Discretebook Fall2015
Discretebook Fall2015
Discretebook Fall2015
Mathematics
An Open Introduction
Oscar Levin
Fall 2015
Discrete
Mathematics
An Open Introduction
Oscar Levin
Fall 2015
Oscar Levin
School of Mathematical Science
University of Northern Colorado
Greely, Co 80639
oscar.levin@unco.edu
http://math.oscarlevin.com/
This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International License.
To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/4.0/.
ISBN-10: 1516921186
ISBN-13: 978-1516921188
Contents iv
Preface vii
0 Introduction 1
0.1 What is Discrete Mathematics? . . . . . . . . . . . . . . . . . . . . . . . . . . 1
0.2 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Relationships Between Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Operations On Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
Venn Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
0.3 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
Surjections, Injections, and Bijections . . . . . . . . . . . . . . . . . . . . . . . 16
Inverse Image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1 Counting 22
1.1 Additive and Multiplicative Principles . . . . . . . . . . . . . . . . . . . . . . 22
Counting With Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
Principle of Inclusion/Exclusion . . . . . . . . . . . . . . . . . . . . . . . . . 27
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
1.2 Binomial Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Pascal’s Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
1.3 Combinations and Permutations . . . . . . . . . . . . . . . . . . . . . . . . . 40
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
1.4 Combinatorial Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Patterns in Pascal’s Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
More Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
1.5 Stars and Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
1.6 Advanced Counting Using PIE . . . . . . . . . . . . . . . . . . . . . . . . . . 61
Counting Derangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
1.7 Counting Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
1.8 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
iv
Contents v
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
2 Sequences 84
2.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
2.2 Arithmetic and Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . 91
Sums of Arithmetic and Geometric Sequences . . . . . . . . . . . . . . . . . 93
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.3 Polynomial Fitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
2.4 Solving Recurrence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
The Characteristic Root Technique . . . . . . . . . . . . . . . . . . . . . . . . 107
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
2.5 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
Stamps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
Formalizing Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
Strong Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
2.6 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
4.3 Coloring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
Coloring in General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
Coloring Edges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
4.4 Euler Paths and Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
Hamilton Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
4.5 Matching in Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
4.6 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
Index 279
This text aims to give an introduction to select topics in discrete mathematics at a level
appropriate for first or second year undergraduate math majors, especially those who
intend to teach middle and high school mathematics. The book began as a set of notes
for the Discrete Mathematics course at the University of Northern Colorado. This course
serves both as a survey of the topics in discrete math and as the “bridge” course for math
majors, as UNC does not offer a separate “introduction to proofs” course. Most students
who take the course plan to teach, although there are a handful of students who will go
on to graduate school or study applied math or computer science. For these students the
current text hopefully is still of interest, but the intent is not to provide a solid mathematical
foundation for computer science, unlike the majority of textbooks on the subject.
Another difference between this text and most other discrete math books is that this
book is intended to be used in a class taught using problem oriented or inquiry based
methods. When I teach the class, I will assign sections for reading after first introducing
them in class by using a mix of group work and class discussion on a few interesting
problems. The text is meant to consolidate what we discover in class and serve as a reference
for students as they master the concepts and techniques covered in the unit. None-the-less,
every attempt has been made to make the text sufficient for self study as well, in a way that
hopefully mimics an inquiry based classroom.
The topics covered in this text were chosen to match the need of the students I teach at
UNC. The main areas of study are combinatorics, sequences, logic and proofs, and graph
theory, in that order. Induction is covered at the end of the chapter on sequences. Most
discrete books put logic first as a preliminary, which certainly has its advantages. However,
I wanted to discuss logic and proofs together, and found that doing both of these before
anything else was overwhelming for my students given that they didn’t yet have context
of other problems in the subject. Also, after spending a couple weeks on proofs, we would
hardly use that at all when covering combinatorics, so much of the progress we made was
quickly lost.
Depending on the speed of the class, it might be possible to include additional material.
In past semesters I have included generating functions (after sequences) and some basic
number theory (either after the logic and proofs chapter or at the very end of the course).
These additional topics are covered in appendix A.
While I (currently) believe this selection and order of topics is optimal, you should feel
free to skip around to what interests you. There are occasionally examples and exercises
that rely on earlier material, but I have tried to keep these to a minimum and usually
can either be skipped or understood without too much additional study. If you are an
instructor, feel free to edit the LATEX source to fit your needs.
vii
Preface viii
Investigate! activities
Sprinkled throughout the sections (usually at the very beginning of a topic) you will find
activities designed to get you acquainted with the topic soon to be discussed. These are
similar (sometimes identical) to group activities I give students to introduce material. You
really should spend some time thinking about, or even working through, these problems
before reading the section. By priming yourself to the types of issues involved in the
material you are about to read, you will better understand what is to come. There are no
solutions provided for these problems, but don’t worry if you can’t solve them or are not
confident in your answers. My hope is that you will take this frustration with you while
you read the proceeding section. By the time you are done with the section, things should
be much clearer.
Examples
I have tried to include the “correct” number of examples. For those examples which include
problems, full solutions are included. Before reading the solution, try to at least have an
understanding of what the problem is asking. Unlike some textbooks, the examples are
not meant to be all inclusive for problems you will see in the exercises. They should not
be used as a blueprint for solving other problems. Instead, use the examples to deepen
our understanding of the concepts and techniques discussed in each section. Then use this
understanding to solve the exercises at the end of each section.
Exercises
You get good at math through practice. Each section concludes with a small number of
exercises meant to solidify concepts and basic skills presented in that section. At the end
of each chapter, a larger collection of similar exercises is included (as a sort of “chapter
review”) which might bridge material of different sections in that chapter. Every exercise
has either a hint, answer or full solution (which in the pdf version of the text can be found
by clicking on the exercises number – clicking on the solution number will bring you back to
the exercise). Readers are encouraged to try these exercises before looking at the solution.
When I teach with this book, I assign these exercises as practice and then use them, or
similar problems, on quizzes and exams.
Homework Problems
Each chapter includes a small number of more involved problems – the type I would assign
as homework to be written up and collected each week. As many of these are problems I
assign, solutions are not included. If you are using this book for self study, consider these
additional Investigate! problems.
Preface ix
Acknowledgments
This book would not exist if not for “Discrete and Combinatorial Mathematics” by Richard
Grassl and Tabitha Mingus. It is the book I learned discrete math out of, and taught out of
the semester before I began writing this text. I wanted to maintain the inquiry based feel
of their book but update, expand and rearrange some of the material. In many ways I see
this current text as a “reboot” of their book.
In Spring 2015, Alees Seehausen, a graduate student at the University of Northern
Colorado, co-taught the Discrete Mathematics course with me and helped develop many
of the Investigate! activities and other problems currently used in the text. She also offered
many suggestions for improvement of the expository text, for which I am quite grateful.
Thanks also to Katie Morrison and Nate Eldredge for their suggestions after using parts of
this text in their class.
Finally, a thank you to the numerous students who have pointed out typos and made
suggestions over the years and a thanks in advance to those who will do so in the future.
Welcome to Discrete Mathematics. If this is your first time encountering the subject, you
will probably find discrete mathematics quite different from other math subjects. You
might not even know what discrete math is! Hopefully this short introduction will shed
some light on what the subject is about and what you can expect as you move forward in
your studies.
1
0.1. What is Discrete Mathematics? 2
Investigate!
Here are a few Discrete Math problems for you to try.
Note: Throughout the text you will see Investigate! activities like this one. Answer the
questions in these as best you can to give yourself a feel for what is coming next.
1. The most popular mathematician in the world is throwing a party for all of his
friends. As a way to kick things off, they decide that everyone should shake
hands. Assuming all 10 people at the party each shake hands with every other
person (but not themselves, obviously) exactly once, how many handshakes
take place?
2. At the warm-up event for Oscar’s All Star Hot Dog Eating Contest, Al ate one
hot dog. Bob then showed him up by eating three hot dogs. Not to be outdone,
Carl ate five. This continued with each contestant eating two more hot dogs
than the previous contestant. How many hot dogs did Zeno (the 26th and final
contestant) eat? How many hot dogs were eaten all together?
3. While walking through a fictional forest, you encounter three trolls. Each is
either a knight, who always tells the truth, or a knave, who always lies. The
trolls will not let you pass until you correctly identify each as either a knight
or a knave. Each troll makes a single statement:
4. Back in the days of yore, five small towns decided they wanted to build roads
directly connecting each pair of towns. While the towns had plenty of money
to build roads as long and as winding as they wished, it was very important
that the roads not intersect with each other (as stop signs had not yet been
invented). Also, tunnels and bridges were not allowed. Is it possible for each
of these towns to build a road to each of the four other towns without creating
any intersections?
Ultimately the best way to learn what discrete math is about is to do it. Let’s get started!
Before we can begin answering more complicated (and fun) problems, we must lay down
some foundation. We start by reviewing sets and functions in the framework of discrete
mathematics.
0.2 Sets
The most fundamental objects we will use in our studies (and really in all of math) are
sets. Much of what follows might be review, but it is very important that you are fluent in
the language of set theory. Most of the notation we use below is standard, although some
might be a little different than what you have seen before.
For us, a set will simply be an unordered collection of objects. Two examples: we could
consider the set of all actors who have played The Doctor on Doctor Who, or the set of natural
numbers between 1 and 10 inclusive. In the first case, Tom Baker is a element (or member)
of the set, while Idris Elba, among many others, is not an element of the set. Also, the two
examples are of different sets. Two sets are equal exactly if they contain the exact same
elements.
Notation
We need some notation to make talking about sets easier. Consider,
A {1, 2, 3} .
This is read, “A is the set containing the elements 1, 2 and 3.” We use curly braces “{ , }”
to enclose elements of a set. Some more notation:
a ∈ { a, b, c } .
The symbol “∈” is read “is in” or “is an element of.” Thus the above means that a is an
element of the set containing the letters a, b, and c. Note that this is a true statement. It
would also be true to say that d is not in that set:
d < { a, b, c } .
Be warned: we write “x ∈ A” when we wish to express that one of the elements of the set
A is x. For example, consider the set,
A {1, b, { x, y, z } , ∅} .
This is a strange set, to be sure. It contains four elements: the number 1, the letter b, the
set { x, y, z }, and the empty set (∅ {}, the set containing no elements). Is x in A? The
answer is no. None of the four elements in A are the letter x, so we must conclude that
x < A. Similarly, if we considered the set B {1, b }, then again B < A. Even though the
elements of B are also elements of A, we cannot say that the set B is one of the things in the
collection A.
0.2. Sets 4
If a set is finite, then we can describe it by simply listing the elements. Infinite sets exists
though, so we need to be able to describe them as well. For instance, if we want A to be the
set of all even natural numbers, would could write,
A {0, 2, 4, 6, . . . } ,
A { x ∈ N : ∃n ∈ N( x 2n )} .
Breaking that down: “x ∈ N” means x is in the set N (the set of natural numbers, starting
with 0), “:” is read “such that” and “∃n ∈ N( x 2n )” is read “there exists an n in the
natural numbers for which x is two times n” (in other words, x is even). Slightly easier
might be,
A { x : x is even} .
Note: sometimes people use | or for the “such that” symbol instead of the colon.
Defining a set using this sort of notation is very useful, although it takes some practice
to read them correctly. It is a way to describe the set of all things that satisfy some condition
(the condition is the logical statement after the “:” symbol). Here are some more examples.
We use the symbols ∧ for “and” and ∨ for “or” (which includes the “or both” for us).
Example:
Describe each of the following sets both in words and by listing out enough elements
to see the pattern.
1. { x : x + 3 ∈ N}. 3. { x : x ∈ N ∨ −x ∈ N}.
2. { x ∈ N : x + 3 ∈ N}. 4. { x : x ∈ N ∧ −x ∈ N}.
Solution:
1. This is the set of all number which are 3 less than a natural number (i.e., that
if you add 3 to them, you get a natural number). The set could also be written
as {−3, −2, −1, 0, 1, 2, . . . } (note that 0 is a natural number, so −3 is in this set
because −3 + 3 0).
2. This is the set of all natural numbers which are 3 less than a natural number.
So here we just have {0, 1, 2, 3 . . . }.
3. This is the set of all integers (positive and negative whole numbers, written
Z). In other words, { . . . , −2, −1, 0, 1, 2, . . . }.
4. Here we want all numbers x such that x and −x are natural numbers. There
is only one: 0. So we have the set {0}.
0.2. Sets 5
We already have a lot of notation, and there is more yet. Below is a handy chart of
symbols. Some of these will be discussed in greater detail as we move forward.
{, } the set containing {1, 2, 3}. The braces enclose the elements of a
set. This is the set which contains the numbers
1, 2, and 3.
: such that { x : x > 2} is the set of all x such that x is
greater than 2.
∈ is an element of 2 ∈ {1, 2, 3} asserts that 2 is one of the elements
in the set {1, 2, 3}. However, 4 < {1, 2, 3}.
⊆ is a subset of A ⊆ B asserts that every element of A is also an
element of B.
⊂ is a proper subset of A ⊂ B asserts that every element of A is also an
element of B, but A , B.
∩ intersection A ∩ B is the set containing all elements which
are elements of both A and B.
∪ union A ∪ B is the set containing all elements which
are elements of A or B or both.
× cross A × B is the set of all ordered pairs ( a, b ) with
a ∈ A and b ∈ B.
\ set difference A \ B is the set containing all elements of A
which are not elements of B.
A complement (of A) A is the set of everything which is not an ele-
ment of A. The A can be any set here.
|A| cardinality (of A) |{4, 5, 6}| 3 because there are 3 elements in
the set. Sometimes we call | A | the size of A.
Logic symbols:
∧ and x ∈ A ∧ x < B means x is both in the set A and
not in the set B.
∨ or x ∈ A ∨ x < B asserts that x is an element of A
or not an element of B, or both.
¬ not Another way to write x < A is ¬x ∈ A.
∀ for all ∀x ( x ≥ 0) claims that every number is greater
than 0.
∃ there exists ∃x ( x < 0) claims that there is a number less
than 0.
0.2. Sets 6
Special sets
∅ The empty set is the set which contains no elements.
U The universe set is the set of all elements.
N The set of natural numbers. That is, N {0, 1, 2, 3 . . . }.
Z The set of integers. That is, Z { . . . , −2, −1, 0, 1, 2, 3, . . . }.
Q The set of rational numbers.
R The set of real numbers.
P (A ) The power set of any set A is the set of all subsets of A.
Investigate!
1. Find the cardinality of each set below.
since these are all ways to write the set containing the first three positive integers (how
we write them doesn’t matter, just what they are).
0.2. Sets 7
What about the sets A {1, 2, 3} and B {1, 2, 3, 4}? Clearly A , B, but notice that
every element of A is also an element of B. Because of this we say that A is a subset of B, or
in symbols A ⊂ B or A ⊆ B. (Both symbols are read “is a subset of.” The difference is that
sometimes we want to say that A is either equal to or a subset of B, in which case we use
⊆. This is analoguous to the difference between < and ≤.)
Example:
Let A {1, 2, 3, 4, 5, 6}, B {2, 4, 6}, C {1, 2, 3} and D {7, 8, 9}. Determine
which of the following are true, false, or meaningless.
1. A ⊂ B. 4. ∅ ∈ A. 7. 3 ∈ C.
2. B ⊂ A. 5. ∅ ⊂ A. 8. 3 ⊂ C.
3. B ∈ C. 6. A < D. 9. {3} ⊂ C.
Solution:
4. False. A has exactly 6 elements, and none of them are the empty set.
9. True. 3 is the only element of the set {3}, and is an element of C, so every
element in {3} is an element of C.
In the example above, B is a subset of A. You might wonder what other sets are subsets
of A. If you collect all these subsets of A into a new set, we get a set of sets. We call the set
of all subsets of A the power set of A, and write it P (A).
Example:
Let A {1, 2, 3}. Find P (A).
0.2. Sets 8
P (A) {∅, {1} , {2} , {3} , {1, 2} , {1, 3} , {2, 3} , {1, 2, 3}} .
Notice that while 2 ∈ A, it is wrong to write 2 ∈ P (A) since none of the elements in
P (A) are numbers! On the other hand, we do have {2} ∈ P (A) because {2} ⊆ A.
What does a subset of P (A) look like? Notice that {2} * P (A) because not
everything in {2} is in P (A). But we do have {{2}} ⊆ P (A). The only element of
{{2}} is the set {2} which is also an element of P (A). We could take the collection
of all subsets of P (A) and call that P (P (A)). Or even the power set of that set of
sets of sets.
Another way to compare sets is by their size. Notice that in the example above, A has 6
elements, B, C, and D all have 3 elements. The size of a set is called the set’s cardinality. We
would write | A | 6, | B | 3, and so on. For sets that have a finite number of elements, the
cardinality of the set is simply the number of elements in the set. Note that the cardinality
of {1, 2, 3, 2, 1} is 3. We do not count repeats (in fact, {1, 2, 3, 2, 1} is exactly the same set
as {1, 2, 3}). There are sets with infinite cardinality, such as N, the set of rational numbers
(written Q), the set of even natural numbers, and the set of real numbers (R). It is possible
to distinguish between different infinite cardinalities, but that is beyond the scope of this
text. For us, a set will either be infinite, or finite; if it is finite, the we can determine its
cardinality by counting elements.
Example:
1. Find the cardinality of A {23, 24, . . . , 37, 38}.
2. Here | B | 3. The three elements are the number 1, the set {2, 3, 4}, and the
empty set.
3. We wrote out the elements of the power set P (C ) above, and there are 8
elements (each of which is a set). So | P (C )| 8.1
1 You
might wonder if there is a relationship between | A | and | P (A)| for all sets A. This is a good question
which we will return to in Chapter 1.
0.2. Sets 9
Operations On Sets
Is it possible to add two sets? Not really, however there is something similar. If we want to
combine two sets to get the collection of objects that are in either set, then we can take the
union of the two sets. Symbolically,
C A ∪ B,
read, “C is the union of A and B,” means that the elements of C are exactly the elements
which are either an element of A or an element of B (or an element of both). For example,
if A {1, 2, 3} and B {2, 3, 4}, then A ∪ B {1, 2, 3, 4}.
The other common operation on sets is intersection. We write,
C A∩B
and say, “C is the intersection of A and B,” when the elements in C are precisely those both
in A and in B. So if A {1, 2, 3} and B {2, 3, 4}, then A ∩ B {2, 3}.
Often when dealing with sets, we will have some understanding as to what “everything”
is. Perhaps we are only concerned with natural numbers. In this case we would say that our
universe is N. Sometimes we denote this universe by U. Given this context, we might wish
to speak of all the elements which are not in a particular set. We say B is the complement of
A, and write,
BA
when B contains every element not contained in A. So if our universe is {1, 2, . . . , 9, 10},
and A {2, 3, 5, 7}, then A {1, 4, 6, 8, 9, 10}.
Of course we can perform more than one operation at a time. For example, consider
A ∩ B.
This is the set of all elements which are both elements of A and not elements of B. What
have we done? We’ve started with A and removed all of the elements which were in B.
Another way to write this is the set difference:
A ∩ B A \ B.
Example:
Let A {1, 2, 3, 4, 5, 6}, B {2, 4, 6}, C {1, 2, 3} and D {7, 8, 9}. If the universe
is U {1, 2, . . . , 10}, find:
0.2. Sets 10
1. A ∪ B. 4. A ∩ D. 7. (D ∩ C ) ∪ A ∩ B.
2. A ∩ B. 5. B ∪ C. 8. ∅ ∪ C.
3. B ∩ C. 6. A \ B. 9. ∅ ∩ C.
Solution:
You might notice that the symbols for union and intersection slightly resemble the logic
symbols for “or” and “and.” This is no accident. What does it mean for x to be an element
of A ∪ B? It means that x is an element of A or x is an element of B (or both). That is,
x ∈A∪B ⇔ x ∈ A ∨ x ∈ B.
Similarly,
x ∈A∩B ⇔ x ∈ A ∧ x ∈ B.
Also,
x∈A ⇔ ¬( x ∈ A).
which says x is an element of the complement of A if x is not an element of A.
There is one more way to combine sets which will be useful for us: the Cartesian product,
A × B. This sounds fancy but is nothing you haven’t seen before. When you graph a
function in calculus, you graph it in the Cartesian plane. This is the set of all ordered pairs
of real numbers ( x, y ). We can do this for any pair of sets, not just the real numbers with
themselves.
Put another way, A × B {( a, b ) : a ∈ A ∧ b ∈ B }. The first coordinate comes from the
first set and the second coordinate comes from the second set. Sometimes we will want to
0.2. Sets 11
take the Cartesian product of a set with itself, and this is fine: A × A {( a, b ) : a, b ∈ A }
(we might also write A2 for this set). Notice that in A × A, we still want all ordered pairs,
not just the ones where the first and second coordinate are the same. We can also take
products of 3 or more sets, getting ordered triples, or quadruples, and so on.
Example:
Let A {1, 2} and B {3, 4, 5}. Find A × B and A × A. How many elements do
you expect to be in B × B?
Solution: A × B {(1, 3) , (1, 4) , (1, 5) , (2, 3) , (2, 4) , (2, 5)}.
A × A A2 {(1, 1) , (1, 2) , (2, 1) , (2, 2)}.
| B × B | 9. There will be 3 pairs with first coordinate 3, three more with first
coordinate 4, and a final three with first coordinate 5.
Venn Diagrams
There is a very nice visual tool we can use to represent operations on sets. Venn diagrams
display sets as intersecting circles. We can shade the region we are talking about when we
carry out an operation. We can also represent cardinality of a particular set by putting the
number in the corresponding region.
A B A B
Each circle represents a set. The rectangle containing the circles represents the universe.
To represent combinations of these sets, we shade the corresponding region. For example,
we could draw A ∩ B as:
A B
A B
0.2 Exercises 12
A B
Notice that the shaded regions above could also be arrived at in another way. We could
have started with all of C, then excluded the region where C and A overlap outside of B.
That region is (A ∩ C ) ∩ B. So the above Venn diagram also represents C ∩ (A ∩ C ) ∩ B .
So using just the picture, we have determined that
( B ∩ C ) ∪ ( C ∩ A ) C ∩ (A ∩ C ) ∩ B .
Exercises
1. Let A {1, 2, 3, 4, 5}, B {3, 4, 5, 6, 7}, and C {2, 3, 5}.
(a) Find A ∩ B.
(b) Find A ∪ B.
(c) Find A \ B.
(d) Find A ∩ (B ∪ C ).
(e) Find A × C.
(f) Is C ⊆ A? Explain.
(g) Is C ⊆ B? Explain.
(a) Find A ∩ B.
(b) Find A ∪ B.
(c) Find B ∩ C.
(d) Find B ∪ C.
3. Find an example of sets A and B such that A ∩ B {3, 5} and A ∪ B {2, 3, 5, 7, 8}.
(c) Find 2Z ∩ 3Z. Describe the set in words, and also in symbols (using a : symbol).
(d) Express { x ∈ Z : ∃y ∈ Z( x 2y ∨ x 3y )} as a union or intersection of two
sets above.
6. Let A2 be the set of all multiples of 2 except for 2. Let A3 be the set of all multiples
of 3 except for 3. And so on, so that A n is the set of all multiple of n except for n, for
any n ≥ 2. Describe (in words) the set A2 ∪ A3 ∪ A4 ∪ · · ·.
7. Draw a Venn diagram to represent each of the following:
(a) A ∪ B
(b) (A ∪ B )
(c) A ∩ (B ∪ C )
(d) (A ∩ B ) ∪ C
(e) A ∩ B ∩ C
(f) (A ∪ B ) \ C
8. Describe a set in terms of A and B which has the following Venn diagram:
A B
0.3 Functions
A function is a rule that assigns each input exactly one output. The set of all inputs for a
function is called the domain. The set of all allowable outputs is called the codomain. For
example, a function might assign each natural number to a natural number from 1 to 5.
In that case, the domain is the natural numbers and the codomain is the set of natural
numbers from 1 to 5. Now it could be that this particular function we are thinking about
assigns each even natural number to the number 2 and each odd natural number to the
number 1. In this case, not all of the codomain is actually used. We would say that the
set {1, 2} is the range of the function. These are the elements in the codomain (allowable
outputs) which are actually outputs for some input.
The key thing that makes a rule actually a function is that there is only one output for
each input. That is, it is important that the rule be a good rule. What output do we assign
to the input 7? There can only be one answer for any particular function.
To specify the name of the function, as well as the domain and codomain, we write
f : X → Y. The function is called f , the domain is the set X, and the codomain is the set
Y. This, however, does not describe the rule. To do that, we say something like this:
This function takes an input x and computes the output by squaring x and then adding
3. In this case, you better hope that X is a set of numbers and Y is a set of numbers which
can be 3 more than squares of numbers from X. It would not work for Y to be negative
numbers here. That would not be a valid function.
The description of the rule can vary greatly. We might just give a list of each output for
each input. You could also describe the function with a table or a graph or in words.
Example:
The following are all examples of functions:
1. f : Z → Z defined by f ( n ) 3n. The domain and codomain are both the set
of integers. However, the range is only the set of integer multiples of 3.
1 2 3
1 2 3
0.3. Functions 15
The arrow diagram used to define the function above can be very helpful in visualizing
functions. We will often be working with functions on finite sets so this kind of picture
is often more useful than a traditional graph of a function. A graph of the function in
example 3 above would look like this:
It would be absolutely WRONG to connect the dots or try to fit them to some curve.
There are only three elements in the domain. A curve suggests that the domain contains
an entire interval of real numbers. Remember, we are not in calculus any more!
It is important to know how to determin if a rule is or is not a function. The arrow
diagrams can help.
Example:
Which of the following diagrams represent a function? Let X {1, 2, 3, 4} and
Y { a, b, c, d }
a c a b c d a b c d
b d
Example:
Which functions are surjective (i.e., onto)?
1. f : Z → Z defined by f ( n ) 3n.
1 2 3
1 2 3
Solution:
1. f is not surjective. There are elements in the codomain which are not in the
range. For example, no n ∈ Z gets mapped to the number 1 (the rule would
say that 31 would be sent to 1, but 13 is not in the domain). In fact, the range of
the function is 3Z (the integer multiples of 3), which is not equal to Z.
To be a function, a map cannot assign a single element of the domain to two or more
different elements of the codomain. However, we have seen that the reverse is permissible.
That is, a function might assign the same element of the codomain to two or more different
elements of the domain. When this does not occur (that is, when each element of the
codomain is assigned to at most one element of the domain) then we say the function
is one-to-one. Again, this terminology makes sense: we are sending at most one element
from the domain to one element from the codomain. One input to one output. The fancy
math term for a one-to-one function is an injection. We call one-to-one functions injective
functions.
In pictures:
Example:
Which functions are injective (i.e., one-to-one)?
1. f : Z → Z defined by f ( n ) 3n.
1 2 3
1 2 3
Solution:
From the examples above, it should be clear that there are functions which are surjective,
injective, both, or neither. In the case when a function is both one-to-one and onto (an
injection and surjection), we say the function is a bijection, or that the function is a bijective
function.
0.3. Functions 18
Inverse Image
When discussing functions, we have notation for talking about an element of the domain
(say x) and its corresponding element in the codomain (we write f ( x )). It would also be
nice to start with some element of the codomain (say y) and talk about which element or
elements (if any) from the domain get sent to it. We could write “those x in the domain
such that f ( x ) y,” but this is a lot of writing. Here is some notation to make our lives
easier.
Suppose f : X → Y is a function. For y ∈ Y (an element of the codomain), we write
f −1 ( y ) to represent the set of all elements in the domain X which get sent to y. That is,
f −1 ( y ) { x ∈ X : f ( x ) y }. We say that f −1 ( y ) is the complete inverse image of y under f .
WARNING: f −1 ( y ) is not an inverse function!!!! Inverse functions only exist for bijec-
tions, but f −1 ( y ) is defined for any function f . The point: f −1 ( y ) is a set, not an element
of the domain.
Example:
Consider the function f : {1, 2, 3, 4, 5, 6} → { a, b, c, d } given by f (1) a, f (2) a,
f (3) b, f (4) c, f (5) c and f (6) c. Find the complete inverse image of each
element in the codomain.
Solution: Remember, we are looking for sets.
f −1 ( a ) {1, 2}
f −1 ( b ) {3}
f −1 ( c ) {4, 5, 6}
f −1 ( d ) ∅.
Example:
Consider the function g : Z → Z defined by g ( n ) n 2 + 1. Find g −1 (1), g −1 (2),
g −1 (3) and g −1 (10).
Solution: To find g −1 (1), we need to find all integers n such that n 2 + 1 1. Clearly
only 0 works, so g −1 (1) {0} (note that even though there is only one element, we
still write it as a set with one element in it).
To find g −1 (2), we need to find all n such that n 2 +1 2. We see g −1 (2) {−1, 1}.
If n 2 + 1 3, then we are looking for an n such that n 2 2. There are no such
integers so g −1 (3) ∅.
Finally, g −1 (10) {−3, 3} because g (−3) 10 and g (3) 10.
0.3 Exercises 19
Since f −1 ( y ) is a set, it makes sense to ask for | f −1 ( y )|, the number of elements in the
domain which map to y.
Example:
Find a function f : {1, 2, 3, 4, 5} → N such that | f −1 (7)| 5.
Solution: There is only one such function. We need five elements of the domain
to map to the number 7 ∈ N. Since there are only five elements in the domain, all
of them must map to 7. So f (1) 7, f (2) 7, f (3) 7, f (4) 7, and f (5) 7.
Function Definitions
• A function is a rule that assigns each element of a set, called the domain, to
exactly one element of a second set, called the codomain.
• The range is a subset of the codomain. It is the set of all elements which are
assigned to at least one element of the domain by the function. That is, the
range is the set of all outputs.
Exercises
1. Write out all functions f : {1, 2, 3} → { a, b }. How many are there? How many are
injective? How many are surjective? How many are both?
2. Write out all functions f : {1, 2} → { a, b, c }. How many are there? How many are
injective? How many are surjective? How many are both?
0.3 Exercises 20
x 1 2 3 4 5
f (x ) 3 2 4 1 2
1 2 3 4 x
5. For each function given below, determine whether or not the function is injective and
whether or not the function is surjective.
(a) f : N → N given by f ( n ) n + 4.
(b) f : Z → Z given by f ( n ) n + 4.
(c) f : Z → Z given by f ( n ) 5n − 8.
n/2 if n is even
(d) f : Z → Z given by f ( n )
(n + 1)/2 if n is odd.
6. Let A {1, 2, 3, . . . , 10}. Consider the function f : P (A) → N given by f (B ) | B |.
That is, f takes a subset of A as an input and outputs the cardinality of that set.
7. Let A { n ∈ N : 0 ≤ n ≤ 999} be the set of all numbers with three or fewer digits.
Define the function f : A → N by f ( abc ) a + b + c, where a, b, and c are the digits
of the number in A. For example, f (253) 2 + 5 + 3 10.
0.3 Exercises 21
n+1 if n is even
12. Consider the function f : Z → Z given by f ( n )
n − 3 if n is odd.
(a) Is f injective? Prove your answer.
(b) Is f surjective? Prove your answer.
13. At the end of the semester a teacher assigns letter grades to each of her students.
Is this a function? If so, what sets make up the domain and codomain, and is the
function injective, surjective, bijective, or neither?
14. In the game of Hearts, four players are each dealt 13 cards from a deck of 52. Is this
a function? If so, what sets make up the domain and codomain, and is the function
injective, surjective, bijective, or neither?
15. Suppose 7 players are playing 5-card stud. Each player initially receives 5 cards from
a deck of 52. Is this a function? If so, what sets make up the domain and codomain,
and is the function injective, surjective, bijective, or neither?
Chapter 1
Counting
One of the first things you learn in mathematics is how to count. Now we want to count
large collections of things quickly and precisely. For example:
• In a group of 10 people, if everyone shakes hands with everyone else exactly once,
how many handshakes took place?
• How many ways can you distribute 10 girl scout cookies to 7 boy scouts?
Before tackling these difficult questions, let’s look at the basics of counting.
Investigate!
1. A restaurant offers 8 appetizers and 14 entrées. How many choices do you
have if:
2. Think about the methods you used to solve question 1. Write down the rules
for these methods.
3. Do your rules work? A standard deck of playing cards has 26 red cards and
12 face cards.
(a) How many ways can you select a card which is either red or a face card?
(b) How many ways can you select a card which is both red and a face card?
(c) How many ways can you select two cards so that the first one is red and
the second one is a face card?
22
1.1. Additive and Multiplicative Principles 23
Consider this rather simple counting problem: at Red Dogs and Donuts, there are 14
varieties of donuts, and 16 types of hot dogs. If you want either a donut or a dog, how
many options do you have? This isn’t too hard, you just add 14 and 16. Will that always
work? What is important here?
Additive Principle
The additive principle states that if event A can occur in m ways, and event B can occur
in n disjoint ways, then the event “A or B” can occur in m + n ways.
It is important that the events be disjoint. For example, a standard deck of 52 cards
contains 26 red cards and 12 face cards. However, the number of ways to select a card
which is either red or a face card is not 26 + 12 38. This is because there are 6 cards which
are both red and face cards.
The additive principle works with more than two events. Say, in addition to your 14
choices for donuts and 16 for dogs, you would also consider eating one of 15 waffles? How
many choices do you have now? You would have 14 + 16 + 15 45 options.
Example:
How many two letter “words” start with either A or B? How many start with one
of the 5 vowels? (A word is just a strings of letters; it doesn’t have to be English, or
even pronounceable.)
Solution: First, how many two letter words start with A? We just need to select the
second letter, which can be accomplished in 26 ways. So there are 26 words starting
with A. There are also 26 words that start with B. To select a word which starts with
either A or B, we can pick the word from the first 26 or the second 26, for a total of
52 words. The additive principle is at work here.
Now what about all the two letter words starting with a vowel? There are 26
starting with A, another 26 starting with E, and so on. We will have 5 groups of
26. So we add 26 to itself 5 times. Of course it would be easier to just multiply
5 · 26. We are really using the additive principle again, just using multiplication as
a shortcut.
Example:
Suppose you are going for some fro-yo. You can pick one of 6 yogurt choices, and
one of 4 toppings. How many choices do you have?
Solution: Break your choices up into disjoint events: A are the choices with the
first topping, B the choices featuring the second topping, and so on. There are four
events; each can occur in 6 ways (one for each yogurt flavor). The events are disjoint,
so the total number of choices is 6 + 6 + 6 + 6 24.
1.1. Additive and Multiplicative Principles 24
Note that in both of the previous examples, when using the additive principle on a
bunch of events all the same size, it is quicker to multiply. This really is the same, and not
just because 6 + 6 + 6 + 6 4 · 6. We can first select the topping in 4 ways (that is, we first
select which of the disjoint events we will take). For each of those first 4 choices, we now
have 6 choices of yogurt. We have:
Multiplicative Principle
The multiplicative principle states that if event A can occur in m ways, and each possi-
bility for A allows for exactly n ways for event B, then the event “A and B” can occur
in m · n ways.
Example:
How many license plates can you make out of three letters followed by three nu-
merical digits?
Solution: Here we have six events: the first letter, the second letter, the third letter,
the first digit, the second digit, and the third digit. The first three events can each
happen in 26 ways; the last three can each happen in 10 ways. So the total number
of license plates will be 26 · 26 · 26 · 10 · 10 · 10, using the multiplicative principle.
Does this make sense? Think about how we would pick a license plate. How
many choices we would have? First, we need to pick the first letter. There are 26
choices. Now for each of those, there are 26 choices for the second letter: 26 second
letters with first letter A, 26 second letters with first letter B, and so on. We add 26
to itself 26 times. Or quicker: there are 26 · 26 choices for the first two letters.
Now for each choice of the first two letters, we have 26 choices for the third
letter. That is, 26 third letters for the first two letters AA, 26 choices for the third
letter after starting AB, and so on. There are 26 · 26 of these 26 third letter choices,
for a total of (26 · 26) · 26 choices for the first three letters. And for each of these
26 · 26 · 26 choices of letters, we have a bunch of choices for the remaining digits.
In fact, there are going to be exactly 1000 choices for the numbers. We can see
this because there are 1000 three-digit numbers (000 through 999). This is 10 choices
for the first digit, 10 for the second, and 10 for the third. The multiplicative principle
says we multiply: 10 · 10 · 10 1000.
All together, there were 263 choices for the three letters, and 103 choices for the
numbers, so we have a total of 263 · 103 choices of license plates.
Careful: “and” doesn’t mean “times.” For example, how many playing cards are both
red and a face card? Not 26 · 12. The answer is 6, and we needed to know something about
cards to answer that question.
Another caution: how many ways can you select two cards, so that the first one is a red
card and the second one is a face card? This looks more like the multiplicative principle
1.1. Additive and Multiplicative Principles 25
(you are counting two separate events) but the answer is not 26·12 here either. The problem
is that while there are 26 ways for the first card to be selected, it is not the case that for each
of those there are 12 ways to select the second card. If the first card was both red and a
face card, then there would be only 11 choices for the second card. The moral of this story
is that the multiplicative principle only works if the events are independent.1
Example:
Suppose you own 9 shirts and 5 pairs of pants.
2. If today is half-naked-day, and you will wear only a shirt or only a pair of
pants, how many choices do you have?
Solution: By now you should agree that the answer to the first question is 9 · 5 45
and the answer to the second question is 9 + 5 14. These are the multiplicative
and additive principles. There are two events: picking a shirt and picking a pair of
pants. The first event can happen in 9 ways and the second event can happen in 5
ways. To get both a shirt and a pair of pants, you multiply. To get just one article of
clothing, you add.
Now look at this using sets. There are two sets, call them S and P. The set S
contains all 9 shirts so | S | 9 while | P | 5, since there are 5 elements in the set
P (namely your 5 pairs of pants). What are we asking in terms of these sets? Well
1 To solve this problem, you could break it into two cases. First, count how many ways there are to select
the two cards when the first card is a red non-face card. Second, count how many ways when the first card is
a red face card. Doing so makes the events in each separate case independent, so the multiplicative principle
can be applied.
1.1. Additive and Multiplicative Principles 26
From this example we can see right away how to rephrase our additive principle in
terms of sets:
Additive Principle (with sets)
Given two sets A and B, if A ∩ B ∅ (that is, if there is no element in common to both
A and B), then
|A ∪ B | |A| + |B |.
This hardly needs a proof. To find A ∪ B, you take everything in A and throw in
everything in B. Since there is no element in both sets already, you will have | A | things and
add | B | new things to it. This is what adding does! Of course, we can easily extend this to
any number of disjoint sets.
From the example above, we see that in order to investigate the multiplicative principle
carefully, we need to consider ordered pairs. We should define this carefully:
Cartesian Product
Given sets A and B, we can form the set A × B {( x, y ) : x ∈ A ∧ y ∈ B } to be the set
of all ordered pairs ( x, y ) where x is an element of A and y is an element of B. We
call A × B the Cartesian product of A and B.
A × B {( a 1 , b 1 ) , ( a 1 , b2 ) , ( a 1 , b3 ) , . . . ( a 1 , b n ) ,
(a2 , b1 ), (a2 , b2 ), (a2 , b3 ), . . . , (a2 , b n ),
(a3 , b1 ), (a3 , b2 ), (a3 , b3 ), . . . , (a3 , b n ),
..
.
(a m , b1 ), (a m , b2 ), (a m , b3 ), . . . , (a m , b n )} .
Notice what we have done here: we made m rows of n pairs, for a total of m · n pairs.
Each row above is really { a i } × B for some a i ∈ A. That is, we fixed the A-element.
Broken up this way, we have
A × B ({ a 1 } × B ) ∪ ({ a 2 } × B ) ∪ ({ a 3 } × B ) ∪ · · · ∪ ({ a m } × B ).
1.1. Additive and Multiplicative Principles 27
So A × B is really the union of m disjoint sets. Each of those sets has n elements in them.
The total (using the additive principle) is n + n + n + · · · + n m · n.
To summarize:
Multiplicative Principle (with sets)
Given two sets A and B, we have | A × B | | A | · | B |.
Principle of Inclusion/Exclusion
While we are thinking about sets, consider what happens to the additive principle when
the sets are NOT disjoint. Suppose we want to find | A ∪ B | and know that | A | 10 and
| B | 8. This is not enough information though. We do not know how many of the 8
elements in B are also elements of A. However, if we also know that | A ∩ B | 6, then we
can say exactly how many elements are in A, and, of those, how many are in B and how
many are not (6 of the 10 elements are in B, so 4 are in A but not in B). We could fill in a
Venn diagram as follows:
A B
4 6 2
|A ∪ B | |A| + |B | − |A ∩ B |.
Example:
An examination in three subjects, Algebra, Biology, and Chemistry, was taken by
41 students. The following table shows how many students failed in each single
subject and in their various combinations:
Subject: A B C AB AC BC ABC
Failed: 12 5 8 2 6 3 1
A B
Now let’s fill in the other intersections. We know A ∩ B contains 2 elements, but
1 element has already been counted. So we should put a 1 in the region where A
and B intersect (but C does not). Similarly, we calculate the cardinality of (A ∩ C ) \ B,
and (B ∩ C ) \ A:
1.1. Additive and Multiplicative Principles 29
A B
1
1
5 2
A B
5 1 1
1
5 2
0
26
C
We found 5 goes in the “A only” region because the entire circle for A needed
to have a total of 12, and 7 were already accounted for.
Thus the number of students who passed all three classes is 26. The number
who failed at least one class is 15.
Note that we can also answer other questions. For example, now many students
failed just Chemistry? None. How many passed Biology but failed both Algebra
and Chemistry? 5.
Could we have solved the problem above in an algebraic way? While the additive
principle generalizes to any number of sets, when we add a third set here, we must be
careful. With two sets, we needed to know the cardinalities of A, B, and A ∩ B in order to
find the cardinality of A ∪ B. With three sets we need more information. There are more
ways the sets can combine. Not surprisingly then, the formula for cardinality of the union
of three non-disjoint sets is more complicated:
|A ∪ B ∪ C | |A| + |B | + |C | − |A ∩ B | − |A ∩ C | − |B ∩ C | + |A ∩ B ∩ C |
To determine how many elements are in at least one of A, B, or C we add up all the
elements in each of those sets. However, when we do that, any element in both A and B
is counted twice. Also, each element in both A and C is counted twice, as are elements in
B and C, so we take each of those out of our sum once. But now what about the elements
1.1 Exercises 30
which are in A ∩ B ∩ C (in all three sets)? We added them in three times, but also removed
them three times. They have not yet been counted. Thus we add those elements back in at
the end.
Returning to our example above, we have | A | 12, | B | 5, | C | 8. We also have
| A ∩ B | 2, | A ∩ C | 6, | B ∩ C | 3, and | A ∩ B ∩ C | 1. Therefore:
| A ∪ B ∪ C | 12 + 5 + 8 − 2 − 6 − 3 + 1 15
This is what we got when we solved the problem using Venn diagrams.
This process of adding in, then taking out, then adding back in, and so on is called
the Principle of Inclusion/Exclusion, or simply PIE. We will return to this counting technique
later to solve for more complicated problems (involving more than 3 sets).
Exercises
1. Your wardrobe consists of 5 shirts, 3 pairs of pants, and 17 bow ties. How many
different outfits can you make?
2. For your college interview, you must wear a tie. You own 3 regular (boring) ties and
5 (cool) bow ties. How many choices do you have for your neck-wear?
3. You realize that the interview is for clown college, so you should probably wear both
a regular tie and a bow tie. How many choices do you have now?
4. For the rest of your outfit, you have 5 shirts, 4 skirts, 3 pants, and 7 dresses. You want
to select either a shirt to wear with a skirt or pants, or just a dress. How many outfits
do you have to choose from?
5. Your Blu-ray collection consists of 9 comedies and 7 horror movies. Give an example
of a question for which the answer is:
(a) 16.
(b) 63.
6. If | A | 10 and | B | 15, what is the largest possible value for | A ∩ B |? What is the
smallest? What are the possible values for | A ∪ B |?
7. If | A | 8 and | B | 5, what is | A ∪ B | + | A ∩ B |?
8. A group of college students were asked about their TV watching habits. Of those
surveyed, 28 students watch The Walking Dead, 19 watch The Blacklist, and 24 watch
Game of Thrones. Additionally, 16 watch The Walking Dead and The Blacklist, 14 watch
The Walking Dead and Game of Thrones, and 10 watch The Blacklist and Game of Thrones.
There are 8 students who watch all three shows. How many students surveyed
watched at least one of the shows?
10. Using the same data as the previous question, describe a set with cardinality 26.
11. Consider all 5 letter “words” made from the letters a through h. (Recall, words are
just strings of letters, not necessarily actual Elglish words.)
Investigate!
In Chess, a rook can move only in straight lines (not diagonally). Fill in each square
of the chess board below with the number of different shortest paths the rook, in the
upper left corner, can take to get to that square. For example, one square is already
filled in. There are six different paths from the rook to the square: DDRR (down
down right right), DRDR, DRRD, RDDR, RDRD and RRDD.
rZ0Z0Z0Z
Z0Z0Z0Z0
0Z 6 Z0Z0Z
Z0Z0Z0Z0
0Z0Z0Z0Z
Z0Z0Z0Z0
0Z0Z0Z0Z
Z0Z0Z0Z0
Subsets
Subsets should be familiar, otherwise read over Section 0.2 again. Suppose we look at the
set A {1, 2, 3, 4, 5}. How many subsets of A contain exactly 3 elements?
First, a simpler question. How many subsets of A are there total? In other words, what
is | P (A)| (the cardinality of the power set of A)? Think about how we would build a subset.
We need to decide, for each of the elements of A, whether or not to include the element
in our subset. So we need to decide “yes” or “no” for the element 1. And for each choice
we make, we need to decide “yes” or “no” for the element 2. And so on. For each of
the 5 elements, we have 2 choices. Therefore the number of subsets is simply 25 (by the
multiplicative principle).
Of those 32 subsets, how many have 3 elements? This is not obvious. Note that we
cannot just use the multiplicative principle. Maybe we want to say we have 2 choices for
the first element, 2 choices for the second, 2 choices for the third, and then only 1 choice for
the other two. But what if we said “no” to one of the first three elements? Then we would
have two choices for the 4th element. What a mess!
Another (bad) idea: we need to pick three elements to be in our subset. There are 5
elements to choose from. So there are 5 choices for the first element, and for each of those 4
choices for the second, and then 3 for the third (last) element. The multiplicative principle
would say then that there are a total of 5 · 4 · 3 60 ways to select the 3 element subset.
But this cannot be correct (60 > 32 for one thing). One of the outcomes we would get from
these choices would be the set {3, 2, 5}, by choosing the element 3 first, then the element
2, then the element 5. Another outcome would be {5, 2, 3} by choosing the element 5 first,
then the element 2, then the element 3. But these are the same set! We can correct this by
dividing the supposed 60 outcomes by the number of different outcomes which count as
the same for each three elements. There happen to be 6 of these. So we expect there to be
10 3-element subsets of A.
Is this right? Well, we could list out all 10 of them, being very systematic in doing so, to
make sure we don’t miss any or list any twice. Or we could try to count how many subsets
of A don’t have 3 elements in them. How many have no elements? Just 1 (the empty set).
How many have 5? Again, just 1. These are the cases in which we say “no” to all elements,
or “yes” to all elements. Okay, what about the subsets which contain a single element?
There are 5 of these. We must say “yes” to exactly one element, and there are 5 to choose
from. This is also the number of subsets containing 4 elements. Those are the ones for
which we must say “no” to exactly one element.
So far we have counted 12 of the 32 subsets. We have not yet counted the subsets with
cardinality 2 and with cardinality 3. There are a total of 20 subsets left to split up between
these two groups. But the number of each must be the same! If we say “yes” to exactly two
elements, that can be accomplished in exactly the same number of ways as the number of
ways we can say “no” to exactly two elements. So the number of 2-element subsets is equal
to the number of 3-element subsets. Together there are 20 of these subsets, so 10 each.
Number of elements: 0 1 2 3 4 5
Number of subsets: 1 5 10 10 5 1
1.2. Binomial Coefficients 33
Bit Strings
“Bit” is short for “binary digit,” so a bit string is a string of binary digits. The binary digits
are simply the numbers 0 and 1. All of the following are bit strings:
The number of bits (0’s or 1’s) in the string is the length of the string; the strings above have
lengths 4, 1, 4, and 10 respectively. We also can ask how many of the bits are 1’s. The
number of 1’s in a bit string is the weight of the string; the weights of the above strings are
2, 0, 4, and 5 respectively.
Bit Strings
• A n-bit string is a bit string of length n. That is, it is a string containing n
symbols, each of which is a bit, either 0 or 1.
For example, the elements of the set B32 are the bit strings 011, 101, and 110. Those are
the only strings containing three bits exactly two of which are 1’s.
The counting questions: How many bit strings have length 5? How many of those have
weight 3? In other words, we are asking for the cardinalities |B5 | and |B53 |.
To find the number of 5-bit strings is straight forward. We have 5 bits, and each can
either be a 0 or a 1. So there are 2 choices for the first bit, 2 choices for the second, and so
on. By the multiplicative principle, there are 2 · 2 · 2 · 2 · 2 25 32 such strings.
Finding the number of 5-bit strings of weight 3 is harder. Think about how such a string
could start. The first bit must be either a 0 or a 1. In the first case (the string starts with a
0), we must then decide on four more bits. To have a total of three 1’s, among those four
remaining bits there must be three 1’s. In other words, we must include all 4-bit strings of
weight 3. In the second case (the string starts with a 1), we still have four bits to choose,
but now only two of them can be 1’s, so we should look at all the 4-bit strings of weight 2.
