonals has an octagon?
An octagon has 8 sides and 8 vertices. Any two vertices determine either a side or a diagonal.
= 28 sides plus diagonals.
28
-
But there a r e 8 sides; hence there a r e
8 = 20 diagonals.
15.7. How many diagonals has a regular polygon with ?z sides?
sides has, also, n vertices. Any two vertices determine either a
’n\
n(n-1)
n ( n - 1)
sides plus diagonals. B u t there
side or a diagonal. Thus there a r e 1
1.2
:
2
/
a r e 71 sides: hence there a r e
The regular polygon with
n(n- 1)
~2
71
n =
722
-n
2
2n
2
-
n2
-
- 3n
2
-
’
3
diagonals
2
15.8. Which regular polygon has the same number of diagonals as sides?
The regular polygon with 71 sides has, by the preceding problem,
we seek the polygon of n sides f o r which
n2
- 3n - n
2
or
n1 - 3 7 ~ =
2n
or
n2
- 5n = 0
nz - 3 n
7
diagonals.
or
Thus
n(n-5) = 0
Since 71 must be a positive integer, the onIy answer is n = 5. In other words, the pentagon is the
only regular polygon with the same number of diagonals a s sides.
15.9. A man is dealt a poker hand (5 cards) from an ordinary playing deck. I n how many
ways can he be dealt (i) a spade flush (5 spades), (ii) an ace high spade flush
(5 spades with the ace), (iii) a flush ( 5 of the same suit), (iv) a n ace high full house
(3 aces with another pair), ( ~ 7 ) a full house (3 of one kind and 2 of another),
(vi) 3 aces (without the fourth ace or another pair), (vii) 3 of a kind (without another
pair), (viii) two pair?
13) - 13 * 1 2 - 1 1 1 0 - 9 = 1287 ways.
The 5 spades can be dealt from the 13 spades in
(i)
( 5 ,/ 1.2*3*4.5
(ii)
If he receives the spade ace, then he can be dealt the other 4 spades from the remaining
(iii)
There a r e 4 suits, and a flush from each suit can be dealt (see (i)) in
flush can be dealt in 4 .
(iv)
(7)
(153)
ways; hence a
= 5148 ways.
The 3 aces can be dealt from the i aces in
(i) = (:>
= 4 ways.
in 12 ways, and the 2 cards of this pair can be dealt in
ways.
The pair can be selected
Thus the ace high full
house can be dealt in 4 12 {4) = 4 12 4 * 3 = 288 ways.
\2
1.2
0-
(v)
One kind can be selected in 13 ways and 3 of this kind can be dealt in
kind can be selected in 12 ways and 2 of this kind can be dealt in
house can be dealt in
13
* 12 *
= 3744 ways
($
(i)
fobserve, by (iii), t h a t a flush is more common than a full house.)
c1
ways; another
ways.
Thus a full
COMBINATIOKS, ORDERED PARTITIOKS
CHAP. 151
(vi)
(:)
kinds in ( y ) ways; and
The 3 aces can be selected from 4 aces in
the remaining 12
167
ways. The 2 other cards can be selected from
1 card in each kind can be dealt in 4 ways.
/
Thus a poker hand of 3 aces can be’ dealt in
(vii)
One kind can be selected in 13 ways: and 3 of this kind can be dealt in
2 other kinds can be selected from the remaining 12 kinds in
(43)
ways. The
ways; and 1 card in each
\ 2 /
kind can be dealt in 4 ways. Thus a poker hand of 3 of a kind can be dealt in
I ” ~ ‘
13‘ ways; and 2 of each kind
\ 2
ways. Another kind can be selected in 11 mays; and 1 of this kind in
(viii) Two kinds (for the pairs) can be selected from the 13 kinds in
can be dealt in
(:I
4 ways. Thus a poker hand of t w o pair can be dealt in
15.10. Consider 4 vowels (including a) and 8 consonants (including b ) .
(i)
How many 5 letter “words” containing 2 different von~elsand 3 different consonants can be formed from the letters?
(ii) How many of them contain b ?
(iii) How many of them begin with b ?
(iv) How many of them begin with a ?
(v) How many of them begin with a and contain b?
(i)
(;)
The 2 vov,els can be selected from the 4 vowels in
ways, and the 3 consonants can be
/8‘
selected from the 8 consonants in
\ 3 1 ways. Furthermore, each 5 letters can be arranged
in a row f a s a “word”) in 5 ! ways. Thus we can form
(:;
(:)
*
5 ! = 6 56
*
120 = 40,320 words
i\
ways. However, since b is one of the consonants, the
‘2
other 2 consonants can be selected from the remaining 7 consonants in
ways. Again each
5 letters can be arranged as a word in 5! ways. Thus we can form
(ii) The 2 vowels can still be selected in
(i) (2‘)
*
*
5 ! = 6 * 21 * 120 = 15,120 Ivords containing b
(z)
(iii) The 2 vowels can be selected in ( 4 ) ways, and the other 2 consonants can be selected in
\2/
ways. The 4 letters can be arranged, following b , in 4 ! ways. Thus we can form
(i) (2’) -
-
4! = 6 21 * 24 = 3024
words beginning with b
168
COMBINATIONS, ORDERED PARTITIONS
[CHAP. 15
( i d The other vowel can be chosen in 3 ways, and the 3 consonants can be chosen in
The 4 letters can be arranged, following a , in 4! ways.
3 . 5 6 . 2 4 = 4032 words beginning with a.
(vi
(z)
Thus we form
-
(:)
3 /8)
\3/
ways.
*
4!
=
The other vowel can be chosen in 3 ways, and the 2 other consonants can be chosen in
3
The 4 letters can be arranged, following a, in 4! ways.
xays.
(7)
2,
- 4! = 3 - 2 1 - 2 4
Thus we can form
= 1612 words beginning with a and containing b .
15.11. How many committees of 5 Tvith a given chairman can be selected from 12 persons?
The chairman can be chosen in 12 ways and, following this, the other 4 on the committee can
be chosen from the 11 remaining in
such committees.
(y)
ways.
Thus there a r e 1 2 .
(121)
= 12.330 = 3960
15.12. Find the number of subsets of a set X containing ?z elements.
Method 1.
The number of subsets of X with
T
subsets of X. The above sum (see Problem 13.30, Page 151) is equal to 2n, i.e. there a r e 2" subsets of X.
Method 2.
There a r e two possibilities for each element of X : either it belongs to the subset o r it doesn't;
hence there a r e
times
2 . 2 . . . . a 2 = 2n
ways t o form a subset of X , i.e. there a r e 2n different subsets of
X.
15.13. In how many ways can a teacher choose one or more students from six eligible
students ?
Method 1.
By the preceding problem, there a r e 2s = 64 subsets of the set consisting of the six students.
However, the empty set must be deleted since one o r more students a r e chosen. Accordingly, there
a r e 26 - 1 = 64 - 1 = 63 ways to choose the students.
Method 2.
Either 1 , 2 . 3 , 4 , 5 or 6 students a r e chosen. Hence the number of choices is
(F)
+
4-
(i) + (:)
+
(i)
+
(z)
=
6
+
15
+
20
+
15
+
6 i1
=
63
15.14. In how many ways can three or more persons be selected from 12 persons?
There a r e 21a - 1 = 4096 - 1 = 4096 ways of choosing one o r more of the twelve persons.
Now there a r e
( y )+ (7)
Hence there a r e 4096-78
= 12
+ 66 = 78 ways of choosing one or two of
= 4018 ways of choosing three or more.
the twelve persons.
CHAP. 151
COMBINATIONS, ORDERED PARTITIONS
PARTITIONS AND CROSS-PARTITIONS
15.15. Which of the following are partitions of
(i)
169
X = {a,b, c, d , e, f , g } ?
[{a.0,c:, "Zj, if,9)]
(ii) [{a,
0,c ) , {c, d , e}, { f , s } ]
(iii) [{a,
b , c ) , ( d , e ) , { f ,9 } ]
(i)
This is not a partition of X since e E X, but e does not belong to any cell.
(ii) This is not a partition of X since c E
A' and belongs to both
(iii) This is a partition of X . Each element of
T
{ a , 6, c)
and {c, d, e } .
belongs t o exactly one cell.
15.16. Let cA = { A , , A , , . . . , A , , L }and 49 = ( B , ,B,, . . . , B J be partitions of the same
set X . Show that the cross-partition C = {Ai n B I : A i E d ,
B j E CB} is also a
partition of X.
Let z E X. Then x belongs to some d L o
and to some Bi,, since 04 and '73 a r e partitions of X .
Thus x E A , o n E , o and so belongs t o a member of the cross-partition.
On the other hand, suppose 3: E A ; , , I ~ I B and
,~
AilnB,il. Then z E Aio and A , , , whence
A . = A i since d is a partition of X . Similarly, B . = B j l . Accordingly, A i O n B j ,= A , l n B , l
'0
1
30
and so C is a partition of X .
13.17. Let X = (1,2,3,4,5,6,7,8}. Find the cross-partition of each pair of partitions of X :
(i) [{I,2,3, 4 } , .(5,6,7,8>1 and [{I,2 , 8 ) , ( 3 , 4 , 5 , 6 , 7 ) ]
(ii) [{I,2 , 3 , 4 ) , { 5 , 6 , 7 , 8 } ] and [il,
2 ) , { 3 , 4 , 5 } , {6,7,8)]
To obtain the cross-paTAition, intersect each set of one partition with each set of the other.
(i)
{1,2,3,4]
C5,6,7,8> n {1,2, 8) = {8)
n {l,2 , 8 } = {1,2} ;
{ 1 , 2 , 3 , 4 ) n{3,4,5,6,7}
= {3,4};
Thus the cross-partition is
{5,6,7,8}n{3,4,5,6,7]
= {5,6,7}
[{l,
Z}, { 3 , 4 } , {8), {5,6,7}].
(ii) (1,2,3,4$ n {l, 2 ) = {1,2} ;
= {3,4};
n {6,7,8> = P ;
{5,6,7,8)
n U , 2 } = 2' )
= (5)
{1,2,3,4}n(3,4,5}
{5,6,7,8)n{3,4,5}
{1,2,3,4:
( 5 , 6, 7, 8) n {6,7,8} = (6, 7,8}
Thus the cross-partition is
[{l, 2 } , {3,4}, {5}, {6,7,8),
81
or, simply,
[{I, a}, {3,4), (51, {6,7,8}]
Remark: We do not distinguish partitions which differ by t h e empty set.
15.18. Let X denote the domain of a proposition P ( p , q, 7') of three variables:
X = {TTT, TTF, TFT, T F F , FTT, F T F , F F T , F F F }
(i) Partition X according to the number of T's appearing in each element.
(ii) For the proposition P
(11- q ) A r , partition X into T ( P ) and ( T ( P ) ) c . (Recall
that 7 ( P ) is the truth set of P.)
)
('T(Q))c.
(iii) For .the proposition Q = - p + ( q A T ) , partition X into 7 ( Q and
(iv) Find the cross-partition of the partitions of X in (ii) and (iii).
(i)
Combine the elements of X with the same number of T's t o form the cells:
[{FFF}, ( T F F , F T F , FFT}, {TTF, T F T , F T T ) , {TTT}]
COMBIKATIONS, ORDERED PARTITIONS
170
iii) and (iii). Construct the t r u t h tables of P = ( p -+ q ) P
f
F
Q
r
P+4
T
and Q = - p
[CHAP. 15
+
(q A r):
(P+Q)AT
T
T
T
T
F
T
T
F
F
T
F
F
F
F
F
F
T
T
T
T
T
F
T
F
F
T
T
T
T
T
F
F
T
F
T
T
I B I
1,'l
T
F
Note t h a t P is t r u e in Cases (lines) 1 , 5 and 7, and Q is t r u e in Cases 1 , 2 , 3 , 4 and 5. Thus
T(P) = {TTT, FTT, F F T )
and
I(Q)
= {TTT, T T F , T F T , T F F , FTT}
and so the required partitions a r e
[{TTT, FTT, FFT}, {TTF, T F T , T F F , F T F , F F F } ]
[{TTT, T T F , TFT, T F F , FTT}, { F T F , F F T , FFF}]
and
(iv) The cross-partition is obtained by intersecting each set of one partition with each set of the
other:
{TTT, FTT, F F T }
n
{TTT, TTF, TFT, T F F , FTT} = {TTT, FTT}
{TTT, F T T , F F T } n { F T F , F F T , F F F } = {FFT}
{TTF, T F T , T F F , F T F , F F F ) n {TTT, T T F , T F T , T F F , FTT} = {TTF, T F T , T F F }
{TTF, T F T , T F F , F T F , FFF] n { F T F , F F T , F F F ) = { F T F , F F F }
Thus t h e cross-partition is
[{TTT, FTT}, {FFT}, {TTF, T F T , TFF}, { F T F , F F F } ]
ORDERED AND UNORDERED PARTITIONS
15.19.In how many ways can 7 toys be divided among 3 chiIdren if the youngest gets
3 toys and each of the others gets 2?
We seek the number of ordered partitions of 7 objects into cells containing 3, 2 and 2 objects,
respectively. By Theorem 15.3, there a r e
"'
-= 210 such partitions.
3! 2 ! 2 !
15.20.There are 12 students in a class. In how many ways can the 12 students take
3 different tests if 4 students are to take each test?
Method 1.
We seek the number of ordered partitions of the 1 2 students into cells containing 4 students
12!
each. By Theorem 15.3, there a r e
= 34,650 such partitions.
Method 2.
There a r e
1' '
(7)
ways to choose 4 students to take the first test; following this, there a r e
ways t o choose 4 students to take the second test. The remaining students take the third test.
Thus. altogether, there a r e
( y ) (t)
*
= 495 70 = 34,650 ways f o r the students to take t h e tests.
171
COMBINATIONS, ORDERED PARTITIONS
CHAP. 151
15.21. In how many ways can 12 students be partitioned into 3 teams, A , , A , and A,, so that
each team contains 4 students?
Method 1.
Observe t h a t each partition { A l , A 2 , A 3 }of the students can be arranged in 3! = 6 ways
as a n ordered partition. Since (see the preceding problem) there a r e
partitions, there a r e 34,650/6 = 5775 (unordered) partitions.
Method 2.
Let A denote one of the students. Then there a r e
(131)
12!
= 34,650 such ordered
ways to choose 3 other students to be
on the same team as A . Now let B denote a student who is not on the same team as A ; then there
are
(:)
ways t o choose 3 students of the remaining students t o be on the same team as B. The
remaining 4 students constitute the third team.
165.35 = 5 7 7 5 ways t o partition the students.
Thus, altogether, there a r e
(Y) *(;I
=
15.22. In how many ways can 6 students be partitioned into (i) 2 teams containing 3 students
each, (ii) 3 teams containing 2 students each?
(i)
Method 1.
61
Since
There ar e 3 ! 3! = 20 ordered partitions into 2 cells containing 3 students each.
each unordered partition determines 2 ! = 2 ordered partitions, there a r e 20/2 = 10 unordered
partitions.
Method 2.
Let A denote one of the students; then there a r e
([) =
1 0 ways to choose 2 other students
to be on the same team as A . The other 3 students constitute the other team. I n other words,
there a r e 10 ways to partition the students.
(ii) Method 1.
61
There a r e
= 90 ordered partitions into 3 cells containing 2 students each. Since
each unordered partition determines 3! = 6 ordered partitions, there a r e 90/6 = 15 unordered
partitions.
&
Method 2.
Let A denote one of the students, Then there a r e 5 ways to choose the other student to
be on the same team as A . Let B denote a student who isn't on the same team as A ; then
there a r e 3 ways to choose another student to be on the same team as B. The remaining
2 students constitute the other team. Thus, altogether, there a r e 5 3 = 15 ways to partition
the students.
-
15.23.In how many ways can a class X with 10 students be partitioned into 4 teams
A,, A,, B,, B,, where A , and A , contain 2 students each and B , and B, contain 3 students
each ?
Method 1.
lo!
There are 2 ! 2 ! 3 ! 3 ! = 25,200 ordered partitions of X into 4 cells containing 2,2,3 and
3 students, respectively. However, each unordered partition { A , ,A,, B1,B,} of X determines
2! 2! = 4 ordered partitions of X . Thus, altogether, there a r e 25,20014 = 6300 unordered partitions.
Method 2.
There a r e
(y)
ways to choose 4 students who will be on the teams A , and A,, and there a re
3 ways in which each 4 students can be partitioned into 2 teams of 2 students each. On the other
hand, there ar e 10 ways (see preceding problem) t o partition the remaining 6 students into 2 teams
containing 3 students each. Thus, altogether, there a r e (:)*3*10
to partition the students.
= 2 1 0 * 3 * 1 0 = 6300 ways
172
COMBINATIONS, ORDERED PARTITIONS
;CHAP. 15
15.24.In how many ways can 9 students be partitioned into three teams containing 4, 3
and 2 students, respectively?
Since all the cells contain different numbers of students, the number of unordered partitions
equals the number of ordered partitions which, by Theorem 15.3, is
- 1260.
&
15.25. (i) In how many ways can a set X containing 10 elements be partitioned into two
cells?
(ii) In how many ways can 10 students be partitioned into two teams?
(i,
Method 1.
Each subset A of X determines a n ordered partition ( A , A c ) of X (where Ac is the complement of A ) , and so there a r e 210 = 1024 such ordered partitions. However, each unordered
partition { A , B } determines two ordered partitions, ( A , B ) and ( B , A ) , and so there a r e
l 0 2 4 / 2 = 512 unordered partitions.
Method 2.
Let j . denote one of the elements in X . Now X - { x } contains 9 elements, and so there a r e
2q = 512 ways of choosing a subset of X - {x} t o be in the same cell as x. In other words,
there a r e 512 such partitions. (Observe t h a t { X , @ ) is one of these partitions.)
(iii
Here we assume t h a t each team must contain at least one student, hence-we do not allow the
partition f o r which one team contains 1 0 students, and the other team none. Accordingly,
there a r e 5 1 2 - 1 = 511 possible partitions.
Supplementary Problems
COMBINATIONS
15.26. A class contains 9 boys and 3 girls. (i) In how many ways can the teacher choose a committee of 4?
(iii How many of them will contain at least one girl? (iii) How many of them will contain exactly
one girl?
15.27.
A woman has 11 close friends. ( i ) In how many ways can she invite 5 of them to dinner? (ii) In
how many ways if two of the friends a r e married and will not attend separately? iiii) In how
many ways if two of them a r e not on speaking terms and will not attend together?
15.28.
A woman has 11 close friends of whom 6 are also women. ii) In how many ways can she invite
3 o r more of them to a p a r t y ? (ii) In how many ways can she invite 3 or more of them if she wants
the same number of men as women fincluding herself)?
15.29.
There a r e 10 points A , R , . , . in a plane, no three on the same line. (i) How many lines a r e
determined by the points? (ii) How many of these lines do not pass through A or E ? (iii) How
many triangles a r e determined by the points? (iv) How many of these triangles contain the
point A ? I W ) How many of these triangles contain the side A B ?
15.30.
A student is t o answer 10 out of 13 questions on a n exam. (i) How many choices has he? iii) How
many if he must answer the first two questions? (iii) How many if he must answer the first or
second question but not both? (iv) How many if he must answer exactly 3 of t h e first 5 questions?
(v) How many if he must answer a t least 3 of the first 5 questions?
15.31.
How inany diagonals h a s a (i) hexagon, (ii) decagon?
15.32. Which regular polygon h as
diagonals a s sides?
15.33.
173
COMBINATIONS, ORDERED PARTITIONS
CHAP. 151
(i) twice as many diagonals as sides,
(ii) three times as many
(i)
How many triangles a r e determined by the vertices of a n octagon?
(ii) How many if the sides of the octagon a re not to be sides of a ny triangle?
15.34. How many triangles ar e determined by the vertices of a regular polygon with
of the polygon ar e not t o be sides of any triangle?
12
sides if the sides
15.35.
A man is dealt a poker hand (5 cards) from a n ordinary playing deck. In how many ways can he
be dealt (i) a straight flush, (ii) four of a kind, (iii) a straight, (iv) a pair of aces, (v) two of a
kind (a p a i r ) ?
15.36.
The English alphabet has 26 letters of which 5 a r e vowels.
(i)
How many 5 letter words containing 3 different consonants and 2 different vowels can be
formed?
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
How
How
How
How
How
How
How
(ix)
(x)
How many of them begin with a and end with b?
How many of them contain the letters a , b and c?
many
many
many
many
many
many
many
of
of
of
of
of
of
of
them contain the letter b ?
them contain the letters b and c ?
them begin with b and contain the letter c?
them begin with b and end with c?
them contain the letters a and b?
them begin with a and contain b?
them begin with b and contain a?
PARTITIONS AND CROSS-PARTITIONS
15.37. Which of t he following are partitions of
(i) [{a,e , Q, Co), {u)]
(ii) [{a,i,u>,
{ e l , (0,ull
(iii) [{a,e ) , Co), Czt)]
35.38.
15.39.
@I
Find the cross-partition of each pair of partitions of X = { 1 , 2 , 3 , 4 , 5 , 6 ) :
.
(i)
[{l, 2 , 3 ) , { 4 , 5 , 6 ) ] and [{I, 3,5), { 2 , 4 , 6 ) ]
(ii)
[{I, 2,3), { 4 , 5 , 6 ) ] and [{I, 21, {3,4, 5), (611
.
Let X denote the domain of a proposition P ( p , q , ~ of
) three variables:
X = {TTT, T T F , T F T , T F F , F T T , F T F , F F T , FFF)
(i)
(ii)
(iii)
(iv)
15.40.
X = (a,e,i,o,u}?
(iv) [ { a ,el, Ci), (01, {a,el, {ull
(v) [ { a ,e , 4, (0,u1,
(vi) [@, { a ) , Ce, u.1,@, {i, 011
Partition X according to the number of F's appearing in each element.
F o r the proposition P = ( p v q ) A T , partition X into T ( P ) and ( 7 ( P ) ) c .
Partition X into T(q) and T ( - q ) .
Find the cross-partition of t h e partitions in (ii) and (iii).
F o r a n y proposition P = P ( p , q,
of P .
. . .),
show tha t [ T ( P ) ,T ( - P ) ] forms a partition of the domain
that is, each
15.41. Let the partition { A , , A , , . . .,A,.} be a refinement of the partition {B,, B,, . . .,Bs},
A i is a subset of some Bj. Describe the cross-partition.
15.42.
Show t h a t a cross-partition is a refinement (see preceding problem) of each of the original partitions.
ORDERED AND UNORDERED PARTITIONS
15.43. I n how many ways can 9 toys be divided evenly among 3 children?
15.44.
In how many ways can 9 students be evenly divided into three teams?
174
COMBINATIONS, ORDERED PARTITIONS
'CHAP. 15
15.45. In how many ways can 1 0 students be divided into three teams, one containing 4 students and
the others 3 1
15.46.
There a r e 12 balls in a n urn. In how many ways can 3 balls be drawn from t h e urn, f o u r times
in succession?
15.47.
In how many ways can a club with 1 2 members be partitioned into three committees containing
5,4 and 3 members respectively?
15.48.
In how many ways can n students be partitioned into two teams containing at least one student?
15.49.
In how many ways can 14 men be partitioned into 6 committees where 2 of the committees contain
3 men and the others 2 ?
15.50. In how many ways can a set X with 3 elements be partitioned into
(ii, three (unordered) cells?
(i) three ordered cells,
15.51.
In how many ways can a set X with 4 elements be partitioned into
(ii) three (unordered) cells?
(i) three ordered cells,
15.52.
In bridge, 13 cards a r e dealt to each of four players who a r e called North, South, E a s t and West.
The distribution of the cards is called a bridge hand. ii) How many bridge hands a r e there?
(ii, In how many of them will one player be dealt all four aces? (iii) In how many of them will
each player be dealt a n ace? (iv) In how many of them will North be dealt 8 spades and South
the other 5 spades? (v) In how many of them will North and South have, together, all four aces?
(Rrmark. Leave answers in factorial notation.)
Answers to Supplementary Problems
15.26.
(i,
15.27.
(i,
( y ) (:)
)
(93) +
( y ) (';)
(7)
= 495,
(ii)
= 462,
(ii)
15.28. (i) 211 - 1 -
15.29.
(i)
('2", =
15.30.
(i)
=
(ii)
15.31.
(i)
(ii)
(i)
or
(iii)
(iii) 2
*
(:)
*(:I
(t:> + ('41)
(130)
= 252
= 252
+ . * . + (")
\ 11
= 120, (iv)
= 36,
= 1981
(v) 8
(7)
= 286
(y)= (y)=
(i)
= 28,
(iii) 3
= 210,
= 1981
-
45,
= 369,
-
- 6 = 9,
165
(ii)
(Y)
-
10 = 35
15.32. Hl?zt. By Problem 15.7, the regular polygon with n sides has ( n 2 - 3n)/2 diagonals.
( i ) (?zL - 3 n ) / 2 = 2% or n = 7, ( i i ) (nL- 3n)/2 = 3n or ?z = 9
15.33.
(i)
= 56, (ii)
(:)
-8
(t)
- 8 = 16
COMBINATIONS, O R D E R E D PARTITIONS
CHAP. 151
175
15.35. (i) 4 - 1 0 = 40, iii) 1 3 - 4 8 = 624, (iii) 10.45 - 40 = 10,200. (We subtract the number of straight
flushes.) (iv) 4 /12 * 43 = 84,480, (v) 13
43 = 1,098,240
(:)(?)
() )
2 \ 3
ti)(:)
(i) =
(:)(:)
15.36. (i)
(ii)
*
5 ! = 1,596,000
(v)
19
5 ! = 228,000
(vi)
4
*
(vii) 4
*
(iii) 19
5!
(iv) 19-(:)
4 ! = 4560
22,800
(i)
(y)
*
(T)
*
3! = 1140
(ix) 4
*
5! = 91,200
(x) 4 * 1 9 . 5 ! = 9120
*
*
3! = 4456
(7)
4 ! = 18,240
fviii) 18,240
15.37. (i) yes, (ii) no, (iii) no, (iv) yes, (v) yes, (vi) yes
15.38. (i) [{1,31, { 2 } , C51, (4, Q],
15.39. (i)
(ii) [C1,2}, {3}, @,C4,5}, (611
[{TTT}, {TTF, TFT, FTT}, {TFF, F T F , FFT}, {FFF}]
(ii) [{TTT, TFT, FTT}, {TTF, T F F , F T F , F F T , FFF}]
(iii) [{TTT, TTF, FTT, FTF}, {TFT, T F F , F F T , FFF}]
(iv) [{TTT, FTT}, {TFT}, {TTF, F T F ) , { T F F , F F T , FFF}]
15.41. { A , , A , ,
. . . , A , } , i.e. the refinement partition.
15.42. A , n B , c A , and A , n B , c B , .
15.43. '!
= 1680
3! 3! 3!
15.44. 1680/3! = 280 or
15.45. lo!
4 ! 3 ! 3 ! 2!
0 -
-
(:)(;)
or
2100
= 280
ti)(;)
= 2100
12!
15.46* 3! 31 3! 3! = 369,600
15.47.
12!
=
5! 4! 3!
15.48. 2n-1
27,720
-1
14!
.-
15'49' 3! 3! 2! 2! 2! 2! 2 ! 4!
15.50. (i)
- 3,153,150
-
33 = 27 (Each element can be placed in any of the three cells.)
(ii) The number of elements in the three cells can be distributed as follows:
(4 [{31, (01, { O } ] , ( b ) [{2}, {1>,{Oil, (c) [{I}, c11, (111
Thus the number of partitions is 1 + 3 + 1 = 5.
15.51. (i) 3 4 = 81. (ii) The number of elements in the three cells can be distributed a s follows: ( a ) [{4},
(01, {O}], ( b ) [{3$, {I}, {O}], (4 [@I, C2}, {O)], (4 [@I, {I}, U}l. Thus the number of partitions
is 1 + 4 + 3 + 6 = 14.
15.52.
52!
48!
(ii) 4 9! 13!48!
13! 13! (iii) 4! 12! 121 12! 12!
(i) 13! 13! 13! 13!
48!
48!
48!
(v) 2 * 9! 13! 13! 13!
2 ' 4 * l o ! 12! 13! 13! + 2 * 3 * ll! ll! 13! 13!
+
iiv) (13)
39!
8 5!8!13!13!
-
48!
2300 11! 13! 13! 13!
Chapter 16
Tree Diagrams
TREE DIAGRAMS
A tree diagram is a device used to enumerate all the logical possibilities of a sequence
of events where each event can occur in a finite number of ways. The construction of
tree diagrams is illustrated in the following examples.
Example 1:
Find the product set A
X
B
X
C where A = ( 1 , 2 } , B = {a, b , c}
and C = { 3 , 4 } .
The tree diagram follows:
(1,a, 3)
(1, a, 4)
3
<<
'<A
'<:
'<:
'<4
3
(1, b, 3)
(1, b, 4)
(1, c, 3)
(1, c, 4)
(2, a, 3)
(2, a, 4)
(2, b , 3)
(2, b, 4)
(2, c, 3)
Observe that the tree is constructed from
:<c left t o right, and
(2, that
c, 4) the number of
branches at each point corresponds to the number of ways the next event can occur.
Example 2:
Mark and Eric a r e to play a tennis tournament. The first person to win two games
in a row or who wins a total of three games wins the tournament. The following
diagram gives the various ways the tournament can occur.
Observe t h a t there a r e 10 endpoints which correspond to the 10 ways that the
tournament can occur:
MM, MEMM, MEMEM, MEMEE, MEE, EhIM, EMEMM, EMEME, E M E E , E E
The path from the beginning of the tree to the endpoint describes who won which
game in the individual tournament.
176
177
T R E E DIAGRAMS
CHAP. 161
Example 3:
A man h a s time t o play roulette at most
five times. A t each play he wins or loses
a dollar. The man begins with one dollar
and will stop playing before the five times
if he loses all his money or if he wins
three dollars, i.e. if he h a s four dollars.
The tree diagram describes the way the
betting can occur. Each number in the
diagram denotes t h e number of dollars
he has at t h a t point.
The betting can occur in 11 different
ways. Note t h a t he will stop betting before the five times a r e up in only three of
the cases.
3-
Solved Problems
16.1. Construct the tree diagram f o r the number of permutations of {a, b, c } .
abc
c
acb
bac
a
bca
The six permutations a r e listed on the right of the diagram.
16.2. Find the product set { 1,2,3} x ( 2 , 4 } x {2,3,4} by constructing the appropriate
tree diagram.
2
6
1
2
3
4
(1,2, 2)
(1, 2, 3)
(1, 2, 4)
(1,
(1,4, 2)
4)
3)
,,/'54<:
245$:
4
(2, 2, 2)
(2, 2, 3)
(2, 2, 4)
\ /2e!
(3, 2, 2)
(3,2,
(3, 2, 4)
3)
-3\4+23
(3, 4, 2)
3)
4
(3,4, 4)
The eighteen elements of the product set a r e listed to the r i g h t of the tree diagram.
178
T R E E DIAGRAMS
[CHAP. 16
16.3. Teams A and B play in a basketball tournament. The team that first wins 3 games
wins the tournament. Find the number of possible ways in which the tournament
can occur.
Construct the appropriate tree diagram:
The tournament can occur in twenty ways:
AAA, AABA, AABBA, AABBB, ABAA, ABABA, ABABB, ABBAA, ABBAB, ABBB
BAAA, BAABA, BAABB, BABAA, BABAB, BABB, BBAAA, BBAAB, BBAB, BBB
16.4. A man is a t the origin on the x-axis
and takes a one unit step either to the
left or to the right. He stops if he
reaches 3 or -3, or if he occupies any
position, other than the origin, more
than once. Find the number of different paths the man can travel.
Construct t h e appropriate tree diagram.
There a r e 14 different paths, each path
corresponding to a n end point of the tree
diagram.
TREE DIAGRAMS
CHAP. 161
179
16.5. Teams A and B play a hockey game consisting of three periods. The number of
goals (points) scored in each period satisfies the following conditions:
(ij
The total number of points scored by both teams in any one period is one
01' two.
(iij The teams are never tied a t the end of any period.
(iii) Neither team is ahead by more than 2 points a t the end of any period.
(iv) Team A is leading a t the end of the first period.
Find the number of ways the scoring can occur if the above conditions a r e satisfied.
In how many ways will team B win?
Note first, by condition (i), t h a t the scoring in a n y one period can occur in only five ways:
B
j
B
j
1jR.l
The number a t the t op is the number of points scored by team A , and the number at t h e bottom
by team B.
Construct the appropriate tree diagram:
Observe t h a t condition (iv) allows only two of the above five scores to occur in t h e first period.
By the tree diagram, the scoring can occur in only 13 ways, and in 4 of these which a r e exhibited
on the right of the diagram, Team B will win.
T R E E DIAGRAMS
180
[CHAP. 16
16.6. Teams A and B play in baseball's world series. (Here, the team that first wins
4 games wins the series.) Find the number of ways the series can occur if (i) A wins
the first game and (ii) the team that wins the second game also wins the fourth game.
Construct the appropriate tree diagram:
Observe t h a t there is only one choice in the fourth game, the winner of the second game. The
series can occur in 15 ways.
16.7. In the following diagram let A , B , . . ., F denote islands, and the lines connecting
them bridges. A man begins at A and walks from island to island. He stops for
lunch when he cannot continue to walk without crossing the same bridge twice.
Find the number of ways that he can take his walk before eating lunch.
Construct the appropriate tree diagram:
C
A
D
D
B
D
F
E
F
There are eleven ways t o take his walk. Observe t h a t he must e a t his lunch at either B , D or
E.
CHAP. 161
181
TREE DIAGRAMS
16.8. Consider the adjacent diagram with nine points A , B , C , R ,
S, T,X,Y ,2. A man begins at X and is allowed t o move
horizontally o r vertically, one step at a time. He stops when
he cannot continue to walk without reaching the same point
more than once. Find the number of ways he can take his
walk, if he first moves from X to R. (By symmetry, the
total number of ways is twice this.)
A-
1
I-1-1
1-1
X-Y-
z
Construct the appropriate tree diagram:
/
‘s/
/
z‘
\
Y\
z
‘Y
T
Y
z
Y
s
Y
z
T
C
C
B
A
There ar e 1 0 different trips. (Note t h at in only 4 of them a re all nine points covered.)
Supplementary Problems
16.9.
Construct the tree diagram f o r t h e number of permutations of { a , b , c , d } .
16.10. Find the product set {a,b , c } X {a,c }
X
{b}
X
{ b , c} by constructing the appropriate tree diagram.
16.11. Teams A and B play in a basketball tournament, The first team tha t wins two games in a row o r a
total of four games wins the tournament. Find the number of ways the tournament can occur.
16-12. A man has time t o play roulette five times. He wins or loses a dollar a t each play. The m a n begins
with two dollars and will stop playing before the five times if he loses all his money or wins three
dollars (i.e. has five dollars). Find the number of ways the playing can occur.
16.13. A man is at t h e origin on t h e x-axis and takes a unit step either t o the left or to the right. H e
stops after 5 steps or if he reaches 3 or -2. Construct the tree diagram to describe all possible
paths the man can travel.
16.14. Teams A and B play in baseball’s world series. (The team that first wins 4 games wins the series.)
Find the number of ways the series can occur if the team t h a t wins the first game also wins the
third game, and the team t h a t wins the second game also wins the fourth game.
16.15. Solve Problem 16.7 if the man (i) begins a t E , (ii) begins at F .
16.16. Solve Problem 16.8 if the man (i) begins a t Y and first moves to S, (ii) begins a t Y and first
moves to 2.
16.17. Solve Problem 16.5 if condition (ii) is replaced by the following: (ii) The game does not end in a tie.
182
TREE DIAGRAMS
[CHAP. 16
Answers to Supplementary Problems
16.12.
16.13. Hint. The tree is essentially the same as the tree of the preceding problem.
16.15. (i)
(ii)
E-
C-
D
D
F-
C
-
D
C
-
D
E
F-
F
\
TREE DIAGRAMS
CHAP. 161
16.16.
(i)
sqB<;
/ R<y
I’
183
‘4
B
C
R
T
X
B
A
z
T
z
R
X
T-=====Z
(ii)
-
2
0
-
2
0
__
__
1
0
__
0
__
1
__
-
0
2
__
-
and
1
0
-
1
__
0
0
~
1
0
~
2
d
Chapter 17
Probability
INTRODUCTION
Probability is the study of random or nondeterministic experiments. If a die is tossed
in the air, then it is certain that the die will come down, but it is not certain that, say,
a 6 will appear. However, suppose we repeat this experiment of tossing a die; let s be
the number of successes, i.e. the number of times a 6 appears, and let n be the number of
tosses. Then it has been empirically observed that the ratio f = sln, called the relative
f r e q u e n c y , becomes stable in the long ru;, i.e. approaches a limit. This stability is the
basis of probability theory.
In probability theory, we define a mathematical model of the above phenomenon by
assigning “probabilities” (or: the limit values of the relative frequencies) to each possible
outcome of an experiment. Furthermore, since the relative frequency of each outcome is
non-negative and the sum of the relative frequencies of all possible outcomes is unity, we
require that our assigned “probabilities” also satisfy these two properties. The reliability
of our mathematical model for a given experiment depends upon the closeness of the
assigned probabilities to the actual relative frequency. This then gives rise to problems
of testing and reliability which form the subject matter of statistics.
Historically, probability theory began with the study of games of chance, such as
roulette and cards. The probability 11 of an event A was defined as follows: if A can occur
in s ways out of a total of n equally likely ways, then
s
p = P(A) = n
For example, in tossing a die an even number can occur in 3 ways out of 6 “equally likely”
ways; hence p = f = 4. This classical definition of probability is essentially circular since
the idea of “equally likely” is the same as that of “with equal probability” which has not
been defined. The modern treatment of probability theory is purely axiomatic. In
particular, we require that the probabilities of our events satisfy the three axioms listed
in Theorem 17.1. However, those spaces consisting of n equally likely elementary events,
called equiprobable spaces, are an important class of finite probability spaces and shall
provide us with many examples.
SAMPLE SPACE AND EVENTS
The set S of all possible outcomes of some given experiment is called the sample space.
A particular outcome, i.e. an element in S, is called a sample point o r sample. An event A
is a set of outcomes or, in other words, a subset of the sample space S. In particular,
the set { a } consisting of a single sample a E S is an event and called an eZewzentury event.
Furthermore, the empty set $2 and S itself are subsets of S and so are events; $3 is sometimes called the impossible event.
184
185
PROBABILITY
CHAP. 171
Since an event is a set, we can combine events to form new events using the various
set operations:
(i) A U B is the event that occurs iff A occurs o r B occurs (or both);
(ii) A n B is the event that occurs iff A occurs and B occurs;
(iii) Ac, the complement of A , also written A, is the event that occurs iff A does not
occur.
Two events A and B are called mutually exclusive if they are disjoint, i.e. A n B = 8.
In other words. A and B are mutually exclusive iff they cannot occur simultaneously.
Example 1.1:
Experiment: Toss a die and observe the number t h a t appears on top.
sample space consists of the six possible numbers:
Then the
S = (1, 2, 3, 4, 5, 6 )
Let A be the event t h a t a n even number occurs, B t h a t a n odd number occurs and
C t h at a prime number occurs:
A = {2,4,6},
B = {1,3,5},
C = {2,3,5}
Then:
A u C = { Z , 3 , 4 , 5, 6} is the event t h a t a n even or a prime number occurs;
B n C = (3, 5)
is the event t h a t a n odd prime number occurs;
Cc = (1, 4, S}
is the event t h a t a prime number does not occur.
Note t h a t A and B a r e mutually exclusive: A n B = 0;in other words, a n even
number and a n odd number cannot occur simultaneously.
Example 1.2:
Experiment Toss a coin 3 times and observe the sequence of heads ( H ) and
tails ( T) t h a t appears. The sample space S consists of eight elements:
S = { HHH , H H T, H TH , H TT, TH H , TH T, TTH, TTT}
Let A be t h e event t h a t two or more heads appear consecutively, and B t h a t all
the tosses are the same:
A = {HHH,HHT,THH}
and
B = {H H H ,TTT}
Then A n B = (HHH} is the elementary event in which only heads appear. The
event t h a t 5 heads appear is the empty set 8.
Example 1.3:
Experiment: Toss a coin until a head appears and then count the number of times
the coin w a s tossed. The sample space of this experiment is S = (1,2,3, . . ., -}.
refers t o the case when a head never appears and so the coin is tossed a n
Here
infinite number of times, This is a n example of a sample space which is co~72tabZy
infinite.
Example 1.4:
Experiment: Let a pencil drop, head first, into a rectangular box and note the point on the bottom of the
box t h at t h e pencil first touches. Here S consists of all
the points on t h e bottom of the box. Let t h e rectangular
ar ea on the r i g h t represent these points. Let A and B
be the events t h a t t h e penciI drops into the corresponding areas illustrated on the right. This is a n example of
a sample space which is not finite nor even countably
infinite, i.e. which is non-denumerable. For theoretical
reasons every subset of this space cannot be considered
to be an event.
I
S
The sample spaces in the preceding two examples, as noted, are not finite. The theory
concerning such sample spaces, especially the non-denumerable case, lies beyond the scope
of this text. Thus, unless otherwise stated, all our sample spaces S shall be finite.
186
PROBABILITY
[CHAP. 17
FINITE PROBABILITY SPACES
Let S be a finite sample space: S = {u,,a,, . . ., a , l ] . A finite probability space is
obtained by assigning to each point a7E S a real number TI,, called the probability of at,
satisfying the following properties:
(i) each p l is non-negative, p i1 0
(ii) the sum of the p l is one, p , + p 2
+
* *
. +pn
= 1.
The probability of any event A , written P ( A ) , is then defined to be the sum of the probabilities of the points in A . For notational convenience we write P(u,) for P ( { a , } ) .
Example 2.1:
Let three coins be tossed and the number of heads observed; then the sample space
is S = { 0 , 1 , 2 , 3 ) . We obtain a probability space by the following assignment
P(0) =
4,
P(1) =
#,
P(2) = Q
P(3) =
and
8
since each probability is non-negative and the sum of the probabilities is 1. Let A
be the event t h a t a t least one head appears and let B be the event t h a t all heads
or all tails appear’
A = (1, 2, 3)
2nd
B = (0, 3)
Then, by definition,
P(A) = P(1)
and
Example 2.2:
+ P(2)
P(B) = P(0)
+ 38 + 18
& +4 =
P ( 3 ) = 38
+ P(3) =
*
=
2
8
Three horses A , B and C a r e in a race; A is twice a s likely to win a s B and 13 is
twice as likely to win a s C. W h a t a r e their respective probabilities of winning,
i.e. P ( A ) ,P ( B ) and P(C)?
Let P ( C ) = p ; since B is twice as likely t o win as C, P ( B ) = 2 p ; and since
A is twice a s likely t o win as B , P ( A ) = 2 P ( B ) = 2 ( 2 p ) = 4p. Now the sum of the
probabilities must be 1: hence
p + Z p f 4 p = l
or
7 p = 1
or
p = +
Accordingly,
P(A) = 4p =
4,
P(B) = 2 p =
+,
P(C) = p =
3
Question: Wha t is the probability t h a t B or C wins, i.e. P ( { B ,C})? By definition
P ( { B ,C}) = P ( B )
+ P(C) = + + 3 = +
EQUIPROBABLE SPACES
Frequently, the physical characteristics of a n experiment suggest that the various
outcomes of the sample space be assigned equal probabilities. Such a finite probability
space S, where each sample point has the same probability, will be called an equiprobable
space. In particular, if S contains n points then the probability of each point is lln.
T
Furthermore, if an event A contains T points then its probability is r * 1- = -.
In other
words,
n n
number of elements in A
=
number of elements in S
or
P(A) =
number of ways that the event A can occur
number of ways that the sample space S can occur
We emphasize that the above formula for P ( A ) can only be used with respect to a n
equiprobable space, and cannot be used in general.
The expression “at random” will be used only with respect to a n equiprobable space;
formally, the statement “choose a point at random from a set S” shall mean that S is a n
equiprobable space, i.e. that each sample point in S has the same probability,
PROBABILITY
CHAP. 171
187
Let a card be selected a t random from a n ordinary deck of 52 cards.
Example 3.1:
Let
A = {the card is a spade}
B = {the card is a face card, i.e. a jack, queen o r king}
and
We compute P ( A ) ,P ( B ) and P ( A n B ) . Since we have a n equiprobable space,
number of spades 13 = 1
= number of cards
52 4
P ( An B ) =
Example 3.2:
P(B) =
number of face cards - 12
number of cards
52
_ -3
-
13
number of spade face cards - 5
number of cards
52
Let 2 items be chosen a t random from a lot containing 12 items of which 4 a re
defective. Let
A = {both items a r e defective}
Find P ( A ) and P ( B ) .
and
B = {both items a r e non-defective)
Now
= 66 ways, the number of ways tha t 2 items can be
chosen from 12 items;
5' can occur in
A can occur in
(;)
= 6 ways, the number of ways tha t 2 defective items
can be chosen from 4 defective items;
B can occur in
(28)
= 28 ways, the number of ways tha t 2 non-defective
items can be chosen from 8 non-defective items.
Accordingly, P ( A ) =
=
and P ( B ) =
= E.
33
THEOREMS ON FINITE PROBABILITY SPACES
The following theorem follows directly from the fact that the probability of an event
is the sum of the probabilities of its points.
Theorem 17.1:
The probability function P defined on the class of all events in a finite
probability space satisfies the following axioms:
[P,] For every event A , 0 6 P ( A ) 1.
[Pal P ( S )= 1.
[P,] If events A and B are mutually exclusive, then
P ( A U B) = P(A) P(B)
+
Using mathematical induction, [PSI can be generalized as follows:
Corollary 17.2: If A , , A,, . . . ,AT are pairwise mutually exclusive events, then
P(AJ
P(A,UA,U * . U A r ) = P ( A , ) P(A,) . * .
+
+
+
Using only the properties of Theorem 17.1, we can prove (see Problems 4.19-4.22).
Theorem 17.3:
If @ is the empty set, and A and B are arbitrary events, then:
(i)
(ii)
(iii)
(iv)
P(P) = 0;
P(Ac) = 1 - P ( A ) ;
P ( A \B) = P ( A ) - P ( A n B ) , i.e. P ( A n B c ) = P(A) - P ( An B ) ;
A c B implies P ( A )L P ( B ) .
Property (ii) of the preceding theorem is very useful in many applications.
Example 4.1:
Refer to Example 3.2, where 2 items a r e chosen at random from 12 items of which
4 a r e defective. Find the probability p t h a t at least one item is defective.
Now
is the complement of
C = {a t least one item is defective}
B = {both items a r e non-defective}
[CHAP. 1 7
PROBABILITY
188
i.e. C = Be; and we found t h a t P ( B ) =
g.
Hence
14
19
p = P ( C ) = P(BC) = 1 - P ( B ) = 1 - 33 = 33
The odds t h at an event with probability p occurs is de$e$ to be the ratio p : (1- p ) .
Th u s t h e odds t h at a t least one item is defective is 5 :33 or 19 : 14 which is read
“19 t o 14”.
Observe that axiom [P,] gives the probability of a union of events in the case that the
events are mutually exclusive, i.e. disjoint. The general formula follows:
Theorem 17.4:
Corollary 17.5:
For any events A and B , P ( A U B ) = P ( A ) + P ( B ) - P ( A n B ) .
For any events A , B and C,
P ( A U B U C ) = P ( A ) P(B) P ( C )- P ( A n B )
- P ( A n C ) - P(BnC) + P ( A n B n C )
Example 4.2:
+
+
Of 100 students, 30 a r e taking mathematics, 20 a r e taking music and 10 a r e taking
both mathematics and music. If a student is selected at random, find the probability p t h a t he is taking mathematics or music.
Let A = {students taking mathematics} and B = {students taking music}.
Then A n B = {students taking mathematics and music} and A U B = {students
taking mathematics o r music}.
30 - 3
2 0 - 1
Since t h e space is equiprobable, then P ( A ) =
- lo, P ( B ) =
- s,
P(AnB)=
=
Thus by Theorem 17.4,
k.
p = P(AuB) = P(A)+P(B)-P(AnB) =
$+i-f =
5
CLASSICAL BIRTHDAY PROBLEM
The classical birthday problem concerns the probability p that n people have distinct
birthdays. In solving this problem, we assume that a person’s birthday can fall on any
day with the same probability.
Since there are n people and 365 different days, there are 365” ways in which the
n people can have their birthdays. On the other hand, if the n persons are to have distinct
birthdays, then the first person can be born on any of the 365 days, the second person can
be born on the remaining 364 days, the third person can be born on the remaining 363 days,
etc. Thus there are 365 * 364.363 . . (365 - n 1) ways the n persons can have distinct
birthdays. Accordingly,
365-364.363 * (365 - n 1) - 365 364 363 . . . 365 - n 1
+
e
P =
365“
+
-.-.-365
- 365
+
365
365
It can be shown that for n s 2 3 , p < $; in other words, amongst 23 or more people it is
more likely that a t least two of them have the same birthday than that they all have
distinct birthdays.
Solved Problems
SAMPLE SPACES AND EVENTS
17.1. Let A and B be events. Find an expression and exhibit the Venn diagram for the
event that: (i) A but not B occurs, i.e. only A occurs; (ii) either A or B, but not
both occurs, i.e. exactly one of the two events occurs.
