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Acknowledgements
Thanks to Daniel Dimijian, Scott Jeffreys, Dan Seabold, C.R. Sincock, Pete Terlecky, Zoran Sunik, and David Wayne for their helpful input during the creation of this book.
CONNECT WITH DR. STEVE WARNER

Pure Mathematics for Beginners

A Rigorous Introduction to Logic, Set Theory, Abstract Algebra, Number Theory, Real Analysis, Topology, Complex Analysis, and Linear Algebra
--------
Dr. Steve Warner

𝑁𝛿⨀(𝑎)
𝛿

𝑁𝜖 (𝐿)
𝜖

© 2018, All Rights Reserved
iii

Table of Contents

Introduction

7

For students

7

For instructors

8

Lesson 1 – Logic: Statements and Truth

9

Statements with Words

9

Statements with Symbols

10

Truth Tables

12

Problem Set 1

16

Lesson 2 – Set Theory: Sets and Subsets

19

Describing Sets

19

Subsets

20

Unions and Intersections

24

Problem Set 2

28

Lesson 3 – Abstract Algebra: Semigroups, Monoids, and Groups

30

Binary Operations and Closure

30

Semigroups and Associativity

32

Monoids and Identity

34

Groups and Inverses

34

Problem Set 3

36

Lesson 4 – Number Theory: The Ring of Integers

38

Rings and Distributivity

38

Divisibility

41

Induction

43

Problem Set 4

48

Lesson 5 – Real Analysis: The Complete Ordered Field of Reals

50

Fields

50

Ordered Rings and Fields

52

Why Isn’t ℚ enough?

56

Completeness

58

Problem Set 5

62

Lesson 6 – Topology: The Topology of ℝ

64

Intervals of Real Numbers

64

Operations on Sets

66

Open and Closed Sets

70

Problem Set 6

76

Lesson 7 – Complex Analysis: The Field of Complex Numbers

78

A Limitation of the Reals

78

The Complex Field

78

Absolute Value and Distance

82

Basic Topology of ℂ

85

Problem Set 7

90

iv

Lesson 8 – Linear Algebra: Vector Spaces

93

Vector Spaces Over Fields

93

Subspaces

98

Bases

101

Problem Set 8

105

Lesson 9 – Logic: Logical Arguments

107

Statements and Substatements

107

Logical Equivalence

108

Validity in Sentential Logic

111

Problem Set 9

116

Lesson 10 – Set Theory: Relations and Functions

118

Relations

118

Equivalence Relations and Partitions

121

Orderings

124

Functions

124

Equinumerosity

130

Problem Set 10

135

Lesson 11 – Abstract Algebra: Structures and Homomorphisms

137

Structures and Substructures

137

Homomorphisms

142

Images and Kernels

146

Normal Subgroups and Ring Ideals

147

Problem Set 11

150

Lesson 12 – Number Theory: Primes, GCD, and LCM

152

Prime Numbers

152

The Division Algorithm

155

GCD and LCM

159

Problem Set 12

167

Lesson 13 – Real Analysis: Limits and Continuity

169

Strips and Rectangles

169

Limits and Continuity

172

Equivalent Definitions of Limits and Continuity

175

Basic Examples

177

Limit and Continuity Theorems

181

Limits Involving Infinity

183

One-sided Limits

185

Problem Set 13

186

Lesson 14 – Topology: Spaces and Homeomorphisms

189

Topological Spaces

189

Bases

192

Types of Topological Spaces

197

Continuous Functions and Homeomorphisms

204

Problem Set 14

210

v

Lesson 15 – Complex Analysis: Complex Valued Functions

212

The Unit Circle

212

Exponential Form of a Complex Number

216

Functions of a Complex Variable

218

Limits and Continuity

223

The Reimann Sphere

228

Problem Set 15

230

Lesson 16 – Linear Algebra: Linear Transformations

234

Linear Transformations

234

Matrices

239

The Matrix of a Linear Transformation

242

Images and Kernels

244

Eigenvalues and Eigenvectors

247

Problem Set 16

253

Index

255

About the Author

259

Books by Dr. Steve Warner

260

vi

INTRODUCTION PURE MATHEMATICS
This book was written to provide a basic but rigorous introduction to pure mathematics, while exposing students to a wide range of mathematical topics in logic, set theory, abstract algebra, number theory, real analysis, topology, complex analysis, and linear algebra.
For students: There are no prerequisites for this book. The content is completely self-contained. Students with a bit of mathematical knowledge may have an easier time getting through some of the material, but no such knowledge is necessary to read this book.
More important than mathematical knowledge is “mathematical maturity.” Although there is no single agreed upon definition of mathematical maturity, one reasonable way to define it is as “one’s ability to analyze, understand, and communicate mathematics.” A student with a higher level of mathematical maturity will be able to move through this book more quickly than a student with a lower level of mathematical maturity.
Whether your level of mathematical maturity is low or high, if you are just starting out in pure mathematics, then you’re in the right place. If you read this book the “right way,” then your level of mathematical maturity will continually be increasing. This increased level of mathematical maturity will not only help you to succeed in advanced math courses, but it will improve your general problem solving and reasoning skills. This will make it easier to improve your performance in college, in your professional life, and on standardized tests such as the SAT, ACT, GRE, and GMAT.
So, what is the “right way” to read this book? Simply reading each lesson from end to end without any further thought and analysis is not the best way to read the book. You will need to put in some effort to have the best chance of absorbing and retaining the material. When a new theorem is presented, don’t just jump right to the proof and read it. Think about what the theorem is saying. Try to describe it in your own words. Do you believe that it is true? If you do believe it, can you give a convincing argument that it is true? If you do not believe that it is true, try to come up with an example that shows it is false, and then figure out why your example does not contradict the theorem. Pick up a pen or pencil. Draw some pictures, come up with your own examples, and try to write your own proof.
You may find that this book goes into more detail than other math books when explaining examples, discussing concepts, and proving theorems. This was done so that any student can read this book, and not just students that are naturally gifted in mathematics. So, it is up to you as the student to try to answer questions before they are answered for you. When a new definition is given, try to think of your own examples before looking at those presented in the book. And when the book provides an example, do not just accept that it satisfies the given definition. Convince yourself. Prove it.
Each lesson is followed by a Problem Set. The problems in each Problem Set have been organized into five levels, Level 1 problems being considered the easiest, and Level 5 problems being considered the most difficult. If you want to get just a small taste of pure mathematics, then you can work on the easier problems. If you want to achieve a deeper understanding of the material, take some time to struggle with the harder problems.
7

For instructors: This book can be used for a wide range of courses. Although the lessons can be taught in the order presented, they do not need to be. The lessons cycle twice among eight subject areas: logic, set theory, abstract algebra, number theory, real analysis, topology, complex analysis, and linear algebra.
Lessons 1 through 8 give only the most basic material in each of these subjects. Therefore, an instructor that wants to give a brief glimpse into a wide variety of topics might want to cover just the first eight lessons in their course.
Lessons 9 through 16 cover material in each subject area that the author believes is fundamental to a deep understanding of that particular subject.
For a first course in higher mathematics, a high-quality curriculum can be created by choosing among the 16 lessons contained in this book.
As an example, an introductory course focusing on logic, set theory, and real analysis might cover Lessons 1, 2, 5, 9, 10, and 13. Lessons 1 and 9 cover basic sentential logic and proof theory, Lessons 2 and 10 cover basic set theory including relations, functions, and equinumerosity, and Lessons 5 and 13 cover basic real analysis up through a rigorous treatment of limits and continuity. The first three lessons are quite basic, while the latter three lessons are at an intermediate level. Instructors that do not like the idea of leaving a topic and then coming back to it later can cover the lessons in the following order without issue: 1, 9, 2, 10, 5, and 13.
As another example, a course focusing on algebraic structures might cover Lessons 2, 3, 4, 5, 10, and 11. As mentioned in the previous paragraph, Lessons 2 and 10 cover basic set theory. In addition, Lessons 3, 4, 5, and 11 cover semigroups, monoids, groups, rings, and fields. Lesson 4, in addition to a preliminary discussion on rings, also covers divisibility and the principle of mathematical induction. Similarly, Lesson 5, in addition to a preliminary discussion on fields, provides a development of the complete ordered field of real numbers. These topics can be included or omitted, as desired. Instructors that would also like to incorporate vector spaces can include part or all of Lesson 8.
The author strongly recommends covering Lesson 2 in any introductory pure math course. This lesson fixes some basic set theoretical notation that is used throughout the book and includes some important exposition to help students develop strong proof writing skills as quickly as possible.
The author welcomes all feedback from instructors. Any suggestions will be considered for future editions of the book. The author would also love to hear about the various courses that are created using these lessons. Feel free to email Dr. Steve Warner with any feedback at
steve@SATPrepGet800.com
8

LESSON 1 – LOGIC STATEMENTS AND TRUTH
Statements with Words
A statement (or proposition) is a sentence that can be true or false, but not both simultaneously.
Example 1.1: “Mary is awake” is a statement because at any given time either Mary is awake or Mary is not awake (also known as Mary being asleep), and Mary cannot be both awake and asleep at the same time.
Example 1.2: The sentence “Wake up!” is not a statement because it cannot be true or false.
An atomic statement expresses a single idea. The statement “Mary is awake” that we discussed above is an example of an atomic statement. Let’s look at a few more examples.
Example 1.3: The following sentences are atomic statements: 1. 17 is a prime number. 2. George Washington was the first president of the United States. 3. 5 > 6. 4. David is left-handed.
Sentences 1 and 2 above are true, and sentence 3 is false. We can’t say for certain whether sentence 4 is true or false without knowing who David is. However, it is either true or false. It follows that each of the four sentences above are atomic statements.
We use logical connectives to form compound statements. The most commonly used logical connectives are “and,” “or,” “if…then,” “if and only if,” and “not.”
Example 1.4: The following sentences are compound statements: 1. 17 is a prime number and 0 = 1. 2. Michael is holding a pen or water is a liquid. 3. If Joanna has a cat, then fish have lungs. 4. Albert Einstein is alive today if and only if 5 + 7 = 12. 5. 16 is not a perfect square.
Sentence 1 above uses the logical connective “and.” Since the statement “0 = 1” is false, it follows that sentence 1 is false. It does not matter that the statement “17 is a prime number” is true. In fact, “T and F” is always F.
Sentence 2 uses the logical connective “or.” Since the statement “water is a liquid” is true, it follows that sentence 2 is true. It does not even matter whether Michael is holding a pen. In fact, “T or T” is always true and “F or T” is always T.
9

It’s worth pausing for a moment to note that in the English language the word “or” has two possible meanings. There is an “inclusive or” and an “exclusive or.” The “inclusive or” is true when both statements are true, whereas the “exclusive or” is false when both statements are true. In mathematics, by default, we always use the “inclusive or” unless we are told to do otherwise. To some extent, this is an arbitrary choice that mathematicians have agreed upon. However, it can be argued that it is the better choice since it is used more often and it is easier to work with. Note that we were assuming use of the “inclusive or” in the last paragraph when we said, “In fact, “T or T” is always true.” See Problem 4 below for more on the “exclusive or.”
Sentence 3 uses the logical connective “if…then.” The statement “fish have lungs” is false. We need to know whether Joanna has a cat in order to figure out the truth value of sentence 3. If Joanna does have a cat, then sentence 3 is false (“if T, then F” is always F). If Joanna does not have a cat, then sentence 3 is true (“if F, then F” is always T).
Sentence 4 uses the logical connective “if and only if.” Since the two atomic statements have different truth values, it follows that sentence 4 is false. In fact, “F if and only if T” is always F.
Sentence 5 uses the logical connective “not.” Since the statement “16 is a perfect square” is true, it follows that sentence 5 is false. In fact, “not T” is always F.
Notes: (1) The logical connectives “and,” “or,” “if…then,” and “if and only if,” are called binary connectives because they join two statements (the prefix “bi” means “two”).
(2) The logical connective “not” is called a unary connective because it is applied to just a single statement (“unary” means “acting on a single element”).
Example 1.5: The following sentences are not statements: 1. Are you happy? 2. Go away! 3. 𝑥 − 5 = 7 4. This sentence is false. 5. This sentence is true.
Sentence 1 above is a question and sentence 2 is a command. Sentence 3 has an unknown variable – it can be turned into a statement by assigning a value to the variable. Sentences 4 and 5 are self-referential (they refer to themselves). They can be neither true nor false. Sentence 4 is called the Liar’s paradox and sentence 5 is called a vacuous affirmation.
Statements with Symbols
We will use letters such as 𝑝, 𝑞, 𝑟, and 𝑠 to denote atomic statements. We sometimes call these letters propositional variables, and we will generally assign a truth value of T (for true) or F (for false) to each propositional variable. Formally, we define a truth assignment of a list of propositional variables to be a choice of T or F for each propositional variable in the list.
10

We use the symbols ∧, ∨, →, ↔, and ¬ for the most common logical connectives. The truth value of a compound statement is determined by the truth values of its atomic parts together with applying the following rules for the connectives.
• 𝑝 ∧ 𝑞 is called the conjunction of 𝑝 and 𝑞. It is pronounced “𝑝 and 𝑞.” 𝑝 ∧ 𝑞 is true when both 𝑝 and 𝑞 are true, and it is false otherwise.
• 𝑝 ∨ 𝑞 is called the disjunction of 𝑝 and 𝑞. It is pronounced “𝑝 or 𝑞.” 𝑝 ∨ 𝑞 is true when 𝑝 or 𝑞 (or both) are true, and it is false when 𝑝 and 𝑞 are both false.
• 𝑝 → 𝑞 is called a conditional or implication. It is pronounced “if 𝑝, then 𝑞” or “𝑝 implies 𝑞.” 𝑝 → 𝑞 is true when 𝑝 is false or 𝑞 is true (or both), and it is false when 𝑝 is true and 𝑞 is false.
• 𝑝 ↔ 𝑞 is called a biconditional. It is pronounced “𝑝 if and only if 𝑞.” 𝑝 ↔ 𝑞 is true when 𝑝 and 𝑞 have the same truth value (both true or both false), and it is false when 𝑝 and 𝑞 have opposite truth values (one true and the other false).
• ¬𝑝 is called the negation of 𝑝. It is pronounced “not 𝑝.” ¬𝑝 is true when 𝑝 is false, and it is false when 𝑝 is true (𝑝 and ¬𝑝 have opposite truth values.)
Example 1.6: Let 𝑝 represent the statement “Fish can swim,” and let 𝑞 represent the statement “7 < 3.” Note that 𝑝 is true and 𝑞 is false.
1. 𝑝 ∧ 𝑞 represents “Fish can swim and 7 < 3.” Since 𝑞 is false, it follows that 𝑝 ∧ 𝑞 is false. 2. 𝑝 ∨ 𝑞 represents “Fish can swim or 7 < 3.” Since 𝑝 is true, it follows that 𝑝 ∨ 𝑞 is true. 3. 𝑝 → 𝑞 represents “If fish can swim, then 7 < 3.” Since 𝑝 is true and 𝑞 is false, 𝑝 → 𝑞 is false. 4. 𝑝 ↔ 𝑞 represents “Fish can swim if and only if 7 < 3.” Since 𝑝 is true and 𝑞 is false, 𝑝 ↔ 𝑞 is
false. 5. ¬𝑞 represents the statement “7 is not less than 3.” This is equivalent to “7 is greater than or
equal to 3,” or equivalently, “7 ≥ 3.” Since 𝑞 is false, ¬𝑞 is true. 6. ¬𝑝 ∨ 𝑞 represents the statement “Fish cannot swim or 7 < 3.” Since ¬𝑝 and 𝑞 are both false,
¬𝑝 ∨ 𝑞 is false. Note that ¬𝑝 ∨ 𝑞 always means (¬𝑝) ∨ 𝑞. In general, without parentheses present, we always apply negation before any of the other connectives. 7. ¬(𝑝 ∨ 𝑞) represents the statement “It is not the case that either fish can swim or 7 < 3.” This can also be stated as “Neither can fish swim nor is 7 less than 3.” Since 𝑝 ∨ 𝑞 is true (see 2 above), ¬(𝑝 ∨ 𝑞) is false.
8. ¬𝑝 ∧ ¬𝑞 represents the statement “Fish cannot swim and 7 is not less than 3.” This statement can also be stated as “Neither can fish swim nor is 7 less than 3.” Since this is the same statement as in 7 above, it should follow that ¬𝑝 ∧ ¬𝑞 is equivalent to ¬(𝑝 ∨ 𝑞). After completing this lesson, you will be able to verify this. For now, let’s observe that since ¬𝑝 is false, it follows that ¬𝑝 ∧ ¬𝑞 is false. This agrees with the truth value we got in 7. (Note: The equivalence of ¬𝑝 ∧ ¬𝑞 with ¬(𝑝 ∨ 𝑞) is one of De Morgan’s laws. These laws will be explored further in Lesson 9. See also Problem 3 below.)
11

Truth Tables
A truth table can be used to display the possible truth values of a compound statement. We start by labelling the columns of the table with the propositional variables that appear in the statement, followed by the statement itself. We then use the rows to run through every possible combination of truth values for the propositional variables followed by the resulting truth values for the compound statement. Let’s look at the truth tables for the five most common logical connectives.

𝒑

𝒒 𝒑∧𝒒

𝐓

𝐓

𝐓

𝐓

𝐅

𝐅

𝐅

𝐓

𝐅

𝐅

𝐅

𝐅

𝒑

𝒒 𝒑∨𝒒

𝐓

𝐓

𝐓

𝐓

𝐅

𝐓

𝐅

𝐓

𝐓

𝐅

𝐅

𝐅

𝒑

𝒒 𝒑→𝒒

𝐓

𝐓

𝐓

𝐓

𝐅

𝐅

𝐅

𝐓

𝐓

𝐅

𝐅

𝐓

𝒑 𝒒 𝒑↔𝒒

𝐓 𝐓

𝐓

𝐓 𝐅

𝐅

𝐅 𝐓

𝐅

𝐅 𝐅

𝐓

𝒑 ¬𝒑 𝐓 𝐅 𝐅 𝐓

We can use these five truth tables to compute the truth values of compound statements involving the five basic logical connectives.
Note: For statements involving just 1 propositional variable (such as ¬𝑝), the truth table requires 2 rows, 1 for each truth assignment of 𝑝 ( T or F ).

For statements involving 2 propositional variables (such as 𝑝 ∧ 𝑞), the truth table requires 2 ⋅ 2 = 4 (or 22 = 4) rows, as there are 4 possible combinations for truth assignments of 𝑝 and 𝑞 ( TT, TF, FT, FF ).
In general, for a statement involving 𝑛 propositional variables, the truth table will require 2𝑛 rows. For example, if we want to build an entire truth table for ¬𝑝 ∨ (¬𝑞 → 𝑟), we will need 23 = 2 ⋅ 2 ⋅ 2 = 8 rows in the truth table. We will create the truth table for this statement in Example 1.8 below (see the third solution).

Example 1.7: If 𝑝 is true and 𝑞 is false, then we can compute the truth value of 𝑝 ∧ 𝑞 by looking at the second row of the truth table for the conjunction.

𝒑

𝒒 𝒑∧𝒒

𝐓

𝐓

𝐓

𝐓

𝐅

𝐅

𝐅

𝐓

𝐅

𝐅

𝐅

𝐅

We see from the highlighted row that 𝑝 ∧ 𝑞 ≡ T ∧ F ≡ 𝐅.

12

Note: Here the symbol ≡ can be read “is logically equivalent to.” So, we see that if 𝑝 is true and 𝑞 is false, then 𝑝 ∧ 𝑞 is logically equivalent to F, or more simply, 𝑝 ∧ 𝑞 is false.

Example 1.8: Let 𝑝, 𝑞, and 𝑟 be propositional variables with 𝑝 and 𝑞 true, and 𝑟 false. Let’s compute the truth value of ¬𝑝 ∨ (¬𝑞 → 𝑟).

Solution: We have ¬𝑝 ∨ (¬𝑞 → 𝑟) ≡ ¬T ∨ (¬T → F) ≡ F ∨ (F → F) ≡ F ∨ T ≡ 𝐓.

Notes: (1) For the first equivalence, we simply replaced the propositional variables by their given truth values. We replaced 𝑝 and 𝑞 by T, and we replaced 𝑟 by F.

(2) For the second equivalence, we used the first row of the truth table for the 𝒑 ¬𝒑

negation (drawn to the right for your convenience).

𝐓 𝐅

We see from the highlighted row that ¬T ≡ F. We applied this result twice.

𝐅 𝐓

(3) For the third equivalence, we used the fourth row of the truth table for the conditional.

𝒑 𝒒 𝒑→𝒒

𝐓 𝐓

𝐓

𝐓 𝐅

𝐅

𝐅 𝐓

𝐓

𝐅 𝐅

𝐓

We see from the highlighted row that F → F ≡ T.

(4) For the last equivalence, we used the third row of the truth table for the disjunction.

𝒑

𝒒 𝒑∨𝒒

𝐓

𝐓

𝐓

𝐓

𝐅

𝐓

𝐅

𝐓

𝐓

𝐅

𝐅

𝐅

We see from the highlighted row that F ∨ T ≡ T.
(5) We can save a little time by immediately replacing the negation of a propositional variable by its truth value (which will be the opposite truth value of the propositional variable). For example, since 𝑝 has truth value T, we can replace ¬𝑝 by F. The faster solution would look like this:
¬𝑝 ∨ (¬𝑞 → 𝑟) ≡ F ∨ (F → F) ≡ F ∨ T ≡ 𝐓.
Quicker solution: Since 𝑞 has truth value T, it follows that ¬𝑞 has truth value F. So, ¬𝑞 → 𝑟 has truth value T. Finally, ¬𝑝 ∨ (¬𝑞 → 𝑟) must then have truth value T.
Notes: (1) Symbolically, we can write the following: ¬𝑝 ∨ (¬𝑞 → 𝑟) ≡ ¬𝑝 ∨ (¬T → 𝑟) ≡ ¬𝑝 ∨ (F → 𝑟) ≡ ¬𝑝 ∨ T ≡ 𝐓

13

(2) We can display this reasoning visually as follows:
¬𝑝 ∨ (¬𝑞 → 𝑟) T
F T
𝐓 The vertical lines have just been included to make sure you see which connective each truth value is written below.

We began by placing a T under the propositional variable 𝑞 to indicate that 𝑞 is true. Since ¬T ≡ F, we then place an F under the negation symbol. Next, since F → 𝑟 ≡ T regardless of the truth value of 𝑟, we place a T under the conditional symbol. Finally, since ¬𝑝 ∨ T ≡ T regardless of the truth value of 𝑝, we place a T under the disjunction symbol. We made this last T bold to indicate that we are finished.