In other words:
|B53 | |B42 | + |B43 | .
This is an example of a recurrence relation. We represented one instance of our counting
problem in terms of two simpler instances of the problem. It holds because the strings in
B53 all have the form 1B42 (that is, a 1 followed by a string from B42 ) or 0B43 . If only we knew
the cardinalities of B42 and B43 . Repeating the same reasoning,
We can keep going down, but this should be good enough. Both B31 and B32 contain 3 bit
strings: we must pick one of the three bits to be a 1 (three ways to do that) or one of the
1.2. Binomial Coefficients 34
three bits to be a 0 (three ways to do that). Also, B33 contains just one string: 111. Thus
|B42 | 6 and |B43 | 4, which puts B53 at a total of 10 strings.
But wait – 32 and 10 were the answers to the counting questions about subsets. Coin-
cidence? Not at all. Each bit string can be thought of as a code for a subset. For the set
A {1, 2, 3, 4, 5}, we would use 5-bit strings, one bit for each element of A. Each bit in the
string is a 0 if its corresponding element of A is not in the subset, and a 1 if the element of
A is in the subset. Remember, deciding the subset amounted to a sequence of five yes/no
votes for the elements of A. Instead of yes, we put a 1; instead of no, we put a 0.
For example, the bit string 11001 represents the subset {1, 2, 5} since the first, second
and fifth bits are 1’s. The subset {3, 5} would be coded by the string 00101. What we really
have here is a bijection from P (A) to B5 .
Now for a subset to contain exactly three elements, the corresponding bit string must
contain exactly three 1’s. In other words, the weight must be 3. Thus counting the number
of 3-element subsets of A is the same as counting the number 5-bit strings of weight 3.
Lattice Paths
The integer lattice is the set of all points in the Cartesian plane for which both the x and y
coordinates are integers. If you like to draw graphs on graph paper, the lattice is the set of
all the intersections of the grid lines.
A lattice path is one of the shortest possible paths connecting two points on the lattice,
moving only horizontally and vertically. For example, here are three possible lattice paths
from the points (0, 0) to (3, 2):
Notice to ensure the path is the shortest possible, each move must be either to the right
or up. Additionally, in this case, each path has length 5 as no matter what order we take
them in, we must take three steps right and two steps up.
The counting question: how many lattice paths are there between (0, 0) and (3, 2)? We
could try to draw all of these, or instead of drawing them, maybe just list which direction
we travel on each of the 5 steps. One path might be RRUUR, or maybe UURRR, or perhaps
RURRU (those correspond to the three paths drawn above). So how many such strings of
R’s and U’s are there?
Notice that each of these strings must contain 5 symbols. Exactly 3 of them must be R’s
(since our destination is 3 units to the right). This seems awfully familiar. In fact, what if
we used 1’s instead of R’s and 0’s instead of U’s? Then we would just have 5-bit strings of
weight 3. There are 10 of those, so there are 10 lattice paths from (0,0) to (3,2).
1.2. Binomial Coefficients 35
The correspondence between bit strings and lattice paths does not stop there. Here is
another way to count lattice paths. Consider the lattice shown below:
A (3,2)
(0,0)
Any lattice path from (0,0) to (3,2) must pass through exactly one of A and B. The point
A is 4 steps away from (0,0) and two of them are right. The number of lattice paths to A is
the same as the number of 4-bit strings of weight 2, namely 6. The point B is 4 steps away
from (0,0), but now 3 of them are right. So the number of paths to point B is the same as
the number of 4-bit strings of weight 3, namely 4. So the total number of paths to (3,2) is
just 6 + 4. This is the same way we calculated the number of 5-bit strings of weight 3. The
point: the exact same recurrence relation exists for bit strings and for lattice paths.
Binomial Coefficients
Binomial coefficients are the coefficients in the expanded version of a binomial, such as
(x + y )5 . What happens when we multiply such a binomial out? We will expand (x + y )n for
various values of n. Each of these are done by multiplying everything out (i.e., FOIL-ing)
and then collecting like terms.
( x + y )1 x + y
(x + y )2 x 2 + 2x y + y 2
(x + y )3 x 3 + 3x 2 y + 3x y 2 + y 3
(x + y )4 x 4 + 4x 3 y + 6x 2 y 2 + 4x y 3 + y 4 .
In fact, there is a quicker way to expand the above binomials. For example, consider
the next one, ( x + y )5 . What we are really doing is multiplying out,
In the expansion, there will be only one x 5 term and one y 5 term. This is because to get an
x 5 , we need to use the x term in each of the copies of the binomial ( x + y ), and similarly for
y 5 . What about x 4 y? To get terms like this, we need to use four x’s and one y, so we need
exactly one of the five binomials to contribute a y. There are 5 choices for this, so there are
5 ways to get x 4 y, so the coefficient of x 4 y is 5. This is also the coefficient for x y 4 for the
same (but opposite) reason: there are 5 ways to pick which of the 5 binomials contribute
the single x. So far we have
(x + y )5 x 5 + 5x 4 y + ? x 3 y 2 + ? x 2 y 3 + 5x y 4 + y 5 .
1.2. Binomial Coefficients 36
We still need the coefficients of x 3 y 2 and x 2 y 3 . In both cases, we need to pick exactly 3 of
the 5 binomials to contribute one variable, the other two to contribute the other. Wait. This
sounds familiar. We have 5 things, each can be one of two things, and we need a total of 3
of one of them. That’s just like taking 5 bits and making sure exactly 3 of them are 1’s. So
the coefficient of x 3 y 2 (and also x 2 y 3 ) will be exactly the same as the number of bit strings
of length 5 and weight 3, which we found earlier to be 10. So we have:
(x + y )5 x 5 + 5x 4 y + 10x 3 y 2 + 10x 2 y 3 + 5x y 4 + y 5 .
These numbers we keep seeing over and over again. They are the number of subsets of a
particular size, the number of bit strings of a particular weight, the number of lattice paths,
and the coefficients of these binomial products. We will call them binomial coefficients. We
n
even have a special symbol for them: k .
Binomial Coefficients
For each integer n ≥ 0 and integer k with 0 ≤ k ≤ n there is a number
!
n
k
The last bullet point is usually taken as the definition of nk . Out of n objects we must
choose k of them, so there are n choose k ways of doing this. Each of our counting problems
above can be viewed in this way:
• How many lattice paths are there from (0,0) to (3,2)? We must choose 3 of the 5 steps
to be towards the right. There are 53 ways to do this, so there are 53 such lattice
paths.
1.2. Binomial Coefficients 37
|Bnk | |Bn−1
k−1 | + | Bk | .
n−1
Remember, this is because we can start the bit string with either a 1 or a 0. In both cases,
we have n − 1 more bits to pick. The strings starting with 1 must contain k − 1 more 1’s,
while the strings starting with 0 still need k more 1’s.
Since |Bnk | nk , the same recurrence relation holds for binomial coefficients:
n
Recurrence relation for k
! ! !
n n−1 n−1
+
k k−1 k
Pascal’s Triangle
n
Let’s arrange the binomial coefficients k into a triangle like follows:
!
0
! 0 !
1 1
! 0 ! 1 !
2 2 2
! 0 ! 1 ! 2 !
3 3 3 3
! 0 ! 1 ! 2 ! 3 !
4 4 4 4 4
0 1 2 3 4
This can continue as far down as we like. The recurrence relation for nk tells us that
each entry in the triangle is the sum of the two entries above it. The entries on the sides of
n
the triangle are always
n
1. This is because 0 1 for all n since there is only one way to pick
0 of n objects and n 1 since there is one way to select all n out of n objects. Using the
recurrence relation, and the fact that the sides ofthe triangle are 1’s, we can easily replace
all the entries above with the correct values of nk . Doing so gives us Pascal’s triangle.
We can use Pascal’s triangle to calculate
binomial coefficients. For example, using the
12
triangle on the next page, we can find 6 924.
Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
1.2 Exercises 39
Exercises
1. Let S {1, 2, 3, 4, 5, 6}
2. Let S {1, 2, 3, 4, 5, 6}
3. You break your piggy-bank to discover lots of pennies and nickels. You start arranging
these in rows of 6 coins.
(a) You find yourself making rows containing an equal number of pennies and
nickels. For fun, you decide to lay out every possible such row. How many
coins will you need?
(b) How many coins would you need to make all possible rows of 6 coins (not
necessarily with equal number of pennies and nickels)?
9. Suppose you are ordering a large pizza from D.P. Dough. You want 3 distinct toppings,
chosen from their list of 11 vegetarian toppings.
(c) How many choices do you have for your pizza if you insist on having pineapple
as one of your toppings?
(d) How do the three questions above relate to each other?
10. Explain why the coefficient of x 5 y 3 the same as the coefficient of x 3 y 5 in the expansion
of ( x + y )8 ?
Investigate!
You have a bunch of chips which come in five different colors: red, blue, green,
purple and yellow.
1. How many different two-chip stacks can you make if the bottom chip must be
red or blue? Explain your answer using both the additive and multiplicative
principles.
2. How many different three-chip stacks can you make if the bottom chip must
be red or blue and the top chip must be green, purple or yellow? How does
this problem relate to the previous one?
3. How many different three-chip stacks are there in which no color is repeated
(but otherwise any colors could be on the top or bottom)? What about four-chip
stacks?
4. Suppose you wanted to take three different colored chips and put them in your
pocket. How many different choices do you have? What if you wanted four
different colored chips? How do these problems relate to the previous one?
We know that we have them all listed above – there are 3 choices for which letter we put
first, then 2 choices for which letter comes next, which leaves only 1 choice for the last letter.
The multiplicative principle says we multiply 3 · 2 · 1.
Example:
How many permutations are there of the letters a, b, c, d, e, f ?
1.3. Combinations and Permutations 41
Solution: We do NOT want to try to list all of these out. However, if we did, we
would need to pick a letter to write down first. There are 6 choices for that letter. For
each choice of first letter, there are 5 choices for the second letter (we cannot repeat
the first letter), and for each of those, there are 4 choices for the third, 3 choices for
the fourth, 2 choices for the fifth and finally only 1 choice for the last letter. So there
are 6 · 5 · 4 · 3 · 2 · 1 720 permutations of the 6 letters. We could also write this as 6!
(read, 6 factorial; in general n! is the product of all natural numbers between 1 and
n.).
Example:
How many 4 letter “words” can you make from the letters a through f, with no
repeated letters?
Solution: This is just like the problem of permuting 4 letters, only now we have
more choices for each letter. For the first letter, there are 6 choices. For each of
those, there are 5 choices for the second letter. Then there are 4 choices for the third
letter, and 3 choices for the last letter. The total number of words is 6 · 5 · 4 · 3 360.
This is not 6! because we never multiplied by 2 and 1. We could start with 6! and
6!
then cancel the 2 and 1, and thus write 2! .
In general, we can ask how many permutations exist of k objects choosing those objects
from a larger collection of n objects. (In the example above, k 4, and n 6.) We write this
number P ( n, k ). From the example above, we see that to compute P ( n, k ) we must apply
the multiplicative principle to k numbers, starting with n and counting backwards. So for
example
P (10, 4) 10 · 9 · 8 · 7.
Notice again that P (10, 4) starts out looking like 10!, but we stop after 7. So
10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 10!
P (10, 4) .
6·5·4·3·2·1 6!
Careful: the factorial in the denominator is not 4! but rather (10 − 4)!.
Permutations
P ( n, k ) is the number of permutations of k out of n objects.
n!
P ( n, k ) .
( n − k )!
arrange them. Once you have selected the k objects, we know there are k! ways to arrange
(permute) them. But how do you select k objects from the n? You have n objects, and you
need to choose k of them. You can do that in nk ways. This gives us another formula for
P ( n, k ): !
n
P ( n, k ) · k!.
k
Now since we have a closed formula for P ( n, k ) already, we can substitute that in:
!
n! n
· k!.
( n − k )! k
n
If we divide both sides by k! we get a closed formula for k .
n
Closed formula for k
!
n n!
.
k (n − k )!k!
Example:
You decide to have a dinner party. Even though you are incredibly popular and
have 14 different friends, you only have enough chairs to invite 6 of them.
2. What if you need to decide not only which friends to invite but also in which
order to invite them in? How many choices do you have then?
Solution:
1. You must simply choose 6 friends from a group of 14. This can be done in 14
6
ways. We can find this number either by using Pascal’s triangle or the closed
14!
formula 8!·6! 3003.
2. Here you must count all the ways you can permute 6 friends chosen from a
8! 2192190.
group of 14. So the answer is P (14, 6), which can be calculated as 14!
1.3 Exercises 43
How are these numbers related? Notice that P (14, 6) is much larger than 14
6 . This
14
makes sense. 6 picks 6 friends, but P (14, 6) arranges the 6 friends as well as picks
them. In fact, we can say exactly how much larger P (14, 6) is. In both counting
problems we choose 6 out of 14 friends. For the first one, we stop there, at 3003
ways. But for the second counting problem, each of those 3003 choices of 6 friends
can be arranged in exactly 6! ways. So now we have 3003 · 6! choices and that is
exactly 2192190.
Alternatively, look at the first problem another way. We want to select 6 out of
14 friends, but we do not care about the order they are selected in. To select 6 out
of 14 friends, we might try this:
14 · 13 · 12 · 11 · 10 · 9.
This is a reasonable guess, since we have 14 choices for the first guest, then 13
for the second, and so on. But the guess is wrong (in fact, that product is exactly
2192190 P (14, 6)). It distinguishes between the different orders in which we could
invite the guests. To correct for this, we could divide by the number of different
arrangements of the 6 guests (so that all of these would count as just one outcome).
There are precisely 6! ways to arrange 6 guests, so the correct answer to the first
question is
14 · 13 · 12 · 11 · 10 · 9
.
6!
Note that another way to write this is
14!
.
8! · 6!
which is what we had originally.
Exercises
1. A pizza parlor offers 10 toppings.
(a) How many 3-topping pizzas could they put on their menu? Assume double
toppings are not allowed.
(b) How many total pizzas are possible, with between zero and ten toppings (but
not double toppings) allowed?
(c) The pizza parlor will list the 10 toppings in two equal-sized columns on their
menu. How many ways can they arrange the toppings in the left column?
2. A combination lock consists of a dial with 40 numbers on it. To open the lock, you
turn the dial to the right until you reach a first number, then to the left until you get
to second number, then to the right again to the third number. The numbers must be
distinct. How many different combinations are possible?
3. Using the digits 2 through 8, find the number of different 5-digit numbers such that:
1.3 Exercises 44
4. How many quadrilaterals can you draw using the dots below as vertices (corners)?
(a) Squares?
(b) Rectangles?
(c) Parallelograms?
(d) Trapezoids?2
(e) Trapezoids that are not parallelograms?
7. How many anagrams are there of the word “assesses” that start with the letter “a”?
(a) You need to divide up into foursomes (groups of 4 people): a first foursome, a
second foursome, and so on. How many ways can you do this?
(b) After all your hard work, you realize that in fact, you want each foursome to
include one of the five Board members. How many ways can you do this?
10. How many different seating arrangements are possible for King Arthur and his 9
knights around their round table?
2 Here, as in calculus, a trapezoid is defined as a quadrilateral with at least one pair of parallel sides. In
particular, parallelograms are trapezoids.
1.4. Combinatorial Proofs 45
Investigate!
1. The Stanley Cup is decided in a best of 7 tournament between two teams. In
how many ways can your team win? Let’s answer this question two ways:
(a) How many of the 7 games does your team need to win? How many ways
can this happen?
(b) What if the tournament goes all 7 games? So you win the last game. How
many ways can the first 6 games go down?
(c) What if the tournament goes just 6 games? How many ways can this
happen? What about 5 games? 4 games?
(d) What are the two different ways to compute the number of ways your
team can win? Write down an equation involving binary coefficients
(that is, nk ’s). What pattern in Pascal’s triangle is this an example of?
2. Generalize. What if the rules changed and you played a best of 9 tournament
(5 wins required)? What if you played an n game tournament with k wins
required to be named champion?
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
There are lots of patterns hidden away in the triangle, enough to fill a reasonably sized
book. Here are just a few of the most obvious ones:
1.4. Combinatorial Proofs 46
2. Any entry not on the border is the sum of the two entries above it.
3. The triangle is symmetric. On any row, entries on the left side are mirrored on the
right side.
4. The sum of all entries on a given row is a power of 2. (You should check this!)
We would like to state these observations in a more precise way, and then prove that
they are correct. Now each entry in Pascal’s triangle is in fact a binomial coefficient. The 1
on the very top of the triangle is 00 . The next row (which we will call row 1, even though
it is not the top-most row) consists of 10 and 11 . Row 4 (the row 1, 4, 6, 4, 1) consists of the
binomial coefficients ! ! ! ! !
4 4 4 4 4
.
0 1 2 3 4
Given this description of the elements in Pascal’s triangle, we can rewrite the above obser-
vations as follows:
n n
1 and 1.
1. 0 n
n n−1 n−1
+
2. k k−1 k .
n n
3. k n−k .
n n n n
+ + +···+ 2n .
4. 0 1 2 n
Each of these are an example of a binomial identity: an identity (i.e., equation) involving
binomial coefficients.
Our goal is to establish these identities. We wish to prove that they hold for all values
of n and k. These proofs can be done in many ways. One option would be to give algebraic
proofs, using the formula for nk :
!
n n!
.
k (n − k )! k!
Here’s how you might do that for the second identity above.
Example:
Give an algebraic proof for the binomial identity
! ! !
n n−1 n−1
+ .
k k−1 k
n
Proof. By the definition of k , we have
! !
n−1 (n − 1)! (n − 1)! n−1 ( n − 1) !
and .
k−1 (n − 1 − (k − 1))!(k − 1)! (n − k )!(k − 1)! k (n − 1 − k )!k!
1.4. Combinatorial Proofs 47
Thus, starting with the right hand side of the equation we are trying to establish:
! !
n−1 n−1 (n − 1)! (n − 1)!
+ +
k−1 k (n − k )!(k − 1)! (n − 1 − k )! k!
(n − 1)!k (n − 1)!(n − k )
+
(n − k )! k! (n − k )! k!
( n − 1 ) !( k + n − k )
(n − k )! k!
n!
(n − k )! k!
!
n
.
k
The second line (where the common denominator is found) works because k ( k−1)!
k! and ( n − k )( n − k − 1)! ( n − k )!. qed
This is certainly a valid proof, but also is entirely useless. Even if you understand the
n
does not tell you why the identity is true. A n−1
proof perfectly, it better
approach
n−1
would be to
explain what k means and then say why that is also what k−1 + k means. Let’s see
Example:
n n
1 and 1.
Explain why 0 n
Solution: What do these binomial coefficients tell us? Well, n0 gives the number
number of n-bit strings with weight n. There is only one string with this property,
the string of all 1’s.
Another way: n0 gives the number of subsets of a set of size n containing 0
elements. There is only one such subset, the empty set. nn gives the number
of subsets containing n elements. The only such subset is the original set (of all
elements).
Example:
n n−1 n−1
+
Explain why k k−1 k .
1.4. Combinatorial Proofs 48
Solution: The easiest way to see this is to consider bit strings. nk is the number
of bit strings of length n containing k 1’s. Of all of these strings, some start with
a 1 and the rest start with a 0. First consider all the bit strings which start with a
1. After the 1, there must be n − 1 more bits (to get the total length up to n) and
exactly k − 1 of them must be 1’s (as we already have one,and we need k total).
How many strings are there like that? There are exactly n−1 k−1 such bit strings, so of
all the length n bit strings containing k 1’s, n−1
k−1 of them start with a 1. Similarly,
n−1
there are k which start with a 0 (we still need n − 1 bits and now k of them must
be 1’s). Since there are n−1
k bit strings containing n − 1 bits with k 1’s, that is the
number of length n bit strings with k 1’s which start with a 0.
Another way: consider the question, how many ways can you select k pizza
toppings from a menu containing n choices? One way to do this is just nk . An-
other way to answer the same question is to first decide whether or not you want
anchovies. If you do want anchovies, you still need to pick k − 1 toppings, now from
just n − 1 choices. That can be done in n−1 k−1 ways. If you do not want anchovies,
then you still need to select k toppings from n − 1 choices (the anchovies are out).
You can do that in n−1
k ways. Since the choices with anchovies are disjoint from
the choices without anchovies, the total choices are n−1 k−1 +
n−1
k . But wait. We
answered the same question
n−1in two different ways, so the two answers must be the
n n−1
same. Thus k k−1 + k .
You can also explain (prove) this identity by counting subsets, or even lattice
paths.
Example:
n n
Prove the binomial identity k n−k .
n
Solution: Why is this true? k counts the number of ways to select k things from
n
n choices. On the other hand, n−k counts the number of ways to select n − k things
from n choices. Are these really the same? Well, what if instead of selecting the
n − k things you choose to exclude them. How many ways are there to choose
n
n − k things to exclude from n choices. Clearly this is n−k as well (it doesn’t matter
whether you include or exclude the things once you have chosen them). And if
you exclude n − k things, then you are including the other k things. So the set of
outcomes should be the same.
Let’s try the pizza counting example like we did above. How many ways are
there to pick k toppings from a list of n choices? On the one hand, the answer
is simply nk . Alternatively, you could make a list of all the toppings you don’t
want. To end up with a pizza containing exactly k toppings, you need to pick n − k
n
toppings to not put on the pizza. You have n−k choices for the toppings you don’t
want. Both of these ways give you a pizza with k toppings, in fact all the ways to
get a pizza with k toppings. Thus these two answers must be the same: nk n−k n
.
1.4. Combinatorial Proofs 49
You can also prove (explain) this identity using bit strings, subsets, or lattice
n
paths. The bit string argument is nice: k counts the number of bit strings of length
n with k 1’s. This is also the number of bit string of length n with k 0’s (just replace
each 1 with a 0 and each 0 with a 1). But
n
if a string of length n has k 0’s, it must
have n − k 1’s. And there are exactly n−k strings of length n with n − k 1’s.
Example:
n n n n
+ + +···+ 2n .
Prove the binomial identity 0 1 2 n
Solution: Let’s do a “pizza proof” again. We need to find a question about pizza
toppings which has 2n as the answer. How about this: If a pizza joint offers n
toppings, how many pizzas can you build using any number of toppings from no
toppings to all toppings, using each topping at most once?
On one hand, the answer is 2n . For each topping you can say “yes” or “no,” so
you have two choices for each topping.
On the other hand, divide the possible pizzas into disjoint groups: the pizzas
with no toppings, the pizzas with one topping, the pizzas with two toppings, etc.
If we want no toppings, there is only one pizza like that (the empty pizza, if you
n
will) but it would be better to think of that number as 0 since we choose 0 of the n
toppings. How many pizzas have 1 topping? We need to choose 1 of the n toppings,
n
so 1 . We have:
n
Pizzas with 0 toppings: 0
n
Pizzas with 1 topping: 1
n
Pizzas with 2 toppings: 2
..
.
n
Pizzas with n toppings: n .
The total number of possible pizzas will be the sum of these, which is exactly the
left hand side of the identity we are trying to prove.
Again, we could have proved the identity using subsets, bit strings, or lattice
paths (although the lattice path argument is a little tricky).
Hopefully this gives some idea of how explanatory proofs of binomial identities can
go. It is worth pointing out that sometimes more traditional proofs are also nice.3 For
example, consider the following rather slick proof of the last identity.
3 Most every binomial identity can be proved using mathematical induction, using the recursive definition
n
for k . We will discuss induction in Section 2.5.
1.4. Combinatorial Proofs 50
More Proofs
The explanatory proofs given in the above examples are typically called combinatorial proofs.
In general, to give a combinatorial proof for a binomial identity, say A B you do the
following:
Since both A and B are the answers to the same question, we must have A B.
The tricky thing is coming up with the question. This is not always obvious, but it gets
easier the more counting problems you solve. You will start to recognize types of answers
as the answers to types of questions. More often what will happen is you will be solving a
counting problem and happen to think up two different ways of finding the answer. Now
you have a binomial identity and the proof is right there. The proof is the problem you just
solved together with your two solutions.
For example, consider this counting question:
How many 10-letter words use exactly four A’s, three B’s, two C’s and one D?
Let’s try to solve this problem. We have 10 spots for letters to go. Four of those need
to be A’s. We can pick the four A-spots in 10
4 ways. Now where can we put the B’s? Well
there are only 6 spots left, we need to pick 3 of them. This can be done in 63 ways. The
two C’s need to go in two of the 3 remaining spots, so we have 32 ways of doing that. That
leaves just one spot of the D, but we could write that 1 choice as 11 . Thus the answer is:
! ! ! !
10 6 3 1
.
4 3 2 1
1.4. Combinatorial Proofs 51
But why stop there? We can find the answer another way too. First let’s decide where to
put the one D: we have 10 spots, we need to choose 1 of them, so this can be done in 10 1
ways. Next, choose one of the 92 ways to place the two C’s. We now have 7 spots left, and
three of them need to be filled with B’s. There are 73 ways to do this. Finally the A’s can
be placed in 44 (that is, only one) ways. So another answer to the question is
! ! ! !
10 9 7 4
.
1 2 3 4
Example:
Prove the identity
n+2
!
1n + 2( n − 1) + 3( n − 2) + · · · + ( n − 1)2 + n1 .
3
Proof. Consider the question “How many 3-element subsets are there of the set
{1, 2, 3, . . . , n + 2}?” We answer this in two ways:
Answer 1: We must select 3 elements from the collection of n + 2 elements. This
can be done in n+2
3 ways.
Answer 2: Break this problem up into cases by what the middle number in the
subset is. Say each subset is { a, b, c } written in increasing order. We count the
number of subsets for each distinct value of b. The smallest possible value of b is 2,
and the largest is n + 1.
When b 2, there are 1 · n subsets: 1 choice for a and n choices (3 through n + 2)
for c.
When b 3, there are 2 · ( n − 1) subsets: 2 choices for a and n − 1 choices for c.
When b 4, there are 3 · ( n − 2) subsets: 3 choices for a and n − 2 choices for c.
1.4. Combinatorial Proofs 52
And so on. When b n + 1, there are n choices for a and only 1 choice for c, so
n · 1 subsets.
Therefore the total number of subsets is
1n + 2( n − 1) + 3( n − 2) + · · · + ( n − 1)2 + n1.
Since Answer 1 and Answer 2 are answers to the same question, they must be
equal. Therefore
n+2
!
1n + 2( n − 1) + 3( n − 2) + · · · + ( n − 1)2 + n1 .
3
qed
Example:
Prove the binomial identity
!2 !2 !2 !2 !
n n n n 2n
+ + +···+ .
0 1 2 n n
Solution: We will give two different proofs of this fact. The first will be very
similar to the previous example (counting subsets). The second proof is a little
slicker, using lattice paths.
Proof. Consider the question: “How many pizzas can you make using n toppings
when there are 2n toppings to choose from?”
Answer 1: There are 2n toppings, from which you must choose n. This can be
done in 2nn ways.
Answer 2: Divide the toppings into two groups of n toppings (perhaps n meats
and n veggies). Any choice of n toppings must include some number from the first
group and some number from the second group. Consider each possible number
of meat toppings
separately:
n n
0 meats: 0 n ,since you need to choose 0 of the n meats and n of the n veggies.
1 meat: n1 n−1
n
, since you need 1 of n meats so n − 1 of n veggies.
n n
2 meats: 2 n−2 . Choose 2 meats and the remaining n − 2 toppings from the n
veggies.
And so on. The last case is n meats, which can be done in nn n0 ways.
This is not quite the left hand side . . . yet. Notice that nn n n
n
0 and n−1 1 and
so on, by the symmetry formula. Thus we do indeed get
!2 !2 !2 !2
n n n n
+ + +···+ .
0 1 2 n
qed
Proof. Consider the question: How many lattice paths are there from (0, 0) to ( n, n )?
Answer 1: Wemust travel 2n units, and n of them must be in the up direction.
Thus there are 2n n paths.
Answer 2: Note that any path from (0, 0) to ( n, n ) must cross the line x + y n.
That is, any path must pass through exactly one of the points: (0, n ), (1, n − 1),
(2, n − 2), . . . , (n, 0). For example, this is what happens in the case n 4:
(0,4) (4,4)
(1,3)
(2,2)
(3,1)
(4,0)
(0,0) x+y4
How many paths pass through (0, n )? To getto that point, you must travel n
units, and 0 of them are to the right, so there are n0 ways to get to (0, n ). From (0, n )
to ( n, n ) takes n steps, and 0 of them are up. So there are n0 ways to get from (0, n )
to ( n, n ). Therefore there are n0 n0 paths from (0, 0) to ( n, n ) through the point
(0, n ).
through (1, n − 1). There are n1 paths to get there (n steps, 1 to the
What about
and n1 paths to complete the journey to ( n, n ) (n steps, 1 up). So there are
right)
n n
1 1 paths from (0, 0) to ( n, n ) through (1, n − 1).
n
In general, to get to ( n, n ) through the point ( k, n − k ) we have k paths
to the
midpoint and then nk paths from the midpoint to ( n, n ). So there are nk nk paths
Exercises
n n−1 n−1
+
1. Prove the identity k k−1 k using a question about subsets.
4. A woman is getting married. She has 15 best friends but can only select 6 of them to
be her bridesmaids, one of which needs to be her maid of honor. How many ways
can she do this?
(a) What if she first selects the 6 bridesmaids, and then selects one of them to be the
maid of honor?
(b) What if she first selects her maid of honor, and then 5 other bridemaids?
15 14
15
(c) Explain why 6 6 5 .
n n−2 n
k
5. Give a combinatorial proof of the identity 2 k−2 k 2 .
6. Consider the bit strings in B62 (bit strings of length 6 and weight 2).
7. Let’s count ternary digit strings, that is, strings in which each digit can be 0, 1, or 2.
8. How many ways are there to rearrange the letters in the word “rearrange”? Answer
this question in at least two different ways to establish a binomial identity.
n
9. Give a combinatorial proof for the identity P ( n, k )
k k!
n+3
! ! ! ! ! ! ! ! !
2 n 3 n−1 4 n−2 n 2
+ + +···+ .
2 2 2 2 2 2 2 2 5
Investigate!
1. Suppose you have some number of identical Rubik’s cubes to distribute to your
friends. Start by creating a single row of the cubes. Now find the number of
different ways you can distribute the cubes provided:
2. Make a conjecture about how many different ways you could distribute 7 cubes
to 4 people. Explain.
3. What if each person were required to get at least one cube? How would your
answers change?
You have 7 cookies to give to 4 kids. How many ways can you do this?
Take a moment to think about how you might solve this problem. You may assume that
it is acceptable to give a kid no cookies. Also, the cookies are all identical and the order in
which you give out the cookies does not matter.
Before solving the problem, here is a wrong answer: you might guess that the answer
should be 47 because for each of the 7 cookies, there are 4 choices of kids to which you
can give the cookie. This is reasonable, but wrong. To see why, consider a few possible
1.5. Stars and Bars 56
outcomes: we could assign the first six cookies to kid A, and the seventh cookie to kid B.
Another outcome would assign the first cookie to kid B and the six remaining cookies to
kid A. Both outcomes are included in the 47 answer. But for our counting problem, both
outcomes are really the same – kid A gets six cookies and kid B gets one cookie.
What do outcomes actually look like? How can we represent them? One approach
would be to write a outcome as a string of four numbers like this:
3112,
which represent the outcome in which the first kid gets 3 cookies, the second and third
kid each get 1 cookie, and the fourth kid gets 2 cookies. Represented this way, the order in
which the numbers occur matters. 1312 is a different outcome, because the first kid gets a
one cookie instead of 3. Each number in the string can be any integer between 0 and 7. But
the answer is not 74 . We need the sum of the numbers to be 7.
Another way we might represent outcomes is to write a string of seven letters:
ABAADCD,
which represents that the first cookie goes to kid A, the second cookie goes to kid B, the
third and fourth cookies go to kid A, and so on. In fact, this outcome is identical to the
previous one – A gets 3 cookies, B and C get 1 each and D gets 2. Each of the seven letters
in the string can be any of the 4 possible letters (one for each kid), but the number of such
strings is not 47 , because here order does not matter. In fact, another way to write the same
outcome is
AAABCDD.
This will be the preferred representation of the outcome. Since we can write the letters in
any order, we might as well write them in alphabetical order for the purposes of counting.
So we will write all the A’s first, then all the B’s, and so on.
Now think about how you could specify such an outcome. All we really need to do is
say when to switch from one letter to the next. In terms of cookies, we need to say after
how many cookies do we stop giving cookies to the first kid and start giving cookies to the
second kid. And then after how many do we switch to the third kid? And after how many
do we switch to the fourth? So yet another way to represent an outcome is like this:
∗ ∗ ∗| ∗ | ∗ | ∗ ∗
Three cookies go to the first kid, then we switch and give one cookie to the second kid, then
switch, one to the third kid, switch, two to the fourth kid. Notice that we need 7 stars and
3 bars – one star for each cookie, and one bar for each switch between kids, so one fewer
bars than there are kids (we don’t need to switch after the last kid – we are done).
Why have we done all of this? Simple: to count the number of ways to distribute 7
cookies to 4 kids, all we need to do is count how many stars and bars charts there are. But a
stars and bars chart is just a string of symbols, some stars and some bars. If instead of stars
and bars we would use 0’s and 1’s, it would just be a bit string. We know how to count
those.
1.5. Stars and Bars 57
Before we get too excited, we should make sure that really any string of (in our case) 7
stars and 3 bars corresponds to a different way to distribute cookies to kids. In particular
consider a string like this:
| ∗ ∗ ∗ || ∗ ∗ ∗ ∗
Does that correspond to a cookie distribution? Yes. It represents the distribution in which
kid A gets 0 cookies (because we switch to kid B before any stars), kid B gets three cookies
(three stars before the next bar), kid C gets 0 cookies (no stars before the next bar) and kid
D gets the remaining 4 cookies. No matter how the stars and bars are arranged, we can
distribute cookies in that way. Also, given any way to distribute cookies, we can represent
that with a stars and bars chart. For example, the distribution in which kid A gets 6 cookies
and kid B gets 1 cookie has the following chart:
∗ ∗ ∗ ∗ ∗ ∗ | ∗ ||
After all that work we are finally ready to count. Each way to distribute cookies
corresponds to a stars and bars chart with 7 stars and 3 bars. So there are 10 symbols, and
we must choose 3 of them to be bars. Thus:
!
10
There are ways to distribute 7 cookies to 4 kids.
3
While we are at it, we can also answer a related question: how many ways are there to
distribute 7 cookies to 4 kids so that each kid gets at least one cookie? What can you say
about the corresponding stars and bars charts? The charts must start and end with at least
one star (so that kids A and D) get cookies, and also no two bars can be adjacent (so that
kids B and C are not skipped). One way to assure this is to only place bars in the spaces
between the stars. With 7 stars, there are 6 spots between the stars, so we must choose 3
6
of those 6 spots to fill with bars. Thus there are 3 ways to distribute 7 cookies to 4 kids
giving at least one cookie to each kid.
Another (and more general) way to approach this modified problem is to first give each
kid one cookie. Now the remaining 3 cookies can be distributed to the 4 kids without
restrictions. So we have 3 stars and
3 bars for a total of 6 symbols, 3 of which must be bars.
6
So again we see that there are 3 ways to distribute the cookies.
Stars and bars can be used in counting problems other than kids and cookies. Here are
a few examples.
Example:
Your favorite mathematical pizza chain offers 10 toppings. How many pizzas can
you make if you are allowed 6 toppings? The order of toppings does not matter
but now you are allowed repeats. So one possible pizza is triple sausage, double
pineapple, and onions.
Solution: We get 6 toppings (counting possible repeats). Represent each of these
toppings as a star. Think of going down the menu one topping at a time: you see
anchovies first, and skip to the next, sausage. You say yes to sausage 3 times (use 3
1.5. Stars and Bars 58
stars), then switch to the next topping on the list. You keep skipping until you get
to pineapple, which you say yes to twice. Another switch and you are at onions.
You say yes once. Then you keep switching until you get to the last topping, never
saying yes again (since you already have said yes 6 times. There are 10 toppings to
choose from, so we must switch from considering one topping to the next 9 times.
These are the bars.
Now that we are confident that we have the right number of stars and bars, we
answer the question simply: there are 6 stars and 9 bars, so 15 symbols. We need to
pick 9 of them to be bars, so there number of pizzas possible is
!
15
.
9
Example:
How many 7 digit phone numbers are there in which the digits are non-increasing?
That is, every digit is less than or equal to the previous one.
Solution: We need to decide on 7 digits so we will use 7 stars. The bars will
represent a switch from each possible single digit number down the next smaller
one. So the phone number 866-5221 is represented by the stars and bars chart
| ∗ || ∗ ∗| ∗ ||| ∗ ∗| ∗ |
There are 10 choices for each digit (0-9) so we must switch between choices 9 times.
We have 7 stars and 9 bars, so the total number of phone numbers is
!
16
.
9
Example:
How many integer solutions are there to the equation
x 1 + x 2 + x3 + x 4 + x 5 13.
(An integer solution to an equation is a solution in which the unknown must have
an integer value.)
Solution: This problem is just like giving 13 cookies to 5 kids. We need to say how
many of the 13 units go to each of the 5 variables. In other words, we have 13 stars
and 4 bars (the bars are like the “+” signs in the equation).
2. Now each variable must be at least 1. So give one unit to each variable to
satisfy that restriction. Now there are 8 stars left, and still 4 bars, so the
number of solutions is 12
4 .
3. Now each variable must be 2 or greater. So before any counting, give each
variable 2 units. We now have 3 remaining stars and 4 bars, so there are 74
solutions.
Exercises
1. A multiset is a collection of objects, just like a set, but can contain an object more than
once (the order of the elements still doesn’t matter). For example, {1, 1, 2, 5, 5, 7} is a
multiset of size 6.
(a) How many sets of size 5 can be made using the 10 numeric digits 0 through 9?
(b) How many multisets of size 5 can be made using the 10 numeric digits 0 through
9?
2. Each of the counting problems below can be solved with stars and bars. For each, say
what outcome the diagram
∗ ∗ ∗| ∗ || ∗ ∗|
represents, if there are the correct number of stars and bars for the problem. Oth-
erwise, say why the diagram does not represent any outcome, and what a correct
diagram would look like.
(a) How many ways are there to select a handful of 6 jellybeans from a jar that
contains 5 different flavors?
(b) How many ways can you distribute 5 identical lollipops to 6 kids?
(c) How many 6-letter words can you make using the 5 vowels?
(d) How many solutions are there to the equation x 1 + x2 + x 3 + x 4 6.
3. After gym class you are tasked with putting the 14 identical dodgeballs away into 5
bins.
(a) How many ways can you do this if there are no restrictions?
(b) How many ways can you do this if each bin must contain at least one dodgeball?
1.5 Exercises 60
4. How many integer solutions are there to the equation x + y + z 8 for which
5. Using the digits 2 through 8, find the number of different 5-digit numbers such that:
(a) Digits cannot be repeated and must be written in increasing order. For example,
23678 is okay, but 32678 is not.
(b) Digits can be repeated and must be written in non-decreasing order. For exam-
ple, 24448 is okay, but 24484 is not.
6. When playing Yahtzee, you roll five regular 6-sided dice. How many different out-
comes are possible from a single roll? The order of the dice does not matter.
7. Your friend tells you she has 7 coins in her hand (just pennies, nickels, dimes and
quarters). If you guess how many of each kind of coin she has, she will give them to
you. If you guess randomly, what is the probability that you will be correct?
9. Solve the three counting problems below. Then say why it makes sense that they all
have the same answer. That is, say how you can interpret them as each other.
10. Conic, your favorite math themed fast food drive-in offers 20 flavors which can be
added to your soda. You have enough money to buy a large soda with 4 added
flavors. How many different soda concoctions can you order if:
(a) you refuse to use any of the flavors more than once?
(b) you refuse repeats but care about the order the flavors are added?
(c) you allow yourself multiple shots of the same flavor?
(d) you allow yourself multiple shots, and care about the order the flavors are
added?
1.6. Advanced Counting Using PIE 61
Investigate!
1. You have 11 identical mini key-lime pies to give to 4 children. However, you
don’t want any kid to get more than 3 pies. How many ways can you distribute
the pies?
(a) How many ways are there to distribute the pies without any restriction?
(b) Let’s get rid of the ways that one or more kid gets too many pies. How
many ways are there to distribute the pies if Al gets too many pies? What
if Bruce gets too many? Or Cat? Or Dent?
(c) What if two kids get too many pies? How many ways can this happen?
Does it matter which two kids you pick to overfeed?
(d) Is it possible that three kids get too many pies? If so, how many ways can
this happen?
(e) How should you combine all the numbers you found above to answer the
original question?
2. Suppose now you have 13 pies and 7 children. No child can have more than 2
pies. How many ways can you distribute the pies?
sets once too often, so we need to add it back in. In terms of cardinality of sets, we have
|A ∪ B ∪ C | |A| + |B | + |C | − |A ∩ B | − |A ∩ C | − |B ∩ C | + |A ∩ B ∩ C |.
Example:
Three kids, Alberto, Bernadette, and Carlos, decide to share 11 cookies. They
wonder how many ways they could split the cookies up provided that none of
them receive more than 4 cookies (someone receiving no cookies is for some reason
acceptable to these kids).
Solution: Without the “no more than 4” restriction, the answer would be 13
2 , using
11 stars and 2 bars (separating the three kids). Now count the number of ways that
one or more of the kids violates the condition, i.e., gets at least 4 cookies.
Let A be the set of outcomes in which Alberto gets more than 4 cookies. Let B
be the set of outcomes in which Bernadette gets more than 4 cookies. Let C be the
set of outcomes in which Carlos gets more than 4 cookies. We then are looking (for
the sake of subtraction) for the size of the set A ∪ B ∪ C. Using PIE, we must find
the sizes of | A |, | B |, | C |, | A ∩ B | and so on. Here is what we find.
8
|A|
. First give Alberto 5 cookies, then distribute the remaining 6 to the
2
three kids without restrictions, using 6 stars and 2 bars.
8
|B |
2 . Just like above, only now Bernadette gets 5 cookies at the start.
8
|C |
2 . Carlos gets 5 cookies first.
3
|A ∩ B |
. Give Alberto and Bernadette 5 cookies each, leaving 1 (star) to
2
distribute to the three kids (2 bars).
3
|A ∩ C |
2 . Alberto and Carlos get 5 cookies first.
3
|B ∩ C |
2 . Bernadette and Carlos get 5 cookies first.
sense now that we see it. The only way to ensure that no kid gets more than 4
cookies is to give two kids 4 cookies and one kid 3; there are three choices for which
kid that should be. We could have found the answer much quicker through this
observation, but the point of the example is to illustrate that PIE works!
1.6. Advanced Counting Using PIE 63
For four or more sets, we do not write down a formula for PIE. Instead, we just think of
the principle: add up all the elements in single sets, then subtract out things you counted
twice (elements in the intersection of a pair of sets), then add back in elements you removed
too often (elements in the intersection of groups of three sets), then take back out elements
you added back in too often (elements in the intersection of groups of four sets), then add
back in, take back out, add back in, etc. This would be very difficult if it wasn’t for the
fact that in these problems, all the cardinalities of the single sets are equal, as are all the
cardinalities of the intersections of two sets, and that of three sets, and so on. Thus we
can group all of these together and multiply by how many different combinations of 1, 2,
3,. . . sets there are.
Example:
How many ways can you distribute 10 cookies to 4 kids so that no kid gets more
than 2 cookies?
Solution: To answer this, we will subtract all the outcomes in which a kid gets 3
or more cookies. How many outcomes are there like that? We can force kid A to
eat 3 or more cookies by giving him 3 cookies before we start. Doing so reduces the
problem to one in which we have 7 cookies to give to 4 kids without any restrictions.
In that case, we have 7 stars (the 7 remaining cookies) and 3 bars (one less than the
10
number of kids) so we can distribute the cookies in 3 ways. Of course we could
choose any
one of the 4 kids to give too many cookies, so it would appear that there
4 10
are 1 3 ways to distribute the cookies giving too many to one kid. But in fact, we
have over counted.
We must get rid of the outcomes in which two kids have too many cookies.
There are 42 ways to select 2 kids to give extra cookies. It takes 6 cookies to do this,
leaving only 4 cookies. So we have 4 stars and still 3 bars.
The remaining 4 cookies
7 4
can thus be distributed in 3 ways (for each of the 2 choices of which 2 kids to
over-feed).
But now we have removed too much. We must add back in all the ways to give
too many cookies to three kids. This uses 9 cookies, leaving only 1 to distribute to
4
the 4 kids using stars and bars, which can be done in 3 ways. We must consider
this outcome for every possible choice of which three kids we over-feed, and there
4
are 3 ways of selecting that set of 3 kids.
Next we would subtract all the ways to give four kids too many cookies, but in
this case, that number is 0.
All together we get that the number of ways to distribute 10 cookies to 4 kids
without giving any kid more than 2 cookies is:
! " ! ! ! ! ! !#
13 4 10 4 7 4 4
− − + 286 − [480 − 210 + 16] 0.
3 1 3 2 3 3 3
1.6. Advanced Counting Using PIE 64
This makes sense: there is NO way to distribute 10 cookies to 4 kids and make sure
that nobody gets more than 2. It is slightly surprising that
! " ! ! ! ! ! !#
13 4 10 4 7 4 4
− +
3 1 3 2 3 3 3
Just so you don’t think that these problems always have easier solutions, consider the
following example.
Example:
Earlier we counted the number of solutions to the equation
x 1 + x 2 + x3 + x 4 + x 5 13.
• Solutions where x 1 > 3 and x 2 > 3: 94 . After you give 4 units to x 1 and
We also need to account for the fact that we could choose any of the five variables in
the place of x 1 above, any pair of variables in the place of x 1 and x2 and so on. It is
because of this that the double counting occurs, so we need to use PIE. All together
we have that the number of solutions with 0 ≤ x i ≤ 3 is
! " ! ! ! ! ! !#
17 5 13 5 9 5 5
− − + 15.
4 1 4 2 4 3 4
1.6. Advanced Counting Using PIE 65
Counting Derangements
Investigate!
For your senior prank, you decide to switch the nameplates on your favorite 5
professors’ doors. So that none of them feel left out, you want to make sure that
all of the nameplates end up on the wrong door. How many ways can this be
accomplished?
but most of these have one or more elements fixed: 123 has all three elements fixed, 132
has the first element fixed (1 is in its original first position), and so on. In fact, the only
derangements of three elements are
If we go up to 4 elements, there are 24 permutations (because we have 4 choices for the first
element, 3 choices for the second, 2 choices for the third leaving only 1 choice for the last).
How many of these are derangements? If you list out all 24 permutations and eliminate
those which are not derangements, you will be left with just 9 derangements. Let’s see how
we can get that number using PIE.
Example:
How many derangements are there of 4 elements?
Solution: We count all permutations, and subtract those which are not derange-
ments. There are 4! 24 permutations of 4 elements. Now for a permutation to
not be a derangement, at least one of the 4 elements must be fixed. There are 41
choices for which single element we fix. Once fixed, we need to find a permutation
of the other three elements. There are 3! permutations on 3 elements. But now we
have counted too many non-derangements, so we must subtract those permutations
4
which fix two elements. There are 2 choices for which two elements we fix, and
then for each pair, 2! permutations of the remaining elements. But this subtracts
4
too many, so add back in permutations which fix 3 elements, all 3 1! of them. Fi-
nally subtract the 44 0! permutations (recall 0! 1) which fix all four elements. All
1.6 Exercises 66
Of course we can use a similar formula to count the derangements on any number
of elements. However, the more elements we have, the longer the formula gets. Here is
another example.
Example:
Five gentlemen attend a party, leaving their hats at the door. At the end of the
party, they hastily grab hats on their way out. How many different ways could this
happen so that none of the gentlemen leave with their own hat?
Solution: We are counting derangements on 5 elements. There are 5! ways for the
gentlemen to grab hats in any order - but many of these permutations will result in
someone getting their own hat. So we subtract all the ways in which one or more
of the men get their own hat. In other words, we subtract the non-derangements.
Doing so requires PIE. Thus the answer is:
" ! ! ! ! ! #
5 5 5 5 5
5! − 4! − 3! + 2! − 1! + 0! .
1 2 3 4 5
Exercises
1. The dollar menu at your favorite tax-free fast food restaurant has 7 items. You have
$10 to spend. How many different meals can you buy if you spend all your money
and:
2. After another gym class you are tasked with putting the 14 identical dodgeballs away
into 5 bins. This time, no bin can hold more than 6 balls. How many ways can you
clean up?
3. Consider the equation x 1 + x 2 + x 3 + x 4 15. How many solutions are there with
2 ≤ x i ≤ 5 for all i ∈ {1, 2, 3, 4}?
4. Suppose you planned on giving 7 gold stars to some of the 13 star students in your
class. Each student can receive at most one star. How many ways can you do this?
Use PIE, and also an easier method, and compare your results.
1.7. Counting Functions 67
5. Based on the previous question, give a combinatorial proof for the identity:
n
n+k−1 n + k − (2j + 1)
! ! ! !
n X n
− (−1) j+1 .
k k j k
j1
7. Ten ladies of a certain age drop off their red hats at the hat check of a museum. As
they are leaving, the hat check attendant gives the hats back randomly. In how many
ways can exactly six of the ladies receive their own hat (and the other four not)?