(i)
Since A but not B occurs, shade the a re a of A outside of B as in Figure (a) below. Note
t h a t Bc, the complement of B , occurs since B does not occur; hence A and Bc occurs. I n other
words, the event is A n B c .
CHAP. 171
PROBABILITY
A but not B occurs.
189
Either A o r B , but not both, occurs.
(b)
(4
(ii) Since A or B but not both occurs, shade the a re a of A and B except where they intersect a s
in Figure ( b ) above. The event is equivalent to A but not B occurs or B but not A occurs.
Now, as in (i), A but not B is the event A n B c , and B but not A is the event BnAc. Thus
the given event is ( A n B c ) u ( B n A c ) .
17.2. Let A , B and C be events. Find an expression and exhibit the Venn diagram for
the event that (i) A and B but not C occurs, (ii) only A occurs.
(i)
Since A and B but not C occurs, shade the intersection of A and B which lies outside of C,
as in Figure ( a ) below. The event is A nBn Cc.
I
1
A and B but not C occurs.
(4
(ii)
Only A occurs.
(b)
Since only A is to occur, shade the area of A which lies outside of B and of C, as in Figure ( b )
above. The event is A nBcn Cc.
17.3. Let a coin and a die be tossed; let the sample space S consist of the twelve elements:
S = {Hl, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
(i) Express explicitly the following events: A = {heads and an even number appears}, B = {a prime number appears}, C = {tails and an odd number appears}.
(ii) Express explicitly the event that: (a) A or B occurs, ( b ) B and C occurs,
( c ) only B occurs.
(iii) Which of the events A , B and C are mutually exclusive?
(i)
To obtain A , choose those elements of S consisting of a n H and a n even number: A =
(H2, H4, H6).
To obtain B, choose those points in S consisting of a prime number: B = (H2, H3, H5, T2, T3, T5).
To obtain C, choose those points in S consisting of a T and a n odd number: C = {Tl,T3, T5).
(ii) (a) A or B = A u B = (H2, H4, H6, H3, H5, T2, T3, T5)
( b ) B and C = B n C = {T3,T5}
of B which do not lie in A or C: B n A c n C c = {H3,H5,T2}.
( c ) Choose those elements
(iii) A and C ar e mutually exclusive since A n C = @.
190
[CHAP. 1 7
PROBABILITY
FINITE PROBABILITY SPACES
17.4. Suppose a sample space S consists of 4 elements: S = {u,, a,, a3,ad}. Which function defines a probability space on S ?
(i) P(al)= +, P(a,) = $, P(a,) = 4, P(a,) = &.
(ii) P(a,) = 4,P(a,) = $, P ( a J = -&, P(a,) =
+.
(iii) P(a,) = 3, P(a,) = d, P(aJ = 6, P(a,) = 6.
(iv) P(al)= 4,P(a,) = t, P ( a , ) = $, P(aJ = 0.
(i)
Since the sum of the values on the sample points is greater than one,
the function does not define a probability space on S.
++$ +i + Q
=
s,
(ii) Since P(a,) = -4, a negative number, the function does not define a probability space on S.
(iii) Since each value is non-negative, and the sum of the values is one,
function does define a probability space on S.
& + & +Q + Q
= 1, the
(iv) The values a r e non-negative and add up to one; hence the function does define a probability
space on S.
17.5. Let S = {ul, a,, a,,a4}, and let P be a probability function on S.
(i)
Find P(a,) if
P(a,) = +, P(a,) = &, P(a,) = +.
(ii) Find P ( u l ) and P(a,) if
(iii) Find P(a,) if
(i)
P(a,) = P(a,) = & and P ( a l ) = 2P(a,).
P({a,,a,}) = b, P((a,,a,}) =
+
and P(a,) = Q.
Let Pia,) = p . Then, f o r P t o be a probability function, the sum of the probabilities on the
& f 8 = 1 o r p = 18
sample points must be one: ?I
+++
(ii) Let P(a,) = p , then P ( a l )= 2 p . Hence 2 p f p - k
P ( a l ) = +.
& +i
= 1 or p =
P(a,) = P({az,u,}) - P(a2) = Q -
(iii) Let P ( a l ) = p .
P ( a J = P({a,,u4}) - P(a,) =
Then p
+ 4+ + + Q = 1 or
p = +, tha t is, P(aJ =
+
4- Q
9.
Thus P(a,)
= & and
= Q
=
Q
4.
17.6. A coin is weighted so that heads is twice as likely to appear as tails.
Find P(T)
and P(H).
Let P I T )= p ; then P(H) = 2 p Now set the sum of the probabilities equal to one: p
or p = $. Thus P(T)= p = Q and P ( H ) = 2 p = 3.
+2p
= 1
17.7. Two men, 7n1 and m,, and three women, w,,w, and w,,are in a chess tournament.
Those of the same sex have equal probabilities of winning, but each man is twice
as likely to win as any woman. (i) Find the probability that a woman wins the
tournament. (ii) If m, and 20, are married, find the probability that one of them
wins the tournament.
Set P(zul)= p ; then P(w2)= P ( w 3 )= p and P ( m l )=P(m,) = 2 p . Next se t the sum of the
probabilities of the five sample points equal to one: p p p 21, 2 p = 1 or p = +,
+ + + +
We seek
ii) P({wl, w,,wn}) and (ii) P({ml,wl}).
Then, by definition,
+ P(WQ)= 3 + + + 3 = 4
P({rnl,w,})= P ( m J + P(W1) = 9 + + = $
P({Wl,
w,,w,})
+
= P(Wl) P(W,)
CHAP. 171
191
PROBABILITY
17.8. Let a die be weighted so that the probability of a number appearing when the die
is tossed is proportional to the given number (e.g. 6 has twice the probability of
appearing as 3). Let A = {even number), B = {prime number), C = {odd number).
(i)
Describe the probability space, i.e. find the probability of each sample point.
(ii) Find P ( A ) , P(B) and P ( C ) .
(iii) Find the probability that: ( a ) an even o r prime number occurs;
prime number occurs; (c) A but not B occurs.
(i)
( b ) an odd
Let P ( 1 ) = p . Then P ( 2 ) = 2 p , P ( 3 ) = 3 p , P ( 4 ) = 4 p , P ( 5 ) = 5 p and P ( 6 )= 6 p . Since the sum
of the probabilities must be one, we obtain p 2 p 3 p 4 p 5 p 613 = 1 o r p = 1 / 2 1 . Thus
1
2
P(1) = 21,
P ( 2 ) = E,
P ( 3 ) $, P(4) =
P(5) =
P(6) = $
+ + + + +
i,
(ii) P ( A ) = P ( ( 2 , 1, 6 ) ) = +,
P ( B ) = P ( ( 2 , 3, 5 ) ) =
g,
&,
P(C) = P({1,3,5}) =
+.
(iii) ( a ) The event that an even o r prime number occurs is A L I B = { 2 , 4 , 6 , 3 , 5 } , o r that 1 does
not occur. Thus P ( A u B ) = 1- P(1) =
g.
( b ) The event t h a t a n odd prime number occurs is
8
P ( { 3 , 5 ) ) = 21.
(c)
BnC =
(3,5}.
Thus
P(BnC) =
The event that A but not B occurs is A n B c = { 4 , 6 ] . Hence P ( A n B c ) = P ( { 4 , 6 ) ) =
8.
17.9. Find the probability p of an event if the odds that it will occur are a : b, that is,
‘‘a to b”.
The odds that an event with probability p occurs is the ratio p : (1- p ) . Hence
a
a
-P
l - p -- - b
or b p = a - a p
or a p + b p = a
or p = a f b
17.10. Find the probability p of an event if the odds that it will occur are “3 to 2”.
i.
2
,= f from which p =
We can also use the formula of the preceding problem t o obtain
1-P
3
3
the answer directly: p =
=
= 5.
5
EQUIPROBABLE SPACES
17.11. Determine the probability p of each event:
(i) an even number appears in the toss of a f a i r die;
(ii) a king appears in drawing a single card from an ordinary deck of 52 cards;
(iii) a t least one tail appears in the toss of three fair coins;
(iv) a white marble appears in drawing a single marble from an urn containing
4 white, 3 red and 5 blue marbles.
(i)
The event can occur in three ways ( a 2 , 4 o r 6) out of 6 equally likely cases; hence p =
(ii) There a r e 4 kings among the 52 cards; hence p =
=
=
i.
3.
(iii) If we consider the coins distinguished, then there a r e 8 equally likely cases: H H H , H H T ,
HTH,HTT,THH,THT,TTH,TTT.Only the first case is not favorable to the given event;
hence p = $.
(iv) There are 4 - 3
+5
= 12 marbles, of which 4 a r e white; hence p =
5 = f.
PROBABILITY
192
[CHAP. 17
17.12. Two cards are drawn a t random from an ordinary deck of 52 cards. Find the probability p that (i) both are spades, (ii) one is a spade and one is a heart.
(i)
There a r e
it2)= 1326 ways to draw 2 cards from 52 cards.
There ar e
(i3)= 78 ways to d r a w 2 spades from 1 3 spades; hence
P
=
number of ways 2 spades can be drawn
number of ways 2 cards can be drawn
- __
78
- 1326
- 3
- 51
(ii) Since there a r e 13 spades and 13 hearts, there a r e 13 * 13 = 169 ways to dra w a spade and a
heart; hence p = 1326 = 102
=.
17.13.Three light bulbs are chosen at random from 15 bulbs of which 5 are defective.
Find the probability p that (i) none is defective, (ii) exactly one is defective, (iii) a t
least one is defective.
There a r e ):'(
(i)
= 455 ways to choose 3 bulbs from the 15 bulbs.
Since there a r e 15 - 5 = 10 non-defective bulbs, there a r e (f,")
= 120 ways to choose 3 nondefective bulbs.
Thus p = L455
?! = %.
91
(ii) There ar e 5 defective bulbs and
(120)
= 45 different pairs of non-defective bulbs; hence there
a r e 5 * 45 = 225 ways to choose 3 bulbs of which one is defective.
22s
Thus p =
45j
45
= 91.
(iii) The event t h a t at least one is defective is the complement of the event t h a t none a re defective
24
which has, by (i), probability 91.
24 - 67
Hence p = 1-91
-
91.
17.14. Two cards are selected a t random from 10 cards numbered 1 to 10. Find the probability p that the sum is odd if (i) the two cards are drawn together, (ii) the two
cards are drawn one after the other without replacement, (iii) the two cards are
drawn one after the other with replacement.
(i)
There a r e
(io)= 45 ways t o
and the other is even.
5.5
5
select 2 cards out of 10. The sum is odd if one number is odd
There a re 5 even numbers and 5 odd numbers; hence there a r e
26 ways of choosing an even and a n odd number.
25
5
Thus p = 45 = 9.
(ii) There ar e 1 0 . 9 = 90 ways t o draw two cards one a fte r the other without replacement.
There a r e 5 5 = 25 ways t o draw a n even number and then a n odd number, and 5 5 = 25
ways to draw an odd number and then a n even number; hence p =
7 - 5 - g.
-
(iii) There ar e 1 0 . 1 0 = 100 ways t o draw two cards one a fte r the other with replacement. As
in (ii), there a r e 5 . 5 = 25 ways t o dra w a n even number and then a n odd number, and
-
25-25
-
5 5 = 25 ways to d r aw an odd number and then a n even number; hence p = 100 -
50 - 1
- 2.
17.15. An ordinary deck of 52 cards is dealt out a t random to 4 persons (called North, South,
East and West) with each person receiving 13 cards. Find the probability p of each
event (answers may be left in combinatorial and factorial notation):
CHAP. 171
193
PROBABILITY
(i) North receives all 4 aces.
(ii) North receives no aces.
(iii) North receives 5 spades, 5 clubs, 2 hearts and 1 diamond.
(iv) North and South receive between them all the spades.
(v) North and South receive between them all the aces and kings.
(vi) Each person receives a n ace.
(i)
North can be dealt 13 cards in (!:)
ways. If he receives the 4 aces, then he can be dealt
the other 9 cards from the remaining 48 cards in (",") ways.
(E)ways.
(ii) North can be dealt 13 cards in
froni the remaining 48 cards in
Thus
If he receives no aces, then he can be deaIt 13 cards
ways. T h u s
ways. He can be dealt 5 spades in (i3) ways, 5 clubs in
(iii) North can be dealt 13 cards in ():
(F) ways, 2 hearts in (i3)ways, and 1 diamond in
(If)
= 13 ways. Hence p =
(iv) North and South can be dealt 26 cards between them in
(ii) ways.
13 13 13 13
(5 )(5 )(2
1 )
If they receive the
13 spades, then they can be dealt 13 other cards from the remaining 39 cards in ):(;
Hence p = (;!)
(3
ways.
/ (!:).
(ii)ways.
(v) North and South can be dealt 26 cards between them in
and the 4 kings, then they can be dealt the other 2 6 - 8
44
52 - 8 = 44 cards in (18)
ways. Thus p = ):;( / (;",.
If they receive the 4 aces
= 18 cards from the remaining
ways (an ordered partition).
(vi) The 52 cards can be distributed among the 4 persons in 1 3 ! l&
The
remaining
48
cards can be distributed
Now each person can get a n ace in 4 ! ways.
among the 4 persons in
481
12! 121 12! 12!
Thus
ways.
48!
4 ! 1 2 ! 12: l e ! IZ!
P =
521
13! 131 13! 131
-
4 ! 134
52 51 * 50 * 49
-
133
17 * 25 * 49
17.16. Six married couples a r e standing in a room.
(i)
If 2 people a r e chosen a t random, find the probability p that (a) they a r e
married, ( b ) one is male and one is female.
(ii) If 4 people are chosen a t random, find the probability p t h a t (a) 2 married
couples are chosen, ( b ) no married couple is among the 4, (c) exactly one
married couple is among the 4.
(iii) If the 12 people are divided into six pairs, find the probability p that (a) each
pair is married, ( b ) each pair contains a male and a female.
194
PROBABILITY
(i)
There a r e
(7)= 66 ways t o choose 2 people from the
6
((1)
[CHAP. 17
12 people.
1
There a r e 6 married couples: hence p = 66 = fi.
( b ) There a r e 6 ways to choose a male and 6 ways to choose a female; hence p =
(ii) There a r e
6 6
6
66
= 11.
(y)= 495 ways t o choose 4 people from the 12 people.
( a ) There are (:)
= 15 ways to choose 2 couples from the 6 couples; hence p
15
1
a = ,%.
1
( b ) The 4 persons come from 4 different couples. There a r e (!) = 15 ways to choose 4 couples
from the 6 couples, and there a r e 2 ways to choose one person from each couple. Hence
p = - 2.2.2.2.15 - 16
495
(c)
- 33.
This event is mutually disjoint from the preceding two events (which are also mutually
16
disjoint) and a t least one of these events must occur. Hence p f $ f = 1 o r p = 33.
(iii) There a r e
in each.
12!
- 12!
2!2!2!2!2!2! -
5
; ways to partition the 12 people into 6 ordered cells with 2 people
((0 The 6 couples can be placed into the 6 ordered cells in 6! ways.
Hence p =
& = &.
( b ) The six men can be placed one each into the 6 cells in 6 ! ways, and the 6 women can be
6: 6 !
16
placed one each into the 6 cells in 6! ways. Hence p = 12!/26= 231.
17.17. If 5 balls are placed in 5 cells a t random, find the probability p that exactly one cell
is empty.
There a r e 53 ways to place the 5 balls in the 5 cells. If exactly one cell is empty, then another
cell contains 2 balls and each of the remaining 3 cells contains one ball. There a r e 5 ways to
select the empty cell, then 4 ways t o select the cell containing 2 balls, and (:) = 10 ways t o select
2 balls to go into this cell. Finally there a r e 3 ! ways to distribute the remaining 3 balls among
103! - 48
the remaining 3 cells. Thus p = j.q.
125'
55
17.18.A class contains 10 men and 20 women of which half the men and half the women
have brown eyes. Find the probability p that a person chosen a t random is a man
or has brown eyes.
Let A = {person is a man} and B = {person has brown eyes}.
Then P ( A ) = lo
=
30
$,
P(Bj =
30
=
1
2'
p = P(AuB) = P(A)
P(AnB)=
30
=
+ P(B)- P ( A n B )
5.
We seek P ( A u B ) .
Thus, by Theorem 17.4,
=
g++-Q
=
3
PROOFS OF THEOREMS
17.19. Prove Theorem 17.3(i): If $3 is the empty set, then P ( @ )= 0.
Let A be a n y event; then A and (3 a r e disjoint and A U $3 = A .
+
Thus P ( A ) = P ( A U @ ) = P ( A ) P(@), from which our result follows.
17.20. Prove Theorem 17.3(ii): For any event A , P ( A C )= 1- P ( A ) .
The sample space S can be decomposed into t h e mutually exclusive events A and Ac:
Thus 1 = P ( S ) = P ( A WAC) = P ( A ) P(Ac), from which our result follows.
S = A UAC.
CHAP. 171
195
PROBABILITY
17.21. Prove Theorem 1'7.3(iii):
= P ( A )- P ( A n B )
P(A\B)
A can be decomposed into t h e mutually exclusive events A \ B and
A n B : A = ( A \B) u ( A n B ) . Thus
P(A) = P(A\ B)
from which our result follows.
+ P ( An B )
A is shaded.
17.22. Prove Theorem 17.3(iv):
If A c B , then P ( A )6 P(B).
If A c B , then B can be decomposed into the mutually exclusive events
A and B\A.
Thus
P ( B ) = P ( A ) P(B\A)
+
The result now follows from t h e f act t h a t P ( B \A)
2
0.
B is shaded.
17.23. Prove Theorem 17.4: For any events A and B,
P ( AU B ) = P ( A ) P ( B ) - P ( A n B )
+
A \B
Note t h a t A u B can be decomposed into the mutually exclusive events
and B : A UB = ( A \B)UB.
Thus, using Theorem 1'7.3(iii), we have
+
P ( A u B ) = P ( A \B)
P(B)
= P ( A ) - P ( An B ) f P ( B )
= P ( A ) P ( B ) - P ( AnB )
~
+
A U B is shaded.
17.24.Prove Corollary 17.5: For any events A , B and C,
P ( A u B u C ) = P ( A ) + P ( B ) + P ( C ) -P ( A n B ) - P ( A n C ) - P ( B n C ) + P ( A n B n C )
Let D = B u C .
Then
P(AnD) = P(AnB)
A n D = A n ( B u C ) = ( A n B ) U ( A n C ) and
+ P(AnC)- P(AnBnAnC) =
Thus
P(AnB)
+ P(AnC)- P ( A n B n C )
+
P(AuBuC) = P(AUD) = P(A) P(D) - P(AnD)
= P ( A ) P ( B ) P ( C ) - P ( B n C ) - [ P ( A n B )+ P ( A n C ) - P ( A n B n C ) ]
= P(A) P(B) P(C) - P ( B n C ) - P ( A n B ) - P(AnC) P ( A n B n C )
+
+
+
+
+
MISCELLANEOUS PROBLEMS
17.25. Let A and B be events with P ( A ) = Q, P ( B ) = & and P ( A n B ) = &. Find (i) P ( AU B ) ,
(ii) &'(Ac) and P ( B c ) , (iii) P ( A c n B c ) , (iv) P ( A c U B c ) , (v) P ( A n B c ) , (vi) P ( B n A c ) .
(i)
P ( A u B ) = P ( A ) + P ( B )- P ( A n B ) = Q
(ii) P(Ac) = 1 - P ( A ) = 1 -
#
= Q
and
+&-$ = Q
P ( B ) = 1- P(B) = 1 - Q = 1
2
(iii) Using De Morgan's Law, ( A u B ) c = ACnBc, we have
= 1-
8
=
8
P(AcuBc) = P ( ( A n B ) c ) = 1 - P ( A n B ) = 1 -
&
=
p
P (A cn B c) = P ( ( A u B ) c ) = 1
- P(AuB)
(iv) Using De Morgan's Law, ( A n B ) c = A C U B Cwe
, have
[CHAP. 1 7
PROBABILITY
196
Equivalently,
P(AcUt7c) = P(Ac)
+ P(Bc) - P(AcnBC)
= P(A)-P(AnB) =
(v, P ( A n B c ) = P(A\B)
(vi) P ( B n A c ) = P ( B ) - P ( A n B ) =
8-$
(iii
Substitute in P ( A u B ) = P ( A ) + P ( B ) - P ( A n B ) to obtain
= .20
+
=
8
4
and P ( A n B ) = Q. Find
4 = ++P(B)-&
& -&
=
Number
1
2
3
4
5
6
Frequency
14
17
20
18
15
16
(iii, P ( A n B c ) = P ( A ) - P ( A n B ) =
number of successes
total number of trials
The relative frequency f =
8
= Q
P(A) = 1-P(Ac) = 1 -
20
2
+-& = 4
(i)
(i) f =
+ 1-
&
=
17.26. Let A and B be events with P ( A U B ) = $, P ( A c ) =
(i) P ( A ) , (ii) P(B),(iii) P ( A n B c ) .
t
6
8
(ii) f = $ = .15
or P(B) =
3.
'
17+18+16 -
(iii) f = 1
0
0 - .51
17+20+15
-
(iv) f = 1
0
0 - .52
17.28. Let S = {ul, a,,. . ., as) and T = {b,, b,, . . ., b,} be finite probability spaces.
Let the number p L j= P(a,)P ( b , ) be assigned to the ordered pair (ai, bj) in the product
set S x T = ((8,t ): s E S, t E T } . Show that the pli define a probability space on
S X T,i.e. that the p , are non-negative and add up t o one. (This is called the product
probability space. We emphasize that this is not the only probability function that
can be defined on the product set S x T.)
Since P(a,),P(b,)
Pll
1
0, for each i and each j , pij = P(uJ P(b,)
+ P12 + . - .+ Plt + P21 + P 2 2 +
=
+ Pzt +
* ' '
0.
Furthermore,
+ Psl + Ps2 + . . * + Pst
+ + P(a1)P(bt) + + P(aJ P(b1) + + P(as)P(b,)
+ P(as)[P(bl)
+ . +P(bt)]
P(al)[P(bl)+ . + P(b,)] +
P(al).l +
+ P(a,) '1 = P(al) + + P(as) = 1
= P(a1)P(bJ
=
* * .
2
* * *
* * *
*
* "
* * *
* * *
* *
a
'
.
CHAP. 171
PROBABILITY
197
Supplementary Problems
SAMPLE SPACES AND EVENTS
17.29. Let A and B be events. Find a n expression and exhibit the Venn diagram for the event t h a t
(i) A or not B occurs, (ii) neither A nor B occurs.
17.30.
Let A , B and C be events. Find an expression and exhibit the Venn diagram f o r the event tha t
(i) exactly one of the three events occurs, (ii) a t least two of the events occurs, (iii) none of
the events occurs, (iv) A o r B , but not C, occurs.
17.31.
Let a penny, a dime and a die be tossed.
(i) Describe a suitable sample space S.
(ii) Express explicitly the following events: A = {two heads and a n even number appears},
B = {a 2 appears}, C = {exactly one head and a prime number appears}.
(iii) Express explicitly t h e event t h a t ( a ) A and B occur, ( b ) only B occurs, ( c ) B o r C occur.
FINITE PROBABILITY SPACES
17.32. Which function defines a probability space on S = {a,,a,,a,}?
4
(i) P(aJ = $, P(ad = g, P b 3 ) = 3
(ii) P ( a l ) = 8, P(a,) = -Q, P(a,) = Q
(iii) P ( a l ) = Q, &‘(a2)= +, &‘(a,) =
(iv) P(al) = 0, P(az)= 5, P(u3)= $
17.33.
Let P be a probability function on S = { a l , a ~ , a 3 } . Find P ( a l ) if (i) P(u2) = Q and P(u3) = 2,
(ii) P ( a l ) = 2 P ( a 2 ) and P(a3) = $, (iii) P({aZ,a3}) = 2 P ( a 1 ) , (iv) P(a3) = 2P(u,) and
P(a2) = 3 P ( a 1 ) .
17.34.
Let A and B be events with P ( A u B ) =
P ( A n Bc).
17.35.
Let A and B be events with P ( A ) =
P(AcuBc) and P ( B n A c ) .
4,
P ( A n B ) = $ and P ( A c ) =
4,P ( A U B ) = 3
and P(B)=
3.
8.
Find P ( A ) , P ( B ) and
Find P ( A n B ) , P(AcnBc),
17.36. A coin is weighted so t h a t heads is three times as likely to appear a s tails. Find P(H) and P(T).
17.37. Three students A , B and C a r e in a swimming race. A and B have the same probability of winning
and each is twice as likely t o win as C. Find the probability t h a t B o r C wins.
17.38.
A die is weighted so t h a t t h e even numbers have the same chance of appearing, the odd numbers
have the same chance of appearing, and each even number is twice as likely to appear a s a n y
odd number. Find the probability t h a t (i) a n even number appears, (ii) a prime number appears,
(iii) a n odd number appears, (iv) an odd prime number appears.
17.39. Find the probability of an event if t h e odds t h a t it will occur a r e (i) 2 to 1, (ii) 5 to 11.
17.40.
I n a swimming race, the odds t h a t A will win a r e 2 to 3 and the odds t h a t B will win a r e 1 to 4.
Find the probability p and t h e odds t h at A or B win the race.
EQUIPROBABLE SPACES
17.41. A class contains 5 freshmen, 4 sophomores, 8 juniors and 3 seniors. A student is chosen at random
to represent the class. Find the probability tha t the student is (i) a sophomore, (ii) a senior,
(iii) a junior o r senior.
17.42. One card is selected at random from 50 cards numbered 1 to 50. Find the probability t h a t the
number on the card is (i) divisible by 5 , (ii) prime, (iii) ends in the digit 2.
17.43.
Of 10 girls in a class, 3 have blue eyes. If two of the girls a r e chosen at random, what is the
probability t h a t ( i ) both have blue eyes, (ii) neither ha s blue eyes, (iii) at least one has blue eyes?
17.44.
Three bolts and three n u t s a r e put in a box.
probability t h a t one is a bolt and one a nut.
If two pa rts a r e chosen at random, find the
198
PROBABILITY
[CHAP. 17
17.45. Ten students, A , B , . . . , a r e in a class. If a committee of 3 is chosen a t random from the class,
find the probability t h a t ( i ) A belongs t o t h e committee, (ii) B belongs t o the committee,
( i i i ' d and B belong to the committee, tivi A or B belongs t o the committee.
17.46. A class consists of 6 girls and 10 boys. If a committee of 3 is chosen at random from the class,
find the probability t h a t (i) 3 boys a r e selected, (ii) exactly 2 boys a r e selected, (iii) a t least one
boy is selected, (iv) exactly 2 girls a r e selected.
17.47. A pair of f a i r dice is tossed.
greater than 1.
Find the probability t h a t the maximum of the two numbers is
17.48. Of 120 students, 60 a r e studying French, 50 a r e studying Spanish, and 20 a r e studying French
and Spanish. If a student is chosen a t random, find the probability t h a t the student (i) is studying
French or Spanish, (ii) is studying neither French nor Spanish.
17.49. Three boys and 3 girls sit in a row. Find the probability t h a t
(iil the boys and girls sit in alternate seats.
17.50. A die is tossed 50 times.
occurrence:
(i) the 3 girls sit together,
The following table gives the six numbers and their frequency of
Find the relative frequency of the event ii) a 4 appears, iii) a n odd number appears, (iii) a prime
number appears.
Answers to Supplementary Problems
17.29. (i) A u B ~ ,(ii) ( A u B ) ~
17.30. (i) (AnBcnCc)u (BnAcnCc)u (CnAcnBc)
(ii) ( A n B )u ( A n C ) U ( B n C )
17.31. (i)
(iii) ( A u B u C).
(iv) ( A u B ) n CC
S = (HH1, HH2, HH3, HH4, HH5, HH6, HT1, HT2, HT3, HT4, HT5, HT6,
TH1, TH2, TH3, TH4, TH5, TH6, TT1, TT2, TT3, TT4, TT5, TT6)
(ii) A = (HH2, HH4, HH6}, B = (HH2, HT2, TH2, TT2}, C = {HTZ, TH2, HT3, TH3, HT5, TH5)
(iii) ( a ) A n B = (HH2)
( b ) B\(AuC)
= (TT2)
(c) B u C = (HH2, HT2, TH2, TT2, HT3, TH3, HT5, TH5}
17.32. (i) no, (ii) no, (iii) yes, (iv) yes
17.33.
17.34.
17.35.
17.36.
17.37.
3
5
17.44.
17.38.
A,
(i) k,
17.45. (i)
17.39.
17.46.
3
17.40. p = 5;
the odds a r e 3 to 2.
17.47. 35
17.41. (i)
I, iii) &,
(iii) 3
11
17.42.
17.43. (i)
A, (ii) &, (iii)
8
(ii)
(ii)
A, (iii) 3 (iv) 15
$,
1
8
27
15
(iii) 28, (iv)
17.48. (i) 4,
3 (ii)
17.49. (i)
i,(ii)
1
17.50. (i)
&,
2, (iii) 8
(ii)
E
Chapter 18
Conditional Probability. Independence
CONDITIONAL PROBABILITY
Let E be an arbitrary event in a sample space S with P ( E )> 0. The probability that
an event A occurs once E has occurred or, in other words, the conditional probability of
A given E , written P ( A ' E ) , is defined as follows:
P ( AnE )
P ( A ' E )=
P(E)
As seen in the adjoining Venn diagram, P(A I E ) in a
certain sense measures the relative probability of A
with respect to the reduced space E .
In particular, if S is an equiprobable space and we let \A]denote the number of elements
in the event A , then
P(AnE) (AnE(
(El
and so
P(A1E) =
P ( A n E ) = --,A n E l
P(E) = P(E)
IEl
IS
IS
In other words,
Theorem 18.1:
If S is an equiprobable space and A and E are events of S, then
number of elements in A n E
number of elements in E
number of ways A and E can occur
P(A1E) =
number of ways E can occur
IE,
or
Example 1.1:
Let a pair of f a i r dice be tossed. If the sum is 6, find the probability t h a t one of
the dice is a 2. In other words, if
E = {sum is 61 = {(1,5), (2,4), (3,3), (4,2), (5,111
A = {a 2 appears on at least one die}
and
find P ( A I E).
Now E consists of five elements and two of them, (2,4) and (4,2), belong to A :
A n E = {(2,4), (4,2)}. Then P ( A I E) =
i.
On t h e other hand, since A consists of eleven elements,
A = ((2, l), 12,2), (2, 3), (2,4), (2,5), ( 2 , 6 ) , ( 1 , 2 ) , ( 3 , 2 ) , (4,2), (5,% (6,211
and S consists of 36 elements, P ( A ) = $.
Example 1.2:
A couple h as two children; the sample space is S = { b b , b g , g b , g g } with probability & f o r each point. Find the probability p t h a t both children a r e boys if
( i ) i t i s known t h a t one of the children is a boy, (ii) it is known t h a t the older child
is a boy.
( i ) Here the reduced space consists of three elements, { b b , b g , g b } ; hence p =
&.
rii) Here the reduced space consists of only two elements, { b b , b g } ; hence p =
4.
199
200
[CHAP. 18
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
MULTIPLICATION THEOREM FOR CONDITIONAL PROBABILITY
If we cross multiply the above equation defining conditional probability and use the
fact that A n E = E n A , we obtain the following useful formula.
Theorem 18.2: P ( E f l A ) = P ( E )P ( A 1 E )
This theorem can be extended by induction as follows:
Corollary 18.3: P ( A l n A z n . . n A , ) = P ( A 1 ) P ( A 2 1 A 1 ) P (1 AA3l n A z )
P(A4 I A l n A e n A 3 ). . P ( A , I A1 n A p n . * * nA,-l)
9
We now apply the above theorem which is called, appropriately, the multiplication
theorem.
Example 2.1:
A lot contains 12 items of which 4 are defective. Three items a r e drawn at random from the lot one a fte r the other. Find the probability p t h a t all three a r e
non-defective.
The probability tha t the first item is non-defective is since 8 of 12 items a r e
non-defective. If the first item is non-defective, then the probability tha t the next
since only 7 of the remaining 11 items a r e non-defective.
item is non-defective is
If the first two items a re non-defective, then the probability t h a t the last item is
non-defective is
since only 6 of the remaining 10 items are now non-defective.
Thus by the multiplication theorem,
= & . $ . " = 11
5
fi
&
10
55
FINITE STOCHASTIC PROCESSES AND TREE DIAGRAMS
A (finite) sequence of experiments in which each experiment has a finite number of
outcomes with given probabilities is called a (finite) stochastic process. A convenient
way of describing such a process and computing the probability of any event is by a tree
diagram as illustrated below; the multiplication theorem of the previous section is used to
compute the probability that the result represented by any given path of the tree does occur.
Example 3.1:
We a r e given three boxes as follows:
Box I has 10 light bulbs of which 4 a r e defective.
Box I1 has 6 light bulbs of which 1 is defective.
Box I11 has 8 light bulbs of which 3 a r e defective.
W e select a box a t random and then dra w a bulb at random. W h a t is the probability p t h a t the bulb is defective?
Here we perform a sequence of two experiments:
(i) select one of the three boxes;
(ii) select a bulb which is either defective (D)or non-defective ( N ) .
The following tree diagram describes this process and gives the probability of each
branch of the tree:
CHAP. 181
CONDITIONAL PROBABILITY. INDEPENDENCE
201
The probability that any particular path of the tree occurs is, by the multiplication theorem, the product of the probabilities of each branch of the path, e.g.,
the probability of selecting box I and then a defective bulb is 3.5
1 2 = IS.
2
Now since there a r e three mutually exclusive paths which lead to a defective
bulb, the sum of the probabilities of these paths is the required probability:
= 13 . 25 + ; . ; + 1 . 33
Example 3.2:
= 113
8
360
A coin, weighted so t h a t P ( H ) = $ and P(T) = $, is tossed. If heads appears,
then a number is selected at random from the numbers 1 through 9; if tails appears, then a number is selected at random from the numbers 1 through 5. Find
the probability p that an even number is selected.
The tree diagram with respective probabilities is
Note t h a t the probability of selecting an even number from the numbers 1
through 9 is 4 since there a r e 4 even numbers out of the 9 numbers, whereas the
probability of selecting a n even number from the numbers 1 through 5 is since
there are 2 even numbers out of the 5 numbers. Two of the paths lead to a n
even number: H E and TE. Thus
p z P(E) = $ . + + ? . * = 55
5
9
Example 3.3:
3
3
135
Consider the stochastic process of t h e preceding example. If a n even number was
selected, w h a t is the probability that tails appeared on the coin? In other words,
we seek the conditional probability of T given E: P(T 1 E).
Now tails and an even number can occur only on the bottom path which has
1 2
2
probability 3.5 =
t h a t is, P ( T n E ) = 5. Furthermore, by the preceding example, P ( E ) =
Thus by the definition of conditional probability,
P ( T n E ) - 2/15 - 9
P ( T / E ) = ___
P(E) - 581135 - 29
&;
%.
~
In other words, we divide the probability of the successful path by the probability
of the reduced sample space.
INDEPENDENCE
An event B is said to be independent of an event A if the probability that B occurs is
not influenced by whether A has or has not occurred. In other words, if the probability
of B equals the conditional probability of B given A : P ( B ) = P(B1A). Now substituting
P(B) for P ( B I A ) in the multiplications theorem P ( A nB) = P ( A )P(B 1 A ) , we obtain
P(AnB) = P(A)P(B)
We formally use the above equation as o u r definition of independence.
1 Definition:]
Events A and B are independent if P ( A n B ) = P ( A ) P ( B ) ;otherwise they
are dependent.
Example 4.1:
Let a f a i r coin be tossed three times; we obtain the equiprobable space
S = {HHH, H H T , HTH, HTT, THH, THT, TTH, TTT}
202
[CHAP. 18
CONDITIONAL PROBABILITY. INDEPENDENCE
Consider the events
B = {second toss is heads)
A = {first toss is heads},
C = {exactly two heads are tossed in a row}
Clearly A and B are independent events; this fact is verified below. On the other
hand, the relationship between A and C o r B and C is not obvious. We claim
that A and C are independent, but that B and C are dependent. We have
P ( A ) = P({HHH, HHT, HTH, HTT}) =
P ( B ) = P({HHH, HHT, THH, THT}) =
P ( C ) = P({HHT, THH)) =
f
$
=
f
=
f
=
Then
P ( A n B ) = P({HHH, HHT}) = 2, P ( A n C ) = P({HHT}) = 8,
P ( B n C ) = P({HHT, THH}) = 2
Accordingly,
P ( A ) P ( B ) = 3.4 = & = P ( A n B ) , and so A and B are independent;
P ( A )P ( C ) = d * $ = 4 = P ( A n C ) , and so A and C a r e independent;
12 * $
P ( B )P(C) =
=
4
P ( B n C ) , and so B and C a r e dependent.
#
Frequently, we will postulate that two events a r e independent, or the experiment itself
will imply that two events are independent.
Example 4.2:
The probability that A hits a target is 2 and the probability that B hits it is
What is the probability that the target will be hit if A and B each shoot a t the
target?
We are given that P ( A ) = 4 and P ( B ) = 5, and we seek P ( A u B ) . Furthermore, the probability that A or B hits the target is not influenced by what the
other does; that is, the event that A hits the target is independent of the event that
B hits the target: P ( A n B ) = P ( A ) P ( B ) . Thus
2.
P ( A uB) =
-
P(A)
+ P ( B ) - P ( A nB )
t+$-'.?
4
5
= P(A)
+ P ( B ) - P ( A )P ( B )
= 11
20
Three events A, B and C are i n d e p e n d e n t if:
P ( A n B ) = P ( A ) P ( B ) , P ( A n C ) = P ( A ) P ( C ) and
i.e. if the events are pairwise independent, and
(ii) P ( A n B n C ) = P ( A ) P ( B ) P ( C )
(i)
P(BnC)= P(B)P(C)
The next example shows that condition (ii) does not follow from condition (i); in other
words, three events may be pairwise independent but not independent themselves.
Example 4.3:
Let a pair of fair coins be tossed; here S = {HH, HT, TH, TT} is an equiprobable
space. Consider the events
A = {heads on the first coin}
= {HH, HT}
B = {heads on the second coin}
= {HH, TH}
C = {heads on exactly one coin} = {HT, TH}
Then P ( A ) = P ( B ) = P ( C ) = $ = f and
P ( A n B ) = P({HH}) = $,
P ( A n C ) = P({HT}) =
t,
P ( B n C ) = ({TH}) = $
Thus condition ii) is satisfied, i.e., the events are pairwise independent.
A n B n C = $3 and so
However,
P ( A n B n C ) = P(@) = 0 # P(A)P(B)P(C)
In other words, condition (ii) is not satisfied and so the three events are not independent.
CONDITIONAL PROBABILITY. INDEPENDENCE
CHAP. 181
203
Solved Problems
CONDITIONAL PROBABILITY IN EQUIPROBABLE SPACES
18.1. A pair of fair dice is thrown. Find the probability p that the sum is 10 or greater if
(i) a 5 appears on the first die, (ii) a 5 appears on at least one of the dice.
ii)
If a 5 appears on the first die, then the reduced sample space is
A = { [ 5 , 1 ) , 15,2), ( 5 , 3 ) , (5,4), ( 5 , 5 ) , ( 5 , 6 ) }
The sum is 10 o r greater on two of the six outcomes: ( 5 , 5 ) , ( 5 , 6 ) . Hence p =
2
1
= 3.
(ii) If a 5 appears on a t least one of the dice, then the reduced sample space has eleven elements:
B = ((5, I ) , (5,2), ( 5 , 3 ) , 15,4), 15,5), ( 5 , 6 ) , ( 1 , 5 ) , ( 2 , 5 ) , 13,5), (4,5),(6,511
3
The sum is 10 or greater on three of the eleven outcomes: ( 5 , 5 ) , ( 5 , 6 ) , (6,5). Hence p = a.
18.2. Three fair coins are tossed. Find the probability p that they are all heads if (i) the
first coin is heads, (ii) one of the coins is heads.
The sample space has eight elements:
(i)
S = {HHH, H H T , HTH, HTT, THH, THT, TTH, TTT}.
If the first coin is heads, the reduced sample space is A = {HHH, HHT, HTH, HTT}. Since
the coins are all heads in 1 of 4 cases, p = $.
(ii) If one of the coins is heads, the reduced sample space is B = {HHH, HHT, HTH, HTT, THH.
THT, TTH}. Since the coins are all heads in 1 of 7 cases, p =
3.
18.3. A pair of fa i r dice is thrown. If the two numbers appearing are different, find the
probability p that (i) the sum is six, (ii) an ace appears, (iii) the sum is 4 or less.
Of the 36 ways the pair of dice can be thrown, 6 will contain the same numbers: ( l , l ) , (2,2),
Thus the reduced sample space will consist of 36 - 6 = 30 elements.
. . ., ( 6 , 6 ) .
(i)
The sum six can appear in 4 ways: (1,5), ( 2 , 4 ) , (4,2), ( 5 , l ) . (We cannot include ( 3 , 3 ) since
4
2
the numbers are the same.) Hence p = 30 = 5.
(ii) An ace can appear in 10 ways: ( 1 , 2 ) , (1,3),
10
p=,=,.
1
. . .,
( 1 , 6 ) and (2, l), ( 3 , 1 ) ,
. . .,
(iii) The sum of 4 or less can occur in 4 ways: (3, I), (1,3), (2, l), ( 1 , 2 ) . Thus p =
( 6 , l ) . Hence
6
$.
18.4. Two digits are selected at random from the digits 1 through 9. If the sum is even,
find the probability p that both numbers are odd.
The sum is even if both numbers a r e even or if both numbers a r e odd. There a r e 4 even
numbers i 2 , 4 , 6 , 8 ) ; hence there are (;) = 6 ways to choose two even numbers. There a r e 5 odd
numbers i l , 3 , 5 , 7 , 9 ) : hence there are
= 1 0 ways to choose two odd numbers. Thus there a r e
6 10 = 16 ways to choose two numbers such t h a t their sum is even: since 10 of these ways occur
10 - 5
when both numbers are odd, p = 16 - 2.
+
(3)
18.5. A man is dealt 4 spade cards from an ordinary deck of 52 cards. If he is given
three more cards, find the probability p that at least one of the additional cards is
also a spade.
CONDITIONAL PROBABILITY. IN D EPEN D EN CE
204
[CHAP. 18
Since he is dealt 4 spades, there a r e 5 2 - 4 = 48 cards remaining of which 1 3 - 4 = 9 a r e
spades. There a r e (",") = 17,296 ways in which he can be dealt three more cards. Since there a r e
48 - 9 = 39 cards which a r e not spades, there a r e
= 9139 ways he can be dealt three cards
which ar e not spades. Thus the probability q t h a t he is not dealt another spade is q = -;
157
hence p = 1 - q = 8
17,296 '
(y)
18.6. Four people, called North, South, East and West, are each dealt 13 cards from an
ordinary deck of 52 cards.
(i) If South has no aces, find the probability p that his partner North has exactly
two aces.
(ii) If North and South together have nine hearts, find the probability p that East
and West each has two hearts.
(i)
There ar e 39 cards, including 4 aces, divided among North,
():
ways t h a t North can be dealt 13 of the 39 cards. There a r e
the four aces, and ):(;
ways he can be dealt 11 cards from the
not aces. Th u s
(;)(::)
- 6.12.13.25.26 p z - 36 * 37 * 38 * 39
(3
E a s t and West.
There a re
(;) ways he can be dealt 2 of
39 - 4 = 35 cards which a r e
-650
2109
ways
(ii) There a r e 26 cards, including 4 hearts, divided among E a s t and West. There a r e (;!)
t h at , say, E a s t can be dealt 13 cards. (We need only analyze East's 1 3 cards since West must
have the remaining cards.) There a r e
ways E a s t can be dealt 2 hearts from 4 hearts,
ways he can be dealt 11 non-hearts from the 26 - 4 = 22 non-hearts. Thus
and );(
(i)
MULTIPLICATION THEOREM
18.7. A class has 12 boys and 4 girls. If three students are selected a t random from the
class, what is the probability p that they are all boys?
The probability t h a t t h e first student selected is a boy is 12/16 since there are 12 boys out of
16 students. If the first student is a boy, then the probability t h a t the second is a boy is 11/15
since there ar e 11 boys left out of 15 students. Finally, if the first two students selected were boys,
then the probability t h a t the third student is a boy is 10114 since there a r e 10 boys left out of
14 students. Thus, by the multiplication theorem, the probability t h a t all three a r e boys is
Another Method.
There ar e
(i6)=
(y)= 220 ways t o select 3 boys out of
560 ways to select 3 students of the 16 students, and
12 boys; hence p =
=
2 6.
A T h i i d M et hod. If the students a re selected one a fte r the other, then there a r e 1 6 * 15 * 14 ways
to select three students, and 12 * 11 * 10 ways to select three boys; hence p = -= $.
18.8. A man is dealt 5 cards one after the other from an ordinary deck of 52 cards. What
is the probability p that they are all spades?
The probability t h a t t h e first card is a spade is 13/52, the second is a spade is 12/51, the third
is a spade is 11/50, the fourth is a spade is 10/49, and the last is a spade is 9/48. (We assumed in
each case t h a t the previous cards were spades.) Thus p =
=
2.
$ 9
$ 0
%* &
-.
CHAP. 181
205
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
18.9. An urn contains 7 red marbles and 3 white marbles. Three marbles are drawn from
the urn one after the other. Find the probability p that the first two are red and
the third is white.
The probability that the first marble is red is 7/10 since there a r e 7 red marbles out of 1 0
marbles. If the first marble is red, then the probability t h a t the second marble is red is 6/9 since
there a r e 6 red marbles remaining out of the 9 marbles. If the first two marbles a r e red, then
the probability t h a t the third marble is white is 3/8 since there a r e 3 white marbles out of the
8 marbles in the urn. Hence, by the multiplication theorem,
&.E..
?,
9
8
=
w
18.10. The students in a class are selected a t random, one after the other, for an examination. Find the probability p that the boys and girls in the class alternate if (i) the
class consists of 4 boys and 3 girls; (ii) the class consists of 3 boys and 3 girls.
(i)
If the boys and girls a r e to alternate, then the first student examined must be a boy. The
probability t h a t the first is a boy is 417. If the first is a boy, then the probability t h a t the
second is a girl is 3/6 since there a r e 3 girls out of 6 students left. Continuing in this manner,
we obtain the probability t h a t the third is a boy is.3/5, t h e fourth is a girl is 2/4, the fifth is a boy is
213, the sixth is a girl is 112, and the last is a boy is 1/1. Thus
p = *.?.3.2.2.1.1
,
6
5
4
3 2
-r
1 - 3 5
(ii) There a r e two mutually exclusive cases: the first pupil is a boy, and the first is a girl. If the
first student is a boy, then by the multiplication theorem the probability pl t h a t the students
alternate is
=
g.;.;.;.;.;
If the first student is a girl, then by the multiplication theorem the probability
the students alternate is
p 2 = 36 . 0~ . $ . $ .2 1 .1 1 = 1
YO
Thus
p = p,+pz
1
= %+& =
p2
that
$.
CONDITIONAL PROBABILITY
18.11. In a certain college, 25% of the students failed mathematics, 15% of the students
failed chemistry, and 10% of the students failed both mathematics and chemistry.
A student is selected at random.
(i) If he failed chemistry, what is the probability that he failed mathematics?
(ii) If he failed mathematics, what is the probability that he failed chemistry?
(iii) What is the probability that he failed mathematics o r chemistry?
Let M = {students who failed mathematics} and C = {students who failed chemistry}; then
P ( M ) = 25% = 2 5 ,
(i)
P ( C ) = 15% = .15,
P ( M n C ) = 10% = . l o
The probability that a student failed mathematics, given t h a t he has failed chemistry is
- .lo - 2
P ( M )C) = P ( M n C ) .15
3
P(C)
~
(ii) The probability t h a t a student failed chemistry, given t h a t he has failed mathematics is
(iii)
P(MuC) = P(M)
+ P(C) - P ( M n C ) =
.25
+ .15 - . l o
= .30 =
3
206
CONDITIONAL PROBABILITY. INDEPENDENCE
[CHAP. 18
18.12. Let A and B be events with P ( A ) = +, P(B) = Q and P ( A n B ) = $. Find (i) P ( A [ B),
iii) P ( B I A ) , (iii) P ( AuB), (iv) P(Ac1 Bc), (v) P(BcI A').
(iii) P ( A U B ) = P ( A )
+ P(B)
1
-
P ( A nB) = 5
+3
1
-
s
-
4 =
riv) First compute P(Bc) and P(AcnBc). P(Bc) = 1 - P ( B ) = 1 - Q = +. By De Morgan's
~ AcnBc; hence P(AcnBc) = P ( ( Ad B ) c ) = 1 - P(A U B ) = 1 - 1
1 2 = 2.
12
Law, ( A u B ) =
18.13. Let A and B be events with P ( A ) = 8, P ( B ) = # and P ( A U B ) = 9. Find P ( A 1 B )
and P ( B I A ) .