(3) Knowing that 𝑞 has truth value T is enough to determine the truth value of ¬𝑝 ∨ (¬𝑞 → 𝑟), as we saw in Note 1 above. It’s okay if you didn’t notice that right away. This kind of reasoning takes a bit of practice and experience.
Truth table solution: An alternative solution is to build the whole truth table of ¬𝑝 ∨ (¬𝑞 → 𝑟) one column at a time. Since there are 3 propositional variables (𝑝, 𝑞, and 𝑟), we will need 23 = 8 rows to get all the possible truth values. We then create a column for each compound statement that appears within the given statement starting with the statements of smallest length and working our way up to the given statement. We will need columns for 𝑝, 𝑞, 𝑟 (the atomic statements), ¬𝑝, ¬𝑞, ¬𝑞 → 𝑟, and finally, the statement itself, ¬𝑝 ∨ (¬𝑞 → 𝑟). Below is the final truth table with the relevant row highlighted and the final answer circled.

𝒑 𝒒 𝒓 ¬𝒑 ¬𝒒 ¬𝒒 → 𝒓 ¬𝒑 ∨ (¬𝒒 → 𝒓)

𝐓𝐓𝐓 𝐅 𝐅

𝐓

𝐓

𝐓𝐓 𝐅 𝐅 𝐅

𝐓

𝐓

𝐓 𝐅 𝐓 𝐅 𝐓

𝐓

𝐓

𝐓 𝐅 𝐅 𝐅 𝐓

𝐅

𝐅

𝐅 𝐓 𝐓 𝐓 𝐅

𝐓

𝐓

𝐅 𝐓 𝐅 𝐓 𝐅

𝐓

𝐓

𝐅 𝐅𝐓𝐓𝐓

𝐓

𝐓

𝐅 𝐅 𝐅𝐓𝐓

𝐅

𝐓

Notes: (1) We fill out the first three columns of the truth table by listing all possible combinations of truth assignments for the propositional variables 𝑝, 𝑞, and 𝑟. Notice how down the first column we have 4 T’s followed by 4 F’s, down the second column we alternate sequences of 2 T’s with 2 F’s, and down the third column we alternate T’s with F’s one at a time. This is a nice systematic way to make sure we get all possible combinations of truth assignments.

14

If you’re having trouble seeing the pattern of T’s and F’s, here is another way to think about it: In the first column, the first half of the rows have a T and the remainder have an F. This gives 4 T’s followed by 4 F’s. For the second column, we take half the number of consecutive T’s in the first column (half of 4 is 2) and then we alternate between 2 T’s and 2 F’s until we fill out the column. For the third column, we take half the number of consecutive T’s in the second column (half of 2 is 1) and then we alternate between 1 T and 1 F until we fill out the column. (2) Since the connective ¬ has the effect of taking the opposite truth value, we generate the entries in the fourth column by taking the opposite of each truth value in the first column. Similarly, we generate the entries in the fifth column by taking the opposite of each truth value in the second column. (3) For the sixth column, we apply the connective → to the fifth and third columns, respectively, and finally, for the last column, we apply the connective ∨ to the fourth and sixth columns, respectively. (4) The original question is asking us to compute the truth value of ¬𝑝 ∨ (¬𝑞 → 𝑟) when 𝑝 and 𝑞 are true, and 𝑟 is false. In terms of the truth table, we are being asked for the entry in the second row and last (seventh) column. Therefore, the answer is 𝐓. (5) This is certainly not the most efficient way to answer the given question. However, building truth tables is not too difficult, and it’s a foolproof way to determine truth values of compound statements.
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Problem Set 1
Full solutions to these problems are available for free download here:
www.SATPrepGet800.com/PMFBXSG
LEVEL 1
1. Determine whether each of the following sentences is an atomic statement, a compound statement, or not a statement at all: (i) I am not going to work today. (ii) What is the meaning of life? (iii) Don’t go away mad. (iv) I watched the television show Parks and Recreation. (v) If pigs have wings, then they can fly. (vi) 3 < – 5 or 38 > 37. (vii) This sentence has five words. (viii) I cannot swim, but I can run fast.
2. What is the negation of each of the following statements: (i) The banana is my favorite fruit. (ii) 7 > – 3. (iii) You are not alone. (iv) The function 𝑓 is differentiable everywhere.
LEVEL 2
3. Let 𝑝 represent the statement “9 is a perfect square,” let 𝑞 represent the statement “Orange is a primary color,” and let 𝑟 represent the statement “A frog is a reptile.” Rewrite each of the following symbolic statements in words, and state the truth value of each statement: (i) 𝑝 ∧ 𝑞 (ii) ¬𝑟 (iii) 𝑝 → 𝑟 (iv) 𝑞 ↔ 𝑟 (v) ¬𝑝 ∧ 𝑞 (vi) ¬(𝑝 ∧ 𝑞) (vii) ¬𝑝 ∨ ¬𝑞 (viii) (𝑝 ∧ 𝑞) → 𝑟
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4. Consider the compound sentence “You can have a cookie or ice cream.” In English this would most likely mean that you can have one or the other but not both. The word “or” used here is generally called an “exclusive or” because it excludes the possibility of both. The disjunction is an “inclusive or.” Using the symbol ⊕ for exclusive or, draw the truth table for this connective.
LEVEL 3
5. Let 𝑝, 𝑞, and 𝑟 represent true statements. Compute the truth value of each of the following compound statements: (i) (𝑝 ∨ 𝑞) ∨ 𝑟 (ii) (𝑝 ∨ 𝑞) ∧ ¬𝑟 (iii) ¬𝑝 → (𝑞 ∨ 𝑟) (iv) ¬(𝑝 ↔ ¬𝑞) ∧ 𝑟 (v) ¬[𝑝 ∧ (¬𝑞 → 𝑟)] (vi) ¬[(¬𝑝 ∨ ¬𝑞) ↔ ¬𝑟] (vii) 𝑝 → (𝑞 → ¬𝑟) (viii) ¬[¬𝑝 → (𝑞 → ¬𝑟)]
6. Using only the logical connectives ¬, ∧, and ∨, produce a statement using the propositional variables 𝑝 and 𝑞 that has the same truth values as 𝑝 ⊕ 𝑞 (this is the “exclusive or” defined in problem 4 above).
LEVEL 4
7. Let 𝑝 represent a true statement. Decide if this is enough information to determine the truth value of each of the following statements. If so, state that truth value.
(i) 𝑝 ∨ 𝑞 (ii) 𝑝 → 𝑞 (iii) ¬𝑝 → ¬(𝑞 ∨ ¬𝑟) (iv) ¬(¬𝑝 ∧ 𝑞) ↔ 𝑝 (v) (𝑝 ↔ 𝑞) ↔ ¬𝑝 (vi) ¬[(¬𝑝 ∧ ¬𝑞) ↔ ¬𝑟] (vii) [(𝑝 ∧ ¬𝑝) → 𝑝] ∧ (𝑝 ∨ ¬𝑝) (viii) 𝑟 → [¬𝑞 → (¬𝑝 → ¬𝑟)]
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8. Assume that the given compound statement is true. Determine the truth value of each propositional variable. (i) 𝑝 ∧ 𝑞 (ii) ¬(𝑝 → 𝑞) (iii) 𝑝 ↔ [¬(𝑝 ∧ 𝑞)] (iv) [𝑝 ∧ (𝑞 ∨ 𝑟)] ∧ ¬𝑟
LEVEL 5
9. Show that [𝑝 ∧ (𝑞 ∨ 𝑟)] ↔ [(𝑝 ∧ 𝑞) ∨ (𝑝 ∧ 𝑟)] is always true. 10. Show that [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠] → [(𝑝 → 𝑟) → 𝑠] is always true.
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LESSON 2 – SET THEORY SETS AND SUBSETS
Describing Sets
A set is simply a collection of “objects.” These objects can be numbers, letters, colors, animals, funny quotes, or just about anything else you can imagine. We will usually refer to the objects in a set as the members or elements of the set.
If a set consists of a small number of elements, we can describe the set simply by listing the elements in the set in curly braces, separating elements by commas.
Example 2.1: 1. {apple, banana} is the set consisting of two elements: apple and banana. 2. {anteater, elephant, egg, trapezoid} is the set consisting of four elements: anteater, elephant, egg, and trapezoid. 3. {2, 4, 6, 8, 10} is the set consisting of five elements: 2, 4, 6, 8, and 10. The elements in this set happen to be numbers.
A set is determined by its elements, and not the order in which the elements are presented. For example, the set {4, 2, 8, 6, 10} is the same as the set {2, 4, 6, 8, 10}.
Also, the set {2, 2, 4, 6, 8, 10, 10, 10} is the same as the set {2, 4, 6, 8, 10}. If we are describing a set by listing its elements, the most natural way to do this is to list each element just once.
We will usually name sets using capital letters such as 𝐴, 𝐵, and 𝐶. For example, we might write 𝐴 = {1, 2, 3}. So, 𝐴 is the set consisting of the elements 1, 2, and 3.
Example 2.2: Consider the sets 𝐴 = {𝑎, 𝑏}, 𝐵 = {𝑏, 𝑎}, 𝐶 = {𝑎, 𝑏, 𝑎}. Then 𝐴, 𝐵, and 𝐶 all represent the same set. We can write 𝐴 = 𝐵 = 𝐶.
We use the symbol ∈ for the membership relation (we will define the term “relation” more carefully in Lesson 10). So, 𝑥 ∈ 𝐴 means “𝑥 is an element of 𝐴,” whereas 𝑥 ∉ 𝐴 means “𝑥 is not an element of 𝐴.”
Example 2.3: Let 𝐴 = {𝑎, 𝑘, 3, ⊡, ⊕}. Then 𝑎 ∈ 𝐴, 𝑘 ∈ 𝐴, 3 ∈ 𝐴, ⊡ ∈ 𝐴, and ⊕ ∈ 𝐴.
If a set consists of many elements, we can use ellipses (…) to help describe the set. For example, the set consisting of the natural numbers between 17 and 5326, inclusive, can be written {17, 18, 19, … ,5325, 5326} (“inclusive” means that we include 17 and 5326). The ellipses between 19 and 5325 are there to indicate that there are elements in the set that we are not explicitly mentioning.
Ellipses can also be used to help describe infinite sets. The set of natural numbers can be written ℕ = {0, 1, 2, 3, … }, and the set of integers can be written ℤ = {… , – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, … }.
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Example 2.4: The odd natural numbers can be written 𝕆 = {1, 3, 5, … }. The even integers can be written 2ℤ = {… , – 6, – 4, – 2, 0, 2, 4, 6, … }. The primes can be written ℙ = {2, 3, 5, 7, 11, 13, 17, … }.
A set can also be described by a certain property 𝑃 that all its elements have in common. In this case, we can use the set-builder notation {𝑥|𝑃(𝑥)} to describe the set. The expression {𝑥|𝑃(𝑥)} can be read “the set of all 𝑥 such that the property 𝑃(𝑥) is true.” Note that the symbol “|” is read as “such that.”
Example 2.5: Let’s look at a few different ways that we can describe the set {2, 4, 6, 8, 10}. We have already seen that reordering and/or repeating elements does not change the set. For example, {2, 2, 6, 4, 10, 8} describes the same set. Here are a few more descriptions using set-builder notation:
• {𝑛 | 𝑛 is an even positive integer less than or equal to 10} • {𝑛 ∈ ℤ | 𝑛 is even, 0 < 𝑛 ≤ 10} • {2𝑘 | 𝑘 = 1, 2, 3, 4, 5} The first expression in the bulleted list can be read “the set of 𝑛 such that 𝑛 is an even positive integer less than or equal to 10.” The second expression can be read “the set of integers 𝑛 such that 𝑛 is even and 𝑛 is between 0 and 10, including 10, but excluding 0. Note that the abbreviation “𝑛 ∈ ℤ” can be read “𝑛 is in the set of integers,” or more succinctly, “𝑛 is an integer.” The third expression can be read “the set of 2𝑘 such that 𝑘 is 1, 2, 3, 4, or 5.”
If 𝐴 is a finite set, we define the cardinality of 𝐴, written |𝐴|, to be the number of elements of 𝐴. For example, |{𝑎, 𝑏}| = 2. In Lesson 10, we will extend the notion of cardinality to also include infinite sets.
Example 2.6: Let 𝐴 = {anteater, egg, trapezoid}, 𝐵 = {2, 3, 3}, and 𝐶 = {17, 18, 19, … , 5325, 5326}. Then |𝐴| = 3, |𝐵| = 2, and |𝐶| = 5310.
Notes: (1) The set 𝐴 has the three elements “anteater,” “egg,” and “trapezoid.”
(2) The set 𝐵 has just two elements: 2 and 3. Remember that {2, 3, 3} = {2, 3}.
(3) The number of consecutive integers from 𝑚 to 𝑛, inclusive, is 𝒏 − 𝒎 + 𝟏. For set 𝐶, we have 𝑚 = 17 and 𝑛 = 5326. Therefore, |𝐶| = 5326 − 17 + 1 = 5310.
(4) I call the formula “𝑛 − 𝑚 + 1” the fence-post formula. If you construct a 3-foot fence by placing a fence-post every foot, then the fence will consist of 4 fence-posts (3 − 0 + 1 = 4).
The empty set is the unique set with no elements. We use the symbol ∅ to denote the empty set (some authors use the symbol { } instead).
Subsets
For two sets 𝐴 and 𝐵, we say that 𝐴 is a subset of 𝐵, written 𝐴 ⊆ 𝐵, if every element of 𝐴 is an element of 𝐵. That is, 𝐴 ⊆ 𝐵 if, for every 𝑥, 𝑥 ∈ 𝐴 implies 𝑥 ∈ 𝐵. Symbolically, we can write ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵).
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Notes: (1) The symbol ∀ is called a universal quantifier, and it is pronounced “For all.”

(2) The logical expression ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) can be translated into English as “For all 𝑥, if 𝑥 is an element of 𝐴, then 𝑥 is an element of 𝐵.”

(3) To show that a set 𝐴 is a subset of a set 𝐵, we need to show that the expression ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) is true. If the set 𝐴 is finite and the elements are listed, we can just check that each element of 𝐴 is also an element of 𝐵. However, if the set 𝐴 is described by a property, say 𝐴 = {𝑥|𝑃(𝑥)}, we may need to craft an argument more carefully. We can begin by taking an arbitrary but specific element 𝑎 from 𝐴 and then arguing that this element 𝑎 is in 𝐵.

What could we possibly mean by an arbitrary but specific element? Aren’t the words “arbitrary” and “specific” antonyms? Well, by arbitrary, we mean that we don’t know which element we are choosing – it’s just some element 𝑎 that satisfies the property 𝑃. So, we are just assuming that 𝑃(𝑎) is true. However, once we choose this element 𝑎, we use this same 𝑎 for the rest of the argument, and that is what we mean by it being specific.

(4) To the right we see a physical representation of 𝐴 ⊆ 𝐵. This figure is called a Venn diagram. These types of diagrams are very useful to help visualize relationships among sets. Notice that set 𝐴 lies completely inside set 𝐵. We assume that all the elements of 𝐴 and 𝐵 lie in some universal set 𝑈.

As an example, let’s let 𝑈 be the set of all species of animals. If we let 𝐴 be the set of species of cats and we let 𝐵 be the set of species of mammals, then we have 𝐴 ⊆ 𝐵 ⊆ 𝑈, and we see that the Venn diagram to the right gives a visual representation of this situation. (Note that every cat is a mammal and every mammal is an animal.)

𝐴⊆𝐵

Let’s try to prove our first theorem using the definition of a subset together with Note 3 above about arbitrary but specific elements.

Theorem 2.1: Every set 𝐴 is a subset of itself.

Before writing the proof, let’s think about our strategy. We want to prove 𝐴 ⊆ 𝐴. In other words, we want to show ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐴). So, we will take an arbitrary but specific 𝑎 ∈ 𝐴 and then argue that 𝑎 ∈ 𝐴. But that’s pretty obvious, isn’t it? In this case, the property describing the set is precisely the conclusion we are looking for. Here are the details.

Proof of Theorem 2.1: Let 𝐴 be a set and let 𝑎 ∈ 𝐴. Then 𝑎 ∈ 𝐴. So, 𝑎 ∈ 𝐴 → 𝑎 ∈ 𝐴 is true. Since 𝑎 was

an arbitrary element of 𝐴, ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐴) is true. Therefore, 𝐴 ⊆ 𝐴.

□

Notes: (1) The proof begins with the opening statement “Let 𝐴 be a set and let 𝑎 ∈ 𝐴.” In general, the opening statement states what is given in the problem and/or fixes any arbitrary but specific objects that we will need.

(2) The proof ends with the closing statement “Therefore, 𝐴 ⊆ 𝐴.” In general, the closing statement states the result.

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(3) Everything between the opening statement and the closing statement is known as the argument.

(4) We place the symbol □ at the end of the proof to indicate that the proof is complete.

(5) Consider the logical statement 𝑝 → 𝑝. This statement is always true (T → T ≡ T and F → F ≡ T). 𝑝 → 𝑝 is an example of a tautology. A tautology is a statement that is true for every possible truth assignment of the propositional variables (see Problems 9 and 10 from Lesson 1 for more examples).

(6) If we let 𝑝 represent the statement 𝑎 ∈ 𝐴, by Note 5, we see that 𝑎 ∈ 𝐴 → 𝑎 ∈ 𝐴 is always true.

Alternate proof of Theorem 2.1: Let 𝐴 be a set and let 𝑎 ∈ 𝐴. Since 𝑝 → 𝑝 is a tautology, we have that

𝑎 ∈ 𝐴 → 𝑎 ∈ 𝐴 is true. Since 𝑎 was arbitrary, ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐴) is true. Therefore, 𝐴 ⊆ 𝐴.

□

Let’s prove another basic but important theorem.

Theorem 2.2: The empty set is a subset of every set.

Analysis: This time we want to prove ∅ ⊆ 𝐴. In other words, we want to show ∀𝑥(𝑥 ∈ ∅ → 𝑥 ∈ 𝐴). Since 𝑥 ∈ ∅ is always false (the empty set has no elements), 𝑥 ∈ ∅ → 𝑥 ∈ 𝐴 is always true.

In general, if 𝑝 is a false statement, then we say that 𝑝 → 𝑞 is vacuously true.

Proof of Theorem 2.2: Let 𝐴 be a set. The statement 𝑥 ∈ ∅ → 𝑥 ∈ 𝐴 is vacuously true for any 𝑥, and

so, ∀𝑥(𝑥 ∈ ∅ → 𝑥 ∈ 𝐴) is true. Therefore, ∅ ⊆ 𝐴.

□

Note: The opening statement is “Let 𝐴 be a set,” the closing statement is “Therefore, ∅ ⊆ 𝐴,” and the argument is everything in between.

Example 2.7: Let 𝐶 = {𝑎, 𝑏, 𝑐}, 𝐷 = {𝑎, 𝑐}, 𝐸 = {𝑏, 𝑐}, 𝐹 = {𝑏, 𝑑}, and 𝐺 = ∅. Then 𝐷 ⊆ 𝐶 and 𝐸 ⊆ 𝐶. Also, since the empty set is a subset of every set, we have 𝐺 ⊆ 𝐶, 𝐺 ⊆ 𝐷, 𝐺 ⊆ 𝐸, 𝐺 ⊆ 𝐹, and 𝐺 ⊆ 𝐺. Every set is a subset of itself, and so, 𝐶 ⊆ 𝐶, 𝐷 ⊆ 𝐷, 𝐸 ⊆ 𝐸, and 𝐹 ⊆ 𝐹.

Note: Below are possible Venn diagrams for this problem. The diagram on the left shows the relationship between the sets 𝐶, 𝐷, 𝐸, and 𝐹. Notice how 𝐷 and 𝐸 are both subsets of 𝐶, whereas 𝐹 is not a subset of 𝐶. Also, notice how 𝐷 and 𝐸 overlap, 𝐸 and 𝐹 overlap, but there is no overlap between 𝐷 and 𝐹 (they have no elements in common). The diagram on the right shows the proper placement of the elements. Here, I chose the universal set to be 𝑈 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔}. This choice for the universal set is somewhat arbitrary. Any set containing {𝑎, 𝑏, 𝑐, 𝑑} would do.

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Example 2.8: The set 𝐴 = {𝑎, 𝑏} has 2 elements and 4 subsets. The subsets of 𝐴 are ∅, {𝑎}, {𝑏}, and {𝑎, 𝑏}.

The set 𝐵 = {𝑎, 𝑏, 𝑐} has 3 elements and 8 subsets. The subsets of 𝐵 are ∅, {𝑎}, {𝑏}, {𝑐}, {𝑎, 𝑏}, {𝑎, 𝑐}, {𝑏, 𝑐}, and {𝑎, 𝑏, 𝑐}.

Let’s draw a tree diagram for the subsets of each of the sets 𝐴 and 𝐵.

{𝑎, 𝑏}

{𝑎, 𝑏, 𝑐}

{𝑎} {𝑏}

{𝑎, 𝑏} {𝑎, 𝑐} {𝑏, 𝑐}

∅

{𝑎} {𝑏} {𝑐}

∅
The tree diagram on the left is for the subsets of the set 𝐴 = {𝑎, 𝑏}. We start by writing the set 𝐴 = {𝑎, 𝑏} at the top. On the next line we write the subsets of cardinality 1 ({𝑎} and {𝑏}). On the line below that we write the subsets of cardinality 0 (just ∅). We draw a line segment between any two sets when the smaller (lower) set is a subset of the larger (higher) set. So, we see that ∅ ⊆ {𝑎}, ∅ ⊆ {𝑏}, {𝑎} ⊆ {𝑎, 𝑏}, and {𝑏} ⊆ {𝑎, 𝑏}. There is actually one more subset relationship, namely ∅ ⊆ {𝑎, 𝑏} (and of course each set displayed is a subset of itself). We didn’t draw a line segment from ∅ to {𝑎, 𝑏} to avoid unnecessary clutter. Instead, we can simply trace the path from ∅ to {𝑎} to {𝑎, 𝑏} (or from ∅ to {𝑏} to {𝑎, 𝑏}). We are using a property called transitivity here (see Theorem 2.3 below).