Investigate!
1. Consider all functions f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5}. How many functions
are there all together? How many of those are injective? Remember, a function
is an injection if every input goes to a different output.
3. Recall that a surjection is a function for which every element of the codomain
is in the range. How many of the functions f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5}
are surjective? Use PIE!
Example:
How many functions f : {1, 2, 3, 4, 5, 6, 7} → { a, b, c, d } are there?4
Solution: We must assign f ( x ) for each x in the domain. We have four choices for
f (1): we could have f (1) a, f (1) b, f (1) c or f (1) d. Similarly, we have four
choices for f (2), and four choices for f (3) and so on. In fact, for each of the seven
elements of the domain, we have four choices, so the number of functions here is
47 .
We might also wonder how many of those functions are injective and how many are
surjective. What would these mean in terms of cookies? The function would be injective if
each kid got no more than one cookie. There is no way to do this, because we have more
cookies than kids. The number of injective functions with this domain and codomain is 0.
On the other hand, a surjective function would be one in which each kid got at least one
cookie. Can we count those? It turns out that counting surjective functions will require
PIE. First back up and consider another simpler example.
Example:
How many functions f : {1, 2, 3, 4, 5} → { a, b, c, d, e } are there? How many of
those functions are injective?
Solution: First we count all the functions with domain {1, 2, 3, 4, 5} and codomain
{ a, b, c, d, e }. For f to be a function, it must assign each element of the domain to
exactly one element of the codomain. In other words, we need to assign one of the
elements of { a, b, c, d, e } to f (1). There are 5 choices for this. Similarly, there are
five choices for f (2). In fact, there are 5 choices for f ( x ) for all five possible values
of x ∈ {1, 2, 3, 4, 5}. Thus the total number of function is 5 · 5 · 5 · 5 · 5 55 .
What about the injective (one-to-one) functions? Again, we have 5 choices for
f (1). However, once we assign a value to f (1), we cannot assign that value to f (2), or
any other f ( x ). So for each of the 5 choices for f (1), we only have 4 choices for f (2),
and then only 3 choices for f (3) and only 2 choices for f (4), leaving only 1 choice
(the last element of the codomain) for f (5). Therefore the number of one-to-one
functions is 5!.
Using the same counting techniques, you should be able to count the total number
of functions from any finite size domain to any finite size codomain, and also count the
injective functions (as long as the size of the codomain is at least as large as the domain,
otherwise there would be none).
So what about surjective functions? In the previous example, there are exactly 5! 120
surjections. We know this because whenever the domain and codomain of a function are
the same finite size, a function is injective if and only if it is surjective. In general, if the
domain and codomain both contain n elements, then the number of surjective functions is
4 The cookies are numbered 1 through 7, the kids are named a through d.
1.7. Counting Functions 69
n!, since this is the number of injective functions. Also, if the codomain is strictly larger
than the domain, then the surjective functions are easy to count: there are 0 of them (since
there will always be elements of the codomain left out of the range). But what if the size of
the codomain is smaller than the size of the domain?
The idea is to count the functions which are not surjective, and then subtract that from
the total number of functions. This works very well when the codomain has two elements
in it:
Example:
How many functions f : {1, 2, 3, 4, 5} → { a, b } are surjective?
Solution: There are 25 functions all together, two choices for where to send each of
the 5 elements of the domain. Now of these, the functions which are not surjective
must exclude one or more elements of the codomain from the range. So first,
consider functions for which a is not in the range. This can only happen one way:
everything gets sent to b. Alternatively, we could exclude b from the range. Then
everything gets sent to a, so there is only one function like this. These are the only
ways in which a function could not be surjective (no function excludes both a and
b from the range) so there are exactly 25 − 2 surjective functions.
When there are three elements in the codomain, there are now three choices for a single
element to exclude from the range. Additionally, we could pick pairs of two elements to
exclude from the range, and we must make sure we don’t over count these. It’s PIE time!
Example:
How many functions f : {1, 2, 3, 4, 5} → { a, b, c } are surjective?
Solution: Again start with the total number of functions: 35 (as each of the five
elements of the domain can go to any of three elements of the codomain). Now we
count the functions which are not surjective.
Start by excluding a from the range. Then we have two choices (b or c) for
where to send each of the five elements of the domain. Thus there are 25 functions
which exclude a from the range. Similarly, there are 25 functions which exclude b,
and another 25 which exclude c. Now have we counted all functions which are not
surjective? Yes, but in fact, we have counted some multiple times. For example,
the function which sends everything to c was one of the 25 functions we counted
when we excluded a from the range, and also one of the 25 functions we counted
when we excluded b from the range. We must subtract out all the functions which
specifically exclude two elements from the range. There is 1 function when we
exclude a and b (everything goes to c), one function when we exclude a and c, and
one function when we exclude b and c.
1.7. Counting Functions 70
We are using PIE: to count the functions which are not surjective, we added up
the functions which exclude a, b, and c separately, then subtracted the functions
which exclude pairs of elements. We would then add back in the functions which
exclude groups of three elements, except that there are no such functions. We find
that the number of functions which are not surjective is
25 + 25 + 25 − 1 − 1 − 1 + 0.
since each of the 25 ’s was the result of choosing 1 of the 3 elements of the codomain
to exclude from the range, each of the three 15 ’s was the result of choosing 2 of the
3 elements of the codomain to exclude. Writing 15 instead of 1 makes sense too: we
have 1 choice of were to send each of the 5 elements of the domain.
Now we can finally count the number of surjective functions:
" ! ! #
3 5 3 5
5
3 − 2 − 1 150.
1 2
You might worry that to count surjective functions when the codomain is larger than
3 elements would be too tedious. We need to use PIE but with more than 3 sets the
formula for PIE is very long. However, we have lucked out. As we saw in the example
above, the number of functions which exclude a single element from the range is the same
no matter which single element is excluded. Similarly, the number of functions which
exclude a pair of elements will be the same for every pair. With larger codomains, we will
see the same behavior with groups of 3, 4, and more elements excluded. So instead of
adding/subtracting each of these, we can simply add or subtract all of them at once, if you
know how many there are. This works just like it did in Section 1.6, only now the size of
the various combinations of sets is a number raised to a power, as opposed to a binomial
coefficient or factorial. Here’s what happens with 4 and 5 elements in the codomain.
Example:
1. How many functions f : {1, 2, 3, 4, 5} → { a, b, c, d } are surjective?
1. There are 45 functions all together; we will subtract the functions which are not
surjective. We could exclude any one of the four elements of the codomain,
and doing so will leave us with 35 functions for each excluded element. This
1.7. Counting Functions 71
counts too many so we subtract the functions which exclude two of the four
elements of the codomain, each pair giving 25 functions. But this excludes
too many, so we add back in the functions which exclude three of the four
elements of the codomain, each triple giving 15 function. There are 41 groups
of functions excluding a single element, 42 groups of functions excluding a
This means that the number of functions which are not surjective is:
! ! !
4 5 4 5 4 5
3 − 2 + 1 .
1 2 3
We can now say that the number of functions which are surjective is:
" ! ! ! #
4 5 4 5 4 5
5
4 − 3 − 2 + 1 .
1 2 3
We took the total number of functions 55 and subtracted all that were not
surjective. There were 51 ways to select a single element from the codomain
to exclude from the range, and for each there were 45 functions. But this
double counts, so we use PIE and subtract functions excluding two elements
from the range: there are 52 choices for the two elements to exclude, and for
each pair, 35 functions. This takes out too many functions,
so we add back
in functions which exclude 3 elements from the range: 53 choices for which
three to exclude, and then 25 functions for each choice of elements. Finally
we take back out the 1 function which excludes 4 elements for each of the 54
choices of 4 elements.
We have seen that counting surjective functions is another nice example of the advanced
use of the Principle of Inclusion/Exclusion first introduced in Section 1.6. Also, counting
injective functions turns out to be equivalent to permutations, and counting all functions
has a solution akin to those counting problems where order matters but repeats are allowed
(like counting the number of words you can make from a given set of letters).
These are not just a few more examples of the techniques we have developed in this
chapter. Quite the opposite: everything we have learned in this chapter are examples of
counting functions!
1.7. Counting Functions 72
Example:
How many 5-letter words can you make using the eight letters a through h? How
many contain no repeated letters?
Solution: By now it should be no surprise that there are 85 words, and P (8, 5)
words without repeated letters. The new piece here is that we are actually counting
functions. For the first problem, we are counting all functions from {1, 2, . . . , 5} to
{ a, b, . . . , h }. The numbers in the domain represent the position of the letter in the
word, the codomain represents the letter that could be assigned to that position. If
we ask for no repeated letters, we are asking for injective functions.
If A and B are any sets with | A | 5 and | B | 8, then the number of functions
f : A → B is 85 and the number of injections is P (8, 5). So if you can represent your
counting problem as a function counting problem, most of the work is done.
Example:
How many subsets are there of {1, 2, . . . , 9}? How many 9-bit strings are there (of
any weight)?
Solution: We saw in Section 1.2 that the answer to both these questions is 29 , as
we can say yes or no (or 0 or 1) to each of the 9 elements in the set (positions in the
bit-string). But 29 also looks like the answer you get from counting functions. In
fact, if you count all functions f : A → B with | A | 9 and | B | 2, this is exactly
what you get.
This makes sense! Let A {1, 2, . . . , 9} and B { y, n }. We are assigning
each element of the set either a yes or a no. Or in the language of bit-strings, we
would take the 9 positions in the bit string as our domain and the set {0, 1} as the
codomain.
So far we have not used a function as a model for binomial coefficients (combinations).
Think for a moment about the relationship between combinations and permutations, say
specifically 93 and P (9, 3). We do have a function model for P (9, 3). This is the number of
injective functions from a set of size 3 (say {1, 2, 3} to a set of size 9 (say {1, 2, . . . , 9}) since
there are 9 choices for where to send the first element of the domain, then only 8 choices
for the second, and 7 choices for the third. For example, the function might look like this:
f (1) 5 f ( 2) 8 f (3) 4.
f (1) 4 f ( 2) 5 f (3) 8.
Now P (9, 3) counts these as different outcomes correctly, but 93 will count these (among
others) as just one outcome. In fact, in terms of functions 93 just counts the number of
1.7 Exercises 73
different ranges possible of injective functions. This should not be a surprise since binomial
coefficients counts subsets, and the range is a possible subset of the codomain.5
While it is possible to interpret combinations as functions, perhaps the better advice is
to instead use combinations (or stars and bars) when functions are not quite the right way
to interpret the counting question.
Example:
You decide to give away your video game collection so to better spend your time
studying advance mathematics. You have 8 games that you will give to 5 different
friends. How many ways can you do this? More importantly, is there a good way
to model this problem using a function?
Solution: What if we did try to use a function to describe the situation? What
should the domain be? It cannot be the 5 friends, because a single element of the
domain cannot be mapped to more than one element of the codomain (and we must
give away all the games). So the domain is the set of 8 games and the codomain
is the set of 5 friends. In this interpretation, assuming that we don’t need to give
at least one game to each friend, we are just looking for functions, not necessarily
injective functions. So the answer would be 58 .
But is this correct? It is if we assume that the 8 games are distinct, or more
importantly, that it matters which game goes to which friend. As it turns out, the
8 games you have are all copies of The Legend of Zelda (it is a great game after all).
Now it really doesn’t make sense to think of this as a function, since the domain is
not a set of 8 objects. Instead we recognize this as a stars and bars problem, and
deduce the answer of 12
8 .
Exercises
1. Write out all functions f : {1, 2, 3} → { a, b }. How many are there? How many are
injective? How many are surjective? How many are both?
2. Write out all functions f : {1, 2} → { a, b, c }. How many are there? How many are
injective? How many are surjective? How many are both?
6. Let A {1, 2, 3, 4, 5}. How many injective functions f : A → A have the property
that for each x ∈ A, f ( x ) , x?
Investigate!
1. Suppose you have a huge box of animal crackers containing plenty of each of
10 different animals. For the counting questions below, carefully examine their
similarities and differences, and then give an answer. The answers are all one
of the following:
10 106 15
P (10, 6) .
6 9
(a) How many animal parades containing 6 crackers can you line up?
(b) How many animal parades of 6 crackers can you line up so that the
animals appear in alphabetical order?
(c) How many ways could you line up 6 different animals in alphabetical
order?
(d) How many ways could you line up 6 different animals if they can come
in any order?
(e) How many ways could you give 6 children one animal cracker each?
(f) How many ways could you give 6 children one animal cracker each so
that no two kids get the same animal?
(g) How many ways could you give out 6 giraffes to 10 kids?
(h) Write
a question about giving animal crackers to kids that has the answer
10
6 .
you practice you start to notice some trends that can help you distinguish between types of
counting problems. Here are some suggestions that you might find helpful when deciding
how to tackle a counting problem and checking whether your solution is correct.
• Remember that you are counting the number of items in some list of outcomes. Write
down part of this list. Write down an element in the middle of the list – how are you
deciding whether your element really is in the list. Could you get this element more
than once using your proposed answer?
• If generating an element on the list involves selecting something (for example, picking
a letter or picking a position to put a letter, etc), can the things you select be repeated?
Remember, permutations and combinations select objects from a set without repeats.
• Does order matter? Be careful here and be sure you know what your answer really
means. We usually say that order matters when you get different outcomes when the
same objects are selected in different orders. Combinations and “Stars & Bars” are
used when order does not matter.
• There are four possibilities when it comes to order and repeats. If order matters and
repeats are allowed, the answer will look like n k . If order matters and repeats are not
allowed, we have P ( n, k ). If order doesn’t matter and repeats are allowed, use stars
n
and bars. If order doesn’t matter and repeats are not allowed, use k . But be careful:
this only applies when you are selecting things, and you should make sure you know
exactly what you are selecting before determining which case you are in.
• Think about how you would represent your counting problem in terms of sets or
functions. We know how to count different sorts of sets and different types of
functions.
• As we saw with combinatorial proofs, you can often solve a counting problem in
more than one way. Do that, and compare your numerical answers. If they don’t
match, something is amiss.
While we have covered many counting techniques, we have really only scratched the
surface of the large subject of enumerative combinatorics. There are mathematicians doing
original research in this area even as you read this. Counting can be really hard.
In the next chapter, we will approach counting questions from a very different direction,
and in doing so, answer infinitely many counting questions at the same time. We will create
sequences of answers to related questions.
Exercises
1. You have 9 presents to give to your 4 kids. How many ways can this be done if:
(a) The presents are identical, and each kid gets at least one present?
(b) The presents are identical, and some kids might get no presents?
1.8 Exercises 76
(c) The presents are unique, and some kids might get no presents?
(d) The presents are unique and each kid gets at least one present?
2. For each of the following counting problems, say whether the answer is
10
A 4 B P (10, 4) C Neither
(a) How many shortest lattice paths are there from (0, 0) to (10, 4)?
(b) If you have 10 bow ties, and you want to select 4 of them for next week, how
many choices do you have?
(c) Suppose you have 10 bow ties and you will wear one on each of the next 4 days.
How many choices do you have?
(d) If you want to wear 4 of your 10 bow ties next week (Monday through Sunday),
how many ways can this be accomplished?
(e) Out of a group of 10 classmates, how many ways can you rank your top 4 friends?
(f) If 10 students come to their professor’s office but only 4 can fit at a time, how
different combinations of 4 students can see the prof first?
(g) How many 4 letter words can be made from the first 10 letters of the alphabet?
(h) How many ways can you make the word “cake” from the first 10 letters of the
alphabet?
(i) How many ways are there to distribute 10 apples among 4 children?
(j) If you have 10 kids (and live in a shoe) and 4 types of cereal, how many ways
can your kids eat breakfast?
(k) How many ways can you arrange exactly 4 ones in a string of 10 binary digits?
(l) You want to select 4 single digit numbers as your lotto picks. How many choices
do you have?
(m) 10 kids want ice-cream. You have 4 varieties. How many ways are there to give
the kids as much ice-cream as they want?
(n) How many 1-1 functions are there from {1, 2, . . . , 10} to { a, b, c, d }?
(o) How many surjective functions are there from {1, 2, . . . , 10} to { a, b, c, d }?
(p) Each of your 10 bow ties match 4 pairs of suspenders. How many outfits can
you make?
(q) After the party, the 10 kids each choose one of 4 party-favors. How many
outcomes?
(r) How many 6-elements subsets are there of the set {1, 2, . . . , 10}
(s) How many ways can you split up 11 kids into 5 teams?
1.8 Exercises 77
3. Recall, you own 3 regular ties and 5 bow ties. You realize that it would be okay to
wear more than two ties to your clown college interview.
(a) You must select some of your ties to wear. Everything is okay, from no ties up
to all ties. How many choices do you have?
(b) If you want to wear at least one regular tie and one bow tie, but are willing to
wear up to all your ties, how many choices do you have for which ties to wear?
(c) How many choices do you have if you wear exactly 2 of the 3 regular ties and 3
of the 5 bow ties?
(d) Once you have selected 2 regular and 3 bow ties, in how many orders could you
put the ties on, assuming you must have one of the three bow ties on top?
4. Give a counting question where the answer is 8 · 3 · 3 · 5. Give another question where
the answer is 8 + 3 + 3 + 5.
5. Consider five digit numbers α a 1 a 2 a 3 a 4 a 5 , with each digit from the set {1, 2, 3, 4}.
8. In a recent small survey of airline passengers, 25 said they had flown American in the
last year, 30 had flown Jet Blue, and 20 had flown Continental. Of those, 10 reported
they had flown on American and Jet Blue, 12 had flown on Jet Blue and Continental,
1.8 Exercises 78
and 7 had flown on American and Continental. 5 passengers had flown on all three
airlines.
How many passengers were surveyed? (Assume the results above make up the entire
survey.)
11. How many 8-letter words contain exactly 5 vowels (a,e,i,o,u)? What if repeated letters
were not allowed?
12. For each of the following, find the number of shortest lattice paths from (0, 0) to (8, 8)
which:
13. You live in Grid-Town on the corner of 2nd and 3rd, and work in a building on the
corner of 10th and 13th. How many routes are there which take you from home to
work and then back home, but by a different route?
14. Give an example of a problem for which P ( n, k ) is the solution. Give another example
of a problem for which nk is the solution.
15. How many 10-bit strings start with 111 or end with 101 or both?
16. How many 10-bit strings of weight 6 start with 111 or end with 101 or both?
17. How many 6 letter words made from the letters a, b, c, d, e, f without repeats do
not contain the sub-word “bad” in (a) consecutive letters? or (b) not-necessarily
consecutive letters (but in order)?
Pn n
2n .
18. Explain using lattice paths why k0 k
!
n
19. Explain the relationship between and P ( n, k ). Be sure to say both how the
k
formulas for each are related, and why that relationship makes sense.
21. Suppose you have 20 one-dollar bills to give out as prizes to your top 5 discrete math
students. How many ways can you do this if:
24. How many functions map {1, 2, 3, 4, 5, 6} onto { a, b, c, d } (i.e., how many surjections
are there)?
25. To thank your math professor for doing such an amazing job all semester, you decide
to bake him (or her) cookies. You know how to make 10 different types of cookies.
(a) If you want to give your professor 4 different types of cookies, how many
different combinations of cookie type can you select? Explain your answer.
(b) To keep things interesting, you decide to make a different number of each type
of cookie. If again you want to select 4 cookie types, how many ways can you
select the cookie types and decide for which there will be the most, second most,
etc. Explain your answer.
(c) You change your mind again. This time you decide you will make a total of 12
cookies. Each cookie could be any one of the 10 types of cookies you know how
to bake (and it’s okay if you leave some types out). How many choices do you
have? Explain.
(d) You realize that the previous plan did not account for presentation. This time,
you once again want to make 12 cookies, each one could be any one of the 10
types of cookies. However, now you plan to shape the cookies into the numerals
1, 2, . . . , 12 (and probably arrange them to make a giant clock, but you haven’t
decided on that yet). How many choices do you have for which types of cookies
to bake into which numerals? Explain.
(e) The only flaw with the last plan is that your professor might not get to sample
all 10 different varieties of cookies. How many choices do you have for which
types of cookies to make into which numerals, given that each type of cookie
should be present at least once? Explain.
1.8 Exercises 80
26. For which of the parts above does it make sense to interpret the counting question
as counting some number of functions? Say what the domain and codomain should
be, and whether you are counting all functions, injections, surjections, or something
else.
Homework Problems
The following are some more involved problems for you to try, which might be assigned
as homework.
1. We usually write numbers in decimal form (or base 10), meaning numbers are com-
posed using 10 different “digits” {0, 1, . . . , 9}. Sometimes though it is useful to write
numbers hexadecimal or base 16. Now there are 16 distinct digits that can be used to
form numbers: {0, 1, . . . , 9, A, B, C, D, E, F}. So for example, a 3 digit hexadecimal
number might be 3B8.
(a) How many 2-digit hexadecimals are there in which the first digit is E or F?
Explain your answer in terms of the additive principle (using either events or
sets).
(b) Explain why your answer to the previous part is correct in terms of the multi-
plicative principle (using either events or sets). Why do both the additive and
multiplicative principles give you the same answer?
(c) How many 3-digit hexadecimals start with a letter (A-F) and end with a numeral
(0-9)? Explain.
(d) How many 3-digit hexadecimals start with a letter (A-F) or end with a numeral
(0-9) (or both)? Explain.
2. For how many three digit numbers (100 to 999) is the sum of the digits even? (For
example, 343 has an even sum of digits: 3 + 4 + 3 10 which is even.) Find the answer
and explain why it is correct in at least two different ways.
3. In a recent survey, 30 students reported whether they liked their potatoes Mashed,
French-fried, or Twice-baked. 15 liked them mashed, 20 liked French fries, and 9
liked twice baked potatoes. Additionally, 12 students liked both mashed and fried
potatoes, 5 liked French fries and twice baked potatoes, 6 liked mashed and baked,
and 3 liked all three styles. How many students hate potatoes? Explain why your
answer is correct.
4. Bonus: The number 735000 factors as 23 · 3 · 54 · 72 . How many divisors does it have?
Explain your answer using the multiplicative principle.
(a) How many subsets of A are there? That is, find | P (A)|. Explain.
(b) How many subsets of A contain exactly 5 elements? Explain.
1.8 Exercises 81
5. How many 9-bit strings (that is, bit strings of length 9) are there which:
6. How many triangles are there with vertices from the points shown below? Note, we
are not allowing degenerate triangles - ones with all three vertices on the same line,
but we do allow non-right triangles. Explain why your answer is correct. (HINT: you
need at exactly two points on either the x- or y-axis, but don’t over-count the right
triangles.)
7. Gridtown USA, besides having excellent donut shoppes, is known for its precisely
laid out grid of streets and avenues. Streets run east-west, and avenues north-south,
for the entire stretch of the town, never curving and never interrupted by parks or
schools or the like.
Suppose you live on the corner of 1st and 1st and work on the corner of 12th and
12th. Thus you must travel 22 blocks to get to work as quickly as possible.
(a) How many different routes can you take to work, assuming you want to get
there as quickly as possible?
(b) Now suppose you want to stop and get a donut on the way to work, from your
favorite donut shoppe on the corner of 8th st and 10th ave. How many routes to
work, via the donut shoppe, can you take (again, ensuring the shortest possible
route)?
(c) Disaster Strikes Gridtown: there is a pothole on 4th avenue between 5th and
6th street. How many routes to work can you take avoiding that unsightly (and
dangerous) stretch of road?
(d) How many routes are there both avoiding the pothole and visiting the donut
shoppe?
1.8 Exercises 82
n!
8. Recall that the formula for P ( n, k ) is . Your task here is to explain why this is
( n − k )!
the right formula.
(a) Suppose you have 12 chips, each a different color. How many different stacks of
5 chips can you make? Explain your answer and why it is the same as using the
formula for P (12, 5).
(b) Using the scenario of the 12 chips again, what does 12! count? What does 7!
count? Explain.
(c) Explain why it makes sense to divide 12! by 7! when computing P (12, 5) (in
terms of the chips).
(d) Does your explanation work for numbers other than 12 and 5? Explain the
formula P ( n, k ) (n−k
n!
)! using the variables n and k.
9. Suppose you own x fezzes and y bow ties. Of course, x and y are both greater than
1.
(a) How many combinations of fez and bow tie can you make? You can wear only
one fez and one bow tie at a time. Explain.
(b) Explain why the answer is also x+y x y
2
− 2 − 2
. (If this is what you claimed the
answer was in part (a), try it again.)
(c) Use your answers to parts (a) and (b) to give a combinatorial proof of the identity
x+y
! ! !
x y
− − xy
2 2 2
11. After a late night of math studying, you and your friends decide to go to your favorite
tax-free fast food Mexican restaurant, Burrito Chime. You decide to order off of the
dollar menu, which has 7 items. Your group has $16 to spend (and will spend all of
it).
(a) How many different orders are possible? Explain. (The order in which the
order is placed does not matter - just which and how many of each item that is
ordered.)
1.8 Exercises 83
(b) How many different orders are possible if you want to get at least one of each
item? Explain.
(c) How many different orders are possible if you don’t get more than 4 of any one
item? Explain. Hint: get rid of the bad orders using PIE.
(a) How many of these functions are strictly increasing? Explain. (A function is
strictly increasing provided if a < b, then f ( a ) < f ( b ).)
(b) How many of the functions are non-decreasing? Explain. (A function is non-
decreasing provided if a < b, then f ( a ) ≤ f ( b ).)
13. The Grinch sneaks into a room with 6 Christmas presents to 6 different people. He
proceeds to switch the name-labels on the presents. How many ways could he do
this if:
(a) No present is allowed to end up with its original label? Explain what each term
in your answer represents.
(b) Exactly 2 presents keep their original labels? Explain.
(c) Exactly 5 presents keep their original labels? Explain.
Chapter 2
Sequences
There is a monastery in Hanoi, as the legend goes, with a great hall containing three tall
pillars. Resting on the first pillar are 64 giant disks (or washers), all different sizes, stacked
from largest to smallest. The monks are charged with the following task: they must move
the entire stack of disks to the third pillar. However, due to the size of the disks, the monks
cannot move more than one at a time. Each disk must be placed on one of the pillars before
the next disk is moved. And because the disks are so heavy and fragile, the monks may
never place a larger disk on top of a smaller disk. When the monks finally complete their
task, the world shall come to an end. Your task: figure out how long before we need to
start worrying about the end of the world.
Investigate!
1. First, let’s find the minimum number of moves required for a smaller number
of disks. Collect some data. Make a table.
3. If the monks were able to move one disk every second without ever stopping,
how long before the world ends?
84
2.1. Basics 85
our question. Of course we will also need to verify that our suspected pattern is correct,
and that this correct pattern really does give us the nth term we think it does, but it is
impossible to prove that your formula is correct without having a formula to start with.
Sequences are also interesting mathematical objects to study in their own right. Let’s
see why.
2.1 Basics
Investigate!
What comes next:
Example:
Can you find the next term in the following sequences?
1. 7, 7, 7, 7, 7, . . . 6. 1, 2, 3, 5, 8, 13, 21, . . .
Solution: No you cannot. You might guess that the next terms are:
2.1. Basics 86
1. 7 4. 64 7. 28 10. 24
2. −3 5. 49 8. 17
3. 4 6. 34 9. −2
In fact, those are the next terms of the sequences I had in mind when I made up the
example, but there is no way to be sure they are correct.
That said, we will often do this. Given the first few terms of a sequence, we can
ask what the pattern in the sequence suggests the next terms are.
Given that no number of initial terms in a sequence is enough to say for certain which
sequence we are dealing with, we need to find another way to specify a sequence. We
consider two ways to do this.
Closed formula
A closed formula for a sequence a 0 , a 1 , a 2 , . . . is a formula for a n using a fixed finite
number of operations on n. This is what you normally think of as a formula in n.
Recursive definition
A recursive definition (sometimes called an inductive definition) for a sequence a 0 , a 1 , a 2 , . . .
consists of a recurrence relation: an equation relating a term of the sequence to previous
terms (terms with smaller index) and an initial condition: a list of a few terms of the
sequence (one less than the number of terms in the recurrence relation).
Example:
Here are a few closed formulas for sequences:
• an n2.
n ( n+1)
• an 2 .
√ n √ −n
1+ 5
2 − 1+2 5
• an 5 .
Note in each case, if you are given n, you can calculate a n directly. Just plug in n.
Here are a few recursive definitions for sequences:
• a n 2a n−1 with a 0 1.
In these cases, if you are given n, you cannot calculate a n directly, you first need to
find a n−1 (or a n−1 and a n−2 ).
Investigate!
You have a large collection of 1 × 1 squares and 1 × 2 dominoes. You want to arrange
these to make a 1 × 15 strip. How many ways can you do this?
1. Start by collecting data. How many length 1 × 1 strips can you make? How
many 1 × 2 strips? How many 1 × 3 strips? And so on.
4. What if I asked you to find the number of 1 × 1000 strips? Would the method
you used to calculate the number fo 1 × 15 strips be helpful?
Example:
Find a 6 in the sequence defined by a n 2a n−1 − a n−2 with a 0 3 and a1 4.
Solution: We know that a 6 2a 5 − a 4 . So to find a 6 we need to find a 5 and a 4 . Well
a 5 2a 4 − a 3 and a4 2a 3 − a 2 ,
a 3 2a 2 − a 1 and a 2 2a 1 − a 0 ,
2.1. Basics 88
a0 3
a1 4
a2 2 · 4 − 3 5
a3 2 · 5 − 4 6
a4 2 · 6 − 5 7
a5 2 · 7 − 6 8
a 6 2 · 8 − 7 9.
Note that now we can guess a closed formula for the nth term of the sequence:
a n n + 3. To be sure this will always work, we could plug in this formula into the
recurrence relation:
Finding closed formulas, or even recursive definitions, for sequences is not trivial. There
is no one method for doing this. Just like in evaluating integrals or solving differential
equations, it is useful to have a bag of tricks you can apply, but sometimes there is no easy
answer.
One useful method is to relate a given sequence to another sequence for which we
already know the closed formula.
Example:
n ( n+1)
Use the formulas Tn 2 and a n 2n to find closed formulas for the following
sequences. Assume the first term is a 1 (not a 0 ).
2. 2, 3, 5, 9, 17, 33,. . .
5. 1, 3, 7, 15, 31, . . .
Solution: Before you say this is impossible, what we are asking for is simply to
find a closed formula which agrees with all of the initial terms of the sequences. Of
course there is no way to read into the mind of the person who wrote the numbers
down, but we can at least do this.
Now the first few terms of Tn , starting with T1 are 1, 3, 6, 10, 15, 21, . . . (these are
the triangular numbers). The first few terms of a n are 2, 4, 8, 16, . . .. Now let’s try
to find formulas for the given sequences.
1. 2, 4, 7, 11, 16, 22, . . . . Note that if subtract 1 from each term, we get the sequence
n ( n+1)
Tn . So this sequence is Tn + 1. Therefore a closed formula is 2 + 1. A
quick check of the first few n confirms we have it right.
3. 2, 6, 12, 20, 30, 42,. . . . Notice that all these terms are even. What happens if
we factor out a 2? We get Tn ! So this sequence has closed formula n ( n + 1).
4. 6, 10, 15, 21, 28, . . . . These are all triangular numbers. However, we are
starting with 6 as our first term instead of as our third term. So if we could
plug in 3 instead of 1 into the formula for Tn , we would be set. Therefore the
(n+2)(n+3)
closed formula is 2 (where n + 3 came from ( n + 2) + 1)
5. 1, 3, 7, 15, 31, . . . . Try adding one to each term and we get powers of 2. You
might guess this because each term is approximately twice the previous term.
Closed formula: 2n − 1.
6. 3, 6, 12, 24, 48, . . . . These numbers are all multiples of 3. Let’s try dividing
each by 3. Doing so gives 1, 2, 4, 8, . . . . Aha. We get the closed formula
3 · 2n−1 .
7. 6, 10, 18, 34, 66, . . . . To get from one term to the next, we almost double
each term. So maybe we can relate this back to 2n . Yes, each term is 2 more
than a power of 2. So we get 2n+1 + 2 (the n + 1 is because the first term is 2
more than 22 , not 21 ). Alternatively, we could have related this sequence to
the second sequence in this example: starting with 3, 5, 9, 17, . . . we see that
this sequence is twice the terms from that sequence (omitting the 2). That
sequence had closed formula 2n−1 + 1. To make the 3 first, we would write
2n + 1. Our sequence would be twice this, so 2(2n + 1), which is the same as
we got before.
2.1 Exercises 90
8. 15, 33, 57, 87, 123, . . . . Try dividing each term by 3. That gives the sequence
5, 11, 19, 29, 41, . . .. Now add one: 12, 20, 30, 42, . . ., which is sequence 3 in
this example, except starting with 6 instead of 2. So let’s start with the formula
for sequence 3: n ( n + 1). To start with the 6, we shift: ( n + 1)( n + 2). But this
is one too many, so subtract 1: ( n + 1)( n + 2) − 1. That gives us our sequence,
but divided by 3. So we want 3(( n + 1)( n + 2) − 1).
Exercises
1. Find the closed formula for each of the following sequences by relating them to a well
know sequence. Assume the first term given is a1 .
3. Consider the three sequences below. For each, find a recursive definition. How are
these sequences related?
4. Write out the first few terms of the sequence given by a 1 3; a n 2a n−1 + 4. Then
find a recursive definition for the sequence 10, 24, 52, 108, . . ..
5. Write out the first few terms of the sequence given by a n n 2 − 3n + 1. Then find a
closed formula for the sequence (starting with a 1 ) 0, 2, 6, 12, 20, . . ..
6. Find a closed formula for the sequence with recursive definition a n 2a n−1 − a n−2
with a 1 1 and a 2 2.
7. Find a recursive definition for the sequence with closed formula a n 3 + 2n. Bonus
points if you can give a recursive definition in which makes use of two previous terms
and no constants.
2.2. Arithmetic and Geometric Sequences 91
Investigate!
For the patterns of dots below, draw the next pattern in the sequence. Then give a
recursive definition and a closed formula for the number of dots in the nth pattern.
n0
n1
n2
n0
n1
n2
n1
n2
n3
n4
Arithmetic Sequences
If the terms of a sequence differ by a constant, we say the sequence is arithmetic.
If the initial term (a 0 ) of the sequence is a and the common difference is d, then
we have,
Recursive definition: a n a n−1 + d with a 0 a.
Closed formula: a n a + dn.
How do we know this? For the recursive definition, we need to specify a 0 . Then
we need to express a n in terms of a n−1 . If we call the first term a, then a 0 a. For the
recurrence relation, by the definition of an arithmetic sequence, the difference between
2.2. Arithmetic and Geometric Sequences 92
a0 a a n a n−1 + d.
To find a closed formula, first try writing out the first write out the sequence using the
recursive definition without simplifying:
a, a + d, a + d + d, a + d + d + d, . . . .
We see that to find the nth term, we need to start with a and then add d a bunch of
times. In fact, add it n times. Thus a n a + dn.
Example:
Find recursive definitions and closed formulas for the sequences below. Assume
the first term listed is a 0 .
1. 2, 5, 8, 11, 14, . . ..
Solution: First we should check that these sequences really are arithmetic. Doing
so will reveal the common difference d.
2. Here the common difference is −7, since we add −7 to 50 to get 43, and so on.
Thus we have a recursive definition of a n a n−1 − 7 with a 0 50. The closed
formula is a n 50 − 7n.
What about sequences like 2, 6, 18, 54, . . .? This is not arithmetic because the difference
between terms is not constant. However, the ratio between successive terms is constant.
We call such sequences geometric.
The recursive definition for the geometric sequence with initial term a and common
ratio r is a n a n · r; a 0 a. To get the next term we multiply the previous term by r. We can
find the closed formula like we did for the arithmetic progression. Write a 0 a, a 1 a · r,
a 2 a 1 · r a · r · r and so on. We must multiply the first term a by r a number of times, n
to be precise. We get a n a · r n .
Geometric Sequences
A sequence is called geometric if the ratio between successive terms is constant. Sup-
pose the initial term a 0 is a and the common ratio is r. Then we have,
Recursive definition: a n ra n−1 with a 0 a.
Closed formula: a n a · r n .
2.2. Arithmetic and Geometric Sequences 93
Example:
Find the recursive and closed formula for the sequences below. Again, the first term
listed is a 0 .
Solution: Again, we should first check that these sequences really are geometric,
and doing so will tell us what r is.
1. 6/3 2, 12/6 2, 24/12 2, etc. Yes, to get from any term to the next, we
multiply by r 2. So the recursive definition is a n 2a n−1 with a 0 3. The
closed formula is a n 3 · 2n .
In the examples and formulas above, we assumed that the initial term was a 0 . If your
sequence starts with a 1 , you can easily find the term that would have been a 0 and use that
in the formula. For example, if we want a formula for the sequence 2, 5, 8, . . . and insist that
2 a 1 , then we can find a 0 −1 (since the sequence is arithmetic with common difference
3, we have a 0 + 3 a 1 ). Then the closed formula will be a n −1 + 3n.
Investigate!
Your neighborhood grocery store has a candy machine full of Skittles.
1. Suppose that the candy machine currently holds exactly 650 Skittles, and every
time someone inserts a quarter, exactly 7 Skittles come out of the machine.
(a) How many Skittles will be left in the machine after 20 quarters have been
inserted?
(b) Will there ever be exactly zero Skittles left in the machine? Explain.
2. What if the candy machine gives 7 Skittles to the first customer who put in
a quarter, 10 to the second, 13 to the third, 16 to the fourth, etc. How many
Skittles has the machine given out after 20 quarters are put into the machine?
3. Now, what if the machine gives 4 Skittles to the first customer, 7 to the second,
12 to the third, 19 to the fourth, etc. How many Skittles has the machine given
out after 20 quarters are put into the machine?
Look at the sequence 1, 3, 6, 10, 15, . . .. These are called the triangular numbers since
they represent the number of dots in an equilateral triangle (think of how you arrange 10
bowling pins: a row of 4 plus a row of 3 plus a row of 2 and a row of 1).
Is this sequence arithmetic? No, since 3 − 1 2 and 6 − 3 3 , 2, so there is no common
difference. Is the sequence geometric? No. 3/1 3 but 6/3 2, so there is no common
ratio. What to do?
Notice that the differences between terms form an arithmetic sequence: 2, 3, 4, 5, 6, . . ..
This says that the nth term of the sequence 1, 3, 6, 10, 15, . . . is the sum of the first n terms in
the sequence 1, 2, 3, 4, 5, . . .. We say that the first sequence is the sequence of partial sums of
the second sequence (partial sums because we are not taking the sum of all infinitely many
terms). If we know how to add up the terms of an arithmetic sequence, we could use this
to find a closed formula for a sequence whose differences are the terms of that arithmetic
sequence.
This should become clearer if we write the triangular numbers like this:
11
31+2
61+2+3
10 1 + 2 + 3 + 4
.. ..
. .
Tn 1 + 2 + 3 + · · · + n.
Consider how we could find the sum of the first 100 positive integers (that is, T100 ).
Instead of adding them in order, we regroup and add 1 + 100 101. The next pair to
combine is 2 + 99 101. Then 3 + 98 101. Keep going. This gives 50 pairs which each
add up to 101, so T100 101 · 50 5050.
n ( n+1)
In general, using this same sort of regrouping, we find that Tn 2 . Incidentally,
this is exactly the same as n+1
2 , which makes sense if you think of the triangular numbers
as counting the number of handshakes that take place at a party with n + 1 people.
The point of all of this is that some sequences, while not arithmetic or geometric, can be
interpreted as the sequence of partial sums of arithmetic and geometric sequences. Luckily
there are methods we can use to compute these sums quickly.
Example:
Find the sum: 2 + 5 + 8 + 11 + 14 + · · · + 470.
Solution: The idea is to mimic how we found the formula for triangular numbers.
If we add the first and last terms, we get 472. The second term and second-to-last
2.2. Arithmetic and Geometric Sequences 95
term also add up to 472. To keep track of everything, we might express this as
follows. Call the sum S. Then,
To find 2S then we add 472 to itself a number of times. What number? We need
to decide how many terms (summands) are in the sum. Since the terms form an
arithmetic sequence, the nth term in the sum (counting 2 as the 0th term) can be
expressed as 2 + 3n. If 2 + 3n 470 then n 156. So n ranges from 0 to 156, giving
157 terms in the sum. This is the number of 472’s in the sum for 2S. Thus
This will work for any sum of arithmetic sequences. Call the sum S. Reverse and add.
This produces a single number added to itself many times. Find the number of times.
Multiply. Divided by 2. Done.
Example:
Find a closed formula for 6 + 10 + 14 + · · · + (4n − 2).
Solution: Again, we have a sum of an arithmetic sequence. We need to know how
many terms are in the sequence. Clearly each term in the sequence has the form
4k − 2 (as evidenced by the last term). For which values of k though? To get 6,
k 2. To get 4n − 2 take k n. So to find the number of terms, we need to know
how many integers are in the range 2, 3, . . . , n. The answer is n − 1. (There are n
numbers from 1 to n, so one less if we start with 2.)
Now use the reverse and add trick:
S 6 + 10 + 14 +···+ 4n − 6 + 4n − 2
+ S 4n − 2 + 4n − 6 + 4n − 10 +···+ 10 + 6
2S 4n + 4 + 4n + 4 + 4n + 4 +···+ 4n + 4 + 4n + 4
(n − 2)(4n + 4)
2S ( n − 2)(4n + 4) so S
2
2.2. Arithmetic and Geometric Sequences 96
The previous example could be used to find the closed formula for 0, 6, 16, 30, . . ., since
it is the sequence of partial sums.
Example:
What is 3 + 6 + 12 + 24 + · · · + 12288?
Solution: Multiply each term by 2, the common ratio. You get 2S 6 + 12 + 24 +
· · · + 24576. Now subtract: 2S − S −3 + 24576 24573. Since 2S − S S, we have
our answer.
To better see what happened in the above example, try writing it this way:
S 3 + 6 + 12 + 24 + · · · + 12288
− 2S 6 + 12 + 24 + · · · + 12288 + 24576
−S 3 + 0 + 0 + 0 + · · · + 0 − 24576
Then divide both sides by −1 and we have the same result for S. The idea is, by
multiplying the sum by the common ratio, each term becomes the next term. We shift over
the sum to get the subtraction to mostly cancel out, leaving just the first term and new last
term.
Example:
Find a closed formula for S ( n ) 2 + 10 + 50 + · · · + 2 · 5n .
Solution: The common ratio is 5. So we have
S 2 + 10 + 50 + · · · + 2 · 5n
− 5S 10 + 50 + · · · + 2 · 5n + 2 · 5n+1
−4S 2 − 2 · 5n+1
2 − 2 · 5n+1
Thus S
−4
Even though this might seem like a new technique, you have probably used it before.
2.2 Exercises 97
Example:
Express 0.464646 . . . as a fraction.
Solution: Let N 0.46464646 . . .. Consider 0.01N. We get:
N 0.4646464 . . .
−0.01N 0.00464646 . . .
0.99N 0.46
So N 9946
. What have we done? We viewed the repeating decimal 0.464646 . . . as a
sum of the geometric sequence 0.46, 0.0046, 0.000046, . . . The common ratio is 0.01.
The only real difference is that we are now computing an infinite geometric sum,
we do not have the extra “last” term to consider. Really, this is the result of taking
a limit as you would in calculus when you compute infinite geometric sums.
P Q
and notation
n
X
To simplify writing out sums, we will use notation like a k . This means add up the a k ’s
k1
where k changes from 1 to n.
Example:
P
Use notation to rewrite the sums:
1. 1 + 2 + 3 + 4 + · · · + 100
2. 1 + 2 + 4 + 8 + · · · + 250
3. 6 + 10 + 14 + · · · + (4n − 2).
Solution:
100
X 50
X n
X
k 3. (4k − 2)
1. k 2. 2
k1 k0 k2
n
Y n
Y
If we want to multiply the a k instead, we would write a k . For example, k n!.
k1 k1
Exercises
1. Consider the sequence 8, 14, 20, 26, . . ..
(b) Find a formula for the nth term of this sequence, assuming a 1 8.
P100
(c) Find the sum of the first 100 terms of the sequence: k1 ak .
3. Find 5 + 7 + 9 + 11 + · · · + 521.
4. Find 5 + 15 + 45 + · · · + 5 · 320 .
230
5. Find 1 − 2
3 + 4
9 −···+ 330
.
6. Find x and y such that 27, x, y, 1 is part of an arithmetic sequence. Then find x and
y so that the sequence is part of a geometric sequence. (Warning: x and y might not
be integers.)
7. Consider the sequence 2, 7, 15, 26, 40, 57, . . . (with a 0 2). By looking at the differ-
ences between terms, express the sequence as a sequence of partial sums. Then find
a closed formula for the sequence by computing the nth partial sum.
P Q
8. Use summation ( ) or product ( ) notation to rewrite the following.
(a) 2 + 4 + 6 + 8 + · · · + 2n.
(b) 1 + 5 + 9 + 13 + · · · + 425.
(c) 1 + 1
2 + 1
3 + 1
4 +···+ 1
50 .
(d) 2 · 4 · 6 · · · · · 2n.
(e) ( 12 )( 23 )( 43 ) · · · ( 100
101 ).
9. Expand the following sums and products. That is, write them out the long way.
100
X
(a) (3 + 4k ).
k1
n
X
(b) 2k .
k0
50
X 1
(c) .
k2
( k 2 − 1)
100
Y k2
(d) .
k2
(k 2 − 1)
n
Y
(e) (2 + 3k ).
k0
2.3. Polynomial Fitting 99
Investigate!
A standard 8 × 8 chessboard contains 64 squares. Actually, this is just the number of
unit squares. How many squares of all sizes are there on a chessboard? Start with
smaller boards: 1 × 1, 2 × 2, 3 × 3, etc. Find a formula for the total number of squares
in an n × n board.
This sequence is not arithmetic (or geometric for that matter), but perhaps it’s sequence of
differences is. For differences we get
4, 9, 16, 25, . . .
Not a huge surprise: one way to count the number of squares in a 4 × 4 chessboard is to
notice that there are 16 squares with side length 1, 9 with side length 2, 4 with side length
3 and 1 with side length 4. So the original sequence is just the sum of squares. Now this
sequence of differences is not arithmetic since it’s sequence of differences (the differences
of the differences of the original sequence) is not constant. In fact, this sequence of second
differences is
5, 7, 9, . . .
which is an arithmetic sequence (with constant difference 2). Notice that our original
sequence had third differences (that is, differences of differences of differences of the original)
constant. We will call such a sequence ∆3 -constant. The sequence 1, 4, 9, 16, . . . has second
differences constant, so it will be a ∆2 -constant sequence. In general, we will say a sequence
is a ∆k -constant sequence if the kth differences are constant.
Now ∆0 -constant sequences are themselves constant, so a closed formula for them is
easy to compute (it’s just the constant). The ∆1 -constant sequences are arithmetic and we
2.3. Polynomial Fitting 100
have a method for finding closed formulas for them as well. Every ∆2 -constant sequence
is the sum of an arithmetic sequence so we can find formulas for these as well. But notice
that the format of the closed formula for a ∆2 -constant sequence is always quadratic. For
example, the square numbers are ∆2 -constant with closed formula a n n 2 . The triangular
n ( n+1)
numbers (also ∆2 -constant) have closed formula a n 2 , which when multiplied out
gives you an n 2 term as well. It appears that every time we increase the complexity of
the sequence, that is, increase the number of differences before we get constants, we also
increase the degree of the polynomial used for the closed formula. We go from constant
to linear to quadratic. The sequence of differences between terms tells us something about
the rate of growth of the sequence. If a sequence is growing at a constant rate, then the
formula for the sequence will be linear. If the sequence is growing at a rate which itself is
growing at a constant rate, then the formula is quadratic. You have seen this elsewhere:
if a function has a constant second derivative (rate of change) then the function must be
quadratic.
This works in general:
Finite Differences
The closed formula for a sequence will be a degree k polynomial if and only if the
sequence is ∆k -constant (i.e., the kth sequence of differences is constant).
This tells us that the sequence 1, 5, 14, 30, 55, . . . will have a cubic (degree 3 polynomial)
for its closed formula.
Now once we know what format the closed formula for a sequence will take, it is much
easier to actually find the closed formula. In the case that the closed formula is a degree k
polynomial, we just need k + 1 data points to “fit” the polynomial to the data.
Example:
Find a formula for the sequence 3, 7, 14, 24, . . .. Assume a 1 3.
Solution: First, check to see if the formula has constant differences at some level.
The sequence of first differences is 4, 7, 10, . . . which is arithmetic, so the sequence
of second differences is constant. The sequence is ∆2 -constant, so the formula for a n
will be a degree 2 polynomial. That is, we know that for some constants a, b, and c,
a n an 2 + bn + c.
Now to find a, b, and c. First, it would be nice to know what a0 is, since plugging
in n 0 simplifies the above formula greatly. In this case, a 0 2 (work backwards
from the sequence of constant differences). Thus
a 0 2 a · 02 + b · 0 + c,
2.3. Polynomial Fitting 101
a1 3 a + b + 2
a 2 7 a4 + b2 + 2.
At this point we have two (linear) equations and two unknowns, so we can solve
the system for a and b (using substitution or elimination or even matrices). We find
a 32 and b −1
2 , so a n 2 n − 2 n + 2.