+
First compute P ( A n B ) using the formula P ( A UB) = P ( A ) P ( B )- P(A n B ) :
_a -
4 + 8 -P(AnB)
P(AnB) =
or
4
and P(d) = P ( e ) = &i, and
18.14. Let S = { a , b , c , d , e } with P(a) = P ( b ) = A, P ( c ) =
let E = { b , c , d } . Find: (i) P ( A IE) where A = {a,b , c } ; (ii) P ( B I E ) where
B = { c , d , e } ; (iii) P ( C I E ) where C = { a , c , e } ; (iv) P ( D IE) where D = { a , e } .
In each case use the definition of conditional probability
+ P(c) + P(d) = &
P ( A n E ) = P ( b ) + P ( c ) = %+; = &,
Here P ( E ) = P ( { b , c , d } ) = P ( b )
(ii
APE = {b,c},
J
-
S
tii) B n E = { c , d } , P ( B n E ) = P(c) t P ( d ) =
f+&
=
i,
P(x E) =
+$
=
P(* nE )
p
.
(El
5.
& 1
(AnE) and P ( A / E ) = P P
(E) - B
~
and P ( B / E ) =
-
2'
P(BnE) - - 5
- a
P ( E ) - % - 6'
~
P(CnE) 1
1
- 5 - (iii) C n E = {c}, P ( C n E ) = P ( c ) = +, and P(C 1 E ) =
P(E)
# 3'
~
P(DnE) - 0
(iv) D n E = $3, P ( D n E ) = P ( @ ) = 0, and P(DI E ) = ___
P(E) - 3 = O.
18.15. Show that the conditional probability function P(* 1 E ) satisfies the three axioms of
Theorem 17.1; that is:
[P,] For any event A, 0 L P ( A 1 E ) f 1.
[P,] For the certain event S , P ( S I E ) = 1.
[P,] If A and B are mutually exclusive, then P ( A uB 1 E ) = P ( A I E ) P ( B E).
+
~
Since A n E c E , P ( A n E ) 5 P ( E ) . Thus P ( A I E ) =
n E ) 4 1 and is also non-negative;
P(E)
that is, 0 5 P(A 1 E ) f 1.
P(E)
Since S n E = E , P(S 1 E ) = P ( S n E ) - __
P(E)
P(E) =
If A n B = 8, then A n B n E = ( A n E ) n ( B n E ) = 8, i.e. A n E and B n E a r e mutually
exclusive events; hence P ( ( A n E )u (BnE))= P ( A n E ) + P ( B n E ) . But ( A n E ) u ( B n E ) =
( A LB)n E. Thus P((Ac:B)n E ) = P ( A n E ) P ( B n E ) . Hence
~
~
+
CHAP. 181
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
207
Remark: Since P(* I E ) satisfies the aKove three axioms, it also satisfies a n y consequence of
these axioms: for example,
P(AUB I E ) = P ( A I E )
+ P(B1E)
-
P ( A i 7 B 1 Ej
18.16. Prove: Let E l ,E,, . . .,En form a partition of a sample space S , and let A be any
event. Then
P ( A ) = P ( E i ) P ( AI E l ) + P(Er)
P ( A I Ez) + . . . + P(En)P ( A 1 En)
Since the E , forin a partition of S, the E, a r e pairwise disjoint and S = E I U E z U . ~ ~ U E , , .
Hence
A = S n A = ( E , u E , u ~ ~ ~ u En,A, ) = ( E l n A ) u ( E , n A ) u . . . u (E,nA)
and the E , n A a r e also pairwise disjoint. Accordingly,
P(A) = P(E,nA)
P(E,nA)
+ - . . + P(E,,nA)
By the multiplication theorem, P ( E , p A ) = P ( E , )P ( A I Et); hence
P(A)
P(E1)P ( A 1 E l ) - P(E,j P ( A 1 Ez)
+ . . - P(E,,)P ( A I En)
18.17.Three machines A, B and C produce respectively 50%) 3 0 5 and 20% of the total
number of items of a factory. The percentages of defective output of these machines
are respectively 3%) 4% and 5 G . If an item is selected at random, what is the
probability 71 that the item is defective?
Let X be the event t h a t a n item is defective.
preceding problem,
P(X) =
=
-
+
Then by the
-
P ( A )P : X I A )
P ( B )P ( X B )
P ( C )P ( X 1 C )
5OCr.3%
30%*4%
205.5%
_
5 0 , 3 + 3o.L + 2 o . z
37 = 3.7%
+
100
100
100
100
+
100
100
-
1000
We can also consider this problem as a stochastic process having
the adjoining tree diagram.
18.18. Prove Bayes’ Theorem: Let E l ,E,, . . . , E l , form a partition of a sample space S ,
and let A be any event. Then, for-each i,
P ( E JP ( A I E l )
P ( E i A) = P(E1)P ( A E l ) + P(E2)P ( A 1 E z ) + . . . + P ( E n )P ( A 1 En)
Using the definitlon of conditional probability, we obtain
P ( E ,nA ) P(E,) P ( A I EL)
P ( E i )P ( A I E l )
P(Ez)P ( A I E 2 j
. . - P(En)P ( A 1 En)
P(A)
Here we used the multiplication theorem, P ( E ,n A ) = P(E,) P ( A 1 Ez),t o obtain the numerator,
and Problem 18.16 t o obtain the denominator.
P(E, A ) =
+
+
18.19. In a certain college, 4% of the men and 1% of the women are taller than 6 feet.
Furthermore, 6 0 5 of the students are women. Now if a student is selected at
random and is taller than 6 feet, what is the probability that the student is a woman?
Let A = {students taller than 6 feet:. We seek P(W 1 A ) , the probability t h a t a student is a
woman given t h a t the student is taller than 6 feet. By Bayes’ theorem,
208
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
[CHAP. 18
18.20. Find P(B A ) if (i) A is a subset of B , (ii) A and I3 are mutually exclusive.
(i)
If A is a subset of B , then whenever A occurs B must occur: hence P ( B 1 A ) = 1. Alternately,
if A is a subset of B then A r B = A ; hence
P(B A ) =
~
P(AnB)
P(A)
-
P(A)
P(A) =
~
(i)
(iii
(ii)
If A and B a r e mutually exclusive. i.e. disjoint, then whenever A occurs B cannot occur; hence
P ( B I A ) = 0. Alternately, if A and B a r e mutually exclusive then A n B = 0; hence
FINITE STOCHASTIC PROCESSES
18.21. A box contains three coins; one coin is fair, one coin is two-headed, and one coin is
weighted so that the probability of heads appearing is
A coin is selected a t
random and tossed. Find the probability p that heads appears.
+.
Construct the tree diagram a s shown in Figure ( a ) below. Note t h a t I refers t o the f a i r coin,
I1 t o the two-headed coin, and I11 t o the weighted coin. Now heads appears along three of the
paths: hence
1.1 + 1.1 + 1.1 = 11
P =
3
2
3
3
3
18
(4
0)
18.22.We are given three urns as follows:
Urn A contains 3 red and 5 white marbles.
Urn B contains 2 red and 1 white marble.
Urn C contains 2 red and 3 white marbles.
An urn is selected at random and a marble is drawn from the urn.
is red, what is the probability that i t came from urn A ?
If the marble
Construct t h e tree diagram a s shown in Figure ( b ) above.
We seek the probability t h a t A was selected, given t h a t the marble is red; t h a t is, P ( A 1 R).
In order t o find P ( A 1 R ) , i t is necessary first t o compute P ( A n R ) and P ( R ) .
The probability t h a t urn A is selected and a red marble drawn is 31 . 83 = g1 ;
P ( A r R )=
Thus
f.
t h a t is,
Since there a r e three paths leading to a redmarble, P ( R ) = 1
* 3 +31 *3 ~ +3 1 5. g= 173
3 8
360'
1
P(A , R ) =
~
P(AnR) - s
P ( R ) - II.?
360
-
45
173
CHAP. 181
209
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
18.23. Box A contains nine cards numbered 1 through 9, and Box B contains five cards
numbered 1 through 5. A box is chosen at random and a card drawn. If the number
is even, find the probability that the card came from Box A .
The tree diagram of t h e process is shown in Figure (a) below.
We seek PIA ' E ) , the probability t h a t A was selected, given t h a t the number is even. The
probability t h a t Box A and a n even number is drawn is 21 - y4 = $; t h a t is, P ( A n E ) = $. Since
there are two paths which lead to an even number, P ( E ) = L*.i-L*
2 9
2
;
= L!4?5 .
Thus
18.24. An urn contains 3 red marbles and 7 white marbles. A marble is drawn from the
urn and a marble of the other color is then put into the urn. A second marble is
drawn from the urn.
0) Find the probability p that the second marble is red.
(ii) If both marbles were of the same color, what is the probability p that they were
both white?
Construct the tree diagram as shown in Figure ( b ) above.
6) Two paths
of the tree lead to a red marble: p =
10 * &
10
&
+
10 * 10
= !?
50
.
ii.
(ii) The probability t h a t both marbles were white is
* $ = The probability t h a t both
marbles were of the same color, i.e. the probability of the reduced sample space, is
lo.lo+-'3
2
7
6
- Hence the conditional probability p =
=
1o 1o
bg.
5/g i.
18.25. We are given two urns as follows:
Urn A contains 3 red and 2 white marbles.
Urn B contains 2 red and 5 white marbles.
An urn is selected at random; a marble is drawn and put into the other urn; then
a marble is drawn from the second urn. Find the probability p that both marbles
drawn are of the same color.
Construct the following tree diagram:
210
CONDITIOKAL PROBABILITY. IISDEPENDENCE
[CHAP. 18
Note t h a t if u r n A is selected and a red marble drawn and put into u r n B , then u r n C has
3 red marbles and 5 white marbles.
Since there a r e f o u r paths which lead to two marbles of the same color,
=
$.;.;
+$.$.2
+ 1.2.2 - L . 5 . L
2
7
2
3
7
=
2
901
1680
INDEPENDENCE
18.26. Let A = event that a family has children of both sexes, and let B = event that a
family has at most one boy. (i) Show that A and B are independent events if a
family has three children. (ii) Show that A and B are dependent events if a family
has two children.
fi)
We have the equiprobable space S = { b b b , b b g , b g b , b g g , g b b , g b g , g g b , ggg). Here
P ( A ) = 68 = 34
A = W g , bgb, b g g , g b b , g b g , g g b )
and so
B = Cbgg, g b g , g g b , s w >
A n B = {bug,gbg,ggb)
Since P ( A ) P ( B )=
Iii)
2-4= #
and so
P(B) =
P(AnB) =
and so
+
z
$
= P ( A n B ) , A and B a r e independent.
We have the equiprobable space 5' = { b b , b g , g b , g g ) .
A = {bg,gb)
and so
B = { b g , gb, gg}
A n B = {bg,gb}
Here
and so
P(B) =
4
2
and so
P(AnB) =
4
P(A) =
Since P ( A ) P ( B )# P ( A n E ) , A and B a r e dependent.
18.27. Prove: If A and B are independent events, then A c and B' are independent events.
Let P ( A ) = x and P ( B ) = y. Then P(Ac) = l - - r and P(Bc) = 1-y.
independent, P ( A n B ) = P ( A )P ( B ) = X U . Furthermore,
P(AuB) = P(A)
By De Morgan's theorem,
+ P(B)- P ( A n B ) = x + y
-
Since A and B a r e
xy
( A u B ) c = AcnBc; hence
P(AcnBc) = P ( ( A u B ) c ) = 1 - P ( A ~ J B =
) 1 -x -y
On the other hand,
P(AC)P(Bc) = (1 - x ) ( l - y) = 1 -
j:
-
y
+
xy
+
Thus P(AcnBc) = P ( A c )P(Bc), and so A' and BC a r e independent.
18.28. The probability that a man will live 10 more years is 4,and the probability that his
wife ill live 10 more years is +. Find the probability that (i) both will be alive in
10 years, (ii) at least one will be alive in 10 years, (iii) neither will be alive in
10 years, (iv) only the wife will be alive in 10 years.
Let A = event t h a t the man is alive in 10 years, and B = event t h a t his wife is alive i n
10 years; then P ( A ) =
and P ( B ) =
t
11)
4.
We seek P ( A n B ) . Since A and B a r e independent, P ( A n B ) = P ( A )P ( B ) =
(ii) We seek P ( A u B ) . P ( A u B ) = P ( A )+ P ( B ) - P ( A n B ) =
&
- + = A.
&++-A= 4.
4
4 3.
iiii) We seek P(AcnBc). Now P(Ac) = 1- P ( A ) = 1- & =
and P(i3c) = 1- P ( B ) = 1 - =
Furthermore, since Ac and Ec a r e independent, P ( A c nSc) = P(AC)P(Bc) = 3
' 42
3 = 1
2 .
Alternately, since ( A U B ) C= ACnBc, P ( A c n B c ) = P ( ( Au B ) ~=
) 1- P ( A L'B) = 1 (iv) We seek P ( A c n B ) . Since P(Ac) = 1- P ( A ) =
P ( A c ) P ( B )= f.
2
4 = 4.
and Ac and B a r e independent, P ( A c n B ) =
CHAP. 181
211
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
18.29. Box A contains 8 items of which 3 are defective, and box B contains 5 items of which
2 are defective. An item is d r a x n at random from each box.
(i) What is the probability p that both items are non-defective?
(ii) What is the probability p that one item is defective and one not?
(iii) If one item is defective and one is not, what is the probability p that the defective item came from box A ?
(i)
The probability of choosing a non-defective item from A is
events a r e independent, p =
= 5
8 ’
and from B is
$.
Since the
fa;
(ii) Method 1.
=
The probability of choosing two defective items is
probability t h a t both a r e non-defective is
i. Hence
p = 1
20
R
&.
From fi) the
= 19.
40
Method 2. The probability p , of choosing a defective item from A and a non-defective item
from B is
The probability p L of choosing a non-defective item from A and a
:*
=
9
1
defective item froin B is
= $. Hence p = pl p , = l o t -4 = ?4! 0 .
&.
+
(iii) Consider the events X = {defective item from A } and Y = {one item is defective and one
and P(Y)=
Hence
non-defective}. We seek P ( X [ Y). By (ii), P ( X n Y ) = p , =
g.
18.30. The probabilities that three men hit a target are respectively Q, & and +. Each
shoots once a t the target. (i) Find the probability p that exactly one of them hits
the target. (ii) If only one hit the target, what is the probability th at it was the
first man ?
Consider the events A = {first man hits the target}, B = {second man hits the target), and
The three events a r e
and P ( C ) =
C = {third man hits the target}; then P ( A ) = +, P ( B ) =
independent, and P(Ac) = g, P ( B c ) = 9 , P(CC) =
a.
t.
(i)
Let E = [exactly one man hits the target}. Then
E = (AnBcnCc)u (AcnBnCc)
In other words, if only one hit the target, then i t was either only the first man, A n B c r C c ,
o r only the second man, AcnBnCc, o r only the third man, A c n B c n C . Since the three events
a r e mutually exclusive,
p = P(E) = P(AnBcnCc)
P(AcnBnCc) P(AcnBcnC)
= P ( A )P(Bc)P(Cr)
P ( A c )P ( B )P(Cc) T P(Ac)P(Bc) P ( C )
1 3 2
5 1 2
5 3 1 = L t 5 - 5 = 31
-
+
+
+
-.-.---.-.__-.-.
6
4
3
6
4
3
6
4
3
12
36
24
72
(ii) We seek P ( A ’ E ) , the probability t h a t the first man h i t the target given t h a t only one man
hit the target. Now A n E = A n Bc n Cc is the event t h a t only the first man h i t the target.
hence
By (i), P ( A r E ) = P ( A n B c n C c ) = $ and P ( E ) =
g;
Supplementary Problems
CONDITIONAL PROBhBILITT IN EQUIPROBABLE SP-ACES
18.31. A die is tossed. If the number is odd, what is the probability t h a t it is prime?
18.32. A letter is chosen a t random from the word “trained”. If i t is a consonant, what i s the probability
t h a t i t is the first letter of t h e word?
[CHAP. 18
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
2 12
18.33.
Three f a i r coins a r e tossed. If both heads and tails appear, determine t h e probability t h a t exactly
one head appears.
18.34.
A pair of dice is tossed. If the numbers appearing a r e different, find the probability t h a t the sum
is even.
18.35.
A man is dealt 5 red cards from a n ordinary deck of 52 cards. W h a t is the probability t h a t they
a r e all of the same suit, i.e. hearts o r diamonds?
18.36. A man is dealt 3 spade cards from a n ordinary deck of 52 cards. If he is given four more cards,
determine the probability t h a t a t least two of the additional cards a r e also spades.
18.37.
18.38.
18.39.
Two different digits a r e selected a t random from the digits 1 through 9.
(i)
If the sum is odd, what is the probability t h a t 2 is one of the numbers selected?
(ii)
If 2 is one of the digits selected, what i s t h e probability t h a t the sum is odd?
Four persons, called North, South, E a s t and West, a r e each dealt 13 cards from a n ordinary deck
of 5 2 cards.
ii)
If South h a s exactly one ace, what is the probability t h a t his p a r t n e r North h a s the other
three aces?
iii)
If North and South together have 10 hearts, what is the probability t h a t neither E a s t nor
West has the other 3 hearts?
A class has 1 0 boys and 5 girls. Three students a r e selected from the class at random, one a f t e r
the other. Find the probability t h a t ( i ) the first two a r e boys and the third is a girl, (ii) the first
and third a r e boys and the second is a girl, (iii) the first and third a r e of the same sex, and the
second is of the opposite sex.
18.40.
In the preceding problem, if the first and third students selected a r e of t h e same sex and the second
student is of the opposite sex, what is the probability t h a t the second student is a girl?
CONDITIONAL PROBABILITY
18.41.
In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both
brown hair and brown eyes. A person is selected at random from the town.
ti)
If he has brown hair, what is the probability t h a t he also has brown eyes?
(ii) If he has brown eyes, what is the probability t h a t he does not have brown h a i r ?
(iii) W h a t is the probability t h a t he has neither brown h a i r nor brown eyes?
a,
(i) P ( A IB),
18.43.
Let S = ( a , b , c , d , e ,f} with P ( a ) =
P(b) =
P(c) =
P(d) = A, P(e) = & and
P ( f )= A
.
Let
A
=
{
a
,
c
,
e
}
,
B
=
{
c
,
d
,
e
,
f
}
and
C
=
{
b
,
c
,
,
f
}
.
Find
(i) P ( A 1 B ) , (ii) P ( B I C),
16
(iii) P ( C ' A c ) , (iv) P(Ac 1 C).
18.44.
In a certain college, 25% of the boys and 10% of the girls a r e studying mathematics. The girls
constitute 60Cb of the student body. If a student is selected at random and i s studying mathematics,
determine the probability t h a t the student is a girl.
and
A,
P(AUB) =
4.
Let A and B be events with P ( A ) =
(ii, P ( B A ) , (iii) P ( A nBc), (iv) P ( A 1 Bc).
A,
P(B) =
&
18.42.
Find
4,
FINITE STOCHASTIC PROCESSES
18.45.
We a r e given two u r n s as follows:
Urn A contains 5 red marbles, 3 white marbles and 8 blue marbles.
Urn B contains 3 red marhles and 5 white marbles.
A f a i r die is tossed; if 3 or 6 appears, a marble is chosen from B , otherwise a marble is chosen
from A . Find the probability t h a t (i) a red marble is chosen, (ii) a white marble is chosen,
(iiii a blue marble is chosen.
CHAP. 181
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
213
18.46.
Refer to the preceding problem. (i) If a red marble is chosen, what is the probability t h a t it came
from urn A ? {iii If a white marble is chosen, what is the probability t h a t a 5 appeared on the die?
18.47.
An u r n contains 5 red marbles and 3 white marbles. A marble is selected a t random from the urn.
discarded, and two marbles of the other color a r e put into the urn. A second marble is then selected
from the urn. Find the probability t h a t i i ) the second marble is red, iii) both marbles a r e of the
same color.
18.48. Refer to the preceding problem. (i) If the second marble is red, what is the probability t h a t the
first marble is red? iii) If both marbles a r e of the same color, what is the probability t h a t they
a r e both white?
18.49.
A box contains three coins, two of them f a i r and one two-headed. A coin is selected at random
and tossed twice. If heads appears both times, what is t h e probability t h a t the coin is two-headed?
18.50. We a r e given two urns as follows:
U r n A contains 5 red marbles and 3 white marbles.
Urn B contains 1 red marble and 2 white marbles.
A f a i r die is tossed; if a 3 o r 6 appears, a marble is drawn from B and put into A and then a marble
is drawn from A : otherwise, a marble is drawn from A and put into B and then a marble is drawn
from B.
(i)
W h a t is the probability t h a t both marbles a r e red?
(ii) What is the probability t h a t both marbles a r e white?
18.51.
Box A contains nine cards numbered 1 through 9, and box B contains five cards numbered 1
through 5. A box is chosen a t random and a card drawn; if the card shows a n even number,
another card is drawn from the same box; if the card shows a n odd number, a card is drawn from
the other box.
(i)
W h a t is the probability t h a t both cards show even numbers?
(ii) If both cards show even numbers, what is the probability t h a t they came f r o m box A ?
(iii) W h a t is the probability t h a t both cards show odd numbers?
18.52. A box contains a f a i r coin and a two-headed coin. A coin i s selected a t random and tossed. If
heads appears, the other coin is tossed: if tails appears, the same coin is tossed.
(i)
Find the probability t h a t heads appears on the second toss.
(ii) If heads appeared on the second toss, find the probability t h a t it also appeared on t h e first toss.
18.53. A box contains three coins, two of them f a i r and one two-headed. A coin is selected at random
and tossed, If heads appears the coin is tossed again; if tails appears, then another coin is selected
from the two remaining coins and tossed.
(i)
Find t h e probability that heads appears twice.
(ii) If the same coin is tossed twice, find the probability t h a t it is the two-headed coin.
(iii) Find the probability t h a t tails appears twice.
18.54.
U r n A contains z red marbles and y white marbles, and u r n B contains z red marbles and
marbles.
(i)
9
white
If a n u r n is selected at random and a marble drawn, what is the probability t h a t t h e marble
is red?
(ii) If a marble is drawn from u r n A and put into u r n B and then a marble is drawn from u r n B ,
what is the probability t h a t t h e second marble is red?
18.55.
A box contains 5 radio tubes of which 2 a r e defective. The tubes a r e tested one a f t e r the other
until the 2 defective tubes a r e discovered. W h a t i s t h e probability t h a t t h e process stopped on t h e
(i) second test, (ii) third test?
18.56.
Refer to the preceding problem. If the process stopped on t h e third test, w h a t is the probability
t h a t the first tube is non-defective?
214
CONDITIONAL PROBABILITY. IN D EPEN D EN CE
[CHAP. 18
INDEPENDENCE
18.57.
Prove: If -4 and B a r e independent, then A and Bc a r e independent and Ac and B a r e independent.
18.58.
and P ( B ) = p . (i) Find p if A and B a r e
Let A and B be events with P ( A ) = $, P ( A u B ) =
mutually exclusive. (ii) Find p if A and B are independent. (iii) Find p if A is a subset of B.
18.59.
Urn A contains 5 red marbles and 3 white marbles, and urn B contains 2 red marbles and
6 white marbles.
+
(i)
If a marble is drawn from each urn, w ha t is the probability tha t they a r e both of the same
color?
(iii
If two marbles a r e drawn from each urn, what is the probability t h a t all four marbles a r e
of the same color?
18.60.
Let three f ai r coins be tossed. Let A = {all heads o r all tails}, B = {at least two heads) and
C = { at most two heads}. Of the pairs ( A , B ) , ( A , C) and ( B , C), which are independent and
which a r e dependent?
18.61.
The probability t h a t A hits a t ar g et is
t and the
probability t h a t B hits a ta rge t is Q.
If each fires twice, what is the probability t h a t the ta rge t will be hit at least once?
(i)
(ii) If each fires once and the target is hit only once, what is the probability t h a t A hit the ta rget?
(iii) If A can fire only twice, how many times must B fire so t h a t there is at least a 90% probability t h at the t ar g et will be hit?
18.62.
Let A and B be independent events with P ( A ) = 8 and P ( A u B ) =
(iii) P(Bc A ) .
'
8.
Find (i) P ( B ) , (ii) P ( A 1 B ) ,
Answers to Supplementary Problems
15
-
18.31.
$
18.40.
%= $
~
21
18.32.
2
18.33.
4
18.34.
+
18.37.
(i)
18.41.
(i) Q , (ii) 3, (iii) fr
18.42.
(i) Q, (ii) $, (iii)
18.43.
(i) $, (ii) 3, (iii) 4, (iv) $
18.44.
Q
18.45.
(i)
+
(ii)
&
(iii)
4
&, (ii) 5
4, (iv) Q
b
8
18.39.
(i)
(ii)
10
- z *13 =
5
9
-.-.--10
15
15
(iii) 91
5
14
9
13
-
15
91
+- = 20
273
a,
(ii)
&
18.46.
(i)
18.47.
( i ) z , (ii) 36
18.48.
(i) 41' (ii) 13
15
91
5
21
41
13
20
3
W
CHAP. 181
CONDITIONAL PROBABILITY. I N D E P E N D E N C E
215
16.49. $
5
16.50. (i) 24
+&
=
61
(ii)
z
,
3
16
+
8
=
81
3il
1206
~
<-+:
6
w e ;
9
Tree diagram for Problem 18.50
1
18.51. (i) 5
16.52. (i)
8,
1
16.53. (i) 12
+1
(ii)
1
-
2
= 15'
(ii) 12
=
2
15
5,
(iii)
4+Q
= Q
8
+ 12 + 5
1
=
f,
2
1
(ii) 3,
(iii) 12
Tree diagram for Problem 18.55
Tree diagram for Problem 18.53
h,
&;
16.55. (i)
(ii)
we must include the case where the three non-defective tubes appear first, since
t h e l a st two tubes must then be t h e defective ones.
18.56.
3
16.60. Only A and B a r e independent.
1
1
1
(i) $, (ii) 9, (iii) 5
18.58. (i) 12, (ii) 9, (iii) 5
18.61.
7
18.59. (i) 16,
(ii)
18.62. (i) 4, (ii) 4, (iii) 3
55
Chapter 19
Independent Trials, Random Variables
INDEPENDENT OR REPEATED TRIALS
We have previously discussed probabiIity spaces which were associated with an experiment repeated a finite number of times, as the tossing of a coin three times. This concept
of repetition is formalized as follows:
[Definition:I Let S* be a finite probability space.
By ?i i n d e p e n d e n t or repeated trials, we
mean the probability space S consisting of ordered n-tuples of elements of S"
with the probability of an ?z-tuple defined to be the product of the probabilities
of its components:
P((s1,s2, . . . , Sn)) = P(s,)P(s2) . . . P(s,)
Example 1.1:
Whenever three horses a, b and c race together, their respective probabilities of
I n other words, S = { u , b, c } with P(a) = +, P ( b ) =
winning a r e +, f and
and P ( c ) = 8. I f the horses race twice, then the sample space of the 2 repeated
trials is
S = {aa, ab, a c , h a , bb, bc, ca, cb, cc}
A.
F o r notational convenience, we have written a c f o r the ordered pair ( a , ~ ) . The
Probability of each point in S is
P(aa) = P(a)P(a) =
4.4 =
$
P(ba) = Q
P(ab) = P(u)P(b) =
+*&
=
4
P(bb) =
&
P(cb) = 3
P ( a c ) = P(u)P(c) =
i.8
=
&
P(bc) =
ilj-
1
P(cc) =
P(ca) =
&
1
&
Thus the probability of c winning the first race and a winning the second race is
P(ca) =
A.
From another point of view, a repeated
trials process is a stochastic process whose
tree diagram has the following properties:
(i) every branch point has the same outcomes;
(ii) the probability is the same f o r each
branch leading to the same outcome. For
example, the tree diagram of the repeated
trials process of the preceding experiment
is as shown in the adjoining figure.
Observe that every branch point has the
outcomes a, b and c, and each branch leading to outcome a has probability &, each
branch leading to b has probability Q, and
each leading to c has probability Q.
216
C
CHAP. 191
217
INDEPENDENT TRIALS, RANDOM VARIABLES
REPEATED TRIALS WITH TWO OUTCOMES
We now investigate repeated trials of an experiment with two outcomes; we call one
of the outcomes success and the other outcome failure. Let p be the probability of success,
and so q = 1 - p is the probability of failure. Frequently we are interested in the number
of successes and not in the order in which they occur. The following theorem applies.
Theorem 19.1:
The probability of exactly k. successes in n repeated trials is denoted and
given by
f ( n ,k , P) = (3P k r k
The probability of at least one success is 1- q".
Here (;) is the binomial coefficient (see Chapter 13).
Example 2.1:
A fair coin is tossed 6 times or, equivalently, six fair coins are tossed; call heads
Then n = 6 and p = q = 4.
a success.
(i)
The probability that exactly two heads occurs (i.e. k = 2) is
f(6,2,4) =
=
($)z(+)4
64
(ii) The probability of getting at least four heads (i.e. k = 4, 5 o r 6) is
f(6,4, #
f
f(6,5,4)
+ f(6,6, Q)
(3)4(g)2+ (E)
=
(#5
(4)+ (i)($J6
5 + 6 + 1 = 11
- 1
64
64
64
32
A, and so the
(iii) The probability of no heads (i.e. all failures) is q6 = (4)6=
probability of a t least one heads is 1 - q 6 = 1 - 614 = a.
64
Example 2.2:
A fair die is tossed 7 times; call a toss a success if a 5 o r a 6 appears. Then n = 7,
p = P({5,6}) = Q and q = 1- p = 2.
3
(i)
The probability that a 5 or a 6 occurs exactly 3 times (i.e. k = 3) is
f(7,3,Q) =
(7) (')3
3
3
(")4
3
1
2187
,187'
(ii) The probability that a 5 or a 6 never occurs (i.e. all failures) is 47 = (Q)' = p.
2059
hence the probability that a 5 or a 6 occurs a t least once is 1- q7 = 2187.
If we treat n and p as constants, then the function defined by f ( n ,k,p ) is called the
binomial distribution since for k = 0,1,2, . . . , n it corresponds t o the successive terms
of the binomial expansion:
( q +p)" = q"
(Y) q n - ' p (;) q"-Zp2 + * + p"
+
= f(n,0,P)
+
a
+ f(n,1,P ) + f(n,2, P) + -
.
+ f ( n n,
, P)
The use of the word distribution will be explained later in the chapter. This distribution
is also called the Bernoulli distribution, and independent trials with two outcomes are
called Bernoulli trials.
RANDOM VARIABLES
Let S be a sample space of some experiment. As noted previously, the outcomes of
the experiment, i.e. the sample points of S , need not be numbers. However, we frequently
wish to assign a specific number to each outcome of the experiment. Such a n assignment
is called a random variable. In other words:
218
I N D E P E N D E N T TRIALS, RANDOM VARIABLES
1 Definition: I
[CHAP. 19
A mndom variable X is a function from a sample space S into the real
numbers.
We shall let R, denote the range of a random variable X : R, = X ( S ) . We shall refer to
R, as the range space.
Remark:
If the sample space S is nondenumerable, then f o r theoretical reasons lying
beyond the scope of this text certain functions defined on S are not random
variables. However, the sample spaces here are finite, and every function defined
on a finite sample space is a random variable.
Example 3.1:
A pair of dice is tossed. The sample space S consists of the 36 ordered pairs of
numbers between 1 and 6:
S = ((1, I), ( 1 , 2 ) , . . . , ( 6 6 ) )
Let X assign to each point in S the sum of the numbers; then X is a random variable with range space
Rx = (2, 3, 4, 5, 6, 7 , 8, 9, 10, 11, 12)
Let Y assign t o each point the maximum of the two numbers; then Y is a random
variable with range space
RY = (1, 2 , 3, 4, 5, 6)
Example 3.2:
A sample of 3 items is selected from a box containing, say, 12 items of which 3
a r e defective. The sample space S consists of the
= 220 different samples of
size 3. Let X denote the number of defective items in the sample; then X is a
random variable with range space R, = { 0 , 1 , 2 , 3 } .
(y)
PROBABILITY SPACE AND DISTRIBUTION O F A RANDOM VARIABLE
Let R, = {x,,x,, . . .,x,} be the range space of a random variable X defined on a
finite sample space S. Then X induces an assignment of probabilities on the range space
R, as follows:
p l = P(x,) = sum of probabilities of points in S whose image is x1
The function assigning p i to xior, in other words, the set of ordered pairs (xl, pl), . . . ,
usually given by a table
...
is called the distribution of the random variable X .
I n the case that S is an equiprobable space, we can easily obtain the distribution of a
random variable from the following result.
Theorem 19.2:
Let S be an equiprobable space. If R, = {xl, . . . ,x,} is the range space
of a random variable X defined on S , then
pi = P(XJ =
Example 4.1:
number of points in S whose image is
number of points in S
xi
A pair of f ai r dice is tossed. W e obtain the equiprobable space '
5 consisting of the
36 ordered pairs of numbers between 1 and 6:
S =
{(LI), ( 1 , 2 ) , . . .,
CHAP. 191
219
I N D E P E N D E N T TRIALS, RANDOM VARIABLES
Let X be t h e random variable which assigns t o each point the sum of the two
numbers; then X has range space
R , = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)
x.
We use Theorem 19.2 t o obtain the distribution of
There is only one point
There a r e two points in S, (1,Z) and
2
(2, l), having image 3; hence P(3) = z.There a r e three points in S , (1,3), (2,2)
3
4
and (3, l),having image 4: hence P ( 4 ) = 36. Similarly, P(5) = x, P(6) = &, etc.
The distribution of X consists of the points in R, with their respective probabilities:
(1,l)whose image is 2: hence P(2) =
Example 4.2:
xi
2
'
36
I
3
-
f6
4
-
'
3 36
&.
5
6
7
1
8
1
36
-5
36
-6
36
1
- 5
36
9
-
6',
10
11
12
3
36
2
-
6',
36
A sample of 3 items is selected a t random from a box containing 12 items of which
3 a r e defective. The sample space S consists of the
= 220 different equally
likely samples of size 3. Let X be the random variable which assigns to each
sample the number of defective items in the sample; then Rx = {0,1,2,3). Again
we use Theorem 19.2 to obtain the distribution of X.
ii')
(i)
There a r e
= 84 samples of size 3 which contain no defective items; hence
84
P ( 0 ) = m. There a r e 3 * (:) = 108 samples of size 3 containing 1 defective item;
There are (;) * 9 = 27 samples of size 3 containing 2 defective
hence P(1) =
2items; hence P ( 2 ) = &. There is only one sample of size 3 containing the 3 defecThe distribution of X follows:
tive items; hence P(3)
g.
=A.
Example 4.3:
+
A coin weighted so t h a t P i H ) = 8 and P(T)=
is tossed three times. The
probabilities of the points in the sample space S = {HHH, HHT, HTH, HTT,
THH, THT, TTH, TTT} a r e as follows:
p ( H H H ) = g.:F.2
3 1 3
z 27
i7
= 6
P ( H H T ) = 2.2.1
a
=
$.;+*+
p(HTH) =
-
p ( H T T ) = +.l.+ = 4
27
P(THH) =
PtTHT) =
P(TTH) =
P(,TTT) =
t.3.g = 4
27
Q.8.Q = &
g.1.2
3
3
&*&*&
= 2
27
=
1
27
Let X be the random variable which assigns t o each point in S the largest
number of successive heads which occurs. Thus,
X(TTT) = 0
X ( H T H ) = 1, X(HTT) = 1, X(THT) = 1, X(TTH) = 1
X(HHT) = 2, X!THH) = 2
X(HHH) = 3
The range space of X is R , = {0,1,2,3}. We obtain P(O),P(1),P ( 2 ) and P(3) as
follows:
P(0) = P(TTT) =
P(1) = P ( H T H )
&
+ P(HTT) + P(THT) + P(TTH)
2
P ( 2 ) = P(HHT) t P ( T H H ) = &-I
$=
P(3) = P(HHH) =
&
=
& + $+& +$
=
10
I N D E P E K D E 9 T T I? IA L S, RAND 0 31 V A R IA B L ES
220
The distribution of X consists of t h e points in the range space of
respective probabilities:
;CHAP. 19
X with their
Now consider repeated trials of an experiment with two outcomes: that of success
with probability p and that of failure with probability Q = 1- p . To each sequence of
n outcomes, let X n assign the number of successes. Then X , 2 is a random variable with
range space RY,,= {0,1,2, . . ., ti). Using Theorem 19.1, v e obtain
PjO) = f ( t z , O , p ) = Q"
P(1) =
f ( i l , 1,p)
=
(:)qrz-1]1
P ( 2 ) = f(i1,2,p) = (;) q"-'p'
..........................
P(H) =
. f ( i r , n, p
) = p"
I n other words, the distribution of X n is the binomial distribution as discussed previously:
EXPECTED VALUE
If the outcomes of an experiment are the numbers T , , x,, . , ., x r which occur with
respective probabilities pl, p 2 , . . , p , , then the expected value is defined t o be
,
E = ~~p~
+ ~ , p ,+
yrpr
In particular, if the xZand p Z come from the distribution of a random variable X , then the
expected value ic, called the expectation of the random variable and is denoted by E ( X ) .
Example 5.1 :
Consider the random variable X of Example 4.2. I t s distribution gives the possible
numbers of defective items in a sample of size 3 with their respective probabilities
of occurrence. Then the expected number of defective items in a sample of size 3,
i.e. the expectation of the random variable X, is
E(X) =
+ 2.sc 3 . L - s3
220
-
o.g+1.E
Example 5.2:
A f a i r coin is tossed 6 times. The possible numbers of heads which can occur
with their respective probabilities a r e as foilom-s:
Note t h a t the above table is t h e binomial distribution with n = 6 and p =
The expected value, i.e., the expected number of heads, is
E
= 0 . k
+ 1-& + 2
Observe t h a t E = np.
- g
+ 3*%
+
4 . 3
+ 5 * & + 6-A
= 3
3.
CHAP. 191
I K D E P E N D E N T TRIALS, RANDOM VARIABLES
22 1
The result in the preceding example holds true in general; namely,
Example 5.3:
xi
0
Pi
qn
1
2
(;)q"-'P
(;)9"-2P2
...
n
...
Pn
The probability is 0.2 t ha t a n item produced by a factory is defective. If 3000
items are made by the factory, then the expected number of defective items is
E = np = (3000)(0.2) = 60.
GAMBLING GAMES
I n a gambling game, the expected value E is considered to be the value of the game
to the player. The game is said to be favorable to the player if E is positive, and tinfavorable if E is negative. If E = 0, the game is fair.
xi
2
5
3
-1
-4
-6
The negative numbers -1, -4 and -6 correspond t o the fa c t t h a t the player loses
if a non-prime number occurs, The expected value of the game is
Thus t h e game is unfavorable to the player since the expected value is negative.
Example 6.2:
*+
A player tosses a f a i r coin. If a head occurs he receives 5 dollars, but if a tail
1* 4 = 3.
occurs he receives only 1 dollar. His expected winnings E = 5
Question: How much money should he pay to play the game so t h a t the game
is f a i r ? W e claim t h at the answer is 3 dollars. For if he pays 3 dollars t o play
the game, then each outcome is decreased by 3 dollars. Thus when a head occurs
he wins 5 - 3 = 2 dollars and when a tail occurs he wins 1 - 3 = -2 dollars,
i.e., he loses 2 dollars; hence the expected value is now E = 2 *
(-2) * $ = 0.
++
The result in the preceding example holds true in general; namely,
Theorem 19.4:
Let E be the expected value of a n experiment. If a constant k is added
to each outcome of the experiment, then the expected value E* of the
new experiment is E k : E" = E k.
+
+
Example 6.3:
A thousand tickets a t $1 each are sold in a lottery in which there is one prize of
$500, four prizes of $100 each, and five prizes of $10 each. The expected winnings
of one ticket are
500-&
loo*&
lo*&
= 0.95
+
+
However, since a ticket costs $1, the expected value of the game is 0.95 - 1.00 =
- $0.05, which is unfavorable to the player.
222
I N D E P E N D E K T TRIALS, RANDOJI VARIABLES
[CHAP. 19
Solved Problems
REPEATED TRIALS WITH TWO OUTCOMES
19.1. Find (i) f ( 5 , 2 , + ) , (ii) f ( 6 , 3 ,i), (iii) f ( 4 , 3 , t ) . Here
f ( n ,k , P ) = (;) Pk (1- P y k
19.2. A fair coin is tossed three times. Find the probability P that there will appear
(i) three heads, (ii) two heads, (iii) one head, (iv) no heads.
Method 1. We obtain the following equiprobable space of eight elements:
S = {HHH, HHT, HTH, HTT, T H H , THT, TTH, TTT]
(i)
Three heads (HHH) occurs only once among the eight sample points; hence P = 4.
(ii) Two heads occurs 3 times IHHT, HTH, and THH); hence P = # .
(iii) One head occurs 3 times (HTT, T H T and TTH); hence P = $ .
(iv) No heads, i.e. three tails (TTT), occurs only once; hence P = $.
Use Theorem 19.1 with .n = 3 and p = q =
Method 2.
(i)
Here k = 3 and P = f ( 3 , 3 , 4 ) =
(i)
(3)O
4.
= 1 . 81 * 1 = 1.
8
(+)z(4) = 3 1.1
4 2 = a.
8
(t)(+)l (4)z = 3 9 4 . 4 = #.
(ii) Here k = 2 and P = f ( 3 , 2 , *) = (;)
(iii) Here k = 1 and
P = f(3,1,3) =
(iv) Here k = 0 and
P = f ( 3 , 0 , 4 ) = (:)
(+)O(4)3
= 1. 1.18 = 1.
8
If 4 items are
(i) 2 are defective, (ii) 3 are
19.3. Suppose that 20% of the items produced by a factory are defective.
chosen a t random, what is the probability P that
defective, (iii) a t least one is defective?
Use Theorem 19.1 with n = 4, p = .2 and q = 1 - p = .8.
(i)
Here k = 2 and
(ii) Here k = 3 and
P = J(4,2, .2) =
(i)(.2)2(.8)2 =
.1536.
P = f(4,3, .2) = (:) (.2)3(.8) = ,0256.
(iii) P = 1 - 44 = 1 - (.8)4 = 1 - .4096 = .5904.
19.4. Team A has probability ;f of winning whenever it plays.
If A plays 4 games, find
the probability that A wins (i) exactly 2 games, (ii) at least 1 game, (iii) more than
half of the games.
Here
12
= 4, p =
and q = 1 - p = $.
CHAP. 191
(i)
I N D E P E N D E N T TRIALS, RANDOM VARIABLES
We seek f ( n , k , p ) when k = 2: f ( 4 , 2 , $ ) =
(i)
=
($)2(&)2
223
&.
(ii) Here 44 =
= $ is the probability t h a t A loses all four games.
probability of winning at least one game.
Then 1- 44 =
is the
(iii) A wins more than half the games if A wins 3 or 4 games. Hence the required probability is
f(4,3,+,
+ f(4,4,9)
= (;)
(9)3
(g) +
(94
g+g = 3
=
27
19.5. A family has 6 children. Find the probability that there are (i) 3 boys and 3 girls,
(ii) fewer boys than girls. Assume that the probability of the birth of either sex i s 3.
(i)
The required probability is
f ( 6 , 3,
4)
=
( 6 ) (1)3
3
2
(1)s = 20 = -5-
16
64
2
(ii) There a r e fewer boys than girls if there a r e 0, 1 or 2 boys. Hence the required probability is
+ f(6,1, +) + f ( 6 , 2 , 4 ) =
.f(6,0,4)
(&6
+
($I5 (4) f (g) (4)4CgP
=
19.6. The probability that a college student does not graduate is .3. Of 5 college students
chosen at random, find the probability that (i) one will not graduate, (ii) three will
not graduate, (iii) a t least one will not graduate.
Here n = 5, p = .3 and q = 1 - p = .7,
(i)
f ( 5 , 1,.3) = (;) (.3) (.7)4 = .36015
(ii) f ( 5 , 3, .3) =
(i)(.3)3(.7)2 =
(iii) 1 - 45 = 1 -
(.7)5
.1323
= 1 - .16807 = .83193
Consider n repeated trials with two outcomes and p = q =
3.
(i) Find the probability of exactly h: successes.
(ii) Find the probability P of more successes than failures if n is (a) odd, ( b ) even.
(i)
The probability of exactly k successes is
f ( n ,k ,
3, =
(T;)
(&)k
= (T;)
(&)"+
(3)n
(ii) Recall t h a t the distribution of the number of successes appears in the expansion of the
binomial ( q p)" =
+
(4+ 4)":
+
1 = (1
2
1
2 )" = (+)n
+ (7) (3)"+ )(;
(4)" +
*
- + (;I
'
($P+ (Y) (&).
+ (4)"
Since the binomial coefficients are symmetrical about the center of the expansion and since
each term is (4)" times a binomial coefficient, the above terms a r e also symmetrical about
the center of the expansion.
(a) If n is odd, then there is an even number of terms in the above expansion and so there
are more successes than failures in exactly the terms in the last half of the expansion.
Thus 1 = P + P or P = Q.
( b ) If n is even, then there is an odd number of terms in the above expansion, with the middle
Now there a r e more successes than failures in the terms to the
term M = (&)(:)".
right of the middle term; hence by the symmetry, 1 = P M P o r P = A(1- M ) .
+ +
224
I N D E P E N D E K T TRIALS, RANDOM VARIABLES
'CHAP. 19
19.8. The probability of team A winning any game is 4. What is the probability P that
A wins more than half of its games if A plays [ i ) 43 games, (ii) 8 games?
(i)
+ since n is odd and 71 = q = +.
of the expansion (4+
i.e. the probability t h a t A wins 4 games and
= g,Then by the preceding problem, P = $(l M ) = + ( l- g)
= z.
By the preceding problem, P =
(ii) The middle term M
loses 4 games, is (:)(&)8
&)8,
-
19.9. How many dice must be thrown so that there is a better than even chance of obtaining
a six?
The probability of not obtaining a six on n dice is
which (Q)" is less than
4:
Hence we seek the smallest
($)rl.
71
for
Thus 4 dice must be thrown.
19.10.The probability of a man hitting a target is 4. (i) If he fires 7 times, what is the
probability P of his hitting the target at least twice? (ii) How many times must he
fire so that the probability of his hitting the target at least once is greater than t ?
(i)
We seek the sum of the probabilities f o r k = 2, 3, 4, 5, 6 and 7 . It is simpler in this case t o
find the sum of t h e probabilities f o r k = 0 and 1,i.e. no hits or 1 hit, and then subtract i t from 1.
f(7,0,$) =
Then
p = 1
2187
16,384
($Ii =
5101
=
16,384
1187
m,
f(7,1,%) =
Ci) ($1
($)6
5103
16,384
~
45-17.
blOZ
(ii) The probability of not hitting the t a r g e t is qn. Thus we seek the smallest n f o r which qn is
less than 1 - $ = Q, where q = 1 - p = 1 - $ = $. Hence compute successive powers of
q until q n < Q is obtained:
(2)l = 34
Q
1.
3'
($)Z
=A
16
Q
A;
d
(9)3
= !7 Q
64
<I
3
1; but ( 9 ) 4 z 81
8
256
In other words, he must fire 4 times.
19.11.Prove Theorem 19.1: The probability of exactly k successes in n repeated trials is
f ( n ,k , p ) = ([)pkq n P k . The probability of at least one success is 1- qn.
The sample space of the 7% repeated trials consists of all ordered n-tuples whose components
a r e either s (success) o r f (failure). The event A o f k successes consists of all ordered n-tuples of
which Is components a r e s and the other n - k components a r e f . The number of n-tuples in the
event A is equal to the number of ways t h a t k letters s can be distributed among t h e ? I components
of a n n-tuple; hence A consists of (E) sample points. Since the probability of each point in A is
p k qll- k , we have P ( A ) = ):( p k q n - k ,
The event B of no successes consists o f the single n-tuple (f,f,, , .,f), i.e. all failures, whose
probability is qn. Accordingly, the probability of at least one success is 1 - qn.
19.12. The mathematics department has 8 graduate assistants who are assigned to the same
office. Each assistant is just as likely to study at home as in the office. How many
desks must there be in the office so that each assistant has a desk at least 90% of
the time?
If there a r e 8 desks, then 0% of the time an assistant will not have a desk at which t o study.
CHAP. 191
INDEPENDENT
TRIALS,RANDOM VARIABLES
225
Suppose theie a r e 7 desks; then an assistant will not have a desk a t which to study if all
8 assistants a r e in the office. The probability t h a t 8 will study in the office is ( $ ) 8 =
= 0.45:.
Thus there will not be enough desks 0 . d c r of the time.
&
Suppose there a r e 6 desks; then an assistant will not have a desk if 8 o r 7 assistants a r e in
the office. The probability t h a t 7 xi11 study in the office is f ( 8 . 7 , i ) =
= 3.2%. Thus there
will not be enough desks 3.2% 0.Ac; = 3.GC; of the time.
+
Suppose there are 5 desks; then a n assistant will not have a desk if 8, 7 o r 6 assistants a r e
in the office. The probability t h a t 6 will study in the office is f ( 8 , 6 ,$) =
= 10.9%. Thus there
will not he enough desks 10.9% 3.2"r 0.1G = 14.5% of the time. In other words, there will be
enough desks only 85.5% of the time.
+
+
Accordingly. 6 desks a r e required.