The tree diagram on the right is for the subsets of 𝐵 = {𝑎, 𝑏, 𝑐}. Observe that from top to bottom we write the subsets of 𝐵 of size 3, then 2, then 1, and then 0. We then draw the appropriate line segments, just as we did for 𝐴 = {𝑎, 𝑏}.

How many subsets does a set of cardinality 𝑛 have? Let’s start by looking at some examples.

Example 2.9: A set with 0 elements must be ∅, and this set has exactly 1 subset (the only subset of the empty set is the empty set itself).

A set with 1 element has 2 subsets, namely ∅ and the set itself.

In the last example, we saw that a set with 2 elements has 4 subsets, and we also saw that a set with 3 elements has 8 subsets.
Do you see the pattern yet? 1 = 20, 2 = 21, 4 = 22, 8 = 23. So, we see that a set with 0 elements has 20 subsets, a set with 1 element has 21 subsets, a set with 2 elements has 22 subsets, and a set with 3 elements has 23 subsets. A reasonable guess would be that a set with 𝑛 elements has 𝟐𝒏 subsets. You will be asked to prove this result later (Problem 12 in Lesson 4). We can also say that if |𝐴| = 𝑛, then |𝒫(𝐴)| = 2𝑛, where 𝒫(𝐴) (pronounced the power set of 𝐴) is the set of all subsets of 𝐴. In set-builder notation, we write 𝒫(𝐴) = {𝐵 | 𝐵 ⊆ 𝐴}.

Let’s get back to the transitivity mentioned above in our discussion of tree diagrams.

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Theorem 2.3: Let 𝐴, 𝐵, and 𝐶 be sets such that 𝐴 ⊆ 𝐵 and 𝐵 ⊆ 𝐶. Then 𝐴 ⊆ 𝐶.

Proof: Suppose that 𝐴, 𝐵, and 𝐶 are sets with 𝐴 ⊆ 𝐵 and 𝐵 ⊆ 𝐶, and let 𝑎 ∈ 𝐴. Since 𝐴 ⊆ 𝐵 and 𝑎 ∈ 𝐴,

it follows that 𝑎 ∈ 𝐵. Since 𝐵 ⊆ 𝐶 and 𝑎 ∈ 𝐵, it follows that 𝑎 ∈ 𝐶. Since 𝑎 was an arbitrary element

of 𝐴, we have shown that every element of 𝐴 is an element of 𝐶. That is, ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) is true.

Therefore, 𝐴 ⊆ 𝐶.

□

Note: To the right we have a Venn diagram illustrating Theorem 2.3.

Theorem 2.3 tells us that the relation ⊆ is transitive. Since ⊆ is transitive, we can write things like 𝐴 ⊆ 𝐵 ⊆ 𝐶 ⊆ 𝐷, and without explicitly saying it, we know that 𝐴 ⊆ 𝐶, 𝐴 ⊆ 𝐷, and 𝐵 ⊆ 𝐷.

Example 2.10: The membership relation ∈ is an example of a relation that is not transitive. For example, let 𝐴 = {0},

𝐵 = {0, 1, {0}}, and 𝐶 = {𝑥, 𝑦, {0, 1, {0}}}. Observe that 𝐴 ∈ 𝐵

𝑨⊆𝑩⊆𝑪

and 𝐵 ∈ 𝐶, but 𝐴 ∉ 𝐶.

Notes: (1) The set 𝐴 has just 1 element, namely 0.

{0} ∈ {0, 1, {0}} ∈ {𝑥, 𝑦, {0, 1, {0}}}

(2) The set 𝐵 has 3 elements, namely 0, 1, and {0}. But wait! 𝐴 = {0}. So, 𝐴 ∈ 𝐵. The set 𝐴 is circled twice in the above image.

(3) The set 𝐶 also has 3 elements, namely, 𝑥, 𝑦, and {0,1, {0}}. But wait! 𝐵 = {0, 1, {0}}. So, 𝐵 ∈ 𝐶. The set 𝐵 has a rectangle around it twice in the above image.

(4) Since 𝐴 ≠ 𝑥, 𝐴 ≠ 𝑦, and 𝐴 ≠ {0, 1, {0}}, we see that 𝐴 ∉ 𝐶.

(5) Is it clear that {0} ∉ 𝐶? {0} is in a set that’s in 𝐶 (namely, 𝐵), but {0} is not itself in 𝐶.

(6) Here is a more basic example showing that ∈ is not transitive: ∅ ∈ {∅} ∈ {{∅}}, but ∅ ∉ {{∅}} The only element of {{∅}} is {∅}.
Unions and Intersections

The union of the sets 𝐴 and 𝐵, written 𝐴 ∪ 𝐵, is the set of elements that are in 𝐴 or 𝐵 (or both). 𝐴 ∪ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵}

The intersection of 𝐴 and 𝐵, written 𝐴 ∩ 𝐵, is the set of elements that are simultaneously in 𝐴 and 𝐵. 𝐴 ∩ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 and 𝑥 ∈ 𝐵}

The following Venn diagrams for the union and intersection of two sets can be useful for visualizing these operations.

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𝑨∪𝑩

𝑨∩𝑩

Example 2.11:
1. Let 𝐴 = {0, 1, 2, 3, 4} and 𝐵 = {3, 4, 5, 6}. Then 𝐴 ∪ 𝐵 = {0, 1, 2, 3, 4, 5, 6} and 𝐴 ∩ 𝐵 = {3, 4}. See the figure below for a visual representation of 𝐴, 𝐵, 𝐴 ∪ 𝐵 and 𝐴 ∩ 𝐵.

2. Recall that the set of natural numbers is ℕ = {0, 1, 2, 3, … } and the set of integers is ℤ = {… , – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, … }. Observe that in this case, we have ℕ ⊆ ℤ. Also, ℕ ∪ ℤ = ℤ and ℕ ∩ ℤ = ℕ.
In fact, whenever 𝐴 and 𝐵 are sets and 𝐵 ⊆ 𝐴, then 𝐴 ∪ 𝐵 = 𝐴 and 𝐴 ∩ 𝐵 = 𝐵. We will prove the first of these two facts in Theorem 2.5. You will be asked to prove the second of these facts in Problem 13 below.
3. Let 𝔼 = {0, 2, 4, 6, … } be the set of even natural numbers and let 𝕆 = {1, 3, 5, 7, … } be the set of odd natural numbers. Then 𝔼 ∪ 𝕆 = {0, 1, 2, 3, 4, 5, 6, 7, … } = ℕ and 𝔼 ∩ 𝕆 = ∅. In general, we say that sets 𝐴 and 𝐵 are disjoint or mutually exclusive if 𝐴 ∩ 𝐵 = ∅. Below is a Venn diagram for disjoint sets.

𝑨∩𝑩=∅
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Let’s prove some theorems involving unions of sets. You will be asked to prove the analogous results for intersections of sets in Problems 11 and 13 below.

Theorem 2.4: If 𝐴 and 𝐵 are sets, then 𝐴 ⊆ 𝐴 ∪ 𝐵.

Before going through the proof, look once more at the Venn diagram above for 𝐴 ∪ 𝐵 and convince yourself that this theorem should be true.

Proof of Theorem 2.4: Suppose that 𝐴 and 𝐵 are sets and let 𝑥 ∈ 𝐴. Then 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵. Therefore,

𝑥 ∈ 𝐴 ∪ 𝐵. Since 𝑥 was an arbitrary element of 𝐴, we have shown that every element of 𝐴 is an element

of 𝐴 ∪ 𝐵. That is, ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐴 ∪ 𝐵) is true. Therefore, 𝐴 ⊆ 𝐴 ∪ 𝐵.

□

Note: Recall from Lesson 1 that if 𝑝 is a true statement, then 𝑝 ∨ 𝑞 (𝑝 or 𝑞) is true no matter what the truth value of 𝑞 is. In the second sentence of the proof above, we are using this fact with 𝑝 being the statement 𝑥 ∈ 𝐴 and 𝑞 being the statement 𝑥 ∈ 𝐵.

We will use this same reasoning in the second paragraph of the next proof as well.

Theorem 2.5: 𝐵 ⊆ 𝐴 if and only if 𝐴 ∪ 𝐵 = 𝐴.

Before going through the proof, it’s a good idea to draw a Venn diagram for 𝐵 ⊆ 𝐴 and convince yourself that this theorem should be true.

Technical note: Let 𝑋 and 𝑌 be sets. The Axiom of Extensionality says that 𝑋 and 𝑌 are the same set if and only if 𝑋 and 𝑌 have precisely the same elements. In symbols, we have
𝑋 = 𝑌 if and only if ∀𝑥(𝑥 ∈ 𝑋 ↔ 𝑥 ∈ 𝑌).
It is easy to verify that 𝑝 ↔ 𝑞 is logically equivalent to (𝑝 → 𝑞) ∧ (𝑞 → 𝑝). To see this, we check that all possible truth assignments for 𝑝 and 𝑞 lead to the same truth value for the two statements. For example, if 𝑝 and 𝑞 are both true, then
𝑝 ↔ 𝑞 ≡ T ↔ T ≡ T and (𝑝 → 𝑞) ∧ (𝑞 → 𝑝) ≡ (T → T) ∧ (T → T) ≡ T ∧ T ≡ T.

The reader should check the other three truth assignments for 𝑝 and 𝑞, or draw the entire truth table for both statements.

Letting 𝑝 be the statement 𝑥 ∈ 𝑋, letting 𝑞 be the statement 𝑥 ∈ 𝑌, and replacing 𝑝 ↔ 𝑞 by the logically equivalent statement (𝑝 → 𝑞) ∧ (𝑞 → 𝑝) gives us
𝑋 = 𝑌 if and only if ∀𝑥((𝑥 ∈ 𝑋 → 𝑥 ∈ 𝑌) ∧ (𝑥 ∈ 𝑌 → 𝑥 ∈ 𝑋)).

It is also true that ∀𝑥(𝑝(𝑥) ∧ 𝑞(𝑥)) is logically equivalent to ∀𝑥(𝑝(𝑥)) ∧ ∀𝑥(𝑞(𝑥)). And so, we have 𝑋 = 𝑌 if and only if ∀𝑥(𝑥 ∈ 𝑋 → 𝑥 ∈ 𝑌) and ∀𝑥(𝑥 ∈ 𝑌 → 𝑥 ∈ 𝑋).

In other words, to show that 𝑋 = 𝑌, we can instead show that 𝑋 ⊆ 𝑌 and 𝑌 ⊆ 𝑋.

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Proof of Theorem 2.5: Suppose that 𝐵 ⊆ 𝐴 and let 𝑥 ∈ 𝐴 ∪ 𝐵. Then 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵. If 𝑥 ∈ 𝐴, then 𝑥 ∈ 𝐴 (trivially). If 𝑥 ∈ 𝐵, then since 𝐵 ⊆ 𝐴, it follows that 𝑥 ∈ 𝐴. Since 𝑥 was an arbitrary element of 𝐴 ∪ 𝐵, we have shown that every element of 𝐴 ∪ 𝐵 is an element of 𝐴. That is, ∀𝑥(𝑥 ∈ 𝐴 ∪ 𝐵 → 𝑥 ∈ 𝐴) is true. Therefore, 𝐴 ∪ 𝐵 ⊆ 𝐴. By Theorem 2.4, 𝐴 ⊆ 𝐴 ∪ 𝐵. Since 𝐴 ∪ 𝐵 ⊆ 𝐴 and 𝐴 ⊆ 𝐴 ∪ 𝐵, it follows that 𝐴 ∪ 𝐵 = 𝐴.

Now, suppose that 𝐴 ∪ 𝐵 = 𝐴 and let 𝑥 ∈ 𝐵. Since 𝑥 ∈ 𝐵, it follows that 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵. Therefore,

𝑥 ∈ 𝐴 ∪ 𝐵. Since 𝐴 ∪ 𝐵 = 𝐴, we have 𝑥 ∈ 𝐴. Since 𝑥 was an arbitrary element of 𝐵, we have shown

that every element of 𝐵 is an element of 𝐴. That is, ∀𝑥(𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴). Therefore, 𝐵 ⊆ 𝐴.

□

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Problem Set 2

LEVEL 1

Full solutions to these problems are available for free download here:
www.SATPrepGet800.com/PMFBXSG

1. Determine whether each of the following statements is true or false: (i) 2 ∈ {2} (ii) 5 ∈ ∅ (iii) ∅ ∈ {1, 2} (iv) 𝑎 ∈ {𝑏, {𝑎}} (v) ∅ ⊆ {1, 2} (vi) {Δ} ⊆ {𝛿, Δ} (vii) {𝑎, 𝑏, 𝑐} ⊆ {𝑎, 𝑏, 𝑐} (viii) {1, 𝑎, {2, 𝑏}} ⊆ {1, 𝑎, 2, 𝑏}

2. Determine the cardinality of each of the following sets: (i) {𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓} (ii) {1, 2, 3, 2, 1} (iii) {1, 2, … , 53} (iv) {5, 6, 7, … , 2076, 2077}

3. Let 𝐴 = {𝑎, 𝑏, Δ, 𝛿} and 𝐵 = {𝑏, 𝑐, 𝛿, 𝛾}. Determine each of the following: (i) 𝐴 ∪ 𝐵 (ii) 𝐴 ∩ 𝐵
LEVEL 2

4. Determine whether each of the following statements is true or false: (i) ∅ ∈ ∅ (ii) ∅ ∈ {∅} (iii) {∅} ∈ ∅ (iv) {∅} ∈ {∅} (v) ∅ ⊆ ∅ (vi) ∅ ⊆ {∅} (vii) {∅} ⊆ ∅ (viii) {∅} ⊆ {∅}