3 2 1
Example:
Find a closed formula for the number of squares on an n × n chessboard.
Solution: We have seen that the sequence 1, 5, 14, 30, 55, . . . is ∆3 -constant, so we
are looking for a degree 3 polynomial. That is,
a n an 3 + bn 2 + cn + d.
We can find d if we know what a 0 is. Working backwards from the third differences,
we find a 0 0 (unsurprisingly, since there are no squares on a 0 × 0 chessboard).
Thus d 0. Now plug in n 1, n 2, and n 3:
1 a + b + c
5 8a + 4b + 2c
14 27a + 9b + 3c.
Not all sequences will have polynomials as their closed formula. We can use the theory
of finite differences to identify these.
Example:
Determine whether the following sequences can be described by a polynomial, and
if so, of what degree.
1. 1, 2, 4, 8, 16, . . .
3. 1, 1, 2, 3, 5, 8, 13, . . .
Solution:
2. The sequence of first differences is 7, 43, 133, 301, 571, . . .. The second dif-
ferences are: 36, 90, 168, 270, . . .. Third difference: 54, 78, 102, . . .. Fourth
differences: 24, 24, . . .. As far as we can tell, this sequence of differences is
constant so the sequence is ∆4 -constant and as such the closed formula is a
degree 4 polynomial.
Exercises
1. Use polynomial fitting to find the formula for the nth term of the following sequences.
Assume the first term is a 0 .
2. Consider the sequence 1, 3, 7, 13, 21, . . .. Explain how you know the closed formula
for the sequence will be quadratic. Then guess the correct formula by comparing this
sequence to the squares 1, 4, 9, 16, . . ..
3. Use a similar technique as in the previous exercise to find a closed formula for the
sequence 2, 11, 34, 77, 146, 247, . . ..
5. Repeat the above assuming this time a n an 2 + bn + c. That is, prove that every
quadratic sequence has arithmetic differences.
6. Can you use polynomial fitting to find the formula for the nth term of the sequence
4, 7, 11, 18, 29, 47, . . . ? Explain why or why not.
7. Will the nth sequence of differences of 2, 6, 18, 54, 162, . . . ever be constant? Explain.
2.4. Solving Recurrence Relations 103
Investigate!
Consider the recurrence relation
a n 5a n−1 − 6a n−2 .
Example:
Find a recurrence relation and initial conditions for 1, 5, 17, 53, 161, 485 . . ..
Solution: Finding the recurrence relation would be easier if we had some context
for the problem (like the Tower of Hanoi, for example). Alas, we have only the
sequence. Remember, the recurrence relation tells you how to get from previous
terms to future terms. What is going on here? We could look at the differences
between terms: 4, 12, 36, 108, . . .. Notice that these are growing by a factor of 3. Is
the original sequence as well? 1 · 3 3, 5 · 3 15, 17 · 3 51 and so on. It appears
that we always end up with 2 less than the next term. Aha!
So a n 3a n−1 + 2 is our recurrence relation and the initial condition is a 0 1.
We are going to try to solve these recurrence relations. By this we mean something
very similar to solving differential equations: we want to find a function of n (a closed
formula) which satisfies the recurrence relation, as well as the initial condition.1 Just like
for differential equations, finding a solution might be tricky, but checking that the solution
is correct is easy.
1 Recurrence relations are sometimes called difference equations since they can describe the difference
between terms and this highlights the relation to differential equations further.
2.4. Solving Recurrence Relations 104
Example:
Check that a n 2n + 1 is a solution to the recurrence relation a n 2a n−1 − 1 with
a 1 3.
Solution: First, it is easy to check the initial condition: a 1 should be 21 +1 according
to our closed formula. But 21 + 1 3, which is what we want. To check that our
proposed solution satisfies the recurrence relation, try plugging it in.
2a n−1 − 1 2(2n−1 + 1) − 1 2n + 2 − 1 2n + 1 a n .
because every third term looks like: 2 + −2 0, and then 3 + −3 0 and so on.
We can use this behavior to solve recurrence relations. Here is an example.
Example:
Solve the recurrence relation a n a n−1 + n with initial term a 0 4.
Solution: To get a feel for the recurrence relation, write out the first few terms of
the sequence: 4, 5, 7, 10, 14, 19, . . .. Look at the difference between terms. a 1 − a 0 1
and a2 − a 1 2 and so on. The key thing here is that the difference between terms is
n. We can write this explicitly: a n − a n−1 n. Of course, we could have arrived at
this conclusion directly from the recurrence relation by subtracting a n−1 from both
sides.
Now use this equation over and over again, changing n each time:
a1 − a0 1
a2 − a1 2
a3 − a2 3
.. ..
. .
a n − a n−1 n.
2.4. Solving Recurrence Relations 105
Add all these equations together. On the right hand side, we get the sum 1 + 2 + 3 +
n ( n+1)
· · · + n. We already know this can be simplified to 2 . What happens on the left
hand side? We get
This sum telescopes. We are left with only the −a 0 from the first equation and the
n ( n+1)
a n from the last equation. Putting this all together we have −a 0 + a n 2 or
n ( n+1)
a n 2 + a 0 . But we know that a 0 4. So the solution to the recurrence relation,
subject to the initial condition is
n ( n + 1)
an + 4.
2
(Now that we know that, we should notice that the sequence is the result of adding
4 to each of the triangular numbers.)
The above example shows a way to solve recurrence relations of the form a n a n−1 + f ( n )
where nk1 f ( k ) has a known closed formula. If you rewrite the recurrence relation as
P
a n − a n−1 f ( n ), and then add up all the different equations with n ranging between 1 and
n, the left hand side will always give you a n − a 0 . The right hand side will be nk1 f ( k ),
P
which is why we need to know the closed formula for that sum.
However, telescoping will not help us with a recursion such as a n 3a n−1 + 2 since the
left hand side will not telescope. You will have −3a n−1 ’s but only one a n−1 . However, we
can still be clever if we use iteration.
We have already seen an example of iteration when we found the closed formula for
arithmetic and geometric sequences. The idea is, we iterate the process of finding the next
term, starting with the known initial condition, up until we have a n . Then we simplify. In
the arithmetic sequence example, we simplified by multiplying d by the number of times
we add it to a when we get to a n , to get from a n a + d + d + d + · · · + d to a n a + dn.
To see how this works, let’s go through the same example we used for telescoping, but
this time use iteration.
Example:
Use iteration to solve the recurrence relation a n a n−1 + n with a 0 4.
Solution: Again, start by writing down the recurrence relation when n 1. This
time, don’t subtract the a n−1 terms to the other side:
a 1 a 0 + 1.
a 2 ( a 0 + 1) + 2.
2.4. Solving Recurrence Relations 106
a 3 (( a 0 + 1) + 2) + 3.
We notice a pattern. Each time, we take the previous term and add the current
index. So
a n (((( a 0 + 1) + 2) + 3) + · · · + n − 1) + n.
Regrouping terms, we notice that a n is just a 0 plus the sum of the integers from 1 to
n. So, since a 0 4,
n ( n + 1)
an 4 + .
2
Of course in this case we still needed to know formula for the sum of 1, . . . , n. Let’s try
iteration with a sequence for which telescoping doesn’t work.
Example:
Solve the recurrence relation a n 3a n−1 + 2 subject to a 0 1.
Solution: Again, we iterate the recurrence relation, building up to the index n.
a 1 3a 0 + 2
a 2 3( a 1 ) + 2 3(3a 0 + 2) + 2
a 3 3[ a 2 ] + 2 3[3(3a 0 + 2) + 2] + 2
.. .. ..
. . .
a n 3( a n−1 ) + 2 3(3(3(3 · · · (3a 0 + 2) + 2) + 2) · · · + 2) + 2.
It is difficult to see what is happening here because we have to distribute all those
3’s. Let’ try again, this time simplifying a bit as we go.
a 1 3a0 + 2
a 2 3( a 1 ) + 2 3(3a 0 + 2) + 2 32 a 0 + 2 · 3 + 2
a 3 3[ a 2 ] + 2 3[32 a 0 + 2 · 3 + 2] + 2 33 a 0 + 2 · 32 + 2 · 3 + 2
.. .. ..
. . .
a n 3( a n−1 ) + 2 3(3n−1 a 0 + 2 · 3n−2 + · · · + 2) + 2
3n a 0 + 2 · 3n−1 + 2 · 3n−2 + · · · + 2 · 3 + 2.
Now we simplify. a 0 1, so we have 3n + (· · · ). Note that all the other terms have a
2 in them. In fact, we have a geometric sum with first term 2 and common ratio 3.
n
We have seen how to simplify 2 + 2 · 3 + 2 · 32 + · · · + 2 · 3n−1 . We get 2−2·3
−2 which
simplifies to 3n − 1. Putting this together with the first 3n term gives our closed
2.4. Solving Recurrence Relations 107
formula:
a n 2 · 3n − 1.
Iteration can be messy, but when the recurrence relation only refers to one previous term
(and maybe some function of n) it can work well. However, trying to iterate a recurrence
relation such as a n 2a n−1 + 3a n−2 will be way too complicated. We would need to keep
track of two sets of previous terms, each of which were expressed by two previous terms,
and so on. The length of the formula would grow exponentially (double each time, in fact).
Luckily there happens to be a method for solving recurrence relations which works very
well on relations like this.
Characteristic Roots
Given a recurrence relation a n + αa n−1 + βa n−2 0, the characteristic polynomial is
x 2 + αx + β
x 2 + αx + β 0.
If r1 and r2 are two distinct roots of the characteristic polynomial (i.e, solutions to the
characteristic equation), then the solution to the recurrence relation is
a n ar1n + br2n ,
Example:
Solve the recurrence relation a n 7a n−1 − 10a n−2 with a0 2 and a 1 3.
Solution: Rewrite the recurrence relation a n − 7a n−1 + 10a n−2 0. Now form the
characteristic equation:
x 2 − 7x + 10 0
and solve for x:
(x − 2)(x − 5) 0
so x 2 and x 5 are the characteristic roots. We therefore know that the solution
to the recurrence relation will have the form
a n a2n + b5n .
To find a and b, plug in n 0 and n 1 to get a system of two equations with two
unknowns:
2 a20 + b50 a + b
3 a21 + b51 2a + 5b
Perhaps the most famous recurrence relation is Fn Fn−1 + Fn−2 , which together with
the initial conditions F0 0 and F1 1 defines the Fibonacci sequence. But notice that
this is precisely the type of recurrence relation on which we can use the characteristic root
2.4. Solving Recurrence Relations 109
technique. When you do, the only thing that changes is that the characteristic equation
does not factor, so you need to use the quadratic formula to find the characteristic roots. In
fact, doing so gives the third most famous irrational number, ϕ, the golden ratio.
Before leaving the characteristic root technique, we should think about what might
happen when you solve the characteristic equation. We have an example above in which
the characteristic polynomial has two distinct roots. These roots can be integers, or perhaps
irrational numbers (requiring the quadratic formula to find them). In these cases, we know
what the solution to the recurrence relation looks like.
However, it is possible for the characteristic polynomial to only have one root. This can
happen if the characteristic polynomial factors as ( x − r )2 . It is still the case that r n would
be a solution to the recurrence relation, but we won’t be able to find solutions for all initial
conditions using the general form a n ar1n + br2n , since we can’t distinguish between r1n
and r2n . We are in luck though:
a n ar n + bnr n
Notice the extra n in bnr n . This allows us to solve for the constants a and b from the
initial conditions.
Example:
Solve the recurrence relation a n 6a n−1 − 9a n−2 with initial conditions a 0 1 and
a 1 4.
Solution: The characteristic polynomial is x 2 − 6x + 9. We solve the characteristic
equation
x 2 − 6x + 9 0
by factoring:
( x − 3) 2 0
so x 3 is the only characteristic root. Therefore we know that the solution to the
recurrence relation has the form
a n a3n + bn3n
a 0 1 a30 + b · 0 · 30 a
a 1 4 a · 3 + b · 1 · 3 3a + 3b.
2.4 Exercises 110
1
a n 3n + n3n .
3
Although we will not consider examples more complicated than these, this character-
istic root technique can be applied to much more complicated recurrence relations. For
example, a n 2a n−1 + a n−2 − 3a n−3 has characteristic polynomial x 3 − 2x 2 − x + 3. Assum-
ing you see how to factor such a degree 3 (or more) polynomial you can easily find the
characteristic roots and as such solve the recurrence relation (the solution would look like
a n ar1n + br2n + cr3n if there were 3 distinct roots). It is also possible to solve recurrence
relations of the form a n αa n−1 + βa n−2 + C for some constant C. It is also possible (and
acceptable) for the characteristic roots to be complex numbers.
Investigate!
1. Think back to the magical candy machine at your neighborhood grocery store.
Suppose that the first time a quarter is put into the machine 1 Skittle comes
out. The second time, 4 Skittles, the third time 16 Skittles, the fourth time 64
Skittles, etc.
(a) Find both a recursive and closed formula for how many Skittles the nth
customer gets.
(b) Check your solution for the closed formula by solving the recurrence
relation using the Characteristic Root technique.
2. You have access to 1 × 1 tiles which come in 2 different colors and 1 × 2 tiles
which come in 3 different colors. We want to figure out how many different
1 × n path designs we can make out of these tiles.
4. Find the solution to the recurrence relation a n 3a n−1 +4a n−2 with initial terms a0 2
and a 1 3.
5. Find the solution to the recurrence relation a n 3a n−1 +4a n−2 with initial terms a0 5
and a 1 8.
7. Solve the recurrence relation a n 3a n−1 + 10a n−2 with initial terms a 0 4 and a 1 1.
2.5 Induction
Mathematical induction is a proof technique, not unlike direct proof or proof by contra-
diction or combinatorial proof.2 In other words, induction is a style of argument we use
to convince ourselves and others that a mathematical statement is always true. Many
mathematical statements can be proved by simply explaining what they mean. Others are
very difficult to prove – in fact, there are relatively simple mathematical statements which
nobody yet knows how to prove. To facilitate the discovery of proofs, it is important to be
familiar with some standard styles of arguments. Induction is one such style. Let’s start
with an example.
Stamps
Investigate!
You need to mail a package, but don’t yet know how much postage you will need.
You have a large supply of 8-cent stamps and 5-cent stamps. Which amounts of
postage can you make exactly using these stamps? Which amounts are impossible
to make?
stamps.” Since each P ( n ) is a statement, it is either true or false. So if we form the sequence
of statements
P (1) , P (2) , P (3) , P (4) , . . .
the sequence will consist of T’s (for true) and F’s (for false). In our particular case the
sequence starts
F, F, F, F, T, F, F, T, F, F, T, F, F, T, . . .
because P (1) , P (2) , P (3) , P (4) are all false (you cannot make 1, 2, 3, or 4 cents of postage)
but P (5) is true (use one 5-cent stamp), and so on.
Let’s think a bit about how we could find the value of P ( n ) for some specific n (the
“value” will be either T or F). How did we find the value of the nth term of a sequence of
numbers? How did we find a n ? There were two ways we could do this: either there was a
closed formula for a n , so we could plug in n into the formula and get our output value, or
we had a recursive definition for the sequence, so we could use the previous terms of the
sequence to compute the nth term. When dealing with sequences of statements, we could
use either of these techniques as well. Maybe there is a way to use n itself to determine
whether we can make n cents of postage. That would be something like a closed formula.
Or instead we could use the previous terms in the sequence (of statements) to determine
whether we can make n cents of postage. That is, if we know the value of P ( n − 1), can we
get from that to the value of P ( n )? That would be something like a recursive definition for
the sequence. Remember, finding recursive definitions for sequences was often easier than
finding closed formulas. The same is true here.
Suppose I told you that P (43) was true (it is). Can you determine from this fact the
value of P (44) (whether it true or false)? Yes you can. Even if we don’t know how exactly
we made 43 cents out of the 5-cent and 8-cent stamps, we do know that there was some
way to do it. What if that way used at least three 5-cent stamps (making 15 cents)? We
could replace those three 5-cent stamps with two 8-cent stamps (making 16 cents). The
total postage has gone up by 1, so we have a way to make 44 cents, so P (44) is true. Of
course, we assumed that we had at least three 5-cent stamps. What if we didn’t? Then we
must have at least three 8-cent stamps (making 24 cents). If we replace those three 8-cent
stamps with five 5-cent stamps (making 25 cents) then again we have bumped up our total
by 1 cent so we can make 44 cents, so P (44) is true.
Notice that we have not said how to make 44 cents, just that we can, on the basis that we
can make 43 cents. How do we know we can make 43 cents? Perhaps because we know we
can make 42 cents, which we know we can do because we know we can make 41 cents, and
so on. It’s a recursion! As with a recursive definition of a numerical sequence, we must
specify our initial value. In this case, the initial value is “P (1) is false.” That’s not good,
since our recurrence relation just says that P ( k + 1) is true if P ( k ) is also true. We need to
start the process with a true P ( k ). So instead, we might want to use “P (31) is true” as the
initial condition.
Putting this all together we arrive at the following fact: it is possible to (exactly) make
any amount of postage greater than 27 cents using just 5-cent and 8-cent stamps.3 In other
words, P ( k ) is true for any k ≥ 28. To prove this, we could do the following:
3 This is not claiming that there are no amounts less than 27 cents which can also be made.
2.5. Induction 113
Suppose we have done this. Then we know that the 28th term of the sequence above
is a T (using step 1, the initial condition or base case), and that every term after the 28th
is T also (using step 2, the recursive part or inductive case). Here is what the proof would
actually look like.
Proof. Let P ( n ) be the statement “it is possible to make exactly n cents of postage using
5-cent and 8-cent stamps.” We will show P ( n ) is true for all n ≥ 28.
First, we show that P (28) is true: 28 4 · 5 + 1 · 8, so we can make 28 cents using four
5-cent stamps and one 8-cent stamp.
Now suppose P ( k ) is true for some arbitrary k ≥ 28. Then it is possible to make k cents
using 5-cent and 8-cent stamps. Note that since k ≥ 28, it cannot be that we use less than
three 5-cent stamps and less than three 8-cent stamps: using two of each would give only 26
cents. Now if we have made k cents using at least three 5-cent stamps, replace three 5-cent
stamps by two 8-cent stamps. This replaces 15 cents of postage with 16 cents, moving from
a total of k cents to k + 1 cents. Thus P ( k + 1) is true. On the other hand, if we have made k
cents using at least three 8-cent stamps, then we can replace three 8-cent stamps with five
5-cent stamps, moving from 24 cents to 25 cents, giving a total of k + 1 cents of postage. So
in this case as well P ( k + 1) is true.
Therefore, by the principle of mathematical induction, P ( n ) is true for all n ≥ 28. qed
Formalizing Proofs
What we did in the stamp example above works for many types of problems. Proof by
induction is useful when trying to prove statements about all natural numbers, or all
natural numbers greater than some fixed first case (like 28 in the example above), and in
some other situations too. In particular, induction should be used when there is some way
to go from one case to the next – when you can see how to always “do one more.”
This is a big idea. Thinking about a problem inductively can give new insight into the
problem. For example, to really understand the stamp problem, you should think about
how any amount of postage (greater than 28 cents) can be made (this is non-inductive
reasoning) and also how the ways in which postage can be made changes as the amount
increases (inductive reasoning). When you are asked to provide a proof by induction, you
are being asked to think about the problem dynamically; how does increasing n change the
problem?
But there is another side to proofs by induction as well. In mathematics, it is not enough
to understand a problem, you must also be able to communicate the problem to others. Like
any discipline, mathematics has standard language and style, allowing mathematicians to
share their ideas efficiently. Proofs by induction have a certain formal style, and being able
to write in this style is important. It allows us to keep our ideas organized and might even
help us with formulating a proof.
Here is the general structure of a proof by mathematical induction:
2.5. Induction 114
1. Base case: Prove that P (0) is true. You do this directly. This is often easy.
2. Inductive case: Prove that P ( k ) → P ( k + 1) for all k ≥ 0. That is, prove that
for any k ≥ 0 if P ( k ) is true, then P ( k + 1) is true as well. This is the proof
of an if. . . then. . . statement, so you can assume P ( k ) is true (P ( k ) is called the
inductive hypothesis). You must then explain why P ( k + 1) is also true, given that
assumption.
Assuming you are successful on both parts above, you can conclude, “Therefore
by the principle of mathematical induction, P ( n ) is true for all n ≥ 0.”
Sometimes the statement P ( n ) will only be true for values of n ≥ 4, for example, or
some other value. In such cases, replace all the 0’s above with 4’s (or the other value).
The other advantage of formalizing inductive proofs is it allows us to verify that the
logic behind this style of argument is valid. Why does induction work? Think of a row
of dominoes set up standing on their edges. We want to argue that in a minute, all the
dominoes will have fallen down. For this to happen, you will need to push the first domino.
That is the base case. It will also have to be that the dominoes are close enough together
that when any particular domino falls, it will cause the next domino to fall. That is the
inductive case. If both of these conditions are met, you push the first domino over and
each domino will cause the next to fall, then all the dominoes will fall.
Induction is powerful! Think how much easier it is to knock over dominoes when you
don’t have to push over each domino yourself. You just start the chain reaction, and the
rely on the relative nearness of the dominoes to take care of the rest.
Think about our study of sequences. It is easier to find recursive definitions for se-
quences than closed formulas. Going from one case to the next is easier than going directly
to a particular case. That is what is so great about induction. Instead of going directly to
the (arbitrary) case for n, we just need to say how to get from one case to the next.
When you are asked to prove a statement by mathematical induction, you should first
think about why the statement is true, using inductive reasoning. Explain why induction
is the right thing to do, and roughly why the inductive case will work. Then, sit down and
write out a careful, formal proof using the structure above.
2.5. Induction 115
Examples
Here are some examples of proof by mathematical induction.
Example:
n ( n+1)
Prove for each natural number n ≥ 1 that 1 + 2 + 3 + · · · + n 2 .
Solution: First, let’s think inductively about this equation. In fact, we know this is
true for other reasons (reverse and add comes to mind). But why might induction
be applicable? The left hand side is adding up the numbers from 1 to n. If we
know how to do that, adding just one more term (n + 1) would not be that hard.
For example, if n 100, suppose we know that the sum of the first 100 numbers is
5050 (so 1 + 2 + 3 + · · · + 100 5050, which is true). Now to find the sum of the first
101 numbers, it makes more sense to just add 101 to 5050, instead of computing the
entire sum again. We would have 1 + 2 + 3 + · · · + 100 + 101 5050 + 101 5151. In
fact, it would always be easy to add just one more term. This is why we should use
induction.
Now the formal proof:
n ( n+2)
Proof. Let P ( n ) be the statement 1 + 2 + 3 + · · · + n 2 . We will show that P ( n )
is true for all natural numbers n ≥ 1.
1(1+1)
Base case: P (1) is the statement 1 2 which is clearly true.
Inductive case: Let k ≥ 1 be a natural number. Assume (for induction) that P ( k )
k ( k+1)
is true. That means 1 + 2 + 3 + · · · + k 2 . We will prove that P ( k + 1) is true as
(k+1)(k+2)
well. That is, we must prove that 1 + 2 + 3 + · · · + k + ( k + 1) 2 . To prove
this equation, start by adding k + 1 to both sides of the inductive hypothesis:
k ( k + 1)
1 + 2 + 3 + · · · + k + ( k + 1) + ( k + 1) .
2
Now, simplifying the right side we get:
k ( k + 1) k ( k + 1) 2( k + 1) k ( k + 1) + 2( k + 1) ( k + 2)( k + 1)
+k+1 + .
2 2 2 2 2
Thus P ( k + 1) is true, so by the principle of mathematical induction P ( n ) is true for
all natural numbers n ≥ 1. qed
Note that in the part of the proof in which we proved P ( k + 1) from P ( k ), we used
the equation P ( k ). This was the inductive hypothesis. Seeing how to use the inductive
hypotheses is usually straight forward when proving a fact about a sum like this. In other
proofs, it can be less obvious where it fits in.
2.5. Induction 116
Example:
Prove that for all n ∈ N, 6n − 1 is a multiple of 5.
Solution: Again, start by understanding the dynamics of the problem. What does
increasing n do? Let’s try with a few examples. If n 1, then yes, 61 − 1 5 is a
multiple of 5. What does incrementing n to 2 look like? We get 62 − 1 35, which
again is a multiple of 5. Next, n 3: but instead of just finding 63 − 1, what did the
increase in n do? We will still subtract 1, but now we are multiplying by another 6
first. Viewed another way, we are multiplying a number which is one more than a
multiple of 5 by 6 (because 62 − 1 is a multiple of 5, so 62 is one more than a multiple
of 5). What do numbers which are one more than a multiple of 5 look like? They
must have last digit 1 or 6. What happens when you multiply such a number by 6?
Depends on the number, but in any case, the last digit of the new number must be
a 6. And then if you subtract 1, you get last digit 5, so a multiple of 5.
The point is, every time we multiply by just one more six, we still get a number
with last digit 6, so subtracting 1 gives us a multiple of 5. Now the formal proof:
Proof. Let P ( n ) be the statement, “6n − 1 is a multiple of 5.” We will prove that P ( n )
is true for all n ∈ N.
Base case: P (0) is true: 60 − 1 0 which is a multiple of 5.
Inductive case: Let k be an arbitrary natural number. Assume, for induction,
that P ( k ) is true. That is, 6k − 1 is a multiple of 5. Then 6k − 1 5j for some integer
j. This means that 6k 5 j + 1. Multiply both sides by 6:
6k+1 − 1 30j + 5.
We had to be a little bit clever (i.e., use some algebra) to locate the 6k − 1 inside of 6k+1 − 1
before we could apply the inductive hypothesis. This is what can make inductive proofs
challenging.
In the two examples above, we started with n 1 or n 0. We can start later if we need
to.
Example:
Prove that n 2 < 2n for all integers n ≥ 5.
2.5. Induction 117
Solution: First, the idea of the argument. What happens when we increase n by 1?
On the left hand side, we increase the base of the square and go to the next square
number. On the right hand side, we increase the power of 2. This means we double
the number. So the question is, how does doubling a number relate to increasing to
the next square? Think about what the difference of two consecutive squares looks
like. We have ( n + 1)2 − n 2 . This factors:
(n + 1)2 − n 2 (n + 1 − n )(n + 1 + n ) 2n + 1.
But doubling the right hand side increases it by 2n , since 2n+1 2n + 2n . When n is
large enough, 2n > 2n + 1.
What we are saying here is that each time n increases, the left hand side grows
by less than the right hand side. So if the left hand side starts smaller (as it does
when n 5), it will never catch up. Now the formal proof:
Proof. Let P ( n ) be the statement n 2 < 2n . We will prove P ( n ) is true for all integers
n ≥ 5.
Base case: P (5) is the statement 52 < 25 . Since 52 25 and 25 32, we see that
P (5) is indeed true.
Inductive case: Let k ≥ 5 be an arbitrary integer. Assume, for induction, that
P ( k ) is true. That is, assume k 2 < 2k . We will prove that P ( k + 1) is true, i.e.,
(k + 1)2 < 2k+1 . To prove such an inequality, start with the left hand side and work
towards the right hand side:
(k + 1)2 k 2 + 2k + 1
< 2k + 2k + 1 . . . by the inductive hypothesis.
< 2k + 2k . . . since 2k + 1 < 2k for k ≥ 5.
2k+1 .
Following the equalities and inequalities through, we get ( k + 1)2 < 2k+1 , in other
words, P ( k + 1). Therefore by the principle of mathematical induction, P ( n ) is true
for all n ≥ 5. qed
The previous example might remind you of the racetrack principle from calculus, which
says that if f ( a ) < g ( a ), and f 0( x ) < g 0( x ) for x > a, then f ( x ) < g ( x ) for x > a. Same
idea: the larger function is increasing at a faster rate than the smaller function, so the
larger function will stay larger. In discrete math, we don’t have derivatives, so we look at
differences. Thus induction is the way to go.
Warning:
With great power, comes great responsibility. Induction isn’t magic. It seems very powerful
to be able to assume P ( k ) is true. After all, we are trying to prove P ( n ) is true and the only
difference is in the variable: k vs. n. Are we assuming that what we want to prove is true?
Not really. We assume P ( k ) is true only for the sake of proving that P ( k + 1) is true.
2.5. Induction 118
Still you might start to believe that you can prove anything with induction. Consider
this incorrect “proof” that every Canadian has the same eye color: Let P ( n ) be the statement
that any n Canadians have the same eye color. P (1) is true, since everyone has the same
eye color as themselves. Now assume P ( k ) is true. That is, assume that in any group of
k Canadians, everyone has the same eye color. Now consider an arbitrary group of k + 1
Canadians. The first k of these must all have the same eye color, since P ( k ) is true. Also,
the last k of these must have the same eye color, since P ( k ) is true. So in fact, everyone
the group must have the same eye color. Thus P ( k + 1) is true. So by the principle of
mathematical induction, P ( n ) is true for all n.
Clearly something went wrong. The problem is that the proof that P ( k ) implies P ( k + 1)
assumes that k ≥ 2. We have only shown P (1) is true. In fact, P (2) is false.
Strong Induction
Investigate!
Start with a square piece of paper. You want to cut this square into smaller squares,
leaving no waste (every piece of paper you end up with must be a square). Obviously
it is possible to cut the square into 4 squares. You can also cut it into 9 squares. It
turns out you can cut the square into 7 squares (although not all the same size).
What other numbers of squares could you end up with?
You have 9 coins which all look identical. However, you know that one of the
coins is counterfeit, and weighs slightly less than the other coins (which all have
the same weight). The weight difference is not great enough to tell by holding
the coins, but luckily you have an old-fashioned balance scale (like the one held
by Lady Justice). What is the smallest number of weighings needed to find the
counterfeit coin?
weigh the coins, you can eliminate 2/3 of them. So using 3 weighings, you could find the
counterfeit coin among up to 27 coins. A forth weighing would be required for 28 to 81
coins.
Conjecture. The minimum number of weighings required to find the counterfeit coin among n
coins is log3 ( n ), rounded up. In other words, if x is the smallest integer such that n ≤ 3x , then the
minimum number of weighings is x.
How do we know this is true? If you have n coins, divide them into 3 equal piles (if
this is not possible, make two piles the same size, and the third pile one more or less as
needed). Put one pile on each side of the scale, leaving one pile off (if the piles were not
equal, weigh the piles with an equal number of coins against each other). If the scales
balance, the counterfeit coin is in the unweighed pile. If the scales tip, the counterfeit coin
is in the light pile. In any event, we have now reduced the number of coins to roughly n/3.
What now? Well obviously we do this again. And again, until we find our bad coin.
Here is where induction comes in: instead of going through the process over and over
again, we can just say that we have reduced the problem to an earlier case, for which we
assume our conjecture holds. For example if n 81, we do the weighing as described,
leaving us with n/3 27 coins. By induction, we already know how to complete the
problem for 27 coins, so we are done.
This “reducing to a simpler case” is a type of induction. But not exactly the same sort of
induction we have encountered so far. In ordinary induction, we assume our result holds
for k, and prove that it holds for the next higher case: k + 1. There is not much difference
in showing that the result holds for the case k assuming that it holds for the previous
case (k − 1). But what we are doing here, is showing that the result holds for the case k,
assuming it works for all smaller cases. Is this okay? In fact, it is. Think of the dominoes:
to prove that they all fall down, it is enough to prove that the first one falls, and that if all
the dominoes up to the kth one have fallen, then the kth one will fall too.
The advantage is that we now have a stronger inductive hypothesis. We can assume
that P (1), P (2), P (3),. . . P ( k ) is true, just to show that P ( k + 1) is true. Previously, we just
assumed P ( k ) for this purpose.
It is slightly easier if we change our variables for strong induction. Here is what the
formal proof would look like:
2. Inductive case: Assume P ( k ) is true for all k < n. Prove that P ( n ) is true.
Proof. Let P ( n ) be the statement, “if x is the smallest natural number such that n ≤ 3x , then
it is possible to find the light coin in no more than x weighings.”
Base case: Consider P (2). Here x is 1, since 2 ≤ 31 . It is possible to find the heavy coin
using just 1 weighing: just put the coins on the scale.
Inductive case: Suppose we have n coins, and assume P ( k ) is true for all k < n. Now
divide the n coins into three piles, equal if possible, otherwise two equal with the third
either one more or one less than the other two. Put two equal piles on each side of the
scales. If the scales balance, then the counterfeit coin is in the unweighed pile. If the scales
tip, the coin is in the pile on the lighter side of the scale. In any event, we now know the
counterfeit coin is one of k n/3 or k n/3 ± 1 coins. What’s more, if x is the smallest
number such that n ≤ 3x , then k ≤ 3x−1 . By the inductive hypothesis, it is now possible to
find the counterfeit coin in x − 1 more weighings. Thus P ( n ) is true.
Therefore, by strong induction, P ( n ) is true for all n ≥ 2. qed
Example:
Prove that any natural number greater than 1 is either prime or can be written as
the product of primes.
Solution: First, the idea: if we take some number n, maybe it is prime. If so, we are
done. If not, then it is composite, so it is the product of two smaller numbers. Each
of these factors is smaller than n (but at least 2), so we can repeat the argument with
these numbers. We have reduced to a smaller case.
Now the formal proof:
Proof. Let P ( n ) be the statement, “n is either prime or can be written as the product
of primes.” We will prove P ( n ) is true for all n ≥ 2.
Base case: P (2) is true because 2 is indeed prime.
Inductive case: assume P ( k ) is true for all k < n. We want to show that P ( n ) is
true. That is, we want to show that n is either prime or is the product of primes. If
n is prime, we are done. If not, then n has more than 2 divisors, so we can write
n m 1 · m2 , with m 1 and m2 less than n (and greater than 1). By the inductive
hypothesis, m 1 and m 2 are each either prime or can be written as the product of
primes. In either case, we have that n is written as the product of primes.
Thus by the strong induction, P ( n ) is true for all n ≥ 2. qed
4 Technically,
strong induction does not require you to prove a separate base case. This is because when
proving the inductive case, you must show that P (0) is true, assuming P ( k ) is true for all k < 0. But this is
not any help so you end up proving P (0) anyway. To be on the safe side, we will always include the base case
separately.
2.5 Exercises 121
Whether you use regular induction or strong induction depends on the statement you
want to prove. If you wanted to be safe, you could always use strong induction. It really
is stronger, so can accomplish everything “weak” induction can. That said, using regular
induction is often easier since there is only one place you can use the induction hypothesis.
There is also something to be said for elegance in proofs. If you can prove a statement using
simpler tools, it is nice to do so.
As a final contrast between the two forms of induction, consider once more the stamp
problem. Regular induction worked by showing how to increase postage by one cent
(either replacing three 5-cent stamps with two 8-cent stamps, or three 8-cent stamps with
five 5-cent stamps). We could give a slightly different proof using strong induction. First,
we could show five base cases: it is possible to make 28, 29, 30, 31, and 32 cents (we would
actually say how each of these is made). Now assume that it is possible to make k cents of
postage for all k < n as long as k ≥ 28. As long as n > 32, this means in particular we can
make k n − 5 cents. Now add a 5-cent stamp to get make n cents.
Exercises
n
X
1. Use induction to prove for all n ∈ N that 2k 2n+1 − 1.
k0
6. What is wrong with the following “proof” of the “fact” that n + 3 n + 7 for all values
of n (besides of course that the thing it is claiming to prove is false)?
7. The proof in the previous problem does not work. But if we modify the “fact,” we
can get a working proof. Prove that n + 3 < n + 7 for all values of n ∈ N. You can do
this proof with algebra (without induction), but the goal of this exercise is to write
out a valid induction proof.
8. Find the flaw in the following “proof” of the “fact” that n < 100 for every n ∈ N.
Proof. Let P ( n ) be the statement n < 100. We will prove P ( n ) is true for all n ∈ N.
First we establish the base case: when n 0, P ( n ) is true, because 0 < 100. Now for
the inductive step, assume P ( k ) is true. That is, k < 100. Now if k < 100, then k is
2.5 Exercises 122
some number, like 80. Of course 80 + 1 81 which is still less than 100. So k + 1 < 100
as well. But this is what P ( k + 1) claims, so we have shown that P ( k ) → P ( k + 1). Thus
by the principle of mathematical induction, P ( n ) is true for all n ∈ N. qed
9. While the above proof does not work (it better not since the statement it is trying to
prove is false!) we can prove something similar. Prove that there is a strictly increasing
sequence a 1 , a 2 , a 3 , . . . of numbers (not necessarily integers) such that a n < 100 for all
n ∈ N. (By strictly increasing we mean a n < a n+1 for all n. So each term must be larger
than the last.)
10. What is wrong with the following “proof” of the “fact” that for all n ∈ N, the number
n 2 + n is odd?
Proof. Let P ( n ) be the statement “n 2 + n is odd.” We will prove that P ( n ) is true for all
n ∈ N. Suppose for induction that P ( k ) is true, that is, that k 2 + k is odd. Now consider
the statement P ( k + 1). Now ( k + 1)2 + ( k + 1) k 2 + 2k + 1 + k + 1 k 2 + k + 2k + 2. By
the inductive hypothesis, k 2 + k is odd, and of course 2k + 2 is even. An odd plus an
even is always odd, so therefore ( k + 1)2 + ( k + 1) is odd. Therefore by the principle
of mathematical induction, P ( n ) is true for all n ∈ N. qed
11. Now give a valid proof (by induction, even though you might be able to do so without
using induction) of the statement, “for all n ∈ N, the number n 2 + n is even.”
12. Prove that there is a sequence of positive real numbers a1 , a 2 , a 3 , . . . such that the
partial sum a 1 + a 2 + a 3 + · · · + a n is strictly less than 2 for all n ∈ N. Hint: think about
how you could define what a k+1 is to make the induction argument work.
13. Prove that every natural number is either a power of 2, or can be written as the sum
of distinct powers of 2.
14. Use induction to prove that if n people all shake hands with each other, that the total
n ( n−1)
number of handshakes is 2 .
15. Suppose that a particular real number x has the property that x + 1
x is an integer.
Prove that x n + x1n is an integer for all natural numbers n.
n !
X n
16. Use induction to prove that 2n . That is, the sum of the nth row of Pascal’s
k
k0
Triangle is 2n .
4 5 6 4+n 5+n
+ + + ··· +
17. Use induction to prove 0 1 2 n n . (This is an example of the
hockey stick theorem.)
18. Use the product rule for logarithms (log( ab ) log( a ) + log( b )) to prove, by induction
on n, that log( a n ) n log( a ), for all natural numbers n ≥ 2.
2.6. Chapter Summary 123
( f1 + f2 + · · · + f n )0 f10 + f20 + · · · + f n0
You may assume the product rule for two functions is true.
Investigate!
Each day your supply of magic chocolate covered espresso beans doubles (each one
splits in half), but then you eat 5 of them. You have 10 at the start of day 0.
1. Write out the first few terms of the sequence. Then give a recursive definition
for the sequence and explain how you know it is correct.
2. Prove, using induction, that the last digit of the number of beans you have on
the nth day is always a 5 for all n ≥ 1.
3. Find a closed formula for the nth term of the sequence and prove it is correct
by induction.
• If the terms of a sequence increase at a polynomial rate (that is, if the differences
between terms form a sequence with a polynomial closed formula), then the sequence
is itself given by a polynomial closed formula (of degree one more than the sequence
of differences).
• If the terms of a sequence increase at an exponential rate, then we expect the closed
formula for the sequence to be exponential. These sequences often have relatively
nice recursive formulas, and the characteristic root technique allows us to find the closed
formula for these sequences.
• If we want to prove that a statement is true for all values of n (greater than some first
small value), and we can describe why the statement being true for n k implies the
statement is true for n k + 1, then the principle of mathematical induction gives us that
the statement is true for all values of n (greater than the base case).
Throughout the chapter we tried to understand why these facts listed above are true.
In part, that is what proofs, by induction or not, attempt to accomplish: they explain
why mathematical truths are in fact truths. As we develop our ability to reason about
mathematics, it is a good idea to make sure that the methods of our reasoning are sound.
The branch of mathematics that deals with deciding whether reasoning is good or not is
mathematical logic, the subject of the next chapter.
Exercises
1. Find 3 + 7 + 11 + · · · + 427.
(a) Find the first 4 terms of the sequence. What sort of sequence is this?
25
X
(b) Find the sum of the first 25 terms. That is, compute ak .
k1
6. Use polynomial fitting to find a closed formula for the sequence: 4, 11, 20, 31, 44, . . .
(assume a 1 4).
7. Suppose the closed formula for a particular sequence is a degree 3 polynomial. What
can you say about the closed formula for:
10. Consider the recurrence relation a n 3a n−1 + 10a n−2 with first two terms a 0 1 and
a 1 2.
(a) Write out the first 5 terms of the sequence defined by this recurrence relation.
(b) Solve the recurrence relation. That is, find a closed formula for a n .
11. Consider the recurrence relation a n 2a n−1 + 8a n−2 , with initial terms a0 1 and
a 1 3.
12. Your magic chocolate bunnies reproduce like rabbits: every large bunny produces 2
new mini bunnies each day, and each day every mini bunny born the previous day
grows into a large bunny. Assume you start with 2 mini bunnies and no bunny ever
dies (or gets eaten).
2.6 Exercises 126
15. Suppose a 0 1, a 1 1 and a n 3a n−1 − 2a n−1 . Prove, using strong induction, that
a n 1 for all n.
16. Prove, using strong induction, that every positive integer can be written as the sum
of distinct powers of 2. For example, 13 1 + 4 + 8 20 + 22 + 23 .
17. Prove using induction that every set containing n elements has 2n different subsets
for any n ≥ 1.
Homework Problems
The following are some more involved problems for you to try, which might be assigned
as homework.
1. For each sequence given below, find a closed formula for a n , the nth term of the
sequence (assume the first terms are a 0 ) by relating it to another sequence for which
you already know the formula. In each case, briefly say how you got your answers.
2. Starting with any rectangle, we can create a new, larger rectangle by attaching a
square to the longer side. For example, if we start with a 2 × 5 rectangle, we would
glue on a 5 × 5 square, forming a 5 × 7 rectangle:
2.6 Exercises 127
5 5
2 5 7
(a) Create a sequence of rectangles using this rule starting with a 1 × 2 rectangle.
Then write out the sequence of perimeters for the rectangles (the first term of the
sequence would be 6, since the perimeter of a 1 × 2 rectangle is 6 - the next term
would be 10).
(b) Repeat the above part this time starting with a 1 × 3 rectangle.
(c) Find recursive formulas for each of the sequences of perimeters you found in
parts (a) and (b). Don’t forget to give the initial conditions as well.
(d) Are the sequences arithmetic? Geometric? If not, are they close to being either
of these (i.e., are the differences or ratios almost constant)? Explain.
3. If you have enough toothpicks, you can make a large triangular grid. Below, are the
triangular grids of size 1 and of size 2. The size 1 grid requires 3 toothpicks, the size
2 grid requires 9 toothpicks.
(a) Let t n be the number of toothpicks required to make a size n triangular grid.
Write out the first 5 terms of the sequence t1 , t2 , . . ..
(b) Find a recursive definition for the sequence. Explain why you are correct.
(c) Is the sequence arithmetic or geometric? If not, is it the sequence of partial sums
of an arithmetic or geometric sequence? Explain why your answer is correct.
(d) Use your results from part (c) to find a closed formula for the sequence. Show
your work.
(a) Find a closed formula for a n , the nth term of the sequence, by writing each term
as a sum of a sequence. Hint: first find a 0 , but ignore it when collapsing the
sum.
(b) Find a closed formula again, this time using either polynomial fitting or the
characteristic root technique (whichever is appropriate). Show your work.
(c) Find a closed formula once again, this time by recognizing the sequence as a
modification to some well known sequence(s). Explain.
2.6 Exercises 128
5. In their down time, ghost pirates enjoy stacking cannonballs in triangular based
pyramids (aka, tetrahedrons), like those pictured here:
Note, in the picture on the right, there are some cannonballs (actually just one) you
cannot see. The next picture would have 4 cannonballs you cannot see. The stacks
are not hollow.
The pirates wonder how many cannonballs would be required to build a pyramid 15
layers high (thus breaking the world cannonball stacking record). Can you help?
(a) Let P ( n ) denote the number of cannonballs needed to create a pyramid n layers
high. So P (1) 1, P (2) 4, and so on. Calculate P (3), P (4) and P (5).
(b) Use polynomial fitting to find a closed formula for P ( n ). Show your work.
(c) Answer the pirate’s question: how many cannonballs do they need to make a
pyramid 15 layers high?
6. Consider the sequences 2, 5, 12, 29, 70, 169, 408, . . . (with a0 2).
7. Let a n be the number of 1 × n tile designs you can make using 1 × 1 squares available
in 4 colors and 1 × 2 dominoes available in 5 colors.
(a) First, find a recurrence relation to describe the problem. Explain why the recur-
rence relation is correct (in the context of the problem).
(b) Write out the first 6 terms of the sequence a 1 , a 2 , . . ..
(c) Solve the recurrence relation. That is, find a closed formula for a n .
(a) Find the general solution to the recurrence relation (beware the repeated root).
(b) Find the solution when a 0 1 and a 1 2.
(c) Find the solution when a 0 1 and a 1 8.
2.6 Exercises 129
9. Zombie Euler and Zombie Cauchy, two famous zombie mathematicians, have just
signed up for Twitter accounts. After one day, Zombie Cauchy has more followers
than Zombie Euler. Each day after that, the number of new followers of Zombie
Cauchy is exactly the same as the number of new followers of Zombie Euler (and
neither lose any followers). Explain how a proof by mathematical induction can show
that on every day after the first day, Zombie Cauchy will have more followers than
Zombie Euler. That is, explain what the base case and inductive case are, and why
they together prove that Zombie Cauchy will have more followers on the 4th day.
10. Find the largest number of points which a football team cannot get exactly using
just 3-point field goals and 7-point touchdowns (ignore the possibilities of safeties,
missed extra points, and two point conversions). Prove your answer is correct by
mathematical induction.
n ( n + 1)(2n + 1)
12 + 22 + 32 + ... + n 2
6
13. Prove, using strong induction, that every natural number is either a Fibonacci number
or can be written as the sum of distinct Fibonacci numbers.
Chapter 3
Logic is the study of consequence. Given a few mathematical statements or facts, we would
like to be able to draw some conclusions. For example, if I told you that a particular real
valued function was continuous on the interval [0, 1], and f (0) −1 and f (1) 5, can we
conclude that there is some point between [0, 1] where the graph of the function crosses
the x-axis? Yes, we can, thanks to the Intermediate Value Theorem from Calculus. Can
we conclude that there is exactly one point? No. Whenever we find an “answer” in math,
we really have a (perhaps hidden) argument. Mathematics is really about proving general
statements (like the Intermediate Value Theorem), and this too is done via an argument,
usually called a proof. We start with some given conditions, the premises of our argument,
and from these we find a consequence of interest, our conclusion.
The problem is, as you no doubt know from arguing with friends, not all arguments are
good arguments. A “bad” argument is one in which the conclusion does not follow from
the premises, i.e., the conclusion is not a consequence of the premises. Logic is the study
of what makes an argument good or bad. In other words, logic aims to determine in which
cases a conclusion is, or is not, a consequence of a set of premises.
By the way, “argument” is actually a technical term in math (and philosophy, another
discipline which studies logic):
Arguments
An argument is a set of statements, one of which is called the conclusion and the rest
of which are called premises. An argument is said to be valid if the conclusion must
be true whenever the premises are all true. An argument is invalid if it is not valid; it
is possible for all the premises to be true and the conclusion to be false.
130
3.1. Propositional Logic 131
Are these arguments valid? Hopefully you agree that the first one is but the second
one is not. Logic tells us why by analyzing the structure of the statements in the argument.
Notice the two arguments above look almost identical. Edith and Florence both eat their
vegetables. In both cases there is a connection between the eating of vegetables and cookies.
But we claim that it is valid to conclude that Edith gets a cookie, but not that Florence does.
The difference must be in the connection between eating vegetables and getting cookies.
We need to be skilled at reading and comprehending these sentences. Do the two sentences
mean the same thing? Unfortunately, in everyday language we are often sloppy, and you
might be tempted to say they are equivalent. But notice that just because Florence must
eat her vegetables, we have not said that doing so would be enough (she might also need
to clean her room, for example). In everyday (non-mathematical) practice, you might be
tempted to say this “other direction” is implied. We don’t ever get that luxury.
Our goal in studying logic is to gain intuition for which arguments are valid and which
are invalid. This will require us to become better at reading and writing mathematics – a
worthy goal in its own right. So let’s get started.
Investigate!
After excavating for weeks, you finally arrive at the burial chamber. The room is
empty except for two large chests. On each is carved a message (strangely in English).
You know exactly one of these messages is true. What should you do?
we will look at the logical form of the statement. First though, let’s back up and make sure
we are very clear on some basics.
Statements
A statement is any declarative sentence which is either true or false.
Example:
These are statements:
• 42 is a perfect square.
• Every even number greater than 2 can be expressed as the sum of two primes.
The reason the last sentence is not a statement is because it contains variables (“that”
and “she”). Unless those are specified, the sentence cannot be true or false, and as such not
a statement. Other examples of this: x + 3 7. Depending on x, this is either true or false,
but as it stands it it neither.
You can build more complicated statements out of simpler ones using logical connectives.