RANDOM VARIABLES
19.13. A pair of fair dice is thrown. Let S be the random variable which assigns to each
point the maximum of the two numbers which occur: X ( ( a ,b ) ) = max (a, b). (i) Find
the distribution of X . (ii) Find the expectation of X .
I
(i)
The sample space S is the equiprobable space consisting of the 36 ordered pairs of numbers
between 1 and 6: S = {(l, l), (1,Z ) , , . , (6,6)}. Since X assigns to each point in S the
maximum of the two numbers in the ordered pair, the range space of X consists of the
numbers from 1 through 6: R , = { l , Z , 3 , 4 , 5 , 6 } .
&.
Each of three points
There is only one point (1,l) whose maximum is 1; hence P ( 1 ) =
Each of five points in S ,
in S, ( 1 , 2 ) , ( 2 , 2 ) and (2, l), has maximum 2; hence P(2) =
Similarly, P ( 4 )
P(5) =
( 1 , 3 ) , ( 2 , 3 ) , 13,3), ( 3,2) and (3, l ) , has 3; hence P(3)
11
and P ( 6 ) = 1 ~ .,
&.
=A.
The distribution of X consists of the points in
I
I
=A,
&
Rx with their respective probabilities:
I
I
I
(ii) To obtain the expectation of X , i.e. the expected value, multiply each number x, by i t s
probability p z and take the sum:
E
:i6+ 2 . A
= 1.-
+ 4-$ +
- 3 . 16
2
5-$
+ 601
36
-
36
= 4.5
19.14. Suppose a random variable X takes on the values -3, -1, 2 and 5 with respective
2 x - 3 x + l x-1
x-2
__ and ___ .
10 ' 10 ' 10
10
expectation of X.
probabilities
__
~
Determine the distribution and
Set the sum of the probabilities equal t o 1, and solve t o find .2: = 3. Then put x = 3 into the
above probabilities and obtain .3, .4, .2 and .1 respectively. The distribution of X is
xi
I
-3
I
-1
5
2
I
I
I
The expectation is obtained by multiplying each value by its probability and taking the sum:
E = (-3)(.3)
+ (-1)(.4) + 2(.2) + 5(.1)
= --.4
226
INDEPENDENT TRIALS, RANDOM VARIABLES
[CHAP. 19
19.15. Four fair coins are tossed. Let X denote the number of heads occurring. Find the
distribution and expectation of the random variable X .
nith
Since this experiment consists of four independent trials, we obtain the binomial distribution
7 1 = 4 and p = $:
P(0) = f(4,0,3) =
P(3) = 1 ( 4 , 3 , 3 ) = 2
1L
6
P(1) = f ( 4 , 1 , + ) =
P(2) = J ( 4 , 2 , 4 ) =
P ( 4 ) = 1'(4,4,3) =
&
&
Thus the distribution is
I
I
I
+
+
+
+
1 - 1A
2 *1A
3 * &. 4 & = 2.
E = 0 1
and the expectation is
16
6
6
Theorem 19.3 which states that the expected value E = ? ~ p= 4
= 2.
This agrees with
4
19.16. A coin weighted so that P(H ) = $ and P ( T )= & is tossed three times.
Let X be the
random variable which denotes the longest string of heads which occurs. Find the
distribution and expectation of X .
The random variable X is defined on the sample space
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The points in S have the following respective probabilities:
P ( H H H ) = 3.8.3
= 27
4 4 . 4
P(HHT) =
3.3.1
4
4
P(THH) =
64
4
P(HTT) = $ * $ * $ =
&
A
P(THT) = 1.3-1 = 8
&
&
P(TTH) =
$ * $ a $
=
P(TTT) =
&.&.&
=
4
64
P(HTH) = $ * $ * $ =
=
$ * $ a $
4
4
64
&
Since X denotes the longest string of heads,
X(TTT) = 0;
X(HTT) = 1, X(HTH) = 1, X(THT) = 1, X(TTH)
X(HHT) = 2, X(THH) = 2;
1;
X(HHH) = 3
Thus the range space of X is R, = {0,1,2,3}. The probability of each number x, in R, is
obtained by summing the probabilities of the points in S whose image is xi:
P ( 0 ) = P(TTT) =
P(1) = P(HTT)
&
P(2) = P(HHT)
+ P(HTH) + P ( T H T ) + P(TTH)
=
+ P(THH)
=
8
P ( 3 ) = P(HHH) =
The distribution of X consists of the numbers in the range space R, with their respective
probabilities:
64
64
64
64
19.17. Prove Theorem 19.2: Let S be an equiprobable space. If R, = {xl, . . , , xt} is the
range space of a random variable X defined on S , then
pi = P ( x , ) =
number of points in S whose image is xi
number of points in S
Let S have n points and let sl, s2,
show that P ( x J = r/n. By definition,
,
. .,s,
be the points in S whose image is x,. We wish to
CHAP. 191
I N D E P E N D E N T TRIALS, RANDOM VARIABLES
227
P ( x i ) = sum of the probabilities of the points in S whose image is xi
= P(Sl) P(s2) . . .
P(s,)
+
+
+
Since S is an equiprobable space, each o l the
points in S has probability lln. Hence
?t
r times
19.18. Let X be a random variable defined on a sample space S and with distribution
Show that the following are equivalent ways of defining the expectation E ( X ) of the
random variable X :
E ( X ) = 2,p1 * * .
x,p, =
X ( s )P ( s )
c
+
+
S E S
Let sil,si2,. , . , sini be the points in S with image x i , t h a t is, X ( s i j ) = x i . Then,
p,
= P ( z J = sum of the probabilities of the points in S whose image is z1
=
p,
=
P(Z2)
P(S1,)
=
+
P(SI*)
+
P(S22)
P(S21)
..............................
p,
= P(rJ =
P(S,l)
t
+
'.*
+
+
' ' '
Z2Pz
=
,z
3=1
P(S1j)
j=1
"2
+ P(s2n2)
=
P(s2j)
j=1
. . . ............................
P(Sr2)
+
'
a
+
*
z
nt
P(S,,t)
=
P(Stj)
j=1
nt
n2
Similarly,
3
=
P(Slnl)
X ( S 2 ; ) P(SZ,),
. . .,
XtP,
=
B
J = lX(s,,) P(Stj)
Since the slJ range over all the points s E S , we obtain
19.19. Let X and Y be random variables defined on the same sample space S. Let X
the random variable on S defined by
(X Y ) ( s ) = X ( s ) Y ( s )
+
+
Prove that E ( X
+ Y ) = E ( X )+ E ( Y ) .
Using the result of the preceding problem, we obtain
E(X
+Y)
=
2
( X + Y ) ( s )P ( s )
2
( X ( S ) P ( S )i
3
X(s)P(s)
ses
=
=
ses
S E S
+
=
2
seS
+
( X ( s ) Y ( s ) )P ( s )
Y(s)P(s))
z Y ( s )P ( s )
seS
=
E(X)
+ E(Y)
+ Y be
228
I N D E P E N D E K T TRIALS, RANDOM VARIABLES
[CHAP. 19
EXPECTED VALUES
19.20. A player tosses two fair coins. He wins $2 if 2 heads occur, and $1 if 1 head occurs.
On the other hand, he loses $3 if no heads occur. Determine the expected value of
the game and if it is favorable to the player.
Note t h at the probability t h at 2 heads occur is f, tha t 2 tails occur is f and tha t 1 head
Thus the probability of winning $2 is $, of winning $1 is
and of losing $3 is
occurs is
Accordingly, E = 2 & 1 * - 3 . i = & = .25.
In other words, the expected value of
the game is 25d in favor of the player.
+.
-
a.
4,
+ 3
19.21. A player tosses two fair coins. He wins $3 if 2 heads occur, and $1 if 1 head occurs.
If the game is to be fair, how much should he lose if no heads occur?
Let .T denote the amount he should lose if no heads occur. We have t h a t the probability of
winning 93 is $, of winning $1 is
and of losing $x is
Since the game is to be fa ir, set E = 0:
a.
3,
E = 3.$+1.1---~.1 = 0
or
2
5 -z -- 0
4
x = 5
or
4
In other words, the game is f a i r if he loses $5 when no heads occur.
19.22. A player tosses two fair coins. He wins $5 if 2 heads occur, $2 if 1 head occurs and
$1 if no heads occur. (i) Find his expected winnings. (ii) How much should he pay
t o play the game if it is to be f a i r ?
(i)
The probability of winning $ 5 is $, of winning $2 is 6, and of winning $1 is $; hence
E = 5
2 * 3 1 4 = 2.50, t h a t is, the expected winnings a r e $2.50.
- t+
+
-
(ii) If he pays $2.50 t o play the game, then the game is fair.
19.23. A box contains 10 items of which 2 are defective. If four items a r e selected from
the box, what is the expected number of defective items in the sample of size 4?
(io)
There ar e
= 210 different samples of size 4 (without replacement). Since there a r e 8 nondefective items in the box, then (:) = 70 different samples contain no defective items. Hence the
=
probability P ( 0 ) of no defective items in the sample is P(0) =
2
(B)
(:)
5.
= 112 different samples contain 1 defective item: hence P(1) =
=
28
= 28 different samples contain the 2 defective items; hence P ( 2 ) =
Thus the expected number of defective items is
E = O * ' +3
l a & +
&.
- 2
-
E.
2m2
15 =
5
*
19.24. A fair die is tossed 300 times. Compute the expected number of sixes.
This is an independent trials experiment with n = 300 and p = Q.
Hence by Theorem 19.3, E = n p = 300.Q = 50.
19.25.Determine the expected number of boys in a family with 4 children, assuming the
sex distribution to be equally probable. What is the probability that the expected
number of boys does occur?
4;
This is an independent (repeated) trials experiment with ?i = 4 and p = hence by Theorem 19.3,
1
2 = 2. In other words, the family can expect to have 2 boys. The probability t h a t
the family has 2 boys is f ( 4 , 2 , 3 ) = 8.
E = np = 4
CHAP. 191
INDEPENDENT TRIALS,RANDOM VARIABLES
229
19.26. Prove Theorem 19.3: For the binomial distribution
the expected value is E = np.
n
E
71
=
k(I;)qn-!
=
i,ib
k=O
n
(li -
l)!
q"-k
since q - 11 = 1.
=
pk-1
11p
(p
+ p)!l-l
zz
np
1111the above summation. n e dropped the term n h e n li = 0 since i t is 0 * q" = 0.)
19.27. A fair coin is tossed until a head or 4 tails occurs.
tosses of the coin.
Find the expected number of
Only one toss occurs if heads occurs the first time, i.e. the ererit H. Two tosses occur if the
first is tails and the second is heads, 1.e. the event TH. Three tosses occur if the first t n o a r e
tails and the third is heads, i.e. the event TTH. Four tosses occur if either T T T H or T T T T occurs.
Hence P 1 1= P I H I=
P ( 2 ) = P ( T H = $, P 31 = P ( T T H ) = +, and
3,
P(4) = P T T T H I - P T T T T ) =
Thus
E = 1.++2.&+3.&+.:.&
=
&+A
=
8
8
19.28.A box contains 8 light bulbs of which 3 are defective. A bulb is selected from the
box and tested until a non-defective bulb is chosen. Find the expected number of
bulbs t o be chosen.
The probability of choosing a non-defective bulb the first time is P(1) = 8. The probability
of first choosing a defective bulb and then a non-defective one is P ( 2 ) = $.f = $. The probability of first selecting two defective bulbs and then a non-defective one is P ( 3 ) =
f=
Lastly, the probability of first selecting the three defective bulbs and then a non-defective one is
P(4) = 3.2.1.5 = 1_
56.
Hence
t.3.
&.
,
= 1.:
E
3-&
L
+
=
4.k
84
56
19.29. Prove Theorem 19.4: Let E be the expected value of an experiment. If a constant h:
is added t o each outcome of the experiment, then the expected value E* of the new
experiment is E + k.
T ~ . ,. . , y r be t h e outcomes of the original experiment with respective probabilities
Let
pl,p,, . . , , p r . Note p, p ,
... 4 = 1. Thus
+ +
E
=
(%l+k)pl -
(/l
=
xlpl
+ kp, -
-
x,p1
+ .'pz +
=
E+k
-
.f$J
' ' .
k)p,
+
+ kp, +
+ xtp, +
k)Pt
f ("t+
"'
...
T
a.,P, + kp,
k(p1 - 112
-
..'
-
Pt)
230
I N D E P E N D E N T TRIALS, RANDOM VARIABLES
[CHAP. 19
Supplementary Problems
REPEATED TRIALS WITH TWO OUTCOMES
19.30.
19.31.
A card is drawn and replaced three times from a n ordinary deck of 52 cards. Find the probability
t h a t iii two hearts a r e drawn, ( i i ) three hearts a r e drawn, (iii, a t least one he a rt is drawn.
19.32.
A baseball player’s batting average is .300. If he comes to bat 4 times, w ha t is the probability
t h a t he will get (i) two hits, (ii) a t least one hit?
19.33.
A box contains 3 red marbles and 2 white marbles. A marble is drawn and replaced three times
from the box. Find the probability t h a t (i) 1 red marble was drawn, (ii) 2 red marbles were
drawn, (iii) a t least one red marble was drawn.
19.34.
Team A has probability Q of winning whenever i t plays. If A plays 4 games, find the probability
t h a t A wins ( i ) 2 games, (ii) a t least 1 game, (iii) more than half of the games.
19.35.
A card is drawn and replaced in an ordinary deck of 52 cards. How many times must a card be
drawn so t h at (i) there is a t least a n even chance of drawing a heart, (ii) the probability of
drawing a heart is greater than $ ?
19.36.
The probability of a man hitting a target is 8. (i) If he fires 5 times, what is the probability of
hitting the target a t least twice? iii) How many times must he fire so t h a t the probability of
hitting the t ar g et at least once is more than 90%.
19.37. Ten typists ar e assigned to the same office. Each typist is a s likely to work at home a s in the office.
How many typewriters must there be in the office so t h a t each typist ha s a typewriter a t least
90r2 of the time?
RANDOM VARIABLES
19.38.
A pair of f ai r dice is thrown. Let X be the random variable which denotes the minimum of the
two numbers which occur. Find the distribution and expectation of X .
19.39.
Five f a i r coins a r e tossed. Let X be the random variable which denotes the number of heads
occurring. Find the distribution and expectation of X.
19.40. A f a i r coin is tossed f o u r times. Let X be the random variable which denotes the longest string
of heads. Find t h e distribution and expectation of X .
+.
19.41.
Solve the preceding problem in the case t h a t the coin is weighted so t h a t P ( H ) = 3 and P ( T ) =
19.42.
Two cards a r e selected from a box which contains five cards numbered 1 , 1 , 2 , 2 and 3. (i) Let X
be the random variable which denotes the sum of the two numbers. Find the distribution and
expectation of X . (ii) Let Y be the random variable which denotes the maximum of the two
numbers. Find the distribution and expectation of Y .
19.43.
Refer to the preceding problem. Let X Y be the random variable defined by ( X Y ) ( s ) =
X ( s ) Y ( s ) , i.e. which denotes the sum of the two numbers added to the maximum number.
F o r example, ( X + Y)(Z,3) = 5 A 3 = 8. Find the distribution and expectation of X
Y . Verify
that E ( X Y ) = E ( X ) E ( Y ) .
+
+
+
+
19.44. A man stands a t the origin. He tosses a coin. If a head occurs, he moves one
if a tail occurs he moves one unit t o the left. (i) If X is the random variable
distance from the origin af t er 4 tosses of the coin, find the distribution and
(ii) If Y is the random variable which denotes his position a fte r 4 tosses of
distribution and expectation of Y .
19.45.
Solve (i) and [ii) of the preceding problem for the case of 5 tosses of the coin.
+
+
unit to the right;
which denotes his
expectation of X .
the coin, find the
CHAP. 191
I N D E P E N D E N T TRIALS, RANDOM VARIABLES
231
EXPECTED VALUE
19.46. A player tosses three f a i r coins. He wins $5 if 3 heads occur, $3 if 2 heads occur, and $1 if only
1 head occurs. On the other hand, he loses $15 if 3 tails occur. Find the value of the game to
the player.
19.47. A player tosses three f a i r coins. He wins $8 if 3 heads occur, $3 if 2 heads occur, and $1 if only
1 head occurs. If the game is to be f ai r, how much should he lose if no heads occur?
19.48. A player tosses three f a i r coins. He wins $8 if 3 heads occur, $3 if 2 heads occur, $1 if 1 head
occurs, but nothing if no heads occur. If the game is to be fa ir, how much should he pay to play
the game?
19.49. A box contains 8 items of which 2 a r e defective. A man selects 3 items from the box. Find the
expected number of defective items he has drawn.
19.50.
Of the bolts produced by a factory, 2 5 a r e defective. In a lot of 8000 bolts from the factory, how
many a r e expected t o be defective?
19.51.
Find the expected number of girls in a family with 6 children, assuming sex distribution t o be
equally probable. Wh at i s the probability t h a t the expected number of girls does occur?
19.52.
A lottery with 500 tickets gives one prize of $100, 3 prizes of $50 each, and 5 prizes of $25 each.
(i) Find the expected winnings of a ticket. fii) If a ticket costs $1, w ha t is the expected value
of t h e game?
19.53. A f a i r coin is tossed until a head or five tails occur. Find the expected number of tosses of the coin.
4
6
19.54.
A coin weighted so t h at P ( H ) = and P ( T ) =
the expected number of tosses of the coin.
19.55.
A plays B in a tournament. The first team
The probability of team A winning an y game is
t o win 2 games in a row o r a total of three games wins the tournament. Find the expected number
of games in the tournament.
is tossed until a head or five tails occurs. Find
4.
19.56. A box contains 1 0 transistors of which 2 a r e defective. A transistor is selected from the box and
tested until a non-defective one is chosen. Find the expected number of transistors to be chosen.
19.57.
Solve the preceding problem in the case t h a t 3 of the 10 items a re defective.
Answers to Supplementary Problems
21
&, (ii) 128,
27
19.30.
(i)
19.31.
(i) &, (ii) &, (iii)
19.32.
(i) 0.2646, (ii) 0.7599
19.33.
(i)
g,(ii) a,(iii)
117
19.34.
(i)
g,(ii) m, (iii)
112
19.35.
(i) 3, (ii) 5
19.36.
(i)
19.37.
7
(iii) 1.r~
54
544
g,(ii) 6
232
I N D E P E N D E N T TRIALS, RANDOM VARIABLES
[CHAP. 19
19.38.
19.39.
c
xi
5
U
1
2
n
Pi
19.40.
19.41.
2
1
xi
0
Pi
1
1
3
l
32
3
16
16
32
O
L
32
3
-
5
4
4
2
-
16
-1
32
E(X)=$
1'6
8
1
E(X)=
-
81
81
81
g
81
19.42.
E ( X ) = 3.6
(i)
19.43.
x
i
3
I
pi
5
I
.1
6
I
.4
7
I
.1
8
1
.2
(ii)
E ( X + Y ) = 5.9
.2
19.44.
19.45.
E ( X ) = 15
19.46.
2%
19.52.
19.47.
$20
19.53.
19.48. $2.50
19.49.
23
19.54.
19.55.
(i) 75d, (ii) - 25@
%
23
19.50.
160
11
19.56. s
19.51.
3,
19.57.
E ( Y ) = 2.3
Chapter 20
Markov Chains
PROBABILITY VECTORS
A row vector u = (u1,u2,. . .,u,,)
is called a probability vector if its components are
non-negative and their sum is 1.
Example 1.1 :
Consider the following vectors:
16 = ( 3 0 -1 I)
7j =
4’ >
4’ ’’
Then:
I
(2, 4,0’ 1)
4
and
zu = (11 0 1)
4’4’
’2
ic
is not a probability vector since its third component is negative;
1:
is not a probability vector since the sum of its components is greater than 1:
zt‘ is a probability vector.
Example 1.2:
Remark:
-
The non-zero vector 2’ = ( 2 , 3 , 5 , 0 , 1 ) is not a probability vector since the sum of
its components is 2 f 3 5 O + 1 = 11. However, since the components of w
a r e non-negative, u has a unique scalar multiple Xu which is a probability vector;
it can be obtained from v by dividing each component of v by the sum of the
’
3
5
1
components of v: & G = (z
11’ fi, fl. 0 , E ) .
+
Since the sum of the components of a probability vector is one, a n arbitrary
probability vector with n components can be represented in terms of n - 1 unknowns as follows:
( X l , x2, . . ., x,1-1, 1 - x1 - * . ’ - x n - 1 )
In particular, arbitrary probability vectors with 2 and 3 components can be
represented, respectively, in the form
(x, 1-x)
and
(x,Y, 1 - x - y )
STOCHASTIC AND REGULAR STOCHASTIC MATRICES
A square matrix P = ( p i J is called a stochastic matrix if each of its rows is a probability vector, i.e. if each entry of P is non-negative and the sum of the entries in every
row is 1.
Example 2.1 :
Consider t h e following matrices:
(ii)
(i)
(iii)
is not a stochastic matrix since the entry in the second row and third column
is negative;
(ii) is not a stochastic matrix since the sum of the entries in t h e second row is
not 1;
(iii) is a stochastic matrix since each row is a probability vector.
233
234
[CHAP. 20
31ARKOV CHAINS
We shall prove (Problem 20.7)
Theorem 20.1:
If A and B are stochastic matrices, then the product AB is a stochastic
matrix. Therefore, in particular, all powers A" are stochastic matrices.
We now define an important class of stochastic matrices whose properties shall be
investigated subsequently.
1-
A stochastic matrix P is said to be regular if all the entries of some power
Pm are positive.
+)
0
Example 2.2:
1
The stochastic matrix A =
is regular since
is positive in every entry.
1
Example 2.3:
Consider the stochastic matrix A =
11
A2 =
0
j4
(+
1
A3
=
+J
0
(3
$1'
Here
0
A4
=
;(
A)
In f act every power A'n will have 1 and 0 in the first row; hence A is not regular.
FIXED POINTS OF SQUARE MATRICES
A non-zero row vector u = (ul,u,,. . . ,u,) is called a fixed point of a square matrix A
if u is left fixed, i.e. is not changed,- when multiplied by A :
uA = u
We do not include the zero vector 0 as a fixed point of a matrix since it is always left fixed
by every matrix A : OA = 0.
(::)
.
Then the vector u = (2, -1) is a fixed point
Example 3.1:
Consider t h e matrix A =
of A . For,
Example 3.2:
Suppose u is a fixed point of a matrix A : zcA = u. W e claim t h a t every scalar
multiple of u , say hti, is also a fixed point of A . For,
(xu)A = k(zcA) = xu
Thus, in particular, the vector 2u = (4, -2)
t h e example above:
is a fixed point of the matrix A of
We state the result of the preceding example as a theorem.
Theorem 20.2:
If u is a fixed vector of a matrix A , then for any real number h # 0 the
scalar multiple XU is also a fixed vector of A .
235
MARKOV CHAINS
CHAP. 201
FIXED POINTS AND REGULAR STOCHASTIC MATRICES
The main relationship between regular stochastic matrices and fixed points is contained
in the next theorem whose proof lies beyond the scope of this text.
Theorem 20.3:
Let P be a regular stochastic matrix.
Then:
(i) P has a unique fixed probability vector t , and the components of t
are all positive;
(ii) the sequence P, P2,P 3 , .. . of powers of P approaches the matrix T
whose rows are each the fixed point t ;
(iii) if p is any probability vector, then the sequence of vectors
pP,pP2,pP3, . . . approaches the fixed point t.
Note: P" approaches T means that each entry of Prl approaches the corresponding
entry of T,and pPn approaches t means that each component of pPn approaches the corresponding component of t.
Example 4.1:
Consider the regular stochastic matrix P =
vector t = (5, 1 - 2) such that tP = t
( +).
0
1
We seek a probability
Multiplying the left side of the above matrix equation, we obtain
Thus t =
1= (Q,+) is the unique fixed probability vector of P. By
Theorem 20.3, the sequence P , P 2 , P 3 , . . . approaches the matrix T whose rows are
each the vector t:
(4, 4)
T
=
(i :)
(
=
*33 *67)
.33 .67
We exhibit some of the powers of P to indicate the above result:
Example 4.2:
Find the unique fixed probability vector of the regular stochastic matrix
0
Method 1.
1
We seek a probability vector t =
0
(2,Y,1 - 2
=
- y)
such that tP = t:
(x,y,l-z-y)
Multiplying the left side of the above matrix equation and then setting corresponding components equal to each other, we obtain the system
236
MARKOV CHAINS
21 - l
2 z -2
l,
Y
[CHAP. 20
3x+y = 1
= x
2 = 1
Z + ~ - + Z - &
= y
z - 3 ~ = -1
or
y = 1-x-y
z+2y
5
or
y = g
= 1
Thus t = (&,$, 8) is the unique fixed probability vector of P.
Method 2.
We first seek a n y fixed vector u = (2,y,z) of the matrix P:
0
1
0
$2
= (z, Y,2 )
or
Z + L Z
2
= x
= y
u = z
We know t h a t the system h a s a non-zero solution; hence we can arbitrarily assign
a value to one of the unknowns. Set z = 2. Then by the first equation x = 1,
and by the third equation y = 2. Thus U. = ( 1 , 2 , 2 ) is a fixed point of P . But
every multiple of u is a fixed point of P ; hence multiply u by 6 to obtain the required
$).
fixed probability vector t = QZL=
(i,?,
MARKOV CHAINS
We now consider a sequence of trials whose outcomes, say, X , , X,, . . . , satisfy the
following two properties:
u2, . . .,am} called the state space
(i) Each outcome belongs to a finite set of outcomes {a,,
of the system; if the outcome on the nth trial is a l , then we say that the system is in
state a?at time n or at the nth step.
(ii) The outcome of any trial depends at most upon the outcome of the immediately preceding trial and not upon any other previous outcome; with each pair of states (a,,aj)
there is given the probability p i j that uj occurs immediately after at occurs.
Such a stochastic process is called a (finite) Alarkov chain. The numbers p,,, called the
transifion probabilities, can be arranged in a matrix
* *
P12
P =
..................
called the transition matrix.
Thus with each state al there corresponds the ith row (pi,,pi2, . . .,pl,J of the transition
matrix P; if the system is in state ui,then this row vector represents the probabilities of
all the possible outcomes of the next trial and so it is a probability vector. Accordingly,
Theorem 20.4:
The transition matrix P of a Markov chain is a stochastic matrix.
Example 5.1:
A man either drives his car or takes a train to work each day. Suppose h e never
takes the train two days in a row; but if he drives to work, then the next day
he is just as likely t o drive again as h e is t o take the train.
The state space of the system is { t (train), d (drive)}. This stochastic process
is a Markov chain since the outcome on any day depends only on what happened
the preceding day. The transition matrix of t h e Markov chain is
t
d
CHAP. 20)
237
SIARKOV CHAINS
The first row of the matrix corresponds to the fa c t t h a t he never takes the
t r ai n two days in a row and so he definitely will drive the day after he takes the
train. The second row of the matrix corresponds t o the fa c t tha t the day a fte r
he drives he will drive o r take the train with equal probability.
Example 5.2:
Three boys A , B and C a re throwing a ball t o each other. A always throws the ball
t o B and B always throws the ball to C; but C is just as likely to throw the ball
to B as to A . Let X, denote t h e n t h person to be thrown t h e ball. The sta te space
of t h e system is ( A ,B , C}. This is a Markov chain since the person throwing the
ball is not influenced by those who previously had the ball. The transition matrix
of the Markov chain is
A
B
C
B
The first row of the matrix corresponds to the fa c t tha t A always throws the
ball to B. The second row corresponds t o the fact tha t B always throws the ball
to C. The last row corresponds to the fa c t tha t
throws the ball t o A or B with
equal probability (and does not throw it to himself).
c
Example 5.3:
A school contains 200 boys and 150 girls. One student is selected after another
t o take a n eye examination. Let X, denote the sex of the n t h student who takes
the examination. The state space of the stochastic process is { m (male), f (female)}.
However, this process is not a Markov chain since, for example, the probability
t h a t the third person is a girl depends not only on the outcome of the second trial
b u t on both the first and second trials.
Example 5.4:
(Random waIk with reflecting barriers.) A man is a t a n integral point on the
x-axis between the origin 0 and, say, the point 5. H e takes a unit step to the
r i g h t with probability p o r to the left with probability g = 1- p , unless he is at
t h e origin where he takes a step to the right to 1 o r at the point 5 where he takes
a step t o the left to 4. Let X, denote his position a fte r n steps. This is a Markov
chain with state space {uo,a,, a,, a,, a,, a,) where a, means t h a t the man is a t the
point i. The transition matrix is
a0
a2
a1
a3
a4
a5
Each row of the matrix, except the first and last, corresponds to the fa c t tha t
the man moves from sta te aito sta te a, + l with probability p or back to state ai-l
with probability q = 1 - p . The first row corresponds to the fa c t t h a t the man
must move from s t ate a. to sta te al,and the la st row that the nian must move
f r o m state a5 to s t at e a,.
HIGHER TRANSITION PROBABILITIES
The entry p l j in the transition matrix P of a Markov chain is the probability that the
system changes from the state a[ to the state a, in one step: a, acl. Question: What is
the probability, denoted by pi;), that the system changes from the state at t o the state alin
exactly n steps:
+ a k , + * * * + ak,-l
a,
U,
-+
+
The next theorem answers this question; here the pi;) are arranged in a matrix
the n-step transition matrix:
called
238
MARKOV CHAINS
Theorem 20.5:
[CHAP. 20
Let P be the transition matrix of a Markov chain process. Then the
= P".
n-step transition matrix is equal to the nth power of P: P(")
Now suppose that, a t some arbitrary time, the probability that the system is in state
ai is 11, ; we denote these probabilities by the probability vector p = (p,, p2, . . . , p,,,) which
is called the probability distribution of the system a t that time. In particular, we shall let
p'0'
= (@', p y ,
. . .)p:;))
denote the initial probability distribution, i.e. the distribution when the process begins,
and we shall let
(n)
= ( P y , P?), * * .> P,,
)
P(ll)
denote the nth s t e p probability distribution, i.e. the distribution after the first n steps.
The following theorem applies.
Theorem 20.6:
Let P be the transition matrix of a Markov chain process. If p = ( p J is
the probability distribution of the system a t some arbitrary time, then
pP is the probability distribution of the system one step later and pPn is
the probability distribution of the system n steps later. In particular,
*(I)
Example 6.1 :
= p<o'p p )= p"'p
f
9
@3)
= p(2'p, . . . )
p(n)
= p(0)pn
Consider the Markov chain of Example 5.1 whose transition matrix is
t
d
Here t is the sta te of taking a tra in to work and d of driving t o work.
ample 4.1,
By
Ex-
Thus t h e probability t h a t the system changes from, say, state t t o state d in exactly
4 steps is
i.e. p $ ) = #. Similarly, p j : ) = #,
=
and p(4) = 11.
8,
dd
16
Now suppose tha t on the first day of work, the man tossed a f a i r die and drove
to work if and only if a 6 appeared. I n other words, p(O) = (&&) is the initial
arobabilitv distribution. Then
is the probability distribution a fte r 4 days, i.e. piq) =
Example 6.2:
and p d( 4 ) =
96'
Consider the Markov chain of Example 5.2 whose transition matrix is
A
B
C
Suppose C was the first person with the ball, i.e. suppose
initial probability distribution. Then
0
p"'
= p'0'P
= (0,0,1)
1
0
p(0)
= ( O , O , 1) is the
CHAP. 201
MARKOV CHAINS
239
Thus, a f t e r three throws, the probability t h a t A has the ball is &, t h a t B h a s the
= 1 and p y ) =
ball is $ and t h a t C has the ball i s $: pa31 = $, ):p
fr.
Example 6.3:
Consider the random walk problem of Example 5.4. Suppose the man began a t the
point 2; find the probability distribution a f t e r 3 steps and a f t e r 4 steps, i.e. ~ ( 3
and pC4).
Now
p(0) =
p“’
p‘2’
p‘3’
p(4)
( O , O , 1,0, 0,O) is t h e initial probability distribution.
Then
= p(0’P = ( 0 , q , 0,p , 0 , 0 )
= p“’P = ( q ! , 0 , 2pq, 0 , p2, 0 )
= p(”P = ( 0 , q L 2p2q, 0 , 3p2q, 0,p 3 )
= p ( 3 ) ~= ( a 3 2 p q 2 , 0 , pq2 2p3q 3 ~ 2 q 2 0, , 3 ~ 3 q
-
+
+
+
+9 , o )
Thus a f t e r 4 steps he is a t , say, the origin with probability q3
+ 2p2q2.
STATIONARY DISTRIBUTION OF REGULAR MARKOV CHAINS
Suppose that a Markov chain is regular, i.e. that its transition matrix P is regular.
By Theorem 20.3, the sequence of n-step transition matrices Pn approaches the matrix T
whose rows are each the unique fixed probability vector t of P ; hence the probability p:;)
that u3occurs for sufficiently large n is independent of the original state a, and it approaches
the component t j of t. In other words,
Theorem 20.7:
Let the transition matrix P of a Markov chain be regular. Then, in the
long run, the probability that any state u3 occurs is approximately equal
to the component t3 of the unique fixed probability vector t of P .
Thus we see that the effect of the initial state or the initial probability distribution of
the process wears off as the number of steps of the process increase. Furthermore, every
sequence of probability distributions approaches the fixed probability vector t of P , called
the stationary distribution of the Markov chain.
Example 7.1:
Consider t h e Markov chain process of Example 5.1 whose transition matrix is
t d
t
P
=
d
(&
0
1
4)
(4,g).
By Example 4.1, the unique fixed probability vector of the above matrix is
Thus, in t h e long run, the man will take t h e train to work of t h e time, and drive
to work the other 8 of the time.
4
Example 7.2:
Consider t h e Markov chain process of Example 5.2 whose transition matrix is
A
P
=
B
C
B
By Example 4.2, the unique fixed probability vector of t h e above matrix is
3, g). Thus, in the long run, A will be thrown the ball 20% of the time, and
B and C 40% of the time.
(&,
)
[CHAP. 20
I I A R K O V CHAINS
240
ABSORBING STATES
A state ut of a Markov chain is called absorbing if the system remains in the state ai
once it enters there. Thus a state azis absorbing if and only if the ith row of the transition
matrix P has a 1 on the main diagonal and zeros everywhere else.
Example 8.1:
Suppose the following matrix is the transition matrix of a Markov chain:
a1
a2
a5
a4
a3
The states a? and a 5 a r e each absorbing, since each of the second and fifth rows
h a s a 1 on the main diagonal.
Example 8.2:
(Random walk with absorbing barriers.) Consider the random walk problem of
Example 5.4, except now we assume t h a t the man remains a t either endpoint whenever he reaches there. This is also a RIarkov chain and t h e transition matrix is
given by
an ai a, a3 a4 a5
a1
“I‘
a4
a5
\
q
o
O
O
p
O
o
O
o
O\
o
o
o
o
q
o
0
0
0
0
0
”I
1
W e call this process a random walk with absorbing barriers, since the a. and
I n this case, phn) denotes t h e probability t h a t the man
reaches the s t a t e a,, on o r before t h e izth step. Similarly, p p ) denotes t h e probability t h a t he reaches the a5 on o r before the n t h step.
a5 a r e absorbing states.
Example 8.3:
A player has, say, z dollars. H e bets one dollar a t a time and wins with probability
p and loses with probability q = 1 - p . The game ends when he loses all his
money, i.e. h a s 0 dollars, or when he wins 3 - x dollars, i.e. h a s N dollars. This
game i s identical to the random walk of the preceding example except t h a t here
t h e absorbing barriers are at 0 and N .
Example 8.4:
A man tosses a f a i r coin until 3 heads occur in a row. Let X , = k if, a t t h e itth
trial, t h e l a s t tail occurred a t t h e (n- k)-th trial, i.e. X , denotes the longest string
of heads ending a t the nth trial. This is a RIarkov chain process with state space
{ao, al, a2, as}, where a, means t h e string of heads h a s length i. The transition
matrix is
a,
u1
a2
a3
a0
a1
a2
a3
+
+
o
o
;i f ;)
Each row, except the last, corresponds to t h e f a c t t h a t a string of heads is
either broken if a tail occurs or i s extended by one if a head occurs. The l a s t
line corresponds t o the f a c t t h a t t h e game ends if three heads a r e tossed in a row.
Note t h a t a3 is a n absorbing state.
CHAP. 201
241
MARKOV CHAINS
Let ui be an absorbing state of a hlarkov chain with transition matrix P. Then, for
i # j , the n-step transition probability p:;) = 0 for every n. Accordingly, every power of
P has a zero entry and so P is not regular. Thus:
Theorem 20.8:
If a stochastic matrix P has a 1 on the main diagonal, then P is not regular
(unless P is a 1x 1 matrix).
Solved Problems
PROBABILITY VECTORS AND STOCHASTIC MATRICES
20.1. Which vectors are probability vectors ?
(i) u = (&, 0 , -&, 4,Q), (ii) v = (4,0, ft, +, h), (iii) w = (4,0 , 0, +,4).
A vector is a probability vector if its components a r e non-negative and their sum is 1.
(i)
ZL
is not a probability vector since its third component is negative.
(ii) v is not a probability vector since the sum of the components is gre a te r tha n 1.
(iii)
PL'
is a probability vector since its components a r e non-negative and their sum is 1.
20.2. MuItiply each vector by the appropriate scalar to form a probability vector:
(i) (2, 1, 0, 2, 3), (ii) (4, 0, 1, 2, 0, 5), (iii) (3, 0, -2, l), (iv) (0, 0, 0, 0, 0).
(i)
+ +
The sum of the components i s 2 - 1 0 3 + 2 = 8; hence multiply the vector, i.e. each
Q,0, $, 8).
component, by & to obtain the probability vector
(ii) The sum of the components is 4
component, by
(t,
+ 0 + 1 + 2 + 0 + 5 = 12;
t o obtain the probability vector
hence multiply the vector, i.e. each
(g, 0, &,Q,0 , A).
(iii) The first component is positive and the third is negative; hence i t is impossible to multiply the
vector by a scalar to form a vector with non-negative components. Thus no scalar multiple
of the vector is a probability vector.
(iv) Every scalar multiple of t h e zero vector is the zero vector whose components add to 0. Thus
no multiple of the zero vector is a probability vector.
20.3. Find a multiple of each vector which is a probability vector:
(i) (g, $, 0, 2, g), (ii) (0, 3, 1, Q , $).
In each case, first multiply each vector by a scalar so that the fractions are eliminated.
(i)
F i r st multiply the vector by 6 to obtain ( 3 , 4 , 0 , 1 2 , 5 ) . Then multiply by 1/(3
& t o obtain (&,&,O,$,&)
which is a probability vector.
+ 4 + 0 + 12 + 5) =
(ii) F i r st multiply the vector by 30 t o obtain (0,20,30,18,25). Then multiply by 1/(0 -t20
20 30 18 25
18 25) =
t o obtain ( 0 , ~3.
' 93'93) which is a probability vector.
+
&
+ 30 +
20.4. Which of the following matrices are stochastic matrices?
(i)
A is not a stochastic matrix since it is not a square matrix.
(ii) B is not a stochastic matrix since the sum of the components in the last row is greater tha n 1.
(iii) C is a stochastic matrix.
(iv) D is not a stochastic matrix since the entry in the first row, second column is negative.
242
MARKOV CHAINS
20.5. Let
A =
(1: 1: 1:)
az bz
[CHAP. 20
be a stochastic matrix and let u =
c,
( U I , U P , U ~ )be
a proba-
bility vector. Show that uA is also a probability vector.
(nt i' :I\
b,
ZLA=
(u1,
u g , 243)
c1
= (ulal
+
+ u3a3,u l b , + u2bZ +
u2a2
+
uzc~
~ $ 3 , ulcl+
~ b 3 ~ 3 )
Since the ui,
a,, bi and cl a r e non-negative and since the products and sums of non-negative
numbers a r e non-negative, the components of wl a r e non-negative a s required. Thus we only need
t o show t h a t t h e sum of the components of uA is 1. Here we use the f a c t t h a t z i 1 + u 2 + u 3 ,
a,
b , cl, u 2 b, c2 and u3 b , c3 a r e each 1:
+ +
+ +
~
1
+
~~
12
+ +
+ +
+
+ ~ 3 b 3+ + +
+ b, + +
+ b, + c,) + + +
+ u2.l + u , * l = + u 2 + UQ = 1
~2 ~ 23 ~ 3~ ~ l b uzb,
l
=
zll(~1
=
Zll*l
~
~~
1
~ 1 )
12
~~ 23
Z L ~ ( U ~b 3
~
3
~ 3 )
u1
If A = (aij) is a stochastic matrix of order n and u = (ul,u,, , . . ,uJ is a
probability vector, then uA is also a probability vector.
20.6. Prove:
I:
The proof is similar t o t h a t of the preceding problem for the case n = 3 :
[::: :1
..'
..*
.................
a12
. . ., u,)
ZLA =
( ~ 1 ~, 2 ,
...
an1
ann
+ uza21+ . - + unanl, ula12+ u2a2,+
= (ulall
*
*
+
*
+ unaR2,. . ., ulaln+ uZa2,,+ - + unarm)
*
9
Since the uiand aZja r e non-negative, the components of uA a r e also non-negative. Thus we
only need to show t h a t t h e sum of the components of uA is 1:
u1a12 u2a22 *
u,an2
*
ulaln uZazn * * *
u,a,,
~~u~~ z ~ ~. u ununl
~ ~
- u,(all+ a12
. ' ' un(anl a,, * * * a,,)
* * *
Ul,)
u,(azl a22 * * * a,,)
+
-
+
2L1'1
+
+
+
u2.l
+
+
+
+ +
* ' .
+
+ - +
+ +
+
+ +
+ + +
+ + + + +
+
Un'l
= u,+u,+
' . . +u, = 1
20.7. Prove Theorem 20.1: If A and B are stochastic matrices, then the product A B is a
stochastic matrix. Therefore, in particular, all powers An are stochastic matrices.
The ith-row si of t h e product matrix A B is obtained by multiplying the ith-row ri of A by
the matrix B: si = riB. Since each ri is a probability vector and B is a stochastic matrix, by
the preceding problem, si is also a probability vector. Hence A B is a stochastic matrix.
20.8. Prove: Let p = ( p l , p 2 , . . .,p,) be a probability vector, and let T be a matrix
whose rows are each the same vector t = ( t l ,t,, . . . , tRJ. Then pT = t .
Using the f act t h a t p1
+ p2 + . . . + p m = 1, we have
i t l t,
* . *
tm\
243
MARKOV CHAINS
CHAP. 201
REGULAR STOCHASTIC MATRICES AND FIXED PROBABILITY VECTORS
20.9. Find the unique fixed probability vector of the regular stochastic matrix A =
What matrix does A n approach?
2
2
We seek a probability vector f = (Y, 1 - 2) such t h a t t A = f:
Multiply the left side of the above matrix equation and then set corresponding components equal
to each other to obtain the two equations
=
$x++-+z
2,
Solve either equation to obtain x = Q . Thus t =
$Z+&-frx
(3,+)
= 1-2
is the required probability vector.
Check the answer by computing the product tA:
The answer checks since t A = t .
The matrix An approaches the matrix T whose rows a r e each the fixed point t:
T =
(f i).
z
20.10. (i) Show that the vector u = ( b , a ) is a fixed point of the general 2 x 2 stochastic
matrix
P =
('I"
lab).
(ii) Use the result of (i) to find the unique fixed probability vector of each of the
following matrices:
1--a
(i)
UP = ( " a ) (
a
1-b
)
= (b-ab+ab, a b f a - a b )
= ( b , a ) = u.
(ii) By (i), z t = ( 1 , i ) i s a fixed point of A . Multiply u by 3 to obtain the fixed point (3,2) of A
which has no fractions. Then multiply ( 3 , 2 ) by 1/(3 2) = 6 to obtain the required unique
fixed probability vector (#, Q).
+
By (i), u = ($, 4) is a fixed point of B. Multiply u by 6 to obtain the fixed point ( 4 , 3 ) ,
t o obtain the required unique fixed probability vector
and then multiply by 1 / ( 4 + 3 ) =
3
cq, $).
Hence (8,3) and the probability vector
By (i), ZL = (.8, .3) is a fixed point of C.
a r e also fixed points of C.
20.11. Find the unique fixed probability vector of the regular stochastic matrix
P
=
0
Method 1. We seek a probability vector t =
(2,y,
1
0
1 - x - y) such t h a t tP = t:
= (Z,Y,1---Y)
0
1
0
( A , 6)
MARKOV CHAINS
246
[CHAP. 20
R L
P
R
=
.8 .2
L (.6
A )
The probability distribution for the first trial is p = ( . 5 , - 5 ) . To compute the probability
distribution for the next step, i.e. the second trial, multiply p by the transition matrix P:
= (.7, .3)
(.5, .5) (*8 *2)
.6
.4
Thus on the second trial he predicts that 70% of the mice will go right and 30% will go left. To
compute the probability distribution for the third trial, multiply that of the second trial by P :
(.7, .3)
(
*8 ‘2)
.4
.6
= (.74, .26)
Thus on the third trial he predicts that 74% of the mice will go right and 26% will g o left.
We assume t h a t t h e probability distribution for the thousandth trial is essentially the
stationary probability distribution of the Markov chain, i.e. the unique fixed probability vector t
of the transition matrix P . By Problem 20.10, u = (.6, .2) is a fixed point of P and SO
t = ($,$) = (.75,.25). Thus he predicts that, on the thousandth trial, 75% of the mice will go to
the right and 25% will go to the left.
20.17. Given the transition matrix
p(O) =
(i)
P =
(~*)
1
0
with initial probability distribution
(i,+).
Define and find: (i) pi:), (ii) p C 3 ’ , (iii) p i 3 ) .
pa:) is the probability of moving from state u2 to state a l in 3 steps. It can be obtained from
the 3-step transition matrix P3; hence first compute P3:
p2
Then p;)
(ii)
=(; l)J =(; ;)
p3
is the entry in the second row first column of
P3:
p:) = 2.
8
is the probability distribution of the system after three steps. It can be obtained by
successively computing p ( 1 ) , p ( 2 ) and then ~ ( 3 ) :
p(3)
However, since the 3-step transition matrix P3 has already been computed in (i), p ( 3 ) can also
be obtained as follows:
(iii) p r ) is the probability t h a t the process is in the state a2 after 3 steps; it is the second component of the 3-step probability distribution ~ ( 3 ) : p?) = L
12‘
247
MARKOV CHAINS
CHAP. 201
0”
20.18. Given the transition matrix P =
2
i) and the initial probability distribution
(4 4
0 1 0
p(O) = (3, O,&). Find: (i)
and p t i ) , (ii) p(4)and p?),
approaches, (iv) the matrix that Pi* approaches.
(i)
‘.
Fi r st compute the 2-step transition matrix P2:
0 1 1
0 1
=
pa
Then p::) =
(ii) To compute
4
”) (i i
1
3
-
-
=
t h e 2-step transition matrix P2 and the initial probability distribution
~ ( ~ use
1 ,
p(0)p2
p(2)
pi41
f
1
0 1 0
0 1 0
a
2
and pi:) = 0, since these numbers refer to the entries in P2.
p(0J:
Since
; ;)(:
(iii) the vector that p(O)P”
=
(1 2
3’ 3’
is the third component of
0)
and
p(4)
= p(2)Pz = (1 14’ 12’
A)
p;) = Q.
p(4),
(iii) By Theorem 20.3, p(o)Pn approaches the unique fixed probability vector t of P. To obtain t ,
first find an y fixed vector u = (x,y, z ) :
’ ”\
4 (O4 3
(%2/*
0
1
3y
= (x,y, z )
= x
=
+x++y+z
or
y
0
Find any non-zero solution of the above system of equations. Set z = 1; then by the third
equation x = 2, and by the first equation y = 4. Thus u = (2,4,1) is a fixed point of P and so
t = (+,+,+). In other words, p(O)Pn approaches ($,$,$).
(iv) Pn approaches the matrix T whose rows a r e each the fixed probability vector of P ; hence
2
Pn approaches
4
(!
- - _
I
1
I
7
r
7
$
.
20.19. A salesman’s territory consists of three cities, A , B and C.
He never sells in the
same city on successive days. If he sells in city A , then the next day he sells in
city B. However, if he sells in either B or C, then the next day he is twice as likely
to sell in city A as in the other city. In the long run, how often does he sell in each
of the cities?
The transition matrix of t h e problem is a s follows:
A
p
=
A
B
C
0
1
0
* V ; )
c
z
3
We seek the unique fixed probability vector t of the matrix P. First find a n y fixed vector
u = (x,y,z):
(E 1 $
0
v,4
(2,
1
0
$y
=
(%
Y,4
01
+ 32
s+$z
=
=
2
y
ly = z
Set, say, z = 1. Then by t h e third equation y=3, a nd by the first equation x =+. Thus
8
u = ( ~ , 3 , 1 ) . Also 3u = (8,9,3) is a fixed vector of P . Multiply 3 2 ~by 1/(8+ 9 3) = $ to
obtain the required fixed probability vector t = (z
5’ 20’ ”)
20
= (.40, .45, J 5 ) . Thus in the long
r u n he sells 407;. of the time in city A , 45% of the time in B and 15% of the time in C .