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5. Determine the cardinality of each of the following sets: (i) {∅, {1, 2, 3}} (ii) {{{∅, {∅}}}}
(iii) {{1,2}, ∅, {∅}, {∅, {∅, 1, 2}}}
(iv) {∅, {∅}, {{∅}}, {∅, {∅}, {{∅}}}}
6. Let 𝑃 = {∅, {∅}} and 𝑄 = {{∅}, {∅, {∅}}}. Determine each of the following: (i) 𝑃 ∪ 𝑄 (ii) 𝑃 ∩ 𝑄
LEVEL 3
7. How many subsets does {𝑎, 𝑏, 𝑐, 𝑑} have? Draw a tree diagram for the subsets of {𝑎, 𝑏, 𝑐, 𝑑}. 8. A set 𝐴 is transitive if ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ⊆ 𝐴) (in words, every element of 𝐴 is also a subset of 𝐴).
Determine if each of the following sets is transitive: (i) ∅ (ii) {∅} (iii) {{∅}} (iv) {∅, {∅}, {{∅}}}
LEVEL 4
9. A relation 𝑅 is reflexive if ∀𝑥(𝑥𝑅𝑥) and symmetric if ∀𝑥∀𝑦(𝑥𝑅𝑦 → 𝑦𝑅𝑥). Show that ⊆ is reflexive, but ∈ is not. Then decide if each of ⊆ and ∈ is symmetric.
10. Let 𝐴, 𝐵, 𝐶, 𝐷, and 𝐸 be sets such that 𝐴 ⊆ 𝐵, 𝐵 ⊆ 𝐶, 𝐶 ⊆ 𝐷, and 𝐷 ⊆ 𝐸. Prove that 𝐴 ⊆ 𝐸. 11. Let 𝐴 and 𝐵 be sets. Prove that 𝐴 ∩ 𝐵 ⊆ 𝐴.
LEVEL 5
12. Let 𝑃(𝑥) be the property 𝑥 ∉ 𝑥. Prove that {𝑥|𝑃(𝑥)} cannot be a set. 13. Prove that 𝐵 ⊆ 𝐴 if and only if 𝐴 ∩ 𝐵 = 𝐵. 14. Let 𝐴 = {𝑎, 𝑏, 𝑐, 𝑑}, 𝐵 = {𝑋 | 𝑋 ⊆ 𝐴 ∧ 𝑑 ∉ 𝑋}, and 𝐶 = {𝑋 | 𝑋 ⊆ 𝐴 ∧ 𝑑 ∈ 𝑋}. Show that there is
a natural one-to-one correspondence between the elements of 𝐵 and the elements of 𝐶. Then generalize this result to a set with 𝑛 + 1 elements for 𝑛 > 0.
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LESSON 3 – ABSTRACT ALGEBRA SEMIGROUPS, MONOIDS, AND GROUPS
Binary Operations and Closure
A binary operation on a set is a rule that combines two elements of the set to produce another element of the set.
Example 3.1: Let 𝑆 = {0, 1}. Multiplication on 𝑆 is a binary operation, whereas addition on 𝑆 is not a binary operation (here we are thinking of multiplication and addition in the “usual” sense, meaning the way we would think of them in elementary school or middle school).
To see that multiplication is a binary operation on 𝑆, observe that 0 ⋅ 0 = 0, 0 ⋅ 1 = 0, 1 ⋅ 0 = 0, and 1 ⋅ 1 = 1. Each of the four computations produces 0 or 1, both of which are in the set 𝑆.
To see that addition is not a binary operation on 𝑆, just note that 1 + 1 = 2, and 2 ∉ 𝑆.
Let’s get a bit more technical and write down the formal definition of a binary operation. The terminology and notation used in this definition will be clarified in the notes below and formalized more rigorously later in Lesson 10.
Formally, a binary operation ⋆ on a set 𝑆 is a function ⋆ ∶ 𝑆 × 𝑆 → 𝑆. So, if 𝑎, 𝑏 ∈ 𝑆, then we have ⋆ (𝑎, 𝑏) ∈ 𝑆. For easier readability, we will usually write ⋆ (𝑎, 𝑏) as 𝑎 ⋆ 𝑏.
Notes: (1) If 𝐴 and 𝐵 are sets, then 𝐴 × 𝐵 is called the Cartesian product of 𝐴 and 𝐵. It consists of the ordered pairs (𝑎, 𝑏), where 𝑎 ∈ 𝐴 and 𝑏 ∈ 𝐵. A function 𝑓: 𝐴 × 𝐵 → 𝐶 takes each such pair (𝑎, 𝑏) to an element 𝑓(𝑎, 𝑏) ∈ 𝐶.
As an example, let 𝐴 = {dog, fish}, 𝐵 = {cat, snake}, 𝐶 = {0, 2, 4, 6, 8}, and define 𝑓: 𝐴 × 𝐵 → 𝐶 by 𝑓(𝑎, 𝑏) = the total number of legs that animals 𝑎 and 𝑏 have. Then we have 𝑓(dog, cat) = 8, 𝑓(dog, snake) = 4, 𝑓(fish, cat) = 4, 𝑓(fish, snake) = 0.
We will look at ordered pairs, cartesian products, and functions in more detail in Lesson 10.
(2) For a binary operation, all three sets 𝐴, 𝐵, and 𝐶 in the expression 𝑓: 𝐴 × 𝐵 → 𝐶 are the same.
As we saw in Example 3.1 above, if we let 𝑆 = {0, 1}, and we let ⋆ be multiplication, then ⋆ is a binary operation on 𝑆. Using function notation, we have ⋆ (0, 0) = 0, ⋆ (0, 1) = 0, ⋆ (1, 0) = 0, and ⋆ (1, 1) = 1.
As stated in the formal definition of a binary operation above, we will usually write the computations as 0 ⋆ 0 = 0, 0 ⋆ 1 = 0, 1 ⋆ 0 = 0, and 1 ⋆ 1 = 1.
We can use symbols other than ⋆ for binary operations. For example, if the operation is multiplication, we would usually use a dot (⋅) for the operation as we did in Example 3.1 above. Similarly, for addition we would usually use +, for subtraction we would usually use −, and so on.
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Recall: ℕ = {0, 1, 2, 3, … } is the set of natural numbers and ℤ = {… , – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, … } is the set of integers.
If 𝐴 is a set of numbers, we let 𝐴+ be the subset of 𝐴 consisting of just the positive numbers from 𝐴. For example, ℤ+ = {1, 2, 3, 4, … }, and in fact, ℕ+ = ℤ+.
Example 3.2:
1. The operation of addition on the set of natural numbers is a binary operation because whenever we add two natural numbers we get another natural number. Here, the set 𝑆 is ℕ and the operation ⋆ is +. Observe that if 𝑎 ∈ ℕ and 𝑏 ∈ ℕ, then 𝑎 + 𝑏 ∈ ℕ. For example, if 𝑎 = 1 and 𝑏 = 2 (both elements of ℕ), then 𝑎 + 𝑏 = 1 + 2 = 3, and 3 ∈ ℕ.
2. The operation of multiplication on the set of positive integers is a binary operation because whenever we multiply two positive integers we get another positive integer. Here, the set 𝑆 is ℤ+ and the operation ⋆ is ⋅. Observe that if 𝑎 ∈ ℤ+ and 𝑏 ∈ ℤ+, then 𝑎 ⋅ 𝑏 ∈ ℤ+. For example, if 𝑎 = 3 and 𝑏 = 5 (both elements of ℤ+), then 𝑎 ⋅ 𝑏 = 3 ⋅ 5 = 15, and 15 ∈ ℤ+.
3. Let 𝑆 = ℤ and define ⋆ by 𝑎 ⋆ 𝑏 = min{𝑎, 𝑏}, where min{𝑎, 𝑏} is the smallest of 𝑎 or 𝑏. Then ⋆ is a binary operation on ℤ. For example, if 𝑎 = – 5 and 𝑏 = 3 (both elements of ℤ), then 𝑎 ⋆ 𝑏 = – 5, and – 5 ∈ ℤ.
4. Subtraction on the set of natural numbers is not a binary operation. To see this, we just need to provide a single counterexample. (A counterexample is an example that is used to prove that a statement is false.) If we let 𝑎 = 1 and 𝑏 = 2 (both elements of ℕ), then we see that 𝑎 − 𝑏 = 1 − 2 is not an element of ℕ.
5. Let 𝑆 = {𝑢, 𝑣, 𝑤} and define ⋆ using the following table:
⋆ 𝑢 𝑣𝑤 𝑢 𝑣 𝑤𝑤 𝑣𝑤𝑢 𝑢 𝑤𝑢𝑣 𝑣
The table given above is called a multiplication table. For 𝑎, 𝑏 ∈ 𝑆, we evaluate 𝑎 ⋆ 𝑏 by taking the entry in the row given by 𝑎 and the column given by 𝑏. For example, 𝑣 ⋆ 𝑤 = 𝑢.
⋆ 𝑢 𝑣𝑤 𝑢 𝑣 𝑤𝑤 𝑣𝑤𝑢 𝑢 𝑤𝑢𝑣 𝑣
⋆ is a binary operation on 𝑆 because the only possible “outputs” are 𝑢, 𝑣, and 𝑤.
Some authors refer to a binary operation ⋆ on a set 𝑆 even when the binary operation is not defined on all pairs of elements 𝑎, 𝑏 ∈ 𝑆. We will always refer to these “false operations” as partial binary operations.
We say that the set 𝑆 is closed under the partial binary operation ⋆ if whenever 𝑎, 𝑏 ∈ 𝑆, we have 𝑎 ⋆ 𝑏 ∈ 𝑆.
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In Example 3.2, part 4 above, we saw that subtraction is a partial binary operation on ℕ that is not a binary operation. In other words, ℕ is not closed under subtraction.
Semigroups and Associativity
Let ⋆ be a binary operation on a set 𝑆. We say that ⋆ is associative in 𝑆 if for all 𝑥, 𝑦, 𝑧 in 𝑆, we have (𝑥 ⋆ 𝑦) ⋆ 𝑧 = 𝑥 ⋆ (𝑦 ⋆ 𝑧)
A semigroup is a pair (𝑆,⋆), where 𝑆 is a set and ⋆ is an associative binary operation on 𝑆.
Example 3.3: 1. (ℕ, +), (ℤ, +), (ℕ, ⋅), and (ℤ, ⋅) are all semigroups. In other words, the operations of addition and multiplication are both associative in ℕ and ℤ. 2. Let 𝑆 = ℤ and define ⋆ by 𝑎 ⋆ 𝑏 = min{𝑎, 𝑏}, where min{𝑎, 𝑏} is the smallest of 𝑎 or 𝑏. Let’s check that ⋆ is associative in ℤ. Let 𝑎, 𝑏, and 𝑐 be elements of ℤ. There are actually 6 cases to consider (see Note 1 below). Let’s go through one of these cases in detail. If we assume that 𝑎 ≤ 𝑏 ≤ 𝑐, then we have (𝑎 ⋆ 𝑏) ⋆ 𝑐 = min{𝑎, 𝑏} ⋆ 𝑐 = 𝑎 ⋆ 𝑐 = min{𝑎, 𝑐} = 𝑎. 𝑎 ⋆ (𝑏 ⋆ 𝑐) = 𝑎 ⋆ min{𝑏, 𝑐} = 𝑎 ⋆ 𝑏 = min{𝑎, 𝑏} = 𝑎. Since both (𝑎 ⋆ 𝑏) ⋆ 𝑐 = 𝑎 and 𝑎 ⋆ (𝑏 ⋆ 𝑐) = 𝑎, we have (𝑎 ⋆ 𝑏) ⋆ 𝑐 = 𝑎 ⋆ (𝑏 ⋆ 𝑐). After checking the other 5 cases, we can say the following: Since 𝑎, 𝑏, and 𝑐 were arbitrary elements from ℤ, we have shown that ⋆ is associative in ℤ. It follows that (ℤ,⋆) is a semigroup. 3. Subtraction is not associative in ℤ. To see this, we just need to provide a single counterexample. If we let 𝑎 = 1, 𝑏 = 2, and 𝑐 = 3, then (𝑎 − 𝑏) − 𝑐 = (1 − 2) − 3 = – 1 − 3 = – 4 and 𝑎 − (𝑏 − 𝑐) = 1 − (2 − 3) = 1 − (– 1) = 1 + 1 = 2. Since – 4 ≠ 2, subtraction is not associative in ℤ. It follows that (ℤ, −) is not a semigroup. Note that (ℕ, −) is also not a semigroup, but for a different reason. Subtraction is not even a binary operation on ℕ (see part 4 in Example 3.2). 4. Let 𝑆 = {𝑢, 𝑣, 𝑤} and define ⋆ using the following table (this is the same table from part 5 in Example 3.2): ⋆ 𝑢 𝑣𝑤 𝑢 𝑣 𝑤𝑤 𝑣𝑤𝑢 𝑢 𝑤𝑢𝑣 𝑣 Notice that (𝑢 ⋆ 𝑣) ⋆ 𝑤 = 𝑤 ⋆ 𝑤 = 𝑣 and 𝑢 ⋆ (𝑣 ⋆ 𝑤) = 𝑢 ⋆ 𝑢 = 𝑣. So, (𝑢 ⋆ 𝑣) ⋆ 𝑤 = 𝑢 ⋆ (𝑣 ⋆ 𝑤). However, this single computation does not show that ⋆ is associative in 𝑆. In fact, we have the following counterexample: (𝑢 ⋆ 𝑤) ⋆ 𝑣 = 𝑤 ⋆ 𝑣 = 𝑣 and 𝑢 ⋆ (𝑤 ⋆ 𝑣) = 𝑢 ⋆ 𝑣 = 𝑤. Thus, (𝑢 ⋆ 𝑤) ⋆ 𝑣 ≠ 𝑢 ⋆ (𝑤 ⋆ 𝑣). So, ⋆ is not associative in 𝑆, and therefore, (𝑆,⋆) is not a semigroup.
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5. Let 2ℤ = {… , – 6, – 4, – 2, 0, 2, 4, 6, … } be the set of even integers. When we multiply two even integers together, we get another even integer (we will prove this in Lesson 4). It follows that multiplication is a binary operation on 2ℤ. Since multiplication is associative in ℤ and 2ℤ ⊆ ℤ, it follows that multiplication is associative in 2ℤ (see Note 2 below). So, (2ℤ, ⋅) is a semigroup.
Notes: (1) In part 2 above, we must prove the result for each of the following 6 cases: 𝑎≤𝑏≤𝑐 𝑎≤𝑐≤𝑏 𝑏≤𝑎≤𝑐 𝑏≤𝑐≤𝑎 𝑐≤𝑎≤𝑏 𝑐≤𝑏≤𝑎
The same basic argument can be used for all these cases. For example, we saw in the solution above that for the first case we get
(𝑎 ⋆ 𝑏) ⋆ 𝑐 = min{𝑎, 𝑏} ⋆ 𝑐 = 𝑎 ⋆ 𝑐 = min{𝑎, 𝑐} = 𝑎. 𝑎 ⋆ (𝑏 ⋆ 𝑐) = 𝑎 ⋆ min{𝑏, 𝑐} = 𝑎 ⋆ 𝑏 = min{𝑎, 𝑏} = 𝑎.
Let’s also do the last case 𝑐 ≤ 𝑏 ≤ 𝑎: (𝑎 ⋆ 𝑏) ⋆ 𝑐 = min{𝑎, 𝑏} ⋆ 𝑐 = 𝑏 ⋆ 𝑐 = min{𝑏, 𝑐} = 𝑐. 𝑎 ⋆ (𝑏 ⋆ 𝑐) = 𝑎 ⋆ min{𝑏, 𝑐} = 𝑎 ⋆ 𝑐 = min{𝑎, 𝑐} = 𝑐.
The reader should verify the other 4 cases to complete the proof.
(2) Associativity is closed downwards. By this, we mean that if ⋆ is associative in a set 𝐴, and 𝐵 ⊆ 𝐴, (𝐵 is a subset of 𝐴) then ⋆ is associative in 𝐵.
The reason for this is that the definition of associativity involves only a universal statement—a statement that describes a property that is true for all elements without mentioning the existence of any new elements. A universal statement begins with the quantifier ∀ (“For all” or “Every”) and never includes the quantifier ∃ (“There exists” or “There is”).
As a simple example, if every object in set 𝐴 is a fruit, and 𝐵 ⊆ 𝐴, then every object in 𝐵 is a fruit. The universal statement we are referring to might be ∀𝑥(𝑃(𝑥)), where 𝑃(𝑥) is the property “𝑥 is a fruit.”
In the case of associativity, the universal statement is ∀𝑥∀𝑦∀𝑧((𝑥 ⋆ 𝑦) ⋆ 𝑧 = 𝑥 ⋆ (𝑦 ⋆ 𝑧)).
Let ⋆ be a binary operation on a set 𝑆. We say that ⋆ is commutative (or Abelian) in 𝑆 if for all 𝑥, 𝑦 in 𝑆, we have 𝑥 ⋆ 𝑦 = 𝑦 ⋆ 𝑥.
Example 3.4: 1. (ℕ, +), (ℤ, +), (ℕ, ⋅), and (ℤ, ⋅) are all commutative semigroups. In other words, the operations of addition and multiplication are both commutative in ℕ and ℤ (in addition to being associative). 2. The semigroup (ℤ,⋆), where ⋆ is defined by 𝑎 ⋆ 𝑏 = min{𝑎, 𝑏} is a commutative semigroup. Let’s check that ⋆ is commutative in ℤ. Let 𝑎 and 𝑏 be elements of ℤ. This time there are just 2 cases to consider (𝑎 ≤ 𝑏 and 𝑏 ≤ 𝑎). Let’s do the first case in detail, and assume that 𝑎 ≤ 𝑏. We then have 𝑎 ⋆ 𝑏 = min{𝑎, 𝑏} = 𝑎 and 𝑏 ⋆ 𝑎 = min{𝑏, 𝑎} = 𝑎. So, 𝑎 ⋆ 𝑏 = 𝑏 ⋆ 𝑎. After verifying the other case (which you should do), we can say that ⋆ is commutative in ℤ.
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3. Define the binary operation ⋆ on ℕ by 𝑎 ⋆ 𝑏 = 𝑎. Then (ℕ,⋆) is a semigroup that is not commutative. For associativity, we have (𝑎 ⋆ 𝑏) ⋆ 𝑐 = 𝑎 ⋆ 𝑐 = 𝑎 and 𝑎 ⋆ (𝑏 ⋆ 𝑐) = 𝑎 ⋆ 𝑏 = 𝑎. Let’s use a counterexample to show that ⋆ is not commutative. Well, 2 ⋆ 5 = 2 and 5 ⋆ 2 = 5.
Note: In part 3 above, the computation 𝑎 ⋆ (𝑏 ⋆ 𝑐) can actually be done in 1 step instead of 2. The way we did it above was to first compute 𝑏 ⋆ 𝑐 = 𝑏, and then to replace 𝑏 ⋆ 𝑐 with 𝑏 to get 𝑎 ⋆ (𝑏 ⋆ 𝑐) = 𝑎 ⋆ 𝑏 = 𝑎. However, the definition of ⋆ says that 𝑎 ⋆ (anything) = 𝑎. In this case, the “anything” is 𝑏 ⋆ 𝑐. So, we have 𝑎 ⋆ (𝑏 ⋆ 𝑐) = 𝑎 just by appealing to the definition of ⋆.
Monoids and Identity
Let (𝑆,⋆) be a semigroup. An element 𝑒 of 𝑆 is called an identity with respect to the binary operation ⋆ if for all 𝑎 ∈ 𝑆, we have 𝑒 ⋆ 𝑎 = 𝑎 ⋆ 𝑒 = 𝑎 A monoid is a semigroup with an identity.
Example 3.5: 1. (ℕ, +) and (ℤ, +) are commutative monoids with identity 0 (when we add 0 to any integer 𝑎, we get 𝑎). (ℕ, ⋅) and (ℤ, ⋅) are commutative monoids with identity 1 (when we multiply any integer 𝑎 by 1, we get 𝑎). 2. The commutative semigroup (ℤ,⋆), where ⋆ is defined by 𝑎 ⋆ 𝑏 = min{𝑎, 𝑏} is not a monoid. To see this, let 𝑎 ∈ ℤ. Then 𝑎 + 1 ∈ ℤ and 𝑎 ⋆ (𝑎 + 1) = 𝑎 ≠ 𝑎 + 1. This shows that 𝑎 is not an identity. Since 𝑎 was an arbitrary element of ℤ, we showed that there is no identity. It follows that (ℤ,⋆) is not a monoid. 3. The noncommutative semigroup (ℕ,⋆), where 𝑎 ⋆ 𝑏 = 𝑎 is also not a monoid. Use the same argument given in 2 above with ℤ replaced by ℕ. 4. (2ℤ, ⋅) is another example of a semigroup that is not a monoid. The identity element of (ℤ, ⋅) is 1, and this element is missing from (2ℤ, ⋅).
Groups and Inverses
Let (𝑀,⋆) be a monoid with identity 𝑒. An element 𝑎 of 𝑀 is called invertible if there is an element 𝑏 ∈ 𝑀 such that 𝑎 ⋆ 𝑏 = 𝑏 ⋆ 𝑎 = 𝑒.
A group is a monoid in which every element is invertible.
Groups appear so often in mathematics that it’s worth taking the time to explicitly spell out the full definition of a group.
A group is a pair (𝐺,⋆) consisting of a set 𝐺 together with a binary operation ⋆ satisfying: (1) (Associativity) For all 𝑥, 𝑦, 𝑧 ∈ 𝐺, (𝑥 ⋆ 𝑦) ⋆ 𝑧 = 𝑥 ⋆ (𝑦 ⋆ 𝑧). (2) (Identity) There exists an element 𝑒 ∈ 𝐺 such that for all 𝑥 ∈ 𝐺, 𝑒 ⋆ 𝑥 = 𝑥 ⋆ 𝑒 = 𝑥. (3) (Inverse) For each 𝑥 ∈ 𝐺, there is 𝑦 ∈ 𝐺 such that 𝑥 ⋆ 𝑦 = 𝑦 ⋆ 𝑥 = 𝑒.
Notes: (1) If 𝑦 ∈ 𝐺 is an inverse of 𝑥 ∈ 𝐺, we will usually write 𝑦 = 𝑥−1.
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(2) Recall that the definition of a binary operation already implies closure. However, many books on groups will mention this property explicitly:
(Closure) For all 𝑥, 𝑦 ∈ 𝐺, 𝑥 ⋆ 𝑦 ∈ 𝐺.

(3) A group is commutative or Abelian if for all 𝑥, 𝑦 ∈ 𝐺, 𝑥 ⋆ 𝑦 = 𝑦 ⋆ 𝑥.

Example 3.6:

1. (ℤ, +) is a commutative group with identity 0. The inverse of any integer 𝑎 is the integer – 𝑎.

2. (ℕ, +) is a commutative monoid that is not a group. For example, the natural number 1 has no inverse in ℕ. In other words, the equation 𝑥 + 1 = 0 has no solution in ℕ.

3. (ℤ, ⋅) is a commutative monoid that is not a group. For example, the integer 2 has no inverse in ℤ. In other words, the equation 2𝑥 = 1 has no solution in ℤ.

4. A rational number is a number of the form 𝑎𝑏, where 𝑎 and 𝑏 are integers and 𝑏 ≠ 0.

We identify rational numbers 𝑎 and 𝑐 whenever 𝑎𝑑 = 𝑏𝑐. For example, 1 and 3 represent the

𝑏

𝑑

2

6

same rational number because 1 ⋅ 6 = 6 and 2 ⋅ 3 = 6.

We

denote

the

set

of

rational

numbers

by

ℚ.

So,

we

have

ℚ

=

{𝑎
𝑏

|

𝑎,

𝑏

∈

ℤ,

𝑏

≠

0}.

In

words,

ℚ is “the set of quotients 𝑎 over 𝑏 such that 𝑎 and 𝑏 are integers and 𝑏 is not zero.”

We

identify

the

rational

number

𝑎 1

with

the

integer

𝑎.

In

this

way,

we

have

ℤ

⊆

ℚ.

We

add

two

rational

numbers

using

the

rule

𝑎 𝑏

+

𝑐 𝑑

=

𝑎⋅𝑑𝑏+⋅𝑑𝑏⋅𝑐.

Note

that

0

=

0 1

is

an

identity

for

(ℚ,

+)

because

𝑎 𝑏

+

0 1

=

𝑎⋅1+𝑏⋅0 𝑏⋅1

=

𝑎 𝑏

and

0 1

+

𝑎 𝑏

=

0⋅𝑏+1⋅𝑎 1⋅𝑏

=

𝑎𝑏.

You will be asked to show in Problem 11 below that (ℚ, +) is a commutative group.

5.

We

multiply

two

rational

numbers

using

the

rule

𝑎 𝑏

⋅

𝑐 𝑑

=

𝑏𝑎⋅⋅𝑑𝑐 .

Note that 1 = 1 is an identity for (ℚ, ⋅) because 𝑎 ⋅ 1 = 𝑎⋅1 = 𝑎 and 1 ⋅ 𝑎 = 1⋅𝑎 = 𝑎.

1

𝑏 1 𝑏⋅1 𝑏

1 𝑏 1⋅𝑏 𝑏

Now,

0⋅𝑎
𝑏

=

0⋅
1

𝑎 𝑏

=

0⋅𝑎 1⋅𝑏

=

0 𝑏

=

0. In particular, when we multiply 0 by any

rational number, we

can never get 1. So, 0 is a rational number with no multiplicative inverse. It follows that (ℚ, ⋅)

is not a group.

However, 0 is the only rational number without a multiplicative inverse. In fact, you will be asked to show in Problem 9 below that (ℚ∗, ⋅) is a commutative group, where ℚ∗ is the set of
rational numbers with 0 removed.

Note: When multiplying two numbers, we sometimes drop the dot (⋅) for easier readability. So, we may

write 𝑥 ⋅ 𝑦 as 𝑥𝑦. We may also use parentheses instead of the dot. For example, we might write 𝑎 ⋅ 𝑐 as
𝑏𝑑

(𝑎)
𝑏

(𝑑𝑐 ),

whereas

we

would

probably

write

𝑎⋅𝑐 𝑏⋅𝑑

as

𝑏𝑎𝑑𝑐 .

We

may

even

use

this

simplified

notation

for

arbitrary group operations. So, we could write 𝑎 ⋆ 𝑏 as 𝑎𝑏. However, we will avoid doing this if it would

lead to confusion. For example, we will not write 𝑎 + 𝑏 as 𝑎𝑏.

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Problem Set 3
Full solutions to these problems are available for free download here:
www.SATPrepGet800.com/PMFBXSG
LEVEL 1
1. For each of the following multiplication tables defined on the set 𝑆 = {𝑎, 𝑏}, determine if each of the following is true or false:

(i) ⋆ defines a binary operation on 𝑆.

(ii) ⋆ is commutative in 𝑆.

(iii) 𝑎 is an identity with respect to ⋆.

(iv) 𝑏 is an identity with respect to ⋆.

I )

⋆

𝑎

𝑏 II ⋆

𝑎

𝑏 III ⋆

𝑎

𝑏

IV ⋆

𝑎

𝑏

𝑎 𝑎 𝑎

𝑎 𝑎 𝑏

𝑎 𝑎 𝑏

𝑎 𝑎 𝑎

𝑏 𝑎 𝑎

𝑏 𝑐 𝑎

𝑏 𝑏 𝑎

𝑏 𝑏 𝑏

2. Show that there are exactly two monoids on the set 𝑆 = {𝑒, 𝑎}, where 𝑒 is the identity. Which of these monoids are groups? Which of these monoids are commutative?

LEVEL 2
3. Let 𝐺 = {𝑒, 𝑎, 𝑏} and let (𝐺,⋆) be a group with identity element 𝑒. Draw a multiplication table for (𝐺,⋆).
4. Prove that in any monoid (𝑀,⋆), the identity element is unique.

LEVEL 3
5. Assume that a group (𝐺,⋆) of order 4 exists with 𝐺 = {𝑒, 𝑎, 𝑏, 𝑐}, where 𝑒 is the identity, 𝑎2 = 𝑏 and 𝑏2 = 𝑒. Construct the table for the operation of such a group.
6. Prove that in any group (𝐺,⋆), each element has a unique inverse.