For example, this is a statement:
Note that we can break this down into two smaller statements. The two shorter statements
are connected by an “and.” We will consider 5 connectives: “and” (Sam is a man and Chris
is a woman), “or” (Sam is a man or Chris is a woman), “if. . . then. . . ” (if Sam is a man,
then Chris is a woman), “if and only if” (Sam is a man if and only if Chris is a woman), and
“not” (Sam is not a man).
Since we rarely care about the content of the individual statements, we can replace
them with variables. By convention, we use capital letters in the middle of the alphabet
for these propositional (or sentential) variables: P, Q, R, S . . .. We also have symbols for the
logical connectives: ∧, ∨, →, ↔, ¬.
1 This is a tricky one. Remember, a sentence is only a statement if it is either true or false. Here, the sentence
is not false, for if it were, it would be true. It is not true, for that would make it false.
3.1. Propositional Logic 133
Logical Connectives
• P ∧ Q means P and Q, called a conjunction.
The logical connectives allow us to construct longer statements out of simpler state-
ments. But the result is still a statement since it is either true or false. The truth value can
be determined by the truth or falsity of the parts, depending on the connectives.
These should be what you would expect, except perhaps for P → Q. Consider the
statement, “if Bob gets a 90 on the final, then Bob will pass the class.” This is definitely an
implication: P is the statement, “Bob gets a 90 on the final,” and Q is the statement, “Bob
will pass the class.” Suppose I made that statement to Bob. In what circumstances would
it be fair to call me a liar? What if Bob really did get a 90 on the final, and he did pass the
class? Then I have not lied; my statement is true. But if Bob did get a 90 on the final and
did not pass the class, then I lied, making the statement false. The tricky case is this: what
if Bob did not get a 90 on the final? Maybe he passes the class, maybe he doesn’t. Did I lie
in either case? I think not. In these last two cases, P was false, and the statement P → Q
was true. In the first case, Q was true, and so was P → Q. So P → Q is true when either P
is false or Q is true.
Perhaps an easier way to look at it is this: P → Q is false in only one case: if P is true
and Q is false. Otherwise, P → Q is true. Admittedly, there are times in English when this
is not how “if. . . , then. . . ” works. However, in mathematics, we define the implication to
work this way.
Truth Tables
Our earlier question about Monopoly is to determine whether the following statement is
true:
3.1. Propositional Logic 134
If you get more doubles than any other player then you will lose, or if you lose
then you must have bought the most properties.
Example:
Make a truth table for the statement ¬P ∨ Q.
Solution: Note that this statement is not ¬(P ∨ Q ), the negation belongs to P alone.
Here is the truth table:
P Q ¬P ¬P ∨ Q
T T F T
T F F F
F T T T
F F T T
We added a column for ¬P to make filling out the last column easier. The entries
in the ¬P column were determined by the entries in the P column. Then to fill in
3.1. Propositional Logic 135
the final column, look only at the column for Q and the column for ¬P and use the
rule for ∨.
You might notice that the final column is identical to the final column in the truth table
for P → Q. Since we listed the possible values for P and Q in the same (in fact, standard)
order, this says that ¬P ∨ Q and P → Q are logically equivalent.
Logical Equivalence
Two (compound) statements P and Q are logically equivalent provided P is true pre-
cisely when Q is true.
To verify that two statements are logically equivalent, you can make a truth table
for each and check whether the columns for the two statements are identical.
Example:
Analyze the statement, “if you get more doubles than any other player you will lose,
or that if you lose you must have bought the most properties,” using truth tables.
Solution: Represent the statement in symbols as (P → Q ) ∨ (Q → R), where P
is the statement “you get more doubles than any other player,” Q is the statement
“you will lose,” and R is the statement “you must have bought the most properties.”
Now make a truth table.
The truth table needs to contain 8 rows in order to account for every possible
combination of truth and falsity among the three statements. Here is the full truth
table:
P Q R P→Q Q→R (P → Q ) ∨ (Q → R )
T T T T T T
T T F T F T
T F T F T T
T F F F T T
F T T T T T
F T F T F T
F F T T T T
F F F T T T
The first three columns are simply a systematic listing of all possible combinations
of T and F for the three statements (do you see how you would list the 16 possible
combinations for four statements?). The next two columns are determined by the
values of P, Q, and R and the definition of implication. Then, the last column is
determined by the values in the previous two columns and the definition of ∨. It is
this final column we care about.
3.1. Propositional Logic 136
Notice that in each of the eight possible cases, the statement in question is true.
So our statement about monopoly is true (regardless of how many properties you
own, how many doubles you roll, or whether you win or lose).
Deductions
Investigate!
Holmes owns two suits: one blue and one brown. He always wears either a blue
suit or white socks. Whenever he wears his blue suit and a blue shirt, he also wears
a blue tie. He never wears the blue suit unless he is also wearing either a blue shirt
or white socks. Whenever he wears white socks, he also wears a blue shirt. Today,
Holmes is wearing a gold tie. What else is he wearing?2
If Edith eats her vegetables, then she can have a cookie. Edith ate her vegetables.
Therefore Edith gets a cookie.
How do we know this is valid? Let’s look at the form of the statements. Let P denote
“Edith eats her vegetables” and Q denote “Edith can have a cookie.” The logical form of
the argument is then:
P→Q
P
∴ Q
This is an example of a deduction rule, an argument form which is always valid. This
one is a particularly famous rule called modus ponens. Are you convinced that it is a valid
deduction rule? If not, consider the following truth table:
P Q P→Q
T T T
T F F
F T T
F F T
2 Adapted from Problem Solving Through Recreational Mathematics by Averbach and Chein, Dover 1999.
3.1. Propositional Logic 137
This is just the truth table for P → Q, but what matters here is that all the lines in the
deduction rule have their own column in the truth table. Remember that an argument is
valid provided the conclusion must be true given that the premises are true. The premises
in this case are P → Q and P. Which rows of the truth table correspond to both of these
being true? P is true in the first two rows, and of those, only the first row has P → Q true
as well. And low-and-behold, in this one case, Q is also true. So if P → Q and P are both
true, we see that Q must be true as well.
Here are a few more examples.
Example:
P→Q
Show that ¬P → Q is a valid deduction rule.
∴ Q
Solution: We make a truth table which contains all the lines of the argument form:
P Q P→Q ¬P ¬P → Q
T T T F T
T F F F T
F T T T T
F F T T F
(we include a column for ¬P just as a step to help getting the column for ¬P → Q).
Now look at all the rows for which both P → Q and ¬P → Q are true. This
happens only in rows 1 and 3. Hey! In those rows Q is true as well, so the argument
form is valid (it is a valid deduction rule).
Example:
(P → R ) ∨ (Q → R )
Decide whether is a valid deduction rule.
∴ (P ∨ Q ) → R
Solution: Let’s make a truth table containing both statements.
is a tautology.
Exercises
1. Consider the statement about a party, “If it’s your birthday or there will be cake, then
there will be cake.”
(a) Translate the above statement into symbols. Clearly state which statement is P
and which is Q.
(b) Make a truth table for the statement.
(c) Assuming the statement is true, what (if anything) can you conclude if there
will be cake?
(d) Assuming the statement is true, what (if anything) can you conclude if there
will not be cake?
(e) Suppose you found out that the statement was a lie. What can you conclude?
2. Suppose P and Q are the statements: P: Jack passed math. Q: Jill passed math.
(a) Translate “Jack and Jill both passed math” into symbols.
(b) Translate “If Jack passed math, then Jill did not” into symbols.
(c) Translate “P ∨ Q” into English.
(d) Translate “¬(P ∧ Q ) → Q” into English.
(e) Suppose you know that if Jack passed math, then so did Jill. What can you
conclude if you know that:
(a) Jill passed math?
(b) Jill did not pass math?
3. Geoff Poshingten is out at a fancy pizza joint, and decides to order a calzone. When
the waiter asks what he would like in it, he replies, “I want either pepperoni or
sausage, and if I have sausage, I must also include quail. Oh, and if I have pepperoni
or quail then I must also have ricotta cheese.”
(b) The waiter knows that Geoff is either a liar or a truth-teller (so either everything
he says is false, or everything is true). Which is it?
(c) What, if anything, can the waiter conclude about the ingredients in Geoff’s
desired calzone?
5. Make a truth table for the statement ¬P ∧ (Q → P ). What can you conclude about P
and Q if you know the statement is true?
Investigate!
Decide which of the following are valid proofs of the following statement.
1. Suppose a and b are odd. That is, a 2k + 1 and b 2m + 1 for some integers
k and m. Then
ab (2k + 1)(2m + 1)
4km + 2k + 2m + 1
2(2km + k + m ) + 1.
Therefore ab is odd.
2. Assume that a or b is even - say it is a (the case where b is even will be identical).
That is, a 2k for some integer k. Then
ab (2k ) b
2( kb ).
Thus ab is even.
3. Suppose that ab is even but a and b are both odd. Namely, ab 2n, a 2k + 1
and b 2j + 1 for some integers n, k, and j. Then
2n (2k + 1)(2j + 1)
2n 4k j + 2k + 2j + 1
1
n 2k j + k + j + .
2
But since 2k j + k + j is an integer, this says that the integer n is equal to a
non-integer, which is impossible.
ab (2k + 1) b
2n 2kb + b
2n − 2kb b
2( n − kb ) b.
No surprise there. Now let’s see how negation plays with conjunctions and disjunctions.
De Morgan’s Laws
Do you believe De Morgan’s laws? If not, make a truth table for each of them. I think
most of us get these right most of the time without thinking about them too hard. If I told
you that I had popcorn and goobers at the movies, but then you found out it was opposite
day (so my statement was false) then you would agree, I hope, that I either did not have
popcorn or did not have goobers (or didn’t have either). You would not insist that I could
not have had either.
I should warn you that often in English, we are sloppy about our and’s, or’s and not’s.
When you write about mathematics, you should be careful and write what you mean. If
you are not sure what to write, rephrasing carefully using De Morgan’s laws can help you
make sure that statement matches your intended meaning.
Here are some rules for implications:
Negation of Implication
This is very important, and not obvious. Implications are tricky. Look again at the truth
table for P → Q:
P Q P→Q
T T T
T F F
F T T
F F T
There is only one way for the implication to be false: P is true and Q is false. Another
way to see that this is true is by using De Morgan’s Laws. We saw earlier that P → Q can
be rephrased as ¬P ∨ Q so we have
But by De Morgan’s laws, ¬(¬P ∨ Q ) is equivalent to ¬¬P ∧ ¬Q. By double negation ¬¬P
is the same as P.
While we are thinking about implications, we should talk about the converse and
contrapositive:
A related, but lesser used, term is the inverse of an implication. The inverse of P → Q is
¬P → ¬Q. Notice that the inverse of an implication is the contrapositive of the converse.
Read that one more time. Good. Since implications and their contrapositives are logically
equivalent, the inverse and converse of an implication are logically equivalent to each other,
but not to the original implication.
Example:
Suppose I tell Sue that if she gets a 93% on her final, she will get an A in the class.
Assuming that what I said is true, what can you conclude in the following cases:
Solution: Note first that whenever P → Q and P are both true statements, Q must
be true as well. For this problem, take P to mean “Sue gets a 93% on her final” and
Q to mean “Sue will get an A in the class.”
(b) You cannot conclude anything. Sue could have gotten the A because she did
extra credit for example. Notice that we do not know that if Sue gets an A, then
she gets a 93% on her final. That is the converse of the original implication, so
it might or might not be true.
(c) The inverse of P → Q is ¬P → ¬Q, which states that if Sue does not get a 93%
on the final then she will not get an A in the class. But this does not follow
from the original implication. Again, we can conclude nothing. Sue could
have done extra credit.
(d) What would happen if Sue does not get an A but did get a 93% on the final.
Then P would be true and Q would be false. But this makes the implication
P → Q false! So it must be that Sue did not get a 93% on the final. Notice
now we have the implication ¬Q → ¬P which is the contrapositive of P → Q.
Since P → Q is assumed to be true, we know ¬Q → ¬P is true as well.
If and only if
P ↔ Q is logically equivalent to (P → Q ) ∧ (Q → P ).
Example: Given an integer n, it is true that n is even if and only if n 2 is even. That is,
if n is even, then n 2 is even, as well as the converse: if n 2 is even, then n is even.
You can think of “if and only if” statements as having two parts: an implication and its
converse. We might say one is the “if” part, and the other is the “only if” part. We also
sometimes say that “if and only if” statements have two directions: a forward direction
(P → Q ) and a backwards direction (P ← Q, which is really just sloppy notation for
Q → P).
Let’s think a little about which part is which. Is P → Q the “if” part or the “only if”
part? Perhaps we should look at an example.
Example:
Suppose it is true that I sing if and only if I’m in the shower. We know this means
that both if I sing, then I’m in the shower, and also the converse, that if I’m in the
3.2. Logical Equivalence 144
shower, then I sing. Let P be the statement, “I sing,” and Q be, “I’m in the shower.”
So P → Q is the statement “if I sing, then I’m in the shower.” Which part of the if
and only if statement is this?
What we are really asking is what is the meaning of “I sing if I’m in the shower”
and “I sing only if I’m in the shower.” When is the first one (the “if” part) false?
When I am in the shower but not singing. That is the same condition on being false
as the statement “if I’m in the shower, then I sing.” So the “if” part is Q → P. On
the other hand, to say, “I sing only if I’m in the shower” is equivalent to saying “if I
sing, then I’m in the shower,” so the only if part is P → Q.
It is not terribly important to know which part is the if or only if part, but this does get
at something very, very important: THERE ARE MANY WAYS TO STATE AN IMPLICA-
TION! The problem is, since these are all different ways of saying the same implication, we
cannot use truth tables to analyze the situation. Instead, we just need good English skills.
Example:
Rephrase the implication, “if I dream, then I am asleep” in as many different ways
as possible. Then do the same for the converse.
Solution: The following are all equivalent to the original implication:
1. I am asleep if I dream.
The following are equivalent to the converse (if I am asleep, then I dream):
1. I dream if I am asleep.
Hopefully you agree with the above example. We include the “necessary and sufficient”
versions because those are common when discussing mathematics. In fact, let’s agree once
and for all what they mean:
3.2 Exercises 145
To be honest, I have trouble with these if I’m not very careful. I find it helps to have an
example in mind:
Example:
Recall from calculus, if a function is differentiable at a point c, then it is continuous
at c, but that the converse of this statement is not true (for example, f ( x ) | x | at
the point 0). Restate this fact using necessary and sufficient language.
Solution: It is true that in order for a function to be differentiable at a point c, it is
necessary for the function to be continuous at c. However, it is not necessary that a
function be differentiable at c for it to be continuous at c.
It is true that to be continuous at a point c, it is sufficient that the function be
differentiable at c. However, it is not the case that being continuous at c is sufficient
for a function to be differentiable at c.
Exercises
1. Determine whether the following two statements are logically equivalent: ¬(P → Q )
and P ∧ ¬Q. Explain how you know you are correct.
3. Consider the statement “If Oscar eats Chinese food, then he drinks milk.”
4. Simplify the following statements (so that negation only appears right before vari-
ables).
(a) ¬(P → ¬Q ).
3.3. Quantifiers and Predicate Logic 146
5. Which of the following statements are equivalent to the implication, “if you win
the lottery, then you will be rich,” and which are equivalent to the converse of the
implication?
(a) Either you win the lottery or else you are not rich.
(b) Either you don’t win the lottery or else you are rich.
(c) You will win the lottery and be rich.
(d) You will be rich if you win the lottery.
(e) You will win the lottery if you are rich.
(f) It is necessary for you to win the lottery to be rich.
(g) It is sufficient to win the lottery to be rich.
(h) You will be rich only if you win the lottery.
(i) Unless you win the lottery, you won’t be rich.
(j) If you are rich, you must have one the lottery.
(k) If you are not rich, then you did not win the lottery.
(l) You will win the lottery if and only if you are rich.
6. Consider the implication, “if you clean your room, then you can watch TV.” Rephrase
the implication in as many ways as possible. Then do the same for the converse.
Investigate!
Consider the statement below. Decide whether any are equivalent to each other, or
whether any imply any others.
So far we have seen how statements can be combined with logical symbols. This is
helpful when trying to understand a complicated mathematical statement since you can
determine under which conditions the complicated statement is true. Additionally, we
have been able to analyze the logical form of arguments to decide which arguments are
valid and which are not. However, the types of statements we have been able to make so
far has be sorely limited. For example, consider a classic argument:
Predicates
We can think of predicates as properties of objects. For example, consider the predicate
E which we will use to mean “is even.” Being even is a property of some numbers, so
E needs to be applied to something. We will adopt the notation E ( x ) to mean x is even.
(Some books would write Ex instead.) Notice that if we put a number in for x, then this
becomes a statement, and as such, can be true or false. So E (2) is true, and E (3) is false. A
predicate is like a function with codomain equal to the set of truth values {T, F }. On the
other hand, E ( x ) is not true or false, since we don’t know what x is. If we have a variable
floating around like that, we say the expression is merely a formula, and not a statement.
Since E (2) is a statement (a proposition), we can apply propositional logic to it. Consider
E (2) ∧ ¬E (3)
which is a true statement, because it is both the case that 2 is even and that 3 is not even.
What we have done here is capture the logical form (using connectives) of the statement
“2 is even and 3 is not” as well as the mathematical content (using predicates).
Notice that we can only assert even-ness of a single number at a time. That is to say,
E is a one-place predicate. There are also predicates which assert a property of two or
more numbers (or other objects) at the same time. Consider the two-place predicate “is less
than.” Perhaps we will use the variable L. Now we can say L(2, 3), which is true because
2 is less than 3. Of course we are already have a symbol for this: 2 < 3. However, what
about “divides evenly into” as a predicate? We can say D (2, 10) is true because 2 divides
evenly into 10, while D (3, 10) is false since there is a remainder when you divide 10 by 3.
Incidentally, there is a standard mathematical symbol for this: 2|10 is read “2 divides 10.”
Predicates can be as complicated and have as many places as we want or need. For
example, we could let R( x, y, z, u, v, w ) be the predicate asserting that x, y, z are distinct
natural numbers whose only common factor is u, the difference between x and y is v and
the difference between y and z is w. This is a silly and most likely useless example, but it
is an example of a predicate. It is true of some ordered lists of six numbers (6-tuples), and
false of others. Additionally, predicates need not have anything to do with numbers: we
could let F ( a, b, c, d ) be the predicate that asserts that a and b are the only two children of
mother c and father d.
Quantifiers
Perhaps the most important reason to use predicate logic is that doing so allows for quan-
tification. We can now express statements like “every natural number is either even or
odd,” and “there is a natural number such that no number is less than it.” Think back
to Calculus and the Mean Value Theorem. It states that for every function f and every
interval ( a, b ), if f is continuous on the interval [ a, b ] and differentiable on the interval
(a, b ), then there exists a number c such that a ≤ c ≤ b and f 0(c )(b − a ) f (b ) − f (a ). Using
the correct predicates and quantifiers, we could express this statement entirely in symbols.
There are two quantifiers we will be interested in: existential and universal.
3.3. Quantifiers and Predicate Logic 149
Quantifiers
• The existential quantifier is ∃ and is read “there exists” or “there is.” For
example,
∃x ( x < 0)
asserts that there is a number less than 0.
• The universal quantifier is ∀ and is read “for all” or “every.” For example,
∀x ( x ≥ 0)
Are the statements ∃x ( x < 0) and ∀x ( x ≥ 0) true? Well, first notice that they cannot
both be true. In fact, they assert exactly the opposite of each other. (Note that x < y ↔
¬( x ≥ y ), although you might wonder what x and y are here, so it might be better to
say ∀x∀y x < y ↔ ¬( x ≥ y ) .) Which one is it though? The answer depends entirely on
our domain of discourse, the universe over which we quantify. Usually, this universe is
clear from the context. If we are only discussing the natural numbers, then ∀x . . . means
“for every natural number x . . ..” On the other hand, in calculus we care about the real
numbers, so it would mean “for every real number x . . ..” If the context is not clear, we
might write ∀x ∈ N . . . to mean “for every natural number x. . . .” Of course, for the two
statements above, the second is true of the natural numbers, the first is true for any universe
which includes negative numbers, such as the integers.3
Some more examples: To say “every natural number is either even or odd,” we would
write, using E and O as the predicates for even and odd respectively:
∀x (E( x ) ∨ O ( x )).
To say “there is a number such that no number is less than it” we would write:
∃x∀y ( y ≥ x ).
Actually, I did a little translation before I wrote that down. The above statement would be
literally read “there is a number such that every number is greater than or equal to it.” This
of course amounts to the same thing. However, if I wanted to have a literal translation, I
could have also written:
∃x¬∃y ( y < x ).
Notice also that saying that there is a number for which no number is smaller is equivalent
to saying that it is not the case that for every number there is a number smaller than it:
¬∀x∃y ( y < x ).
In other words, to move a negation symbol past a quantifier, you must switch the
quantifier. This can be done multiple times:
We also know how to move negation symbols through other connectives (using De Mor-
gan’s Laws) so it is always possible to rewrite a statement so that the only negation symbols
that appear are right in front of a predicate. This hints at the possibility of having a stan-
dard form for all predicate statements. To get this, however, we must also understand how
to move quantifiers through connectives.
Before we get too excited, note that we only need to worry about two connectives: ∧
and ∨. This is because we can rewrite p → q as ¬p ∨ q (they are logically equivalent) and
p ↔ q as ( p ∧ q ) ∨ (¬p ∧ ¬q ) (also logically equivalent).
Consider an example to see what can happen:
Example:
Let E be the predicate for being even, and O for being odd. Consider:
∃xE( x ) ∧ ∃xO ( x ) ,
which says that there is a number which is even and a number which is odd. This
is of course true. However there is no number which is both even and odd, so
∃x (E( x ) ∧ O ( x ))
∀x (E( x ) ∧ O ( x ))
3.3. Quantifiers and Predicate Logic 151
∃x∃y (E ( x ) ∧ O ( y )).
The same thing works with ∨ and for ∀ with either connective. As long as there is
no repeat in quantified variables we can move the quantifiers outside of conjunctions and
disjunctions.
A warning though: you cannot do this for →, at least not directly.4 Let’s see what
happens.
Example:
Consider
∀xP ( x ) → ∃yQ ( y )
for some predicates P and Q. This sentence is not the same as
∀x∃y (P ( x ) → Q ( y )).
¬∀xP ( x ) ∨ ∃yQ ( y ).
Before we move the quantifiers out, we must move the ∀x past the negation sign,
which switches it to a ∃x:
∃x¬P ( x ) ∨ ∃yQ ( y ).
Then we can finish by writing,
∃x∃y (¬P ( x ) ∨ Q ( y ))
or equivalently
∃x∃y (P ( x ) → Q ( y )).
Exercises
1. Translate into symbols. Use E( x ) for “x is even” and O ( x ) for “x is odd.”
(a) ∀x (E ( x ) → E ( x + 2)).
(b) ∀x∃y (sin( x ) y ).
(c) ∀y∃x (sin( x ) y ).
(d) ∀x∀y ( x 3 y 3 → x y ).
3. Simplify the statements (so negation appears only directly next to predicates).
4. Suppose P ( x ) is some predicate for which the statement ∀xP ( x ) is true. Is it also the
case that ∃xP ( x ) is true? In other words, is the statement ∀xP ( x ) → ∃xP ( x ) always
true? Is the converse always true? Explain.
5. For each of the statements below, give a universe of discourse for which the statement
is true, and a universe for which the statement is false.
(a) ∀x∃y ( y 2 x ).
(b) ∀x∀y∃z ( x < z < y ).
(c) ∃x∀y∀z ( y < z → y ≤ x ≤ z ) Hint: universes need not be infinite.
6. Can you switch the order of quantifiers? For example, consider the two statements:
Depending on what the predicate P ( x, y ) is, each statement might be true or false.
Give an example of a predicate for which the first statement is true and the second
false of the natural numbers. Can you also give an example of a (different) predicate
for which the first statement is false and the second true? Explain.
3.4. Proofs 153
3.4 Proofs
Investigate!
Suppose you have a collection of 5-cent stamps and 8-cent stamps. We saw earlier that
it is possible to make any amount of postage greater than 27 cents using combinations
of both these types of stamps. But, let’s ask some other questions:
1. What amounts of postage can you make if you only use an even number of
both types of stamps? Prove your answer.
2. Suppose you made an even amount of postage. Prove that you used an even
number of at least one of the types of stamps.
3. Suppose you made exactly 72 cents of postage. Prove that you used at least 6
of one type of stamp.
Proof. Suppose this were not the case. That is, suppose there are only finitely many primes.
Then there must be a last, largest prime, call it p. Consider the number
N p! + 1 ( p · ( p − 1) · · · · 3 · 2 · 1) + 1.
Now N is certainly larger than p. Also, N is not divisible by any number less than or equal
to p, since every number less than or equal to p divides p!. Thus the prime factorization of
N contains prime numbers (possibly just N itself) all greater than p. So p is not the largest
prime, a contradiction. Therefore there are infinitely many primes. qed
This proof is an example of a proof by contradiction, one of the standard styles of math-
ematical proof. First and foremost, the proof is an argument. It contains sequence of
3.4. Proofs 154
statements, the last being the conclusion which follows from the previous statements. The
argument is valid so the conclusion must be true if the premises are true. Let’s go through
the proof line by line.
3. Let N p! + 1.
basically just notation, although this is the inspired part of the proof; looking at p! + 1 is the
key insight.
4. N is larger than p.
by the definition of p!
8. This is a contradiction.
from line 2 and line 7: the largest prime is p and there is a prime larger than p.
We should say a bit more about the last line. Up through line 8, we have a valid argument
with the premise “there are only finitely many primes” and the conclusion “there is a prime
larger than the largest prime.” This is a valid argument as each line follows from previous
lines. So if the premises are true, then the conclusion must be true. However, the conclusion
is NOT true. The only way out: the premise must be false.
The sort of line-by-line analysis we did above is a great way to really understand what
is going on. Whenever you come across a proof in a textbook, you really should make
sure you understand what each line is saying and why it is true. Additionally, it is equally
important to understand the overall structure of the proof. This is where using tools from
logic is helpful. Luckily there are a relatively small number of standard proof styles that
keep showing up again and again. Being familiar with these can help understand proof,
as well as give ideas of how to write your own.
3.4. Proofs 155
Direct Proof
The simplest (from a logic perspective) style of proof is a direct proof. Often all that is
required to prove something is a systematic explanation of what everything means. Direct
proofs are especially useful when proving implications. The general format to prove P → Q
is this:
Example:
Prove: For all integers n, if n is even, then n 2 is even.
Solution: The format of the proof with be this: Let n be an arbitrary integer.
Assume that n is even. Explain explain explain. Therefore n 2 is even.
To fill in the details, we will basically just explain what it means for n to be even,
and then see what that means for n 2 . Here is a complete proof.
Example:
Prove: For all integers a, b, and c, if a | b and b | c then a | c. Here x | y, read “x divides
y” means that y is a multiple of x (so x will divide into y without remainder).
Solution: Even before we know what the divides symbol means, we can set up
a direct proof for this statement. It will go something like this: Let a, b, and c be
arbitrary integers. Assume that a | b and b | c. Dot dot dot. Therefore a | c.
How do we connect the dots? We need to say what our hypothesis (a | b and b | c)
really means and why this gives us what the conclusion (a | c) really means. Another
way to say that a | b is to say that b ka for some integer k (that is, that b is a multiple
of a). What are we going for? That c la, for some integer l (because we want c to
be a multiple of a). Here is the complete proof.
and c jb. Combining these (through substitution) we get that c jka. But jk is
an integer, so this says that c is a multiple of a. Therefore a | c. qed
Proof by Contrapositive
Recall that an implication P → Q is logically equivalent to its contrapositive ¬Q → ¬P.
There are plenty of examples of statements which are hard to prove directly, but whose
contrapositive can easily be proved directly. This is all that proof by contrapositive does.
It gives a direct proof of the contrapositive of the implication. This is enough because the
contrapositive is logically equivalent to the original implication.
The skeleton of the proof of P → Q by contrapositive will always look roughly like this:
As before, if there are variables and quantifiers, we set them to be arbitrary elements of
our domain. Here are a couple examples.
Example:
Is the statement “for all integers n, if n 2 is even, then n is even” true?
Solution: This is the converse of the statement we proved above using a direct
proof. From trying a few examples, this statement definitely appears this is true.
So let’s prove it.
A direct proof of this statement would require fixing an arbitrary n and assuming
that n 2 is even. But it is not at all clear how this would allow us to conclude anything
about n. Just because n 2 2k does not in itself suggest how we could write n as a
multiple of 2.
Try something else: write the contrapositive of the statement. We get, for all
integers n, if n is odd then n 2 is odd. This looks much more promising. Our proof
will look something like this:
Let n be an arbitrary integer. Suppose that n is not even. This means that. . . . In
other words. . . . But this is the same as saying . . . . Therefore n 2 is not even.
Now we fill in the details.
Example:
Prove: for all integers a and b, if a + b is odd, then a is odd or b is odd.
3.4. Proofs 157
Solution: The problem with trying a direct proof is that it will be hard to separate
a and b from knowing something about a + b. On the other hand, if we know
something about a and b separately, then combining them might give us information
about a + b. Our proof will have the following format:
Let a and b be integers. Assume that a and b are both even. la la la. Therefore
a + b is even.
Now for a complete proof.
Proof. Let a and b be integers. Assume that a and b are even. Then a 2k and
b 2l for some integers k and l. Now a + b 2k + 2l 2( k + 1). Since k + l is an
integer, we see that a + b is even, completing the proof. qed
Note that our assumption that a and b are even is really the negation of a or b is
odd. We used De Morgan’s law here.
We have seen how to prove some statements in the form of implications: either directly
or by contrapositive. Some statements aren’t written as implications to begin with.
Example:
Consider the statement, for every prime number p, either p 2 or p is odd. We can
rephrase this: for every prime number p, if p , 2, then p is odd. Now try to prove
it.
Proof by Contradiction
There
√ might be statements which really cannot be rephrased as implications. For example,
“ 2 is irrational.” In this case, it is hard to know where to start. What can we assume?
Well, say we want to prove the statement P. What if we could prove that ¬P → Q where
Q was false? If this implication is true, and Q is false, what can we say about ¬P? It must
be false as well, which makes P true!
This is why proof by contradiction works. If we can prove that ¬P leads to a contradic-
tion, then the only conclusion is that ¬P is false, so P is true. That’s what we wanted to
prove. In other words, if it is impossible for P to be false, P must be true.
Here are a couple examples of proofs by contradiction.
Example:
√
Prove that 2 is irrational.
3.4. Proofs 158
√
Proof. Suppose not. Then 2 is equal to a fraction ba . Without loss of generality,
assume ba is in lowest terms (otherwise reduce the fraction). So,
a2
2
b2
2b 2 a 2
Thus a 2 is even, and as such a is even. So a 2k for some integer k, and a 2 4k 2 .
We then have,
2b 2 4k 2
b 2 2k 2
a
Thus b 2 is even, and as such b is even. Since a is also even, we see that b is not in
√
lowest terms, a contradiction. Thus 2 is irrational. qed
Example:
Prove: there are no integers x and y such that x 2 4y + 2.
Proof. We proceed by contradiction. So suppose there are integers x and y such that
x 2 4y +2 2(2y +1). So x 2 is even. We have seen that this implies that x is even. So
x 2k for some integer k. Then x 2 4k 2 . This in turn gives 2k 2 (2y + 1). But 2k 2
is even, and 2y + 1 is odd, so these cannot be equal. Thus we have a contradiction,
so there must not be any integers x and y such that x 2 4y + 2. qed
Example:
The Pigeonhole Principle: if more than n pigeons fly into n pigeon holes, then at
least one pigeon hole will contain at least two pigeons. Prove this!
Proof. Suppose, contrary to stipulation, that each of the pigeon holes contain at
most one pigeon. Then at most, there will be n pigeons. But we assumed that there
are more than n pigeons, so this is impossible. Thus there must be a pigeon hole
with more than one pigeon. qed
n 1 2 3 4 5 6 7
n 2 − n + 41 41 43 47 53 61 71 83
So far we have gotten only primes. You might be tempted to conjecture, “For all positive
integers n, the number n 2 − n + 41 is prime.” If you wanted to prove this, you would need
to use a direct proof, a proof by contrapositive, or another style of proof, but certainly it is
not enough to give even 7 examples. In fact, we can prove this conjecture is false by proving
its negation: “There is a positive integer n such that n 2 − n + 41 is not prime.” Since this is
an existential statement, it suffices to show that there does indeed exist such a number.
In fact, we can quickly see that n 41 will give 412 which is certainly not prime. You
might say that this is a counterexample to the conjecture that n 2 − n + 41 is always prime.
Since so many statements in mathematics are universal, making their negations existential,
we can often prove that a statement is false (if it is) by providing a counterexample.
Example:
Above we proved, “for all integers a and b, if a + b is odd, then a is odd or b is odd.”
Is the converse true?
Solution: The converse is the statement, “for all integers a and b, if a is odd or b
is odd, then a + b is odd.” This is false! How do we prove it is false? We need to
prove the negation of the converse. Let’s look at the symbols. The converse is
5 This is not to say that looking at examples is a waste of time. Doing so will often give you an idea of how
to write a proof. But the examples do not belong in the proof.
3.4. Proofs 160
Proof by Cases
We could go on and on and on about different proof styles (we haven’t even mentioned
induction or combinatorial proofs here), but instead we will end with one final useful
technique: proof by cases. The idea is to prove that P is true by proving that Q → P and
¬Q → P for some statement Q. So no matter what, whether or not Q is true, we know that
P is true. In fact, we could generalize this. Suppose we want to prove P. We know that
at least one of the statements Q1 , Q2 , . . . , Q n are true. If we can show that Q1 → P and
Q 2 → P and so on all the way to Q n → P, then we can conclude P. The key thing is that
we want to be sure that one of our cases (the Q i ’s) must be true no matter what.
If that last paragraph was confusing, hopefully an example will make things better.
Example:
Prove: for any integer n, the number ( n 3 − n ) is even.
Solution: It is hard to know where to start this, because we don’t know much of
anything about n. We might be able to prove that n 3 − n is even if we knew that n
was even. In fact, we could probably prove that n 3 − n was even if n was odd. But
since n must either be even or odd, this will be enough. Here’s the proof.
n 3 − n 8k 3 − 2k
2(4k 2 − k ) ,
Exercises
1. Consider the statement “for all integers a and b, if a + b is even, then a and b are even”
(a) Prove the statement. What sort of proof are you using?
(b) Is the converse true? Prove or disprove.
√
3. Prove that 3 is irrational.
4. Consider the statement: for all integers a and b, if a is even and b is a multiple of 3,
then ab is a multiple of 6.
(a) Prove the statement. What sort of proof are you using?
(b) State the converse. Is it true? Prove or disprove.
5. Prove the statement: For all integers n, if 5n is odd, then n is odd. Clearly state the
style of proof you are using.
7. The game TENZI comes with 40 six-sided dice (each numbered 1 to 6). Suppose you
roll all 40 dice. Prove that there will be at least seven dice that land on the same
number.
8. How many dice would you have to roll before you were guaranteed that some four
of them would all match or all be different? Prove your answer.
11. Prove that every prime number greater than 3 is either one more or one less than a
multiple of 6. Hint: prove the contrapositive by cases.
3.5. Chapter Summary 162
12. For each of the statements below, say what method of proof you should use to prove
them. Then say how the proof starts and how it ends. Bonus points for filling in the
middle.
(a) There are no integers x and y such that x is a prime greater than 5 and x 6y +3.
(b) For all integers n, if n is a multiple of 3, then n can be written as the sum of
consecutive integers.
(c) For all integers a and b, if a 2 + b 2 is odd, then a or b is odd.
Every even number greater than 2 can be written as the sum of two primes.
We are not going to try to prove the statement here, but we can at least say what a proof
might look like, based on the logical form of the statement. Perhaps we should write the
statement in an equivalent way which better highlights the quantifiers and connectives:
For all integers n, if n is even and greater than 2, then there exists integers p
and q such that p and q are prime and n p + q.
What would a direct proof look like? Since the statement starts with a universal
quantifier, we would start by, “Let n be an arbitrary integer." The rest of the statement is an
implication. In a direct proof we assume the “if” part, so the next line would be, “Assume
n is greater than 2 and is even.” I have no idea what comes next, but eventually, we would
3.5 Exercises 163
need to find two prime numbers p and q (depending on n) and explain how we know that
n p + q.
Or maybe we try a proof by contradiction. To do this, we first assume the negation of
the statement we want to prove. What is the negation? From what we have studied we
should be able to see that it is,
There is an integer n such that n is even and greater than 2, but for all integers
p and q, either p or q is not prime or n , p + q.
Could this statement be true? A proof by contradiction would start by assuming it was
and eventually conclude with a contradiction, proving that our assumption of truth was
incorrect. And if you can find such a contradiction, you will have proved the most famous
open problem in mathematics. Good luck.
Exercises
1. Complete a truth table for the statement ¬P → (Q ∧ R).
2. Suppose you know that the statement “if Peter is not tall, then Quincy is fat and Robert
is skinny” is false. What, if anything, can you conclude about Peter and Robert if you
know that Quincy is indeed fat? Explain (you may reference problem 1).
P→Q
P→R
∴ P → ( Q ∧ R ).
5. Write the negation, converse and contrapositive for each of the statements below.
(a) If the power goes off, then the food will spoil.
(b) If the door is closed, then the light is off.
(c) ∀x ( x < 1 → x 2 < 1).
(d) For all natural numbers n, if n is prime, then n is solitary.
(e) For all functions f , if f is differentiable, then f is continuous.
(f) For all integers a and b, if a · b is even, then a and b are even.
(g) For every integer x and every integer y there is an integer n such that if x > 0
then nx > y.
(h) For all real numbers x and y, if x y 0 then x 0 or y 0.
(i) For every student in Math 228, if they do not understand implications, then they
will fail the exam.
3.5 Exercises 164
6. Consider the statement: for all integers n, if n is even and n ≤ 7 then n is negative or
n ∈ {0, 2, 4, 6}.
(a) Explain what the statement says in words. Is this statement true? Be sure to
state what you are taking the universe of discourse to be.
(b) Write the converse of the statement, both in words and in symbols. Is the
converse true?
(c) Write the contrapositive of the statement, both in words and in symbols. Is the
contrapositive true?
(d) Write the negation of the statement, both in words and in symbols. Is the
negation true?
8. Write each of the following statements in the form, “if . . . , then . . . .” Careful, some
of the statements might be false (which is alright for the purposes of this question).
10. Consider the statement: for all integers n, if n is odd, then 7n is odd.
(a) Prove the statement. What sort of proof are you using?
(b) Prove the converse. What sort of proof are you using?
11. Suppose you break your piggy bank and scoop up a handful of 22 coins (pennies,
nickels, dimes and quarters).
3.5 Exercises 165
(a) Prove that you must have at least 6 coins of a single denomination.
(b) Suppose you have an odd number of pennies. Prove that you must have an odd
number of at least one of the other types of coins.
(c) How many coins would you need to scoop up to be sure that you either had 4
coins that were all the same or 4 coins that were all different? Prove your answer.
Homework Problems
The following are some more involved problems for you to try, which might be assigned
as homework.
1. Your “friend” has shown you a “proof” he wrote to show that 1 3. Here is the
proof:
What is going on here? Is your friends argument valid? Is the argument a proof of
the claim 1 3? Carefully explain using what we know about logic. Hint: What
implication follows from the given proof?
2. Tommy Flanagan was telling you what he ate yesterday afternoon. He tells you, “I
had either popcorn or raisins. Also, if I had cucumber sandwiches, then I had soda.
But I didn’t drink soda or tea.” Of course you know that Tommy is the worlds worst
liar, and everything he says is false. What did Tommy eat?
Justify your answer by writing all of Tommy’s statements using sentence variables
(P, Q, R, S, T), taking their negations, and using these to deduce what Tommy actually
ate.
3. Use De Morgan’s Laws, and any other logical equivalence facts you know to simplify
the following statements. Show all your steps. Your final statements should have
negations only appear directly next to the sentence variables or predicates (P, Q, E ( x ),
etc.), and no double negations. It would be a good idea to use only conjunctions,
disjunctions, and negations.
4. Can you chain implications together? That is, if P → Q and Q → R, does that means
the P → R? Can you chain more implications together? Let’s find out:
3.5 Exercises 166
P→Q
(a) Prove that the following is a valid deduction rule: Q→R
∴ P→R
(b) Prove that the following is a valid deduction rule for any n ≥ 2:
P1 → P2
P2 → P3
..
.
Pn−1 → Pn
∴ P1 → Pn .
I suggest you don’t go through the trouble of writing out a 2n row truth table.
Instead, you should use part (a) and mathematical induction.
(a) Explain what this statement says in words. Is the statement true?
(b) State the contrapositive of the original statement. Do so both in words and in
symbols.
(c) State the converse of the original statement. Is the converse true?
(d) State the negation of the original statement. Do so both in words and in symbols
(simplifying as much as possible).
Write out the beginning and end of the argument if you were to prove the statement,
(a) Directly
(b) By contrapositive
(c) By contradiction
You do not need to provide details for the proofs (since you do not know what solitary
means). However, make sure that you provide the first few and last few lines of the
proofs so that we can see that logical structure you would follow.
7. A standard deck of 52 cards consists of 4 suites (hearts, diamonds, spades and clubs)
each containing 13 different values (Ace, 2, 3, . . . , 10, J, Q, K). If you draw some
number of cards at random you might or might not have a pair (two cards with the
same value) or three cards all of the same suit. However, if you draw enough cards,
you will be guaranteed to have these. For each of the following, find the smallest
number of cards you would need to draw to be guaranteed having the specified cards.
Prove your answers.
8. Suppose you are at a party with 19 of your closest friends (so including you, there are
20 people there). Explain why there must be least two people at the party who are
friends with the same number of people at the party. Assume friendship is always
reciprocated.
9. Suppose you have an n × n chessboard but your dog has eaten one of the corner
squares. Can you still cover the remaining squares with dominoes? What needs to
be true about n? Give necessary and sufficient conditions (that is, say exactly which
values of n work and which do not work). Prove your answers.
10. Bonus: What if your n × n chessboard is missing two opposite corners? Prove that no
matter what n is, you will not be able to cover the remaining squares with dominoes.
Chapter 4
Graph Theory
Investigate!
In the time of Euler, in the town of Königsberg in Prussia, there was a river containing
two islands. The islands were connected to the banks of the river by seven bridges
(as seen below). The bridges were very beautiful, and on their days off, townspeople
would spend time walking over the bridges. As time passed, a question arose: was
it possible to plan a walk so that you cross each bridge once and only once? Euler
was able to answer this question. Are you?
168
169
There is an obvious connection between these two problems. Any path in the dot and
line drawing corresponds exactly to a path over the bridges of Königsberg.
Pictures like the dot and line drawing are called graphs. Graphs are made up of a
collection of dots called vertices and lines connecting those dots called edges. When two
vertices are connected by an edge, we say they are adjacent. The nice thing about looking at
graphs instead of pictures of rivers, islands and bridges is that we now have a mathematical
object to study: a set containing of a set of vertices and a set of edges (in fact, we can take the
set of edges to be a set of two element subsets from the set of vertices). We have distilled the
“important” parts of the bridge picture for the purposes of the problem. It does not matter
how big the islands are, what the bridges are made out of, if the river contains alligators,
etc. All that matters is which land masses are connected to which other land masses, and
how many times. This was the great insight that Euler had.
We will return to the question of finding paths through graphs later. But first, here are
a few other situations you can represent with graphs.
Example:
Al, Bob, Cam, Dan, and Euclid are all members of the social networking website
Facebook. The site allows members to be “friends” with each other. It turns out
that Al and Cam are friends, as are Bob and Dan. Euclid is friends with everyone.
Represent this situation with a graph.
Solution: Each person will be represented by a vertex and each friendship will be
represented by an edge. That is, there will be an edge between two vertices if and
only if the people represented by those vertices are friends. We get the following
graph:
A B
E
C D
Example:
Each of three houses must be connected to each of three utilities. Is it possible to
do this without any of the utility lines crossing?
Solution: We will answer this question later. For now, notice how we would ask
this question in the context of graph theory. We are really asking whether it is
possible to redraw the graph below without any edges crossing (except at vertices).
4.1. Basics 170
4.1 Basics
Investigate!
1. Which (if any) of the graphs below are the same?
f b c f c b f v1 v2 v3
b d
a c e a d e a e d v6 v5 v4
3. Actually, all the graphs we have seen above are just drawings of graphs. A
graph is really an abstract mathematical object consisting of two sets V and E
where E is a set of 2-element subsets of V.
Are the graphs below the same or different?
Graph 1: V { a, b, c, d, e }, E {{ a, b } , { a, c } , { a, d } , { a, e } , { b, c } , { d, e }}.
Graph 2: V { v 1 , v 2 , v 3 , v4 , v 5 },
E {{ v 1 , v 3 } , { v 1 , v 5 } , { v2 , v 4 } , { v 2 , v 5 } , { v 3 , v5 } , { v 4 , v 5 }}.
{{ a, b } , { a, c } , { b, c } , { b, d } , { c, d }} .
We could have described the graph in words as follows: we have four vertices, a, b, c, and
d, and a is adjacent to b and c, b is adjacent to both c and d, and c is also adjacent to d. One
way to draw this graph is this:
a b
c d
4.1. Basics 171
However we could also have drawn the graph differently. For example either of these:
a d
c b a b c d
We should be careful about what it means for two graphs to be “the same.” Remember,
as mathematical objects, graphs are just sets. Two sets are equal if they have the exact same
members. However, even if two graphs are not equal, they might be basically the same.
Graphs that are basically the same (but perhaps not equal) are called isomorphic. We will
give a precise definition of this term after a quick example:
Example:
Consider the graphs:
G1 {V1 , E1 } where V1 { a, b, c } and E1 {{ a, b } , { a, c } , { b, c }};
G2 {V2 , E2 } where V2 { u, v, w } and E2 {{ u, v } , { u, w } , { v, w }}.
Are these graphs the same?
Solution: The two graphs are NOT equal. It is enough to notice that V1 , V2 since
a ∈ V1 but a < V2 . However, both of these graphs consist of three vertices with
edges connecting every pair of vertices. We can draw them as follows:
a u
b c v w
Clearly we want to say these graphs are basically the same, so while they are not
equal, they will be isomorphic. The reason is we can rename the vertices of one
graph and get the second graph as the result.
Intuitively, graphs are isomorphic if they are basically the same, or better yet, if they are
the same except for the names of the vertices. To make the concept of renaming vertices
precise, we give the following definitions:
Isomorphic Graphs
An isomorphism between two graphs G1 and G2 is a bijection f : V1 → V2 between the
vertices of the graphs such that if { a, b } is an edge in G1 then { f ( a ) , f ( b )} is an edge
in G2 .
Two graphs are isomorphic if there is an isomorphism between them. In this case
we write G1 G2 .
4.1. Basics 172
Example:
Decide whether the graphs G1 {V1 , E1 } and G2 {V2 , E2 } are equal or isomor-
phic.
V1 { a, b, c, d }, E1 {{ a, b } , { a, c } , { a, d } , { c, d }}
V2 { a, b, c, d }, E2 {{ a, b } , { a, c } , { b, c } , { c, d }}
Solution: The graphs are NOT equal, since { a, d } ∈ E1 but { a, d } < E2 . However,
since both graphs contain the same number of vertices and same number of edges,
they might be isomorphic (this is not enough in most cases, but it is a good start).
We can try to build an isomorphism. How about we say f ( a ) b, f ( b ) c,
f ( c ) d and f ( d ) a. This is definitely a bijection, but to make sure that the
function is an isomorphism, we must make sure it respects the edge relation. In G1 ,
vertices a and b are connected by an edge. In G2 , f ( a ) b and f ( b ) c are
connected by an edge. So far, so good, but we must check the other three edges.
The edges { a, c } in G1 corresponds to { f ( a ) , f ( c )} { b, d }, but here we have a
problem. There is no edge between b and d in G2 . Thus f is NOT an isomorphism.
Not all hope is lost, however. Just because f is not an isomorphism does not
mean that there is no isomorphism at all. We can try again. At this point it might be
helpful to draw the graphs to see how they should match up. Alternatively, notice
that in G1 , the vertex a is adjacent to every other vertex. In G2 , there is also a vertex
with this property: c. So build the bijection g : V1 → V2 by defining g ( a ) c to
start with. Next, where should we send b? In G1 , the vertex b is only adjacent to
vertex a. There is exactly one vertex like this in G2 , namely d. So let g ( b ) d. As
for the last two, in this example, we have a free choice: let g ( c ) b and g ( d ) a
(switching these would be fine as well).
We should check that this really is an isomorphism. It is definitely a bijection.
We must make sure that the edges are respected. The four edges in G1 are
{ a, b } , { a, c } , { a, d } , { c, d }
{ g (a ), g (b )} , { g (a ), g (c )} , { g (a ), g (d )} , { g (c ), g (d )}
{ c, d } , { c, b } , { c, a } , { b, a }
which are precisely the edges in G2 . Thus g is an isomorphism, so G1 G2
Sometimes we will talk about a graph with a special name (like K n or the Peterson graph)
or perhaps draw a graph without any labels. In this case we are really referring to all graphs
4.1. Basics 173
isomorphic to any copy of that particular graph. A collection of isomorphic graphs is often
called an isomorphism class.1
Back to some basic graph theory definitions. Notice that the graphs above have the
property that no pair of vertices is connected more than once, and no vertex is connected to
itself. Graphs like these are sometimes called simple, although we will just call them graphs.