+
248
MARKOV CHAINS
[CHAP. 20
20.20. There are 2 white marbles in urn A and 3 red marbles in urn B . At each step of
the process a marble is selected from each urn and the two marbles selected are
interchanged. Let the state a,of the system be the number i of red marbles in urn A .
(i) Find the transition matrix P . (ii) What is the probability that there are 2 red
marbles in urn A after 3 steps? (iii) In the long run, what is the probability that
there are 2 red marbles in urn A ?
(i)
There a r e three states, ao,
and a2 described by the following diagrams:
a,
B
A
B
A
A
a1
a0
B
a2
If the system is in state a,, then a white marble must be selected from urn A and a red
marble from u r n B , so the system must move to state a,. Accordingly, the first row of the
transition matrix is (0,1,0).
Suppose the system is in state a,. It can move to state a. if and only if a red marble is
selected from u r n A and a white marble from urn B; the probability of t h a t happening is
1.1 = 1. Thus p l o = Q. The system can move from state al to a, if and only if a white
2 3
6
marble is selected from u r n A and a red marble from urn B ; the probability of t h a t happening
Accordingly, the probability t h a t the system remains in state
is 12 . 23 = 1
a . Thus p12=
Thus the second row of the transition matrix is (Q,
(Note
a , is p , , = 1- Q - Q =
t h a t p , , can also be obtained from the f a c t t h a t the system remains in the state a , if either
a white marble is drawn from each urn, probability 4 - i = Q, o r a red marble is drawn
from each urn, probability
* -g =
thus p11 = & f =
4.
4.
4,g).
4
4;
4 4.)
Now suppose the system is in state a2. A red marble must be drawn from urn A . If a
red marble is selected from u rn B , probability Q, then the system remains in state a,; and if
a white marble is selected from urn B , probability $, then the system moves to state al. Note
t h at the system can never move from sta te a2 t o the state uo. Thus the third row of the
Tha t is,
transition matrix is (0,+, i).
a, a1 a2
a0
P
=
a1
a2
(; ;)
1
3
3
(ii) The system began i n state ao, i.e. p(O) = (l,O,O). Thus:
p"'
= pco'p = ( 0 , 1, O),
p(2)
= pcup =
(1
6 , ~,
1
g),
p'3'
= p(2)P = (L
12' 36' ")
18
5
Accordingly, t h e probability t h a t there a r e 2 red marbles in urn A a f t e r 3 steps is g .
(iii, We seek the unique fixed probability vector t of the transition matrix P . F i r s t find a n y fixed
vector u = (x,y,z):
Qy = x
0 1 0
(5,Y, 2)
J
Q Q 4
(o g
= (x,Y, 4
or
x+&y+Qz = y
&y+Qz
= z
Set, say, x = l . Then by the first equation y = 6 , and by the third equation x = 3 . Hence
3) = & to obtain the required unique fixed probazi = (1,6,3). Multiply u by 1/(1+6
bility vector t = (.1,.6, .3). Thus, in the long run, 30% of the time there will be 2 red marbles
in u r n A .
Note that the long r u n probability distribution is the same as if the five marbles were
placed in a n u r n and 2 were selected a t random t o put into urn A .
+
20.21. A player has $2. He bets $1 at a time and wins with probability 8. He stops playing
if he loses the $2 or wins $4. (i) m'hat is the probability that he has lost his money
at the end of, a t most, 5 plays? (ii) What is the probability that the game lasts
more than 7 plays?
249
MARKOV CHAINS
CHAP. 201
This is a random walk with absorbing barriers at 0 and 6 (see Examples 8.2 and 8.3). The
transition matrix is
::/ 4 *
a2
1
0
0
0
0
a
P
=
~
a
as
a
with initial probability distribution
(i)
,
o
I
3
0
0
a6
0
0
0
+
o
3
u5
0
0
\
0
0 ‘
o
o
+
o
’
o
i
I
0
o
p(0)
o
0
4
6
g
a4
a3
0
o
0
0
+
o
0
O
o
0
+
0
0
+
0
o
+
,
0
1
/
= (0, 0 , 1, O , O , 0,O) since he began with $2.
We seek p;5), the probability t h at the system is in sta te a. a fte r five steps.
5th step probability distribution ~ ( 5 ) :
+, 0, 4, 0, 0, 0 )
p“’
p‘2)
p‘3’
= p‘0’P = (0,
p‘4’ = p‘3’P
= p “ ’ P = (&O, *, 0, *, 0, 0 )
= p‘2’P = (*,
0, #, 0, &, 0)
p‘5’ = p‘4’P
A)
= (#, 0 , A, 0, *, 0,
= (#,
0,&, 0 1 L)
A,
4,
Thus p i s ) , the probability t h at he ha s no money a f t e r 5 plays, is
(ii) Compute
p(7):
p(6)
= p(5)p=
(29,
0,
64
A,0’ 13 0’ 1)
8 ’
64’
Compute the
’ 8’
16
8.
p ( 6 ) P = (2
2 0 27 0 13 i)
6 4 ’ 6 4 ’ ’ 128’ ’ 128’ 6
p(7)
The probability that t h e g ame lasts more tha n 7 plays, i.e. t h a t the system i s not in state
a6 af t er 7 steps, i s
=
&+ +
a, or
20.22. Consider repeated tosses of a fair die.
g.
Let Xn be the maximum of the numbers
occurring in the first n trials.
(i) Find the transition matrix P of the Markov chain. Is the matrix regular?
(ii) Find p C 1 ) the
,
probability distribution after the first toss.
(iii) Find p ( z ) and p C 3 ) .
(i)
The state space of t h e Markov chain is { l , 2,3,4,6,6}.
1
P
=
2
3
The transition matrix is
4
5
6
o o o + g g
0
0
0
0
g
Q
We obtain, f o r example, the third row of the matrix as follows. Suppose the system is in
state 3, i.e. the maximum of t h e numbers occurring on the first n tria ls is 3. Then the
3
system remains in state 3 if a 1, 2, o r 3 occurs on the (n 1)-st trial; hence p33 = 6. On the other
hand, the system moves to s t at e 4 , 5 or 6, respectively, if a 4 , 5 o r 6 occurs on the (n4-1)-st trial;
hence p34 = p,, = p S 6 = Q. The system can never move to state 1 or 2 since a 3 h a s occurred
on one of the trials; hence p31= p 3 2 = 0. Thus the third row of the transition matrix is
The other rows a r e obtained similarly.
(0, O,:,
+
i,i,i).
The matrix is not regular since s ta te 6 is absorbing, i.e. there is a 1 on the main diagonal
in row 6.
(ii) On the first toss of the die, the state of the system XI is the number occurring; hence
p“’ = (Q, Q, Q, Q, Q,
9).
(iii)
#2’
= p“’ p = (23 s I 2 “)* p ‘ 3 ) =
369 361 36’ 36, 3 6 ’ 36
p(2’
p = (L
119 37 61 91).
216’ 216’ 216’ 216’ 216’ 216
250
MARKOV CHAINS
[CHAP. 20
20.23. Two boys b , and b, and two girls gl and g, are throwing a ball from one to the other.
Each boy throws the ball t o the other boy with probability 8 and to each girl with
probability 4. On the other hand, each girl throws the ball to each boy with
probability 4 and never t o the other girl. In the long run, how often does each
receive the ball?
This is a Markov chain with state space { b l , b,, g,, g 2 } and transition matrix
bl
P
=
b2
[;1
b,
g1
b2
gz
0
8
t
z
91
'4
92
;)
h
We seek a fixed vector u = (2,u,z , w)of P : ( 2 ,y, x , w)P =
components of UPequal to u t o obtain the system
+ y + + z + a 2w = 5
+5++z+$w =
u,x , w).
(2,
Set the corresponding
u
hx+&y = x
$x+&y = w
We seek any non-zero solution. Set, say, x = 1; then w = 1, 5 = 2 and y = 2. Thus u = ( 2 , 2 , 1 , 1 )
and so the unique fixed probability of P is t = ($,Q,&,&). Thus, in the long run, each boy
receives the ball Q of the time and each girl Q of the time.
20.24. Prove Theorem 20.6: Let P = (p,,) be the transition matrix of a Markov chain.
If p = ( p J is the probability distribution of the system at some arbitrary time k ,
then pP is the probability distribution of the system one step later, i.e. at time k 1;
hence pPn is the probability distribution of the system n steps later, i.e. at time
) p ( l )P , . . . and also pen) = p ( 0 ) P R .
k n. In particular, pc1)= p(O)P,p C 2 =
+
+
Suppose the state space is { a l ,u2, . . .,am}. The probability that the system is in state a, a t
time k and then in state a, a t time k 1 is the product pjp,,. Thus the probability that the system
is in state at a t time k f 1 is the sum
m
+
PlPlt
+~
2 ~ 2 1 +
Thus the probabiilty distribution at time k
+ Pmpmi
+ 1 is
m
=
jg,P ~ P , ,
m
\
P*
However, this vector is precisely the product o f the vector p = (pi)by the matrix
P = ( p i j ) : p* = p P .
20.25. Prove Theorem 20.5: Let P be the transition matrix of a Markov chain. Then the
n-step transition matrix is equal to the nth power of P:
Pn)
= Pn.
Suppose the system is in state ai at, say, time k. We seek the probability 13:;) that the system
is in state aj a t time k + n . Now the probability distribution of the system at time k , since the
system is in state ai,is the vector ei = (0,. . ., 0,1,0,. . ., 0) which has a 1 at the ith position
and zeros everywhere else. By the preceding problem, the probability distribution a t time k + n .
is the product eiPn. But ei Pn is the ith row of the matrix Pll. Thus P!?) i s the jth component
of the ith row of Pn, and so P ( n ) = Pn.
MISCELLANEOUS PROBLEMS
20.26. The transition probabilities of a Markov chain can be represented by a diagram,
called a transition diagram, where a positive probability p i j is denoted by an arrow
from the state ui to the state a j . Find the transition matrix of each of the following
transition diagrams:
CHAP. 201
(i)
MARKOV CHAINS
251
Note first t h a t the s t at e space is {al,az ,u3$ and so the transition matrix is of the form
a1
P
a2
a3
=
a3
The ith row of the matrix is obtained by finding those arrows which emanate from a, in the
diagram; the number attached to the arrow from a, to aJ is the j t h component of the ith row.
Thus the transition matrix is
a, a2 a 3
P
=
(ii) The state space is {a1,a2,a3,a4}.The transition matrix is
a1
P
a2
a4
a3
=
a3
20.27. Suppose the transition matrix of a Markov chain is as follows:
a1
a2
a3
a4
0
0
a4
Is the Markov chain regular?
Note t h a t once the system enters the state a , o r the state a2, then i t can never move to
s t a t e a3 or state a4, i.e. t h e system remains in the sta te subspace { a l , u Z } . Thus, in particular,
pi;) = 0 f o r every n and so every power Pn will contain a zero entry. Hence P is not regular.
20.28. Prove: An rn x m matrix A = (alJ)has a non-zero fixed point if and only if the
determinant of the matrix A - I is zero, Here I is the identity matrix.
The matrix A = (aij) h as a non-zero fixed point if and only if there exists a non-zero vector
u = (q,
xz, . . . , x m ) such t h a t
all
a12
am1
am2
...
...................
* "
252
MARKOV CHAINS
or
(allxl
[CHAP. 20
+ a21x2+ ... + umlzm,u12x1+ a,,%, + ... + um2xm, . . . . almxl+ a 2 m +~ 2... + ammxm)
=
(x1, x 2 ,
. . . . xm)
or, equivalently, if there exists a non-zero solution to the homogeneous system
+ u21x2 + ... + arnlzm =
a12x1 + uX2x2+ ... + amzxm =
allxl
..................................
aln,sl
+ aXmz2+
*.*
+ ammxm=
+ a21x2 + ... +
+ ( u * 2 - l)s, + +
(all-l)xl
z1
z2
a1221
* * .
amlxm
= 0
am2xm
= 0
.............................................
almXl +
a 2 m ~ 2 + ... + (amm-l ) x n L =
x,
0
Now m homogeneous equations in ?n unknowns have a non-zero solution if and only if the
determinant of the matrix of coefficients is zero. But the matrix of coefficients of the above
homogeneous system of equations is the transpose of the matrix
all a12
A-I
10
* *.
...
=
.................
am1 am2
’ * ‘
amm
all-
* * *
...
1
a12
. * *
...
..........................
..........
0 0 *..
aml
am,
...
amm-l
Since the determinant of a matrix is equal to the determinant of its transpose, A has a non-zero
fixed point if and only if det(A - I ) = 0.
20.29. Prove: Every stochastic matrix P = ( p J has a non-zero fixed point,
Since the rows of the stochastic matrix P a r e probability vectors, the sum of its column
vectors is
Accordingly, the sum of t h e columns of the matrix P - Z is
Thus the columns of t h e matrix P - I a r e dependent (see Page 114). B u t the determinant of a
matrix is zero iff its columns a r e dependent; hence de t ( P - I ) = 0 and so, by the preceding
problem, P - Z has a non-zero fixed point.
20.30. Suppose m points on a circle are numbered respectively 1, 2, , . .,m in a counterclockwise direction. A particle performs a “random walk” on the circle; it moves one
step counterclockwise with probability p or one step clockwise with probability
g = 1 - p . Find the transition matrix of this Markov chain.
The state space is {1,2, . . . . m}. The diagram to the right below can be used to obtain the
transition matrix which appears to the left below.
1
1
m-1
2
3
4
...
rn-2
m-1
P
0
0
...
0
0
0
P
0
0
0
...
...
0
P
0
0
Q
...................................................
...
0
0
0
P
0
0
0
0
...
m
m-2
ni- 1
0
Q
CHAP. 201
MARKOV CHAINS
253
Supplementary Problems
PROBABILITY VECTORS AND STOCHASTIC MATRICES
20.31. Which vectors are probability vectors?
(i)
20.33.
(5,4, -d., 4)
(ii)
(4, 0, Q, 4.9)
(iii)
(A,
4, Q, 0, $1.
Which matrices are stochastic?
REGULAR STOCHASTIC MATRICES AND FIXED PROBABILITY VECTORS
20.34.
Find the unique fixed probability vector of each matrix:
20.35.
(i) Find the unique fixed probability vector t of
P =
0
(iii) What vector does
(ii) What matrix does Pn approach?
20.36.
Find the unique fixed probability vector t of each matrix:
20.37.
(i)
Find the unique fixed probability vector t of
P =
(ii) What matrix does Pn approach?
(iii) What vector does (t,
0, +, &)Pn approach?
:I.
(;:
1
0
(a,$, +)Pn approach?
1; ;,” ; ) *
\ 0 + 0 12
I
(iv) What vector does (+, O,O, &)Pn approach?
20.38.
(i)
Given that
(ii) Given that
t = ( I , O , &) is a fixed point of a stochastic matrix P , is P regular?
t = (t,f,f,&)
is a fixed point of a stochastic matrix P , is P regular?
20.39.
Which of the stochastic matrices are regular?
20.40.
Show that (cf
+ ce + de, uf + b f + ae, ad + bd + bc)
P
=
(
1-a-b
c
e
is a fixed point of the matrix
a
1-c-d
f
b
d
1-e-f
)
254
MARKOV CHAINS
[CHAP. 20
MARKOV CHAINS
20.41. A man’s smoking habits a r e as follows. If he smokes filter cigarettes one week, he switches to
non-filter cigarettes the next week with probability .2. On the other hand, the probability that
he smokes non-filter cigarettes two weeks in succession is .7. In the long run, how often does he
smoke filter cigarettes?
20.42.
A gambler’s luck follows a pattern. If he wins a game, the probability of winning the next game
i s .6. However, if he loses a game, the probability of losing the next game is .7. There is a n
even chance t h a t t h e gambler wins the first game.
(i)
What is the probability t h a t he wins the second game?
(ii) Wh at is the probability t h a t he wins the third game?
(iii) In the long run, how often will he win?
F o r a Markov chain, t h e transition matrix is P =
(ii)
p ( 0 ) = (&,$).
(iv) p i 2 ) ; (v) the vector p(O)Pn approaches;
1
20.43.
Find: (i) $:I;
(vi) the matrix Pn approaches.
20.44.
(ii) p i ; ) ; (iii)
For a Markov chain, t h e transition matrix is
~(2);
1
P =
and the initial probability disZ
tribution is p ( O ) = (T,z,O).
1 1
20.45.
with initial probability distribution
Find (i) p i t ) , (ii) p;:),
T
Z
(iii) P ( ~ ) , (iv) p y ) .
Each year a man trades his car f o r a new car. If he ha s a Buick, he trades it for a Plymouth.
If he has a Plymouth, he trades i t for a Ford. However, if he has a Ford, he is just a s likely to
trade i t f o r a new Ford as t o trade i t f o r a Buick or a Plymouth. In 1955 he bought his first car
which was a Ford.
(i)
Find the probability t h a t he ha s a
( d ) 1958 Ford.
(a) 1957 Ford,
( b ) 1957 Buick,
(c) 1958 Plymouth,
(ii) In the long run, how often will he have a Ford?
20.46.
There a r e 2 white marbles in u r n A and 4 red marbles in urn B. A t each step of the process a
marble is selected f r o m each urn, and the two marbles selected a r e interchanged. Let X , be the
number of red marbles in u r n A af t er n interchanges. (i) Find the transition matrix P. (ii) W hat
i s the probability t h a t there a r e 2 red marbles in urn A a fte r 3 steps? (iii) In the long run, what
i s the probability t h a t there a r e 2 red marbles in u r n A ?
20.47.
Solve the preceding problem in the case t h a t there a r e 3 white marbles in u r n A and 3 red marbles
in u r n B.
20.48.
A f a i r coin is tossed until 3 heads occur in a row. Let X, be the length of the sequence of heads
ending at the n t h trial. (See Example 8.4.) W h a t is the probability that there a r e a t least
8 tosses of the coin?
20.49.
A player has 3 dollars. A t each play of a game, he loses one
dollar with probability 3 b u t wins two dollars with probability
He stops playing if he h a s lost his 3 dollars or he ha s won
at least 3 dollars.
&.
(i)
Find the transition matrix of the Markov chain.
(ii) Wh at is the probability t h a t there a r e at least 4 plays to
the game?
20.50.
The diagram on t h e r i g h t shows f o ur compartments with doors
leading from one to another. A y o u s e in a n y compartment is
equally likely to pass through each of the doors of the compartment. Find the transition matrix of the Markov chain.
CHAP. 201
MARKOV CHAINS
255
20.51. Find the transition matrix corresponding to each transition diagram:
20.52. Draw a transition diagram for each transition matrix:
a1
a2
Answers to Supplementary Problems
20.31. Only (iii).
20.32. (i)
(3/13,0, 2/13,5/13,3/13)
(ii) (8/18,2/18,0, 1/18,3/18,0, 4/18)
(iii) (4/45,24/45,6/45,0, 3/45,8/45)
20.33. Only (ii) and (iv).
20.34. (i) (6/11,5/11), (ii) (10/19,9/19), (iii) (5/13,8/13), (iv)
20.35. (i) t = (4/13,8/13,1/13),
20.36. (i) t = (2/9,6/9,1/9),
(z,Q)
(iii) t = (4/13,8/13,1/13)
(ii) t = (5/15,6/15,4/15)
20.37. (i) t = (2/11,4/11,1/11,4/11), (iii) t ,
(iV) t
20.39. Only (iii).
20.41. 60% of the time.
20.42. (i) 9/20, (ii) 87/200, (iii) 3/7 of the time.
20.43. (i) 9/16, (ii) 3/8, (iii) (37/64,27/64j, (iv) 27/64, (v) (.6,.4),
20.44. (i) 3/8, (ii) l/2, (iii) (7/16,2/16,7/16), (iv) 7/16
20.45. (i) ( a ) 4/9,
( b ) 1/9, ( c ) 7/27, ( d ) 16/27. (ii) 50% of the time
0
1
0
(ii) 3/8
(iii) 215
(vi)
(::::)
256
MARKOV CHAINS
1
1 0
10
20.47.
(i) P =
o\
1)
(ii) 32/81
o a a +
\o
0 1
(iii) 9/20
o/
1 0 0 0 0 0 0
20.52.
(i)
~
0
o
%
0
o
~
o
0
~
3
t
a
0
0
o
(ii)
[CHAP. 20
Chapter 21
Inequalities
THE REAL LINE
The set of real numbers, denoted by R, plays a dominant role in mathematics. We
assume the reader is familiar with the geometric representation of R by means of the
points on a straight line. As shown below, a point, called the origin, is chosen t o represent 0 and another point, usually to the right of 0, to represent 1. Then there is a natural
way to pair off the points on the line and the real numbers, i.e. each point will represent
a unique real number and each real number will be represented by a unique point. For
this reason we refer to R as the real line and use the words point and number interchangeably.
--
I
-4
Il
-3
4
1,
-2
I
-1
I
I
I
I
I
0
1
2
3
4
The real line R
POSITIVE NUMBERS
Those numbers to the right of 0 on the real line R, i.e. on the same side as 1, are the
positive numbers; those numbers to the left of 0 are the negative numbers. The set of
positive numbers can be completely described by the following axioms:
[P,] If a E R,then exactly one of the following is true: a is positive; a = 0; -a is positive.
rPP] If a, b E R are positive, then their sum a + b and their product a - b are also positive.
It follows that a is positive if and only if -a is negative.
Example 1.1:
We show, using only [P,] and [PJ, that the real number 1 is positive.
By [PI], either 1 or -1 is positive. If -1 is positive then, by [P2], the product
(-l)(-1) = 1 is positive. But this contradicts [P,] which states t h a t 1 and -1
cannot both be positive. Hence the assumption that -1 is positive is false and
so 1 is positive.
Example 1.2:
The real number -2 is negative. For, by the preceding example, 1 is positive and
so, by [P2], the sum 1 1 = 2 is positive; hence -2 is not positive, i.e. -2 is
negative.
+
ORDER
An order relation in R is defined using the concept of positiveness.
1-1
The real number a is less than the real number b, written a
difference b - a is positive.
257
< b,
if the
258
[CHAP. 21
INEQUALITIES
The following notation is also used:
a > b , read a is greater than b,
u 6 b , read a is less than o r equal to b,
a 2 b, read a is greater than or equal to b,
means b < a
means a < b or a = b
means b 6 a
Geometrically speaking,
a < b means a is to the left of b on the real line R
a > b means a is to the right o f b on the real line R
< 5,
2
Example 2.2:
A real number 3: is positive iff x > 0, and z is negative iff z
x2 is always greater than 0: x2 > 0.
Example 2.3:
The notation 2 < x < 5 means 2
2 and 5 on the real line.
-6
f
-3,
4
We refer to the relations <, >,
6
4, 5
> -8
Example 2.1:
and
< 0.
For z
f
0,
< x and also z < 5; hence x will lie between
as inequalities in order to distinguish them
< and > as strict inequalities. We
now state basic properties about inequalities which shall be used throughout.
1
from the equality relation = . We also shall refer to
Theorem 21.1:
Let a, b and c be real numbers.
(i) The sense o f an inequality is not changed if the same real number
is added to both sides:
If a < b , then a + c < b + c .
If a 6 b, then u + c 6 b + c .
(ii) The sense of an inequality is not changed if both sides are multiplied
by the same positive real number:
If a < b and c > 0, then ac < be.
If a 4 b and c > 0, then ac 4 be.
(iii) The sense of an inequality is reversed if both sides are multiplied by
the same negative number:
If a < b and c < 0, then ac > be.
If a 6 b and c < 0. then ac be.
(FINITE) INTERVALS
Let a and b be real numbers such that a < b . Then the set of all real numbers x
satisfying
a < x < b is called the open interval from a to b
a 6 x 4 b is called the closed interval from a to b
a < x 6 b is called the open-closed interval from a to b
a L x < b is called the closed-open interval from a to b
The points a and b are called the endpoints of the interval. Observe that a closed interval
contains both its endpoints, an open interval contains neither endpoint, and a n open-closed
and a closed-open interval contains exactly one of its endpoints:
"
1
a
b
open interval: a
"
<x <b
c
-
a
b
closed interval:
(L
5
x 6b
CHAP. 211
INEQUALITIES
-
"
a
259
-
b
<x
-
-
U
b
d
<b
The open-closed and closed-open intervals are also referred to as being half-open (or:
half -closed).
open-closed interval: a
f
b
closed-open interval: a
6 2
INFINITE INTERVALS
Let a be any real number. Then the set of all real numbers x satisfying x
x > a or x A a is called an infinite i n t e m n l .
< a, x 6 a,
LINEAR INEQUALITIES I N ONE UNKNOWN
Every linear inequality in one unknown x can be reduced to the form
a x < b, a x ' b ,
an:> b or a x k b
If a Z 0 , then both sides of the inequality can be multiplied by l l a with the sense of the
inequality reversed if a is negative. Thus the inequality can be further reduced to the f o r m
x < c , x g c , n . > c or x z c
whose solution set is an infinite interval.
Example 3.1 :
+
Consider the inequality 5~ 7
posing), we obtain
5 2 - 22
Multiplying both sides by
x
Example 3.2:
5
1
+ 1.
-
Adding - 2 2 - 7 to both sides (or: trans-
7
or
3 x 5 -6
+ (or: dividing both sides by 3) we finally obtain
6
Consider the inequality 22
22
22
-
4
-2
+ 3 < 4 2 + 9.
42
< 9-3
:
-2
0
2
Transposing, we obtain
or
<
-22
6
-3
AIultiplying both sides of the inequality by
(or: dividing both sides by - 2 ) and
is negative, we finally obtain
reversing t h e inequality since
-+
-
x > -3
u
-3
0
3
r
ABSOLUTE VALUE
The absolute value of a real number x, written 1x1, is defined by
=
1x1
{
x if
-n:
X A O
if x < O
that is, if x is non-negative then
= x, and if x is negative then 1x1 = -x. Thus the
absolute value of every real number is non-negative: 1x1 2 0 f o r every x E R.
Geometrically speaking, the absolute value of x is the distance between the point x
on the real line and the origin, i.e. the point 0. Furthermore, the distance between any
two points a, b E R is la- b( = Ib - a ' .
,-fil
= fi
Example 4.1:
,-21 = 2, 171 = 7,
Example 4.2:
' 3 - 81 = 1-51 = 5 and I8 - 3'l - 151 = 5
1-7,
= 7,
260
IN EQ U A LITIES
[CHAP. 21
The statement 1x1 < 5 can be interpreted to mean t h a t the distance between x
and t h e origin is less than 5; hence z must lie between -5 and 5 on the real line.
I n other words,
Example 4.3:
Ix' < 5 and -5 < x
--
have identical meaning and, similarly,
1x1
f
5
and
- 5 5 2 6
-
-
<5
5
-5
n
-5
n
5
5
have identical meaning.
The graph of the function f(x) = 1x/, i.e. the absolute value function, lies entirely in
the upper half plane since f ( x )2 0 for every x E R:
-2
--4
t
2
4
t
Graph of f(z) = 1x1
The central facts about the absolute value function are the following:
Let a and b be any real numbers. Then:
(i) la1 0, and la1 = 0 iff a = 0
(ii) -la1
a la1
(iii) lab1 = /a1* lbl
Theorem 21.2:
(iv) la
(v) la
+ bl
+ bl
6
2
+
la1 lbl
la1 - Ibl
Solved Problems
INEQUALITIES
21.1. Write each statement in notational form:
(iv) a is greater than b.
(i) a is less than b.
(ii) a is not greater than b.
(v) a is not less than b.
(iii) a is less than or equal to b.
(vi) a is not greater than or equal t o b.
A vertical or slant line through a symbol designates the opposite meaning of the symbol,
(i) a
< b,
(ii) a
4 b,
(iii) a
5
b,
(iv) a
> b, (v) a 4
b,
(vi) a
b
21.2. Rewrite the following geometric relationships between the given real numbers using
the inequality notation:
(iii) x lies between -3 and 7.
(i) y lies to the right of 8.
(ii) x lies to the left of -3.
(iv) w lies between 5 and 1.
Recall t h a t a < b means a lies to the left of 8 on the real line:
(i) y
(ii) z
> 8 or
< -3.
8 < g.
< z and x < 7 or, more concisely,
1 < w < 5.
(iii) -3
(iv)
-3
< 2 < 7.
CHAP. 211
261
I NE Q U A LITIES
21.3. Describe and diagram each of the following intervals:
(i) 2 < x < 4, (ii) -1 6 x A 2, (iii) -3 < x 4 1, (iv) -4
(vi) x 2.
(i)
All numbers greater than 2 and less tha n 4:
6
x < -1,
4
b
r
l2
2
0
-1,
(ii) All numbers greater t h an o r equal to -1 and less tha n or equal to 2:
--
(iii) All numbers greater t h an -3 and less tha n or equal to 1:
21.4.
'
'
-
(vi) All numbers less t h an or equal to 2 :
4,
I
I
,
2
0
Solve and diagram the solution set of each inequality:
(i) 4x - 3 L 2x 3, (ii) 3x 2 < 6n: - 7.
+
(i) Transpose to obtain:
or:
+
4x - 2x
22
Divide by 2 :
3
6
4
4
+3
(ii) Transpose t o obtain:
or:
32 - 6 x
-3x
-
4,
x'3
<
<
-7
-
-3
0
-3
0
2
-9
x > 3
Divide by -3:
3
-
r
L
3
Note t h a t the inequality is reversed since -3 is negative.
21.5.
Solve each inequality and diagram its solution set:
(i) 3%-1
4n:+2, (ii) x - 3 > 1 + 3 x , (iii) 2 % - 3 4 5x-9.
(i)
Transpose to obtain:
3x
- 42
-x
or:
Divide by -1:
x
2
3
4
2
' -3
(ii) Transpose to obtain:
or:
Divide by - 2 :
x - 32 > 1
(iii) Transpose t o obtain:
22 - 5% 6 -9
-2x
2
or :
-32:
Divide by -3:
+1
>
<
f
4,
I
0
3
-2
0
2
-2
0
2
-3
+3
4
-2
4,
-
+3
--
-6
2 2 2
In each case the inequality was reversed since division w a s by a negative number.
21.6. Solve:
4% +
x-2
-< 2x-&
3
Multiply both sides by 12:
or:
Transpose:
or:
Divide by -14:
+4(~-2)<
+ 42 - 8 <
6% + 4x - 2 4 ~<
62
6%
2 4~1
2 4~ 1
8-1
-14~< 7
5
> -Q
*
,
,
2
0
*
-2
_
'
1
-2
2
0
'
'
0
-4
I
-2
,
-2
(iv) All numbers greater t h an o r equal to -4 and less tha n -1:
(v) All numbers greater t h an -1:
,
> -1,
(v) x
'
2
262
[CHAP. 21
INEQUALITIES
21.7. Solve each inequality, i.e. rewrite the inequality so that x is alone between the
inequality signs:
12, (iv) -6 A -2s 4 4.
(i) 3 L x - 4 5 8, (ii) -1 6 x + 3 6 2, (iii) -9 6 3 x
(i)
Add 4 to each side t o obtain 7
f
z
12.
6
(ii) Add -3 to each side t o obtain -4 5 x 4 -1.
to obtain -3
(iii) Divide each side by 3 (or: multiply by
5)
x
4.
(iv) Divide each side by -2 (or: multiply by -+) and reverse the inequalities to obtain -2
(i)
<
Add 5 to each side to obtain:
Divide each side by 2:
4 < 2 < 6
(ii) Add -3 t o each side to obtain: -10 f -2x 6 2.
Divide each side by -2 and reverse inequalities:
ABSOLUTE VALUE
21.9. Evaluate: (i) 13 - 51, (ii) 1-3
+5
21.10. Evaluate:
(i)
= 121 = 2
+
(iii) 1-3
-
6
x
5
5.
(iv) 13 - 71
51,
-
1-51.
= 1-61+l21
+ 1-1 - 51 - 1-81
13 - 61 - 1-2 + 41 - 1-2 - 31
(iv)
= 6+2 = 8
= 1-31 - 1-31 = 3 - 3 = 0
= 4 1-61 - 1-81 = 4 6
(iv) 13- 61 - 1-2+4/ - 1-2 - 31 = 3 - 2 - 5 = -4
(ij)
-1
(iii) 4
(i) 12 - 81 '3 - 11
(ii) 12 - 51 - 14 - 71
12-81+'3-11
- 2%
(iii) 1-3 - 51 = 1-81 = 8
(iv) 13 - 71 - 1-51 = 1-41 - 1-51 = 4 - 5 = -1
(i) 13 - 51 = '-21 = 2
(ii) 1-3
+ 51,
3.
< 12
8
22
x
+ 3 A 5.
< 2x - 5 < 7, (ii) -7
21.8. Solve each inequality: (i) 3
f
12-51 - 14-71
(iii) 4
+
+
+ 1-1 - 51 - 1-81
-
8 = 2
21.11. Rewrite without the absolute value sign:
(i) 1x1 6 3, (ii) (x-21 < 5 , (iii) (2x-31 5 7.
(i)
-3 5 x 5 3
or
(ii) - 5 < 2 - 2 < 5
(iii) -7
f
22 - 3
5
7
or
-3<x<7
-4
5
22
5
10
or
-2
21.12. Rewrite using a n absolute value sign: (i) -2
f
x
L
5
x
5
6 , (ii) 4
< x < 10.
F i r s t rewrite t h e inequality so t h a t a number and its negative appear at the ends of the
inequality.
(i)
Add -2 to each side to obtain -4
(ii) Add -7 to each side to obtain -3
6
x -2
44
which is equivalent to
IX
-21
6
< x - 7 < 3 which is equivalent to I$-- 71 <
4.
3.
PROOFS
21.13. Prove: If a < b and b < c, then a < c.
By definition, a < b and b < c means b - a and c - b a r e positive. By [P2],the sum
positive numbers ( b - a) + (c - b ) = c - a is positive. Thus, by definition, a < c.
of two
CHAP. 211
INEQUALITIES
< b, then a + c < b + c.
(ii) If a < b and c is positive, then ac < be.
(iii) If a < b and c is negative, then ac > be.
21.14. Prove Theorem 21.1:
(i)
By definition, a
263
If a
(i)
<b
means b - a is positive. B u t
( b + ~ ) - ( a + ~=) b - a -
Hence ( b
+ c) - (a + c )
<
is positive and so a + c
b
+ c.
(ii) By definition, a < b means b - a is positive. But c is also positive; hence by [P2]the product
c ( b - a ) = b c - ac is positive. Accordingly, a c < bc.
(iii) By definition, a < b means b - a is positive. By [P,], if c is negative then -c is positive;
hence by [P2] t h e product ( b - a)(-c) = a c - bc is also positive. Thus, by definition,
bc < ac or, equivalently, ac > bc.
+ bl
21.15. Prove Theorem 21.2(iv): la
Method 1.
Since la[ = ? a ,
--/a1 f a
5
L
la\; also,
-(l al
+
la+ bl
Therefore
la/
-161
bl)
6
+ JbJ.
f
'a:
f
lb[.
+b
6
+ Ibl I
=
a
f
b
Then, adding,
+ Ib/
la1 + 161
la1
Method 2.
Now ab
f
d
21.16. Prove:
implies 2ab
6
2 (a1lbl, and so
+ 2ab + b2
+ 2 la( lb( + b2 = la12 + 2 lb( + [bl2 = (la1 + lW
m = la + bl and so, by the square root of the above, 10 + bl 6 la1 4-lbl.
(a+ b)i =
But
lab1 = la1 Ibl
a?
5
(a- b /
la1
a2
+ Ibl.
+
Using the result of t h e preceding problem, we have la- bl = ( a (--b)\
21.17. Prove:
la - c J
6
]a- bJ
3 is a
la1
+ 1-41= la/ + \ b / .
+ ] b- el.
l~-~c/
21.18. Prove that
6
= !(a-b)
+ (b-C)[
f
la-bl
+ /b-c/
positive number.
By [P,], either
or 8 i s positive. Suppose -3 is positive and so, by [P2],the sum
= -1 i s also positive. But by Example 1.1, the number 1 is positive and not -1.
T h u s we have a contradiction, and so & is positive.
-4
(-4)+ (-4)
21.19. Prove that the product a . b of a positive number a and a negative number b is
negative.
If b is negative then, by [PI],-b is positive; hence by [P2]the product a (-b) is also positive.
B u t a (-b) = - ( a b). Thus -(a * b ) is positive and so, by [P,], a * b is negative.
-
21.20. Prove that the product a - b of negative numbers a and b is positive.
(-a)
If a and b a r e negative then, by [PI], --a and -b are positive. Hence by [P2],the product
B u t a . b = (-a) (-b), a nd so a *b is positive.
* (-b) is positive.
-
264
INEQUALITIES
21.21. Prove: If a and b are positive, then a
iff a2 < b2.
< bz; hence
<b
a and b a r e positive, a2 < ab and ab
< b. Since
Suppose a
[CHAP. 21
+
a2 < bz,
On the other hand, suppose a2 < b2. Then b 2 - a2 = ( b a ) ( b - a ) is positive. Since a and b
a r e positive, the sum b a is positive; hence b - a is positive or else the product ( b a ) ( b- a)
would be negative. Thus, by definition, a < b.
+
+
21.22. Prove that the sum of a positive number a and its reciprocal l l a is greater than or
equal t o 2: if a > 0, then a l l a 2 2.
+
If a = 1, then lla = 1 and so a + lla = 1 + l = 2.
a - 1 # 0 and so
(u-1)* > 0
or
a2- 2 a + 1
>
On the other hand, if
or
0
az+ 1
a # 1, then
> 2a
Since a is positive, we can divide both sides of the inequality by a to obtain a+ lla
> 2.
Supplementary Problems
INEQUALITIES
21.23.
Write each statement in notational form:
(i)
x is not greater t h an y.
(iii) m is not less tha n n.
(ii) Y is greater t h an or equal t o t.
21.24.
(iv) u is less tha n w.
Using inequality notation, rewrite each geometric relationship between the given real numbers:
(i)
x lies to the right of y.
(iii) x lies between 4 and - 6 .
(ii) r lies to the left of t .
21.25.
(iv) y lies between -2 and -5.
Describe and diagram t h e following intervals:
(i) -1
<x
6
(ii) 1
3,
+
f
x
6 4,
f
x < 0, (iv) x
f
1.
> 4 x + 5.
21.26.
Solve: (i) 6% 5
21.27.
Solve: (i) 5 - 2
21.28.
2-1
Solve: - 1 5
-1
3
2
2
'
21.29.
Solve, i.e. rewrite so t h a t x is alone between the inequality signs:
<
2x -6,
(ii) 2x - 3
(ii) 3%
+2
2
62+ 11.
+
(i) -2
21.30.
6
2x -7,
(iii) -2
Solve:
f
x -3
(i) 4
6
f
< x +2 <
4, (ii) -5
-2x
f
10, (ii) -1
1, (iii) -12
< 2 s - 3 < 5,
< 42 <
(iii) -3
ABSOLUTE VALUES
21.31.
Evaluate:
(i, 14 - 71, (ii) 17 - 41, (iii) 14 f 71, (iv) 1-4 - 71.
21.33.
Evaluate:
(i) I 1-31 - 1-91
I,
(ii) ! 12 - 61 - 11 - 91 I.
-8.
5 - 2x
6
7.
CHAP. 211
INEQUALITIES
265
21.34. Rewrite without the absolute value sign:
(i) 1x1
1%
8, (ii)
5
- 31
+
<
8, (iii) 12x 41
8, (iv) 12-2x1
<
8.
21.35. Rewrite using the absolute value sign:
<x<
(i) -3
8, (iii) -7
<
< b and c is positive, then
cla
x
9, (ii) 2
<
x
(iv) -1
-1,
f
2x+3
5.
f
THEOREMS ON INEQUALITIES
21.36.
Prove:
If
21.37.
Prove:
If a
a
<b
and c
< d,
then a
> c/b.
+ c < b +d.
6 ( a+ b ) / 2 .
21.38. Prove:
If a and b are positive, then
21.39. Prove:
If
21.40. Prove:
For any real numbers a, b and c, (i) 2ab
a<
b
where
d
Q,
f
6 , c and d are positive, then a + c
b S d
~
f
a2
+ b2
<
c*
d
and
(ii) ab
THEOREMS ON ABSOLUTE VALUES
21.41. Prove: (i) (--a1 = 1a1, (ii) u2 = Ia12, (iii) 'a]
= @, (iv) 1x1 < a iff -a
21.42.
+ ac + be
f
< x < a.
Prove Theorem 21,2(iii): lab/ = /a] lbl.
21.43. Prove Theorem 21.2(v): ]a1- /bl
21.44. Prove:
la'
-
lbl
5 la
+ bl.
ja- bl.
6
Answers to Supplementary Problems
21.23. (i) x
4 y,
21.24. (i) x
> y or
(ii) r
y
21.25. (i)
(i) x
f
21.27.
(i) x
> 4,
-6
< x < 4, (iv) -5 < y < -2
(iii)
(ii) x
4
2
0
(ii) .T
f
x
6
c
I
-4
6
-3
f
21.31. (i) 3,
(ii) 3,
(i) -2,
21.33. (i) 6,
(i) -8
x
f
-2,
< x < -1,
(ii) 1 < x
< -2
(5)-3 < x
< 4, (iii) -1
f
x
f
4
(iii) 11, (iv) 11
(ii) 7,
(iii) 6,
(iv) 3
(ii) 4
f
21.35. (i) 1%-31
x
f
<
8,
(ii) -5
< x < 11, (iii) -6
6, (ii) 'x-51
f
3,
L
x
6
(iii) ' ~ $ 4 1< 3,
2,
(iv) -3
<x <5
(iv) 12%+1l
f
0
-2
< -4
(ii) -7
7,
21.30. (i) -5
21.34.
< w.
(iv) u
7
21.29. (i) 1
21.32.
4: n,
< x, (ii) T < t, (iii)
-- ,
-2
-3,
21.26.
2
t , (iii) m
r
-4
21.28. x
2
3
1
2
I
4
+ + c2.
u2
b2
Chapter 22
Points, lines and Hyperplanes
We assume the reader is familiar with the
Cartesian plane, denoted by R2, as shown on
the right. Here each point P represents an
ordered pair of real numbers (a,b ) and vice
versa; the vertical line through the point P
meets the horizontal axis (x axis) a t a, and the
horizontal line through P meets the vertical
axis (y axis) a t b. Thus in discussing the Cartesian plane R2 we use the words point and
b
*P
x axis
a
DISTANCE BETWEEN POINTS
The distance d between two points 11 = (x,,y,) and q = (x2,y2) is given by
d =
d(.,
- x,)2
+ (y, -
y,)2
For example, the distance between p = (3, -1) and q = (7,2) is
d =
d(7-3)'
3. (2 -(-l))' = d16
+9
= 5
We shall denote the distance between p and q by d ( p , q ) .
INCLINATION AND SLOPE OF A LINE
The inclination B of a line I is the angle that I makes with the horizontal; that is, 6 = 0"
if 1 is horizontal or B is the smallest positive angle measured counterclockwise from the
positive end of the x axis to the line 1. The range of 0 is given by 0" 4 8 < 180".
The slope 172 of a line is defined as the tangent of its angle t? of inclination: 7n = tan8.
In particular, the slope of the line passing
through two points p = (x,,y,) and q = (x2,y,)
is given by
!I*
- 1-11
tv = t a n 0 =
~
x, - x,
The slope of a vertical line, i.e. when x2=x1,
is not defined.
Two distinct lines I, and I , are parallel if and only if their slopes, say nz, and m2
respectively, are equal: m, = nz,. On the other hand, the lines are perpendicular if and
only if the slope of one is the negative reciprocal of the other: m, = -Urn2 or 7n,?n, = -1.
266
CHAP. 221
267
POINTS, LINES A N D HYPERPLANES
LINES AND LIKEAR EQUATIONS
Every line 1 in the Cartesian plane R' can be represented by a linear equation of the
form
+ by = c , where u and b a r e not both zero
(1)
i.e. each point on 1 is a solution to ( 1 ) and each solution to ( 1 ) is a point on 1. Conversely,
every equation of the form ( 1 ) represents a line in the plane. Thus, in discussing the
Cartesian plane, we sometimes use the terms line and linear equation synonymously.
(See Chapter 11.)
a.15
HORIZONTAL AND VERTICAL LINES
The equation of a horizontal line. i.e. a line parallel to the n: axis, is of the form
y = 1;
where k is the point a t which the line intersects the
y axis. In particular, the equation of the :I' axis is
y = 0.
x=a
The equation of a certical line, i.e. a line parallel
to the y axis, is of the form
y=b
r = k
a
where lc is the point a t which the line intersects the
3' axis.
In particular, the equation of the y axis is
z'= 0.
-
POINT-SLOPE FORRl
A line is completely determined once its direction and a point on the line are known.
The equation of the line having slope 02 and passing through the point (xl,yl) is
y - y1 = m(.u - X I )
Example 1.1:
Example 1.2:
F i n d th e equation of the line h av i n g slope 3 an d passing t h r o u g h t h e point ( 5 , -21.
Substitute in t h e above f o r m u l a to obtain y 2 = 31, - 5) or y = 3% - 1 7 .
+
F in d t h e equation of t h e line passing t h r o u g h the two points (3,2) a n d 15,-6).
Yz-YI
-G-2
112 = ____ - -4.
T h u s t h e equation
xq-xl
5-3
of the line is y - 2 = -4( 7. - 3 ) or 4 2 y = 1-1.
T h e slope m of t h e line is
+
SLOPE-INTERCEPT FORM
The equation of the line having slope vz and intersecting the y axis at k , i.e. having
?/-intercept k , is
y = 023' + k
Thus if the equation of a line 1 is a?
- by
2 / =
= c, where b f 0, then
a
--.r
11
and so m = - a / b
Example 2.1:
Example 2.2:
+
C
-
13
is the slope of 1.
The slope of t h e line I! = -2.).
F i n d th e slope 7n of t h e line
= 2,~X - - 33. , hence ) ) I = 3.
1/
+8
is -2.
2.x - 3y = 5.
Rewrite t h e equation in t h e f o r m
268
POINTS, L I N E S AND H Y P E R P L A N E S
[CHAP. 22
PARALLEL LINES. DISTANCE BETWEEN A POINT AND A LINE
Let I be the line ax by = c. The set of all
lines parallel to I is represented by the equation
+
ax
+ by
= h.
(2)
i.e. every value for k determines a line parallel
to 1 and every line parallel to l is of the above
form f o r some k . We call k a p a m m e t e r , and
the set (2) a one parameter family of lines.
The distance d between a point 11 = ( x ~ , y ~ )
and the line 1 is given by
d =
lax,
+ byl
c1
VGTi7
I n particular, the distance
Example 3.1:
-
between the origin (0,O) and the line Z is
Find the line I parallel to t h e line 2r - 51/ = 7 and passing through the point
( 3 , 4 ) . The line 1 belongs to the family
2x
Iz
5y =
-
Since the point 1 3 , l ) belongs t o 1 it must satisfy the equation of I:
2.3
-
5.4
=
k
k = -14
or
Thus t h e equation of I is 2%-55y = -14.
Example 3.2:
Let 1 be the line 3.c
d
-
4y = 2. The distance d between the point (2, - G ) and 1 is
=
13.2+4*(--6)-21
The distance p between t h e origin and 1 is
EUCLIDEAN m-SPACE
Let ic and v be vectors with
112
-
JIm
p
1-201
5
=
121
- 2
= ___ - 5'
l/m
components of real numbers:
u = (ul,u,, . . . ,uIn) and v = (vl,v2, . . ., v,)
Recall (see Chapter 9) the following operations of vector addition, scalar multiplication, and
dot product :
u + v = ( 2 1 , 4G,, 26, + v*, . . ., un,
+ v,J
ku
U'
v
= (k'u,, ku2, . . . , ku,),
=
21,2',
+
1L2V2
+ .- +
*
f o r any scalar k
Ui,,7jli,
The set of all such vectors with the above operations is called Euclidean m-space o r
m dimensional Euclidean space o r , simply, m-space, and will be denoted by R".
The distance between the above vectors u and v, written d(u,v), is defined by
CHAP. 221
POINTS, LINES AND HYPERPLANES
The norm (or: l e n g t h of the vector
ilull =
269
written ]lull, is defined by
ZL,
-\/... = dzq + u;+ . + Uf,
* *
Observe that d(u,v) = Iju - ~111.
The vectors u and 1: are said to be orthogonal (or: perpendicular) if their dot product
is zero: u-v= 0.
The set of vectors of the form 10, . . . , 0, Z L ~ ,0, . . ., 0) is called the xi axis. Note that
the zero vector, also called the origin, belongs to every axis.
Example 4.1 :
Consider the following vectors in R4: u = (1, - 2 , 4 , 1 ) and
ZL
+v
+
= (1 3 , -2 - 1, -1 - 5 , 1+ 0 )
7u =
U.W
= (3,1, -5,O).
Then:
(4, -1, -1, 1)
(7, -14, 28, 7 )
= 1.3
+
+ 4.(-5) + 1 . 0 = 3
+ (1+ 2 ) 2 + (-5 4)2 + (0 - 1 ) 2
(-2)*1
diu,w) = d ( 3- 1 ) 2
Example 4.2:
=
2)
-
We frequently identify a point in,
say, R2 with t h e arrow from the
origin to the given point. Then
the arrow corresponding to the
sum p q is t h e diagonal of the
parallelogram determined by the
arrows corresponding to the points
p and q as shown on the right.
Furthermore, llpll, the norm of p,
is precisely the length of the a r row to p .