36

LEVEL 4
7. Let (𝐺,⋆) be a group with 𝑎, 𝑏 ∈ 𝐺, and let 𝑎−1 and 𝑏−1 be the inverses of 𝑎 and 𝑏, respectively. Prove (i) (𝑎 ⋆ 𝑏)−1 = 𝑏−1 ⋆ 𝑎−1. (ii) the inverse of 𝑎−1 is 𝑎.
8. Let (𝐺,⋆) be a group such that 𝑎2 = 𝑒 for all 𝑎 𝐺. Prove that (𝐺,⋆) is commutative.
9. Prove that (ℚ∗, ⋅) is a commutative group.
LEVEL 5
10. Prove that there are exactly two groups of order 4, up to renaming the elements. 11. Show that (ℚ, +) is a commutative group. 12. Let 𝑆 = {𝑎, 𝑏}, where 𝑎 ≠ 𝑏. How many binary operations are there on 𝑆? How many semigroups
are there of the form (𝑆,⋆), up to renaming the elements?
37

LESSON 4 – NUMBER THEORY THE RING OF INTEGERS
Rings and Distributivity
Before giving the general definition of a ring, let’s look at an important example.
Example 4.1: Recall that ℤ = {… , – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, … } is the set of integers. Let’s go over some of the properties of addition and multiplication on this set.
1. ℤ is closed under addition. In other words, whenever we add two integers, we get another integer. For example, 2 and 3 are integers, and we have 2 + 3 = 5, which is also an integer. As another example, – 8 and 6 are integers, and so is – 8 + 6 = – 2.
2. Addition is commutative in ℤ. In other words, when we add two integers, it does not matter which one comes first. For example, 2 + 3 = 5 and 3 + 2 = 5. So, we see that 2 + 3 = 3 + 2. As another example, – 8 + 6 = – 2 and 6 + (– 8) = – 2. So, we see that – 8 + 6 = 6 + (– 8).
3. Addition is associative in ℤ. In other words, when we add three integers, it doesn’t matter if we begin by adding the first two or the last two integers. For example, (2 + 3) + 4 = 5 + 4 = 9 and 2 + (3 + 4) = 2 + 7 = 9. So, (2 + 3) + 4 = 2 + (3 + 4). As another example, we have (– 8 + 6) + (– 5) = – 2 + (– 5) = – 7 and – 8 + (6 + (– 5)) = – 8 + 1 = – 7. So, we see that (– 8 + 6) + (– 5) = – 8 + (6 + (– 5)).
4. ℤ has an identity for addition, namely 0. Whenever we add 0 to another integer, the result is that same integer. For example, we have 0 + 3 = 3 and 3 + 0 = 3. As another example, 0 + (– 5) = – 5 and (– 5) + 0 = – 5.
5. Every integer has an additive inverse. This is an integer that we add to the original integer to get 0 (the additive identity). For example, the additive inverse of 5 is – 5 because we have 5 + (– 5) = 0 and – 5 + 5 = 0. Notice that the same two equations also show that the inverse of – 5 is 5. We can say that 5 and – 5 are additive inverses of each other.
We can summarize the five properties above by saying that (ℤ, +) is a commutative group.
6. ℤ is closed under multiplication. In other words, whenever we multiply two integers, we get another integer. For example, 2 and 3 are integers, and we have 2 ⋅ 3 = 6, which is also an integer. As another example, – 3 and – 4 are integers, and so is (– 3)(– 4) = 12.
7. Multiplication is commutative in ℤ. In other words, when we multiply two integers, it does not matter which one comes first. For example, 2 ⋅ 3 = 6 and 3 ⋅ 2 = 6. So, 2 ⋅ 3 = 3 ⋅ 2. As another example, – 8 ⋅ 6 = – 48 and 6(– 8) = – 48. So, we see that – 8 ⋅ 6 = 6(– 8).
8. Multiplication is associative in ℤ. In other words, when we multiply three integers, it doesn’t matter if we begin by multiplying the first two or the last two integers. For example, (2 ⋅ 3) ⋅ 4 = 6 ⋅ 4 = 24 and 2 ⋅ (3 ⋅ 4) = 2 ⋅ 12 = 24. So, (2 ⋅ 3) ⋅ 4 = 2 ⋅ (3 ⋅ 4). As another example, (– 5 ⋅ 2) ⋅ (– 6) = −10 ⋅ (– 6) = 60 and – 5 ⋅ (2 ⋅ (– 6)) = – 5 ⋅ (– 12) = 60. So, we see that (– 5 ⋅ 2) ⋅ (– 6) = – 5 ⋅ (2 ⋅ (– 6)).
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9. ℤ has an identity for multiplication, namely 1. Whenever we multiply 1 by another integer, the result is that same integer. For example, we have 1 ⋅ 3 = 3 and 3 ⋅ 1 = 3. As another example 1 ⋅ (– 5) = – 5 and (– 5) ⋅ 1 = – 5.
We can summarize the four properties above by saying that (ℤ, ⋅) is a commutative monoid.
10. Multiplication is distributive over addition in ℤ. This means that whenever 𝑘, 𝑚, and 𝑛 are integers, we have 𝑘 ⋅ (𝑚 + 𝑛) = 𝑘 ⋅ 𝑚 + 𝑘 ⋅ 𝑛. For example, 4 ⋅ (2 + 1) = 4 ⋅ 3 = 12 and 4 ⋅ 2 + 4 ⋅ 1 = 8 + 4 = 12. So, 4 ⋅ (2 + 1) = 4 ⋅ 2 + 4 ⋅ 1. As another example, we have – 2 ⋅ ((– 1) + 3) = – 2(2) = – 4 and – 2 ⋅ (– 1) + (– 2) ⋅ 3 = 2 − 6 = – 4. Therefore, we see that – 2 ⋅ ((– 1) + 3) = – 2 ⋅ (– 1) + (– 2) ⋅ 3.
Notes: (1) Since the properties listed in 1 through 10 above are satisfied, we say that (ℤ, +, ⋅) is a ring. We will give the formal definition of a ring below.
(2) Observe that a ring consists of (i) a set (in this case ℤ), and (ii) two binary operations on the set called addition and multiplication.
(3) (ℤ, +) is a commutative group and (ℤ, ⋅) is a commutative monoid. The distributive property is the only property mentioned that requires both addition and multiplication.
(4) We see that ℤ is missing one nice property—the inverse property for multiplication. For example, 2 has no multiplicative inverse in ℤ. There is no integer 𝑛 such that 2 ⋅ 𝑛 = 1. So, the linear equation 2𝑛 − 1 = 0 has no solution in ℤ.
(5) If we replace ℤ by the set of natural numbers ℕ = {0, 1, 2, … }, then all the properties mentioned above are satisfied except property 5—the inverse property for addition. For example, 1 has no additive inverse in ℕ. There is no natural number 𝑛 such that 𝑛 + 1 = 0.
(6) ℤ actually satisfies two distributive properties. Left distributivity says that whenever 𝑘, 𝑚, and 𝑛 are integers, we have 𝑘 ⋅ (𝑚 + 𝑛) = 𝑘 ⋅ 𝑚 + 𝑘 ⋅ 𝑛. Right distributivity says that whenever 𝑘, 𝑚, and 𝑛 are integers, we have (𝑚 + 𝑛) ⋅ 𝑘 = 𝑚 ⋅ 𝑘 + 𝑛 ⋅ 𝑘. Since multiplication is commutative in ℤ, left distributivity and right distributivity are equivalent. (7) Let’s show that left distributivity together with commutativity of multiplication in ℤ implies right distributivity in ℤ. If we assume that we have left distributivity and commutativity of multiplication, then for integers 𝑘, 𝑚, and 𝑛, we have (𝑚 + 𝑛) ⋅ 𝑘 = 𝑘(𝑚 + 𝑛) = 𝑘 ⋅ 𝑚 + 𝑘 ⋅ 𝑛 = 𝑚 ⋅ 𝑘 + 𝑛 ⋅ 𝑘.
We are now ready to give the more general definition of a ring.
A ring is a triple (𝑅, +, ⋅), where 𝑅 is a set and + and ⋅ are binary operations on 𝑅 satisfying (1) (𝑅, +) is a commutative group. (2) (𝑅, ⋅) is a monoid. (3) Multiplication is distributive over addition in 𝑅. That is, for all 𝑥, 𝑦, 𝑧 ∈ 𝑅, we have 𝑥 ⋅ (𝑦 + 𝑧) = 𝑥 ⋅ 𝑦 + 𝑥 ⋅ 𝑧 and (𝑦 + 𝑧) ⋅ 𝑥 = 𝑦 ⋅ 𝑥 + 𝑧 ⋅ 𝑥.
39

Recall: The symbol ∈ is used for membership in a set. Specifically, the statement 𝑎 ∈ 𝑆 can be read as “𝑎 is a member of the set 𝑆,” or more simply as “𝑎 is in 𝑆.” For example, 2 ∈ ℕ means “2 is in the set of natural numbers,” or more simply, “2 is a natural number.”
We will always refer to the operation + as addition and the operation ⋅ as multiplication. We will also adjust our notation accordingly. For example, we will refer to the identity for + as 0, and the additive inverse of an element 𝑥 ∈ 𝑅 as – 𝑥. Also, we will refer to the identity for ⋅ as 1, and the multiplicative inverse of an element 𝑥 ∈ 𝑅 (if it exists) as 𝑥–1 or 𝑥1.
Notes: (1) Recall from Lesson 3 that (𝑅, +) a commutative group means the following: • (Closure) For all 𝑥, 𝑦 ∈ 𝑅, 𝑥 + 𝑦 ∈ 𝑅. • (Associativity) For all 𝑥, 𝑦, 𝑧 ∈ 𝑅, (𝑥 + 𝑦) + 𝑧 = 𝑥 + (𝑦 + 𝑧). • (Commutativity) For all 𝑥, 𝑦 ∈ 𝑅, 𝑥 + 𝑦 = 𝑦 + 𝑥. • (Identity) There exists an element 0 ∈ 𝑅 such that for all 𝑥 ∈ 𝑅, 0 + 𝑥 = 𝑥 + 0 = 𝑥. • (Inverse) For each 𝑥 ∈ 𝑅, there is – 𝑥 ∈ 𝑅 such that 𝑥 + (– 𝑥) = (– 𝑥) + 𝑥 = 0.
(2) Recall from Lesson 3 that (𝑅, ⋅) a monoid means the following: • (Closure) For all 𝑥, 𝑦 ∈ 𝑅, 𝑥 ⋅ 𝑦 ∈ 𝑅. • (Associativity) For all 𝑥, 𝑦, 𝑧 ∈ 𝑅, (𝑥 ⋅ 𝑦) ⋅ 𝑧 = 𝑥 ⋅ (𝑦 ⋅ 𝑧). • (Identity) There exists an element 1 ∈ 𝑅 such that for all 𝑥 ∈ 𝑅, 1 ⋅ 𝑥 = 𝑥 ⋅ 1 = 𝑥.
(3) Although commutativity of multiplication is not required for the definition of a ring, our most important example (the ring of integers) satisfies this condition. When multiplication is commutative in 𝑅, we call the ring a commutative ring. In this case we have the following additional property:
• (Commutativity) For all 𝑥, 𝑦 ∈ 𝑅, 𝑥 ⋅ 𝑦 = 𝑦 ⋅ 𝑥. (4) Observe that we have two distributive properties in the definition for a ring. The first property is called left distributivity and the second is called right distributivity.
(5) In a commutative ring, left distributivity implies right distributivity and vice versa. In this case, the distributive property simplifies to
• (Distributivity) For all 𝑥, 𝑦, 𝑧 ∈ 𝑅, 𝑥 ⋅ (𝑦 + 𝑧) = 𝑥 ⋅ 𝑦 + 𝑥 ⋅ 𝑧
(6) Some authors leave out the multiplicative identity property in the definition of a ring and call such a ring a unital ring or a ring with identity. Since we are mostly concerned with the ring of integers, we will adopt the convention that a ring has a multiplicative identity. If we do not wish to assume that 𝑅 has a multiplicative identity, then we will call the structure “almost a ring” or rng (note the missing “i”).
(7) The properties that define a ring are called the ring axioms. In general, an axiom is a statement that is assumed to be true. So, the ring axioms are the statements that are given to be true in all rings. There are many other statements that are true in rings. However, any additional statements need to be proved using the axioms.
40

Example 4.2:

1. (ℤ, +, ⋅) is a commutative ring with additive identity 0 and multiplicative identity 1. The additive inverse of an integer 𝑎 is the integer – 𝑎. This is the ring we will be focusing most of our attention on. See Example 4.1 for more details.

2. (ℕ, +, ⋅) is not a ring because (ℕ, +) is not a group. The only group property that fails is the additive inverse property. For example, the natural number 1 has no additive inverse. That is, 𝑛 + 1 = 0 has no solution in ℕ. Note that (ℕ, ⋅) is a commutative monoid and the distributive property holds in ℕ. Therefore, (ℕ, +, ⋅) misses being a commutative ring by just that one property. (ℕ, +, ⋅) is an example of a structure called a semiring.

3.

Recall

from

Example

3.6

(4

and

5)

that

the

set

of

rational

numbers

is

ℚ

=

{𝑎
𝑏

|

𝑎,

𝑏

∈

ℤ,

𝑏

≠

0}

and we define addition and multiplication on ℚ by 𝑎 + 𝑐 = 𝑎𝑑+𝑏𝑐 and 𝑎 ⋅ 𝑐 = 𝑎𝑐.

𝑏𝑑

𝑏𝑑

𝑏 𝑑 𝑏𝑑

(ℚ, +, ⋅) is a commutative ring with additive identity 0 = 0 and multiplicative identity 1 = 1.

The

additive

inverse

of

a

rational

number

𝑎 𝑏

is

the

rational

1
number

–𝑏𝑎.

1

ℚ has one additional property not required in the definition of a ring. Every nonzero element

of

ℚ

has

a

multiplicative

inverse.

The

inverse

of

the

nonzero

rational

number

𝑎 𝑏

is

the

rational

number 𝑏. This is easy to verify: 𝑎 ⋅ 𝑏 = 𝑎𝑏 = 𝑎𝑏 = 1 = 1 and 𝑏 ⋅ 𝑎 = 𝑏𝑎 = 𝑎𝑏 = 1 = 1. So,

𝑎

𝑏 𝑎 𝑏𝑎 𝑎𝑏 1

𝑎 𝑏 𝑎𝑏 𝑎𝑏 1

(ℚ∗, ⋅) is a commutative group, where ℚ∗ is the set of nonzero rational numbers.

If we replace the condition “(𝑅, ⋅) is a monoid” in the definition of a ring (condition 2) with the condition (𝑅∗, ⋅) is a commutative group, we get a structure called a field. By the remarks in
the last paragraph, we see that (ℚ, +, ⋅) is a field.

Technical note: The definition of semiring has one additional property: 0 ⋅ 𝑥 = 𝑥 ⋅ 0 = 0. Without the additive inverse property this new property does not follow from the others, and so, it must be listed explicitly.

Divisibility
An integer 𝑎 is called even if there is another integer 𝑏 such that 𝑎 = 2𝑏.

Example 4.3:

1. 6 is even because 6 = 2 ⋅ 3.

2. – 14 is even because – 14 = 2 ⋅ (– 7).

3. We can write 1 = 2 ⋅ 12, but this does not show that 1 is even (and as we all know, it is not). In

the

definition

of

even,

it

is

very

important

that

𝑏

is

an

integer.

The

problem

here

is

that

1 2

is

not

an integer, and so, it cannot be used as a value for 𝑏 in the definition of even.

We define the sum of integers 𝑎 and 𝑏 to be 𝑎 + 𝑏. We define the product of 𝑎 and 𝑏 to be 𝑎 ⋅ 𝑏.

Theorem 4.1: The sum of two even integers is even.

41

Strategy: Before writing the proof, let’s think about our strategy. We need to start with two arbitrary but specific even integers. Let’s call them 𝑚 and 𝑛. Notice that we need to give them different names because there is no reason that they need to have the same value.

When we try to add 𝑚 and 𝑛, we get 𝑚 + 𝑛. Hmmm…I see no reason yet why the expression 𝑚 + 𝑛 should represent an even integer.

The problem is that we haven’t yet used the definition of even. If we invoke the definition, we get integers 𝑗 and 𝑘 such that 𝑚 = 2𝑗 and 𝑛 = 2𝑘.

Now, when we add 𝑚 and 𝑛, we get 𝑚 + 𝑛 = 2𝑗 + 2𝑘.

Is it clear that 2𝑗 + 2𝑘 represents an even integer? Nope…not yet. To be even, our final expression needs to have the form 2𝑏, where 𝑏 is an integer.

Here is where we use the fact that (ℤ, +, ⋅) is a ring. Specifically, we use the distributive property to rewrite 2𝑗 + 2𝑘 as 2(𝑗 + 𝑘).

It looks like we’ve done it. We just need to verify one more thing: is 𝑗 + 𝑘 an integer? Once again, we can use the fact that (ℤ, +, ⋅) is a ring to verify this. Specifically, we use the fact that + is a binary operation on ℤ.

I think we’re now ready to write the proof.

Proof of Theorem 4.1: Let 𝑚 and 𝑛 be even integers. Then there are integers 𝑗 and 𝑘 such that 𝑚 = 2𝑗

and 𝑛 = 2𝑘. So, 𝑚 + 𝑛 = 2𝑗 + 2𝑘 = 2(𝑗 + 𝑘) because multiplication is distributive over addition in ℤ.

Since ℤ is closed under addition, 𝑗 + 𝑘 ∈ ℤ. Therefore, 𝑚 + 𝑛 is even.

□

The property of being even is a special case of the more general notion of divisibility.

An integer 𝑎 is divisible by an integer 𝑘, written 𝑘|𝑎, if there is another integer 𝑏 such that 𝑎 = 𝑘𝑏. We also say that 𝑘 is a factor of 𝑎, 𝑘 is a divisor of 𝑎, 𝑘 divides 𝑎, or 𝑎 is a multiple of 𝑘.

Example 4.4: 1. Note that being divisible by 2 is the same as being even. 2. 18 is divisible by 3 because 18 = 3 ⋅ 6. 3. – 56 is divisible by 7 because – 56 = 7 ⋅ (– 8).

Theorem 4.2: The product of two integers that are each divisible by 𝑘 is also divisible by 𝑘.

Proof: Let 𝑚 and 𝑛 be integers that are divisible by 𝑘. Then there are integers 𝑏 and 𝑐 such that 𝑚 = 𝑘𝑏 and 𝑛 = 𝑘𝑐. So, 𝑚 ⋅ 𝑛 = (𝑘 ⋅ 𝑏) ⋅ (𝑘 ⋅ 𝑐) = 𝑘 ⋅ (𝑏 ⋅ (𝑘 ⋅ 𝑐)) because multiplication is associative in ℤ. Since ℤ is closed under multiplication, 𝑏 ⋅ (𝑘 ⋅ 𝑐) ∈ ℤ. Thus, 𝑚 ⋅ 𝑛 is divisible by 𝑘. □

Notes: (1) If you’re confused about how associativity was used here, it might help to make the substitution 𝑢 = (𝑘 ⋅ 𝑐). Then we have (𝑘 ⋅ 𝑏) ⋅ (𝑘 ⋅ 𝑐) = (𝑘 ⋅ 𝑏) ⋅ 𝑢 = 𝑘 ⋅ (𝑏 ⋅ 𝑢) = 𝑘(𝑏 ⋅ (𝑘 ⋅ 𝑐)).

42

(2) Although it may seem tempting to simplify 𝑘 ⋅ (𝑏 ⋅ (𝑘 ⋅ 𝑐)) further, it is unnecessary. The definition of divisibility by 𝑘 requires us only to generate an expression of the form 𝑘 times some integer, and that’s what we have done.

(3) If the generality of the proof confuses you, try replacing 𝑘 by a specific integer. For example, if we let 𝑘 = 2, we have 𝑚 = 2𝑏, 𝑛 = 2𝑐, and therefore 𝑚 ⋅ 𝑛 = (2𝑏) ⋅ (2𝑐) = 2(𝑏 ⋅ (2𝑐)). Is it clear that this final expression is even (divisible by 2)?
(4) It’s worth noting that the product 𝑚 ⋅ 𝑛 is actually divisible by 𝑘2. Indeed, we have 𝑚 ⋅ 𝑛 = 𝑘 ⋅ (𝑏 ⋅ (𝑘 ⋅ 𝑐)) = 𝑘 ⋅ ((𝑏 ⋅ 𝑘) ⋅ 𝑐) = 𝑘 ⋅ ((𝑘 ⋅ 𝑏) ⋅ 𝑐) = 𝑘 ⋅ (𝑘 ⋅ (𝑏 ⋅ 𝑐)) = 𝑘2(𝑏 ⋅ 𝑐)

Induction
The Well Ordering Principle says that every nonempty subset of natural numbers has a least element.

For example, the least element of ℕ itself is 0.

Theorem 4.3 (The Principle of Mathematical Induction): Let 𝑆 be a set of natural numbers such that (i) 0 ∈ 𝑆 and (ii) for all 𝑘 ∈ ℕ, 𝑘 ∈ 𝑆 → 𝑘 + 1 ∈ 𝑆. Then 𝑆 = ℕ.

Notes: (1) The Principle of Mathematical Induction works like a chain reaction. We know that 0 ∈ 𝑆 (this is condition (i)). Substituting 0 in for 𝑘 in the expression “𝑘 ∈ 𝑆 → 𝑘 + 1 ∈ 𝑆” (condition (ii)) gives us 0 ∈ 𝑆 → 1 ∈ 𝑆. So, we have that 0 is in the set 𝑆, and “if 0 is in the set 𝑆, then 1 is in the set 𝑆.” So, 1 ∈ 𝑆 must also be true.

(2) In terms of Lesson 1 on Sentential Logic, if we let 𝑝 be the statement 0 ∈ 𝑆 and 𝑞 the statement 1 ∈ 𝑆, then we are given that 𝑝 ∧ (𝑝 → 𝑞) is true. Observe that the only way that this statement can be true is if 𝑞 is also true. Indeed, we must have both 𝑝 ≡ T and 𝑝 → 𝑞 ≡ T. If 𝑞 were false, then we would have 𝑝 → 𝑞 ≡ T → F ≡ F. So, we must have 𝑞 ≡ T.

(3) Now that we showed 1 ∈ 𝑆 is true (from Note 1 above), we can substitute 1 for 𝑘 in the expression “𝑘 ∈ 𝑆 → 𝑘 + 1 ∈ 𝑆” (condition (ii)) to get 1 ∈ 𝑆 → 2 ∈ 𝑆. So, we have 1 ∈ 𝑆 ∧ (1 ∈ 𝑆 → 2 ∈ 𝑆) is true. So, 2 ∈ 𝑆 must also be true.

(4) In general, we get the following chain reaction:
0∈𝑆→1∈𝑆→2∈𝑆→3∈𝑆→⋯
I hope that the “argument” presented in Notes 1 through 4 above convinces you that the Principle of Mathematical Induction should be true. Now let’s give a proof using the Well Ordering Principle. Proofs involving the Well Ordering Principle are generally done by contradiction.

Proof of Theorem 4.3: Let 𝑆 be a set of natural numbers such that 0 ∈ 𝑆 (condition (i)), and such that

whenever 𝑘 ∈ 𝑆, 𝑘 + 1 ∈ 𝑆 (condition (ii)). Assume toward contradiction that 𝑆 ≠ ℕ. Let

𝐴 = {𝑘 ∈ ℕ | 𝑘 ∉ 𝑆} (so, 𝐴 is the set of natural numbers not in 𝑆). Since 𝑆 ≠ ℕ, 𝐴 is nonempty. So, by

the Well Ordering Principle, 𝐴 has a least element, let’s call it 𝑎. 𝑎 ≠ 0 because 0 ∈ 𝑆 and 𝑎 ∉ 𝑆. So,

𝑎 − 1 ∈ ℕ. Letting 𝑘 = 𝑎 − 1, we have 𝑎 − 1 ∈ 𝑆 → 𝑘 ∈ 𝑆 → 𝑘 + 1 ∈ 𝑆 → (𝑎 − 1) + 1 ∈ 𝑆 → 𝑎 ∈ 𝑆.

But 𝑎 ∈ 𝐴, which means that 𝑎 ∉ 𝑆. This is a contradiction, and so, 𝑆 = ℕ.