This is because our definition for a graph says that the edges form a set of 2-element subsets
of the vertices. Remember that it doesn’t make sense to say a set contains an element more
than once. So no pair of vertices can be connected by an edge more than once. Also, since
each edge must be a set containing two vertices, we cannot have a single vertex connected
to itself by an edge.
That said, there are times we want to consider double (or more) edges and single edge
loops. For example, the “graph” we drew for the Bridges of Königsberg problem had
double edges because there really are two bridges connecting a particular island to the
near shore. We will call these objects multigraphs. This is a good name: a multiset is a set in
which we are allowed to include a single element multiple times.
The graphs above are also connected: you can get from any vertex to any other vertex
by following some path of edges. A graph that is not connected can be thought of as two
separate graphs drawn close together. Unless otherwise stated, we will assume all our
graphs are connected.
Vertices in a graph do not always have edges between them. If we add all possible
edges, then the resulting graph is called complete. That is, a graph is complete if every
pair of vertices is connected by an edge. Since a graph is determined completely by which
vertices are adjacent to which other vertices, there is only one complete graph with a given
number of vertices. We give these a special name: K n is the complete graph on n vertices.
Each vertex in K n is adjacent to n − 1 other vertices. We call the number of edges
emanating from a given vertex the degree of that vertex. So every vertex in K n has degree
n − 1. How many edges does K n have? One might think the answer should be n ( n − 1),
since we count n − 1 edges n times (once for each vertex). However, each edge is incident
to 2 vertices, so we counted every edge exactly twice. Thus there are n ( n − 1)/2 edges in
K n . Alternatively, we can say there are n2 edges, since to draw an edge we must choose 2
of the n vertices.
In general, if we know the degrees of all the vertices in a graph, we can find the number
of edges. The sum of the degrees of all vertices will always be twice the number of edges,
since each edge adds to the degree of two vertices. Notice this means that the sum of the
degrees of all vertices in any graph must be even!
Example:
At a recent math seminar, 9 mathematicians greeted each other by shaking hands.
Is it possible that each mathematician shook hands with exactly 7 people at the
seminar?
1 This is not unlike geometry, where we might have more than one copy of a particular triangle. There
instead of isomorphic we say congruent.
4.1. Basics 174
Solution: It seems like this should be possible. Each mathematician chooses one
person to not shake hands with. But this cannot happen. We are asking whether a
graph with 9 vertices can have each vertex have degree 7. If such a graph existed,
the sum of the degrees of the vertices would be 9 · 7 63. This would be twice
the number of edges (handshakes) resulting in a graph with 31.5 edges. That is
impossible. Thus at least one (in fact an odd number) of the mathematicians must
have shaken hands with an even number of people at the seminar.
One final definition: we say a graph is bipartite if the vertices can be divided into two
sets, A and B, with no two vertices in A adjacent and no two vertices in B adjacent. The
vertices in A can be adjacent to some or all of the vertices in B. If each vertex in A is adjacent
to all the vertices in B, then the graph is a complete bipartite graph, and gets a special name:
K m,n , where | A | m and | B | n. The graph in the houses and utilities puzzle is K3,3 .
Named Graphs
Some graphs are used more than others, and get special names.
K5 K2,3 C6 P6
4.1. Basics 175
Exercises
1. If 10 people each shake hands with each other, how many handshakes took place?
What does this question have to do with graph theory?
3. Is it possible for two different (non-isomorphic) graphs to have the same number of
vertices and the same number of edges? What if the degrees of the vertices in the
two graphs are the same (so both graphs have vertices with degrees 1, 2, 2, 3, and 4,
for example)? Draw two such graphs or explain why not.
4. Are the two graphs below equal? Are they isomorphic? If they are isomorphic, give
the isomorphism.
Graph 1: V { a, b, c, d, e }, E {{ a, b } , { a, c } , { a, e } , { b, d } , { b, e } , { c, e }}.
a
e b.
Graph 2:
d c
7. For each of the following, try to give two different unlabeled graphs with the given
properties, or explain why doing so is impossible.
(a) Two different trees with the same number of vertices and the same number of
edges. A tree is a connected graph with no cycles.
(b) Two different graphs with 8 vertices all of degree 2.
(c) Two different graphs with 5 vertices all of degree 4.
(d) Two different graphs with 5 vertices all of degree 3.
4.2. Planar Graphs 177
Investigate!
When a connected graph can be drawn without any edges crossing, it is called planar.
When a planar graph is drawn in this way, it divides the plane into regions called
faces.
1. Draw, if possible, two different planar graphs with the same number of vertices,
edges, and faces.
2. Draw, if possible, two different planar graphs with the same number of vertices
and edges, but a different number of faces.
The graphs are the same, so if one is planar, the other must be too. However, the original
drawing of the graph was not a planar representation of the graph.
When a planar graph is drawn without edges crossing, the edges and vertices of the
graph divide the plane into regions. We will call each region a face. The graph above
has 3 faces (yes, we do include the “outside” region as a face). The number of faces does
not change no matter how you draw the graph (as long as you do so without the edges
crossing), so it makes sense to ascribe the number of faces as a property of the planar graph.
A warning: you can only count faces when the graph is drawn in a planar way. For
example, consider these two representations of the same graph:
4.2. Planar Graphs 178
If you try to count faces using the graph on the left, you might say there are 5 faces
(including the outside). But drawing the graph with a planar representation shows that in
fact there are only 4 faces.
There is a connection between the number of vertices (v), the number of edges (e) and
the number of faces ( f ) in any connected planar graph. This relationship is called Euler’s
formula.
Euler’s Formula for Planar Graphs
For any (connected) planar graph with v vertices, e edges and f faces, we have
v−e+ f 2
Why is Euler’s formula true? One way to convince yourself of its validity is to draw a
planar graph step by step. Start with the graph P2 :
Any connected graph (besides just a single isolated vertex) must contain this subgraph.
Now build up to your graph by adding edges and vertices. Each step will consist of either
adding a new vertex connected by a new edge to part of your graph (so creating a new
“spike”) or by connecting two vertices already in the graph with a new edge (completing
a circuit).
What do these “moves” do? When adding the spike, the number of edges increases by
1, the number of vertices increases by one, and the number of faces remains the same. But
this means that v − e + f does not change. Completing a circuit adds one edge, adds one
face, and keeps the number of vertices the same. So again, v − e + f does not change.
Since we can build any graph using a combination of these two moves, and doing so
never changes the quantity v − e + f , that quantity will be the same for all graphs. But
notice that our starting graph P2 has v 2, e 1 and f 1, so v − e + f 2. This argument
is essentially a proof by induction. A good exercises would be to rewrite it as a formal
induction proof.
4.2. Planar Graphs 179
Non-planar Graphs
Investigate!
1. For the complete graphs K n , we would like to be able to say something about
the number of vertices, edges, and (if the graph is planar) faces. Let’s first
consider K 3
5. What about complete bipartite graphs? How many vertices, edges, and faces
(if it were planar) does K 7,4 have?
If you try to redraw this without edges crossing, you quickly get into trouble. There
seems to be one edge too many. In fact, we can prove that no matter how you draw it, K 5
will always have edges crossing.
Proof. The proof is by contradiction. So assume that K5 is planar. Then the graph must
satisfy Euler’s formula for planar graphs. K 5 has 5 vertices and 10 edges, so we get
5 − 10 + f 2
which says that if the graph is drawn without any edges crossing, there would be f 7
faces.
Now consider how many edges surround each face. Each face must be surrounded by
at least 3 edges. Let B be the total number of boundaries around all the faces in the graph.
4.2. Planar Graphs 180
Thus we have that B ≥ 3 f . But also B 2e, since each edge is used as a boundary exactly
twice. Putting this together we get
3 f ≤ 2e
But this is impossible, since we have already determined that f 7 and e 10, and 21 6≤ 20.
This is a contradiction so in fact K 5 is not planar. qed
Proving that K 3,3 is not planar answers the houses and utilities puzzle: it is not possible
to connect each of three houses to each of three utilities without the lines crossing.
Proof. Again, we proceed by contradiction. Suppose K3,3 were planar. Then by Euler’s
formula there will be 5 faces, since v 6, e 9, and 6 − 9 + f 2.
How many boundaries surround these 5 faces? Let B be this number. Since each edge
is used as a boundary twice, we have B 2e. Also, B ≥ 4 f since each face is surrounded
by 4 or more boundaries. We know this is true because K 3,3 is bipartite, so does not contain
any 3-edge cycles. Thus
4 f ≤ 2e.
But this would say that 20 ≤ 18, which is clearly false. Thus K3,3 is not planar. qed
Note the similarities and differences in these proofs. Both are proofs by contradiction,
and both start with using Euler’s formula to derive the (supposed) number of faces in the
graph. Then we find a relationship between the number of faces and the number of edges
based on how many edges surround each face. This is the only difference. In the proof
for K 5 , we got 3 f ≤ 2e and for K 3,3 we go 4 f ≤ 2e. The coefficient of f is the key. It is
the smallest number of edges which could surround any face. If some number of edges
surround a face, then these edges form a cycle. So that number is the size of the smallest
cycle in the graph.
In general, if we let g be the size of the smallest cycle in a graph (g stands for girth,
which is the technical term for this) then for any planar graph we have g f ≤ 2e. When this
disagrees with Euler’s formula, we know for sure that the graph cannot be planar.
Polyhedra
Investigate!
A cube is an example of a convex polyhedron. It contains 6 identical squares for
its faces, 8 vertices, and 12 edges. The cube is a regular polyhedron (also known as a
4.2. Planar Graphs 181
Platonic solid) because each face is an identical regular polygon and each vertex joins
an equal number of faces.
There are exactly four other regular polyhedra: the tetrahedron, octahedron,
dodecahedron, and icosahedron with 4, 8, 12 and 20 faces respectively. How many
vertices and edges do each of these have?
In fact, every convex polyhedron can be projected onto the plane without edges crossing.
Think of placing the polyhedron inside a sphere, with a light at the center of the sphere.
The edges and vertices of the polyhedron cast a shadow onto the interior of the sphere. You
can then cut a hole in the sphere in the middle of one of the projected faces and “stretch” the
sphere to lay down flat on the plane. The face that was punctured becomes the “outside”
face of the planar graph.
The point is, we can apply what we know about graphs (in particular planar graphs) to
convex polyhedra. Since every convex polyhedron can be represented as a planar graph,
we see that Euler’s formula for planar graphs holds for all convex polyhedra as well. We
also can apply the same sort of reasoning we use for graphs in other contexts to convex
polyhedra. For example, we know that there is no convex polyhedron with 11 vertices all
of degree 3, as this would make 33/2 edges.
Example:
Is there a convex polyhedron consisting of three triangles and six pentagons? What
about three triangles, six pentagons and five heptagons (7-sided polygons)?
2 An alternative definition for convex is that the internal angle formed by any two faces must be less than
180 deg.
4.2. Planar Graphs 182
Solution: How many edges would such polyhedra have? For the first proposed
polyhedron, the triangles would contribute a total of 9 edges, and the pentagons
would contribute 30. However, this counts each edge twice (as each edge borders
exactly two faces), giving 39/2 edges, an impossibility. There is no such polyhedron.
The second polyhedron does not have this obstacle. The extra 35 edges con-
tributed by the heptagons give a total of 74/2 = 37 edges. So far so good. Now how
many vertices does this supposed polyhedron have? We can use Euler’s formula.
There are 14 faces, so we have v − 37 + 14 2 or equivalently v 25. But now use
the vertices to count the edges again. Each vertex must have degree at least three
(that is, each vertex joins at least three faces), so the sum of the degrees is at least 75.
Since the sum of the degrees must be exactly twice the number of edges, this says
that there are strictly more than 37 edges. Again, there is no such polyhedron.
To conclude this application of planar graphs, consider the regular polyhedra. Above
we claimed there are only five. How do we know this is true? We can prove it using graph
theory.
Proof. Recall that a regular polyhedron has all of its faces identical regular polygons, and
that each vertex has the same degree. Consider the cases, broken up by what the regular
polygon might be.
Case 1: Each face is a triangle. Let f be the number of faces. There are then 3 f /2 edges.
Using Euler’s formula we have v − 3 f /2 + f 2 so v 2 + f /2. Now each vertex has the
same degree, say k. So the number of edges is also kv/2. Putting this together gives
3f k (2 + f /2)
e
2 2
which says
6f
k
4+ f
6f
We need k and f to both be positive integers. Note that 4+ f is an increasing function for
positive f , and has a horizontal asymptote at 6. Thus the only possible values for k are 3,
4, and 5. Each of these are possible. To get k 3, we need f 4 (this is the tetrahedron).
For k 4 we take f 8 (the octahedron). For k 5 take f 20 (the icosahedron). Thus
there are exactly three regular polyhedra with triangles for faces.
Case 2: Each face is a square. Now we have e 4 f /2 2 f . Using Euler’s formula we
get v 2 + f , and counting edges using the degree k of each vertex gives us
k (2 + f )
e 2f
2
Solving for k gives
4f 8f
k
2+ f 4+2f
4.2 Exercises 183
This is again an increasing function, but this time the horizontal asymptote is at k 4, so
the only possible value that k could take is 3. This produces 6 faces, and we have a cube.
There is only one regular polyhedron with square faces.
Case 3: Each face is a pentagon. We perform the same calculation as above, this time
getting e 5 f /2 so v 2 + 3 f /2. Then
5f k (2 + 3 f /2)
e
2 2
so
10 f
k
4+3f
Now the horizontal asymptote is at 10
3 . This is less than 4, so we can only hope of making
k 3. We can do so by using 12 pentagons, getting the dodecahedron. This is the only
regular polyhedron with pentagons as faces.
Case 4: Each face is an n-gon with n ≥ 6. Following the same procedure as above, we
deduce that
2n f
k
4 + ( n − 2) f
2n
which will be increasing to a horizontal asymptote of n−2 . When n 6, this asymptote is
at k 3. Any larger value of n will give an even smaller asymptote. Therefore no regular
polyhedra exist with faces larger than pentagons.3 qed
Exercises
1. Is it possible for a planar graph to have 6 vertices, 10 edges and 5 faces? Explain.
2. The graph G has 6 vertices with degrees 2, 2, 3, 4, 4, 5. How many edges does G have?
Could G be planar? If so, how many faces would it have.
3. If a graph has 10 vertices and 10 edges and contains an Euler circuit, must it be planar?
How many faces would it have?
4. I’m think of a polyhedron containing 12 faces. Seven are triangles and four are
squares. The polyhedron has 11 vertices other than those around the mystery face.
How many sides does the last face have?
5. Prove Euler’s formula using induction on the number of edges in the graph.
6. Euler’s formula (v − e + f 2) holds for all connected planar graphs. What if a graph
is not connected? Suppose a planar graph has two components. What is the value of
v − e + f now? What if it has k components?
7. Prove that any planar graph with v vertices and e edges satisfies e ≤ 3v − 6.
8. Prove that any planar graph must have a vertex of degree 5 or less.
3 Notice
that you can tile the plane with hexagons. This is an infinite planar graph; each vertex has degree
3. These infinitely many hexagons correspond to the limit as f → ∞ to make k 3.
4.3. Coloring 184
4.3 Coloring
Investigate!
Mapmakers in the fictional land of Euleria have drawn the borders of the various
dukedoms of the land. To make the map pretty, they wish to color each region.
Adjacent regions must be colored differently, but it is perfectly fine to color two
distant regions with the same color. What is the fewest colors the mapmakers can
use and still accomplish this task?
Given any map of countries, states, counties, etc., how many colors are needed to
color each region on the map so that neighboring regions are colored differently?
Actual map makers usually use around seven colors. For one thing, they require watery
regions to be a specific color, and with a lot of colors it is easier to find a permissible coloring.
We want to know whether there is a smaller palette that will work for any map.
How is this related to graph theory? Well, if we place a vertex in the center of each
region (say in the capital of each state) and then connect two vertices if their states share a
border, we get a graph. Coloring regions on the map corresponds to coloring the vertices
of the graph. Since neighboring regions cannot be colored the same, our graph cannot have
vertices colored the same when those vertices are adjacent.
In general, given any graph G, a coloring of the vertices is called (not surprisingly) a
vertex coloring. If the vertex coloring has the property that adjacent vertices are colored
differently, then the coloring is called proper. Every graph has a proper vertex coloring. For
4.3. Coloring 185
example, you could color every vertex with a different color. But often you can do better.
The smallest number of colors needed to get a proper vertex coloring is called the chromatic
number of the graph, written χ (G).
Example:
Find the chromatic number of the graphs below.
Solution: The graph on the left is K6 . The only way to properly color the graph is
to give every vertex a different color (since every vertex is adjacent to every other
vertex). Thus the chromatic number is 6.
The middle graph can be properly colored with just 3 colors (Red, Blue, and
Green). For example:
G R
R B G
There is no way to color it with just two colors, since there are three vertices
mutually adjacent (i.e., a triangle). Thus the chromatic number is 3.
The graph on the right is just K2,3 . As with all bipartite graphs, this graph has
chromatic number 2. Color the vertices on the top row red and the vertices on the
bottom row blue.
It appears that there is no limit to how large chromatic numbers can get. It should not
come as a surprise that K n has chromatic number n. So how could there possibly be an
answer to the original map coloring question? If the chromatic number of graph can be
arbitrarily large, then it seems like there would be no upper bound to the number of colors
needed for any map. But there is.
The key observation is that while it is true that for any number n, there is a graph with
chromatic number n, only some graphs arrive as representations of maps. If you convert a
map to a graph, the edges between vertices correspond to borders between the countries.
So you should be able to connect vertices in such a way where the edges do not cross. In
other words, the graphs representing maps are all planar!
So the question is, what is the largest chromatic number of any planar graph? The
answer is one of the best known theorems of mathematics:
4.3. Coloring 186
Theorem (The Four Color Theorem). If G is a planar graph, then the chromatic number of G is
less than or equal to 4. Thus any map can be properly colored with 4 or fewer colors.
We will not prove this theorem. Really. Even though the theorem is easy to state
and understand, the proof is not. In fact, there is currently no “easy” known proof
of the theorem. The current best proof still requires powerful computers to check an
unavoidable set of 633 reducible configurations. The idea is that every graph must contain one
of these reducible configurations (this fact also needs to be checked by a computer) and
that reducible configurations can, in fact, be colored in 4 or fewer colors.
Coloring in General
Investigate!
The math department plans to offer 10 classes next semester. Some classes cannot
run at the same time (perhaps they are taught by the same professor, or are required
for seniors).
How many different time slots are needed to teach these classes (and which
should be taught at the same time)? More importantly, how could we use graph
coloring to answer this question?
Example:
Radio stations broadcast their signal at certain frequencies. However, there are a
limited number of frequencies to choose from, so nationwide many stations use the
same frequency. This works because the stations are far enough apart that their
signals will not interfere; no one radio could pick them up at the same time.
Suppose 10 new radio stations are to be set up in a currently unpopulated (by
radio stations) region. The radio stations that are close enough to each other to
cause interference are recorded in the table below. What is the fewest number of
frequencies the stations could use.
KQEA KQEB KQEC KQED KQEE KQEF KQEG KQEH KQEI KQEJ
KQEA x x x x
KQEB x x
KQEC x x x x
KQED x x x x
KQEE x x
KQEF x x x x x
KQEG x x x x
KQEH x x
KQEI x x x
KQEJ x x x x x
Solution: Represent the problem as a graph with vertices as the stations and
edges when two stations are close enough to cause interference. We are looking for
the chromatic number of the graph. Vertices that are colored identically represent
stations that can have the same frequency.
This graph has chromatic number 5. A proper 5-coloring is shown on the right.
Notice that the graph contains a copy of the complete graph K 5 so no fewer than 5
colors can be used.
KQEA R
KQEJ KQEB P G
KQEI KQEC B B
KQEH KQED R B
KQEG Y
KQEE G
KQEF G
In the example above, the chromatic number was 5, but this is not a counter-example
to the Four Color Theorem, since the graph representing the radio stations is not planar.
It would be nice to have some quick way to find the chromatic number of a (possibly non-
4.3. Coloring 188
planar) graph. It turns out nobody knows whether an efficient algorithm for computing
chromatic numbers exists.
While we might not be able to find the exact chromatic number of graph easily, we can
often give a reasonable range for the chromatic number. In other words, we can give upper
and lower bounds for chromatic number.
This is actually not very difficult: for every graph G, the chromatic number of G is at
least 1 and at most the number of vertices of G.
What? You want better bounds on the chromatic number? Well you are in luck.
A clique in a graph is a set of vertices all of which are pairwise adjacent. In other words,
a clique of size n is just a copy of the complete graph K n . We define the clique number of a
graph to be the largest n for which the graph contains a clique of size n. Any clique of size
n cannot be colored with fewer than n colors, so we have a nice lower bound:
There are times when the chromatic number of G is equal to the clique number. These
graphs have a special name – they are called perfect. If you know that a graph is perfect,
then finding the chromatic number is simply a matter of searching for the largest clique.4
However, not all graphs are perfect.
For an upper bound, we can improve on “the number of vertices” by looking to the
degrees of vertices. Let ∆(G ) be the largest degree of any vertex in the graph G. One
reasonable guess for an upper bound on the chromatic number is χ (G ) ≤ ∆(G) + 1. Why
is this reasonable? Starting with any vertex, it together with all of its neighbors can always
be colored in ∆(G ) + 1 colors, since at most we are talking about ∆(G) + 1 vertices in this set.
Now fan out! At any point, if you consider an already colored vertex, some of its neighbors
might be colored, some might not. But no matter what, that vertex and its neighbors could
all be colored distinctly, since there are at most ∆(G) neighbors, plus the one vertex being
considered.
In fact, there are examples of graphs for which χ (G ) ∆(G ) + 1. For any n, the complete
graph K n has chromatic number n, but ∆(K n ) n −1 (since every vertex is adjacent to every
other vertex). Additionally, any odd cycle will have chromatic number 3, but the degree of
every vertex in a cycle is 2. It turns out that these are the only two types of examples where
we get equality, a result known as Brooks’ Theorem.
Theorem (Brooks’ Theorem). Any graph G with maximal degree ∆(G ) satisfies χ (G ) ≤ ∆(G ),
unless G is a complete graph or an odd cycle, in which case χ (G ) ∆(G ) + 1.
The proof of this theorem is just complicated enough that we will not present it here
(although you are asked to prove a special case in the exercises). The adventurous reader
is encouraged to find a book on graph theory for suggestions on how to prove the theorem.
4 There are special classes of graphs which can be proved to be perfect. One such class is the set of chordal
graphs, which have the property that every cycle in the graph contains a chord – an edge between two vertices
in of the cycle which are not adjacent in the cycle.
4.3. Coloring 189
Coloring Edges
The chromatic number of a graph tells us about coloring vertices, but we could also ask
about coloring edges. Just like with vertex coloring, we might insist that edges that are
adjacent must be colored differently. Here, we are thinking of two edges as being adjacent
if they are incident to the same vertex. The least number of colors required to properly
color the edges of a graph G is called the chromatic index of G, written χ0(G ).
Example:
Six friends decide to spend the afternoon playing chess. Everyone will play every-
one else once. They have plenty of chess sets but nobody wants to play more than
one game at a time. Games will last an hour (thanks to their handy chess clocks).
How many hours will the tournament last?
Solution: Represent each player with a vertex and put an edge between two players
if they will play each other. In this case, we get the graph K 6 :
We must color the edges; each color represents a different hour. Since different
edges incident to the same vertex will be colored differently, no player will be
playing two different games (edges) at the same time. Thus we need to know the
chromatic index of K6 .
Notice that for sure χ0(K 6 ) ≥ 5, since there is a vertex of degree 5. It turns out 5
colors is enough (find such a coloring). Thus the friends will play for 5 hours.
Interestingly, if one of the friends in the above example left, the remaining 5 chess-letes
would still need 5 hours: the chromatic index of K5 is also 5.
In general, what can we say about chromatic index? Certainly χ0(G ) ≥ ∆(G). But how
much higher could it be? Only a little higher.
Theorem (Vizing’s Theorem). For any graph G, the chromatic index χ0(G) is either ∆(G ) or
∆(G ) + 1.
At first this theorem makes it seem like chromatic index might not be very interesting.
However, deciding which case a graph is in is not always easy. Graphs for which χ0(G)
∆(G ) are called class 1, while the others are called class 2. Bipartite graphs always satisfy
χ0(G) ∆(G), so are class 1 (this was proved by König in 1916, decades before Vizing
proved his theorem in 1964). In 1965 Vizing proved that all planar graphs with ∆(G ) ≥ 8
are of class 1, but this does not hold for all planar graphs with 2 ≤ ∆(G ) ≤ 5. Vizing
conjectured that all planar graphs with ∆(G ) 6 or ∆(G ) 7 are class 1; the ∆(G) 7 case
was proved in 2001 by Sanders and Zhao; the ∆(G) 6 case is still open.
4.3 Exercises 190
There is another interesting way we might consider coloring edges, quite different from
what we have discussed so far. What if we colored every edge of a graph either red or
blue. Can we do so without, say, creating a monochromatic triangle (i.e., an all red or all blue
triangle)? Certainly for some graphs the answer is yes. Try doing so for K4 . What about
K5 ? K 6 ? How far can we go?
The answer to the above problem is known and is a fun problem to do as an exercise.
We could extend the question in a variety of ways. What if we had three colors? What if
we were trying to avoid other graphs. The surprising fact is that very little is known about
these questions. For example, we know that you need to go up to K17 in order to force
a monochromatic triangle using three colors, but nobody knows how big you need to go
with more colors. Similarly, we know that using two colors K18 is the smallest graph that
forces a monochromatic copy of K4 , but the best we have to force a monochromatic K 5 is
a range, somewhere from K 43 to K49 . If you are interested in these sorts of questions, this
area of graph theory is called Ramsey theory. Check it out.
Exercises
1. What is the smallest number of colors you need to properly color the vertices of K 4,5 ?
That is, find the chromatic number of the graph.
2. Draw a graph with chromatic number 6 (i.e., which requires 6 colors to properly color
the vertices). Could your graph be planar? Explain.
4. What is the smallest number of colors that can be used to color the vertices of a cube
so that no two adjacent vertices are colored identically?
5. Not all graphs are perfect. Give an example of a graph with chromatic number 4
that does not contain a copy of K 4 . That is, there should be no 4 vertices all pairwise
adjacent.
6. Prove by induction on vertices that any graph G which contains at least one vertex of
degree less than ∆(G) (the maximal degree of all vertices in G) has chromatic number
at most ∆(G ).
7. You have a set of magnetic alphabet letters (one of each of the 26 letters in the
alphabet) that you need to put into boxes. For obvious reasons, you don’t want to
put two consecutive letters in the same box. What is the fewest number of boxes you
need (assuming the boxes are able to hold as many letters as they need to)?
4.4. Euler Paths and Circuits 191
8. Prove that if you color every edge of K 6 either red or blue, you are guaranteed a
monochromatic triangle.
Investigate!
An Euler path, in a graph or multigraph, is a path which uses every edge exactly
once. An Euler circuit is an Euler path which starts and stops at the same vertex.
Our goal is to find a quick way to check whether a graph (or multigraph) has an
Euler path or circuit.
1. Which of the graphs below have Euler paths? Which have Euler circuits?
2. List the degrees of each vertex of the graphs above. Is there a connection
between degrees and the existence of Euler paths and circuits?
3. Is it possible for a graph with a degree 1 vertex to have an Euler circuit? If so,
draw one. If not, explain why not. What about an Euler path?
4. What if every vertex of the graph has degree 2. Is there an Euler path? An
Euler circuit? Draw some graphs.
5. Below is part of a graph. Even though you can only see some of the vertices,
can you deduce whether the graph will have an Euler path or circuit?
trace along every edge exactly once and end up where you started), then the path is called
an Euler circuit. Of course if a graph is not connected, there is no hope of finding such a
path or circuit. For the rest of this section, assume all the graphs discussed are connected.
The bridges of Königsberg problem is really a question about the existence of Euler
paths. There will be a route that crosses every bridge exactly once if and only if the graph
below has an Euler path:
This graph is small enough that we could actually check every possible path and in
doing so convince ourselves that there is no Euler path (let alone an Euler circuit). On
small graphs which do have an Euler path, it is usually not difficult to find one. Our goal
is to find a quick way to check whether a graph has an Euler path or circuit, even if the
graph is quite large.
One way to guarantee that a graph does not have an Euler circuit is to include a “spike,”
a vertex of degree 1.
The vertex a has degree 1, and if you try to make an Euler circuit, you see that you will
get stuck at the vertex. It is a dead end. That is, unless you start there. But then there is
no way to return, so there is no hope of finding an Euler circuit. There is however an Euler
path. It starts at the vertex a, then loops around the triangle. You will end at the vertex of
degree 3.
You run into a similar problem whenever you have a vertex of any odd degree. If you
start at such a vertex, you will not be able to end there (after traversing every edge exactly
once). After using one edge to leave the starting vertex, you will be left with an even
number of edges emanating from the vertex. Half of these could be used for returning to
the vertex, the other half for leaving. So you return, then leave. Return, then leave. The
only way to use up all the edges is to use the last one by leaving the vertex. On the other
hand, if you have a vertex with odd degree that you do not start a path at, then you will
eventually get stuck at that vertex. The path will use pairs of edges incident to the vertex
to arrive and leave again. Eventually all but one of these edges will be used up, leaving
only an edge to arrive by, and none to leave again.
What all this says is that if a graph has an Euler path and two vertices with odd degree,
then the Euler path must start at one of the odd degree vertices and end at the other. In such
a situation, every other vertex must have an even degree since we need an equal number of
edges to get to those vertices as to leave them. How could we have an Euler circuit? The
graph could not have any odd degree vertex as an Euler path would have to start there or
4.4 Exercises 193
end there, but not both. Thus for a graph to have an Euler circuit, all vertices must have
even degree.
The converse is also true: if all the vertices of a graph have even degree, then the graph
has an Euler circuit, and if there are exactly two vertices with odd degree, the graph has an
Euler path. To prove this is a little tricky, but the basic idea is that you will never get stuck
because there is an “outbound” edge for every “inbound” edge at every vertex. If you try
to make an Euler path and miss some edges, you will always be able to “splice in” a circuit
using the edges you previously missed.
Euler Paths and Circuits
• A graph has an Euler circuit if and only if the degree of every vertex is even.
• A graph has an Euler path if and only if there are at most two vertices with odd
degree.
Since the bridges of Königsberg graph has all four vertices with odd degree, there is no
Euler path through the graph.
Hamilton Paths
Suppose you wanted to tour Königsberg in such a way where you visit each land mass (the
two islands and both banks) exactly once. This can be done. In graph theory terms, we are
asking whether there is a path which visits every vertex exactly once. Such a path is called
a Hamilton path (or Hamiltonian path). It appears that finding Hamilton paths would be
easier because graphs often have more edges than vertices, so there are fewer requirements
to be met. However, nobody knows whether this is true. There is no known simple test
for whether a graph has a Hamilton path. For small graphs this is not a problem, but as
the size of the graph grows, it gets harder and harder to check wither there is a Hamilton
path. In fact, this is an example of a question which as far as we know is too difficult for
computers to solve; it is an example of a problem which is NP-complete.
Exercises
1. You and your friends want to tour the southwest by car. You will visit the nine
states below, with the following rather odd rule: you must cross each border between
neighboring states exactly once (so, for example, you must cross the Colorado-Utah
border exactly once). Can you do it? If so, does it matter where you start your road
trip? What fact about graph theory solves this problem?
4.5. Matching in Bipartite Graphs 194
2. Which of the following graphs contain an Euler path? Which contain an Euler circuit?
(a) K4
(b) K5 .
(c) K 5,7
(d) K 2,7
(e) C 7
(f) P7
4. For which m and n does the graph K m,n contain an Euler path? An Euler circuit?
Explain.
5. A bridge builder has come to Königsberg and would like to add bridges so that it is
possible to travel over every bridge exactly once. How many bridges must be built?
Investigate!
Given a bipartite graph, a matching is a subset of the edges for which every vertex
belongs to exactly one of the edges. Our goal in this activity is to discover some
criterion for when a bipartite graph has a matching.
2. Not all bipartite graphs have matchings. Draw as many fundamentally differ-
ent examples of bipartite graphs which do NOT have matchings. Your goal is
to find all the possible obstructions to a graph having a matching. Write down
the necessary conditions for a graph to have a matching (that is, fill in the blank:
If a graph has a matching, then ). Then ask yourself whether
these conditions are sufficient (is it true that if , then the graph
has a matching?).
is nothing left for the second student to like, so it is very much as if the second student
has degree 0. Or what if three students like only two topics between them. Again, after
assigning one student a topic, we reduce this down to the previous case of two students
liking only one topic. We can continue this way with more and more students.
It should be clear at this point that if there is every a group of n students who as a
group like n − 1 or fewer topics, then no matching is possible. This is true for any value of
n, and any group of n students.
To make this more graph-theoretic, say you have a set S ⊆ A of vertices. Define N (S )
to be the set of all the neighbors of vertices in S. That is, N (S ) contains all the vertices (in B)
which are adjacent to at least one of the vertices in S. (In the student/topic graph, N (S ) is
the set of topics liked by the students of S.) Our discussion above can be summarized as
follows:
Matching Condition
If a bipartite graph G { A, B } has a matching of A, then
| N (S)| ≥ | S |
for all S ⊆ A.
Is the converse true? Suppose G satisfies the matching condition | N (S )| ≥ | S | for all
S ⊆ A (every set of vertices has at least as many neighbors than vertices in the set). Does
that mean that there is a matching? Surprisingly, yes. The obvious necessary condition is
also sufficient.6 This is a theorem first proved by Philip Hall in 1935.7
Theorem (Hall’s Marriage Theorem). Let G be a bipartite graph with sets A and B. Then G has
a matching of A if and only if
| N (S)| ≥ | S |
for all S ⊆ A.
There are quite a few different proofs of this theorem – a quick internet search will get
you started.
In addition to its application to marriage and student presentation topics, matchings
have applications all over the place. We conclude with one such example.
Example:
Suppose you deal 52 regular playing cards into 13 piles of 4 cards each. Prove that
you can always select one card from each pile to get one of each of the 13 card values
Ace, 2, 3, . . . , 10, Jack, Queen, and King.
6 This happens often in graph theory. If you can avoid the obvious counter-examples, you often get what
you want.
7 There is also an infinite version of the theorem which was proved by Marshal Hall, Jr. The name is a
coincidence though as the two Halls are not related.
4.5. Matching in Bipartite Graphs 197
Solution: Doing this directly would be difficult, but we can use the matching con-
dition to help. Construct a graph G with 13 vertices in the set A, each representing
one of the 13 card values, and 13 vertices in the set B, each representing one of the
13 piles. Draw an edge between a vertex a ∈ A to a vertex b ∈ B if a card with value
a is in the pile b. Notice that we are just looking for a matching of A; each value
needs to be found in the piles exactly once.
We will have a matching if the matching condition holds. Given any set of card
values (a set S ⊆ A) we must show that | N (S )| ≥ | S |. That is, the number of piles that
contain those values is at least the number of different values. But what if it wasn’t?
Say | S | k. If | N (S )| < k, then we would have fewer than 4k different cards in those
piles (since each pile contains 4 cards). But there are 4k cards with the k different
values, so at least one of these cards must be in another pile, a contradiction. Thus
the matching condition holds, so there is a matching, as required.
Investigate!
The two richest families in Westeros have decided to enter into an alliance by mar-
riage. The first family has 10 sons, the second has 10 girls. The ages of the kids in the
two families match up. To avoid impropriety, the families insist that each child must
marry someone either their own age, or someone one position younger or older. In
fact, the graph representing agreeable marriages looks like this:
1. How many marriage arrangements are possible if we insist that there are
exactly 6 boys marry girls not their own age?
2. Could you generalize the previous answer to arrive at the total number of
marriage arrangements?
3. How do you know you are correct? Try counting in a different way. Look at
smaller family sizes and get a sequence.
• Can the graph be drawn in the plane without edges crossing? If so, how many regions
does this drawing divide the plane into?
• Is it possible to color the vertices of the graph so that related vertices have different
colors using a small number of colors? How many colors are needed?
• Is it possible to trace over every edge of a graph exactly once without lifting up your
pencil? What other sorts of “paths” might a graph posses?
• Can you find subgraphs with certain properties? For example, when does a (bipartite)
graph contain a subgraph in which all vertices are only related to one other vertex?
Not surprisingly, these questions are often related to each other. For example, the
chromatic number of a graph cannot be greater than 4 when the graph is planar. Whether
the graph has an Euler path depends on how many vertices each vertex is adjacent to
(and whether those numbers are always even or not). Even the existence of matchings in
bipartite graphs can be proved using paths.
Exercises
1. Which (if any) of the graphs below are the same? Which are different? Explain.
2. Which of the graphs in the previous question contain Euler paths or circuits? Which
of the graphs are planar?
4.6 Exercises 199
4. Draw a graph which does not have an Euler path and is also not planar.
6. At a school dance, 6 girls and 4 boys take turns dancing (as couples) with each other.
(a) How many couples danced if every girl dances with every boy?
(b) How many couples danced if everyone danced with everyone else (regardless
of gender)?
(c) Explain what graphs can be used to represent these situations.
8. Your friend has challenged you to create a convex polyhedron containing 9 triangles
and 6 pentagons.
(a) Is it possible to build such a polyhedron using only these shapes? Explain.
(b) You decide to also include one heptagon (seven-sided polygon). How many
vertices does your new convex polyhedron contain?
(c) Assuming you are successful in building your new 16-faced polyhedron, could
every vertex be the joining of the same number of faces? Could each vertex join
either 3 or 4 faces? If so, how many of each type of vertex would there be?
9. Is there a convex polyhedron which requires 5 colors to properly color the vertices of
the polyhedron? Explain.
10. How many edges does the graph K n,n have? For which values of n does the graph
contain an Euler circuit? For which values of n is the graph planar?
11. The graph G has 6 vertices with degrees 1, 2, 2, 3, 3, 5. How many edges does G have?
If G was planar how many faces would it have? Does G have an Euler path?
12. What is the smallest number of colors you need to properly color the vertices of K7 .
Can you say whether K 7 is planar based on your answer?
13. What is the smallest number of colors you need to properly color the vertices of K 3,4 ?
Can you say whether K 3,4 is planar based on your answer?
(a) Suppose you color each pentagon with one of three colors. Prove that there
must be two adjacent pentagons colored identically.
(b) What if you use four colors?
(c) What if instead of a dodecahedron you colored the faces of a cube?
15. If a planar graph G with 7 vertices divides the plane into 8 regions, how many edges
must G have?
17. For each part below, say whether the statement is true or false. Explain why the true
statements are true, and give counterexamples for the false statements.
18. Find a matching of the bipartite graphs below or explain why no matching exists.
19. Consider the statement “If a graph is planar, then it has an Euler path.”
Homework Problems
The following are some more involved problems for you to try, which might be assigned
as homework.
G1 : V1 { a, b, c, d, e , f , g }
E1 {{ a, b } , { a, d } , { b, c } , { b, d } , { b, e } , { b, f } , { c, g } , { d, e } , { e , f } , { f , g }}.
G2 : V2 { v 1 , v 2 , v 3 , v 4 , v 5 , v6 , v7 },
E2 {{ v 1 , v 4 } , { v 1 , v 5 } , { v1 , v 7 } , { v 2 , v 3 } , { v 2 , v6 } , { v 3 , v 5 } , { v 3 , v 7 } ,
{ v4 , v5 } , { v5 , v6 } , { v5 , v7 }}
x a b c d e f g
f (x ) v4 v5 v1 v6 v2 v3 v7
2. Recall that we proved that every (connected) planar graph with v vertices, e edges
and f faces satisfied v − e + f 2. We also saw that every convex polyhedron could
be represented as a planar graph.
3. Prove that the Petersen graph (below) is not planar. Hint: what is the length of the
shortest cycle?
(a) Edward wants to give a tour of his new pad to a lady-mouse-friend. Is it possible
for them to walk through every doorway exactly once? If so, in which rooms
must they begin and end the tour? Explain.
(b) Is it possible to tour the house visiting each room exactly once (not necessarily
using every doorway)? Explain.
(c) After a few mouse-years, Edward decides to remodel. He would like to add
some new doors between the rooms he has. Of course, he cannot add any doors
to the exterior of the house. Is it possible for each room to have an odd number
of doors? Explain.
(a) What is the smallest number of cars you need if all the relationships were strictly
heterosexual? Represent an example of such a situation with a graph. What
kind of graph do you get?
(b) Because a number of these friends dated there are also conflicts between friends
of the same gender, listed below. Now what is the smallest number of conflict-
free cars they could take to the cabin?
Friend A B C D E F G H I J
Conflicts with BEJ ADG HJ BF AI DJ B CI EHJ ACFI
7. We say that a graph has a Hamilton path if there is a path which visits each vertex
exactly once (you do not need to use every edge in the path).
(a) Suppose a graph has a Hamilton path. What is the maximum number of vertices
of degree one the graph can have? Explain why your answer is correct.
(b) Find a graph which does not have a Hamilton path even though no vertex has
degree one. Explain why your example works.
(b) The chromatic number of C n is two when n is even. What goes wrong when n
is odd?
(c) Prove that your procedure from part (a) always works for any tree.
(d) Now, prove using induction that every tree has chromatic number 2.
9. BONUS: King Arthur has hired a consultant, Sir Cumference, to plan a seating
arrangement for Arthur and nine of his knights to sit around the round table. This
would not be a difficult task, were it not for the jealousies and petty rivalries that
exist among the knights. Arthur insists that Lancelot should sit on his right and
that Kay should sit on his left; Bedivere refuses to sit next to anyone but Lionel and
Tristan; Gawain won’t sit next to Tristan, Lancelot, or Lionel; Gareth won’t sit next
to Galahad, Lancelot, or Kay; Perceval objects to sitting next to Galahad, Lancelot, or
Lionel; Tristan refuses to sit next to Lancelot, Perceval or Kay; Galahad will sit next
to anyone except Gawain and Kay; and Lionel will not sit next to Galahad. The other
two knights are not particular about whom they sit next to.
4.6 Exercises 204
Help Sir Cumference find a suitable seating arrangement and explain what this
problem has to do with graphs.8
8 from Problem Solving through Recreational Mathematics by Averbach and Chein, Dover 1999
Appendix A
Additional Topics
c0 + c1 x + c2 x 2 + c3 x 3 + c4 x 4 + c5 x 5 + · · · .
When viewed in the context of generating functions, we call such a power series a generating
series. The generating series generates the sequence
c0 , c1 , c2 , c3 , c4 , c5 , . . . .
In other words, the sequence generated by a generating series is simply the sequence of
coefficients of the infinite polynomial.
Example:
5
What sequence is represented by the generating series 3 + 8x 2 + x 3 + x7 + 100x 6 + · · · ?
Solution: We just read off the coefficients of each x n term. So a 0 3 since the
coefficient of x 0 is 3 (x 0 1 so this is the constant term). What is a 1 ? It is NOT 8,
since 8 is the coefficient of x 2 , so 8 is the term a 2 of the sequence. To find a1 we need
to look for the coefficient of x 1 which in this case is 0. So a 1 0. Continuing, we
205
A.1. Generating Functions 206
1
3, 0, 8, 1, , 100, . . .
7
Note that when discussing generating functions, we always start our sequence with
a0 .
Now you might very naturally ask why we would do such a thing. One reason is that
encoding a sequence with a power series helps us keep track of which term is which in the
sequence. For example, if we write the sequence 1, 3, 4, 6, 9, . . . , 24, 41, . . . it is impossible
to determine which term 24 is (even if we agreed that the first term was supposed to be
a 0 ). However, if we wrote the generating series instead, we would have 1 + 3x + 4x 2 + 6x 3 +
9x 4 + · · · + 24x 17 + 41x 18 + · · · . Now it is clear that 24 is the 17th term of the sequence (that
is, a 17 24). Of course to get this benefit we could have displayed our sequence in any
number of ways, perhaps 1 0 3 1 4 2 6 3 9 4 · · · 24 17 41 18 · · · , but we do not do this. The
reason is that the generating series looks like an ordinary power series (although we are
interpreting it differently) so we can do things with it that we ordinarily do with power
series such as write down what it converges to.
2 3 x4 n
For example, from calculus we know that the power series 1+x+ x2 + x6 + 24 +· · ·+ xn! +· · ·
converges to the function e x . So we can use e x as a way of talking about the sequence of
coefficients of the power series for e x . When we write down a nice compact function which
has an infinite power series that we view as a generating series, then we call that function
a generating function. In this example, we would say
1 1 1 1
1, 1, , , , . . . , , . . . has generating function e x
2 6 24 n!
S 1 + x + x2 + x3 + · · ·
−xS x + x2 + x3 + x4 + · · ·
(1 − x ) S 1
You might remember from calculus that this is only true on the interval of convergence for
the power series, in this case when | x | < 1. That is true for us, but we don’t care. We are
never going to plug anything in for x, so as long as there is some value of x for which the
generating function and generating series agree, we are happy. And in this case we are
happy.
1, 1, 1, . . .
1
The generating function for 1, 1, 1, 1, 1, 1, . . . is
1−x
Let’s use this basic generating function to find generating functions for more sequences.
What if we replace x by −x. We get
1
1 − x + x 2 − x 3 + · · · which generates 1, −1, 1, −1, . . .
1+x
If we replace x by 3x we get
1
1 + 3x + 9x 2 + 27x 3 + · · · which generates 1, 3, 9, 27, . . .
1 − 3x
1
By replacing the x in 1−x we can get generating functions for a variety of sequences,
but not all. For example, you cannot plug in anything for x to get the generating function
for 2, 2, 2, 2, . . .. However, we are not lost yet. Notice that each term of 2, 2, 2, 2, . . . is the
result of multiplying the terms of 1, 1, 1, 1, . . . by the constant 2. So multiply the generating
function by 2 as well.
2
2 + 2x + 2x 2 + 2x 3 + · · · which generates 2, 2, 2, 2, . . .
1−x
Similarly, to find the generating function for the sequence 3, 9, 27, 81, . . ., we note that
this sequence is the result of multiplying each term of 1, 3, 9, 27, . . . by 3. Since we have the
generating function for 1, 3, 9, 27, . . . we can say
3
3 · 1 + 3 · 3x + 3 · 9x 2 + 3 · 27x 3 + · · · which generates 3, 9, 27, 81, . . .
1 − 3x
What about the sequence 2, 4, 10, 28, 82, . . .? Here the terms are always 1 more than
powers of 3. That is, we have added the sequences 1, 1, 1, 1, . . . and 1, 3, 9, 27, . . . term
by term. Therefore we can get a generating function by adding the respective generating
functions:
How could we get 0, 1, 0, 1, 0, 1, . . .? Start with the previous sequence and shift it over
by 1. But how do you do this? To see how shifting works, let’s first try to get the generating
function for the sequence 0, 1, 3, 9, 27, . . .. We know that 1−3x 1
1 + 3x + 9x 2 + 27x 3 + · · · .
To get the zero out front, we need the generating series to look like x + 3x 2 + 9x 3 + 27x 4 +
· · · (so there is no constant term). Multiplying by x has this effect. So the generating
function for 0, 1, 3, 9, 27, . . . is 1−3x
x
. This will also work to get the generating function for
0, 1, 0, 1, 0, 1, . . .:
x
x + x 3 + x 5 + · · · which generates 0, 1, 0, 1, 0, 1 . . .
1 − x2
What would happen if we add the sequences 1, 0, 1, 0, 1, 0, . . . and 0, 1, 0, 1, 0, 1, . . .
term by term? We should get 1, 1, 1, 1, 1, 1 . . .. What happens when we add the generating
functions? It works (try it)!
1 x 1
2
+ 2
.
1−x 1−x 1−x
1 1
Here’s a sneaky one: what happens if you take the derivative of 1−x ? We get (1−x )2
. On the
other hand, if we differentiate term by term in the power series, we get (1+x+x 2 +x 3 +· · · )0
1 + 2x + 3x 2 + 4x 3 + · · · which is the generating series for 1, 2, 3, 4, . . .. This says
1, 2, 3, . . .
1
The generating function for 1, 2, 3, 4, 5, . . . is .
(1 − x )2
Differencing
1
We have seen how to find generating functions from 1−x using multiplication (by a constant
or by x), substitution, addition, and differentiation. To use each of these, you must notice
a way to transform the sequence 1, 1, 1, 1, 1 . . . into your desired sequence. This is not
always easy. It is also not really the way we have analyzed sequences. One thing we have
considered often is the sequence of differences between terms of a sequence. This will
turn out to be helpful in finding generating functions as well. The sequence of differences
is often simpler than the original sequence. So if we know a generating function for the
differences, we would like to use this to find a generating function for the original sequence.
For example, consider the sequence 2, 4, 10, 28, 82, . . .. How could we move to the
sequence of first differences: 2, 6, 18, 54, . . .? We want to subtract 2 from the 4, 4 from the
10, 10 from the 28, and so on. So if we subtract (term by term) the sequence 0, 2, 4, 10, 28, . . .
from 2, 4, 10, 28 . . ., we will be set. We can get the generating function for 0, 2, 4, 10, 28, . . .
from the generating function for 2, 4, 10, 28 . . . by multiplying by x. Use A to represent the
generating function for 2, 4, 10, 28, 82, . . . Then:
A.1. Generating Functions 209
While we don’t get exactly the sequence of differences, we do get something close. In
this particular case, we already know the generating function A (we found it in the previous
section) but most of the time we will use this differencing technique to find A: if we have
the generating function for the sequence of differences, we can then solve for A.