-
2 - 20
=
+0
=
-19
a
+
The sum p
+$+
BOUNDED SETS
A subset A of Rm is bounded iff there exists a
real number m > 0 such that llull < nz for every
point u E A . In the case of the plane R2, this says
that the points of A lie inside of the circle of
radius rn centered a t the origin, as indicated on
the right.
-
+ q.
-m
A is bounded.
HYPERPLANES
Consider a linear equation in the m unknowns x,, . . ., xm:
CIXl
+
C2X,
+
* * *
+ CmXm
= b
(3)
Using the notation of the dot product of RI'~,the above equation can be written in the f o r m
C'X
= b
where c = (c1,c2, . . .,cJ and x = (x1,z2,. . .,x,).
that c # 0, i.e. that the ci in ( 3 ) are not all zero.
Definition:
Unless otherwise stated, we assume
The set of all points in Rm which satisfy the equation c x = b, i.e. the set of
all solutions of the linear equation (3), is called a hyperplane.
270
POINTS, L I N E S AND H Y P E R P L A K E S
Example 5.1:
The hyperplanes i n the Cartesian plane R’ a r e the s t r a i g h t
lines a s discussed a t the beginning of this chapter.
Example 5.2:
The hyperplanes in 3-space RJ,
i.e. in solid geometry, correspond t o the planes a s illustrated on the right. I n
particular, the equation of t h e
x y p l a n e is z = 0, the ?z-plane
is y=O, and the ?/:-plane is
x = 0.
[CHAP. 22
Y
Hyperplane in
R3.
Observe that the “dimension” of a hyperplane in R’ is one, and in R3 is two; that is,
we can identify the points of a hyperplane in R2 with the points on the real line R and
we can identify the points on a hyperplane of R3 with the points in R2 with the further
property that distances between corresponding points are equal. In each case the dimension
of the hyperplane is one less than the dimension of the Ivhole space. In fact, this property
completely characterizes hyperplanes in general; namely,
Theorem 22.1:
A subset A of R”‘ is a hyperplane if and only if there exists a one-to-one
correspondence between A and Ril’-i such that distances between corresponding points are equal.
The proof of this theorem is beyond the scope of this text.
PARALLEL HYPERPLANES. DISTANCE BETWEEN A POINT
AND A HYPERPLANE
Let N be the hyperplane c , ~ , C J , t . - . cliplII = b . Then
+
+
el,?-, + C _J , _
+ . . + ClnXnl
*
= k
is the equation of the family of all hyperplanes parallel to a, i.e. which do not have a
point in common with a. The distance d between the point 71. = (ul,u2,. . ., un,) and N is
given by
1 c , l ~ lt c,u,
. . . e,,LzL;,,- 2, 1
+
+
a =
dcf+ c; +
* * *
+ c;l
I n other words, the distance between u and the hyperplane c * R: = b is
d
=
1c-u - bl
llcll
I n particular, the distance
between the origin, i.e. the zero vector, and
p
P
Example 6.1:
=
Ibl
dcf+ c; + - - - + C t
-
~r
is
-P I
ICIl
I n solid geometry, i.e. R3, t h e family of hyperplanes parallel t o the r p p l a n e , i.e.
t o z = O , is represented by x = k where k is the point at which the hyperplane
intersects the i axis. ( S e e d i a g r a m be1ow.r
Y
Two members of the family z = k.
271
POINTS, LINES A N D HYPERPLANES
CHAP. 221
Example 6.2:
Let iy be the hyperplane 3 2
point (1,-4,-2,3) and a is
d
-
4y
+ x - 2w
1
7 in R4. The distance d between the
+ 1*(-2) - 2 ' 3
+ (-4)2 + 1 2 + (-2)Z
i 3 . 1 - 4.(-4)
=
d3'
The distance p between
and the origin is
a
p
-7
-
4
~
fi
171
=
d32
+ (-4)' + 1 2 + (-2)2
7
- -
fi'
Solved Problems
DISTANCE I N THE PLANE
22.1. Find the distance d between
(i) ( - 3 , 5 ) and (6,4), (ii) ( 5 , -2) and (-1, -4).
d =
In each case use the formula
(i)
+ (4-5)2 =
d ( - l - 5 ) * + (-4+2)2
d = d ( 6 + 3)?
(ii) d =
f
+
~ ( X ' - X ~ ) ~ (yz-yJ'.
?
=
=
d
m
fi
=
fi
22.2. Show that the points p = (--1,4), q = ( 2 , l ) and r = (3,5) are the vertices of an
isosceles triangle.
Compute the distance between each pair of points:
d(p,q)
=
d(2+ 1)' + (1-4)2
d(q,r)
=
d ( 3 - 2)'
+ ( 5 - 1)2
=
&m
=
fi
=
d r n
=
m
Since d ( p , T ) = d ( q , T ) , t h e triangle is isosceles.
22.3. Show that the points p = (4,-2), q = (-1,3) and r = (1,4) are the vertices of a
right triangle, and find its area.
Compute the distance between each pair of points:
Since d ( q , T ) Z
+ d(p,
+ (3+2)2
d(p,q)
= l/-l- 4)'
d(p,r)
=
d(q,r)
= d(1+1)"(4-3)2
~ ) 2=
d(1- 4 ) 2 + (4 + 2)2
d ( p , q)2,
=
d r n
dw-5
=
m
=
rn
=
=
6
=
the triangle is a right triangle.
The area of the right triangle is A =
+(a)(fi)
= 2.
22.4. Find the point equidistant from the points a = (-4,3), b = ( 5 , 6 ) and c = (4,-1).
Since d ( p , a) = d ( p , b ) ,
Let p = (x,y) be the required point. Then d ( p , a ) = d ( p , 6 ) = d ( p , c).
d(z+ 4)2 + (y-3)Z
=
d ( z - 5 ) 2 + (y-
6)2
or
3z
+y
= 6
Since d ( p , a ) = d ( p , c ) ,
d ( z +4)2
+ (y-3)2
=
d ( x - 4)2 + (y + 1 ) 2
or
2%- y = -1
Solve the two linear equations simultaneously to obtain x = 1 and y = 3.
required point.
Thus p = (1,3) is the
2'72
POINTS, LIN ES AND H Y P E R P L A N E S
Show that the points p = (1,-2), q = (3,2) and
the same straight line.
1'
[CHAP. 22
= (6,8) are collinear, i.e. lie on
Method 1.
Compute the distance between each p'air of points:
d ( p , q ) = d ( 3 - 1)Z
+ (2+2)2
= lLi-TT6
= d(6-1)2
+ (8+2)2
=
d(p,r)
Since d ( p , T ) = d ( p , q )
+ d(y,
T),
d
m
= &G
= 2 6
=
= 5
6
the points lie on the same straight line.
Method 2.
Compute the slope m, of the line passing through p and y, and the slope m2 of the line passing
through p and r:
m2 = 8 - (-2) - lo = 2
ml = 2 - (-2) = 4 = 2,
- 5
2
6 - 1
3-1
~
~
Since m, = m 2 , the three points lie on the same line.
LINES
22.6. Plot the following lines:
(i) x = -4, x = -1, x = 3 and x = 5;
(i)
(ii) y = 4, y = 1, y = -2, and y = -3.
Each line is vertical and meets the x axis at its respective value of x. See diagram below.
(ii) Each line is horizontal and meets the
~jaxis
at its respective value of y. See diagram below.
y=4
1 5
y=l
,
I
I
-5
t
*
5
y=-2
y=-3
r
t
-5
22.7. Find the equation of the line (i) parallel to the x axis and passing through the point
(-2, - 5 ) , (ii) parallel to the y axis and passing through the point (3,4).
(i)
If the line is parallel t o the z axis, then i t is horizontal and its equation is of the form y = k.
Since i t passes through the point (-2,-5), k must equal the y-component of the point. Thus
the required equation is y = - 5 .
fii)
If the line is parallel to the u axis, then i t is vertical and its equation is of the form 2 = k.
Since it passes through the point (3,4), k must equal the 2-component of the point. Thus
the required equation is r = 3 .
22.8. Plot the following lines: (ij 2%- 3y = 6, (ii) 4x t 3y = 6, (iii) 3x
+ 2y = 0.
In each case find at least two points on the line, preferably the intercepts, and as a check
a third point. The line drawn through these points is the required line.
273
POINTS, LINES A N D HYPERPLANES
CHAP. 221
-- 5
22.9. Find the equation of the line (i) passing through (3,2) with slope -1, (ii) passing
through (-4,1) with slope 4, (iii) passing through (-2, -5) with slope 2.
In each case use the formula y
point.
(i)
y - 2 = -l(.z-3)
1= i(xt4)
(ii) y
-
(iii) y
+5
= 2(2
+ 2)
-
y1 =
or
x
+y
or
x
-
or
22 - y = 1
m(.r - xl) where m is the slope and (xl, yl) the given
= 5
2y = -6
22.10. Find the equation of the line passing through each pair of points:
(i) (1,3)and (4, -5); (ii) (-2, -6) and ( 3 , l ) .
Y2 - Y1
I n each case, first find the slope nt of the line using the formula m = ___
and then
2 2 - $1
find the equation of the line using the point-slope formula.
8
(i) m = - --5-- -3
The equation of the line is Y - 3 = --(X
- 1) or 8~ 376 = 17.
4-1
3'
3
11-6-7
7
- (ii) m = 3 f 2
5 ' The equation of the line is y - 1 = -5( x - 3 ) o r 72 - 5 y = 16.
+
22.11. Find the equation of the line passing through each point and parallel to the given line:
(i) (3, -2) and 5% 4y = -5; (ii) (1,6) and 3x - 7y = 4.
+
(i)
+
The line belongs t o the family 6n: 4y = k. Substitute (3,-2) into the equation to obtain
5 . 3 i 4 * (-2) = k or k = 7. Thus the required line is 52 41/ = 7.
+
(ii) The line belongs t o the family 32 - 71/ = k . Substitute (1,6) into the equation to obtain
Thus the required line is 32 - 7y = -39.
3 * 1 - 7 * 6 = k or k = -39.
22.12. Let ax + by = c be the equation of a line 1. Show that
bx
-
ay = k
(4
is the equation of the family of lines perpendicular t o 1.
If b = 0, then 1 is vertical and ( 1 ) is the equation of the family of horizontal lines; hence ( 1 ) is
the family of lines perpendicular to I .
274
[CHAP. 22
POINTS, LINES AND H Y P E R P L A N E S
On the other hand, if a = 0 then 1 is horizontal and ( 1 ) is the equation of the family of vertical
lines and so is the family of lines perpendicular to 1.
Lastly, suppose b # 0 and a # 0. Then the slope of 2 is - a / b and ( 1 ) is the equation of the
family of lines with slope b l a . But ( - a l b ) ( b / a ) = -1; hence, again, ( 1 ) is the family of lines
perpendicular to 1.
22.13. Find the equation of the line passing through the given point and perpendicular to
the given line: (i) (2,5) and 32 4y = 9; (5) (3, -1) and 2x - 49 = 7.
+
The line belongs to the family 4r -3y = k. Substitute ( 2 , 5 ) into the equation t o obtain
8 - 15 = k or k = -7.
Thus the required line is 45.- 3 y = -7.
ii)
(ii) The line belongs to the family 4x - 2y = k. Substitute (3, -1) into the equation t o obtain
12 - 2 = k or k = 10. Thus the required line is 4r 2 y = 10 o r 2 2 + y = 5 .
+
22.14. Determine the distance d of each line from the origin:
(i) 3%- 4 =
~ 8, (ii) 5 x + 12y = -4.
Use the formula d =
lei
(i) d =
d r n '
_-
181
(rn
(ii) d =
5'
4
- 13'
1-41
@Ti5
22.15. Determine the distance d between each point and line:
(i) (5,3) and 5x-12y = 6; (ii) (-3,-2) and 4 x - 5y = -8.
lax1
Use the formula d =
+ by1 - C I
&Fiz
.
EUCLIDEAN m-SPACE
22.16. Find the distance d(u, v) and the dot product u * v for each pair of vectors:
(i) u = (3,-2,0,5) and 2) = (6,2,-3,-4);
(ii) u = ( 2 , 3 , - 5 , - 7 , l ) and v = (5,3,-2,-4,-1);
(iii) ZL = (4,-1,6,5) and v = (1,-4,2,0,7).
Use the formulas
d ( u , v) =
d ( v l - u1)2+
(iii) The distance d ( u , w ) and the dot product
different spaces: u E R4 and v E R5.
22.17. Find the norm of each vector:
In each case use t h e formula
ZL
*
+ (wnL-
U"U
+
and
U*
w =
UlW,
+
* *
. + umvm.
a re not defined since the vectors belong to
= (2,-1,3,5);
v = (3,-2,-1).
dut + ui + . . . + u$
~ l u
= ~
22.18. Let u = (1,-3, 2, 4) and v = (3, 5, -1, -2).
(i) u
u,,,)2
Find:
v ; (ii) 5u; (iii) 2u - 371; (iv) 'urn v ; (v) ~ l u land
~ 11v11.
POINTS, LINES AND H Y P E R P L A K E S
CHAP. 221
273
(i) u + v = ( l + 3 , - 3 + 5 , 2 - 1 , 4 - 2 ) = ( 4 , 2 , 1 , 2 )
(ii) 5u = (5 1, 5 * (-3), 5 2, 5 . 4 ) = (5, -15, 10, 20)
(iii) 2 u - 3v = (2, -6, 4, 8) (-9, -15, 3 , 6 ) = (-7, -21, 7 , 14)
(iv) U . L ' = 3 - 15 - 2 - 8 = -22
-
(v)
1 1 ~ 1=
-
+
=
d1+9+4+16
m.
,Jvll = d 9 + 2 5 + 1 + 4
=
a.
22.19. For each pair of vectors, determine li so that the vectors are orthogonal:
(i) (3, -2, k , 5 ) and (2k, 5, -3, -1); (ii) ( 2 , 3 k , -4,1,5) and (6, -1,3,7,2k).
In each case, set the product of the vectors equal to zero and solve f o r k.
+ (-2)
(i) 3 2k
(ii) 2 . 6
A
*
5
3k*(-1)
--
+5
k (-3)
(-1) = 0
+ 7 + 5*2k = 0
- (-4).3
or
6k
-
10 - 3k
or
12
-
3k
-
12
-
5 = 0
k =5
or
+ 7 + 10k = 0
or
k = -1
22.20. Determine which of the following subsets of R" are bounded: (i) A consists of all
probability vectors; (ii) B consists of the points on the x, axis; (iii) C is finite;
(iv) D consists of all points with positive components.
Recall t h a t X C R ~i s bounded if there exists a real number T ~ Z> 0 such t h a t IIu(, m f o r
f
every u E X .
(i) A is bounded since IIuI/ 1 f o r every probability vector u .
(ii) B is not bounded; f o r if rn is a n y positive real number, then ( m
and has norm greater t h a n rn.
f
+ 1, 0 , 0 , . . ., 0)
belongs t o B
(iii) C is bounded. Choose rn to be the maximum norm of the vectors in C; then rn satisfies the
required property.
(iv) D is not bounded.
HYPERPLANES
22.21. Find the distance d of each hyperplane from the origin:
(i) 2 ~ - 3 y + 5 x = 8; (ii) 2 ~ + 7 y - z + f z u = -4.
bi where the hyperplane is c x = b .
In each case use the formula d = -
I
(i) d =
181
d22
-
8
+ (-3)Z + 52 6 .
~
ell
iii) d =
1-41
d4+49+1+4
-~4
-
6'
22.22. Determine the distance d between each point and hyperplane: (i) (4,0,-2, -3) and
x - 6.y 5 x 22.9 = 3; (ii) ( 2 , - 1 , 3 , 7 , -2, -4) and 3x, - 5 x , - 2x, 4x, X, - 3x, = -9.
+
+ +
In each case use the formula d =
(i) d =
'4
+ 0 - 10 - 6 - 31
dl+36+25+4
- 15
-~
m'
~ * lt
1~
i
+
b ' where u i s the point and c * x = b the hyperplane.
iii) cl = / 6 + 5 - 6 + 2 8 - 2 + 1 2 + 9 1
~/9+25+4+16+1+9
-
52
a
-
13
2 '
22.23. Find the equation of the hyperplane passing through the given point and parallel to
the given hyperplane:
(i) (3,-2,1, -4) and 2 x + 5 y - 6 2 - 2 ~ : = 8;
(ii) ( 1 , - 2 , 3 , 5 ) and 4 ~ - 5 y + 2 x + w = 11.
The hyperplane belongs to the family 2 x f 5 y - 6 2 - 2 ~ = k . Substitute (3,-2,1,-4) into
the equation to obtain 6 - 1 0 - 6 - 8 = k or k = -2.
The required hyperplane is
2~ 5~ - 62 - 2 %=~ -2.
(ii) The hyperplane belongs to the family 4 x - 5 y + 2 x + w
= k . Substitute ( 1 , - 2 , 3 , 5 ) into
the equation to obtain 4 + 1 0 + 6 + 5 = k o r k = 25. The required hyperplane is
4 ~ - 5 5 ~ 1 2 z + u=
: 25.
(i)
+
276
POINTS, LINES AND H Y P E R P L A K E S
[CHAP. 22
22.24. Show that the equation of the hyperplane in Rm which intersects the xi axis at
aj f 0 is
2,
2,
-+-+
. . . $X 2 1 1
a1
a2
am
(Y
+
Let clxl i
czxz+
c,x,
= b be the equation of a. Substituting each x,-intercept
(0,. . ., 0 , a,, 0 , . . ., 0) into the equation, we obtain the homogeneous system
ale1
a2cz
-b
= 0
-b
= 0
-b = 0
a,c,
where the ci and b a r e the unknowns. The system is in echelon form and we can arbitrarily assign
a value to b . Set b = 1; then c i = l / n i . Thus the equation of a is as claimed.
22.25. Find the equation of the hyperplane in R4 which intersects the axis at -2, 4,3 and -6,
respectively.
By the preceding problem, the equation is
x1
-2
xq
+3
+=
+4
3
-6
22
5
1
6x1
or
-
35, - 4~~
1
2x4 = -12
22.26. Find the equation of the hyperplane 01 in R4 which passes through the points
(1,3,-4, a), (@,I,6, -3), (O,O, 2 , 5 ) and (O,O,3,7).
Let a x + b y + c z + d w = e be the equation of 01. Substituting each point into the equation,
we obtain the following homogeneous system of linear equation where a , b , c , d and e a r e the
unknowns:
a+3b-4c+2d-e
= 0
a+3b-4c+2d-e
= 0
I
b+6c-3d-e
2c+5d-e
3c+7d-e
= 0
= 0
= 0
Or
I
b+Gc-3d-e
= 0
2c+5d-e
= 0
d - e
= 0
The system has been reduced to echelon form on the right. There is only one free variable, i.e. we
can arbitrarily assign a value to e . Set e = 1. Then d = 1 by equation four, c = -2 by equation
three, b = 1 6 by equation two, and a = - 4 1 by equation one. Thus the required equation is
- 4 l ~ + l G y - 2 ~ +=~ 1.
Remark: The hyperplane a is uniquely determined since there is only one free variable in the
homogeneous system. I n other words, every value of e determines a different equation but all
these equations a r e multiples of each other and so they determine the same hyperplane a.
22.27. Show that there is more than one hyperplane in R4 which passes through the points
(1,6, -1,4), (@,I,2,2), (0,1,1,6) and (0,1,3,-2)-
+
Let ax b y + c z + d w = e be the general equation of the hyperplanes passing through the
given points. Substitute each point into the general equation and then reduce the system of
equations to echelon form as follows:
~ + 6 b - c+4d-e
b+2c+2d-e
10
= 0
bS c + 6 d - e = 0
b+3c-2d-e
= 0
a+Gb-
or
cf4d-e
b+2c+2d-e
= 0
= 0
C-4d
1 0
C-4d
= 0
or
a+6b- c+4d-e
b+2c+2d-e
C-4d
= 0
= 0
= O
The system has two free variables, e and d ; t h a t is, we can arbitrarily assign values t o both e and d
and obtain a n equation of a hyperplane passing through the given points. Accordingly, all the
equations will not be multiples of each other and so there is more than one hyperplane passing
through the given points.
CHAP. 221
POINTS, L I N E S A N D H Y P E R P L A N E S
277
22.28. Show that there is a t least one hyperplane in Rm passing through any m arbitrary
points.
+
+
+
Substituting each of the points into the equation clxl c2x2 * * *
z,c
= b , we obtain
a system of n z homogeneous linear equations in the m 1 unknowns, el, . . . , c, and b . Thus the
system has a t least one free variable and so has a non-zero solution. The non-zero solution
determines a hyperplane passing through the given points.
+
22.29. Show that any r < m arbitrary points in Rttzdo not uniquely determine a hyperplane,
i.e. that there is more than one hyperplane passing through the r points.
+
+
+
Substituting each of the r points into the equation clxl c2x2
c,x,
= b, we obtain
a system of r homogeneous equations in the m 1 unknowns, el, . . ., c, and b. Since r < m, the
system has a t least two f r ee variables. Accordingly, all the solutions of the system will not be
multiples of each other and so they determine more tha n one hyperplane.
+
1
D
2
2
-1
3
2
*
a
*
3
-1
=
1(-2+2)
- 2(4+3)
+ 3(4+3)
= 0 - 14
+ 21
= 7
2
Substitute in the second equation to obtain z = 2, and then substitute in the first equation t o find
x = 1. Thus the hyperplanes intersect in the unique point (1, -3,2).
22.31. Determine whether the given hyperplanes in R3 intersect in a unique point:
(i) x + 3 ~ - 2 x = 1, 2 x - y + x
(ii) x + y + x = 1, 2 x + y - x
= 2 and 4 x + 5 y - 3 x
= 4;
= 6 and 3 x - 3 y + x = -5.
In each case, compute t h e determinant D of the matrix of coefficients:
(i)
D
=
1
3
2
-1
4
5
-2
1
= l ( 3 - 5 ) - 3(-6-4)
- 2(10+4)
=
-3
Since D = @ , the hyperplanes do not intersect in a unique point.
1 3 -3
11
Since D # @ ,the hyperplanes intersect in a unique point.
-2
+ 30 - 28
= 0
278
POINTS, LIN ES AND HYPERPLANES
[CHAP. 22
Show that iii arbitrary hyperplanes u1 x = b,, ti, * x = b,, . . ., urn*
x = bm in Rrn
intersect i ; ~a unique point if and only if the ??z rectors P L ~ , U.~. ,.,urnare (linearly)
independent.
The system of
linear equations in m unknowns has a unique solution if and only if the
detemiinarit of the matrix A of coefficients is not zero. However, the determinant of A is not zero
if and only if the rows of A ar e independent vectors. The result now follows from the fa c t that
the rows of A a r e the vectors z i l .
MISCELLANEOUS PROBLEMS
22.33. Prove the Cauchy-Schwarz Inequality: For any pair of vectors u = (ul, . . .,urn)
and v =
. . .,vJ in Rm,
(til,
c Iulvll
m
11/
III
*
6
6
1 ull
1
110
2=1
If 7~ = 0 o r w = 0, then the inequality reduces to 0 6 0 0 and is therefore true. Thus we need
only consider the case in which z( # 0 and w # 0, i.e. in which I J Z L I I # 0 and IIwIJ # 0. Furthermore,
f
IU.
+ ...
= IuIz.I
w1
1ZtlnlJnll
Iulwl(
6
+
.'.
+
m
~z~l,lwm/ =
(u,7J,l
a=1
Thus we need only prove the second inequality.
Now for any real numbers x , y E R. 0
f
2xy
Set
Y
=
?(,I/
z ~ l l and
y =
1~ /
t
(z - Y ) ~= r 2- 2ey
f
22
+
+ y2
or, equivalently,
(1)
y2
in ( 1 ) to obtain, for any
1,
But, b y definition of the norm of a vector, ((all = 2 a ; = 2 ( z L i i 2 and (IwI(=
Thus summing ( 2 ) with respect t o i and using Iztiwil = izti' l w i l , we have
2 v?
=
2 iv[l2.
t h a t is,
hIultiplyinp both sides by
IIL'
1 w'l, we obtain the required inequality.
22.34. Prove Minkowski's Inequality: For any pair of vectors u = (ul,
. . .,urn) and
ZI = (vl, . . .,vrn) in R",
I ' Z L + vll 6 I(u(I ~
~ v ~ ~ .
+
If
7i
+
2)
+%I/
1
= 0, t h e inequality clearly holds. Thus we need only consider the case in which
# 0.
Now
7(1
+ v,I
IIu
6
IU,~ + Iwl~
+w / p
f o r a ny real numbers
=
I: + w , ) 2
f
2 I?(,
Hence
B IU, + w,/2 = 2 Ik + 4 1% + w,1
+ Iv,l) = B I Z f , + vtl ILL,/ + B 1% + wtl Iw,l
=
(ZL,
+ w11
wul E R.
(IZL,i
But by the Cauchy-Schwarz Inequality (see preceding problem),
~ ~ u , + w , 1 I ~f, l I Z L + V I I
llz~ll
and
~ l z ~ , f wp,Il ~
(l~+vl\lI~11
CHAP. 221
POINTS, L I N E S AND H Y P E R P L A N E S
279
SuppIementary Problems
DISTANCE BETWEEN POINTS
22.35. Compute the distance between each pair of points:
(iii) (4,Oi and (-2, -71;
(iv) (-3, -51 and 14, -11.
(i) (5,-1)
and ( 2 , G ) ; (ii) ( 2 , 2 ) and ( 5 , -5);
22.36. Determine k so t h a t the distance between (1,3) and (2k, 7) is 5.
22.37.
Find the point equidistant from ( 1 , 2 ) , i - l , 4 ) and (5,3).
22.38. Determine li so t h a t the points (2, l ) ,(-2, k ) and 16,4) a r e collinear, i.e. belong to t h e same straight
line.
22.39.
Show t h a t the points 16,5), (2, -5) and (-1,2) a r e the vertices of a right triangle, and find its area.
LINES
22.40. Plot the following lines:
(i) x = -5,
.t
= -2, x = 1, x = 4; (ii) y = - 4 , y = -1, y = 2,
ZJ
= 6.
22.41. Find the equation of the line:
(i) parallel to the J- axis and passing through the point (3, -4);
(ii) parallel to the u axis and passing through the point (-2,l);
(iii) horizontal and passing through the point 15, -7);
(iv) vertical and passing through the point (6,-3).
22.42.
Plot the following lines: (i) 3 2 - 5y = 15 and 2x
(iii) 2 x + 5 y = 0 and 4 x - 3 y = 0.
22.43. Find the slope of each line:
(v) ?J = 7 .
22.44.
22.45.
Find the equation of
(i) passing through
(ii) passing through
(iii) passing through
(iv) passing through
(i) y = -2x i 8;
+ 3y = 12;
iii) 2 2 - 5 y
+
(ii) 42 - 3y = -6
= 14;
and 5% 2 ~ 1= 8 ;
(iv) x = -2;
[iii) z = 3 y - 7 ;
the line:
( 3,- 2) with slope &;
(-5,O) and ( 3 , 5 ) ;
(1, -3) and parallel to 2 s - 3y = 7;
(2, -1) and perpendicular to 3 s 4y = 6.
+
Determine the distance of each line from the origin:
(i) 4x 31/ = -8; (ii) 2x 5y = 3; (iii) 12s - 5y = 26.
+
+
22.46. Determine the distance between each point and line:
(i) (2, -1) and 6s - 8y = 15; (ii) ( 3 , l ) and 3: - 3y = -6;
+
(iii) (-2,4) and 12% 5y = 3.
EUCLIDEAN m-SPACE
22.47. Let u = (1,- 2 , 5 ) and v = (3,1, -2). Find:
(i) u v ; (ii) 2 2 ~- 5v; (iii) 2 6 . v; (iv) I'dI; (v) d(u,w).
+
22.48. Let u = ( 2 , 1 , - 3 , 0,4) and v = (5, -3, -1,2,7). Find:
(i) u + v ; (ii) 32~-21;; (iii) u - v ; (iv) I;v:l; (v) d(u,v), i.e. IIu-dI.
22.49.
F o r each pair of vectors determine the value of k so t h a t the vectors a r e orthogonal:
(ii) ( 1 , 7 ,k + 2 , -2) and ( 3 , k , - 3 , k ) .
(i) (5,k,-4,2) and (1, - 3,2,2k) ;
22.50. Determine the value of k so t h a t the distance between ( 2 , k , 1 , - 4 ) and ( 3 , - 1 , 6 , - 3 )
22.51.
Show t h a t a hyperplane is not bounded.
22.52.
Show t h a t a set A is bounded if and only if there exists a real number
-m* 6 u,5 177 for every component of any vector u E A .
m*
HYPERPLANES
22.53. Calculate the distance of each hyperplane from the origin, i.e. the zero vector:
(i) 2x - y 32 4u: = 7; (ii) 4x - 12y - 32 = -10; (iii) 3% 2y 22 - w = -6.
+
+ +
is 6.
>0
such t h a t
280
22.54.
POINTS, LIN ES AND HYPERPLANES
Calculate the distance between the point and the hyperplane:
(i) 12, - 3 , l ) and 3 x + 4 y - l 2 x = 8; (ii) (5,-2,1,-3)
and r
and x + 8 y - 4 x = -7.
22.55.
Find the equation o f the hyperplane in R3 which:
(i) contains (2,-3,-5) and is parallel t o the xx plane;
(ii) intersects t h e x, y and x axes at 2, -3 and 4 respectively;
(iii) contains (1,-2,2), (0, 1 , 3 ) and (0,2, -1);
(iv) contains (1,-5,4) and is parallel to 3%-7y-6x
= 3.
22.56.
Find the equation o f the hyperplane in R4 which:
(i) contains the xl, x2 and x4 axes;
(ii) contains (1,5,-4,3) and is parallel to xl-3x2-4x3-2x1
(iii) contains (1,1,1,l), ( O , l , -3, l), (O,O,1,4) and (O,O, 2, -1);
(iv) intersects t h e axes at 3, -1,2, and G, respectively.
22.57.
Determine whether t h e given hyperplanes in
(i)
1
.v+3y+5x = 1
52
+ y+
12"-
22.58.
x = 2;
(ii)
= 7
y-22
I
2$2y+x
32
- 2y
[CHAP. 22
+ 72/-5x+5w
= 1; (iii) (4,1,-2)
= -1;
intersect in a unique point:
2-4y-2x
= 1
= 1
R3
-x = 2;
52+2y+x = 4
(iii)
1
+ +
22 2y
\3x- y
-
x = 2.
x = o
Show t h a t the intersection of r < ? n hyperplanes in Rm is either empty or contains a n infinite
number of points.
Answers to Supplementary Problems
22.35.
(i)
m, (ii) m, (iii) m, (iv) a
22.36. k = 2 o r k = -1
22.37.
(2.3, 5.3)
22.38. k = -2
22.39. 29
22.41.
(i) y = -4,
22.43.
(i) -2,
22.44.
(i) 2 - 4 y
22.45.
(i) 8/5, (ii) 3 / 6 9 ,
22.46.
(i)
22.47.
22.48.
[ii) x = -2,
(ii) 2/5,
(iii)
/ n ,
(iv) x = 6
(iii) 1/3, (iv) not defined,
= 11, (ii) 5x-8y
4, (ii) G
= -7,
= -25,
(v) 0
(iii) 2%-33y = 11, (iv) 4 2 - 3 y
(iii) 2
(iii) 7/13
+ v = (4, -1,3); (ii) 2 u - 5w = (-13, -9,20); (iii) u v = -9; (iv)
(i) u + w = (7, -2, -4,2,11); (ii) 3u - 2w = (-4,9, -7, -4, -2); (iii) u
(i) u
2 a ; (v) d(2(,2)) =
a
22.49.
(i) k = 3; (ii) k = 3/2
22.50.
k = 2 or k = -4
22.53.
(i) 7/m, (ii) 10/13, (iii) 6/m= 2 / f i
22.54.
(i) 2,
22.55.
(i) y = -3;
22.56.
(i) x3 = 0; (ii) x1 - 3x2 - 433 - 22, = -4;
xq = 6
22.57.
(i) No.
(ii) 3,
= 11
= 6;
(v) d(u, w) = &i
IIU/I
w = 38; (iv) ;1w11 =
fi =
(iii) 3
(ii) 6 % - 4y f 3 x = 12; (iii) 13xl-4y
(ii) No.
(iii) Yes.
+x
= 7; (iv) 32 - 7y- 6x = 14
(iii) 20x1 - 23x2 - 52, - x4 = -9;
(iv) 2x1 - 62, f 3x3
+
Chapter 23
Convex Sets and linear Inequalities
LINE SEGMENTS AND CONVEX SETS
a2, . . .,am) and Q = (bl, b,,
The line segment joining two points P = (ul,
Rm, denoted by P&, is the set of points
tP
+ (1- t)Q =
(ta, + (1- t ) h , ta2
+ (1- t ) h ,
,
. .,
ta,,,
+ (1- t ) b j , l ) ,
. . ., b,,)
where 0 L t
f
in
1
A subset X of Rm is convex iff the line segment joining any two points P and Q in X is
Q cX.
contained in X : P,Q E X implies P
Example 1.1.
The rectangular and elliptical ar eas below ar e each convex since the line segment joining a ny two
points P and Q, in either figure, is contained in the figure. On the other hand, the U-shaped a re a to the
right below is not convex, since t h e line segment joining the points P and Q also has points lying outside
of the figure.
Example 1.2.
Let X and Y be an y two convex sets in R7n. We show tha t
the intersection X n Y is also convex. For let P and Q be any
two points in X n Y ; then P , Q E X and P, Q E Y. Since X
and Y a r e each convex, the line segment &P i s contained in both
X and Y : P&cX and P&cY. Hence &P is contained in the
intersection: &P c X n Y . Thus X n Y is convex.
In fact,
Theorem 23.1:
The intersection of any number of convex sets is convex.
We next state a theorem about line segments which will be used throughout.
Theorem 23.2:
Let f be a linear function defined over a line segment PQ.
point A E P&, either
(i) f ( P )6 f(A) 6 f ( Q )
In particular, if f ( P ) = f ( Q )
in P&.
or
(ii) f(Q)
Then for every
f(A) f(p)
then f has the same value at every point
In other words, the value of a linear function f at any point of a line segment lies
between the values of f a t the end points.
Remark:
A function f defined over Rm is said t o be linear if it is of the form
f = f(x1, X2) . . . ,Xm)
281
=
ClXl
+ czxz +
'
..
+
Cn1Xm
282
CONVEX S E T S AND L I N E A R I N E Q U A L I T I E S
LINEAR INEQUALITIES
The linear equation a,x,
+ a,x,+
* *
;CHAP. 23
. + amxm = b partitions R" into 3 subsets:
(i) those points which satisfy the given equation;
(ii) those points which satisfy the inequality a,x, a,zz
(iii) and those points which satisfy the inequality a,x, a,xz
+ - + amxnl < b ;
+ + - + a,llxnL> b.
+
The first set, as defined previously, is a hyperplane; the other two sets are called open half
spaces.
Accordingly, the solution set or, simply, solution (or: graph) of the linear inequality
(or: a l x l ULT? . * + amxnl 2 b )
al x l + a2xz + - + a,,,~,,,6 Z,
= b, called the boundconsists of the points of the hyperplane a,x, + a,x, + - . + a n l ~ m
i n g hyperplane, and of one of the open half spaces; such a set is termed a closed half space
or, simply, a hal f space.
An important property of half spaces follows.
- -
+
9
Theorem 23.3: Half spaces are convex.
I n particular, the solution of the linear
inequality
ax b y
c (or: ax+ by 2 c )
is called a hal f plane. Geometrically speaking,
the half plane consists of all of the points lying
on the bounding line ax by = c and on one side
of the line.
+
+
5
Example 2.1.
-6,
To d r a w the g r a p h of the inequality 2 r - 3 y
first plot the line 2 x - 3 3 ~ = -6.
Then choose a n arbit r a r y point t h a t doesn't belong to the line, say ( O , O ) , and
test it to see if i t belongs t o t h e solution set. Since (0,O)
does not satisfy the given inequality, the points on the
other side of the line belong to the solution set as shown
on the right.
f
Graph of 2%- 3y
f
-6
POLYHEDRAL CONVEX SETS AND EXTREME POINTS
The intersection X of a finite number of (closed) half spaces in R" is called a polyhedral
convex set. A point P E X is called an e x t r e m e point or c o m e r point of X if P is the
intersection of 7n of the bounding hyperplanes determining X . Observe that Theorem 23.1
and Theorem 23.3 justify the use of the term convex; that is, half spaces are convex and
so their intersection is also convex.
Theorem 23.4:
The solution of a system of linear inequalities
a11n.1
a21r1
+
+
a12xz
a2252
+
+
*
. . -a1,,>:c,,1
6
bl
-a2ii,:c,n
6
bz
* ' '
................................
is a polyhedral convex set.
The theorem follows from the fact that the solution to each inequality is a half space
and that the solution of the system is the intersection of the individual solutions.
CHAP. 231
Remark:
283
CONVEX S E T S A N D L I N E A R I N E Q U A L I T I E S
Since the sense of an inequality can be changed by multipIying che inequality
by, say, -1, there is no loss in generality in assuming the inequalities in
Theorem 23.4 are of the same given sense.
Example 3.1.
The bounded polyhedral convex set in the figure
on the r i g h t is the solution of the following system
of five linear inequalities:
x + u + x
f 2
2 - - y
2
0
l'x
x
'
0
2 ' 0
The points A , B , C, D, E and 0 in the diagram
a r e the corner points, i.e. extreme points of the
solution; each point is the intersection of three of
the bounding planes. For example, D is the intersection of the planes x y x = 2, x - y = 0 and
x = 1; solving the three equations simultaneously.
we obtain D =
1). Similarly we can obtain
A = (2,0,0), B = (1,l , O ) , C = ( l , O , l ) , E = ( O , O , 1)
and 0 = ( O , O , O ) .
Y
+ +
(a,+,
Remark:
Although there are (i) = 10 different ways to select 3 of the 5 given bounding
planes in the above example, four of the ten sets of equations will not determine
an extreme point. In general, a set of m bounding hyperplanes of a polyhedral
convex set X in R" does not determine an extreme point of X if and only if
one of the following holds:
(i)
its solution is empty, i.e. the equations are inconsistent;
(ii) its solution is infinite, i.e. the equations are dependent;
(iii) its solution is unique but does not belong to X .
POLYGONAL CONVEX SETS AND CONVEX POLYGONS
In the special case of the plane R*, a polyhedral convex set X is called a polugonal
convex set and, if it is bounded, a convex polygon. Furthermore, a point P E X is a n
extreme point of X iff it is the intersection of two of the bounding lines of X.
z=1
-i
= 4
y = -1
E
x-y
B = (3, -l), the solution of
- -5
[CHAP. 23
CONVEX S E T S A N D L I N E A R I N E Q U A L I T I E S
284
Although there a r e
(i)= 6 different ways to select 2 of the 4 bounding lines,
E = (1,-3),
p + 2 y = 4
t h e solution of
F = (6, -1), t h e solution of
lx = 1
, does not satisfy
z -y
= 4
, does not satisfy y
= -1
Jx-y
\y
5
L
4
-1
I n other words, E and F do not belong t o the convex polygon.
Example 4.2.
Consider the following f o u r systems of inequalities:
x+2y
f
4
x-y
5
0
y 22
-1
x t 2 y
f
x - y L
y
4
x f 2 y
5
4
z+2y
0
2--y
='
0
x-y
y
6
-1
-1
(ii)
(i)
y
4
g
0
f
-1
(iv)
(iii)
+
In each case the bounding lines of the given half planes are x 2y = 4, x - y = 0 and y = -1.
are
= 3 different ways to select 2 of the 3 bounding lines. Furthermore,
(i)
x-7J
A = (-1, -l), is the solution of
= 0
y = -1
x+21/ = 4
B = (6,-l),
is the solution of
4 4
C = (3,s)'
is the solution of
-5
1
y = -1
-i
x+2y = 4
x-y
= 0
There
CHAP. 231
CONVEX S E T S AND LINEAR INEQUALITIES
I
285
/
(iii)
(iv)
W e now discuss the solution of each system which is sketched above.
Here the solutions a r e bounded and hence called a convex polygon. All three points A , B and C a r e
its extreme points.
Here the solutions a r e not bounded. Only the points A and C a r e its extreme points.
x-y 6 0
, t h a t is, without the inequality
Here the solutions a r e the same as the solutions to
+
x 2y 6 4. In this case we say t h a t the inequality x
point A is the only extreme point.
1
+ 2y
--1
f
4 is redztndant (or: superfluous). The
Here the solution is empty, and we say t h a t the system is inconsistent. There a r e no extreme points.
LINEAR FUNCTIONS ON POLYHEDRAL CONVEX SETS
We seek the maximum o r minimum value of a linear function
f =
+
+
~ 1 x 1 ~ 2 x 2
*
*
+ CmXm
where the unknowns are subject to the following constraints:
UllXl
UZlXl
+
a12x2
+
az2x2
+
ar2x2
+
+
’ ’ ‘
* *
’
+
UlniXrn
6
+ UZrnXrn
I
+ amxm
6
.................................
U,lXl
+
*
*.
61
ba
b,
That is, we seek the maximum or minimum value of a linear function over a polyhedral
convex set. The following fundamental theorem applies:
Theorem 23.5:
Let f be a linear function defined over a bounded polyhedral set X in R”.
Then f takes on its maximum (and minimum) value a t a n extreme point
of x.
Thus the solution of the above problem consists of two steps:
(i) Find the extreme points of X .
(ii) Evaluate the function f at each extreme point.
The maximum of the values of f in step (ii) is the maximum value of f over the entire set
and the minimum value of f in step (ii) is the minimum value of f over X .
X,
286
CONVEX SETS AND LIN EA R IN EQ U A LITIES
[CHAP. 23
Example 5.1.
Find the maximum and minimum value of the function f = 2 r i 5y over the convex polygon X
given by
2+2y L 4
2-y 5 4
x t l
y 2 -1
The extreme points of X a r e as follows (see Example 4.1):
A = (1,-1),
B = (3, -l), C = (4,O) and
D = (1,;)
Evaluate f a t each of these points:
f ( A ) = f(1,-1) = 2 . 1 - 5 . 1 = -3
j ( B ) = f(3,-1) = 2 . 3 - 5 . 1 = 1
f(C) = f(4,O) = 2 . 4
f ( D ) = f'(l,i) = 2 . 1
+ 5.0
+5.;
= 8
= 9.5
Thus the maximum value of f over the entire convex polygon X is 9.5 and occurs at D, and the minimum
value i s -3 and occurs at A .
We shall prove Theorem 23.5 f o r the case m = 2 as a solved problem (see Problem 23.17).
However, it is instructive to show here how this result follows from geometric considerations in the case of the preceding example.
We can consider the equation f = 2% 5y
as a one parameter family of lines in the plane
R2 (see adjoining diagram). We seek the maximum (or: minimum) value of f with the condition that the line f = 2 x + 5 y intersects the
given convex polygon X . As indicated by the
diagram, starting with lines of the family with
large values of f , the family will first intersect
X at the point D. Similarly, starting with
lines having small values of f, the family will
first intersect X at the point A . In other
words, the linear function f = 2 x f 5 y will
take on its maximum and minimum values over
The family f = 2x + 5y and X a r e plotted.
X at the points D and A, respectively.
+
Example 5.2.
Find the maximum value of the function f = x
+ 2y 4-82
x+y+z
x-y
over the polyhedral convex set
X given
by
2
0
6
2
z " 1
2 2 0
2 ' 0
The extreme points of X are (see Example 3.1, Page 283).
A = (2,0,0), B = (1,1,0), C = (1,0,1), D = ( + , + , I ) , E = ( O , O , l ) ,
0 = (O,O,O)
Evaluate f a t each of t h e six extreme points:
f ( A ) = 2, f ( B ) = 3, f(C) = 9,
f ( D ) = 9.5, f ( E ) = 8, f ( 0 ) = 0
Thus, by Theorem 23.5, t h e maximum value of f over the entire polyhedral convex set
and is 9.5.
X
occurs a t
Consider now a function f of the form
f =
+
+
~1x1 ~ 2 x 2
* * *
+
CmXm
+d
over a polyhedral convex set X in Rm. The function f differs from the linear function
D
287
CONVEX S E T S AND LINEAR I N E Q U A L I T I E S
CHAP. 231
g = c1x1
+
c2x2
+
* *
.
+
CmX,n
a t every point of X by the fixed amount d ; hence f takes on its maximum (and minimum)
value a t the same point as g. In other words,
Theorem 23.6:
A function of the form
f =
ClXl
+
CZXZ
+
* * *
+
C,nXm
+d
defined over a bounded polyhedral set X in Rm takes on its maximum
(and minimum) value a t an extreme point of X.
Example 5.3.
Find the minimum value of the function
f = 2z-4411+3z+2
over t h e bounded polyhedral convex set X of the preceding example.
Evaluate f a t each of the six extreme points of X :
+ +0+2
+0+2
2+0 +3
2
f(A) = f(2,0,0) = 4 0
f(B) = f(l,l,O) = 2 - 4
f(C) = f ( l , O , 1) =
T
= 6
= 0
f ( D ) = f(+,+,1) = 1 - 2
f ( E ) = f ( O , O , 1) = 0 4 0
= 7
f(0) = f ( O , O , 0 ) = 0
+3+2
+3 +2
+0 +0 +2
= 4
= 5
= 2
Thus t h e minimum value of f over X occurs a t the point B and is 0.
Remark: If a polyhedral convex set X is not bounded, then a linear function f may or
may not take on a maximum (and minimum) value.
However, the following theorem always applies (see Problem 23.9):
Theorem 23.7:
Let f be a linear function defined over a polyhedral convex set X. Then
(i) f will take on a maximum (minimum) value over X if and only if the
values of f are bounded above (below) and (ii) the maximum (minimum)
value will always occur at an extreme point of X.
Solved Problems
LINEAR INEQUALITIES IN TWO UNKNOWNS
23.1. Graph each inequality:
(i) 2x + y 4 -4,
(ii) 4% - 3 y
-6,
(iii) 2x - 3y 6 0.
In each case, first plot t h e line bounding
the half plane by finding at least t w o points on
t h e line; then choose a point which doesn't belong t o the line and test i t in t h e given inequality.
(i)
+
On the line 22 y = -4: if R: = 0, I/ = -4:
if y = O , z = - 2 .
Draw t h e line through
(0,-4) and ( - 2 , O ) . Test, say, the origin
(0,O). Since (0,O) does not satisfy 21
y 5 -4,
i.e. 0 =f -4, shade the side of
the line which does not contain the origin.
+
(i) Graph of 2 x + y
6
-4.
288
CONVEX SETS AND LINEAR INEQUALITIES
[CHAP. 23
(ii) On the line 42 - 3y = -6: if x = 0, y = 2; if y = 0, z = -1.5. Draw the line through (0,2)
and (-1.5,O). Test the origin (0,O). Since (0,O) satisfies the given inequality 4x - 3y 2 -6,
shade the side of the line containing the origin.
(iii) On the line 22 - 3y = 0: if z = 0, y = 0; if z = 3, y = 2. Draw the line through (0,O) and
( 3 , 2 ) . Since the origin belongs t o the line, test the point (2,O). Since (2,O) does not satisfy
the given inequality 22 - 3y 6 0, shade the other side of the line.
-5
(ii) Graph of 4 2 - 3 y
5
23.2. Graph each inequality: (i) x
(i)
1
(iii) Graph of 22 - 3y
-6.
2, (ii) x
1, (iii) y
4
0.
-1.
First plot the line x = 2 , the vertical line meeting the x axis a t 2. Then shade the area to
the left of the line since the x-coordinates of these points are less than 2.
(ii) First plot the line z = 1, the vertical line meeting the x axis a t 1. Then shade the area to the
right of this line since the x-coordinates of these points are greater than 1.
(iii) First plot the line y = -1, the horizontal line meeting the y axis a t -1. Then shade the area
above the line since the y-coordinate of these points are greater than -1.
(i) Graph of
2 4 2.
23.3. Graph each inequality: (i) 1
(i)
(ii) Graph of x
x
4,
(ii) -2
1.
(iii) Graph of y
2
-1.
4 y 4 1.
Recall that 1 5 x 6 4 means 1 fz and 3 5 4 . Plot the vertical lines s = l and x = 4 ; then
shade the area between the two lines as shown in Fig. (i) below.
(ii) Plot the horizontal lines y = -2 and y = 1; then shade the area between the lines as shown in
Fig. 6.i) below.
289
CONVEX SETS AND LINEAR INEQUALITIES
CHAP. 231
y = -2
-4
(i) Graph of 1 f z
5
(ii) Graph of -2
4.
f
y
5
1.
23.4. Plot the solution of each pair of inequalities:
(i) 2%- 39
6
6
2x+y
6
4
(ii) 2x
+ 5y
10
X
L
l
(iii) -1
x
x-y
6
3
6
2
In each case, plot each half-plane with different slanted lines; the cross-hatched area is the
required solution.
22 -3376
Graphsof 22-3y
f
6 and 2 z + y
f
4.
Solution of
2z+y
5
6
f
4
(ii)
Graphs of 2 2 + 5y
10 and z
2
1.