□

43

Note: The proof given here is a proof by contradiction. A proof by contradiction works as follows: 1. We assume the negation of what we are trying to prove. 2. We use a logically valid argument to derive a statement which is false. 3. Since the argument was logically valid, the only possible error is our original assumption. Therefore, the negation of our original assumption must be true.
In this problem we are trying to prove that 𝑆 = ℕ. The negation of this statement is that 𝑆 ≠ ℕ, and so that is what we assume.
We then define a set 𝐴 which contains elements of ℕ that are not in 𝑆. In reality, this set is empty (because the conclusion of the theorem is 𝑆 = ℕ). However, our (wrong!) assumption that 𝑆 ≠ ℕ tells us that this set 𝐴 actually has something in it. Saying that 𝐴 has something in it is an example of a false statement that was derived from a logically valid argument. This false statement occurred not because of an error in our logic, but because we started with an incorrect assumption (𝑆 ≠ ℕ).
The Well Ordering Principle then allows us to pick out the least element of this set 𝐴. Note that we can do this because 𝐴 is a subset of ℕ. This wouldn’t work if we knew only that 𝐴 was a subset of ℤ, as ℤ does not satisfy the Well Ordering Principle (for example, ℤ itself has no least element).
Again, although the argument that 𝐴 has a least element is logically valid, 𝐴 does not actually have any elements at all. We are working from the (wrong!) assumption that 𝑆 ≠ ℕ.
Once we have our hands on this least element 𝑎, we can get our contradiction. What can this least element 𝑎 be? Well 𝑎 was chosen to not be in 𝑆, so 𝑎 cannot be 0 (because 0 is in 𝑆). Also, we know that 𝑎 − 1 ∈ 𝑆 (because 𝑎 is the least element not in 𝑆). But condition (ii) then forces 𝑎 to be in 𝑆 (because 𝑎 = (𝑎 − 1) + 1).
So, we wind up with 𝑎 ∈ 𝑆, contradicting the fact that 𝑎 is the least element not in 𝑆.
The Principle of Mathematical Induction is often written in the following way:
(⋆) Let 𝑃(𝑛) be a statement and suppose that (i) 𝑃(0) is true and (ii) for all 𝑘 ∈ ℕ, 𝑃(𝑘) → 𝑃(𝑘 + 1). Then 𝑃(𝑛) is true for all 𝑛 ∈ ℕ.
In Problem 9 below, you will be asked to show that statement (⋆) is equivalent to Theorem 4.3.
There are essentially two steps involved in a proof by mathematical induction. The first step is to prove that 𝑃(0) is true (this is called the base case), and the second step is to assume that 𝑃(𝑘) is true, and use this to show that 𝑃(𝑘 + 1) is true (this is called the inductive step). While doing the inductive step, the statement “𝑃(𝑘) is true” is often referred to as the inductive hypothesis.
Subtraction in ℤ: For 𝑥, 𝑦 ∈ ℤ, we define the difference 𝑥 − 𝑦 to be equal to the sum 𝑥 + (– 𝑦). For example, 𝑛2 − 𝑛 = 𝑛2 + (– 𝑛) (where 𝑛2 is defined to be the product 𝑛 ⋅ 𝑛).
Example 4.5: Let’s use the Principle of Mathematical Induction to prove that for all natural numbers 𝑛, 𝑛2 − 𝑛 is even.
44

Base Case (𝑘 = 0): 02 − 0 = 0 = 2 ⋅ 0. So, 02 − 0 is even. Inductive Step: Let 𝑘 ∈ ℕ and assume that 𝑘2 − 𝑘 is even. Then 𝑘2 − 𝑘 = 2𝑏 for some integer 𝑏. Now,
(𝑘 + 1)2 − (𝑘 + 1) = (𝑘 + 1)[(𝑘 + 1) − 1] = (𝑘 + 1)[𝑘 + (1 − 1)] = (𝑘 + 1)(𝑘 + 0) = (𝑘 + 1) ⋅ 𝑘 = 𝑘2 + 𝑘 = (𝑘2 − 𝑘) + 2𝑘 = 2𝑏 + 2𝑘 = 2(𝑏 + 𝑘).

Here we used the fact that (ℤ, +, ⋅) is a ring. Since ℤ is closed under addition, 𝑏 + 𝑘 ∈ ℤ. Therefore, (𝑘 + 1)2 − (𝑘 + 1) is even.

By the Principle of Mathematical Induction, 𝑛2 − 𝑛 is even for all 𝑛 ∈ ℕ.

□

Notes: (1) Instead of listing every property that we used at each step, we simply stated that all the computations we made were allowed because (ℤ, +, ⋅) is a ring. We will discuss the property we used at each step in the notes below.
(2) We first used left distributivity to rewrite (𝑘 + 1)2 − (𝑘 + 1) as (𝑘 + 1)[(𝑘 + 1) − 1]. If you have trouble seeing this, try working backwards, and making the substitutions 𝑥 = (𝑘 + 1), 𝑦 = (𝑘 + 1), and 𝑧 = – 1. We then have
(𝑘 + 1)[(𝑘 + 1) − 1] = (𝑘 + 1)[(𝑘 + 1) + (– 1)] = 𝑥(𝑦 + 𝑧) = 𝑥𝑦 + 𝑥𝑧 = (𝑘 + 1)(𝑘 + 1) + (𝑘 + 1)(– 1) = (𝑘 + 1)2 + (– 1)(𝑘 + 1) = (𝑘 + 1)2 − (𝑘 + 1).

Notice how we also used commutativity of multiplication for the second to last equality.

(3) For the second algebraic step, we used associativity of addition to write (𝑘 + 1) − 1 = (𝑘 + 1) + (– 1) = 𝑘 + (1 + (– 1)) = 𝑘 + (1 − 1).

(4) For the third algebraic step, we used the inverse property for addition to write 1 − 1 = 1 + (– 1) = 0.

(5) For the fourth algebraic step, we used the additive identity property to write 𝑘 + 0 = 𝑘.

(6) For the fifth algebraic step, we used right distributivity and the multiplicative identity property to write (𝑘 + 1) ⋅ 𝑘 = 𝑘 ⋅ 𝑘 + 1 ⋅ 𝑘 = 𝑘2 + 𝑘.

(7) For the sixth algebraic step, we used what I call the “Standard Advanced Calculus Trick.” I sometimes abbreviate this as SACT. The trick is simple. If you need something to appear, just put it in. Then correct it by performing the opposite of what you just did.
In this case, in order to use the inductive hypothesis, we need 𝑘2 − 𝑘 to appear, but unfortunately, we have 𝑘2 + 𝑘 instead. Using SACT, I do the following:
• I simply put in what I need (and exactly where I need it): 𝑘2 − 𝒌 + 𝑘.
• Now, I undo the damage by performing the reverse operation: 𝑘2 − 𝑘 + 𝒌 + 𝑘.
• Finally, I leave the part I need as is, and simplify the rest: (𝑘2 − 𝑘) + 2𝑘

45

(8) For the seventh step, we simply replaced 𝑘2 − 𝑘 by 2𝑏. We established that these two quantities were equal in the second sentence of the inductive step.

(9) For the last step, we used left distributivity to write 2𝑏 + 2𝑘 as 2(𝑏 + 𝑘).

Sometimes a statement involving the natural numbers may be false for 0, but true from some natural number on. In this case, we can still use induction. We just need to adjust the base case.

Example 4.6: Let’s use the Principle of Mathematical Induction to prove that 𝑛2 > 2𝑛 + 1 for all natural numbers 𝑛 ≥ 3.

Base Case (𝑘 = 3): 32 = 9 and 2 ⋅ 3 + 1 = 6 + 1 = 7. So, 32 > 2 ⋅ 3 + 1. Inductive Step: Let 𝑘 ∈ ℕ with 𝑘 ≥ 3 and assume that 𝑘2 > 2𝑘 + 1. Then we have
(𝑘 + 1)2 = (𝑘 + 1)(𝑘 + 1) = (𝑘 + 1)𝑘 + (𝑘 + 1)(1) = 𝑘2 + 𝑘 + 𝑘 + 1 > (2𝑘 + 1) + 𝑘 + 𝑘 + 1 = 2𝑘 + 2 + 𝑘 + 𝑘 = 2(𝑘 + 1) + 𝑘 + 𝑘 ≥ 2(𝑘 + 1) + 1 (because 𝑘 + 𝑘 ≥ 3 + 3 = 6 ≥ 1).

By the Principle of Mathematical Induction, 𝑛2 > 2𝑛 + 1 for all 𝑛 ∈ ℕ with 𝑛 ≥ 3.

□

Notes: (1) If we have a sequence of equations and inequalities of the form =, ≥, and > (with at least one inequality symbol appearing), beginning with 𝑎 and ending with 𝑏, then the final result is 𝑎 > 𝑏 if > appears at least once and 𝑎 ≥ 𝑏 otherwise.

For example, if 𝑎 = 𝑗 = ℎ = 𝑚 > 𝑛 = 𝑝 = 𝑞 ≥ 𝑏, then 𝑎 > 𝑏. The sequence that appears in the solution above has this form.
(𝑘 + 1)2 = (𝑘 + 1)(𝑘 + 1) = (𝑘 + 1)𝑘 + (𝑘 + 1)(1) = 𝑘2 + 𝑘 + 𝑘 + 1 > (2𝑘 + 1) + 𝑘 + 𝑘 + 1
= 2𝑘 + 2 + 𝑘 + 𝑘 = 2(𝑘 + 1) + 𝑘 + 𝑘 ≥ 2(𝑘 + 1) + 1 (2) By definition, 𝑥2 = 𝑥 ⋅ 𝑥. We used this in the first equality in the inductive step to write (𝑘 + 1)2 as (𝑘 + 1)(𝑘 + 1).

(3) For the second equality in the inductive step, we used left distributivity to write (𝑘 + 1)(𝑘 + 1) as (𝑘 + 1)𝑘 + (𝑘 + 1)(1). If you have trouble seeing this, you can make a substitution like we did in Note 2 following Example 4.5.

(4) For the third equality in the inductive step, we used right distributivity to write (𝑘 + 1)𝑘 as 𝑘 ⋅ 𝑘 + 1 ⋅ 𝑘 = 𝑘2 + 𝑘. We also used the multiplicative identity property to write (𝑘 + 1)(1) = 𝑘 + 1.

(5) Associativity of addition is being used when we write the expression 𝑘2 + 𝑘 + 𝑘 + 1. Notice the lack of parentheses. Technically speaking, we should have written (𝑘2 + 𝑘) + (𝑘 + 1) and then taken another step to rewrite this as 𝑘2 + (𝑘 + (𝑘 + 1)). However, since we have associativity, we can
simply drop all those parentheses.

(6) The inequality “𝑘2 + 𝑘 + 𝑘 + 1 > (2𝑘 + 1) + 𝑘 + 𝑘 + 1” was attained by using the inductive hypothesis “𝑘2 > 2𝑘 + 1.”

46

(7) The dedicated reader should verify that the remaining equalities in the proof are valid by determining which ring properties were used at each step.

Example 4.7: Let’s use the Principle of Mathematical Induction to prove that for every natural number 𝑛, there is a natural number 𝑗 such that 𝑛 = 2𝑗 or 𝑛 = 2𝑗 + 1.

Base Case (𝑘 = 0): 0 = 2 ⋅ 0

Inductive Step: Suppose that 𝑘 ∈ ℕ and there is 𝑗 ∈ ℕ such that 𝑘 = 2𝑗 or 𝑘 = 2𝑗 + 1. If 𝑘 = 2𝑗, then 𝑘 + 1 = 2𝑗 + 1. If 𝑘 = 2𝑗 + 1, then 𝑘 + 1 = (2𝑗 + 1) + 1 = 2𝑗 + (1 + 1) = 2𝑗 + 2 = 2(𝑗 + 1). Here we used the fact that (ℕ, +, ⋅) is a semiring (more specifically, we used associativity of addition in ℕ and distributivity of multiplication over addition in ℕ). Since ℕ is closed under addition, 𝑗 + 1 ∈ ℕ.

By the Principle of Mathematical Induction, for every natural number 𝑛, there is a natural number 𝑗

such that 𝑛 = 2𝑗 or 𝑛 = 2𝑗 + 1.

□

Notes: (1) We can now prove the analogous result for the integers: “For every integer 𝑛, there is an integer 𝑗 such that 𝑛 = 2𝑗 or 𝑛 = 2𝑗 + 1.”

We already proved the result for 𝑛 ≥ 0. If 𝑛 < 0, then – 𝑛 > 0, and so there is a natural number 𝑗 such that – 𝑛 = 2𝑗 or – 𝑛 = 2𝑗 + 1. If – 𝑛 = 2𝑗, then 𝑛 = 2(– 𝑗) (and since 𝑗 ∈ ℕ, – 𝑗 ∈ ℤ). If – 𝑛 = 2𝑗 + 1, then 𝑛 = – (2𝑗 + 1) = – 2𝑗 − 1 = – 2𝑗 − 1 − 1 + 1 (SACT) = – 2𝑗 − 2 + 1 = 2(– 𝑗 − 1) + 1. Here we used the fact that (ℤ, +, ⋅) is a ring. Since ℤ is closed under addition, – 𝑗 − 1 = – 𝑗 + (– 1) ∈ ℤ.

(2) If there is an integer 𝑗 such that 𝑛 = 2𝑗, we say that 𝑛 is even. If there is an integer 𝑗 such that 𝑛 = 2𝑗 + 1, we say that 𝑛 is odd.

(3) An integer 𝒏 cannot be both even and odd. Indeed, if 𝑛 = 2𝑗 and 𝑛 = 2𝑘 + 1, then 2𝑗 = 2𝑘 + 1. So, we have
2(𝑗 − 𝑘) = 2𝑗 − 2𝑘 = (2𝑘 + 1) − 2𝑘 = 2𝑘 + (1 − 2𝑘) = 2𝑘 + (– 2𝑘 + 1) = (2𝑘 − 2𝑘) + 1 = 0 + 1 = 1.
So, 2(𝑗 − 𝑘) = 1. But 2 does not have a multiplicative inverse in ℤ, and so, this is a contradiction.

Theorem 4.4: The product of two odd integers is odd.

Proof: Let 𝑚 and 𝑛 be odd integers. Then there are integers 𝑗 and 𝑘 such that 𝑚 = 2𝑗 + 1 and 𝑛 = 2𝑘 + 1. So,
𝑚 ⋅ 𝑛 = (2𝑗 + 1) ⋅ (2𝑘 + 1) = (2𝑗 + 1)(2𝑘) + (2𝑗 + 1)(1) = (2𝑘)(2𝑗 + 1) + (2𝑗 + 1) = ((2𝑘)(2𝑗) + 2𝑘) + (2𝑗 + 1) = (2(𝑘(2𝑗)) + 2𝑘) + (2𝑗 + 1) = 2(𝑘(2𝑗) + 𝑘) + (2𝑗 + 1)
= (2(𝑘(2𝑗) + 𝑘) + 2𝑗) + 1 = 2((𝑘(2𝑗) + 𝑘) + 𝑗) + 1.

Here we used the fact that (ℤ, +, ⋅) is a ring. (Which properties did we use?) Since ℤ is closed under

addition and multiplication, we have (𝑘(2𝑗) + 𝑘) + 𝑗 ∈ ℤ. Therefore, 𝑚𝑛 is odd.

□

47

Problem Set 4

Full solutions to these problems are available for free download here:
www.SATPrepGet800.com/PMFBXSG
LEVEL 1
1. The addition and multiplication tables below are defined on the set 𝑆 = {0, 1}. Show that (𝑆, +, ⋅) does not define a ring.

+0 1 0 0 1 1 1 0

⋅

0

1

0 1 0

1 0 1

2. Let 𝑆 = {0, 1} and define addition (+) and multiplication (⋅) so that (𝑆, +, ⋅) is a ring. Assume that 0 is the additive identity in 𝑆 and 1 is the multiplicative identity in 𝑆. Draw the tables for addition and multiplication and verify that with these tables, (𝑆, +, ⋅) is a ring.

LEVEL 2
3. Use the Principle of Mathematical Induction to prove the following: (i) 2𝑛 > 𝑛 for all natural numbers 𝑛 ≥ 1. (ii) 0 + 1 + 2 + ⋯ + 𝑛 = 𝑛(𝑛+1) for all natural numbers.
2
(iii) 𝑛! > 2𝑛 for all natural numbers 𝑛 ≥ 4 (where 𝑛! = 1 ⋅ 2 ⋯ 𝑛 for all natural numbers 𝑛 ≥ 1).
(iv) 2𝑛 ≥ 𝑛2 for all natural numbers 𝑛 ≥ 4.
4. Show that the sum of three integers that are divisible by 5 is divisible by 5.
LEVEL 3
5. Prove that if 𝑎, 𝑏, 𝑐 ∈ ℤ with 𝑎|𝑏 and 𝑏|𝑐, then 𝑎|𝑐.
6. Prove that 𝑛3 − 𝑛 is divisible by 3 for all natural numbers 𝑛.
LEVEL 4
7. Prove that if 𝑎, 𝑏, 𝑐, 𝑑, 𝑒 ∈ ℤ with 𝑎|𝑏 and 𝑎|𝑐, then 𝑎|(𝑑𝑏 + 𝑒𝑐).
8. Prove that 3𝑛 − 1 is even for all natural numbers 𝑛.

48

9. Show that Theorem 4.3 (the Principle of Mathematical Induction) is equivalent to the following statement: (⋆) Let 𝑃(𝑛) be a statement and suppose that (i) 𝑃(0) is true and (ii) for all 𝑘 ∈ ℕ, 𝑃(𝑘) → 𝑃(𝑘 + 1). Then 𝑃(𝑛) is true for all 𝑛 ∈ ℕ.
LEVEL 5
10. The Principle of Strong Induction is the following statement: (⋆⋆) Let 𝑃(𝑛) be a statement and suppose that (i) 𝑃(0) is true and (ii) for all 𝑘 ∈ ℕ, ∀𝑗 ≤ 𝑘 (𝑃(𝑗)) → 𝑃(𝑘 + 1). Then 𝑃(𝑛) is true for all 𝑛 ∈ ℕ.
Use the Principle of Mathematical Induction to prove the Principle of Strong Induction. 11. Show that (ℚ, +, ⋅) is a field. 12. Use the Principle of Mathematical Induction to prove that for every 𝑛 ∈ ℕ, if 𝑆 is a set with
|𝑆| = 𝑛, then 𝑆 has 2𝑛 subsets. (Hint: Use Problem 14 from Lesson 2.)
49

LESSON 5 – REAL ANALYSIS THE COMPLETE ORDERED FIELD OF REALS
Fields
Let’s review the number systems we have discussed so far. The set ℕ = {0, 1, 2, 3, … } is the set of natural numbers and the structure (ℕ, +, ⋅) is a semiring. The set ℤ = {… , – 3, – 2, – 1, 0, 1, 2, 3, … } is the set of integers and the structure (ℤ, +, ⋅) is a ring. The set ℚ = {𝑎 |𝑎 ∈ ℤ, 𝑏 ∈ ℤ∗} is the set of rational numbers and the structure (ℚ, +, ⋅) is a field.
𝑏
And now let’s formally introduce the notion of a field (and we will review the definitions of ring and semiring in the notes below).
A field is a triple (𝐹, +, ⋅), where 𝐹 is a set and + and ⋅ are binary operations on 𝐹 satisfying (1) (𝐹, +) is a commutative group. (2) (𝐹∗, ⋅) is a commutative group. (3) ⋅ is distributive over + in 𝐹. That is, for all 𝑥, 𝑦, 𝑧 ∈ 𝐹, we have 𝑥 ⋅ (𝑦 + 𝑧) = 𝑥 ⋅ 𝑦 + 𝑥 ⋅ 𝑧 and (𝑦 + 𝑧) ⋅ 𝑥 = 𝑦 ⋅ 𝑥 + 𝑧 ⋅ 𝑥. (4) 0 ≠ 1.
We will refer to the operation + as addition, the operation ⋅ as multiplication, the additive identity as 0, the multiplicative identity as 1, the additive inverse of an element 𝑥 ∈ 𝐹 as – 𝑥, and the multiplicative inverse of an element 𝑥 ∈ 𝐹 as 𝑥−1. We will often abbreviate 𝑥 ⋅ 𝑦 as 𝑥𝑦.
Notes: (1) Recall from Lesson 3 that (𝐹, +) a commutative group means the following: • (Closure) For all 𝑥, 𝑦 ∈ 𝐹, 𝑥 + 𝑦 ∈ 𝐹. • (Associativity) For all 𝑥, 𝑦, 𝑧 ∈ 𝐹, (𝑥 + 𝑦) + 𝑧 = 𝑥 + (𝑦 + 𝑧). • (Commutativity) For all 𝑥, 𝑦 ∈ 𝐹, 𝑥 + 𝑦 = 𝑦 + 𝑥. • (Identity) There exists an element 0 ∈ 𝐹 such that for all 𝑥 ∈ 𝐹, 0 + 𝑥 = 𝑥 + 0 = 𝑥. • (Inverse) For each 𝑥 ∈ 𝐹, there is – 𝑥 ∈ 𝐹 such that 𝑥 + (– 𝑥) = (– 𝑥) + 𝑥 = 0.
(2) Similarly, (𝐹∗, ⋅) a commutative group means the following: • (Closure) For all 𝑥, 𝑦 ∈ 𝐹∗, 𝑥𝑦 ∈ 𝐹∗. • (Associativity) For all 𝑥, 𝑦, 𝑧 ∈ 𝐹∗, (𝑥𝑦)𝑧 = 𝑥(𝑦𝑧). • (Commutativity) For all 𝑥, 𝑦 ∈ 𝐹∗, 𝑥𝑦 = 𝑦𝑥. • (Identity) There exists an element 1 ∈ 𝐹∗ such that for all 𝑥 ∈ 𝐹∗, 1𝑥 = 𝑥 ⋅ 1 = 𝑥. • (Inverse) For each 𝑥 ∈ 𝐹∗, there is 𝑥−1 ∈ 𝐹∗ such that 𝑥𝑥−1 = 𝑥−1𝑥 = 1.
50

(3) Recall that 𝐹∗ is the set of nonzero elements of 𝐹. We can write 𝐹∗ = {𝑥 ∈ 𝐹 | 𝑥 ≠ 0} (pronounced “the set of 𝑥 in 𝐹 such that 𝑥 is not equal to 0”) or 𝐹∗ = 𝐹 ∖ {0} (pronounced “𝐹 with 0 removed”).

(4) The properties that define a field are called the field axioms. These are the statements that are given to be true in all fields. There are many other statements that are true in fields. However, any additional statements need to be proved using the axioms.
(5) If we replace the condition that “(𝐹∗, ⋅) is a commutative group” by “(𝐹, ⋅) is a monoid,” then the resulting structure is called a ring. The most well-known example of a ring is ℤ, the ring of integers. See Lesson 4 for details about ℤ and rings in general.

We also do not require 0 and 1 to be distinct in the definition of a ring. If 0 = 1, we get the zero ring, which consists of just one element, namely 0 (Why?). The operations of addition and multiplication are defined by 0 + 0 = 0 and 0 ⋅ 0 = 0. The reader may want to verify that the zero ring is in fact a ring.

The main difference between a ring and a field is that in a ring, there can be nonzero elements that do not have multiplicative inverses. For example, in ℤ, 2 has no multiplicative inverse. So, the equation 2𝑥 = 1 has no solution.

(6) If we also replace “(𝐹, +) is a commutative group” by “(𝐹, +) is a commutative monoid,” then the resulting structure is a semiring. The most well-known example of a semiring is ℕ, the semiring of natural numbers.

The main difference between a semiring and a ring is that in a semiring, there can be elements that do not have additive inverses. For example, in ℕ, 1 has no additive inverse. Thus, the equation 𝑥 + 1 = 0 has no solution.