Example:
Find a generating function for 1, 3, 5, 7, 9, . . ..
Solution: Notice that the sequence of differences is constant. We know how to
find the generating function for any constant sequence. So denote the generating
function for 1, 3, 5, 7, 9, . . . by A. We have
A 1 + 3x + 5x 2 + 7x 3 + 9x 4 + · · ·
−xA 0 + x + 3x 2 + 5x 3 + 7x 4 + 9x 5 + · · ·
(1 − x )A 1 + 2x + 2x 2 + 2x 3 + 2x 4 + · · ·
2x
We know that 2x + 2x 2 + 2x 3 + 2x 4 + · · · . Thus
1−x
2x
(1 − x ) A 1 + .
1−x
Now solve for A:
1 2x 1+x
A + .
1 − x (1 − x )2 (1 − x )2
Does this makes sense? Before we simplified the two fractions into one, we were
adding the generating function for the sequence 1, 1, 1, 1, . . . to the generating func-
tion for the sequence 0, 2, 4, 6, 8, 10, . . . (remember (1−x
1
)2
generates 1, 2, 3, 4, 5, . . .,
multiplying by 2x shifts it over, putting the zero out front, and doubles each term).
If we add these term by term, we get the correct sequence 1, 3, 5, 7, 9, . . ..
Now that we have a generating function for the odd numbers, we can use that to find
the generating function for the squares:
Example:
Find the generating function for 1, 4, 9, 16, . . .. Note we take 1 a0 .
A.1. Generating Functions 210
Solution: Again we call the generating function for the sequence A. Using differ-
encing:
A 1 + 4x + 9x 2 + 16x 3 + · · ·
−xA 0 + x + 4x 2 + 9x 3 + 16x 4 + · · ·
(1 − x )A 1 + 3x + 5x 2 + 7x 3 + · · ·
1+x 1+x
Since 1 + 3x + 5x 2 + 7x 3 + · · · 2
we have A .
(1 − x ) (1 − x )3
In each of the examples above, we found the difference between consecutive terms
which gave us a sequence of differences for which we knew a generating function. We
can generalize this to more complicated relationships between terms of the sequence. For
example, if we know that the sequence satisfies the recurrence relation a n 3a n−1 − 2a n−2 ?
In other words, if we take a term of the sequence and subtract 3 times the previous term
and then add 2 times the term before that, we get 0 (since a n − 3a n−1 + 2a n−2 0). That
will hold for all but the first two terms of the sequence. So after the first two terms, the
sequence of results of these calculations would be a sequence of 0’s, for which we definitely
know a generating function.
Example:
The sequence 1, 3, 7, 15, 31, 63, . . . satisfies the recurrence relation a n 3a n−1 −2a n−2 .
Find the generating function for the sequence.
Solution: Call the generating function for the sequence A. We have
A 1 + 3x + 7x 2 + 15x 3 + 31x 4 + · · · + a n x n + · · ·
−3xA 0 − 3x − 9x 2 − 21x 3 − 45x 4 − · · · − 3a n−1 x n − · · ·
+ 2x 2 A 0 + 0x + 2x 2 + 6x 3 + 14x 4 + · · · + 2a n−2 x n + · · ·
(1 − 3x + 2x 2 )A 1
We multiplied A by −3x which shifts every term over one spot and multiplies
them by −3. On the third line, we multiplied A by 2x 2 , which shifted every term
over two spots and multiplied them by 2. When we add up the corresponding
terms, we are taking each term, subtracting 3 times the previous term, and adding
2 times the term before that. This will happen for each term after a 1 because
a n − 3a n−1 + 2a n−2 0. In general, we might have two terms from the beginning of
the generating series, although in this case the second term happens to be 0 as well.
Now we just need to solve for A:
1
A .
1 − 3x + 2x 2
A.1. Generating Functions 211
AB a0 b0 + ( a 0 b1 + a 1 b0 ) x + ( a 0 b2 + a 1 b1 + a 2 b0 ) x 2 + ( a 0 b3 + a 1 b2 + a 2 b 1 + a3 b0 ) x 3 + · · ·
Example:
“Multiply” the sequence 1, 2, 3, 4, . . . by the sequence 1, 2, 4, 8, 16, . . ..
Solution: The new constant term is just 1 · 1. The next term will be 1 · 2 + 2 · 1 4.
The next term: 1 · 4 + 2 · 2 + 3 · 1 11. One more: 1 · 8 + 2 · 4 + 3 · 2 + 4 · 1 28. The
resulting sequence is
1, 4, 11, 28, 57, . . .
Since the generating function for 1, 2, 3, 4, . . . is 1
(1−x )2
and the generating function
for 1, 2, 4, 8, 16, . . . is 1−2x
1
, we have that the generating function for 1, 4, 11, 28, 57, . . .
1
is (1−x )2 (1−2x )
Consider the special case when you multiply a sequence by 1, 1, 1, . . .. For example,
multiply 1, 1, 1, . . . by 1, 2, 3, 4, 5 . . .. The first term is 1 · 1 1. Then 1 · 2 + 1 · 1 3.
Then 1 · 3 + 1 · 2 + 1 · 1 6. The next term will be 10. We are getting the triangular
numbers. More precisely, we get the sequence of partial sums of 1, 2, 3, 4, 5, . . .. In terms
of generating functions, we take 1−x 1
(generating 1, 1, 1, 1, 1 . . .) and multiply it by (1−x
1
)2
(generating 1, 2, 3, 4, 5, . . .) and this give (1−x
1
)3
. This should not be a surprise as we found
the same generating function for the triangular numbers earlier.
The point is, if you need to find a generating function for the sum of the first n terms
of a particular sequence, and you know the generating function for that sequence, you can
1
multiply it by 1−x . To go back from the sequence of partial sums to the original sequence,
you look at the sequence of differences. When you get the sequence of differences you end
1 1
up multiplying by 1 − x, or equivalently, dividing by 1−x . Multiplying by 1−x gives partial
1
sums, dividing by 1−x gives differences.
Example:
Solve the recurrence relation a n 3a n−1 − 2a n−2 with initial conditions a 0 1 and
a 1 3.
Solution: We saw in an example above that this recurrence relation gives the
1
sequence 1, 3, 7, 15, 31, 63, . . . which has generating function . We did
1 − 3x + 2x 2
this by calling the generating function A and then computing A − 3xA + 2x A which
2
1 1
.
1 − 3x + 2x 2 (1 − x )(1 − 2x )
Partial fraction decomposition tells us that we can write this faction as the sum
of two fractions (we decompose the given fraction):
1 a b
+ for some constants a and b.
(1 − x )(1 − 2x ) 1 − x 1 − 2x
To find a and b we add the two decomposed fractions using a common denominator.
This gives
1 a (1 − 2x ) + b (1 − x )
.
(1 − x )(1 − 2x ) (1 − x )(1 − 2x )
so
1 a (1 − 2x ) + b (1 − x ).
This must be true for all values of x. If x 1, then the equation becomes 1 −a so
a −1. When x 21 we get 1 b/2 so b 2. This tells us that we can decompose
the fraction like this:
1 −1 2
+ .
(1 − x )(1 − 2x ) 1 − x 1 − 2x
This completes the partial fraction decomposition. Notice that these two fractions
are generating functions we know. In fact, we should be able to expand each of
them.
−1
−1 − x − x 2 − x 3 − x 4 − · · · which generates − 1, −1, −1, −1, −1, . . . .
1−x
2
2 + 4x + 8x 2 + 16x 3 + 32x 4 + · · · which generates 2, 4, 8, 16, 32, . . . .
1 − 2x
A.1 Exercises 213
We can give a closed formula for the nth term of each of these sequences. The first
is just a n −1. The second is a n 2n+1 . The sequence we are interested in is just
the sum of these, so the solution to the recurrence relation is
a n 2n+1 − 1
We can now add generating functions to our list of methods for solving recurrence
relations.
Exercises
1. Find the generating function for each of the following sequences by relating them
back to a sequence with known generating function.
(a) 4, 4, 4, 4, 4, . . ..
(b) 2, 4, 6, 8, 10, . . ..
(c) 0, 0, 0, 2, 4, 6, 8, 10, . . ..
(d) 1, 5, 25, 125, . . ..
(e) 1, −3, 9, −27, 81, . . ..
(f) 1, 0, 5, 0, 25, 0, 125, 0, . . ..
(g) 0, 1, 0, 0, 2, 0, 0, 3, 0, 0, 4, 0, 0, 5, . . ..
4. Use differencing to find the generating function for 4, 5, 7, 10, 14, 19, 25, . . ..
A.1 Exercises 214
5. Find a generating function for the sequence with recurrence relation a n 3a n−1 − a n−2
with initial terms a 0 1 and a1 5.
6. Use the recurrence relation for the Fibonacci numbers to find the generating function
for the Fibonacci sequence.
7. Use multiplication to find the generating function for the sequence of partial sums
of Fibonacci numbers, S0 , S1 , S2 , . . . where S0 F0 , S1 F0 + F1 , S2 F0 + F1 + F2 ,
S3 F0 + F1 + F2 + F3 and so on.
8. Find the generating function for the sequence with closed formula a n 2(5n ) +7(−3)n .
9. Find a closed formula for the nth term of the sequence with generating function
3x 1
+ .
1 − 4x 1 − x
2 x
10. Find a 7 for the sequence with generating function · .
(1 − x )2 1 − x − x 2
1
11. Explain how we know that is the generating function for 1, 2, 3, 4, . . ..
(1 − x )2
12. Starting with the generating function for 1, 2, 3, 4, . . ., find a generating function for
each of the following sequences.
(a) 1, 0, 2, 0, 3, 0, 4, . . ..
(b) 1, −2, 3, −4, 5, −6, . . ..
(c) 0, 3, 6, 9, 12, 15, 18, . . ..
(d) 0, 3, 9, 18, 30, 45, 63, . . .. (Hint: relate this sequence to the previous one.)
1
13. You may assume that 1, 1, 2, 3, 5, 8, . . . has generating function (because it
1 − x − x2
does). Use this fact to find the sequence generated by each of the following generating
functions.
x2
(a) 1−x−x 2
.
1
(b) 1−x 2 −x 4
.
1
(c) 1−3x−9x 2
.
1
(d) (1−x−x 2 )(1−x )
.
14. Find the generating function for the sequence 1, −2, 4, −8, 16, . . ..
16. Suppose A is the generating function for the sequence 3, 5, 9, 15, 23, 33, . . ..
(a) Find a generating function (in terms of A) for the sequence of differences between
terms.
A.2. Introduction to Number Theory 215
(b) Write the sequence of differences between terms and find a generating function
for it (without referencing A).
(c) Use your answers to parts (a) and (b) to find the generating function for the
original sequence.
Which amounts of postage can be made exactly using just 5-cent and 8-cent
stamps?
We were able to prove that any amount greater than 27 cents could be made. You might
wonder what would happen if we changed the denomination of the stamps. What if we
instead had 4- and 9-cent stamps? Would there be some amount after which all amounts
would be possible? Well, again, we could replace two 4-cent stamps with a 9-cent stamp,
or three 9-cent stamps with seven 4-cent stamps. In each case we can create one more cent
of postage. Using this as the inductive case would allow us to prove that any amount of
postage greater than 23 cents can be made.
What if we had 2-cent and 4-cent stamps. Here it looks less promising. If we take some
number of 2-cent stamps and some number of 4-cent stamps, what can we say about the
total? Could it ever be odd? Doesn’t look like it.
Why does 5 and 8 work, 4 and 9 work, but 2 and 4 not work? What is it about these
numbers? If I gave you a pair of numbers, could you tell me right away if they would work
or not? We will answer these questions, and more, after first investigating some simpler
properties of numbers themselves.
Divisibility
It is easy to add and multiply natural numbers. If we extend our focus to all integers, then
subtraction is also easy (we need the negative numbers so we can subtract any number from
any other number, even larger from smaller). Division is the first operation that presents
A.2. Introduction to Number Theory 216
a challenge. If we wanted to extend our set of numbers so any division would be possible
(maybe excluding division by 0) we would need to look at the rational numbers (the set
of all numbers which can be written as fractions). This would be going too far, so we will
refuse this option.
In fact, it is a good thing that not every number can be divided by other numbers. This
helps us understand the structure of the natural numbers and opens the door to many
interesting questions and applications.
If given numbers a and b, it is possible that a ÷ b gives a whole number. In this case,
we say that b divides a, in symbols, we write b | a. If this holds, then b is a divisor or factor
of a, and a is a multiple of b. In other words, if b | a, then a bk for some integer k (this is
saying a is some multiple of b).
m|n “m divides n”
provided n ÷ m is an integer. Thus the following assertions mean the same thing:
1. m | n
4. n is a multiple of m.
Example:
Decide whether each of the statements below are true or false.
1. 4 | 20 4. 5 | 0 7. −3 | 12
2. 20 | 4 5. 7 | 7 8. 8 | 12
3. 0 | 5 6. 1 | 37 9. 1642 | 136299
Solution:
1. True. 4 “goes into” 20 five times without remainder. In other words, 20÷4 5,
an integer. We could also justify this by saying that 20 is a multiple of 4:
20 4 · 5.
A.2. Introduction to Number Theory 217
7. True. Negative numbers work just fine for the divisibility relation. Here
12 −3 · 4. It is also true that 3 | −12 and that −3 | −12.
8. False. Both 8 and 12 are divisible by 4, but this does not mean that 12 is
divisible by 8.
This last example raises a question: how might one decide whether m | n? Of course, if
you had a trusted calculator, you could ask it for the value of n ÷ m. If it spits out anything
other than an integer, you know m - n. This seems a little like cheating though: we don’t
have division, so should we really use division to check divisibility?
While we don’t really know how to divide, we do know how to multiply. We might try
multiplying m by larger and larger numbers until we get close to n. How close? Well, we
want to be sure that if we multiply m by the next larger integer, we go over n.
For example, let’s try this to decide whether 1642 | 136299. Start finding multiples of
1642:
1642 · 2 3284 1642 · 3 4926 1642 · 4 6568 ···
All of these are well less than 136299. I suppose we can jump ahead a bit:
1642 · 83 136286
Is this the best we can do? How far are we from our desired 136299? If we subtract, we get
136299 − 136286 13. So we know we cannot go up to 84, that will be too much. In other
words, we have found that
136299 83 · 1642 + 13
Since 13 < 1642, we can now safely say that 1642 - 136299.
It turns out that the process we went through above can be repeated for any pair of
numbers. We can always write the number a as some multiple of the number b plus some
remainder. We know this because we know about division with remainder from elementary
school. This is just a way of saying it using multiplication. Due to the procedural nature
that can be used to find the remainder, this fact is usually called the division algorithm:
A.2. Introduction to Number Theory 218
a qb + r
The idea is that we can always take a large enough multiple of b so that the remainder
r is as small as possible. We do allow the possibility of r 0, in which case we have b | a.
Remainder Classes
The division algorithm tells us that there are only b possible remainders when dividing by
b. If we fix this divisor, we can group integers by the remainder. Each group is called a
remainder class modulo b (or sometimes residue class).
Example:
Describe the remainder classes modulo 5.
Solution: We want to classify numbers by what their remainder would be when di-
vided by 5. From the division algorithm, we know there will be exactly 5 remainder
classes, because there are only 5 choices for what r could be (0 ≤ r < 5).
First consider r 0. Here we are looking for all the numbers divisible by 5 since
a 5q + 0. In other words, the multiples of 5. We get the infinite set
There are three more to go. The remainder classes for 2, 3, and 4 are, respectively
Note that in the example above, every integer is in exactly one remainder class. The
technical way to say this is that the remainder classes modulo b form a partition of the
integers.1 The most important fact about partitions, is that it is possible to define an
equivalence relation from a partition: this is a relationship between pairs of numbers which
acts in all the important ways like the “equals” relationship.2
All fun technical language aside, the idea is really simple. If two numbers belong to the
same remainder class, then in some way, they are the same. That is, they are the same up
to division by b. In the case where b 5 above, the numbers 8 and 23, while not the same
number, are the same when it comes to dividing by 5, because both have remainder 3.
It matters what the divisor is: 8 and 23 are the same up to division by 5, but not up to
division by 7, since 8 has remainder of 1 when divided by 7 while 23 has a remainder of 2.
With all this in mind, let’s introduce some notation. We want to say that 8 and 23 are
basically the same, even though they are not equal. It would be wrong to say 8 23.
Instead, we write 8 ≡ 23. But this is not always true. It works if we are thinking division
by 5, so we need to denote that somehow. What we will actually write is this:
8 ≡ 23 (mod 5)
which is read, “8 is congruent to 23 modulo 5” (or just “mod 5”). Of course then we could
observe that
8 . 23 (mod 7)
Congruence Modulo n
We say a is congruent to b modulo n, and write,
a ≡ b (mod n )
provided a and b have the same remainder when divided by n. In other words,
provided a and b belong to the same remainder class modulo n.
Many books define congruence modulo n slightly differently. They say that a ≡ b
(mod n ) if and only if n | a − b. In other words, two numbers are congruent modulo n,
if their difference is a multiple of n. So which definition is correct? Turns out, it doesn’t
matter: they are equivalent.
To see why, consider two numbers a and b which are congruent modulo n. Then a and
b have the same remainder when divided by n. We have
a q1 n + r b q2 n + r
Here the two r’s really are the same. Consider what we get when we take the difference of
a and b:
a − b q1 n + r − (q2 n + r ) q1 n − q2 n (q1 − q2 )n
1 It is possible to develop a mathematical theory of partitions, prove statements about all partitions in
general and then apply those observations to our case here.
2 Again, there is a mathematical theory of equivalence relations which applies in many more instances than
the one we look at here.
A.2. Introduction to Number Theory 220
So a − b is a multiple of n, or equivalently, n | a − b.
On the other hand, if we assume first that n | a − b, so a − b kn, then consider what
happens if we divide each term by n. Dividing a by n will leave some remainder, as will
dividing b by n. However, dividing kn by n will leave 0 remainder. So the remainders on
the left hand side must cancel out. That is, the remainders must be the same.
Thus we have:
Congruence and Divisibility
For any integers a, b, and n, we have
It will also be useful to switch back and forth between congruences and regular equa-
tions. The above fact helps with this. We know that a ≡ b (mod n ) if and only if n | a − b, if
and only if a − b kn for some integer k. Rearranging that equation, we get a b + kn. In
other words, if a and b are congruent modulo n, then a is b more than some multiple of n.
This conforms with our earlier observation that all the numbers in a particular remainder
class are the same amount larger than the multiples of n.
Properties of Congruence
We said earlier that congruence modulo n behaves, in many important ways, the same
way equality does. Specifically, we could prove that congruence modulo n is an equivalence
relation, which would require checking the following three facts:
1. a ≡ a (mod n ).
You should take a minute to convince yourself that each of the properties above actu-
ally hold of congruence. Try explaining each using both the remainder and divisibility
definitions.
A.2. Introduction to Number Theory 221
Next, consider how congruence behaves when doing basic arithmetic. We already
know that if you subtract two congruent numbers, the result will be congruent to 0 (be a
multiple of n). What if we add something congruent to 1 to something congruent to 2?
Will we get something congruent to 3?
1. a + c ≡ b + d (mod n ).
2. a − c ≡ b − d (mod n ).
3. ac ≡ bd (mod n ).
The above facts might be written a little strangely, but the idea is simple. If we have
a true congruence, and we add the same thing to both sides, the result is still a true
congruence. This sounds like we are saying:
Of course this is true as well, it is the special case where c d. But what we have
works in more generality. Think of congruence as being “basically equal.” If we have two
numbers which are basically equal, and we add basically the same thing to both sides, the
result will be basically equal.
This seems reasonable. Is it really true? Let’s prove the first fact:
a + c b + d + kn + jn.
Example:
Find the remainder of 3491 divided by 9.
Solution: We could do long division, but there is another way. We want to find
x such that x ≡ 3491 (mod 9). Now 3491 3000 + 400 + 90 + 1. Of course 90 ≡ 0
(mod 9), so we can replace the 90 in the sum with 0. Why is this okay? We are
A.2. Introduction to Number Theory 222
Next, note that 400 4 · 100, and 100 ≡ 1 (mod 9) (since 9 | 99). So we can in fact
replace the 400 with simply a 4. Again, we are appealing to our claim that we can
replace congruent elements, but we are really appealing to property 3 about the
arithmetic of congruence: we know 100 ≡ 1 (mod 9), so if we multiply both sides
by 4, we get 400 ≡ 4 (mod 9).
Similarly, we can replace 3000 with 3, since 1000 1 + 999 ≡ 1 (mod 9). So our
original congruence becomes
x ≡3+4+0+1 (mod 9)
x ≡ 8 (mod 9).
Therefore 3491 divided by 9 has remainder 8.
The above example should convince you that the well known divisibility test for 9 is
true: the sum of the digits of a number is divisible by 9 if and only if the original number
is divisible by 9. In fact, we now know something more: any number is congruent to the
sum of its digits, modulo 9.3
Let’s try another:
Example:
Find the remainder when 3123 is divided by 7.
Solution: Of course, we are working with congruence because we want to find the
smallest positive x such that x ≡ 3123 (mod 7). Now first write 3123 (33 )41 . We
have:
3123 2741 ≡ 641 (mod 7) ,
since 27 ≡ 6 (mod 7). Notice further that 62 36 is congruent to 1 modulo 7. Thus
we can simplify further:
3 This works for 3 as well, but definitely not for any modulus in general.
A.2. Introduction to Number Theory 223
In the above example, we are using the fact that if a ≡ b (mod n ), then a p ≡ b p (mod n ).
This is just applying property 3 a bunch of times.
So far we have seen how to add, subtract and multiply with congruences. What about
division? There is a reason we have waited to discuss it. It turns out that we cannot simply
divide. In other words, even if ad ≡ bd (mod n ), we do not know that a ≡ b (mod n ).
Consider, for example:
18 ≡ 42 (mod 8).
This is true. Now 18 and 42 are both divisible by 6. However,
While this doesn’t work, note that 3 ≡ 7 (mod 4). We cannot divide 8 by 6, but we can
divide 8 by the greatest common factor of 8 and 6. Will this always happen?
Suppose ad ≡ bd (mod n ). In other words, we have ad bd + kn for some integer k.
Of course ad is divisible by d, as is bd. So kn must also be divisible by d. Now if n and d
have no common factors (other than 1), then we must have d | k. But in general, if we try
to divide kn by d, we don’t know that we will get an integer multiple of n. Some of the n
might get divided as well. To be safe, let’s divide as much of n as we can. Take the largest
factor of both d and n, and cancel that out from n. The rest of the factors of d will come
from k, no problem.
We will call the largest factor of both d and n the gcd( d, n ), for greatest common divisor.
In our example above, gcd(6, 8) 2 since the greatest divisor common to 6 and 8 is 2.
Example:
Simplify the following congruences using division: (a) 24 ≡ 39 (mod 5) and (b)
24 ≡ 39 (mod 15).
Solution: (a) Both 24 and 39 are divisible by 3, and 3 and 5 have no common factors,
so we get
8 ≡ 13 (mod 5).
(b) Again, we can divide by 3. However, doing so blindly gives us 8 ≡ 13
(mod 15) which is no longer true. Instead, we must also divide the modulus 15 by
the greatest common factor of 3 and 15, which is 3. Again we get
8 ≡ 13 (mod 5).
A.2. Introduction to Number Theory 224
Solving Congruences
Now that we have some algebraic rules to govern congruence relations, we can attempt to
solve for an unknown in a congruence. For example, is there a value of x that satisfies,
3x + 2 ≡ 4 (mod 5) ,
and if so, what is it?
In this example, since the modulus is small, we could simply try every possible value
for x. There are really only 5 to consider, since any integer that satisfied the congruence
could be replaced with any other integer it was congruent to modulo 5. Here, when x 4
we get 3x + 2 14 which is indeed congruent to 4 modulo 5. This means that x 9 and
x 14 and x 19 and so on will each also be a solution because as we saw above, replacing
any number in a congruence with a congruent number does not change the truth of the
congruence.
So in this example, simply compute 3x + 2 for values of x ∈ {0, 1, 2, 3, 4}. This gives 2,
5, 8, 11, and 14 respectively, for which only 14 is congruent to 4.
Let’s also see how you could solve this using our rules for the algebra of congruences.
Such an approach would be much simpler than the trial and error tactic if the modulus
was larger. First, we know we can subtract 2 from both sides:
3x ≡ 2 (mod 5).
Then to divide both sides by 3, we first add 0 to both sides. Of course, on the right hand
side, we want that 0 to be a 10 (yes, 10 really is 0 since they are congruent modulo 5). This
gives,
3x ≡ 12 (mod 5).
Now divide both sides by 3. Since gcd(3, 5) 1, we do not need to change the modulus:
x≡4 (mod 5).
Notice that this in fact gives the general solution: not only can x 4, but x can be any number
which is congruent to 4. We can leave it like this, or write “x 4 + 5k for any integer k.”
Example:
Solve the following congruences for x.
1. 7x ≡ 12 (mod 13).
Solution:
1. All we need to do here is divide both sides by 7. We add 13 to the right hand
side repeatedly until we get a multiple of 7 (adding 13 is the same as adding
A.2. Introduction to Number Theory 225
0, so this is legal). We get 25, 38, 51, 64, 77 – got it. So we have:
7x ≡ 12 (mod 13)
7x ≡ 77 (mod 13)
x ≡ 11 (mod 13).
2. Here, since we have numbers larger than the modulus, we can reduce them
prior to applying any algebra. We have 84 ≡ 9, 38 ≡ 8 and 79 ≡ 4. Thus,
6x ≡ 9 (mod 14).
We could now divide both sides by 3, or try to increase 9 by a multiple of 14
to get a multiple of 6. If we divide by 3, we get,
2x ≡ 3 (mod 14).
The last congruence above illustrates the way in which congruences might not have
solutions. We could have seen this immediately in fact. Look at the original congruence:
20x 23 + 14k,
or equivalently 20x − 14k 23. We can easily see there will be no solution to this equation
in integers. The left hand side will always be even, but the right hand side is odd. A similar
problem would occur if the right hand side was divisible by any number the left hand side
was not.
So in general, given the congruence
ax ≡ b (mod n ),
if a and n are divisible by a number which b is not divisible by, then there will be no
solutions. In fact, we really only need to check one divisor of a and n: the greatest common
divisor. Thus, a more compact way to say this is:
Diophantine Equations
An equation in two or more variables is called a Diophantine equation if only integers
solutions are of interest. A linear Diophantine equation takes the form a1 x 1 + a 2 x x +
· · · + a n x n b for constants a 1 , . . . , a n , b.
A solution to a Diophantine equation is a solution to the equation consisting only
of integers.
We have the tools we need to solve linear Diophantine equations. We will consider, as
a main example, the equation
51x + 87y 123.
The general strategy will be to convert the equation to a congruence, then solve that
congruence.4 Let’s work this particular example to see how this might go.
4 This is certainly not the only way to proceed. A more common technique would be to apply the Euclidean
algorithm. Our way can be a little faster, and is presented here primarily for variety.
A.2. Introduction to Number Theory 227
First, check if perhaps there are no solutions because a divisor of 51 and 87 is not a
divisor of 123. Really, we just need to check whether gcd(51, 87) | 123. This greatest
common divisor is 3, and yes 3 | 123. At this point, we might as well factor out this greatest
common divisor. So instead, we will solve:
Now observe that if there are going to be solutions, then for those values of x and y,
the two sides of the equation must have the same remainder as each other, no matter what
we divide by. In particular, if we divide both sides by 17, we must get the same remainder.
Thus we can safely write
17x + 29y ≡ 41 (mod 17).
We choose 17 because 17x will have remainder 0. This will allow us to reduce the congru-
ence to just one variable. We could have also moved to a congruence modulo 29, although
there is usually a good reason to select the smaller choice, as this will allow us to reduce
the other coefficient. In our case, we reduce the congruence as follows:
Now at this point we know y 2 + 17k will work for any integer k. If we haven’t made
a mistake, we should be able to plug this back into our original Diophantine equation to
find x:
1. Divide both sides of the equation by gcd( a, b ) (if this does not leave the right hand
side as an integer, there are no solutions). Let’s assume that ax + b y c has already
been reduced in this way.
2. Pick the smaller of a and b (here, assume it is b), and convert to a congruence modulo
b:
ax + b y ≡ c (mod b ).
This will reduce to a congruence with one variable, x:
ax ≡ c (mod b ).
A.2. Introduction to Number Theory 228
3. Solve the congruence as we did in the previous section. Write your solution as an
equation, such as,
x n + kb
4. Plug this into the original Diophantine equation, and solve for y.
Example:
How can you make $6.37 using just 5-cent and 8-cent stamps? What is the smallest
and largest number of stamps you could use?
Solution: First, we need a Diophantine equation. We will work in numbers of
cents. Let x be the number of 5-cent stamps, and y be the number of 8-cent stamps.
We have:
5x + 8y 637.
Convert to a congruence and solve:
8y ≡ 367 (mod 5)
3y ≡ 2 (mod 5)
3y ≡ 12 (mod 5)
y ≡ 4 (mod 5).
Using this method, as long as you can solve linear congruences in one variable, you can
solve linear Diophantine equations of two variables. There are times though that solving
the linear congruence is a lot of work. For example, suppose you need to solve,
You could keep adding 51 to the right side until you get a multiple of 13: You would get 57,
108, 159, 210, 261, 312, and 312 is the first of these that is divisible by 13. This works, but is
really too much work. Instead we could convert back to a Diophantine equation:
13x 6 + 51k
Now solve this like we have in this section. Write it as a congruence modulo 13:
Of course you could do this switching back and forth between congruences and Dio-
phantine equations as many times as you like. If you only used this technique, you would
essentially replicate the Euclidean algorithm, a more standard way to solve Diophantine
equations.
Exercises
1. Suppose a, b, and c are integers. Prove that if a | b, then a | bc.
(a) 2.
(b) 5.
(c) 7.
(d) 9.
A.2 Exercises 230
6. Determine which of the following congruences have solutions, and find any solutions
(between 0 and the modulus) by trial and error.
8. I’m thinking of a number. If you multiply my number by 7, add 5, and divide the
result by 11, you will be left with a remainder of 2. What remainder would you get
if you divided my original number by 11?
9. Solve the following linear Diophantine equations, using modular arithmetic (describe
the general solutions).
10. You have a 13 oz. bottle and a 20 oz. bottle, with which you wish to measure exactly
2 oz. However, you have a limited supply of water. If any water enters either bottle
and then gets dumped out, it is gone forever. What is the least amount of water you
can start with and still complete the task?
Appendix B
Solutions to Exercises
0.2.7. (a) A ∪ B:
231
Solutions for 0.2 232
A B
(b) (A ∪ B ):
A B
(c) A ∩ (B ∪ C ):
A B
(d) (A ∩ B ) ∪ C:
A B
(e) A ∩ B ∩ C:
A B
C
Solutions for 0.3 233
(f) (A ∪ B ) \ C:
A B
0.2.8. For example, A ∪ B ∩ (A ∩ B ). Note that A ∩ B would almost work, but it also contains
the area outside of both circles.
0.2.12. {2, 3, 5}, {1, 2, 3, 5}, {2, 3, 4, 5}, {2, 3, 5, 6}, {1, 2, 3, 4, 5}, {1, 2, 3, 5, 6}, {2, 3, 4, 5, 6},
and {1, 2, 3, 4, 5, 6}.
0.2.15. No. There must be 5 elements in common to both sets. Since there are 10 distinct
elements all together in A and B, this means that between A and B, there must be 5
elements which they do not have in common (some in A but not in B, some in B but
not in A). But 5 is odd, so to have | A | | B |, we would need 7.5 elements in each set,
which is impossible.
0.2.16. If R is the set of red cards and F is the set of face cards, we want to find | R ∪ F |. This
is not simply | R | + | F | because there are 6 cards which are both red and a face card;
| R ∩ F | 6. We find | R ∪ F | 32.
0.3.2. There are 9 functions. You have a choice of three outputs for f (1), and for each,
you have three choices for the output f (2). Of these functions, 6 are injective, 0 are
surjective, and 0 are both.
Solutions for 0.3 234
0.3.3. (a) f is not injective, since f (2) f (5) - two different inputs have the same output.
(b) f is surjective, since every element of the codomain is an element of the range.
0.3.6. (a) f is not injective. To prove this, we must simply find two different elements of
the domain which map to the same element of the codomain. Since f ({1}) 1
and f ({2}) 1, we see that f is not injective.
(b) f is not surjective. The largest subset of A is A itself, and | A | 10. So no natural
number greater than 10 will ever be an output.
(c) f −1 (1) {{1} , {2} , {3} , . . . {10}} (the set of all the singleton subsets of A).
(d) f −1 (0) {∅}. Note, it would be wrong to write f −1 (0) ∅ - that would claim
that there is no input which has 0 as an output.
(e) f −1 (12) ∅, since there are no subsets of A with cardinality 12.
0.3.7. (a) f −1 (3) {003, 030, 300, 012, 021, 102, 201, 120, 210, 111}
(b) f −1 (28) ∅ (since the largest sum of three digits is 9 + 9 + 9 27)
(c) Part (a) proves that f is not injective – the output 3 is assigned to 10 different
inputs.
(d) Part (b) proves that f is not surjective – there is an element of the codomain (28)
which is not assigned to any inputs.
0.3.8. (a) | f −1 (3)| ≤ 1. In other words, either f −1 (3) is the emptyset or is a set containing
exactly one element. Injective functions cannot have two elements from the
domain both map to 3.
(b) | f −1 (3)| ≥ 1. In other words, f −1 (3) is a set containing at least one element,
possibly more. Surjective functions must have something map to 3.
(c) | f −1 (3)| 1. There is exactly one element from X which gets mapped to 3, so
f −1 (3) is the set containing that one element.
0.3.9. X can really be any set, as long as f ( x ) 0 or f ( x ) 1 for every x ∈ X. For example,
X N and f ( n ) 0 works.
0.3.10. (a) | X | ≤ | Y |. Otherwise two or more of the elements of X would need to map to
the same element of Y.
(b) | X | ≥ | Y |. Otherwise there would be one or more elements of Y which were
never an output.
Solutions for 0.3 235
(c) | X | | Y |. This is the only way for both of the above to occur.
(b) f is surjective.
0.3.13. Yes, this is a function, if you choose the domain and codomain correctly. The domain
will be the set of students, and the codomain will be the set of possible grades. The
function is almost certainly not injective, because it is likely that two students will get
the same grade. The function might be surjective – it will be if there is at least one
student who gets each grade.
0.3.14. Yes, as long as the set of cards is the domain and the set of players is the codomain.
The function is not injective because multiple cards go to each player. It is surjective
since all players get cards.
0.3.15. This cannot be a function. If the domain were the set of cards, then it is not a
function because not every card gets dealt to a player. If the domain were the set
of players, it would not be a function because a single player would get mapped to
multiple cards. Since this is not a function, it doesn’t make sense to say whether it is
injective/surjective/bijective.
Solutions for 1.1 236
1.1.2. 8.
1.1.3. 15.
1.1.4. 5 · (4 + 3) + 7 42.
1.1.5. (a) 16 is the number of choices you have if you want to watch one movie, either a
comedy or horror flick.
(b) 63 is the number of choices you have if you will watch two movies, first a comedy
and then a horror.
1.1.7. | A ∪ B | + | A ∩ B | 13.
1.1.8. 39.
6
15.
1.2.2. (a) 4
3
3. We need to select 1 of the 3 remaining elements to be in the subset.
(b) 1
6
15. All subsets of cardinality 4 must contain at least one odd number.
(c) 4
3
3. Select 1 of the 3 even numbers. The remaining three odd numbers of S
(d) 1
must all be in the set.
1.2.3. (a) We can think of each row as a 6-bit string of weight 3 (since of the 6 coins,
we require 3 to be pennies). Thus there are 63 20 rows possible. Each row
requires 6 coins, so if we want to make all the rows at the same time, we will
need 120 coins (60 of each).
6 6 6 6 6 6 6
(b) Now there are 26 64 rows possible, which is also + + + + + +
0 1 2 3 4 5 6 .
Thus we need 6 · 64 384 coins (192 of each).
10 10 10 10 10
+ + + + 386.
1.2.4. 6 7 8 9 10
10
+ 10 10 10 10
7 + 8 + 9 + 10 386. This is the same as the previous question, since we
1.2.5. 6
can think of each subset as a 10-bit string with a 1 representing that we include that
element in the subset.
1.2.6. To get an x 12 , we must pick 12of the 15 factors to contribute an x, leaving the other 3
to contribute a 2. There are 15
12 ways to select these 12 factors. So the term containing
12 15 12 3 12 15 3
an x will be 12 x 2 . In other words the coefficient of x is 12 2 .
14 15
+ 29 .
1.2.7. Use the binomial theorem. 9 6
14
1.2.8. (a) 7 .
6 8
(b) 2 5 .
14
6 8
(c) 7 − 2 5 .
11
165, since you have to select a 3-element subset of the set of 11 toppings.
1.2.9. (a) 3
10
120, since you must select 3 of the 10 non-pineapple toppings.
(b) 3
10
45, since you must select 2 of the remaining 10 non-pineapple toppings to
(c) 2
have in addition to the pineapple.
(d) 165 120 + 45, which makes sense because every 3-topping pizza either has
pineapple or does not have pineapple as a topping.
x. But 85 83 , because we could just as easily have picked 5 out of the 8 factors to
contribute a y.
Solutions for 1.3 238
1.3.2. Despite its name, we are not looking for a combination here. The order in which the
three numbers appears matters. There are P (40, 3) 40 · 39 · 38 different possibilities
for the “combination”.
1.3.3. (a) This is just the multiplicative principle. There are 7 digits which we can select
for each of the 5 positions, so we have 75 such numbers.
(b) Now we have 7 choices for the first number, 6 for the second, etc. So there are
7 · 6 · 5 · 4 · 3 P (7, 5) such numbers.
(c) To build such a number we simply must select 5 different digits. After doing so,
there will only be one way to arrange them into a 5-digit number. Thus there
are 75 such numbers.
(d) The permutation is in part (b), while the combination is in part (c). At first this
seems backwards, since usually we use combinations for when order does not
matter. Here it looks like in part (c) that order does matter. The better way to
distinguish between combinations and permutations is to ask whether we are
counting different arrangements as different outcomes. In part (c), there is only
one arrangement of any set of 5 digits, while in part (b) each set of 5 digits gives
5! different outcomes.
7 7
1.3.4. 2 2 .
1.3.5. (a) 5 (you need to skip one dot on the top and on the bottom).
7
(b) 2 . Once you select the two dots on the top, the bottom two are determined.
(c) This is tricky since you need to worry about running out of space. One way7to
7
count: break into cases by the location of the top left corner. You get 2 + ( 2 −
1) + ( 72 − 3) + ( 72 − 6) + ( 72 − 10) + ( 72 − 15).
1.3.6. Since there are 15 different letters, we have 15 choices for the first letter, 14 for the
next, and so on. Thus there are 15! anagrams.
1.3.7. After the first letter, we must rearrange the remaining 7 letters. There are only two
letters, so this is really just a bit-string question. Thus there 72 anagrams starting
with “a”.
Solutions for 1.4 239
1.3.8. First, decide where to put the “a”s. There are 7 positions, and we must choose 3 of
them to fill with an “a”. This can be done in 73 ways. The remaining 4 spots all get
a different letter, so there are 4! ways to finish off the anagram. This gives a total of
7
3 · 4! anagrams. Strangely enough, this is 840, which is also equal to P (7, 4). To get
the answer that way, start by picking one of the 7 positions to be filled by the “n”, one
of the remaining 6 positions to be filled by the “g”, one of the remaining 5 positions
to be filled by the “r”, one of the remaining 4 positions to be filled by the “m” and
then put “a”s in the remaining 3 positions.
20 16 12 8 4
1.3.9. (a) 4 4 4 4 4 .
5! 15
12 9 6 3
(b) 3 3 3 3 3 .
1.3.10. 9! (there are 10 people seated around the table, but it does not matter where King
Arthur sits, only who sits to his left, two seats to his left, and so on).
1: You must choose k out of n elements to put in the set, which can be done
Answer
n
in k ways.
Answer 2: First count the number of k-element subsets of {1, 2, . . . , n } which contain
the number n. We must choose k − 1 of the n − 1 other element to include in this
set. Thus there are n−1 k−1 such subsets. We have not yet counted all the k-element
subsets of {1, 2, . . . , n } though. In fact, we have missed exactly those subsets which
do NOT contain n. To form one of these subsets, we need to choose k of the other
n − 1 elements, so this can be done in n−1 k ways. Thus the answer to the question is
n−1
n−1
k−1 + k . qed
1.4.2. Proof. Question: How many 2-letter words start with a, b, or c and end with either y
or z?
Answer 1: There are two words that start with a, two that start with b, two that start
with c, for a total of 2 + 2 + 2.
Answer 2: There are three choices for the first letter and two choices for the second
letter, for a total of 3 · 2.
Since the two answers are both answers to the same question, they are equal. Thus
2 + 2 + 2 3 · 2. qed
n+1
!
1+2+3+···+ n
2
qed
15
1.4.4. (a) She has 6 ways to select the 6 bridesmaids, and then for each way, has 6 choices
for the maid of honor. Thus she has 15
6 6 choices.
(b) She has 15 choices for who will be her maid of honor. Then she needs to select
5 of the remaining 14 friends to be bridesmaids, which she can do in 14
5 ways.
14
Thus she has 15 5 choices.
(c) We have answered the question (how many wedding parties can the bride choose
from) in two ways. The first way gives the left hand side of the identity and
the second way gives the right hand side of the identity. Therefore the identity
holds.
1.4.5. Proof. Question: You have a large container filled with ping-pong balls, all with a
different number of them. You must select k of the balls, putting two of them in a jar
and the others in a box. How many ways can you do this?
Answer 1: First select 2 of the n balls to put in the jar, then select k −2 of the remaining
n − 2 balls to put in the box. The first task can be completed in n2 different ways, the
second task in n−2 n n−2
k−2 ways. Thus there are 2 k−2 ways to select the balls.
Answer 2: First select k balls from the n in the container. Then pick 2 of the k balls
you picked to put in
n k
k − 2 in the box. The nfirst
the jar, placing the remaining k task can
be completed in k ways, the second task in 2 ways. Thus there are k 2 ways to
select the balls.
Since both answers count the same thing, they must be equal. qed
5
1.4.6. (a) After the 1, we need to find a 5-bit string with one 1. There are 1 ways to do
this.
4
(b) 1 (we need to pick 1 of the remaining 4 slots to be the second 1).
3
(c) 1
2
(d) Yes. We still need strings starting with 0001 (there are 1 of these) and strings
starting 00001 (there is only 11 1 of these).
6
(e) 2
Solutions for 1.4 241
1.4.7. (a) 3n , since there are 3 choices for each of the n digits.
n
(b) 1, since all the digits need to be 2’s. However, we might write this as 0 .
There are n1 places to put the
(c) non-2 digit. That digit can be either a 0 or a 1, so
there are 2 n1 such strings.
(d) We must choose two slots to fill with 0’s or 1’s. There are n2 ways to do that.
Once the slots are picked, we have two choices for the first slot (0 or 1) and two
2 n
choices for the second slot (0 or 1). So there are a total of 2 2 such strings.
(e) There are nk ways to pick which slots don’t have the 2’s. Then those slots can
(placing the “g” in the last spot). This gives the answer
! ! ! ! !
9 6 4 2 1
.
3 2 2 1 1
Alternatively, we could select the positions of the letters in the opposite order, which
would give an answer ! ! ! ! !
9 8 7 5 3
.
1 1 2 2 3
(where the 3 “r”s go in the remaining 3 spots). These two expressions are equal:
! ! ! ! ! ! ! ! ! !
9 6 4 2 1 9 8 7 5 3
.
3 2 2 1 1 1 1 2 2 3
1.4.9. Proof. Question: How many k-letter words can you make using n different letters
without repeating any letter?
Solutions for 1.5 242
Answer 1: There are n choices for the first letter, n − 1 choices for the second letter,
n − 2 choices for the third letter, and so on until n − ( k − 1) choices for the kth letter
(since k − 1 letters have already been assigned at that point). The product of these
n!
numbers can be written (n−k )! which is P ( n, k ).
Answer
n
2: First pick k letters to be in the word from the n choices. This can be done
in k ways. Now arrange those letters into a word. There are k choices for the first
letter, k − 1 choices for the second, and so on, for a total of k! arrangements of the k
n
letters. Thus the total number of words is k k!. qed
1.4.10. Proof. Question: How many 5-element subsets are there of the set {1, 2, . . . , n + 3}.
n+3
Answer 1: We choose 5 out of the n + 3 elements, so
5 .
Answer 2: Break this up into cases by what the “middle” (third smallest) element
of the 5 element subset is. The smallest this could be is a 3. In that case, we
have 22 choices for the numbers below it, and n2 choices for the numbers above it.
Alternatively, the middle number could be a 4. In this case there are 32 choices for
1.5.2. (a) You take 3 strawberry, 1 lime, 0 licorice, 2 blueberry and 0 bubblegum.
(b) This is backwards. We don’t want the stars to represent the kids because the
kids are not identical, but the stars are. Instead we should use 5 stars (for the
lollipops) and use 5 bars to switch between the 6 kids. For example, **||***|||
would represent the outcome with the first kid getting 2 lollipops, the third kid
getting 3, and the rest of the kids getting none.
(c) This is the word AAAEOO.
(d) This doesn’t represent a solution. Each star should represent one of the 6 units
that add up to 6, and the bars should switch between the different variables. We
have one too many bars.
Solutions for 1.6 243
18
1.5.3. (a) 4 . Each outcome can be represented by a sequence of 14 stars and 4 bars.
13
(b) 4 . First put one ball in each bin. This leaves 9 stars and 4 bars.
7
1.5.4. (a) 2 . After each variable gets 1 star for free, we are left with 5 stars and 2 bars.
10
(b) 2 . We have 8 stars and 2 bars.
19
2 . This problem is equivalent to finding the number of solutions to x + y +z
0 0 0
(c)
17 where x 0, y 0 and z 0 are non-negative. (In fact, we really just do a substitution.
Let x x 0 − 3, y y 0 − 3 and z z 0 − 3).
1.5.5. (a) This is 75 . We simply choose five of the seven digits and once chosen put them
in increasing order.
(b) This requires stars and bars. Use a star to represent each of the 5 digits in the
number, and use their position relative to the bars to say what numeral fills that
spot. So we will have 5 stars and 6 bars, giving 11
6 .
10
1.5.6. 5 . We have 5 stars (the five dice) and 5 bars (the five switches between the numbers
1-6).
1.5.7. We must figure out how many different combinations of 7 coins are possible. Let a
star represent each coin, and a bar represent switching between type of coin. So for
example **|*||**** represents 2 pennies, one nickel, no dimes and 4 quarters. The
number of such star and bar diagrams (with 7 stars and 3 bars) is 103 120. Thus
you have a 1 in 120 chance of guessing correctly.
18
. Distribute 10 units to the variables before finding all solutions to x 10 +x 20 +x 30 +x 40
1.5.8. 3
15 in non-negative integers.
16
(b) 6 .
f g
16 7 13 7 10 7 7
+
(c) 6 − 1 6 − 2 6 3 6 .
18
f 5 4g
− 1 11 5
1.6.2. 4 4 − 2 4 . Subtract all the distributions for which one or more bins contain
7 or more balls.
1.6.3. The easiest way to solve this is to instead count the solutions to y1 + y2 + y3 + y4 7
with 0 ≤ y i ≤ 3. By taking x i y i + 2, each solution to this new equation corresponds
to exactly one solution to the original equation.
Now all the ways to distribute the 7 units to the four y i variables can be found using
stars and bars, specifically 7 stars and 3 bars, so 103 . But this includes the ways that
one or more y i variables can be assigned more than 3 units. So subtract, using PIE.
We get ! ! !
10 4 6
− .
3 1 3
Note that this is the final answer because it is not possible to have two variables both
get 4 units.
1.6.5. The question is, how many ways can you distribute k cookies to n kids so that each
kid gets at most one cookie. On one hand, the answer is just nk since you must choose
k kids to get a cookie. Alternatively, we can use stars and bars with PIE, which is
how we get the right hand side of the identity. Note that lots of the terms on the right
hand side will be zero, as soon as n + k − (2 j + 1) drops below k.
1.6.6. The 9 derangements are: 2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321.
10
f g
4! − 41 3! − 42 2! + 43 1! − 44 0! . We choose 6 of the 10 ladies to get their own
1.6.7. 6
hat, and the multiply by the number of ways the remaining hats can be deranged.
1.7.2. There are 9 functions – you have a choice of three outputs for f (1), and for each,
you have three choices for the output f (2). Of these functions, 6 are injective, 0 are
surjective, and 0 are both.
1.7.3. (a) 64 1296, since there are six choices of where to send each of the 4 elements of
the domain.
(b) P (6, 4) 6 · 5 · 4 · 3 360, since outputs cannot be repeated.
(c) None.
(d) There are 5 · 63 functions for which f (1) , a and another 5 · 63 functions for
which f (2) , b. There are 52 · 62 functions for which both f (1) , a and f (2) , b.