Solution of
{
22
+ 5y
f
2’1
10
(IJ
1 . 1 1 ~p ~
i u b cllci
u - Y,
vrlVllYl
...vvl---D
---- ..
-. -.
the left of the line since the 2-coordinates of these points a r e less than 2.
L ~ L G
YIIG
I____
~~
(ii) First plot the line x = 1, the vertical line meeting the s axis at 1. Then shade the area to the
right of this line since the s-coordinates of these points a r e greater than 1.
(iii) F i r s t plot the line y=-1, the horizontal line meeting the y axis at -1. Then shade the area
above the line since the y-coordinate of these points a r e greater than -1.
(i) Graph of x
6
2.
23.3. Graph each inequality:
(i)
(ii) Graph of z
(i) 1 4 x
4 4,
(ii) -2
5
(iii) Graph of y
1.
2
-1.
4 y 4 1.
Recall t h a t 1 6 x 5 4 means 1 5 x and x r 4 . Plot the vertical lines x = l and x = 4 ; then
shade the area between the two lines as shown in Fig. (i) below.
(ii) Plot the horizontal lines y = -2 and y = 1; then shade the area between the lines a s shown in
Fig. (ii) below.
/
/
l
/
i
l
-5
Each half plane is plotted.
The required solution.
Observe that the solution of the system is not bounded.
(ii)
Each half plane is plotted.
Here the solution of the system is bounded.
The required solution.
CHAP. 231
CONVEX S E T S A N D LIN EA R IN EQ U A LITIES
291
LINEAR FUNCTIONS OVER POLYHEDRAL CONVEX SETS
23.6. Find the maximum and minimum of each
function over the convex polygon X
shown in the diagram on the right:
f = 2x +52J
8-
A = (0,2)
B = (4, -1)
c = (5,l)
D = (3,5)
g = x-22J
h = -3xf2J
-2
-2 -
Evaluate each function at each extreme,
i.e. corner point:
f(A) = t(0,2) = 0
f ( B ) = t(4,-1)
+ 10
= 10
= 8-5 = 3
+5
6 + 25
X is shaded.
y ( A ) = 0 - 4 = -4
g(B) = 4
+2
h(A) = 0
+2
1 = -13
h ( B ) = -12
-
+1=
f ( C ) = t ( 5 , l i = 10
= 15
y(C) = 5 - 2 = 3
h ( C ) = -15
f ( D ) = t(3,5) =
= 31
~ ( 0=) 3
h ( 0 ) = -9
-
= 2
= 6
10 = -7
Thus:
maximum of f over X is 31 and occurs at the point D
maximum of g over X is 6 and occurs a t the point B
maximum of h over X is 2 and occurs a t the point A
and
minimum of i over X is 3 and occurs a t the point B
minimum of g over X is -7 and occurs a t the point D
minimum of lz over X is -14 and occurs a t the point C.
+5
-14
= -4
over the convex polygon X of the
23.7. Find the maximum of the function f = 2 x - y
preceding example.
Evaluate t' a t each of t h e four extreme points of X :
f ( A ) = i ( 0 , 2 ) = -2
f(C) = f ( 5 , l ) = 9
f ( B ) = f(4, -1) = 9
f ( D ) = f(3,5) = 1
Thus the maximum of f over X is 9 and occurs
at two of the corner points, B = (4,-1) and
C = ( 5 , l ) : hence by Theorem 23.2 it will occur
at every point on the line segment joining
B and C.
The above can be interpreted geometrically:
the member of the family of parallel lines determined by the equation f = 2%- y which has
the maximum value of f and intersects X ,
passes through both B and C and so contains
every point on the line segment joining B and
C, as shown in the diagram.
The diagram also indicates t h a t f takes on
its minimum value at t h e point A .
23.8. Find the extreme points of the polygonal convex set X determined by the system
3,r+2y
x-y
G
6
0
x G 4
Method 1 (Algebraic).
= 3 ways to select 2 of the 3 bounding lines:
There a r e
(i)
23.5.
Plot the solution of each system of inequalities:
(i) 3x + 2 y
6
(ii) ,T + 4y L 4
x-y
4
n: ‘1 -1
$ 5 2
y 1 0
(i)
Each half plane is plotted.
The required solution.
Observe t h a t the solution of the system is not bounded.
(ii)
Each half plane is plotted.
The required solution,
Here the solution of the system is bounded.
__.~.~
iiiinimum value of
It
over
X.
X and the family g
X f
76.
X and the family h =
%
- 276.
293
CONVEX S E T S A N D LINEAR INEQUALITIES
CHAP. 231
23.10. Find the maximum and minimum of the function f = x - 2y 4 over the convex polygon
X given by
xsy 5 4
+
Ic
+ 2y + -2
x-y
2
-2
x g 3
To obtain the extreme points of X, plot X as
shown in the diagram. The points A , B . C and D a r e
the corner points of X . The coordinates of each
point can be obtained by finding the intersection of
the appropriate pair of bounding lines:
A = (-2,O)
x + 2 y = -2
is the solution of
x-
x + 2 y = -2
{
{x+: 1 ,"
B = (3, -%) is the solution of
C = (3,l)
is the solution of
D = (1,3)
is the solution of
y = -2
2 = 3
i
x + y = 4
x-y
= -2
Next evaluate the function f at each extreme point:
f(A) = --2'0+4
f(B) = 3
= 2
+5 +4
f(C) = 3 - 2 1 - 4
= 12
+4
f ( D ) = 1- 6
= 5
= -1
Thus the maximum of f over X is 12 and occurs at B, and the minimum is -1 and occurs a t
D.
23.11. Find the extreme points of the polyhedral convex set X given by
2 +
y + x g 3
4
0
2x+2y+2
x-y
X k O
2'0
A point p E R3 is an extreme point of X iff it is the intersection of three of the bounding
= 1 0 ways to choose 3 of the 5 bounding planes:
planes of X. There a r e
(g)
i
= 3
= 4
z = o
= 3
= 4
z = o
x + y + z = 3
x+y+x
x + y + z
2x+2y+z
= 4
2x+2y+x
2x+2y+x
x-y
= 0
ii)
x+y+z
2-y
= 3
= 0
x = o
(v)
(iii)
(ii)
1
= 3
2x+2y+x
x = o
2-y
x + y + z
x = o
(vi)
x--y
= 4
= 0
= 0
x = o
(iv)
2 = 0
(vii)
x-ty+z = 3
294
CHAP. 23
CONVEX SETS AND L I N E A R I N E Q U A L I T I E S
I
I
2.'.
2y - z = 4
.t -
II'-y
x
g = 0
z
7
I
0
=
= oO
2 - 0
Each system except (iii) yields a unique solution:
A,
= (0, 3, 0)
(vii) A ,
= (0, 0, 4)
(vi)
!iii)
No solution, i.e. system is inconsistent.
(viii) A s
= (1, 1,0 )
IIV)
A , = (0, 0 , 3 )
(IX)
A,
= (0, 2, 0)
(x)
A i o = (0, 0, 0)
rv)
A, =
3 3
(2' -2 '
0)
To find the extreme points of X, test each of the above
points in the given inequalities. The following a r e not
extreme points of X:
A = ($, f , 0) does not satisfy 2.x
+ 2u - z
A , = (0, 3,O) does not satisfy 2a
i
A ; = ( O , O , 4) does not s a t ~ s f y .L
2u
x
y-z
L
='
4
4
3
The other six points do satisfy the given inequalities
and a r e the extreme points of X. The adjacent diagram
sho\\s the position of these points. Note t h a t we obtained the extreme points purely algebraically.
23.12. Find the maximum and minimum values of the function f = 4 x - 2 y - x
over the
bounded polyhedral convex set X of the preceding problem.
Evaluate t' a t each of the six extreme points of X:
t'(A1)
1
f(+,+,2)
f ( A J =- f(0,1,2)
t(A4)
1
f(O,O,3)
= 2 - 1- 2
!(A,)
-1
= f(l,l,O)
+0
+0
0+0+0
= 4 -2
2
-
2 = -4
J(A,) = f ( 0 , 2 , 0 ) = 0 - 4
0 -0
-
3
f(A1,) = f(0,0 , O )
= 0
-
-3
=z
= 2
-4
z=
0
The maximum value of f over X is 2 and occurs a t the point A s . The minimum value of f over X
is -4 and occurs at the extreme points A , and A , and so also occurs at each point on the line
segment joining A , to A,.
MISCELLANEOLIS PROBLEMS
23.13. Prove: Let f be a linear function defined over R":
f(X1, x 2 ,
. . . , X!!!) =
ClXl
+
C2Xr
+. .+
*
Then f o r any vectors P,Q E R"' and any real number k ,
(i) f ( p+ Q) = f(P)+ f(Q),
(ii) f ( k P ) = k
f(0
CnlXrn
295
CONVEX SETS AND LINEAR IN EQ U A LITIES
CHAP. 231
23.14. Prove: Let f be a linear function over R”’ with f ( P )6 f ( Q ) . Then for 0 6 t
f(p)L f ( t P
+ (1
Since t and ( 1
-
~
t)&)
1,
f(&).
a r e each non-negative, t ( P )
f )
6
tj(P)
5
and
tf(Qi
f ( Q ) implies
(1- t ) f ( P )
(1- t ) f ( Q j
f
Hence, using the iesult of the preceding problem, we have
f(P)
t f(P)
-
+ (1 - t ) f ( P )
t t ( P ) f (1 tj f ( Q ) = t ( t P -- (1- t ) Q )
~
t f ( Q ) + (1- t ) t ( Q ) = f ( Q )
t ( f P (1- t ) Q ) = t f ( P ) + (1 - t ) fCQ)
and
23.15. Prove Theorem 23.2: Let f be a linear function defined over a line segment p& in R”‘.
Then for every point A E P&, either (i) f ( P )A f ( A )6 f ( Q ) or (ii) f(Q) A f ( A ) f ( P ) .
In particular, if f ( P )= f ( & ) then f’ has the same value a t every point in P&.
The proof folloivs from the preceding problem and the fa c t t h a t A = f P f (1- t ) Q for some
t between 0 and 1.
23.16. Prove Theorem 23.3: Let X be a half space
convex: P , Q E X implies &
P CX.
+ * . . + unIsm4 b.
a,x,
Then X is
Now X consists of all the points A E R”’ such t h a t
f(A)
Hence if P , Q
E
5
b
X , then f(Pj
6
t(P)
6
f(A)
f = alxl
where
+
~
2
f%
. . ~. + a,,,r,,,
&
b and f ( & ) f b. Thus, by Theorem 23.2, if A E P
f
f(Q)
aE
b
f(Q)
or
6
f(A)
5
f(P)
f
then either
b
In either case T”A) 5 b and so W c X .
23.17. Prove Theorem 23.5 (for the case w i = 2): Let f be a linear function defined over a
convex polygon X . Then f takes on its maximum (and minimum) value at a corner
point of X .
Suppose t h at t’ takes on its largest corner value at the
corner point P . Let A be any other point in X. We w a n t to
prove t h a t f ( A )5 j ( P ) .
Since X is bounded, the line passing through the points
P and A will intersect a bounding line of X a t a point Q such
t h a t A lies between P and Q: A E PQ.
Furthermore, Q must
lie on the line segment joining two corners, say, U and V :
Q
E
P
\
UT.
Suppose f ( P ) < f ( A ) ; then by Theorem 23.2, f ( P ) <
f ( A ) f ( Q ) . That is, f ( A ) lies between f ( P ) and f ( Q ) . On
t h e other hand, by Theorem 23.2, either i ( Q ! f ( U ) or
f(&) f(V’). Combining the various inequalities, we finally
have
f(p)
f(A)
f(Q) f ( l 7 o r f ( P ) < f ( A )
)V
/
f
f
f
4
f(Q)
fW)
However, both cases contradict the assumption t h a t f ( P ) w a s the largest corner value. Accordingly,
f(P)< f ( A ! IS impossible and so f ( A ) t ( P ) .
f
23.18. Prove Theorem 23.5: Let f be a linear function defined over a bounded polyhedral
convex set X in R”. Then f takes on its maximum (and minimum) value a t a n
extreme point of X .
The proof is by induction on m. If m = 1, then X is a line segment and by Theorem 23.2
f takes on its nlasimum (and minimunii value a t a n extreme point, i.e. a n end point.
CONVEX SETS AND LINEAR INEQUALITIES
296
'CHAP. 23
We assume the theorem holds f o r a polygonal convex set in Rm-1. Let A E X. Since X is
bounded, the point A belongs to some line segment p& where P and Q each belongs to a bounding
half space of X . By Theorem 23.2, f ( A ) must lie between f ( P ) and f(Q) and so f must take on its
inaximum land minimum) value a t a point on a bounding half space.
Kow every point B E X on a bounding half space of X belongs to the intersection Y of X and
a hyperplane H . However, Y is a bounded polyhedral convex set of H and H is isomorphic to
R1"-1.
Thus by induction f takes on its maximum value a t a n extreme point of Y which is also
an extreme point of X.
Supplementary Problems
LINEAR INEQUALITIES IN TWO UNKNON'NS
23.19.
Graph each inequality:
(i) x - 2y
23.20.
Graph each inequality:
(i) x
23.21.
Graph each inequality:
(i) -2
23.22.
Plot the solution to each system:
ii) 2.c - y 6 4
(ii) x
y
x 5 1
x-2y
23.23.
f
f
4, (ii) 3x
3, iii) x
f
x
f
+
2,
f
2
2
+y
(iii) y
-4,
(ii) 0
5
x
2
5
6
2,
iiv) y
1-
2
0,
(iv) 3x - 5y
-15.
-3.
3.
(iv) -1 5 x 6 4
- 2 f y 5 1
(iii) 2 2 4- 3y
6
- 1 f y 6 2
5
2
Plot the solution to each system:
ii) x - y e 2
(ii, 3 r 1 2y 6 6
2 x f y 2 -2
2 y ' l
u s 1
y
(iii) 2x + 5y
6,
f
(iii)
x - 2y 2 -4
- 2 5 x 5 1
-1
y
2
-3
LINEAR FUNCTIONS OVER POLYHEDRAL CONVEX SETS
23.24.
A = (-1,l)
B
(1,-2)
Find the maximum and minimum values of each
function over the convex polygon X shown in
the diagram on the right: (i) f = 3 x f 2 y ;
f i i ) .g = --z
2y - 4; (iii) h = x - 2y.
C = (4, -1)
D = (5,2)
+
23.25.
,
Find the maximum and minimum values of
over the convex polygon
f = 3r-y
x+2y
x-y
x
f
f
%
4
1
-1
23.26.
Find the maximum and minimum values of the function f = 2 2
given by
2-y 2 0
3x+y 2 3
x r 4
23.27.
Draw the convex polygon
X given by
x
+y
x-y
6
+y +3
over the convex polygon
3
-3
y ' 0
x 2 -1
2 4 2
Find the corner points of
over X .
X
2
and the maximum and minimum values of the function f = 22 - y
CHAP. 231
23.28.
CONVEX SETS AND L I N E A R I N E Q U A L I T I E S
Draw the convex polygon X given by
s+2y
x-y
297
6
2
0
2 ' 4
y r o
Find the corner points of
over X.
23.29.
X and the maximum and minimum values of the function f = 2 x - 3 y
Find the extreme points of the following bounded polyhedral convex set X in R3:
3x+3y+x 5 6
x-y 2 0
2'0
y ' 0
2 ' 0
2 5 3
Find the maximum and minimum values of the function f = x
23.30.
Draw the polygonal convex set X given by
zf2y
2-y
+ 2y - x
over X .
'4
2
-2
2 ' 2
Find the maximum and minimum values (if they exist) of each function over X
f = 2x++,
g = 22-yy,
h = x-3y
Answers to Supplementary Problems
(ii)
(iii)
(ii)
(iii)
23.20.
;CHAP. 23
C O N V E X SETS AND LINEAR I N E Q C A L I T I E S
298
23.21.
41
-4
(ii)
23.22.
\
-4
23.23.
\
(ii)
(iii)
(ii)
(iii)
4i
i i'i
23.24.
(i) max f = 19, min f = -1;
23.25.
max f = 5, min f = -5.5
23.26.
max f = 15, niin f = 2
i,ii) max g = -1,
min g = -10;
(iii) rnax h = 6, min h = -3
23.27.
A =
B =
c =
D =
E =
(-1, 0)
(2, 0 )
(2, 1)
( 0 , 3)
(-1, 2)
Maxi = 4, a t B.
M i n f = -4, a t E.
CHAP. 231
299
CONVEX S E T S A N D L I N E A R I N E Q U A L I T I E S
23.28.
A = (0, 0)
B = (4, 0)
c = (4, 1)
D = (2, 2)
Maxf = 8, a t B.
Min f = -2, a t D .
23.29. The corner points are: (O,O,O),(2,0,0), ( 1 , 1 , 0 ) ,(0,0,3), (1,0,3),
(4,$ , 3 ) .
Max f = 4, min i = -3.
23.30.
Maxf = 5 a t A . niin t’ does not exist.
Max g and niin g do not exist.
Min h = -6 a t 8 . max h does not exist.
Chapter 24
linear Programming
LINEAR PROGRAMMING PROBLEMS
Linear programming problems are concerned with finding the maximum or minimum
of a linear function
f = C I X I C Z X Z + * * . + CntXrn
+
called the objective function, over a polyhedral convex set X which includes the condition
that the unknowns (variables) are non-negative:
21 2
0,
X
Z
0 ,~ . . .,
X,n 2
0
Each point in X is called a feasible solution of the problem, and a point in X a t which
f takes on its maximum (or minimum) value is called an o p t i m m solution.
Example 1.1. (Rlaximum Problem.)
A tailor has the following materials available: 16 sq. yd. cotton, 11 sq. yd. silk, 15 sq. yd. wool.
A suit requires the following: 2 sq. yd. cotton, 1 sq. yd. silk, 1 sq. yd. wool. A gown requires the following:
1 sq. yd. cotton, 2 sq. yd. silk, 3 sq. yd. wool. I f a suit sells for $30 and a gown f o r $50, how many
of each garment should the tailor make t o obtain t h e maximum amount of money?
The following table summarizes the above facts:
Let
5
be the number of suits the tailor makes and y t h e number of gowns. We seek to maximize
+
f = 30% 50y
subject to
2x
+y
L_
16,
x
+ 2y
f
11,
x i - 3y
15,
r 2 0, y
%
0
Observe t h a t the inequality 2 x + y 6 16 corresponds to t h e f a c t t h a t 2 sq. yd. of cotton is required f or
each suit and 1 sq. yd. f o r each gown, but only 16 sq. yd. is available. The inequalities x + 2 y f 11 and
x + 3 y f 15 correspond to the analogous facts about t h e silk and wool,
Fig. 24-1 is the g r a p h of t h e convex polygon X given by
t h e above inequalities. The corner points of X a r e
0 = ( O , O ) , A = (8,0), B = (7,2), C = (3,4), D = (0,5)
The values of t h e objective function f = 302 50y at t h e
+
corner points a r e
f(0) = 0 , J ( A )= 240, f ( B ) = 310, f(C) = 290, f ( D ) = 250
Thus the maximum of f over X is 310 and occurs at t h e
point B = ( 7 , 2 ) . I n other words, the maximum amount of
money the tailor can g e t is $310, and this will happen if he
makes 7 suits and 2 gowns.
300
Fig. 24-1
CHAP. 241
301
LINEAR PROGRAMMING
Example 1.2. (Minimum Problem.)
A person requires 10, 1 2 and 12 units of chemicals A , B and C, respectively, for his garden.
product contains 5 , 2 and 1 units of A, B and C, respectively, per j a r ; and a dry product
1, 2 and 4 units of A , B and C, respectively, per carton. If the liquid product sells f o r $3 per
the d r y product sells for $2 per carton, how many of each should he purchase to minimize the
meet the requirements?
The following table summarizes the above facts:
Units per j a r
Units per carton
Units needed
Chemical A
Chemical B
Chemical C
5
2
1
1
2
10
12
12
co s t
$3
$2
4
A liquid
contains
j a r and
cost and
Let x denote the number of j a r s bought and y the number of cartons. We seek to minimize the function
g = 32
2y
subject to the conditions
12
5x y 2 10, 22 2y 2 12, z 4 y 2 12
P
and t h e f a c t t h a t z and ?j a r e non-negative, z
0, u 2 0.
+
+
+
+
Fig. 24-2 is the g r ap h of t h e polygonal convex set T of
this minimum linear programming problem. The corner
points of T a r e
8
P = (0, lo), Q = (1,5), R = (4,2), S = (12,O)
N t h o u g h T is not bounded, the objective function y = 32 2y
is never negative over T and so takes on a minimum value
a t a corner point of T . Since
g ( P ) = 20, g ( & ) = 13, g(R)= 16, g ( S ) = 36
4
+
4
8
12
16
Fig. 24-2
the minimum value of g over T is 13 and occurs a t Q = (1,5).
In other words, he should buy 1 j a r and 5 cartons.
DUAL PROBLEMS
Every maximum linear programming problem has associated with it a corresponding
minimum problem called its dual and, conversely, every minimum linear programming
problem has associated with it a corresponding maximum problem called its dual. The
original problem is termed the prinzal problem. To be more specific, we define each of
the following problems to be the dual of the other:
(i) Maximum problem:
Maximize f =
0, xz
and
(ii) Minimum problem:
Minimize g =
subject to
+
CIXI
b1zu1
+
+ . + CmXm
~ 2 x 2
0,
*
....
Xm
10.
+- +
b 2 ~ 2
* *
bkWk
+
+ + aklwii
+ a 2 m 2 + . . . + ak~wk 2 cz
..............................
almzuI+
+ + akmwk
allwl
a12wl
alzc2
* * *
~
* ~*
2
2 c1
~
2
Cm
302
[CHAP. 24
L I N E A R PROGRAMMING
Let us write the coefficients of the above inequalities in the form of a matrix as follows
with the coefficients of the objective function as the last row:
(i) Maximum problem:
(ii) Minimum problem:
In each case the matrix of coefficients of the dual problem can be obtained by taking the
transpose of the matrix of coefficients of the primal problem. (Note that there is no last
entry in the last row and last column of either matrix.)
Example 2.1.
Find the dual of the maximum problem of Example 1.1. Write the matrix of coefficients of the given
(primal) problem and then take its transpose:
Primal problem
30
3
15
50
*
Dual problem
16 11 15
Thus the dual minimum problem is as follows:
Minimize g = 16u
subject to
+v f w
+ 2v -I- 3w
2u
u
and
+ l l v + 15w
ZL 2
0, v
2
30
2
50
0.
0, w
Example 2.2.
Find the dual of t h e minimum problem of Example 1.2. Write the matrix of coefficients of the
given (primal) problem and then take i t s transpose:
Primal problem
1 4
12
3
*
2
Dual problem
10 12 1 2
Thus the dual maximum problem is as follows:
+ 12v + 12w
5u + 2v + w
3
Maximize f = 10%
subject t o
f
ZL+2V+4W
and
21 2
0, v
2
6
0, w
2
5
0.
In general, a linear programming problem need not have an optimum solution. However, the following theorem, which is the main result in the theory of linear programming,
applies:
CHAP. 241
303
LINEAR PROGRAMMING
Duality Theorem:
The objective function f of a maximum linear programming problem
takes on a maximum value if and only if the objective function g of
the corresponding dual problem takes on a minimum value and, in
this case, m a x f = min g. Furthermore, if P and Q are feasible
solutions such that f ( P ) = g(&), then P and Q are optimum solutions
of their corresponding problems.
The proof of this theorem lies beyond the scope of this text.
Example 2.3.
Consider the maximum problem of Example 2.2. The
graph of the polyhedral convex set X given by the inequalities of t h a t problem appears on the right. The corner points a r e
P = (lOil9, 0, 7/19)
Q = ( 3 / 5 , 0,0)
R = (114, 7/8, 0)
s = 10,1, 0 )
T = (0,0 , 3)
0 = (0,0,0 )
+
2)
+
The objective function t' = lox 12u 12w takes on its
maximum value a t the point R and m a x f = 13. On the
other hand, min g = 13 for the corresponding miniinum
problem appearing in Example 1.2, as predicted by the
Duality Theorem.
U
MATRIX NOTATION
Let the matrix A , the column vectors b and x, and the row vectors c and w be defined
as follows:
A =
u":
1:: :::,):I
........*.......
akl
ah2
' ' '
b =
p, [;:I
x =
b k /
akm
c =
'
20
=
( C l , c2,
. . ., e m )
. . ., W k )
(201, '202,
Xm
Then the maximum and minimum linear programming problems (1) in the preceding section can be rewritten in the following short compact form:
(ii) Minimum problem:
(i) Maximum problem:
Minimize g = wb
Maximize f = cx
subject to A x
and
X
Remark:
b
0.
subject to zuA
and
20
c
(1)
0.
If ZL and v are vectors, then zc v shall mean that each component of u is less
than or equal to the corresponding component of v.
INTRODUCTION TO T H E SIMPLEX METHOD
Let X be the polyhedral convex set, i.e. set of feasible solutions, of the above maximum
linear programming problem. An optimum solution can be found by evaluating the
objective function f a t each corner point of X. However, if the number of unknowns and
inequalities is large, then i t can be a long and tedious task to find all the corners of X .
An alternate way of finding an optimum solution is the following:
304
LINEAR PROGRAMMING
[CHAP. 24
(1) Select a corner of X as a starting point.
(2) Choose an edge through this corner such that f increases in value along the edge.
(3) Proceed along the chosen edge to the next corner.
(4) Repeat steps (2) and (3) until a n optimum solution is obtained, i.e. a corner is
reached such that f does not increase in value along any of its edges.
The above process is accomplished mathematically by the so-called simplex method.
Observe, first of all, that the maximum linear programming problem (I), Page 301, is
equivalent to the following problem which contains only linear equalities and not linear
inequalities; the variables xm 1, . . , , rFl1
t lc are called "slack" variables:
Maximize f =
subject to
. . . + c , x m + O ~ x , l t ~ + O ~ ~ m +.z. +. + O * x m t h
clx1+cLx3+
+
~ ~ 1 1 U
~ U1
a21X1
+ . . . + almXm +
X~
+ a2222 +
~
. + azmxm
+
~
1
= bl
Xnit2
= bz
-
+
................................................
and
aklXl
+
x1 A
0,
ahax2
c
' '
'
+
0, . . .,
XL
f
akmXm
Xm
0,
Xmtl
= bk
Zmtk
0, . . . .
Xm+k %
(2)
0.
The algorithm of the simplex method consists of four steps (see Page 307) which correspond,
respectively, to the above steps. For the sake of clarity we shall explain each step separately.
Remark:
In order to avoid special cases, we shall assume the following additional conditions on our linear programming problems:
(i) Positive condition: Each component of b is non-negative ( b % 0) and each
row of A contains at least one positive entry.
(ii) N o n - d e g e n e r a c y condition: If r is the rank of A , i.e. the number of independent rows (or columns) of A , then c is not a linear combination of less
than T rows of A and Z, is not a linear combination of less than r columns
of A .
INITIAL SIMPLEX TABLEAU
The initial simplex tableau of the maximum linear programming problem (1) or the
equivalent problem ( 2 ) is as follows:
P1
Pa
* ' *
Pm
..................
akl
I
akz
" '
akm
1
............
0 0
I
" '
1 I
I
I
J
indicators
Observe that the rows of the tableau, excluding the last, are the coefficients of the linear
equalities in ( 2 ) . The columns of the coefficients of the non-slack variables are labeled Pi,
respectively; their significance will appear later. The entries in the last row, excluding
the last entry, are called i n d i c a t o r s and are the negatives of the coefficients of the objective
function f in ( 2 ) . The entry in the lower right hand corner is zero; i t corresponds to the
value of the objective function f at the origin.
CHAP. 241
305
LINEAR PROGRAMMING
Although the simplex method is stated in terms of the maximum linear programming problem, the method will give the optimum solution to both the
maximum problem and to its dual minimum problem.
Remark:
Example 3.1.
Maximize j = 30.1.
subject to
+y
+ 22/
22:
6
2:
5
16
11
6
15
+ 3y
3:
and
Initial simplex tableau:
- 50y
Pl
15
-30
0, u ’ O
n
Example 3.2.
Maximize f = 1 0 ~ 12v
~
subject to
and
-50
0
0
0
0
Initial simplex tableau:
+ + 12w
5 u + 2w + w
3
+ 2v 4w ‘2
5
ti
pi?
Pl
pz
p3
5
2
1
1
0
3
T
Z L ~ O ,w
2 0 ,
W k O
Example 3.3.
Initial simplex tableau:
Maximize f = x
2y
subject to
+ 8x
z
.r.+y+x
1
f
2
pz
p3
l o
1
-1
‘0
-z+y
and
r
PI
3 - 5 0 , y z o , 75’0
PIVOT ENTRY O F A SIMPLEX TABLEAU
A pivot entry or simply a pivot of a simplex tableau is obtained as follows:
(i) Arbitrarily select a column which contains a negative indicator.
(ii) Divide each positive entry in this column into the corresponding element in the
last column.
The entry which yields the smallest quotient in step (ii) is the pivot; its column is called
the pivotal column and its row the pivotal row. If the tableau has no negative indicator,
then it is a terminal tableau and has no pivot. Properly interpreted, the terminal tableau
yields a solution to our problem. The procedure which follows represents a method of
passing to such a terminal configuration.
Example 4.1.
Consider the following simplex tableaux:
Pl
I
1
p2
2 1 1
0
0 1 1 2 1
PI
pz
0
0
p3
1 I
1
1 @ 0 1 - 1
-30 -50
,
0
0
0
0
0
2
0
-1
0
0
0
0
1
0 1
1
1
1
7
1
0
(ii)
1
1
2 1
5
1
6
306
0
1
0
0
0
-
0
1
1 1
40 -10
4
290
f iv)
(iii)
(i)
[CHAP. 24
LINEAR PROGRAMMING
Either the first column o r the second column can be chosen a s the pivotal column. In the first case
2 is the pivot since 312 is less than 1 2 / 1 and 5/3. In the second case 1 is the pivot since 5/1 is less
than 1212. We need not consider the quotient 3/(-1) since -1 is not positive.
(ii) Here the second column must be chosen a s the pivotal column since i t is the only one with a negative
indicator. Furthermore, 1 is the pivot since 2/1 is less tha n S/2.
(iii) The fifth column is t h e pivotal column and 3 is the pivot since 6/3 is less tha n 4/1.
(iv) There is no pivotal column and no pivot since there is no negative indicator. This tableau is a
terminal tableau.
CALCULATING THE NEW SIMPLEX TABLEAU
Let D be a simplex tableau with rows R, whose pivot appears in the rth row and
sth column; say D is the matrix (&) and so d,, is the pivot. A new tableau D* with rows
RF is calculated from D (using appropriate “elementary row operations”) so that 1 appears
in the original pivot position and zeros elsewhere in that column. This is accomplished
as follows:
(i) First divide each entry of the pivotal row R, of D by the pivot d,, to obtain the corresponding row R: of D*::
1
R: = -I?,
drs
If the pivotal column of D is labeled P,, then this row of D* is labeled (or relabeled) Pi.
(ii) Every other row €2: of D* is obtained by adding the appropriate multiple of the above
row R: to the row Ri of D:
R: = -disRT -k R,
If one of these rows of D is labeled Pi, then the corresponding row of D* is given the
same label; otherwise the row is not labeled.
Example 5.1.
Consider the following tableau with a pivot entry circled:
R,:
R,:
R$
I
R,:
1 - 8 - 6 1 0
I
0
I
0 1 0 1
We remark t h a t the circled 2 i s a pivot entry since i t appears in a column with a negative indicator and
412 is less than both 22/1 and 914. We now calculate the new tableau step by step:
(i)
Divide each entry of the pivotal row R , by the pivot 2 t o obtain the new row RZ; label
label of the pivotal column, i.e. PI:
Rl
with the
CHAP. 241
L I N E A R PROGRAMMING
307
2
R4
+
(ii) RF is obtained by multiplying the above row R: by -1 and adding i t t o R,: RT = -R,*
R,;
R: is obtained by multiplying R,* by -4 and adding i t to R3: R; = -4R*
R3; RZ is obtained by
multiplying RZ by 8 and adding i t t o R,: RT = SR; R4. These rows do not change labels. Thus
the completed new tableau is
PI
p,
J
-
+
The intermediate tableau appearing in the preceding example was illustrated separately
from the completed tableau only f o r the sake of clarity; in general, we calculate the new
tableau directly from the given tableau. We remark that if we choose the second column
as the pivotal column and so obtain a different pivot, then we would calculate a different
new tableau. This corresponds to the fact that in moving from corner to corner there
may be more than one edge on which the objective function f increases in value; hence
there is more than one way to finally reach the optimum solution.
The simplex method begins with the initial tableau and then calculates new tableaux
one after the other until a terminal tableau is obtained. We remark that in each step a t
most one row, the pivotal row, changes its label; the columns keep the same label throughout.
Example 5.2.
Consider the following tableau with a pivot entry circled:
Pl
p2
Pl
(*I
0
0
0 -10
4
0
0
16
Divide each entry of the pivotal row by the pivot 2 to obtain the row
o l o l g a
The new tableau is
Pl
p2
1 0
0
~
0
1
(**)
45 1 2 1 1
It is calculated as follows:
(i) t h e first row is obtained by multiplying (**) by -3 and adding the result t o the first row of (*);
(ii) t h e second row is obtained by multiplying (**) by fr and adding the result t o the second row of (*);
it retains the label PI;
(iii) t h e third row is (**); it i s labeled P,, the label of the pivotal column of (*);
(iv) the fourth row is obtained by multiplying ( * c ) by 10 and adding the result t o the fourth row of (*).
Note t h a t t h e new tableau is a terminal tableau, i.e. it ha s no negative indicators.
308
L I N E A R PROGRAMMING
[CHAP. 24
INTERPRETING T H E TERMINAL TABLEAU
Suppose after performing the algorithm of the simplex method on the linear programming problem ( I ) , Page 301, we obtain the following terminal tableau:
...
some labeled
with Pi
Then:
(i)
.............
a
* ... *
* 4: . . . *
...
.............
* a ...
I
I
I
I
4,
...
(I,
q
k
I
v
I
v, the entry in the lower right hand corner of the tableau, is the maximum
value of f (and so by the duality theorem is the minimum value of g:
m a x f = m i n g = v);
(ii) the point Q = (q,, qL, . . . . (I,) is the optimum solution of the minimum problem;
its components are the last entries in the columns corresponding to the slack
variables;
(iii) the point P = ( p l , p l , . . . . p,,,), where
1the last entry of a row labeled Pt
), 0 if
no row is labeled P,
is the optimum solution of the maximum problem.
'I1
=
Thus a necessary condition (and so a check) that the calculations are correct is that
f ( P )= g ( Q ) = v.
Example 6.1.
Find the optimum solution Q = ( q l , . . . . q k ) of the minimum problem, the optimum solution
P = (p,, . . ~,p,?,)
of the maximum problem, and v = max f = min g according to each of the following
terminal tableaux:
PI p2 p,
PI P , P ,
:::
*
::
*
*
*
(i)
(i)
Q = ( 3 , 4 , 5 ) , P = ( 0 , 2 , 6 ) , w = max f = min g = 19.
no r o w is labeled P,.
:*
*
(ii)
Note tha t the first component of P is 0 since
(ii) Q = ( 2 , 5 ) , P = (0,7,0), w = max f = min g = 11. Note tha t the first and third components of P
a r e each 0 since no row is labeled P , o r P,.
ALGORITHM OF T H E SIMPLEX METHOD
The algorithm of the simplex method consists of four steps which correspond, respectively, to the steps on Page 304:
(1) Construct the initial tableau.
(2) Find and circle a pivot entry in the tableau.
(3) Calculate the new tableau from the given tableau and circled pivot.
(4)Repeat steps (2) and (3) until a terminal tableau is obtained.
The following examples illustrate the use of this algorithm.
CHAP. 241
309
L I N E A R PROGRAMMING
Example 7.1.
Applying the simplex method to Example 1.1, Page 300, we obtain the following sequence of tableaux:
Pl
-30
p2
-50
0
0
, 0
-1
0
0
Pl
p2
0
0
0
40 -10
290
(iii)
Thus, by the terminal tableau (iv), the optimum solution is P = (7,2) since 7 and 2 a r e the last entries
in the rows labeled Pl and P , respectively, and the maximum value of f is 310. This agrees with the
previous solution of the problem. Observe t h at the numbers 0, 250, 290 and 310 which a ppe a r respectively
in the lower right hand corners of t h e tableaux correspond precisely to the values of the objective
function f a t the points 0, D , G and B in Fig. 24-1.
Example 7.2.
We solve the problem in Example 1.2, Page 301, by the simplex method; the first tableau is the
initial tableau of the dual maximum problem:
I ; : I : :I:)
Pl
0
-10 -12
-12
0
0
p2
p3
0
12
PZ
0
-4
12
(ii)
0
Remark:
0
9
1
1
5
13
The rows and columns of the tableaux in the simplex method are labeled so
that we can obtain the optimum solution of the maximum problem from the
terminal tableau. However, if only the optimum solution of the minimum
problem is required, then we can dispense with the labels. For example, the
labels in the preceding example were not needed.
LI N E A R PRO GRAM PVII N G
310
[CHAP. 24
Solved Problems
DUALITY
24.1. Find the dual of each problem.
ii)
Maximize f = 3n: t 5 y
subject to
2x - y + 32
and
X k O ,
+ 22
subject to
6
8
4
+ 6y + 72
x + 2y + 52 2 4
(ii) Minimize g = 22
2x-y+2x
3x+5y+x
x+2y+4x 4
2/20, 220.
and
y”0,
X k O ,
3
1
A
2
220.
In each case form the matrix of coefficients of the given problem with the objective function
as the last row, and then form t h e transpose matrix.
(i)
Coefficient matrix:
3
2 - 1
Transpose:
2
6
1
Dual problem:
Minimize g = 621
3
subject to
6
8
*
and
(ii)
Coefficient matrix:
Transpose:
1
2
+ 8v
+
+
2u v
-u+2v
3u 4v
2
?=
2
3
5
2
uk0, vk0.
Dual problem:
Maximize f = 4 2 ~ 3v
3
subject to
+ +w
u + 2v + 3w
2
2u - v + 5tu
6
f
f
6
7
5u+2v+w
*
7
f
uso, v s o , wso.
and
24.2. Find the dual of each problem.
(i) Maximize f = 5 x
-y
+ 22
7x + 2y - x
subject to
L
8
4
subject to
x-39-28 A
~ x - Y + ~6z 5
and
ii)
x”0,
21’0,
2
6r-4s-
and
rZ0, sk0,
F i r s t multiply the second inequality by -1 to change its sense to
7
2 - 1
[-:-:
5 -1
8
Transpose:
7 -1
3
-j [-: : -;-j
5
7
7
2
tho.
6.
subject to
Transpose:
9
-6
2
7u - v -t 3w
2u+3v-w
- ~ t t2 Gw~
5.
Dual problem:
Maximize f = 72
subject to
7 -2
*
and
+ 5w
1
5
-1
2
2
2
u50, v t o , w t o .
and
F i r s t multiply the second inequality by -1 to change its sense to
9 - 3
L
Dual problem:
Minimize g = 8zc - 4v
8 -4
Coefficient matrix:
3t
X L O .
Coefficient matrix:
iii)
+ 5s + t
9r - 3s + 5t
(ii) Minimize y = 2r
- 2y
9x - Gy
- 3 x + 4y
5x t 3y
f
5
f
z20, y t o .
2
5
1
L I N E A R PROGRAMMING
CHAP. 241
311
SIMPLEX METHOD
24.3. Find the initial tableau of each problem.
(ii) Minimize y = 2r
(i) Maximize f = 5% f 2 y - x
subject to
x - 3y f 4 x
+ +
6
8
5
2 y 62 L
2 ~ + 7 y -f~ 9
3x
and
(i)
X L O , y’0,
subject to
+ 5s + t
9 r t 2s + 4 t
2
67’+3~-2t
6
8
w[
and
,220.
The initial tableau is of the form
with the columns of A labeled Pi.
p
i
Pl pz
p3
0
0
1 1 9
1 1 0
0
0 1 0 1
7 - 1
2
-5-2
(ii) Find the dual maximum problem and then construct the initial tableau as i n (i).
PI
pz
24.4. Find a pivot entry of each of the initial tableaux in the preceding problem.
In each case, (1) select any column with a negative indicator a s the pivotal column,
(2) divide each positive entry in this column into the corresponding entry in the l a s t column, and
(3) choose a s the pivot t h a t e n t r y which yields the smallest quotient in 12).
(i)
We can choose either the first o r second column as the pivotal column.
( a ) If the first column is chosen, f o r m the quotients 8/1, 513 and 912. Since 5/3 is the smallest
number of the three, the entry 3 in the second row is the pivot.
( b ) If the second column is chosen, form the quotients 512 and 917. Since 9/7 is the smaller
of the two numbers, the entry 7 in the third row is the pivot. Note t h a t the quotient
8 / ( - 3 ) is not considered since -3 is not a positive number.
(ii) We can choose either the first o r the second column as the pivotal column. I n the first column,
9 is the pivot since 2/9 i s the smallest of the numbers 2/9, 5/2 and 114. In the second column,
6 is the pivot since 2/6 i s the smaller of the numbers 216 and 513.
2
1 1 1
0
0
7
1111
Coefficientmatrix:
Transpose:
2
3
1
Dual problem:
Maximize f = 4u
2
subject to
\2
6
7
ZL
6
Zu-v+fw
5u+2v+w
*
and
24.2.
+ 3v + w
+ 2v + 32u 2
U h O ,
6
7
6
f
?.‘so, w“0.
Find the dual of each problem.
(i)
Maximize f = 5% - 3
‘ix + 2y - x
subject to
(ii) Minimize y = 2 r + 5s + t
8
x-3y-22 1 4
3 x - y + 6 ~6 5
and
(i)
+ 22
+
subject to
_L
9 r - 3s 5t
6r - 4s - 3t
and
T
~
O
First multiply the second inequality by -1 to change its Sense to
Transpose:
7 - 1
3
5 -1
8 -4
5
Coefficient matrix:
+
u”0, vso,
Transpose:
7
9
-6
2
7
-2
Dual problem:
Maximize f = 7 2
*
and
3
1 obtain
1
1
t and add t h e result 1to the0 second
row
0
z 1to 6
1fj
z1 s
4
Multiply ( 1 ) by 7 and add t h e result oto theo lasto row$to obtain
tableau is
Thus the calculated
Pl
p2
W S O .
k.
subject to
Multiply ( 1 ) by
t”0.
Dual problem:
Minimize g = 8u - 4v 3. 5w
subject to
7u- v
3w
5
2 u + 3 v - w 2 -1
-~+2v+6w 2 2
First multiply t h e second inequality by -1 to change its Sense to
5
7
2
6,
and
9 - 3
f
x--”o.
x”0, y ” 0 ,
Coefficient matrix:
2 -1
8
(ii)
s”0,
,
1
- 276
9% - 6y
-32+4y
5x+3y
z”0,
f
5
f
y10.
2
5
1
CHAP. 241
313
L I N E A R PROGRAMMING
24.7. Using the pivot circled in the following tableau, calculate the new tableau and then
find a pivot entry of it.
pi
I
p,
2
3 ' 1
0
0
3
8
1
o I 1 : l
'
0
Divide each entry in t h e pivotal row, i.e. the third row, by the pivot 1 to obtain the (same) row
1
2
0
0
1
3
(1)
Multiply ( 1 ) by -2 and add the result t o the first row to obtain
0 - 1
1
0
-
1
2
Multiply ( 1 ) by -3 and add the result to the second row to obtain
0
0
2
1 - 3
2
Multiply ( 1 ) by 2 and add the result t o the last row to obtain
0 - 6
Thus the new tableau i s
Pl
0
2
0
6
pz
where the third row is labeled P I , the label of the original pivotal column.
The second column of this tableau has the only negative indicator and so must be chosen as
the pivotal column. Since 2/2 is less t h an 3 / 2 , the 2 in the second r o w is the pivot.
24.8. Using the tableau and the pivot obtained in the preceding problem, calculate the
new tableau.
Divide each entry of t h e pivotal row, i.e. the second row, by the piTiot 2 to obtain the row
0
1
0
1 - 3
2
Then the new tableau i s
Pl
1
p2
PZ
Pl
0
0
2
0
0
2
2
-1
4
3 - 7
12
(1)
314
24.9.
LINEAR PROGRAMMING
[CHAP. 21
In each of the following terminal tableaux, find the optimum solution Q = (ql, . . ., q,)
of the minimum problem, the optimum solution P = (pl, . . .,p,) of the maximum
problem, and v = m a x f = m i n g .
Pl
I
:
;
Pl p ,
p2 p3
p3
Pl
*
*
*
*
*
*
*
*
*
*
*
7
1
p3
2
12
*
*
8:
a:
(i)
p3
*
*
*
3
4
(ii)
Recall that: (1) the components of Q a r e the la st entries, respectively, in the columns corresponding t o the slack variables; (2) the component p i of P is 0 if no row is labeled P , or the
last entry in a row labeled P i ; ( 3 ) v is the entry in the lower right hand corner of the tableau.
(i) Q = (7,1, 2), P = (5,0,3), v = 12. Note t h a t the second component of P is 0 since no row
is labeled P2.
(ii) Q = ( 3 , 4 ) , P = (0,0:2), v = 5 . Note t h a t the first two components of P a r e 0 since no row
is labeled P , or P2.
24.10. Solve the following linear programming problem by the simplex method.
Maximize f = 4 x - 2 y - x
subjectto
and
+ +
x-y
3
4
0
y’0,
x”0.
x y x
2x+2y+x
X k O ,
f
A
We obtain the following sequence of tableaux:
Pl
p,
-4
2
Pl
p3
p2
p3
0
(i)
(ii)
Pl
p2
p3
(iii)
Thus, by the terminal tableau, the maximum of f is 2 and occurs at t h e point P = (1,1,0).
(See Problem 23.11, Page 293.)
24.11. Solve the following linear programming problem by the simplex method.
Maximize f = 2x-32j + x
subject to
and
3x + 6 y + x
4x+2y+x
x-y+x
x s o , y’0,
A
6
4
3
2’0.
CHAP. 241
315
LIiYEAR PROGRAMMING
4
1
2
We obtain the following sequence of tabIeaux:
Pl
p,
p3
3
6
1
1
1
0
4
2
1 1 0
1 - 1 0
0
0
3 - 1
-2
p,
Pl
0
1
0
0
0
Pl
6
1
4
1
1
3
0
0
P3
p3
P3
0 -2
0
-1
1
-
0
2
3
3
:i
0-1
4
s3
3
3
1
0
1
3
3
2
3
'
0
0
0
1
0
5
(iii)
Thus, by the terminal tableau, the maximum value of t' is 1013 and occurs at the point
P = (1/3,0,8/3).
24.12. Solve the following linear programming problem by the simplex method.
Minimize g = 6 x + 5 y + 2 x
+ +
subject to
.P
3y 2x -I 5
2x+2y+x 3 2
4,2:-2y+3x '1 -1
and
c r o , y'0,
zro.
We obtain the following sequence of tableaux.
Pl
-6
pz
-2
p3
Pl
0
0
0
0
p,
p3
0
(ii)
(i)
Pl
p3
p2
- . .
Pt
1
2
2
.
2
2
1
2
2
(iii)
Note first t h a t ( I ) IS the tableau of the dual maximum problem. Note also t h a t the third I'OW
of (iii) i s relabeled P , , the label of the pivotal column of the precedlng tabIeau (ii). By the terminal
tableau, the minimum value of g IS 5 and occurs at the point Q = (0,0,5/2).
LINEAR PROGRAMMING WORD PROBLEMS
24.13. A tailor has 80 sq. yd. of cotton material and 120 sq. yd. of wool material. A suit
requires 1 sq. yd. of cotton and 3 sq. yd. of wool, and a dress requires 2 sq. yd. of
each. How many of each garment should the tailor make to maximize his income if
316
[CHAP. 24
LINEAR PROGRAMMING
(i) a suit and dress each sells for $30, (ii) a suit sells for $40 and a dress for $20,
(iii) a suit sells for $30 and a dress for $20?
Construct the following table which summarizes the given facts:
Suit
I
Dress
~
Available
Cotton
1
2
80
Wool
3
2
120
(ii) Here the objective function is g = 4 0 z + 2 0 y .
I
Evaluate g a t the corner points:
g ( 0 ) = 0, g ( A ) = 800, g(B)= 1400, g ( C ) = 1600
The maximum value of g occurs a t C = (40,O). Thus he should make 40 suits (and no
dresses), for which he will receive $1600.
(iii) Here the objective function is
It
= 3@x+20y. Evaluate h at the corner points:
h ( 0 ) = 0, h ( A ) = 800, h(B) = 1200, h ( C ) = 1200
The maximum value of h occurs a t both B = (20,30) and C = (40,O). Thus the tailor should
make either 20 suits and 30 dresses o r only 40 suits; in either case he will receive $1200.
(Actually, any point on the line segment joining B and C is an optimum solution.)
24.14. A truck rental company has two types of trucks; type A has 20 ft3 of refrigerated
space and 40 f t 3 of non-refrigerated, and type B has 30 ft3 each of refrigerated and
non-refrigerated space. A food plant must ship 900 f t 3 of refrigerated produce and
1200 f t 3 of non-refrigerated produce. How many trucks of each type should the
plant rent so as to minimize cost if (i) truck A rents for 30e per mile and B for 40@
per mile, (ii) truck A rents for 30c per mile and B for 50C per mile, (iii) truck A
rents for 2OC per mile and B for 30c per mile?