Technical note: For a semiring, we include one additional axiom: For all 𝑥 ∈ 𝐹, 0 ⋅ 𝑥 = 𝑥 ⋅ 0 = 0.

(7) Every field is a commutative ring. Although this is not too hard to show (you will be asked to show
this in Problem 6 below), it is worth observing that this is not completely obvious. For example, if (𝐹, +, ⋅) is a ring, then since (𝐹, ⋅) is a monoid with identity 1, it follows that 1 ⋅ 0 = 0 ⋅ 1 = 0. However, in the definition of a field given above, this property of 0 is not given as an axiom. We are given that (𝐹∗, ⋅) is a commutative group, and so, it follows that 1 is an identity for 𝐹∗. But 0 ∉ 𝐹∗, and so, 1 ⋅ 0 = 0 ⋅ 1 = 0 needs to be proved.

Similarly, in the definition of a field given above, 0 is excluded from associativity and commutativity. These need to be checked.

(8) You were asked to verify that (ℚ, +, ⋅) is a field in Problems 9 and 11 from Lesson 3 and Problem 11 from Lesson 4.

Subtraction

and

Division:

If

𝑎,

𝑏

∈

𝐹,

we

define

𝑎

−

𝑏

=

𝑎

+

(–

𝑏)

and

for

𝑏

≠

0,

𝑎 𝑏

=

𝑎𝑏−1.

51

Ordered Rings and Fields
We say that a ring (𝑅, +, ⋅) is ordered if there is a nonempty subset 𝑃 of 𝑅, called the set of positive elements of 𝑅, satisfying the following three properties:
(1) If 𝑎, 𝑏 ∈ 𝑃, then 𝑎 + 𝑏 ∈ 𝑃. (2) If 𝑎, 𝑏 ∈ 𝑃, then 𝑎𝑏 ∈ 𝑃. (3) If 𝑎 ∈ 𝑅, then exactly one of the following holds: 𝑎 ∈ 𝑃, 𝑎 = 0, or – 𝑎 ∈ 𝑃.

Note: If 𝑎 ∈ 𝑃, we say that 𝑎 is positive and if – 𝑎 ∈ 𝑃, we say that 𝑎 is negative.

Also, we define 𝑅+ = 𝑃 and 𝑅– = {𝑎 ∈ 𝑅 | – 𝑎 ∈ 𝑃}.

Example 5.1: Let 𝑅 = ℤ and let 𝑃ℤ = {1, 2, 3, … }. It’s easy to see that properties (1), (2), and (3) are satisfied. It follows that (ℤ, +, ⋅) is an ordered ring.

Theorem 5.1: (ℚ, +, ⋅) is an ordered field.

Note: The proof of this result is a bit technical, but I am including it for completeness. The student just starting out in pure mathematics can feel free to just accept this result and skip the proof.

Recall: (1) Rational numbers have the form 𝑚, where 𝑚 and 𝑛 are integers and 𝑛 ≠ 0.
𝑛

(2)

Two

rational

numbers

𝑚 𝑛

and

𝑝 𝑞

are

equal

if

and

only

if

𝑚𝑞

=

𝑛𝑝.

(3)

For

rational

numbers

𝑚 𝑛

and

𝑝𝑞,

we

define

addition

and

multiplication

by

𝑚+𝑝
𝑛𝑞

= 𝑚𝑞+𝑛𝑝
𝑛𝑞

and

𝑚 ⋅ 𝑝 = 𝑚𝑝.

𝑛 𝑞 𝑛𝑞

(4) The additive inverse of

𝑚 𝑛

is

–

𝑚 𝑛

=

–𝑛𝑚.

Analysis: Before writing out the proof in detail, let’s think about how we would go about it. First of all,

we already know from Problem 11 in Lesson 4 that (ℚ, +, ⋅) is a field. So, we need only show that it is

ordered. To do this, we need to come up with a set 𝑃 of positive elements from ℚ. The natural choice

would be to take the set of quotients whose numerator (number on the top) and denominator (number

on the bottom) are both positive integers. In other words, we will let 𝑃ℚ be the set of all the rational

numbers

Since

–𝑚 –𝑛

of =

𝑚𝑛th(ebfeocramus𝑚e𝑛 ,(w– h𝑚e)r𝑛e

𝑚 =

and 𝑛 are both elements of 𝑃ℤ (as defined in (– 𝑛)𝑚), we must automatically be including

Example 5.1 above). all quotients whose

numerator and denominator are both negative integers as well.

With this definition of 𝑃ℚ, it is straightforward to verify properties (1) and (2) of an ordered field.

To verify property (3), we need to check three things.
(i) For any rational number 𝑎, 𝑎 is positive, zero, or negative (𝑎 ∈ 𝑃ℚ, 𝑎 = 0, or – 𝑎 ∈ 𝑃ℚ). We will show this by assuming 𝑎 ∉ 𝑃ℚ and 𝑎 ≠ 0, and then proving that we must have – 𝑎 ∈ 𝑃ℚ.

52

(ii) For any rational number 𝑎, 𝑎 cannot be both positive and negative. We will show this by assuming 𝑎 ∈ 𝑃ℚ and – 𝑎 ∈ 𝑃ℚ, and then deriving a contradiction.
(iii) A positive or negative rational number is not zero, and a rational number that is zero is not positive or negative. This is straightforward to check.

Let’s write out the details.

Proof of Theorem 5.1: By Problem 11 from Lesson 4, (ℚ, +, ⋅) is a field.

Let 𝐹

= ℚ and let 𝑃ℚ

= {𝑥 ∈ ℚ | 𝑥 = 𝑚
𝑛

with 𝑚, 𝑛 ∈ 𝑃ℤ}. Let 𝑎, 𝑏 ∈

𝑃ℚ. Then there are 𝑚, 𝑛, 𝑝, 𝑞

∈ 𝑃ℤ

with

𝑎

=𝑚
𝑛

and

𝑏 = 𝑝.
𝑞

We

have

𝑎+𝑏

= 𝑚 + 𝑝 = 𝑚𝑞+𝑛𝑝.

𝑛𝑞

𝑛𝑞

Since

𝑃ℤ

satisfies

(2)

above,

we

have

𝑚𝑞, 𝑛𝑝, 𝑛𝑞 ∈ 𝑃ℤ. Since 𝑃ℤ satisfies (1) above, we have 𝑚𝑞 + 𝑛𝑝 ∈ 𝑃ℤ. Therefore, 𝑎 + 𝑏 ∈ 𝑃ℚ and (1)

holds.

Also,

we

have

𝑎𝑏

=

𝑚 𝑛

⋅

𝑝 𝑞

=

𝑚𝑝.
𝑛𝑞

Since

𝑃ℤ

satisfies

(2)

above,

we

have

𝑚𝑝,

𝑛𝑞

∈

𝑃ℤ,

and

therefore,

𝑎𝑏 ∈ 𝑃ℚ and (2) holds.

Now,

suppose

𝑎

∉

𝑃ℚ

and

𝑎

≠

0.

Since

𝑎

∈

ℚ,

there

are

𝑚

∈

ℤ

and

𝑛

∈

ℤ∗

such

that

𝑎

=

𝑚.
𝑛

But

𝑎 ≠ 0, and so, we must have 𝑚 ∈ ℤ∗. Since 𝑎 ∉ 𝑃ℚ, either 𝑚 ∉ 𝑃ℤ or 𝑛 ∉ 𝑃ℤ (or both). If both 𝑚 ∉ 𝑃ℤ

and

𝑛

∉

𝑃ℤ,

then

we

have

𝑎

=

𝑚 𝑛

=

–𝑚 –𝑛

(because

𝑚(–

𝑛)

=

𝑛(–

𝑚)).

Then

–

𝑚,

–

𝑛

∈

𝑃ℤ,

and

so,

𝑎

∈

𝑃ℚ,

contrary to our assumption that 𝑎 ∉ 𝑃ℚ. If 𝑚 ∉ 𝑃ℤ and 𝑛 ∈ 𝑃ℤ, then – 𝑚 ∈ 𝑃ℤ, and therefore,

–

𝑎

=

–𝑚 𝑛

∈

𝑃ℚ.

If

𝑚

∈

𝑃ℤ

and

𝑛

∉

𝑃ℤ,

then

–

𝑛

∈

𝑃ℤ,

and

therefore,

–

𝑎

=

–𝑚 𝑛

=

𝑚 –𝑛

∈

𝑃ℚ.

So,

at

least

one

of 𝑎 ∈ 𝑃, 𝑎 = 0, or – 𝑎 ∈ 𝑃 holds.

If

𝑎 ∈ 𝑃ℚ

and

– 𝑎 ∈ 𝑃ℚ,

then

𝑎=𝑚
𝑛

and

–

𝑎

=

𝑝 𝑞

with

𝑚, 𝑛, 𝑝, 𝑞 ∈ 𝑃ℤ.

We

can

also

write

–𝑎

as

–

𝑎

=

–𝑛𝑚.

So,

–𝑚 𝑛

=

𝑝,
𝑞

and

thus,

(–

𝑚)𝑞

=

𝑛𝑝.

Since

𝑛,

𝑝

∈

𝑃ℤ,

we

have

𝑛𝑝

∈

𝑃ℤ.

Since

(–

𝑚)𝑞

=

𝑛𝑝,

we

must have (– 𝑚)𝑞 ∈ 𝑃ℤ. But – 𝑚 ∉ 𝑃ℤ, and so, – (– 𝑚) ∈ 𝑃ℤ. Since we also have 𝑞 ∈ 𝑃ℤ, we must have

– (– 𝑚)𝑞 ∈ 𝑃ℤ. But then by (3) for 𝑃ℤ, (– 𝑚)𝑞 ∉ 𝑃ℤ. This contradiction shows that we cannot have both

𝑎 ∈ 𝑃ℚ and – 𝑎 ∈ 𝑃ℚ.

If

𝑎

∈

𝑃ℚ,

then

𝑎

=

𝑚 𝑛

with

𝑚,

𝑛

∈

𝑃ℤ.

So,

𝑚

≠

0,

and

therefore,

𝑎

≠

0.

If

–

𝑎

∈

𝑃ℚ,

then

–

𝑎

=

𝑚 𝑛

with

𝑚, 𝑛 ∈ 𝑃ℤ. If 𝑎 = 0, then – 𝑎 = 0, and so, 𝑚 = 0. But 𝑚 ∈ 𝑃ℤ, and so, 𝑚 ≠ 0. Thus, 𝑎 ≠ 0.

If

𝑎

=

0,

then

we

have

0

=

0 1

∉

𝑃ℚ

and

–

0

=

–0 1

=

0 1

∉

𝑃ℚ.

It follows that (ℚ, +, ⋅) is an ordered field.

□

If (𝑅, +, ⋅) is an ordered ring and 𝑃 is the set of positive elements from the ring, we will write 𝑎 > 0 instead of 𝑎 ∈ 𝑃 and 𝑎 < 0 instead of – 𝑎 ∈ 𝑃. If 𝑎 − 𝑏 > 0, we will write 𝑎 > 𝑏 or 𝑏 < 𝑎.

We write 𝑎 ≥ 0 if 𝑎 ∈ 𝑃 or 𝑎 = 0, we write 𝑎 ≤ 0 if – 𝑎 ∈ 𝑃 or 𝑎 = 0, and we write 𝑎 ≥ 𝑏 or 𝑏 ≤ 𝑎 if 𝑎 − 𝑏 ≥ 0.

We may use the notation (𝑅, ≤) for an ordered ring, where ≤ is the relation defined in the last paragraph. Note that + and ⋅ aren’t explicitly mentioned, but of course they are still part of the ring.

53

In the future, we may just use the name of the set for the whole structure when there is no danger of confusion. For example, we may refer to the ring 𝑅 or the ordered field 𝐹 instead of the ring (𝑅, +, ⋅) or the ordered field (𝐹, ≤).

Fields are particularly nice to work with because all the arithmetic and algebra we’ve learned through

the years can be used in fields. For example, in the field of rational numbers, we can solve the equation

2𝑥 = 1. The multiplicative inverse property allows us to do this. Indeed, the multiplicative inverse of 2

is

12,

and

therefore,

𝑥

=

1 2

is

a

solution

to

the

given

equation.

Compare

this

to

the

ring

of

integers.

If

we

restrict ourselves to the integers, then the equation 2𝑥 = 1 has no solution.

Working with ordered fields is very nice as well. In the problem set below, you will be asked to derive some additional properties of fields and ordered fields that follow from the axioms. We will prove a few of these properties now as examples.
Theorem 5.2: Let (𝐹, ≤) be an ordered field. Then for all 𝑥 ∈ 𝐹∗, 𝑥 ⋅ 𝑥 > 0.

Proof: There are two cases to consider: (i) If 𝑥 > 0, then 𝑥 ⋅ 𝑥 > 0 by property (2) of an ordered field. (ii) If 𝑥 < 0, then – 𝑥 > 0, and so, (– 𝑥)(– 𝑥) > 0, again by property (2) of an ordered field. Now, using Problem 3 (parts (vi) and (vii)) in the problem set below, together with commutativity and associativity of multiplication, and the multiplicative identity property, we have
(– 𝑥)(– 𝑥) = (– 1𝑥)(– 1𝑥) = (– 1)(– 1)𝑥 ⋅ 𝑥 = 1(𝑥 ⋅ 𝑥) = 𝑥 ⋅ 𝑥.

So, again we have 𝑥 ⋅ 𝑥 > 0.

□

Theorem 5.3: Every ordered field (𝐹, ≤) contains a copy of the natural numbers. Specifically, 𝐹 contains a subset ℕ = {𝑛 | 𝑛 ∈ ℕ} such that for all 𝑛, 𝑚 ∈ ℕ, we have 𝑛 + 𝑚 = 𝑛 + 𝑚, 𝑛 ⋅ 𝑚 = 𝑛 ⋅ 𝑚, and 𝑛 < 𝑚 ↔ 𝑛 < 𝑚.

Proof: Let (𝐹, ≤) be an ordered field. By the definition of a field, 0, 1 ∈ 𝐹 and 0 ≠ 1.

We let 0 = 0 and 𝑛 = 1 + 1 + ⋯ + 1, where 1 appears 𝑛 times. Let ℕ = {𝑛 | 𝑛 ∈ ℕ}. Then ℕ ⊆ 𝐹.

We first prove by induction on 𝑚 that for all 𝑛, 𝑚 ∈ ℕ, 𝑛 + 𝑚 = 𝑛 + 𝑚.

Base case (𝑘 = 0): 𝑛 + 0 = 𝑛 = 𝑛 + 0 = 𝑛 + 0.

Inductive step: Suppose that 𝑛 + 𝑘 = 𝑛 + 𝑘. Then we have 𝑛 + (𝑘 + 1) = (𝑛 + 𝑘) + 1 = 𝑛 + 𝑘 + 1 = (𝑛 + 𝑘) + 1 = 𝑛 + (𝑘 + 1) = 𝑛 + 𝑘 + 1.

By the Principle of Mathematical Induction, for all natural numbers 𝑚, 𝑛 + 𝑚 = 𝑛 + 𝑚.

Similarly, we prove by induction on 𝑚 that for all 𝑛, 𝑚 ∈ ℕ, 𝑛 ⋅ 𝑚 = 𝑛 ⋅ 𝑚.

Base case (𝑘 = 0): 𝑛 ⋅ 0 = 0 = 𝑛 ⋅ 0.

54

Inductive step: Suppose that 𝑛 ⋅ 𝑘 = 𝑛 ⋅ 𝑘. Then we have 𝑛 ⋅ (𝑘 + 1) = 𝑛𝑘 + 𝑛 = 𝑛𝑘 + 𝑛 = 𝑛 ⋅ 𝑘 + 𝑛 = 𝑛(𝑘 + 1) = 𝑛(𝑘 + 1) = 𝑛(𝑘 + 1).

By the Principle of Mathematical Induction, for all natural numbers 𝑚, 𝑛 ⋅ 𝑚 = 𝑛 ⋅ 𝑚. We now wish to prove that for all 𝑛, 𝑚 ∈ ℕ, 𝑛 < 𝑚 ↔ 𝑛 < 𝑚.

We first note that for all 𝑛 ∈ ℕ, 𝑛 + 1 > 𝑛 because 𝑛 + 1 − 𝑛 = 𝑛 + 1 − 𝑛 = 1 = 1 ⋅ 1 > 0 by Theorem 5.2.
We now prove by induction on 𝑛 that for all 𝑛 ∈ ℕ with 𝑛 > 0 that 𝑛 > 0.

Base case (𝑘 = 1): 1 = 1 = 1 ⋅ 1 > 0 by Theorem 5.2.

Inductive step: Assume that 𝑘 > 0. Then 𝑘 + 1 = 𝑘 + 1 = 𝑘 + 1 > 0. Here we have used Order Property 1 together with 𝑘 > 0 and 1 > 0.

By the Principle of Mathematical Induction, for all natural numbers 𝑛 with 𝑛 > 0, we have 𝑛 > 0.

Conversely, if 𝑛 > 0, then 𝑛 ≠ 0 (because 0 = 0). Since 𝑛 is defined only for 𝑛 ≥ 0, we have 𝑛 > 0.

So, we have shown that for 𝑛 ∈ ℕ, 𝑛 > 0 if and only if 𝑛 > 0.

Next, note that if 𝑛 < 𝑚, then 𝑚 = (𝑚 − 𝑛) + 𝑛 = 𝑚 − 𝑛 + 𝑛. It follows that 𝑚 − 𝑛 = 𝑚 − 𝑛.

Finally, we have 𝑛 < 𝑚 ↔ 𝑚 − 𝑛 > 0 ↔ 𝑚 − 𝑛 > 0 ↔ 𝑚 − 𝑛 > 0 ↔ 𝑚 > 𝑛 ↔ 𝑛 < 𝑚.

□

Notes: (1) The function that sends 𝑛 ∈ ℕ to 𝑛 ∈ ℕ is called an isomorphism. It has the following properties: (i) 𝑛 + 𝑚 = 𝑛 + 𝑚, (ii) 𝑛 ⋅ 𝑚 = 𝑛 ⋅ 𝑚, and (iii) 𝑛 < 𝑚 if and only if 𝑛 < 𝑚. The function gives a one-to-one correspondence between the elements of ℕ and the elements of ℕ .

So, when we say that every field contains a “copy” of the natural numbers, we mean that there is a subset ℕ of the field so that (ℕ, ≤) is isomorphic to (ℕ, ≤) (note that addition and multiplication are preserved as well, even though they’re not explicitly mentioned in the notation).

(2) We will formally introduce isomorphisms in Lesson 11.

Theorem

5.4:

Let

(𝐹,

≤)

be

an

ordered

field

and

let

𝑥

∈

𝐹

with

𝑥

>

0.

Then

1 𝑥

>

0.

Proof:

Since

𝑥

≠

0,

1 𝑥

=

𝑥−1

exists

and

is

nonzero.

55

Assume

toward

contradiction

that

1 𝑥

<

0.

Then

–

1 𝑥

>

0.

Using

Problem

3

(part

(vi))

from

the

problem

set below, together with commutativity and associativity of multiplication, the multiplicative inverse

property,

and

the

multiplicative

identity

property,

𝑥 (– 1) = 𝑥(– 1)𝑥−1 = – 1𝑥𝑥−1 = – 1 ⋅ 1 = – 1.
𝑥

Since

𝑥

>

0

and

–

1 𝑥

>

0,

we

have

–

1

=

𝑥

(–

1𝑥)

>

0.

So,

1

≯

0.

But

by

Theorem

5.2,

1

=

1

⋅

1

>

0.

This

is

a

contradiction.

Therefore,

1 𝑥

>

0.

□

Why Isn’t ℚ Enough?
At first glance, it would appear that the ordered field of rational numbers would be sufficient to solve all “real world” problems. However, a long time ago, a group of people called the Pythagoreans showed that this was not the case. The problem was first discovered when applying the now well-known Pythagorean Theorem.

Theorem 5.5 (Pythagorean Theorem): In a right triangle with legs of lengths 𝑎 and 𝑏, and a hypotenuse of length 𝑐, 𝑐2 = 𝑎2 + 𝑏2.

The picture to the right shows a right triangle. The vertical and horizontal segments (labeled 𝑎 and 𝑏, respectively) are called the legs of the right triangle, and the side opposite the right angle (labeled 𝑐) is called the hypotenuse of the right triangle.

There are many ways to prove the Pythagorean Theorem. Here, we will

provide a simple geometric argument. For the proof we will want to

recall that the area of a square with side length 𝑠 is 𝐴 = 𝑠2, and the area of a triangle with base 𝑏 and

height

ℎ

is

𝐴

=

1 2

𝑏ℎ.

Notice

that

in

our

right

triangle

drawn

here,

the

base

is

labeled

𝑏

(how

convenient),

and

the

height

is

labeled

𝑎.

So,

the

area

of

this

right

triangle

is

𝐴

=

1 2

𝑏𝑎

=

1 2

𝑎𝑏.

Proof of Theorem 5.5: We draw 2 squares, each of side length 𝑎 + 𝑏, by rearranging 4 copies of the given triangle in 2 different ways:

56

We can get the area of each of these squares by adding the areas of all the figures that comprise each square.

The square on the left consists of 4 copies of the given right triangle, a square of side length 𝑎 and a

square

of

side

length

𝑏.

It

follows

that

the

area

of

this

square

is

4

⋅

1 2

𝑎𝑏

+

𝑎2

+

𝑏2

=

2𝑎𝑏

+

𝑎2

+

𝑏2.

The square on the right consists of 4 copies of the given right triangle, and a square of side length 𝑐. It

follows

that

the

area

of

this

square

is

4

⋅

1 2

𝑎𝑏

+

𝑐2

=

2𝑎𝑏

+

𝑐2.

Since the areas of both squares of side length 𝑎 + 𝑏 are equal (both areas are equal to (𝑎 + 𝑏)2), 2𝑎𝑏 + 𝑎2 + 𝑏2 = 2𝑎𝑏 + 𝑐2. Cancelling 2𝑎𝑏 from each side of this equation yields 𝑎2 + 𝑏2 = 𝑐2. □

Question: In a right triangle where both legs have length 1, what is the length of the hypotenuse?

Let’s try to answer this question. If we let 𝑐 be the length of the hypotenuse of the triangle, then by the Pythagorean Theorem, we have 𝑐2 = 12 + 12 = 1 + 1 = 2. Since 𝑐2 = 𝑐 ⋅ 𝑐, we need to find a number
with the property that when you multiply that number by itself you get 2. The Pythagoreans showed
that if we use only numbers in ℚ, then no such number exists.