So the total number of functions for which f (1) , a or f (2) , b or both is
5 · 63 + 5 · 63 − 52 · 62 1260
(h) Neither. Actually, “k” is the 11th letter of the alphabet, so the answer is 0. If “k”
was among the first 10 letters, there would only be 1 way - write it down.
9 13
(i) Neither. Either 3 (if every kid gets an apple) or 3 (if appleless kids are
allowed).
10
(j) Neither. Note that this could not be 4 since the 10 things and 4 things are from
different groups. 410 .
10
(k) 4 - don’t be fooled by the “arrange” in there - you are picking 4 out of 10 spots
to put the 1’s.
10
(l) 4 (assuming order is irrelevant).
(m) Neither. 1610 (each kid chooses yes or no to 4 varieties).
(n) Neither. 0.
4 4 4
(o) Neither. 410 − [ 310 − 210 + 110 ].
1 2 3
(p) Neither. 10 · 4.
(q) Neither. 410 .
10 10
(r) 4 (which is the same as 6 ).
(s) Neither.
If all the kids were identical, and you wanted no empty teams, it would
be 10
4 . Instead, this will be the same as the number of surjective functions from
a set of size 11 to a set of size 5.
10
(t) 4 .
10
(u) 4 .
(v) Neither. 4!.
(w) Neither. It’s 10
4 if you won’t repeat any choices. Ifrepetition is allowed, then
this becomes x 1 + x 2 + · · · + x 10 4, which has 13
9 solutions in non-negative
integers.
(x) Neither. Since repetition of cookie type is allowed, the answer is 104 . Without
repetition, you would have P (10, 4).
10 9 9
+
(y) 4 since that is equal to 4 3 .
(z) Neither. It will be a complicated (possibly PIE) counting problem.
1.8.3. (a) 28 256. You have two choices for each tie: wear it or don’t.
(b) You have 7 choices for regular ties (the 8 choices less the “no regular tie” option)
and 31 choices for bow ties (32 total minus the “no bow tie” option). Thus total
you have 7 · 31 217.
3 5
30.
(c) 2 3
(d) Select one of the 3 bow ties to go on top. There are then 4 choices for the next
tie, 3 for the tie after that, and so on. Thus 3 · 4! 72.
Solutions for 1.8 247
1.8.4. You own 8 purple bow ties, 3 red bow ties, 3 blue bow ties and 5 green bow ties.
How many ways can you select one of each color bow tie to take with you on a trip?
8 · 3 · 3 · 5. How many choices do you have for a single bow tie to wear tomorrow?
8 + 3 + 3 + 5.
1.8.5. (a) 45 .
(b) 44 · 2 (choose any digits for the first four digits - then pick either an even or an
odd last digit to make the sum even).
(c) We need 3 or more even digits. 3 even digits: 53 23 22 . 4 even digits: 5
24 2. 5
4
even digits: 55 25 . So all together: 53 23 22 + 54 24 2 + 55 25 .
1.8.7. 215. Use PIE: 100 + 83 + 71 − 16 − 14 − 11 + 2 215 or a Venn diagram. To find out how
many numbers are divisible by 6 and 7, for example, take 500/42 and round down.
1.8.8. 51.
1.8.9. (a) 28 .
8
(b) 5 .
8
(c) 5 .
(d) There is a bijection between subsets and bit strings: a 1 means that element in
is the subset, a 0 means that element is not in the subset. To get a subset of an
8 element set we have a 8-bit string. To make sure the subset contains exactly 5
elements, there must be 5 1’s, so the weight must be 5.
13 17
+
1.8.10. 10 8 .
8 8
55 213 . Without repeats:
1.8.11. With repeated letters allowed: 5 5 5!P (21, 3).
5 11
1.8.12. (a) 2 6 .
16
12 4
(b) 8 − 7 1 .
5 11
12 4 5 7 4
2 6 + 5 3
(c) − 2 3 3 .
18
18
1.8.13. 8 8 −1 .
1.8.14. A test had n questions on it, of which you must answer any k questions. How
many choices do you have as to what order you answer the questions on the test?
P ( n, k ). When grading
the test, how many different combinations of question might
the professor see? nk .
1.8.15. 27 + 27 − 24 .
7 7 4
+
1.8.16. 3 4 − 1 .
Solutions for 1.8 248
6
1.8.17. (a) 6! − 4 · 3!. (b) 6! − 3 3!.
1.8.18. 2n is the number of lattice paths which have length n, since for each step you can go
up or right. Such a path would end along the line x + y n. So you will end at (0, n ),
or (1, n − 1) or (2,n −2) or
. . . orn (n, 0). Counting the paths to each of these points
n n n
separately, give 0 , 1 , 2 , . . . , n (each time choosing which of the n steps to be to
the right).
n
1.8.19. Hint: give a combinatorial proof for the identity P ( n, k )
k k!.
1.8.20. Of your n bow ties, you decide to give k away to charity. How many ways can you
do this? On one hand, you can choose k of the n bow ties to give away in nk ways.
Alternatively, you can choose which bow ties to keep. You must keep
n
n − k of them if
you will give k away, so you can choose the bow ties to keep in n−k ways. This gives
a combinatorial proof for the identity.
1.8.22. (a) 54 + 54 − 53 .
(b) 4 · 54 + 5 · 4 · 53 − 4 · 4 · 53 .
(c) 5! − [4! + 4! − 3!].
f g
5 5 5 5 5
3! + 1! +
(d) 5! − 1 4! − 2 3 2! − 4 5 0! .
5
f
4 4 4 4
g
2! +
1.8.23. 1 4! − 1 3! − 2 3 1! − 4 0! .
f 6g
4 4 4
1.8.24. 46 − 36 − 26 +
1 2 3 1 .
10
1.8.25. (a) 4 . You need to choose 4 of the 10 cookie types. Order doesn’t matter.
(b) P (10, 4) 10 · 9 · 8 · 7. You are choosing and arranging 4 out of 10 cookies. Order
matters now.
21
(c) 9 . You must switch between cookie type 9 times as you make your 12 cookies.
The cookies are the stars, the switches between cookie types are the bars.
(d) 1012 . You have 10 choices for the “1” cookie, 10 choices for the “2” cookie, and
so on.
f g
(e) 1012 − 10 10 12 10 12
1 9 − 2 8 + · · · − 10 0
12
. We must use PIE to remove all the ways
in which one or more cookie type is not selected.
1.8.26. (a) You are giving your professor 4 types of cookies coming from 10 different types
of cookies. This does not lend itself well to a function interpretation. We could
say that the domain contains the 4 types you will give your professor and the
Solutions for 2.1 249
codomain contains the 10 you can choose from, but then counting injections
would be too much (it doesn’t matter if you pick type 3 first and type 2 second,
or the other way around, just that you pick those two types).
(b) We want to consider injective functions from the set {most, second most, second
least, least} to the set of 10 cookie types. We want injections because we cannot
pick the same type of cookie to give most and least of (for example).
(c) This is not a good problem to interpret as a function. The problem is that
the domain would have to be the 12 cookies you bake, but these elements are
indistinguishable (there is not a first cookie, second cookie, etc.).
(d) The domain should be the 12 shapes, the codomain the 10 types of cookies. Since
we can use the same type for different shapes, we are interested in counting all
functions here.
(e) Here we insist that each type of cookie be given at least once, so now we are
asking for the number of surjections of those functions counted in the previous
part.
2.2.3. 68117.
5−5·321
2.2.4. −2 .
31
1+ 231
3
2.2.5. 5/3 .
100
X
2.2.9. (a) (3 + 4k ) 7 + 11 + 15 + · · · + 403.
k1
n
X
(b) 2k 1 + 2 + 4 + 8 + · · · + 2n .
k0
50
X 1 1 1 1 1
(c) 1 + + + + · · · + .
k2
( k 2 − 1) 3 8 15 2499
Solutions for 2.3 251
100
Y k2 4 9 16 10000
(d) 2
· · ··· .
k2
(k − 1) 3 8 15 9999
Yn
(e) (2 + 3k ) (2)(5)(8)(11)(14) · · · (2 + 3n ).
k0
2.3.2. The first differences are 2, 4, 6, 8, . . ., and the second differences are 2, 2, 2, . . .. Thus
the original sequence is ∆2 -constant, so can be fit to a quadratic.
Call the original sequence a n . Consider a n − n 2 . This gives 0, −1, −2, −3, . . .. That
sequence has closed formula 1 − n (starting at n 1) so we have a n − n 2 1 − n or
equivalently a n n 2 − n + 1.
2.3.3. This is a ∆3 -constant sequence. If we subtract off n 3 , we are left with 1, 3, 7, 13, 21, . . .,
the sequence from the previous question. Thus here the closed formula is n 3 +n 2 −n+1.
2.3.4. a n−1 ( n −1)2 +3( n −1) +4 n 2 + n +2. Thus a n − a n−1 2n +2. Note that this is linear
(arithmetic). We can check that we are correct. The sequence a n is 4, 8, 14, 22, 32, . . .
and the sequence of differences is thus 4, 6, 8, 10, . . . which agrees with 2n + 2 (if we
start at n 1).
2.3.6. No. The sequence of differences is the same as the original sequence so no differences
will be constant.
2.3.7. No. The sequence is geometric, and in fact has closed formula 2 · 3n . This is an
exponential function, which is not equal to any polynomial of any degree. If the
nth sequence of differences was constant, then the closed formula for the original
sequence would be a degree n polynomial.
2.4.3. We claim a n 4n works. Plug it in: 4n 3(4n−1 ) + 4(4n−2 ). This works - just simplify
the right hand side.
2.4.5. a n 13 n
54 + 12 n
5 (−1) .
2.4.6. The general solution is a n a + bn where a and b depend on the initial conditions.
(a) a n 1 + n.
(b) For example, we could have a 0 21 and a 1 22.
(c) For every x - take a 0 x − 9 and a 1 x − 8.
2.4.7. a n 19
7 (−2)
n + 79 5n .
2 · 2k+1 − 1
2k+2 − 1
2.5.2. Proof. Let P ( n ) be the statement “7n − 1 is a multiple of 6.” We will show P ( n ) is true
for all n ∈ N.
First we establish the base case, P (0). Since 70 − 1 0, and 0 is a multiple of 6, P (0) is
true.
Now for the inductive case. Assume P ( k ) holds for an arbitrary k ∈ N. That is, 7k − 1
is a multiple of 6, or in other words, 7k − 1 6j for some integer j. Now consider
Solutions for 2.5 253
7k+1 − 1:
7k+1 − 1 7k+1 − 7 + 6 by cleverness: −1 −7 + 6
7( 7 − 1) + 6
k
factor out a 7 from the first two terms
7(6j ) + 6 by the inductive hypothesis
6(7j + 1) factor out a 6
( k + 1)2 by factoring
2.5.5. Proof. Let P ( n ) be the statement 2n < n!. We will show P ( n ) is true for all n ≥ 4. First,
we check the base case and see that yes, 24 < 4! (as 16 < 24) so P (4) is true. Now for
the inductive case. Assume P ( k ) is true for an arbitrary k ≥ 4. That is, 2k < k!. Now
consider P ( k + 1): 2k+1 < ( k + 1)!. To prove this, we start with the left side and work
to the right side.
2k+1 2 · 2k
< 2 · k! by the inductive hypothesis
< ( k + 1) · k! since k + 1 > 2
( k + 1) !
Solutions for 2.5 254
Therefore 2k+1 < ( k + 1)! so we have established P ( k + 1). Thus by the principle of
mathematical induction P ( n ) is true for all n ≥ 4. qed
2.5.6. The only problem is that we never established the base case. Of course, when n 0,
0 + 3 , 0 + 7.
2.5.7. Proof. Let P ( n ) be the statement that n + 3 < n + 7. We will prove that P ( n ) is true
for all n ∈ N. First, note that the base case holds: 0 + 3 < 0 + 7. Now assume for
induction that P ( k ) is true. That is, k + 3 < k + 7. We must show that P ( k + 1) is
true. Now since k + 3 < k + 7, add 1 to both sides. This gives k + 3 + 1 < k + 7 + 1.
Regrouping ( k + 1) + 3 < ( k + 1) + 7. But this is simply P ( k + 1). Thus by the principle
of mathematical induction P ( n ) is true for all n ∈ N. qed
2.5.8. The problem here is that while P (0) is true, and while P ( k ) → P ( k + 1) for some values
of k, there is at least one value of k (namely k 99) when that implication fails. For a
valid proof by induction, P ( k ) → P ( k + 1) must be true for all values of k greater than
or equal to the base case.
2.5.10. We once again failed to establish the base case: when n 0, n 2 + n 0 which is even,
not odd.
2.5.11. Proof. Let P ( n ) be the statement “n 2 + n is even.” We will prove that P ( n ) is true for
all n ∈ N. First the base case: when n 0, we have n 2 + n 0 which is even, so P (0)
is true. Now suppose for induction that P ( k ) is true, that is, that k 2 + k is even. Now
consider the statement P ( k +1). Now ( k +1)2 + ( k +1) k 2 +2k +1+k +1 k 2 +k +2k +2.
By the inductive hypothesis, k 2 + k is even, and of course 2k + 2 is even. An even
plus an even is always even, so therefore ( k + 1)2 + ( k + 1) is even. Therefore by the
principle of mathematical induction, P ( n ) is true for all n ∈ N. qed
2.5.12. Further hint: the idea is to define the sequence so that a n is less than the distance
between the previous partial sum and 2. That way when you add it into the next
partial sum, the partial sum is still less than 2. You could do this ahead of time, or use
a clever P ( n ) in the induction proof. Let P ( n ) be the statement, “there is a sequence
of positive real numbers a 1 , a2 , a 3 , . . . , a n such that a 1 + a 2 + a 3 + · · · + a n < 2.” The
base case should be easy (just pick a 1 < 2). For the inductive case, you know that
a1 + a 2 + · · · + a k < 2 so you just need to argue that you can find some a k+1 small
enough to have a 1 + a 2 + · · · + a k + a k+1 < 2.
Solutions for 2.5 255
2.5.13. The base case should be easy; 0 is a power of 2. For the inductive case, you actually
want to use strong induction. Suppose k is either a power of 2 or can be written as
the sum of distinct powers of 2, for any k < n. Now if n is a power of 2, we are done.
If not, subtract the largest power of 2 from n possible. You get n − 2x , which is a
smaller number, in fact smaller than both n and 2x . Thus n − 2x is either a power of
2 or can be written as the sum of distinct powers of 2, but none of them are going to
be 2x , so the together with 2x we have written n as the sum of distinct powers of 2.
2.5.14. If n 2, this should work out (so their’s your base case). If we assume it works for
k ( k−1)
k people (that the number of handshakes is 2 , what happens if a k + 1st person
shows up. How many new handshakes take place? Now make this into a formal
induction argument.
Note, we have already proven this without using induction, but this is fun too.
2.5.15. When n 0, we get x 0 + x10 2 and when n 1, x + x1 is an integer, so the base case
holds. Now assume the result holds for all natural numbers n < k. In particular,
we know that x k−1 + x k−1
1
and x + x1 are both integers. Thus their product is also an
integer. But,
1 1 x k−1 x 1
x k−1 + x+ xk + + k−1 + k
x k−1 x x x x
1 1
x k + k + x k−2 + k−2
x x
Note also that x k−2 + x k−2
1
is an integer by the induction hypothesis, so we can conclude
that x + x k is an integer.
k 1
2.5.16. Here’s the idea: since every entry in Pascal’s Triangle is the sum of the two entries
above it, we can get the k + 1st row by adding up all the pairs of entry from the kth
row. But doing this uses each entry on the kth row twice. Thus each time we drop
to the next row, we double the total. Of course, row 0 has sum 1 20 (the base case).
Now try to make this precise with a formal induction proof. You will use the fact
that nk n−1
n−1
k−1 + k for the inductive case.
2.5.17. To see why this works, try it on a copy of Pascal’s triangle. We are adding up the
entries along a diagonal, starting with the 1 on the left hand side of the 4th row.
Suppose we add up the first 5 entries on this diagonal. The claim is that the sum is
the entry below and to the left of the last of these 5 entries. Note that if this is true,
and we instead add up the first 6 entries, we will need to add the entry one spot to the
right of the previous sum. But these two together give the entry below them, which
is below and left of the last of the 6 entries on the diagonal.
If you follow that, you can see what is going on. But it is not a great proof. A formal
induction proof is needed:
4 5 6 4+k 5+k
+ + +···+
for the inductive case, suppose P ( k ) is true. That is, 0 1 2 k k .
If we add 4+k+1
k+1 to both sides, we get
2.5.18. The idea here is that if we take the logarithm of a n , we can increase n by 1 if we
multiply by another a (inside the logarithm). This results in adding 1 more log( a ) to
the total.
Proof. Let P ( n ) be the statement log( a n ) n log( a ). The base case, P (2) is true,
because log( a 2 ) log( a · a ) log( a ) + log( a ) 2 log( a ), by the product rule for
logarithms.
Now assume, for induction, that P ( k ) is true. That is, log( a k ) k log( a ). Consider
log( a k+1 ). We have
with the last equality due to the inductive hypothesis. But this simplifies to ( k +
1) log( a ), establishing P ( k + 1).
Therefore by the principle of mathematical induction, P ( n ) is true for all n ≥ 2. qed
2.5.19. Hint: You are allowed to assume the base case. For the inductive case, group all
but the last function together as one sum of functions, then apply the usual sum of
derivatives rule, and then the inductive hypothesis.
2.5.20. Hint: For the inductive step, we know by the product rule for two functions that
Then use the inductive hypothesis on the first summand, and distribute.
Solutions for 2.6 257
2.6.6. a n n 2 + 4n − 1.
2.6.7. (a) The sequence of partial sums will be a degree 4 polynomial (its sequence of
differences will be the original sequence).
(b) The sequence of second differences will be a degree 1 polynomial - an arithmetic
sequence.
2.6.8. (a) 4, 6, 10, 16, 26, 42, . . ..
(b) No, taking differences gives the original sequence back, so the differences will
never be constant.
2.6.9. b n ( n + 3) n.
2.6.10. (a) 1, 2, 16, 68, 364, . . ..
(b) a n 73 (−2)n + 74 5n .
2.6.11. (a) a2 14. a 3 52.
(b) a n 16 (−2)n + 65 4n .
2.6.12. (a) On the first day, your 2 mini bunnies become 2 large bunnies. On day 2, your
two large bunnies produce 4 mini bunnies. On day 3, you have 4 mini bunnies
(produced by your 2 large bunnies) plus 6 large bunnies (your original 2 plus
the 4 newly matured bunnies). On day 4, you will have 12 mini bunnies (2 for
each of the 6 large bunnies) plus 10 large bunnies (your previous 6 plus the 4
newly matured). The sequence of total bunnies is 2, 2, 6, 10, 22, 42 . . . starting
with a0 2 and a 1 2.
Solutions for 2.6 258
(b) a n a n−1 + 2a n−2 . This is because the number of bunnies is equal to the number
of bunnies you had the previous day (both mini and large) plus 2 times the
number you had the day before that (since all bunnies you had 2 days ago are
now large and producing 2 new bunnies each).
(c) Using the characteristic root technique, we find a n a2n + b (−1)n , and we can
find a and b to give a n 34 2n + 23 (−1)n .
2.6.15. Hint: there are two base cases P (0) and P (1). Then, for the inductive case, assume
P ( k ) is true for all k < n. This allows you to assume a n−1 1 and a n−2 1. Apply the
recurrence relation.
2.6.16. Note that 1 20 ; this is your base case. Now suppose k can be written as the sum
of distinct powers of 2 for all 1 ≤ k < n. We can then write n as the sum of distinct
powers of 2 as follows: subtract the largest power of 2 less than n from n. That is,
write n 2 j + k for the largest possible j. But k is now less than n, and also less
than 2 j , so write k as the sum of distinct powers of 2 (we can do so by the inductive
hypothesis). Thus n can be written as the sum of distinct powers of 2 for all n ≥ 1.
2.6.17. Let P ( n ) be the statement, “every set containing n elements has 2n different subsets.”
We will show P ( n ) is true for all n ≥ 1.
Base case: Any set with 1 element { a } has exactly 2 subsets: the empty set and the
set itself. Thus the number of subsets is 2 21 . Thus P (1) is true.
Inductive case: Suppose P ( k ) is true for some arbitrary k ≥ 1. Thus every set
containing exactly k elements has 2k different subsets. Now consider a set containing
k + 1 elements: A { a 1 , a 2 , . . . , a k , a k+1 }. Any subset of A must either contain a k+1 or
not. In other words, a subset of A is just a subset of { a 1 , a 2 , . . . , a k } with or without
a k+1 . Thus there are 2k subsets of A which contain a k+1 and another 2k subsets of A
which do not contain a k+1 . This gives a total of 2k + 2k 2 · 2k 2k+1 subsets of A. But
our choice of A was arbitrary, so this works for any subset containing k + 1 elements,
so P ( k + 1) is true.
Solutions for 3.1 259
3.1.2. (a) P ∧ Q.
(b) P → ¬Q.
(c) Jack passed math or Jill passed math (or both).
(d) If Jack and Jill did not both pass math, then Jill did.
(e) (a) Nothing else.
(b) Jack did not pass math either.
3.1.3. (a) Three statements: P ∨ S, S → Q, (P ∨ Q ) → R. You could also connect the first
two with a ∧.
(b) He cannot be lying about all three sentences, so he is telling the truth.
(c) No matter what, Geoff wants ricotta. If he doesn’t have quail, then he must have
pepperoni but not sausage.
P Q (P ∨ Q ) → (P ∧ Q )
T T T
3.1.4. T F F
F T F
F F T
P Q ¬P ∧ (Q → P )
T T F
3.1.5. T F F If the statement is true, then both P and Q are false.
F T F
F F T
3.1.6. Hint: Like above, only now you will need 8 rows instead of just 4.
3.1.7. The rule is valid. To see this, make a truth table which contains P ∨ Q and ¬P (and
P and Q of course). Look at the truth value of Q in each of the rows that have P ∨ Q
and ¬P true.
3.1.8. The deduction rule is valid. Again, make a truth table containing the premises and
conclusion. Look at the rows for which the premises are true.
Solutions for 3.2 260
3.1.9. The rule is NOT valid. If you make a truth table containing the premises and
conclusion, there will be a row with both premises true but the conclusion false. For
example, if P and Q are false and R is true, then P ∧ Q is false, so (P ∧ Q ) → R is true.
Also ¬P is true, so ¬P ∨ ¬Q is true. However, ¬R is false.
3.2.2. Again, make two truth tables. The statements are logically equivalent.
3.2.4. (a) P ∧ Q.
(b) (P ∧ Q ) ∨ (Q ∨ ¬R).
(c) F. Or (P ∧ Q ) ∧ (R ∧ ¬R ).
(d) Either Sam is a woman and Chris is a man, or Chris is a woman.
(a) converse.
(b) implication.
(c) neither.
(d) implication.
(e) converse.
(f) converse.
(g) implication.
(h) converse.
(i) converse.
(j) converse (in fact, this IS the converse).
(k) implication (the statement is the contrapositive of the implication).
(l) neither.
3.2.6. There are many answers. It helps to assume that the statement is true and the converse
is NOT true. Think about what that means in the real world and then start saying it
in different ways. Some ideas: use necessary and sufficient language, use “only if,”
consider negations, use “or else” language.
Solutions for 3.3 261
3.3.4. If P ( x ) is true of every x, then in particular it is true of x 0 (or any fixed element
of the universe). So then there is definitely some x (namely 0) for which P ( x ) holds.
Thus ∀xP ( x ) → ∃xP ( x ) is always true.
The converse is not always true though. Consider the predicate x 5. So P ( x ) is true
if and only if x 5. Certainly it is true that ∃xP ( x ) (since we can take x 5), but false
that ∀xP ( x ).
3.3.5. (a) This says that everything has a square root (every element is the square of
something). This is true of the positive real numbers, and also of the complex
numbers. It is false of the natural numbers though, as for x 2 there is no
natural number y such that y 2 2.
(b) This asserts that between every pair of numbers there is some number strictly
between them. This is true of the rationals (and reals) but false of the integers.
If x 1 and y 2, then there is nothing we can take for z.
(c) Here we are saying that something is between every pair of numbers. For almost
every universe, this is false. In fact, if the universe contains {1, 2, 3, 4}, then no
matter what we take x to be, there will be a pair that x is NOT between. However,
the set {1, 2, 3} as our universe makes the statement true. Let x 2. Then no
matter what y and z we pick, if y < z, then 2 is between them.
Solutions for 3.4 262
3.3.6. Let P ( x, y ) be the predicate x < y. It is true that for all x there is some y greater than
it (since there are infinitely many numbers). However, there is not a natural number
y which is greater than every number x.
We cannot do the reverse of this though. If there is some y for which every x satisfies
P ( x, y ), then certainly for every x there is some y which satisfies P ( x, y ). The first is
saying we can find one y that works for every x. The second allows different y’s to
work for different x’s, but there is nothing preventing us from using the same y that
works for every x.
(b) The converse is false. That is, there is an integer n such that 8n is even but n is
odd. For example, consider n 3. Then 8n 24 which is even but n 3 is odd.
√ √
3.4.3. Proof. Suppose 3 were rational. Then 3 ba for some integers a and b , 0. Without
loss of generality, assume ba is reduced. Now
a2
3
b2
b23 a2
So a 2 is a multiple of 3. This can only happen if a is a multiple of 3, so a 3k for some
integer k. Then we have
b 2 3 9k 2
b 2 3k 2
So b 2 is a multiple of 3, making b a multiple of 3 as well. But this contradicts our
assumption that ba is in lowest terms. qed
Solutions for 3.4 263
ab (2k )(3j ) 6( k j )
(b) The converse is: for all integers a and b, if ab is a multiple of 6, then a is even
and b is a multiple of 3. This is false. Consider a 3 and b 10. Then ab 30
which is a multiple of 6, but a is not even and b is not divisible by 3.
Proof. Let n be an arbitrary integer, and suppose n is even. Then n 2k for some
integer k. Thus 5n 5 · 2k 10k 2(5k ). Since 5k is an integer, we see that 5n must
be even. This completes the proof. qed
3.4.6. Proof. Suppose, contrary to stipulation, that there are integers a, b and c such that
a 2 + b 2 c 2 but a and b are both odd. Then a 2k + 1 and b 2j + 1 for some integers
k and j. We then have
Proof. Suppose that each number only came up 6 or fewer times. So there are at most
six 1’s, six 2’s, and so on. That’s a total of 36 dice, so you must not have rolled all 40
dice. qed
3.4.8. We can have 9 dice without any four matching or all being different: three 1’s, three
2’s, three 3’s. We will prove that whenever you roll 10 dice, you will always get four
matching or all being different.
Proof. Suppose you roll 10 dice, but that there are NOT four matching rolls. This
means at most, there are three of any given value. If we only had three different
values, that would be only 9 dice, so there must be 4 different values, giving 4 dice
that are all different. qed
Equivalently,
7b 10a
But this is impossible as any power of 7 will be odd while any power of 10 will be
even. qed
3.4.10. Proof. Suppose there were integers x and y such that x 2 4y + 3. Now x 2 must be
odd, since 4y + 3 is odd. Since x 2 is odd, we know x must be odd as well. So x 2k + 1
for some integer k. Then x 2 4k 2 + 4k + 1 4( k 2 + k ) + 1. Therefore we have,
4( k 2 + k ) + 1 4y + 3
which implies
4( k 2 + k ) 4y + 2
and therefore
2( k 2 + k ) 2y + 1.
But this is a contradiction – the left hand side is even while the right hand side is
odd. qed
3.4.11. We must prove that for any integer greater than 3, if the integer is prime, then it is
one more or one less than a multiple of 6. The contrapositive of this statement is that
if a number greater than 3 is not one more or one less than a multiple of 6, then it is
not prime.
3.4.12. (a) Proof by contradiction. Start of proof: Assume, for the sake of contradiction,
that there are integers x and y such that x is a prime greater than 5 and x 6y +3.
End of proof: . . . this is a contradiction, so there are no such integers.
(b) Direct proof. Start of proof: Let n be an integers. Assume n is a multiple of 3.
End of proof: Therefore n can be written as the sum of consecutive integers.
(c) Proof by contrapositive. Start of proof: Let a and b be integers. Assume that a
and b are even. End of proof: Therefore a 2 + b 2 is even.
Solutions for 3.5 265
3.5.3. Yes. To see this, make a truth table for each statement and compare.
3.5.4. Make a truth table that includes all three statements in the argument:
P Q R P→Q P→R P → (Q ∧ R )
T T T T T T
T T F T F F
T F T F T F
T F F F F F
F T T T T T
F T F T T T
F F T T T T
F F F T T T
Notice that in every row for which both P → Q and P → R is true, so is P → (Q ∧ R).
Therefore, whenever the premises of the argument are true, so is the conclusion. In
other words, the deduction rule is valid.
3.5.5. (a) Negation: The power goes off and the food does not spoil.
Converse: If the food spoils, then the power went off.
Contrapositive: If the food does not spoil, then the power did not go off.
(b) Negation: The door is closed and the light is on.
Converse: If the light is off then the door is closed.
Contrapositive: If the light is on then the door is open.
(c) Negation: ∃x ( x < 1 ∧ x 2 ≥ 1)
Converse: ∀x ( x 2 < 1 → x < 1)
Contrapositive: ∀x ( x 2 ≥ 1 → x ≥ 1).
(d) Negation: There is a natural number n which is prime but not solitary.
Converse: For all natural numbers n, if n is solitary, then n is prime.
Contrapositive: For all natural numbers n, if n is not solitary then n is not prime.
(e) Negation: There is a function which is differentiable and not continuous.
Converse: For all functions f , if f is continuous then f is differentiable.
Solutions for 3.5 266
3.5.6. (a) The statement is true. If n is an even integer less than or equal to 7, then the
only way it could not be negative is if n was equal to 0, 2, 4, or 6.
(b) There is an integer n such that n is even and n ≤ 7 but n is not negative and
n < {0, 2, 4, 6}. This is false, since the original statement is true.
(c) For all integers n, if n is not negative and n < {0, 2, 4, 6} then n is odd or n > 7.
This is true, since the contrapositive is equivalent to the original statement
(which is true).
(d) For all integers n, if n is negative or n ∈ {0, 2, 4, 6} then n is even and n ≤ 7.
This is false. n −3 is a counterexample.
3.5.7. (a) For any number x, if it is the case that adding any number to x gives that number
back, then multiplying any number by x will give 0. This is true (of the integers
or the reals). The “if” part only holds if x 0, and in that case, anything times
x will be 0.
(b) The converse in words is this: for any number x, if everything times x is zero,
then everything added to x gives itself. Or in symbols: ∀x (∀z ( x · z 0) →
∀y ( x + y y )). The converse is true: the only number which when multiplied
by any other number gives 0 is x 0. And if x 0, then x + y y.
(c) The contrapositive in words is: for any number x, if there is some number which
when multiplied by x does not give zero, then there is some number which
when added to x does not give that number. In symbols: ∀x (∃z ( x · z , 0) →
Solutions for 3.5 267
(b) The converse is: for all integers n if 7n is odd, then n is odd. We will prove this
by contrapositive.
Proof. Let n be an integer. Assume n is not odd. Then n 2k for some integer
k. So 7n 14k 2(7k ) which is to say 7n is even. Therefore 7n is not odd. qed
3.5.11. (a) Suppose you only had 5 coins of each denomination. This means you have 5
pennies, 5 nickels, 5 dimes and 5 quarters. This is a total of 20 coins. But you
have more than 20 coins, so you must have more than 5 of at least one type.
(b) Suppose you have 22 coins, including 2k nickels, 2j dimes, and 2l quarters (so
an even number of each of these three types of coins). The number of pennies
you have will then be
22 − 2k − 2j − 2l 2(11 − k − j − l )
Solutions for 4.1 268
But this says that the number of pennies is also even (it is 2 times an integer).
Thus we have established the contrapositive of the statement, “If you have an
odd number of pennies then you have an odd number of at least one other coin
type.”
(c) You need 10 coins. You could have 3 pennies, 3 nickels, and 3 dimes. The 10th
coin must either be a quarter, giving you 4 coins that are all different, or else a
4th penny, nickel or dime. To prove this, assume you don’t have 4 coins that are
all the same or all different. In particular, this says that you only have 3 coin
types, and each of those types can only contain 3 coins, for a total of 9 coins,
which is less than 10.
4.1.2. It is possible for everyone to be friends with exactly 2 people. You could arrange the
5 people in a circle and say that everyone is friends with the two people on either
side of them (so you get the graph C 5 ). However, it is not possible for everyone to
be friends with 3 people. That would lead to a graph with an odd number of odd
degree vertices which is impossible since the sum of the degrees must be even.
4.1.3. Yes. For example, both graphs below contain 6 vertices, 7 edges, and have degrees
(2,2,2,2,3,3).
4.1.4. The graphs are not equal. For example, graph 1 has an edge { a, b } but graph 2 does
not have that edge. They are isomorphic. One possible isomorphism is f : G1 → G2
defined by f ( a ) d, f ( b ) c, f ( c ) e, f ( d ) b, f ( e ) a.
4.1.5. Three of the graphs are bipartite. The one which is not is C 7 (second from the right).
(b) This is not possible if we require the graphs to be connected. If not, we could
take C 8 as one graph and two copies of C 4 as the other.
(c) Not possible. If you have a graph with 5 vertices all of degree 4, then every
vertex must be adjacent to every other vertex. This is the graph K 5 .
(d) This is not possible. In fact, there is not even one graph with this property (such
a graph would have 5 · 3/2 7.5 edges).
4.2.2. G has 10 edges. It could be planar, and then it would have 6 faces.
4.2.3. Yes. According to Euler’s formula it would have 2 faces. It does. The only such graph
is C10 .
4.2.4. Say the last polyhedron has n edges, and also n vertices. The total number of edges
the polyhedron has then is (7 · 3 + 4 · 4 + n )/2 (37 + n )/2. In particular, we know
the last face must have an odd number of edges. By Euler’s formula, we have
v − (37 + n )/2 + 12 2, so v (17 + n )/2. But we also know that v 11 + n. Putting
these together we get n 5, so the last face is a pentagon.
4.2.5. Proof. Let P ( n ) be the statement, “every planar graph containing n edges satisfies
v − n + f 2.” We will show P ( n ) is true for all n ≥ 0.
Base case: there is only one graph with zero edges, namely a single isolated vertex.
In this case v 1, f 1 and e 0, so Euler’s formula holds.
Inductive case: Suppose P ( k ) is true for some arbitrary k ≥ 0. Now consider an
arbitrary graph containing k + 1 edges (and v vertices and f faces). No matter what
this graph looks like, we can remove a single edge to get a graph with k edges which
we can apply the inductive hypothesis to. There are two possibilities. First, the
edge we remove might be incident to a degree 1 vertex. In this case, also remove that
vertex. The smaller graph will now satisfy v −1−k + f 2 by the induction hypothesis
(removing the edge and vertex did not reduce the number of faces). Adding the edge
and vertex back gives v − ( k + 1) + f 2, as required. The second case is that the edge
we remove is incident to vertices of degree greater than one. In this case, removing
the edge will keep the number of vertices the same but reduce the number of faces
by one. So by the inductive hypothesis we will have v − k + f − 1 2. Adding the
edge back will give v − ( k + 1) + f 2 as needed.
Therefore, by the principle of mathematical induction, Euler’s formula holds for all
planar graphs. qed
Solutions for 4.3 270
4.2.6. Say the first component has v 1 vertices, e1 edges and f1 faces. The second graph has
v2 vertices, e2 edges and f2 faces. Thinking of each of these separately, we have
v1 − e1 + f1 2,
v 2 − e2 + f2 2.
Adding these two equations gives
v−e+ f 4
(since the graph has v v 1 + v 2 vertices, etc). However, the two components have
one common face (the outside of one of them must be contained in one of the faces
of the other) so in fact we get
v − e + f 3.
In general, a planar graph with k components will satisfy v − e + f 1 + k.
4.2.7. Proof. We know in any planar graph the number of faces f satisfies 3 f ≤ 2e since each
face is bounded by at least three edges, but each edge borders two faces. Combine
this with Euler’s formula:
v−e+ f 2
2e
v−e+ ≥2
3
3v − e ≥ 6
3v − 6 ≥ e.
qed
4.2.8. Proof. Suppose this were not the case. Then there would be a graph with v vertices,
each with degree 6 or more. At a minimum then, there would be 6v/2 3v edges,
so e ≥ 3v. By the previous exercise, we also have that e ≤ 3v − 6. But these two facts
are contradictory. qed
4.3.2. For example, K 6 . If the chromatic number is 6, then the graph is not planar; the
4-color theorem states that all planar graphs can be colored with 4 or fewer colors.
4.3.3. The chromatic numbers are 2, 3, 4, 5, and 3 respectively from left to right.
4.3.4. The cube can be represented as a planar graph and colored with two colors as follows:
Solutions for 4.3 271
R B
B R
R B
B R
Since it would be impossible to color the vertices with a single color, we see that the
cube has chromatic number 2 (it is bipartite).
4.3.5. The wheel graph below has this property. The outside of the wheel forms an odd
cycle, so requires 3 colors, the center of the wheel must be different than all the
outside vertices.
4.3.6. Proof. Let G be a graph with n vertices, maximal degree ∆(G ) and at least one vertex
of degree less than ∆(G ). Assume for the sake of induction that all graphs G0 with
fewer than n vertices and a vertex of degree less than ∆(G0) have chromatic number
less than ∆(G0).
Find a vertex of G with degree less than ∆(G ) and remove it. This forms a subgraph G0
which has n − 1 vertices. Also, since we removed edges, we know that ∆(G0) ≤ ∆(G).
If these maximal degrees are equal, then G0 must also have a vertex of degree less
than ∆(G0), since at least one of its vertices had one more edge in G. In this case, we
can apply our inductive hypothesis to produce a coloring of the vertices of G0 using
just ∆(G ) colors. If ∆(G0) < ∆(G ), then we also can easily find a proper coloring of
the vertices of G0 using just ∆(G ) colors by starting with any vertex and coloring it
and all of its neighbors differently, and then fanning out. Thus G0 has a proper vertex
coloring using just ∆(G )-many colors.
Now move back to G. Put the removed vertex back into the graph. Since it is adjacent
to at most ∆(G ) −1 other vertices, there will be one of the ∆(G) colors that is not present
among its neighbors, which we could use to color the newly inserted vertex. qed
4.3.7. If we drew a graph with each letter representing a vertex, and each edge connecting
two letters that were consecutive in the alphabet, we would have a graph containing
two vertices of degree 1 (A and Z) and the remaining 24 vertices all of degree 2 (for
example, D would be adjacent to both C and E). By Brooks’ theorem, this graph
has chromatic number at most 2, as that is the maximal degree in the graph and the
graph is not a complete graph or odd cycle. Thus only two boxes are needed.
4.3.8. Proof. Start with a single vertex of K6 , call it v 0 . There are five edges incident to v0 ,
and by the pigeonhole principle, three of these must be colored identically. Without
Solutions for 4.4 272
loss of generality, say these edges are ( v 0 , v 1 ), ( v0 , v 2 ) and ( v 0 , v3 ), and are colored
red. Consider the edges ( v1 , v 2 ), ( v 2 , v 3 ), and ( v 3 , v 1 ). If any of these are colored red,
we would have a monochromatic red triangle (it plus the two edges incident to v 0 ).
If they are all colored blue, then we have a monochromatic blue triangle (those three
edges). Either way, we are guaranteed to monochromatic triangle.
qed
4.4.3. When n is odd, K n contains an Euler circuit. This is because every vertex has degree
n − 1, so an odd n results in all degrees being even.
4.4.4. If both m and n are even, then K m,n has an Euler circuit. When both are odd, there is
no Euler path or circuit. If one is 2 and the other is odd, then there is an Euler path
but not an Euler circuit.
4.4.5. If we build one bridge, we can have an Euler path. Two bridges must be built for an
Euler circuit.
4.4.6. We are looking for a Hamiltonian cycle, and this graph does have one:
Solutions for 4.6 273
4.6.2. The first (and third) graphs contain an Euler path. All the graphs are planar.
4.6.6. (a) There were 24 couples: 6 choices for the girl and 4 choices for the boy.
10
(b) There were 45 couples: 2 since we must choose two of the 10 people to dance
together.
(c) For part (a), we are counting the number of edges in K 4,6 . In part (b) we count
the edges of K10 .
4.6.7. Yes, as long as n is even. If n were odd, then corresponding graph would have an
odd number of odd degree vertices, which is impossible.
4.6.8. (a) No. The 9 triangles each contribute 3 edges, and the 6 pentagons contribute 5
edges. This gives a total of 57, which is exactly twice the number of edges, since
each edge borders exactly 2 faces. But 57 is odd, so this is impossible.
(b) Now adding up all the edges of all the 16 polygons gives a total of 64, meaning
there would be 32 edges in the polyhedron. We can then use Euler’s formula
v − e + f 2 to deduce that there must be 18 vertices.
(c) If you add up all the vertices from each polygon separately, we get a total of 64.
This is not divisible by 3, so it cannot be that each vertex belongs to exactly 3
faces. Could they all belong to 4 faces? That would mean there were 64/4 16
vertices, but we know from Euler’s formula that there must be 18 vertices. We
can write 64 3x + 4y and solve for x and y (as integers). We get that there
Solutions for 4.6 274
must be 10 vertices with degree 4 and 8 with degree 3. (Note the number of
faces joined at a vertex is equal to its degree in graph theoretic terms.)
4.6.9. No. Every polyhedron can be represented as a planar graph, and the Four Color
Theorem says that every planar graph has chromatic number at most 4.
4.6.10. K n,n has n 2 edges. The graph will have an Euler circuit when n is even. The graph
will be planar only when n < 3.
4.6.11. G has 8 edges (since the sum of the degrees is 16). If G is planar, then it will have 4
faces (since 6 − 8 + 4 2). G does not have an Euler path since there are more than 2
vertices of odd degree.
4.6.12. 7 colors. Thus K 7 is not planar (by the contrapositive of the Four Color Theorem).
4.6.13. The chromatic number of K3,4 is 2, since the graph is bipartite. You cannot say whether
the graph is planar based on this coloring (the converse of the Four Color Theorem
is not true). In fact, the graph is not planar, since it contains K3,3 as a subgraph.
4.6.14. For all these questions, we are really coloring the vertices of a graph. You get the
graph by first drawing a planar representation of the polyhedron and then taking
its planar dual: put a vertex in the center of each face (including the outside) and
connect two vertices if their faces share an edge.
(a) Since the planar dual of a dodecahedron contains a 5-wheel, it’s chromatic
number is at least 4. Alternatively, suppose you could color the faces using 3
colors without any two adjacent faces colored the same. Take any face and color
it blue. The 5 pentagons bordering this blue pentagon cannot be colored blue.
Color the first one red. Its two neighbors (adjacent to the blue pentagon) get
colored green. The remaining 2 cannot be blue or green, but also cannot both
be red since they are adjacent to each other. Thus a 4th color is needed.
(b) The planar dual of the dodecahedron is itself a planar graph. Thus by the 4-
color theorem, it can be colored using only 4 colors without two adjacent vertices
(corresponding to the faces of the polyhedron) being colored identically.
(c) The cube can be properly 3-colored. Color the “top” and “bottom” red, the
“front” and “back” blue, and the “left” and “right” green.
4.6.16. (a) The graph does have an Euler path, but not an Euler circuit. There are exactly
two vertices with odd degree. The path starts at one and ends at the other.
(b) The graph is planar. Even though as it is drawn edges cross, it is easy to redraw
it without edges crossing.
(c) The graph is not bipartite (there is an odd cycle), nor complete.
(d) The chromatic number of the graph is 3.
Solutions for A.1 275
4.6.18. The first and third graphs have a matching, shown in bold (there are other matchings
as well). The middle graph does not have a matching. If you look at the three circled
vertices, you see that they only have two neighbors, which violates the matching
condition | N (S )| ≥ S (the three circled vertices form the set S).
A.1.2. (a) 0, 4, 4, 4, 4, 4, . . ..
(b) 1, 4, 16, 64, 256, . . ..
(c) 0, 1, −1, 1, −1, 1, −1, . . ..
(d) 0, 3, −6, 9, −12, 15, −18, . . ..
(e) 1, 3, 6, 9, 12, 15, . . ..
1 2
A.1.3. (a) The second derivative of is which expands to 2 + 6x + 12x 2 +
1−x (1 − x )3
20x 3 + 30x 4 + · · · . Dividing by 2 gives the generating function for the triangular
numbers.
(b) Compute A − xA and you get 1 + 2x + 3x 2 + 4x 3 + · · · which can be written as
1
. Solving for A gives the correct generating function.
(1 − x )2
(c) The triangular numbers are the sum of the first n numbers 1, 2, 3, 4, . . .. To
1
get the sequence of partial sums, we multiply by 1−x . This gives the correct
generating function again.
A.1.10. Hint: you should “multiply” the two sequences. Answer: 158.
1
A.1.11. Starting with 1−x 1 + x + x 2 + x 3 + · · · , we can take derivatives of both sides, given
1
(1−x )2
1 + 2x + 3x 2 + · · · . By the definition of generating functions, this says that
1
generates the sequence 1, 2, 3. . . . You can also find this using differencing or by
(1−x )2
multiplying.
1
A.1.12. (a) (1−x 2 )2
.
1
(b) (1+x )2
.
3x
(c) (1−x )2
.
3x
(d) (1−x )3
. (partial sums).
Solutions for A.2 277
A.1.13. (a) 0, 0, 1, 1, 2, 3, 5, 8, . . ..
(b) 1, 0, 1, 0, 2, 0, 3, 0, 5, 0, 8, 0, . . ..
(c) 1, 3, 18, 81, 405, . . ..
(d) 1, 2, 4, 7, 12, 20, . . ..
1
A.1.14. 1+2x .
x3
A.1.15. (1−x )2
+ 1
1−x .
A.2.2. Proof. Assume a | b and a | c. This means that b and c are both multiples of a, so
b am and c an for integers m and n. Then b + c am + an a ( m + n ), so b + c
is a multiple of a, or equivalently, a | b + c. Similarly, b − c am − an a ( m − n ), so
b − c is a multiple of a, which is to say a | b − c. qed
A.2.3. { . . . , −8, −4, 0, 4, 8, 12, . . . }, { . . . , −7, −3, 1, 5, 9, 13, . . . }, { . . . , −6, −2, 2, 6, 10, 14, . . . },
and { . . . , −5, −1, 3, 7, 11, 15, . . . }.
a − c b + kn − ( d + jn ) b − d + ( k − j ) n.
A.2.6. For all of these, just plug in all integers between 0 and the modulus to see which, if
any, work.
Solutions for A.2 278
(a) No solutions.
(b) x 3.
(c) x 2, x 5, x 8.
(d) No solutions.
(e) No solutions.
(f) x 3.
A.2.8. We must solve 7x+5 ≡ 2 (mod 11). This gives x ≡ 9 (mod 11). In general, x 9+11k,
but when you divide any such x by 11, the remainder will be 9.
A.2.10. First, solve the Diophantine equation 13x + 20y 2. The general solution is x
−6 − 20k and y 4 + 13k. Now if k 0, this correspond to filling the 20 oz. bottle 4
times, and emptying the 13 oz. bottle 6 times, which would require 80 oz. of water.
Increasing k would require considerably more water. Perhaps k −1 would be
better? Then we would have x −6 + 20 14 and y 4 − 13 −11, which describes
the solution where we fill the 13 oz. bottle 14 times, and empty the 20 oz. bottle 11
times. This would require 182 oz. of water. Thus the most efficient procedure is to
repeatedly fill the 20 oz bottle, emptying it into the 13 oz bottle, and discarding full
13 oz. bottles, which requires 80 oz. of water.
Index
279
Index 280
statement, 133
subgraph, 176
subset, 7
sufficient condition, 146
summation notation, 98
surjection, 16
syllogism, 148
tautology, 137
telescoping, 105
tetrahedron, 183
Tower of Hanoi, 85
tree, 176
triangular numbers, 90, 95
truth table, 134
truth value, 134
union, 5, 9
universal quantifier, 5, 150
valid, 131
Venn diagram, 11, 28
vertex coloring, 176, 185
vertices, 170
Vizing’s Theorem, 190
weight, of a string, 34
words, 24
List of Symbols
∩ intersection, 5
∪ union, 5, 9
\ set minus, 5, 9
∈ element of, 3, 5
: such that, 4, 5
{, } braces to enclose elements of a set, 5
⊆ subset of, 5
⊂ proper subset of, 5
× Cartesian product, 5
|A| cardinality (size) of A, 5
A complement of A, 5
282
List of Symbols 283
! factorial, 40
n
k
number of combinations of k out of n objects; n choose k, 36, 42
P ( n, k ) number of permutations of k out of n objects, 41
isomorphic to, 171
Kn complete graph on n vertices, 173, 174
K m,n complete bipartite graph with groups of size m and n, 174
Cn cycle with n vertices, 174
Pn path with n vertices, 174
χ (G ) the chromatic number of G, 184
χ0 ( G ) the chromatic index of G, 188
∆(G ) the maximum degree in G, 188
N (S ) set of neighbors of vertices in S, 196
m|n m divides n, 216
≡ congruent (modulo n, perhaps), 219
(mod n ) modulo (or mod) n, 219