Construct the following table which summarizes the given facts:
Type A
Type B
Requirements
Non-ref rigerated
1200
Let x denote the number of trucks of type A and y the number of type B t h a t the plant rents.
Then x and y a r e subject to the following conditions:
20x
+ 30y
2
900,
40%
+ 30y
5
1200,
x
2
0,
0
CHAP. 241
317
LINEAR PROGRAMMING
The set of feasible solutions is drawn on the right; its
corner points a r e
40
P = (0,40), Q = (15, 20), R = (45,O)
(i)
Here t h e objective function to minimize is f =
30s 4011. Evaluate f at the corner points:
+
f ( P ) = f ( 0 , 4 0 ) = 1600, f ( Q ) = f(15,ZO) = 1250,
f ( R )= f(45,O) = 1350
30
20
10
The minimum value of f occurs a t Q = (15,ZO).
Thus the plant should r en t 15 trucks of type A and
20 of B at a total cost of $12.50 per mile.
(ii) Here the objective function is
g
10
20
30
40
50
= 3 0 x f 5 0 y . Evaluate g a t the corner points:
g(P)= 2000, g(&) = 1450, g ( R ) = 1350
occurs at R = (45,O). Thus the plant should re nt 45 trucks of type A
The minimum value of g
(and none of type B ) at a total cost of $13.50 per mile.
(iii) Here the objective function is h = 202
+ 30y.
Evaluate h at the corner points:
h(P) = 1200, h(Q) = 900, h(R) = 900
The minimum value of h occurs at both Q = (15,ZO) and R = (45,O). Thus the plant can
r e n t either 15 trucks of type A and 20 of B o r only 45 of type A a t a total cost of $9.00
per mile.
24.15. A company owns two mines: mine A produces 1 ton of high grade ore, 3 tons of
medium grade ore and 5 tons of low grade ore each day; and mine B produces 2 tons
of each of the three grades of ore each day. The company needs 80 tons of high
grade ore, 160 tons of medium grade ore and 200 tons of low grade ore. How many
days should each mine be operated if it costs $200 per day to work each mine?
1
Construct the following table which summarizes the given facts:
Mine B
Requirements
High
Medium
grade
g r ad e
Low g r ad e
200
Let x denote the number of days mine A is open and y the number of days B is open. Then
x and y a r e subject to t h e following conditions:
x 2y
80, 3% 2g 5 160, 52 2y 2 200, 2 5 0 , 2( 2 0
+
+
+
We wish t o minimize t h e function g = 2002+200y subject to the above conditions.
Method 1.
Draw the set of feasible solutions as on the’right.
The corner points are: p = (0, loo), Q = (20,50), R =
(40,20), S = (80,O).
Evaluate the objective function g = 200%4-2 0 0 ~at
t h e corner points: g ( P ) = 20,000, g ( Q ) = 14,000, g(R)=
12,000, g(S) = 16,000. The minimum value of g occurs
at R = (40,ZO). Thus t h e company should keep mine A
open 40 days and mine B open 20 days f or a total cost
of $12,000.
Method 2.
Consider t h e dual problem:
+ 160v + 2OOw
u + 3w + 5w
200
2u + 2w + 2w
200
Maximize f = 80%
subject to
6
and
u20,w+o,
W’O.
318
[CHAP. 24
L I N E A R PROGRAMMING
Using the simplex method we obtain the following sequence of tableaux:
1
-80 -120
0
r-a
40
1
40
20
1
1
I
0
;
100
0
40
8,000
I
12,000
(iii)
Thus the minimum value of g is 12,000 and occurs a t the point (40, 20). (Note t h a t labels were
not necessary since we were only solving the minimum problem.)
24.16. Suppose, in the preceding problem, the production cost at mine A is $250 per day
and at mine B is $150 per day. How should the production schedule be changed?
+
In this case, the objective function to minimize is g = 2502 150y subject to the same set
of feasible solutions, i.e. conditions. Evaluate g at the corner points of the set of feasible solutions:
g ( P ) = g(0,lOO) = 15,000, g ( Q ) = g(20, 50) = 12,500, g ( R ) = g(40,20) = 13,000,
g ( S ) = g(80,O) = 20,000
The minimum value of g occurs at Q = (20,50). Thus the company should now keep mine A open
20 days and mine B open 50 days f o r a total cost of $12,500.
Supplementary Problems
DUALITY
24.17.
Find the dual of each problem.
(i) hlaximize
subject to
and
f =
- 2y
52
+ 32
+ 22
2%+ 7y + 52
32
-
4y
2’0,
yA0,
(ii) Minimize
f
9
f
11
subject to
32+9y+x
2 2 0 .
and
24.18.
+ 3y + 72
6% - 2y + 6x
3
2 2 + 3y - 4x ’ -2
g = 122
’8
2 2 0 , 2/20, x”0.
Find the dual of each problem.
+ 216 - 2x
(i) Maximize f = 32
subject to
42
+ 3y
5% - 2y
-
22
f
11
+
72
2
2
f
15
x+y+3z
and
2’0,
(ii)
y’0,
Minimize
subject to
+ 7s + 2t
7r + 3s + 5 t
9
g =
T
4r -i- 8s -
t
10
2’0.
SIMPLEX METHOD
24.19.
Find the initial tableau and all possible pivot entries f o r each of the problems given in Problem 24.17.
CHAP. 241
24.20.
Calculate the new tableau from each tableau and circled pivot.
0)
pi
p,
-3-4
2
0
1
0
6
0
0
1
2 1 0
0
0 1 0
3
4
p,
p,
p,
Pl
IT
:I
$
*
*
$
*
*
*
*
3
1
Solve by the simplex method:
2
5
9
Maximize f = 9z
subject t o
and
24.23.
PI
I n each of the following terminal tableaux, find the optimum solution Q = (q,, . . . , q k ) of the
minimum problem, the optimum solution P = (pl, . . ., p,)
of the maximum problem, and
ZI = max f = rnin g.
P z I ;
21.22.
(ii)
p,
5 - 5
24.21.
319
L I N E A R PROGRAMMING
Solve by the simplex method:
+ 2y + 52
+ 3y - 5 x
y + 3x
3x + y - 2x
2s
2.1:
-
X’O,
y’0,
12
f
6
f
3
2
2’0.
+ 5y + x
+y +x 3
2 2 + 2y - ’ 2
x + 2 y ’4
Minimize
g = 4x
subject t o
.?:
1
2
and
2’0,
y’0,
x 10.
LINEAR PROGRAMMING WORD PROBLEMS
24.24. A carpenter has 90, 80 and 50 running feet of plywood, pine and birch, respectively. Product A
requires 2 , 1 and 1 running feet of plywood, pine and birch respectively: and product B requires
1, 2 and 1 running feet of plywood, pine and birch respectively. ti) If A sells f o r $12 and B sells
f o r $10, how many of each should he make to obtain the maximum gross income? (ii) If B sells
f o r $12 and A sells f o r $10, how many of each should he make?
24.25.
Suppose t h a t 8, 1 2 and 9 units of proteins, carbohydrates and f a t s respectively a r e t h e minimum
weekly requirements f o r a person. Food A contains 2, 6 and 1 units of proteins, carbohydrates
and f a t s respectively per pound; and food B contains 1, 1 and 3 units respectively per pound.
(i) If A costs 85c per pound and B costs 40r per pound, how many pounds of each should he buy
per week t o minimize his cost and still meet his minimum requirements? (ii) If the price of A
drops to 75( per pound while B still sells a t do6 per pound, how many pounds of each should he
then buy?
24.26.
A baker has 150. 90 and 150 units of ingredients A , B and C respectively. A loaf of bread requires
1, 1 and 2 units of A , B and C respectively; and a cake requires 5 , 2 and 1 units of A , B and C
respectively. ( i ) If a loaf of bread sells f o r 35C an$ a cake f o r 80<,how many of each should he
bake and what would his gross income be? (ii) If t h e price of bread is increased to 45C per loaf
and a cake still sells f o r 80$, how many of each should he then bake and what would be the
gross income?
24.27.
A manufacturer h a s 240, 370 and 180 pounds of wood, plastic and steel respectively. Product A
requires 1, 3 and 2 pounds of wood, plastic and steel respectively; and product B requires 3, 4 and
1 pounds of wood, plastic and steel respectively. (i) If A sells f o r $4 and B for $6, how many
of each should be made to obtain the maximum gross income? (ii) If the price of B drops t o $5
and A remains a t $4, how many of each should then be made t o obtain the maximum gross income?
320
L I K E A R PROGRAMMIKG
[CHAP. 24
24.28. An oil company requires 9,000, 12,000 and 26,000 barrels of high, medium, and low grade oil,
respectively. It owns two oil refineries, say A and B. A produces 100, 300 and 400 barrels of high,
medium and low grade oil respectively per day; and B produces 200, 100 and 300 barrels of high,
medium and low grade oil respectively per day. (i) If each refinery costs $200 per day t o operate,
how many days should each be r u n to minimize the cost and meet the requirements? (ii) If the
cost a t A rises to $300 per day and B remains at $200 per day, how many days should each be
operated?
MISCELLANEOUS PROBLEMS
24.29. Solve Problem 24.16 by t h e simplex method.
24.30.
Show t h a t the maximum linear programming problem (I), Page 301, is equivalent t o the problem
(Z), Page 304.
Answers to Supplementary Problems
24.17. (i) Minimize
g = 9u
subject to
3u
+ llv
+ 2v
2
-4u+7v
22
2 u + 5v
24.18. (i) Minimize
subject to
and
(i)
P,
g = llu
4 u - 5v
3u 2v
-2U - 7V
v
GZL
-221
2a
+ 15w
+ to
’0,
+ 8w
+ 2v + 3w
+ 3v + 9w
57L - 4v -t w
and
+ + to
+ 3ZC+
u 5 0,
P,
-
.f = 3u - 2v
subject t o
5
-2
3
u s o , V’O.
and
24.19.
2
(ii) Maximize
2
3
2
5
-2
2
24. 2
(ii) Maximize
7.7: - 4y
32 - 8y
5x
and
P3
2
0, w
5
0.
f = 9% - 1Oy
subject t o
zc 2 0.
0, v
f
12
3
7
f
f
+y
L_
f
6
1
7
2
s 2 0 , y’0.
(ii)
P I
P,
P,
0
-5
2 - 3
0
0
0
24.20.
P,
24.21. (i) Q = (3,1,5), P = ( & , & , O ) ,
24.22.
v = 9.
(ii) Q = (3,1), P = ( 0 , 0 , 7 , 0 ) , w = 11.
The rnaximuin value of f is 49 and occurs at P = (0,12,5).
24.23. The minimum value of g is 11 and occurs at the point Q = ( 0 , 2 , 1 ) .
24.24.
(i) 40 of A and 10 of B f o r $580.
24.25.
(i) 1 of A and 6 of B f o r $3.26.
24.26.
(i) 50 loaves of bread and 20 cakes for $33.50.
24.27.
(i) 30 of A and 70 of B f o r $540.
24.28.
(i) A r u n 50 days and B r u n 20 days.
iii) 20 of A and 30 of B f o r $560.
(ii) 3 of A and 2 of B f o r $3.05.
(ii) 70 loaves of bread and 10 cakes for $39.50.
(ii) 70 of A and 40 of B f o r $480.
(ii)
A r u n 20 days and B r u n 60 days.
Chapter 25
Theory of Games
INTRODUCTION TO MATRIX GAMES
Every matrix A defines a game as follows:
(1) There are two players, one called R and the other C.
(2) A p l a y of the game consists of R’s choosing a row of A and, simultaneously, C’s
choosing a column of A .
(3) After each play of the game, R receives from C an amount equal to the entry
in the chosen row and C O ~ L I I I ~ I Ia, negative entry denoting a payment from R to C.
Example 1.1.
Consider the following matrix games:
I
:
[
(i)
(ii)
I n the first game, if R consistently plays the first row, hoping to win a n amount 3 or 4, then C can play
the second column and YO win a n amount 1. However, if C consistently plays the second column then
R can play the second row and so win a n amount 1. Thus we see t h a t if either player should consistently
play a particular row or column, the other player can take advantage of t h a t fact.
Now in the second game, player R can guarantee winning 1 or more by consistently playing row 2.
Player C can then minimize his loss by playing column 1. Thus in this game i t is “best” if R plays a
given row and C plays a given column.
On the other hand, suppose a competitive situation is as follows:
(i) there a r e two people (or: organizations, companies, countries,
(ii) each person has a finite number of courses of action;
(iii) the two people choose a course of action simultaneously;
. . .);
(iv) for each pair (i, j ) of choices, there is a payment b , , from one given player to
the other.
Then the above competitive situation is equivalent to the matrix game determined by the
matrix B = ( b l J ) .
Example 1.2.
Each of the two players R and C sin~ultaneouslyshows 1 or 2 fingers. If the sum of t h e fingers
shown is even, R wins the sum f r o m C ; if the sum is odd, R loses the sum t o C. The matrix of this game
is a s follows:
Player C
1
2
Player R
2
We shall show later (Example 3.1) t h a t this game is “favorable” t o player C.
321
322
THEORY O F GAMES
[CHAP. 25
We remark that matrix games and the above competitive games which can be represented as matrix games are termed two-person zero sum games. The zero sum corresponds
to the fact that the sum of the winnings and losses of the players after each play is zero.
There also exists a theory f o r n-person games and non-zero sum games but that lies beyond
the scope of this text.
STRATEGIES
By a strategll for R in a matrix game, we mean a decision by R to play the various
rows with a given probability distribution, say to play row 1 with probability pl, to play
row 2 with probability p 2 , . . . . This strategy for R is formally denoted by the probability
vector p = (pl, p,, . . . ,pk). F o r example, if the matrix game has two rows and R tosses a
coin to decide which row to play, then his strategy is the probability vector p =
3).
(a,
Similarly, by a s t r a t e g y for C we mean a decision by C t o play the various columns
with a given probability distribution, say to play column 1 with probability q,, to play
column 2 with probability qs, . . . . This strategy f o r C is formally denoted by the probability vector (I = (q,,qs,. . .,q,).
A strategy which contains a 1 as a component and 0 everywhere else, i.e. where R
decides to play consistently a given row or C decides to play consistently a given column,
is called a p u r e s t r a t e g y ; otherwise it is called a m i x e d s t r a t e g y .
OPTIMUM STRATEGIES AND THE VALUE OF A GAME
F o r simplicity, we shall first consider a 2 X 2 matrix game, say
A
=
m
Suppose player R adopts the strategy 11 = ( p l , p 2 ) and player C adopts the strategy
q = (q,,q,). Then R plays row 1 with probability pl and C plays column 1 with probability ql, and so the entry a will occur with probability p,q,. Similarly, b will occur with
probability p,qz, c with probability pLql and d with probability p 2 q 2 . Hence the expected
winning (see Page 220) of R is
where qt, At and p t denote the transpose of g, A and p , respectively
Similarly, if A is any k x n z matrix game and player R adopts the strategy p =
(p,, . . .,p k ) and C adopts the strategy Q = (ql, . . ,qJ, then the expected winnings of R
is given by
E ( p ,q ) = p A qt = q A t p t
.
Now suppose that v1 is the maximum number f o r which there is a strategy pa for R
such that the expected winnings of II is at least v1 f o r any strategy adopted by C:
E(po,q) 2
v1,
for every strategy q of C
(11
323
THEORY O F G A M E S
CHAP. 251
Since V I is the maximum of such numbers. the strategy p" is in a certain sense a "best"
strategy that R can adopt; we call p o an optinzunz strategy for R. We show in Problem 25.9,
Page 233, that ( I ) is equivalent to the condition
p"A
2? ( V l , 2'1,
. . ., Vl)
On the other hand, suppose v2 is the minimum number for which there is a strategy qo
for player C such that the expected loss of C is not greater than l j 2 f o r any strategy
adopted by R:
E( p, q") 4 vL, for every strategy 11 f o r R
(2)
Then the strategy q0 is in a certain sense a "best" strategy that C can adopt; we call
qo a n optimum s t m t e y y for C . Similarly. ( 2 ) can be shown to be equivalent to the condition
qo A'
6
( ~ r 2'2,
,
. . ., 71.)
The fundamental theorem on game theory by von Neumann states that V I and v z exist f o r
any matrix game and are equal to each other: v = V I = vz. This number u is called the
value of the game, and the game is said t o be f a i r if v=O. We observe that (1) and ( 2 )
imply that v1 6 E(po,qo) 6 v z ; hence by von Xeumann's theorem we have the equality
relation
We summarize and formally define the terms introduced above:
FDefinition:]
In a matrix game A , a strategy p o f o r player R is an optimum strategy, a
strategy qo f o r player C is an optii?zumstrategy, and a number v is the calzie
of the game if P O , qo and 21 are related as follows:
(i) p o A
1
( v , ~ ., . .,v);
(ii) @ A t
L
( v , ~ !. ,. .,v)
If v = 0, the game is said to be f a i r .
We point out that there may be more than one pair of strategies for which (i) and (ii) hold;
that is, a player may have more than one optimum strategy.
We remark that the solution of a matrix game means finding optimum strategies for
the players and the value of the game. Before we give the solution to an arbitrary matrix
game by means of the simplex method developed in the preceding chapter, we shall discuss
two special cases which occur frequently and which offer simple solutions: strictly
determined games and 2 x 2 games.
STRICTLY DETERMINED GAMES
A matrix game is called strictly d e t e i w i i l e d if the matrix has an entry which is a
minimum in its row and a maximum in its column; such a n entry is called a "saddle point".
The following theorem applies.
Theorem 25.1:
Let 2' be a saddle point of a strictly determined game. Then an optimum
strategy f o r R is always t o play the row containing v, a n optimum strategy
f o r C is always to play the column containing o, and v is the value of
the game.
Thus in a strictly determined game a pure strategy f o r each player is an optimum strategy.
Example 2.1.
Consider the following game:
I
I
3
-4
I
0
,
2
1 - 4 1 - 3 1
-1
3
324
THEORY O F GAMES
[CHAP. 25
If we circle the minimum entry in each row and put a square about the maximum e ntry in each column,
we obtain
A
=
The point 1 is both circled and “boxed” and so is a saddle point of A . Thus w = 1 is the value of the game,
and optimum strategies po for R and qo for C are as follows:
pa = (O,l,Oj
and
qo = ( O , O , 1 , O j
T h a t is, R consistently plays row 2, and C consistently plays column 3.
We now show t h at po, q 0 and w do satisfy the required properties to be optimum strategies and the
value of the game:
3
0 -4 -3
(i)
pOA
= (O,l,O)(f
1
3
)!2
= (2,3,1,2j
(1,1,1,1) = (w,v,v,wj
-1
2
We remark that the proof of Theorem 25.1 is similar to the proof above f o r the special case.
2 x 2 MATRIX GAMES
+ll
Consider the matrix game
A =
If A is strictly determined, then the solution is indicated in the preceding section. Thus
we need only consider the case in which A is non-strictly determined. We first state a
useful criterion to determine whether or not a 2 X 2 matrix game is strictly determined.
Lemma 25.2:
The above matrix game A is non-strictly determined if and only if each
of the entries on one of the diagonals is greater than each of the entries
on the other diagonal:
(i) a,d > b and a,d > c ; or (ii) b, c > a and b,c > d.
The following theorem applies.
Theorem 25.3:
Suppose the above matrix game A is non-strictly determined.
Then
p o = (xl,xz) is an optimum strategy for player R, qo = (y,,y2) is a n
optimum strategy for player C, and v is the value of the game where
d-c
x, = a + d - b - c ’
II‘, =
d-b
y1= a+d-b-c’
and
2,=
zJ2
a-b
a+d-b-c
a-c
= a+d-b-c
ad - bc
~ + d - b - ~
325
THEORY O F GAMES
CHAP. 251
Note that the five quotients above have the same denominator, and that the numerator of
the formula f o r v is the determinant of the matrix A .
Example 3.1.
Consider the following matrix game (see Example 1.2):
Using the above formula, we can obtain the value
v
2:
of the game:
-
2 . 4 - (-3).(-3)
=
-1
12
Thus the game is not f a i r and is in favor of the column player C. Optimum strategies p o f o r R and qo
for C a r e as follows: po = (L
1 2 ' 1A)
2
and qo =
2 f 4 - 3 f 3
(&,A).
RECESSIVE ROWS AND COLUMNS
Let A be a matrix game. Suppose A contains a row r asuch that r l L r , for some other
row r,. Recall that r l L r l means that every entry of r z is less than or equal t o the corresponding entry of T,. Then r z is called a yecessive row and r, is said to dominate it.
Clearly, player R would always rather play row r, than row rl since he is guaranteed to
win the same o r a greater amount in every possible play of the game. Accordingly, the
recessive row r? can be omitted from the game.
On the other hand, suppose A contains a column Ca such that c , h c 3 for some other
column cj. Then cl is called a recessive coLumri and c, is said to dominate it. For analogous
reasons player C would always rather play column c, than column el. Hence the recessive
column c, can be omitted from the game.
We emphasize that a recessive row contains numbers smaller than those of another
row, whereas a recessive column contains numbers larger than those of another column.
Example 4.1.
.
Consider the game A =
Note that ( - 5 , - 3 , l )
6
(2,-1,2),
i.e. every entry
-2
in the first row is f the corresponding entry in the second row. Thus the first row is recessive and can
be omitted from the game and the game may be reduced to
1 2 1 - 1
I
2
Now observe t h a t the third column is recessive since each entry is
second column. Thus the game may be reduced to the 2 X 2 game
A*
=
Fl
2
the corresponding entry in the
The solution to A* can be found by using the formulas in Theorem 25.3 and is
Thus the solution to the original game A is
326
THEORY O F GAMES
JCHAP. 25
SOLUTIOS OF A MATRIX GAME BY THE SIMPLEX METHOD
Suppose we are given a matrix game A . We assume that A is non-strictly determined
and does not contain any recessive rows or columns. We obtain the solution to A as follows.
(1) Add a sufficiently large number h. to every entry of A to form, say, the following matrix
game which has only positive entries:
I
A'
1
1
x
1
..
ult'!
nil a22 . . . a2,!(
................
/all
\
ah1
a12
\
*
1
ah!,,1
ak2
(The purpose of this step is to guarantee that the value of the new matrix game is
positive.)
(2) Form the following initial tableau:
PI Pr
*.
P,,,
'
. . 0
0 1 ... 0
a21
a.2
U2r,1
.............................
0 0 - .. 1
a k l
ak2
'
ahnl
a11
a12
. . . al,,, 1 0
1
'
1
* * *
1
' '
-1
-1
...
-1 I 0 0
~~
* ' .
0
0
~
~
This tableau corresponds to the linear programming problem:
Maximize f =
subject t o
and
cc'i
- xz + - . + T , ( ~
+ - . + ui,,,Tl,l 1
+ .+
1
............................
+
+ +
1
+
ali?-l
~~12x2
*
f
ar13'1 t a22xz
* *
ahi.1'1
nk2x2
* '
s.1'0,
X S S O ,
Cllr,!5m
'
...,
6
ahnis)'itr
X,,,'O
(3) Solve the above linear programming problem by the simplex method. Let P be the
optimum solution to the maximum problem, Q the optimum solution to the dual minimum
problem, and 2' = f ( P )= g ( Q ) the entry in the lower right hand corner of the
terminal tableau. Set
1
1
1
(i) p o = -Q
(ii) qo = --+P
-k
(iii) 2: = v*
V*
'
Then p o is an optimum strategy f o r player R in game A , qo is an optimum strategy f o r
player C, and c is the value of the game.
We remark that the games ,4 and A have the same optimum strategies for their
respective players and that their values differ by the added constant k ; that is, p o is an
optimum strategy f o r player R in game A * , 4" is an optimum strategy for player C in
game A*, and 2' 4h. = 1/v* is the value of the matrix game A
4 .
Example 5.1.
Consider the mati I X game
THEORY O F GAMES
CHAP. 25!
327
Add 3 to each entry of A t o form the matrix game
The initial simplex tableau and successive tableaux a r e as follows:
(iii)
Thus P = C O , & , i ) , Q =
(i,d)
and z- =
4; hence
l/v* = 81'3.
Then for the original game A ,
p o = (8/3)& = (2/3, 1/3) is a n optimum strategy f o r player R,
qo = ( 8 / 3 ) P = (0, 1/3, 2/3) is a n optimum strategy f o r player C,
v = 8/3 - 3 = -1/3
is the value of the game.
Observe t h a t the game is favorable t o the column player C.
ExampIe 5.2.
Consider the matrix game A shown below:
A
=
;
-1
-1
Adding 2 t o each entry of A , we obtain the matrix game A'
on the right.
The initial tableau and t h e successive tableaux a r e as follows:
P,
P1
P,
P,
4
3
1
1
0
0
1
4
1
0
1
0 1 1
0
0
0
-1
-1
-1
~~
(iii)
1
0
P,
P,
328
THEORY O F GAMES
[CHAP. 2 3
Thus P = (1/22,3/22,9/22), Q = (3/22,1/22,9/22) and v* = 13/22; hence l/w* = 22/13.
f o r the original game A ,
po = (22/13)Q = (3/13, 1/13, 9/13)
is a n optimum strategy for R ,
qo
= (22/13)P = (1/13, 3/13, 9/13)
is a n optimum strategy f o r C,
w
= 22/13 - 2 = -4/13
Accordingly,
is the value of the game.
2 x m AND m x 2 MATRIX GAMES
Consider the 2 x m matrix game
Suppose we apply the simplex method t o the above game. Note that the optimum solution
of the corresponding maximum linear programming problem will contain at most two
non-zero components; hence so will an optimum strategy f o r the column player C. In
other words, the 2 x m matrix game A can be reduced to a 2 x 2 game A*.
I n the next example we show a way of solving such a 2 x nz game A and, in particular,
how to reduce the game to a 2 x 2 game A X . A similar method holds f o r any m X 2 matrix
game.
Example 6.1.
Consider the following matrix game (see Example 5.1):
-1
3
A
Let
p0
=
(2,1-2)
=
1
(-2
-:)
denote an optimum strategy for R and let v be the value of the game. Then
or
32 - 2 ( 1 - 2 )
5
w
-x+(l-x)
5
v
f
v
O-(l-x)
L 5x-2
V
or
v
f
V
L
Let us plot the three linear inequalities in w and x on a
coordinate plane where 0 x 1, as shown in the diagram.
The shaded region depotes all possible values of v; however,
R chooses his optimum strategy so t h a t v is a maximum.
This occurs at the point W which is the intersection of
v = -2x+ 1 and u = x - 1; t h a t is, where x = Q and
v = -&. Thus t h e value of the game is -$ and the optimum
strategy f o r R is ($,Q).
f
+1
-22
2-1
f
To find a n optimum strategy f o r C, we can set qo =
(y, z, 1 - y - x )
and use the condition qoAt 5 (w,v, v) =
-Q, -Q). However, a simpler way is as follows. Observe that t h e two lines intersecting at W a r e determined
by only the second and third columns of A . Omit the other
column and reduce the game t o t h e 2 X 2 niatrix game
4
2
1
(-4,
A*
=
El
-1
-2
-3
-4
By Theorem 25.3, the solution t o A* i s
V'
=
-9, po'
=
(3, Q)
and
q0'
= (I 2 )
3' 3
CHAP. 251
329
THEO.RY O F G A M E S
An optimum strategy f o r the column player C in the matrix game A is then
v = v’ and po = PO‘, as expected.)
q0
= (O,&,$).
(Note tha t
Remark: In the diagram above, the three hounding lines intersect the line E = 0 at -2, 1 and -1
respectively, and the line r = l at 3, -1 and 0, respectively. Observe tha t these numbers correspond t o
the components of the r o w of the original matrix A .
SUMMARY
I n finding a solution to a matrix game, the reader should follow the following steps:
(1)Test to see if the game is strictly determined.
(2) Eliminate all recessive rows and columns.
(3) I n the case of a 2 x 2 game, use the formulas in Theorem 25.3.
(4) In the case of a 2
Example 6.1.
X
m game or an ni X 2 game, reduce the game to a 2 X 2 game as in
( 5 ) Use the simplex method in all other cases.
Solved Problems
MATRIX GAMES
25.1. Which of the following games are strictly determined? For the strictly determined
games, find the value v of the game and find an optimum strategy po for the row
player R and an optimum strategy qo for the column player C.
1
I
-8
0 1
-1
-3
(i)
3 1 - 1
7
3
4
(iii)
(ii)
In each case, circle the minimum entries in each row and put a box around the maximum
entries in each column as follows:
@
1
(i)
(i)
mi01 1 I
-3
(iii)
(ii)
The -1 in row 2 and column 2 is a saddle point; hence
z1
= -1,
p0
= (O,l,O),
q0
= (0,1,0).
(ii) There is no saddle point; hence the game is non-strictly determined. Note t h a t both 3’s in
the first column a r e “boxed”, since both a r e maximum entries in the column.
(iii) The 2 in row 2 and column 3 is a saddle point; hence v = 2, p o = (0,1, O),
QO
= (0,0,1,0).
330
25.2.
THEORY O F GAMES
[CHAP. 25
Find the solution to each of the following 2 x 2 games.
(ii)
po
(iv)
If the game is non-strictly determined, use the following formulas of Theorem 25.3, where
( y l , x2) and 40 = (ul, ur).
a-b
d - b
d -c
XI = 0 ‘ d x2 = a + c l - b - c ’
a+d-b-c’
b -c’
Llq
(ii
(iii)
a-e
= a f d - b - ~ ’
.u=
The game is non-strictly determined. F i r s t compute
ad - bc
a + d - - b - ~
(L
td -b -c = 2
+ 2 - 0 + 1 = 5.
Then
That is, (g, +) is an optimum strategy f o r R and [g,Q I is a n optimum strategy for C. The
2.2-0.(-1)
- L
value of the game is ti =
5
- 5 . Observe t h a t the game I S not Pair and is in
favor of t h e TOW player R.
lii)
This game is strictly determined with saddle point -1. Thus w = -1; and p @ = ( 1 , O )
qo = ( 0 , l), i.e. R should always play row 1 and C should always play column 2.
and
(iii) The game is non-strictly determined, F i r s t compute a - d - b - c = 1- 5 43 ’1 = 10. Then
.yl = 8/10 = .8,
x2 = . 2 , y1 = .6, yz = .4. Tha t is, (A,.2) is a n optimum strategy for R and
1.5- (-3)*(-1)
- .2.
i.6, .4) is an optimum strategy f o r C. The value of the game is t =
10
The game is not f ai r and, since v is positive, the game is in favor of the row player R.
+
iivj The game is non-strictly determined. First compute a d - b - c = -3 - 3 - 2 - 5 = -13.
Then x1 = 5/13, y 2 = 8/13, y 1 = 8/13, y, = 5/13. Thus (&,&) is a n optimum strategy f o r R
(-3).(-3)-2.5
- 1
and
is a n optimum strategy f o r C. The value of the game is L’ =
- 13
13 *
The game is not f ai r and is in favor of the row player R.
(A,&)
25.3. Find a solution to the matrix game
First circle each row minimum and box each column maximum:
Observe t h at there is no saddle point and so the game is non-strictly determined. We now te st f o r
recessive rows and columns. Note t h a t the third column is recessive since each entry is larger
than the corresponding entry in the second column; hence the third column can be omitted to
obtain the game
-5
CHAP. 251
331
THEORY OF G A M E S
Note t h a t the second row is recessive since each entry is smaller than the corresponding entry in
the first row. Thus the second row can be omitted to obtain the game
I
1 1 - 3 1
Using the formulas in Theorem 25.3, the solution to the above 2
v' = -13/11,
= (2
11' 5
1 1 ) (Io' =
,o'
t
2 game is
- f i
(11'
2 -)
11
Thus a solution to the original game is
v = -13/11,
25.4.
pa =
Find a solution t o the matrix game
-
(fi.O,,),
4
(I"
-1
A =
1
Set p o = (:r, 1 - x ) .
Then p o A
(-;-;-;-;)
(x,1-2:)
or
-2%
+1
2
=
(5,h. 0 )
5
-3
1
-2
-2
is equivalent to
5 iil, I ' , c , i ' j
(V,V,C,")
v
8 ~ - 32
3x-2
%
v
-7x+5
5
v
Plot t h e above four linear inequalities a s shown in the
diagram. Note t h a t t h e maximum L' satisfying the
inequalities occurs a t t h e intersection of the lines
-2x+ 1 = 2; and 3 ~ - 2 = v, t h a t is, a t the point
x=
and L' = -L.
Thus v = -8 is the value of the
game, and p o = (2, 8) is a n optimum strategy f o r
player R.
Since the two lines determining L* come from the
first and third columns of t h e matrix game A , omit the
other columns t o reduce A t o the 2 X 2 matrix game
A :1 -
F
l
.
1
1
A solution of A " , by Theorem
-2
(g, 0 , b, 0 )
Thus q 0 =
in game A .
25.5.
I
is a n optimum strategy f o r player C
Note t h a t v = v' and po 1P O ' , as expected.)
Find a solution t o the matrix game
or
A =
47-1
-3.r
-1
23.
El.
4
v
f
21
f
v
5
332
THEORY O F GAMES
[CHAP. 25
Plot the above three inequalities (where O‘x‘l)
as
shown in the diagram. W e seek the miximum v satisfying the given inequalities; note t h at i t occurs a t the intersection of the lines -3% 1 = v and 2x = v, i.e. at the
and v = %. Thus v =
is the value of
point r = 1
5
the game and qo = (&,Q) i s an optimum strategy f o r
player C.
Since the two lines determining v come from the
second and third rows of A , omit the other row to obtain
+
the 2
X
+
Fl.
2 matrix game A:: =
A solution
of A’?,by Theorem 23.5, is
3’ = 2 ,
5
= (2
5 ) qo’
5’5
’
= (I,
+)
5
mi.
Thus po = (0,$, 8) is a n optimum strategy for player R .
(Note t h at u = v’ and qo = q o ’ , as expected.)
25.6.
Find a solution to the matrix game
-1
The reader should first verify t h a t the game is non-strictly determined and tha t there a re no
recessive rows or columns. We find a solution by the simplex method. F i r s t add 4 to each entry
t o obtain the game
3
4
The initial and subsequent tableaux follow:
Pl
p2
Pl
p3
pz
p3
4
(i)
(ii)
(iii)
(Renzayk: Since the l as t row in (iii) has no negative indicators, the tableau is a terminal one
and so all of its entries need not be computed.)
Thus P = (L
li’ 5
17’ O ) , Q = (0’ 1 7 ’ ”)
17
and ~ 4 :
hence 1/~:,: !?, Accordingly, po = ( $ ) Q =
9 3
( 0 ,i
,~
40 = ( )
y ) P,
= (i
5
0
)
and
2: = 17 - 4 = -5
5’5’
5
5’
=&;
25.7. Two players R and C simultaneously show 2 or 3 fingers. If the sum of the fingers
shown is even, then R wins the sum from C; if the sum is odd, then R loses the sum
to C. Find optimum strategies f o r the players and to whom the game is favorable.
THEORY OF GAMES
CHAP. 251
The matrix of the game is
2
333
3
The game is non-strictly determined; hence the formulas in Theorem 25.3, Page 234, apply. First
compute a d - b - c = 4 6 5 5 = 20. Then
+
+ + +
(g,
&), is an optimum strategy f o r R and (&,%) is a n optimum strategy f o r C. The value
T h a t is,
= 4 . 6 - - ( - 5 )20' ( - 5 )
- -L.
of the game is
2 0 , and, since i t is negative, the game is favorable t o C .
25.8. Two players R and C match pennies.
If the coins match, then R wins; if the coins
do not match then C wins. Determine optimum strategies for the players and the
value of the game.
The matrix of the game is
H
T
The game is non-strictly determined. Using the formulas in Theorem 25.3, Page 234, we obtain
Thus (g, &) is an optimum strategy for both R and C. The value of
x1 = x, = y1 = y2 =
=
1.1 - ( - l ) . ( - i )
= 0: hence the game is fair.
the game is
4
4.
THEOREMS
25.9. Let p o be a given strategy for player R in a matrix game A . Show that the following
conditions are equivalent:
v, for every strategy q for C; (ii) p o A 2 (v,v, . . . , v).
(i) p o A qt
Assume that iii) holds. Then, for any strategy q f o r C,
pOA qt
2
(v,v, . . ., v) qt = v
On the other hand, assume t h a t (i) holds and t h a t pOA = ( a l , a 2 , . . ., a k ) .
strategy q = (1,0, . . .,O), we have
Choosing the pure
pOA qt = (al,aL, . . ., ak)qt = a , s v
Similarly, a2 2 w, . . ., ak 5 v. I n other words, @ A
2
(v, v, . . ., w).
25.10. Let p = ( a l , a r ,. . ., ak) and p* = ( b l ,bz, . . ., bk) be probability vectors. Show that
each point on the line segment joining p and p * , i.e. t p -t (1- t)pa where 0 g t 4 1,
is also a probability vector.
Observe that
tp
+ (1- t ) b z , . . ., t a L + (1- t ) b k )
Since t , 1 - f, the a , and the b, are all non-negative, ta, + (1- t ) b , is also non-negative
every i. Furthermore,
f a , + (1- t ) b , + ta2 + (1- t ) b , +
+ tak + (1- t)bh
. . + (1- t ) b k
= ta, + ta, + . + tah + (1 t ) b , + (1 t ) b z
+ (1- t)p"
= (tal
+ (1 - t ) b l ,
ta2
* ' '
*.
-
-
*
= t ( a , + a 2 + ~ ~ ~ + a ~ ) + ( 1 - t ) ( b l + b , + ~ ~=~ +
t +b (~ l)- t ) = 1
for
334
T H E O R Y O F GAMES
'CHAP. 2 6
25.11. Let p o and p be optimum strategies for player R in a matrix game A with value c .
Show that p' = t p o + (1- t ) p , where 0 t 6 1, is also an optimum strategy for R.
p A
By the pieceding problem, p is a probability vector, hence we need only show t h a t
(L,I ,
,v). We a r e given t h a t
pOA
Thus
(L,
I!,
.v)
p A
and
p'A == (tpo - i l - t ) p ) A
tpOA
2
4
+ (1
t ) A~
-
t ( v , v ,. . , l
= (v,w,. .
,v)
( L , L,
(1 - t ) ( C , w, . . ., w )
)
,L)
25.12. Show that if a and b are saddle points of a matrix A , then a = b.
If n and b a r e in the sanie row, then each is a minimum of the row and so a = ti. Similarly
if u and b a r e in the same column, then n and b a r e maxima of the column and so a = b.
Now assume t h a t a and b appear in different columns and different rows as illustrated below:
*:
:,:
::i
:::
\:,:
\*
*
:L
*
::.
.,.
.'.
Q
r;;l
:::
f
:,:
pJ
:.:
*
*
Q
k.;
Q
*
*
@
:
:
p
J
:.k
*
*\
:I
:,:
:.:
.::
::
:i
i'
*I
i::
Since a is a saddle point, a G d and c f a ; and since b is a saddle point, b s c and d E b.
Thus
a ' d s b ' c 5 a
Accordingly, the equality relation holds above:
25.13. Show that the game A =
a = d = b = c.
is strictly determined.
Observe t h a t each a is a minimum of the first row. If a
and if u 1 c then the second a is a saddle point.
b then the first a is a saddle point,
Thus n e are left with the case t h a t b > a and c > a, i.e. where b and c a r e maxima of their
respective columns. Now if b f then b is a saddle point, and if c b then c is a saddle point.
(3
Accordingly, in all cases A has a saddle point and so A is strictly determined.
25.14. Prove Lemma 25.2: The matrix game A =
is non-strictly determined
if and only if each of the entries on one of the diagonals is greater than each of the
entries on the other diagonal:
(i) a,d > b and a,d > c; or (ii) b,c > a and b,c > d.
If ( i ) or l i i ) holds, then there is no saddle point for A and so A is non-strictly determined.
Now suppose A is non-strictly determined.
By t h e preceding problem, a # b . We consider
two cases.
Case (1): a < b.
If a c then a is a saddle point; hence IL < c. We need only show t h a t
and d < c. Suppose d 2 b: then d > c o r else d is a saddle point. But a < c and d > c implies
c is a saddle point. Accordingly, d % b leads to a contradiction, and so d < b. Similarly, d < c.
d
<b
Case ( 2 ) : n
> b.
The proof is similar t o t h e proof in Case (1).
CHAP. 251
THEORY O F GAMES
335
+l+l
25.15. Prove Theorem 25.3: Suppose the matrix game A =
is not strictly
determined. Then
a-c
a-b
a-b
a-c
?Jo = ( a + d - b - c ' a + d - b - c
ad - bc
u = a+d-b-c
are respectively an optimum strategy f o r R, a n optimum strategy for C and the
value of the game.
Set
p0
= (x,1-x)
and q0 = (y,1-y),
+c
( b - d ) ~+ d
or
(a-c)z
2
5
Then pOA
v
v
and
2
(w,w) and qoAf 6 (w,w) is equivalent to
(a- b)y
+b
(c-d)y
4- d
f
t?
f
v
Solving the above inequalities as equations f o r the unknowns 2 , y and v, we obtain the required
result.
We remark t h a t a b - c - d # 0 by the preceding problem, i.e. by Lemma 25.2. We also
note t h a t in the case of a 2 X 2 non-strictly determined matrix game A , we have the following
stronger result: poA = (w,v) and qoAt = ( t i , w),
+
25.16. Let A be a matrix game with optimum strategy p o f o r R, optimum strategy qo f o r C,
and value v ; and let k be a positive number. Show that f o r the game kA, p o is an
optimum strategy for R, qo is an optimum strategy for C, and kv is the value of
the game.
We a r e given t h at pOA
(w,w, . ., c) and qOA (w,W , . . ., c). Then
pO(kA) = k(p0A) 2 k(w,v, . . ., w) = (kv,kv, . . . , kv)
qO(kA)t = qOkAt = k(q0At) f k ( ~ W,, . . ., W ) = (lit],kv, . . ., kw)
,
f
which was to be shown.
Supplementary Problems
MATRIX GAMES
25.17.
Find a solution t o each game, i.e. an optimum strategy po for R , an optimum strategy qo f o r C,
and t h e value of the game.
(9
25.18.
Find a solution to each game.
(4
25.19.
-4
-
-3
-4
Find a solution t o each game.
(ii)
(iii)
-1
4
-2
336
25.20.
THEORY O F GAMES
[CHAP. 25
Find a solution to each game.
(i)
-1
.
Find a solution to the game if iii a < 1, (ii) 1 < a
< 3, (iii) a > 3.
25.21.
Consider the game
25.22.
Each of two players R and C h a s a dime and a quarter. They each show a coin simultaneously.
If the coins ar e the same, R wins C's coin; if the coins a r e different, C wins R's coin. Represent
the game as a matrix game and find a solution.
25.23.
Each of two players R and C has a penny, nickel and dime. They each show a coin simultaneously.
If the total amount of money shown is even, R wins C's coin; if i t is odd, C wins R's coin.
Represent the game as a matrix and find a solution.
THEOREMS
25.24. Suppose every en t r y in a matrix game A is increased by a n amount k. Show tha t the value of
the game also increases by k , but t h a t the optimum strategies remain the same.
25.25.
Show t h at if every entry in a matrix game is positive, then the value of the game is positive.
Answers to Supplementary Problems
(i) p o = (0, l), q0 = (0,l), w = 1; (ii) p o = (l,O), q0 = (0,I), v = 2;
= ( l , O ) , v = 1.
(iii)
25.18.
(i) po = (,7,.3), qo = (.4,.6), w = .2; (ii) p o = (3/4,1/4),
(iiij p o = (7/8,1/8), qo = (5/8,3/8), v = 3/8.
= (5/12,7/12),
25.19.
(i) p o = (5/8,3/8), qo = (3/8,0,5/8), w = 1/8;
(ii) PO = (5/7,2/7), q0 = (5/7,2/7,0), v = 3/7;
(iii) P O = (2/7,0,5/7), q 0 = (4/7,3/7), w = 13/7.
25.21.
(i)
25.17.
p0
can be any strategy,
yo
25.22.
p o = (l,O), q 0 = (O,l), w = 1;
(ii) p o = (0,1), q0 = (0,1), w = a ;
=I;
p o = (5/7, 2 / 7 ) , q' =
5
1
25.23.
qo
= 0.
10
5
10
(i,g), w
; po = ( l O / l l , 0, l / i l ) , yo = (1/2, 0, 1/2),
-10
j
-10
I
10
w = 0.
v = 1/4;
INDEX
De Morgan’s laws, 11
Degenerate equations, 107
Dependent vectors, 114
Detachment, law of, 26
Determinants, 127
Disjoint sets, 37
Disjunction, 2
exclusive, 3, 1 5
Distance, 266
Distribution, 218
binomial, 217
Domain of a function, 66
Dot product, 84
Dual problem, 301
Duality theorem, 303
Absolute value, 259
Absorbing states, 240
Algebra of propositions, 11
Algebra of sets, 38
Algorithm of the simplex method, 308
Angle of inclination, 266
Argument, 26
Bayes’ theorem, 207
Biconditional statement, 19
Binomial coefficients, 141
theorem, 142
Binomial distribution, 217
Birthday problem, 188
Bounded sets, 269
Bounding hyperplane, 282
Echelon form, 109
Element, 35
Elementary event, 184
Empty set, 37
Equations, linear, 104
system of, 107
Equiprobable space, 186
Equivalence class, 60
Equivalence relation, 59
Equivalent, logically, 10
Euclidean space, 268
Event, 184
Exclusive disjunction, 3, 1 5
Expected value, 220
of a game, 221
Extreme point, 282
C player, 321
Cartesian product, 50
plane, 266
Cauchy-Schwarz inequality, 278
Cell, 60
Class, 37
equivalence, 60
Closed half space, 282
Closed interval, 258
Co-domain of a function, 66
Coefficients,
binomial, 141
of equations, 106
Collection, 37
Column vector, 81
Combinations, 161
Complement of a set, 37
Components of vectors, 81, 83
Composition function, 68
Compound statements, 1
Conclusion of an argument, 26
Conditional probability, 199
Conditional statement, 18
Conjunction, 1
Consistent equations, 109
Constant, 106
Contradiction, 10
Contrapositive, 19
Converse statement, 19
Convex sets, 281
polygonal, 283
polyhedral, 282
Corner point, 282
Counting, fundamental principle of, 152
Cross-partitions, 162
Factorial, 141
F a i r game, 221, 323
Fallacy, 26
False statement, 1
Family, 37
of lines, 268
Feasible solution, 300
Finite set, 36
Fixed point, 234
Frequency, relative, 184
Function, 66
composition of, 68
identity, 70
inverse, 70
linear, 67, 281
objective, 300
polynomial, 68
337
INDEX
Games, gambling, 221
f a i r , 221
Games, matrix, 321
f a i r , 323
strictly determined, 323
value of, 323
Graph of a function, 67
Half space, 282
plane, 282
Homogeneous equations, 111
Hyperplane, 269
Identity function, 70
Image, 66
Implication, 28
Inclination of a line, 266
Inconsistent equations, 109
inequalities, 285
Independent events, 201
trials, 216
Independent vectors, 114
Inequalities, 258
linear, 259
Infinite set, 36
Initial simplex tableau, 304
Inner product, 84
Intersection of sets, 37
Intervals, 258
Inverse,
function, 70
matrix, 132
relation, 59
statement, 19
Invertible matrix, 131
Joint denial, 16
Line,
in the plane, 267
real, 257
segment, 281
Linear,
equation, 1 0 4
function, 67, 281
inequality, 259
programming, 300
space, 83
Logical implication, 28
Logically equivalent, 10
Mapping, 66
Markov chain, 236
Matrix, 90, 112
addition of, 90
game, 321
multiplication of, 92
non-singular, 133
regular, 234
scalar multiplication of, 91
singular, 133
square, 93
stochastic, 233
Matrix (cont.)
transition, 236
transpose, 94
Member, 35
Minkowski’s inequality, 278
Mixed strategy, 322
Multinoniial coefficients, 144
Mutually exclusive events, 185
Negation, 3
Negative numbers, 257
New simplex tableau, 306
Non-singular matrix, 133
Norm of a vector, 269
Null set, 36
Numbers, real, 256
Objective function, 300
Odds, 188
One-to-one function, 69
Onto function, 69
Open half space, 282
Optimum solution, 300
Optimum strategy, 322
Order, 257
Ordered pair, 50
Ordered partition, 163
Origin, 257
Pair, ordered, 50
Parallel lines, 266
Parameter, 268
Partitions, 60, 162
ordered, 163
Pascal’s triangle, 143
Permutation, 152
Perpendicular lines, 266
Pivot, 305
Pivotal column, 305
row, 305
Plane, Cartesian, 266
Plane, half, 282
Play of a game, 321
Players, R and C, 321
Point, 266
corner, 282
extreme, 282
fixed, 234
saddle, 323
Polygonal convex set, 283
Polyhedral convex set, 282
Polynomial, 68
Premises, 26
Primal problem, 301
Probability, 184
conditional, 199
distribution, 238
vector, 233
Product set, 50
Proper subset, 36
Proposition, 3
P u r e strategy, 322
Quotient set, 60