Theorem 5.6: There does not exist a rational number 𝑎 such that 𝑎2 = 2.

Analysis: We will prove this Theorem by assuming that there is a rational number 𝑎 such that 𝑎2 = 2,

and

arguing

until

we

reach

a

contradiction.

A

first

attempt

at

a

proof

would

be

to

let

𝑎

=

𝑚 𝑛

∈

ℚ

satisfy

(𝑚)2
𝑛

=

2.

It

follows

that

𝒎𝟐

=

𝟐𝒏𝟐

(𝑚𝑛22

=

𝑚⋅𝑚 𝑛⋅𝑛

=

𝑚⋅𝑚
𝑛𝑛

=

(𝑚)2
𝑛

and

2

=

2 1

⇒

𝑚2 𝑛2

=

2 1

⇒

𝑚2

=

2𝑛2),

showing that 𝒎𝟐 is even. We will then use this information to show that both 𝑚 and 𝑛 are even (at

this point, you may want to try to use the two statements in bold to prove this yourself).

Now, in our first attempt, the fact that 𝑚 and 𝑛 both turned out to be even did not produce a contradiction. However, we can modify the beginning of the argument to make this happen.

Remember

that

every

rational

number

has

infinitely

many

representations.

For

example,

6 12

is

the

same

rational

number

as

2 4

(because

6

⋅

4

=

12

⋅

2).

Notice

that

in

both

representations,

the

numerator

(number on the top) and the denominator (number on the bottom) are even. However, they are both

equivalent to 12, which has the property that the numerator is not even.

In Problem 9 below, you will be asked to show that every rational number can be written in the form 𝑚𝑛 , where at least one of 𝑚 or 𝑛 is not even. We can now adjust our argument to get the desired contradiction.

Proof of Theorem 5.6: Assume, toward contradiction, that there is a rational number 𝑎 such that 𝑎2 = 2. Since 𝑎 is a rational number, there are 𝑚 ∈ ℤ and 𝑛 ∈ ℤ∗, not both even, so that 𝑎 = 𝑚𝑛 .

57

So,

we

have

𝑚2 𝑛2

=

𝑚⋅𝑚 𝑛⋅𝑛

=

𝑚 𝑛

⋅

𝑚 𝑛

=

𝑎

⋅

𝑎

=

𝑎2

=

2

=

2.
1

Thus,

𝑚2

⋅

1

=

𝑛2

⋅

2.

So,

𝑚2

=

2𝑛2.

Therefore,

𝑚2 is even. If 𝑚 were odd, then by Theorem 4.4 (from Lesson 4), 𝑚2 = 𝑚 ⋅ 𝑚 would be odd. So, 𝒎 is

even.

Since 𝑚 is even, there is 𝑘 ∈ ℤ such that 𝑚 = 2𝑘. Replacing 𝑚 by 2𝑘 in the equation 𝑚2 = 2𝑛2 gives us 2𝑛2 = 𝑚2 = (2𝑘)2 = (2𝑘)(2𝑘) = 2(𝑘(2𝑘)). So, 𝑛2 = 𝑘(2𝑘) = (𝑘 ⋅ 2)𝑘 = (2𝑘)𝑘 = 2(𝑘 ⋅ 𝑘). So, we see that 𝑛2 is even, and again by Theorem 4.4, 𝒏 is even.

So, we have 𝑚 even and 𝑛 even, contrary to our original assumption that 𝑚 and 𝑛 are not both even.

Therefore, there is no rational number 𝑎 such that 𝑎2 = 2.

□

So, the big question is, “Is there an ordered field 𝐹 with 𝐹 containing ℚ and 𝑎 ∈ 𝐹 such that 𝑎2 = 2?” Spoiler Alert! There is! We call it ℝ, the ordered field of real numbers.

Completeness
Let (𝐹, ≤) be an ordered field and let 𝑆 be a nonempty subset of 𝐹. We say that 𝑆 is bounded above if there is 𝑀 ∈ 𝐹 such that for all 𝑠 ∈ 𝑆, 𝑠 ≤ 𝑀. Each such number 𝑀 is called an upper bound of 𝑆.

In words, an upper bound of a set 𝑆 is simply an element from the field that is at least as big as every element in 𝑆.

Similarly, we say that 𝑆 is bounded below if there is 𝐾 ∈ 𝐹 such that for all 𝑠 ∈ 𝑆, 𝐾 ≤ 𝑠. Each such number 𝐾 is called a lower bound of 𝑆.

In words, a lower bound of a set 𝑆 is simply an element from the field that is no bigger than any element in 𝑆.

We will say that 𝑆 is bounded if it is both bounded above and bounded below. Otherwise 𝑆 is unbounded.

A least upper bound of a set 𝑆 is an upper bound that is smaller than any other upper bound of 𝑆, and a greatest lower bound of 𝑆 is a lower bound that is larger than any other lower bound of 𝑆.

Example 5.2: Let (𝐹, ≤) be an ordered field with ℚ ⊆ 𝐹.

Note: The only two examples of 𝐹 that we are interested in right now are ℚ (the set of rational numbers) and ℝ (the set of real numbers). Although we haven’t finished defining the real numbers, you probably have some intuition as to what they look like—after all, this is the number system you have used throughout high school. As you look at the set in each example below, think about what it looks like as a subset of ℚ and as a subset of ℝ.

1. 𝑆 = {1, 2, 3, 4, 5} is bounded.
5 is an upper bound of 𝑆, as is any number larger than 5. The number 5 is special in the sense that there are no upper bounds smaller than it. So, 5 is the least upper bound of 𝑆.

58

Similarly, 1 is a lower bound of 𝑆, as is any number smaller than 1. The number 1 is the greatest lower bound of 𝑆 because there are no lower bounds larger than it.

Notice that the least upper bound and greatest lower bound of 𝑆 are inside the set 𝑆 itself. This will always happen when the set 𝑆 is finite.

2. 𝑇 = {𝑥 ∈ 𝐹 | – 2 < 𝑥 ≤ 2} is also bounded. Any number greater than or equal to 2 is an upper bound of 𝑇, and any number less than or equal to – 2 is a lower bound of 𝑇.

2 is the least upper bound of 𝑇 and – 2 is the greatest lower bound of 𝑇.

Note that the least upper bound of 𝑇 is in 𝑇, whereas the greatest lower bound of 𝑇 is not in 𝑇.

3. 𝑈 = {𝑥 ∈ 𝐹 | 𝑥 < – 3} is bounded above by any number greater than or equal to – 3, and – 3 is the least upper bound of 𝑈. The set 𝑈 is not bounded below, and therefore, 𝑈 is unbounded.

4. 𝑉 = {𝑥 ∈ 𝐹 | 𝑥2 < 2} is bounded above by 2. To see this, note that if 𝑥 > 2, then 𝑥2 > 4 ≥ 2, and therefore, 𝑥 ∉ 𝑉. Any number greater than 2 is also an upper bound.

Is 2 the least upper bound of 𝑉? It’s not! For example, 3 is also an upper bound. Indeed, if

2

𝑥

>

32,

then

𝑥2

>

9 4

≥

2

(the

reader

should

verify

that

for

all

𝑎,

𝑏

∈

ℝ+,

𝑎

>

𝑏

→

𝑎2

>

𝑏2).

Does 𝑉 have a least upper bound? A moment’s thought might lead you to suspect that a least upper bound 𝑀 would satisfy 𝑀2 = 2. And it turns out that you are right! (Proving this,
however, is quite difficult). Clearly, this least upper bound 𝑀 is not in the set 𝑉. The big question
is “Does 𝑀 exist at all?”

Well, if 𝐹 = ℚ, then by Theorem 5.6, 𝑀 does not exist in 𝐹. In this case, 𝑉 is an example of a set which is bounded above in ℚ, but has no least upper bound in ℚ.

So, if we want an ordered field 𝐹 containing ℚ where 𝑀 does exist, we can insist that 𝐹 has the property that any set which is bounded above in 𝐹 has a least upper bound in 𝐹. It turns out that there is exactly one such ordered field (up to renaming the elements) and we call it the ordered field of real numbers, ℝ.

Many authors use the term supremum for “least upper bound” and infimum for “greatest lower bound,” and they may write sup 𝐴 and inf 𝐴 for the supremum and infimum of a set 𝐴, respectively (if they exist).

In the examples above, we stated the least upper bound and greatest lower bound of the sets 𝑆, 𝑇, 𝑈, and 𝑉 without proof. Intuitively, it seems reasonable that those numbers are correct. Let’s do one of the examples carefully.

Theorem 5.7: Let 𝑈 = {𝑥 ∈ 𝐹 | 𝑥 < – 3}. Then sup 𝑈 = – 3.

Analysis: We need to show that – 3 is an upper bound of 𝑈, and that any number less than – 3 is not an upper bound of 𝑈. That – 3 is an upper bound of 𝑈 follows immediately from the definition of 𝑈.

The harder part of the argument is showing that a number less than – 3 is not an upper bound of 𝑈. However, conceptually it’s not hard to see that this is true. If 𝑎 < – 3, we simply need to find some number 𝑥 between 𝑎 and – 3. Here is a picture of the situation.

59

𝑥

Notice that 𝑎 can be very close to – 3 and we don’t know exactly what 𝑎 is—we know only that it’s less

than – 3. So, we need to be careful how we choose 𝑥. The most natural choice for 𝑥 would be to go

midway between 𝑎 and – 3. In other words, we can take the average of 𝑎 and – 3. So, we will let

𝑥

=

1 2

(𝑎

+

(–

3)).

Then

we

just

need

to

verify

that

𝑎

<

𝑥

and

that

𝑥

∈

𝑈

(that

is,

𝑥

<

–

3).

Proof of Theorem 5.7: If 𝑥 ∈ 𝑈, then 𝑥 < – 3 by definition, and so, – 3 is an upper bound of 𝑈.

Suppose that 𝑎 < – 3 (or equivalently, – 𝑎 − 3 > 0). We want to show that 𝑎 is not an upper bound of 𝑈. To do this, we let 𝑥 = 1 (𝑎 − 3) = 2−1(𝑎 + (– 3)). 𝑥 ∈ 𝐹 because 𝐹 is closed under addition and
2
multiplication, and the multiplicative inverse property holds in 𝐹∗. We will show that 𝑎 < 𝑥 < – 3.

1

1

1

1

1

1

𝑥 − 𝑎 = 2 (𝑎 − 3) − 𝑎 = 2 (𝑎 − 3) − 2 (2𝑎) = 2 (𝑎 − 3 − 2𝑎) = 2 (𝑎 − 2𝑎 − 3) = 2 (– 𝑎 − 3).

Since

1 2

>

0

(by

Theorem

5.4)

and

–

𝑎

−

3

>

0,

it

follows

that

𝑥

−

𝑎

>

0,

and

therefore,

𝑥

>

𝑎.

–

3

−

𝑥

=

–

3

−

1 2

(𝑎

−

3)

=

1 2

(–

6)

−

1 2

𝑎

+

1 2

⋅

3

=

1 2

(–

6

−

𝑎

+

3)

=

1 2

(–

𝑎

−

3).

Again,

since

1 2

>

0

and

–

𝑎

−

3

>

0,

it

follows

that

–

3

−

𝑥

>

0,

and

therefore,

𝑥

<

–

3.

Thus,

𝑥

∈

𝑈.

So, we found an element 𝑥 ∈ 𝑈 (because 𝑥 < – 3) with 𝑎 < 𝑥. This shows that 𝑎 is not an upper bound

of 𝑈. It follows that – 3 = sup 𝑈.

□

An ordered field (𝐹, ≤) has the Completeness Property if every nonempty subset of 𝐹 that is bounded above in 𝐹 has a least upper bound in 𝐹. In this case, we say that (𝐹, ≤) is a complete ordered field.

Theorem 5.8: There is exactly one complete ordered field (up to renaming the elements).

The proof of Theorem 5.8 is quite long and requires some machinery that we haven’t yet developed. We will therefore accept it as true for the purpose of this book, and we let ℝ be the unique complete ordered field guaranteed to exist by the theorem.

We will finish this section by proving two useful theorems about the complete ordered field ℝ.

Theorem 5.9 (The Archimedean Property of ℝ): For every 𝑥 ∈ ℝ, there is 𝑛 ∈ ℕ such that 𝑛 > 𝑥.

In other words, the Archimedean Property says that the set of natural numbers is unbounded in the reals. In particular, the set of natural numbers is not bounded from above in the set of real numbers.

60

We will prove this theorem by contradiction using the Completeness Property of the reals. If we (wrongly) assume that the set of natural numbers is bounded from above, then the Completeness Property of the reals gives us a least upper bound 𝑥. Since 𝑥 is a least upper bound, 𝑥 − 1 is not an upper bound. Do you see the problem yet? If 𝑥 − 1 < 𝑛 ∈ ℕ, then 𝑥 < 𝑛 + 1. But then 𝑥 is not an upper bound for the set of natural numbers, contrary to our assumption. Let’s write out the details.

Proof: Suppose toward contradiction that ℕ is bounded from above. By the Completeness Property of

ℝ, 𝑥 = sup ℕ exists. Since 𝑥 − 1 is not an upper bound for ℕ, there is 𝑛 ∈ ℕ such that 𝑥 − 1 < 𝑛. Then

we have 𝑥 = 𝑥 + (– 1 + 1) = (𝑥 − 1) + 1 < 𝑛 + 1. Since ℕ is closed under addition, 𝑛 + 1 ∈ ℕ. So, 𝑥

is not an upper bound for ℕ, contradicting the fact that 𝑥 = sup ℕ. It follows that ℕ is not bounded

from above. So, for every 𝑥 ∈ ℝ, there is 𝑛 ∈ ℕ such that 𝑛 > 𝑥.

□

Theorem 5.10 (The Density Theorem): If 𝑥, 𝑦 ∈ ℝ with 𝑥 < 𝑦, then there is 𝑞 ∈ ℚ with 𝑥 < 𝑞 < 𝑦.

In other words, the Density Theorem says that between any two real numbers we can always find a rational number. We say that ℚ is dense in ℝ.

To help understand the proof, let’s first run a simple simulation using a specific example. Let’s let

𝑥

=

16 3

and

𝑦

=

137.

We

begin

by

subtracting

to

get

𝑦

−

𝑥

=

13.

This

is

the

distance

between

𝑥

and

𝑦.

We

wish to find a natural number 𝑛 such that 1 is smaller than this distance. In other words, we want

𝑛

1 𝑛

<

13,

or

equivalently,

𝑛

>

3.

So,

we

can

let

𝑛

be

any

natural

number

greater

than

3,

say

𝑛

=

4.

We

now want to “shift” 1 = 1 to the right to get a rational number between 𝑥 and 𝑦. We can do this as

𝑛4

follows.

We

multiply

𝑛

times

𝑥

to

get

𝑛𝑥

=

4

⋅

16 3

=

634.

We

then

let

𝑚

be

the

least

integer

greater

than

𝑛𝑥.

So,

𝑚

=

66 3

=

22.

Finally,

we

let

𝑞

=

𝑚 𝑛

=

22 4

=

121.

And

we

did

it!

Indeed,

we

have

16 3

<

11 2

<

137.

The

reader should confirm that these inequalities hold. Let’s write out the details of the proof.

Proof: Let’s first consider the case where 0 ≤ 𝑥 < 𝑦. Let 𝑧 = 𝑦 − 𝑥 = 𝑦 + (– 𝑥). Since ℝ has the

additive inverse property and is closed under addition, 𝑧 ∈ ℝ. Also, 𝑧 > 0. By the Archimedean

Property, there is 𝑛 ∈ ℕ such that 𝑛 > 1𝑧. Using Problem 5 (part (v)) in the problem set below, we have

1 𝑛

<

𝑧.

By

the

Archimedean

Property

once

again,

there

is

𝑚

∈

ℕ

such

that

𝑚

>

𝑛𝑥.

Therefore,

𝑚 𝑛

>

𝑥

(Check

this!).

So,

{𝑚

∈

ℕ

|

𝑚 𝑛

>

𝑥}

≠

∅.

By

the

Well

Ordering

Principle,

{𝑚

∈

ℕ

|

𝑚 𝑛

>

𝑥}

has

a

least

element, let’s call it 𝑘. Since 𝑘 > 0, (because 𝑥 ≥ 0 and 𝑛 > 0) and 𝑘 is the least natural number such

that

𝑘 𝑛

>

𝑥,

it

follows

that

𝑘

−

1

∈

ℕ

and

𝑘−1 𝑛

≤

𝑥,

or

equivalently,

𝑘 𝑛

−

1 𝑛

≤

𝑥.

Therefore,

we

have

𝑘 𝑛

≤

𝑥

+

1 𝑛

<

𝑥

+

𝑧

=

𝑥

+

(𝑦

−

𝑥)

=

𝑦.

Thus,

𝑥

<

𝑘 𝑛

<

𝑦.

Since

𝑘,

𝑛

∈

ℕ,

we

have

𝑘 𝑛

∈

ℚ.

Now, we consider the case where 𝑥 < 0 and 𝑥 < 𝑦. By the Archimedean Property, there is 𝑡 ∈ ℕ such that 𝑡 > – 𝑥. Then, we have 0 < 𝑥 + 𝑡 < 𝑦 + 𝑡. So, 𝑥 + 𝑡 and 𝑦 + 𝑡 satisfy the first case above. Thus, there is 𝑞 ∈ ℚ with 𝑥 + 𝑡 < 𝑞 < 𝑦 + 𝑡. It follows that 𝑥 < 𝑞 − 𝑡 < 𝑦. Since 𝑡 ∈ ℕ, – 𝑡 ∈ ℤ. Since ℤ ⊆ ℚ, – 𝑡 ∈ ℚ. So, we have 𝑞, – 𝑡 ∈ ℚ. Since ℚ is closed under addition, 𝑞 − 𝑡 = 𝑞 + (– 𝑡) ∈ ℚ. □

61

Problem Set 5

LEVEL 1

Full solutions to these problems are available for free download here:
www.SATPrepGet800.com/PMFBXSG

1. The addition and multiplication tables below are defined on the set 𝑆 = {0, 1, 2}. Show that (𝑆, +, ⋅) does not define a field.

+0 1 2 0 0 1 2 1 1 2 0 2 2 0 1

⋅

0

1

2

0 0 0 0

1 0 1 2

2 0 2 2

2. Let 𝐹 = {0, 1}, where 0 ≠ 1. Show that there is exactly one field (𝐹, +, ⋅), where 0 is the additive identity and 1 is the multiplicative identity.
LEVEL 2

3. Let (𝐹, +, ⋅) be a field. Prove each of the following: (i) If 𝑎, 𝑏 ∈ 𝐹 with 𝑎 + 𝑏 = 𝑏, then 𝑎 = 0. (ii) If 𝑎 ∈ 𝐹, 𝑏 ∈ 𝐹∗, and 𝑎𝑏 = 𝑏, then 𝑎 = 1. (iii) If 𝑎 ∈ 𝐹, then 𝑎 ⋅ 0 = 0. (iv) If 𝑎 ∈ 𝐹∗, 𝑏 ∈ 𝐹, and 𝑎𝑏 = 1, then 𝑏 = 1.
𝑎
(v) If 𝑎, 𝑏 ∈ 𝐹 and 𝑎𝑏 = 0, then 𝑎 = 0 or 𝑏 = 0. (vi) If 𝑎 ∈ 𝐹, then – 𝑎 = – 1𝑎 (vii) (– 1)(– 1) = 1.

4. Let (𝐹, +, ⋅) be a field with ℕ ⊆ 𝐹. Prove that ℚ ⊆ 𝐹.
LEVEL 3

5. Let (𝐹, ≤) be an ordered field. Prove each of the following:

(i) If 𝑎, 𝑏 ∈ 𝐹, exactly one of the following holds: 𝑎 < 𝑏, 𝑎 = 𝑏, or 𝑎 > 𝑏.

(ii) If 𝑎, 𝑏 ∈ 𝐹, 𝑎 ≤ 𝑏, and 𝑏 ≤ 𝑎, then 𝑎 = 𝑏.

(iii) If 𝑎, 𝑏, 𝑐 ∈ 𝐹, 𝑎 < 𝑏, and 𝑏 < 𝑐, then 𝑎 < 𝑐.

(iv) If 𝑎, 𝑏, 𝑐 ∈ 𝐹, 𝑎 ≤ 𝑏, and 𝑏 ≤ 𝑐, then 𝑎 ≤ 𝑐.

(v)

If

𝑎,

𝑏

∈

𝐹+

and

𝑎

>

𝑏,

then

1 𝑎

<

𝑏1.

(vi) If 𝑎, 𝑏 ∈ 𝐹, then 𝑎 > 𝑏 if and only if – 𝑎 < – 𝑏.

(vii) If 𝑎, 𝑏 ∈ 𝐹, then 𝑎 ≥ 𝑏 if and only if – 𝑎 ≤ – 𝑏.

62

6. Let (𝐹, +, ⋅) be a field. Show that (𝐹, ⋅) is a commutative monoid.
LEVEL 4
7. Prove that there is no smallest positive real number. 8. Let 𝑎 be a nonnegative real number. Prove that 𝑎 = 0 if and only if 𝑎 is less than every positive
real number. (Note: 𝑎 nonnegative means that 𝑎 is positive or zero.) 9. Prove that every rational number can be written in the form 𝑚𝑛 , where 𝑚 ∈ ℤ, 𝑛 ∈ ℤ∗, and at least
one of 𝑚 or 𝑛 is not even.
LEVEL 5
10. Show that every nonempty set of real numbers that is bounded below has a greatest lower bound in ℝ.
11. Show that between any two real numbers there is a real number that is not rational. 12. Let 𝑇 = {𝑥 ∈ 𝐹 | – 2 < 𝑥 ≤ 2}. Prove sup 𝑇 = 2 and inf 𝑇 = – 2.
CHALLENGE PROBLEM
13. Let 𝑉 = {𝑥 ∈ 𝐹 | 𝑥2 < 2} and let 𝑎 = sup 𝑉. Prove that 𝑎2 = 2.
63

LESSON 6 – TOPOLOGY THE TOPOLOGY OF ℝ

Intervals of Real Numbers
A set 𝐼 of real numbers is called an interval if any real number that lies between two numbers in 𝐼 is also in 𝐼. Symbolically, we can write
∀𝑥, 𝑦 ∈ 𝐼 ∀𝑧 ∈ ℝ (𝑥 < 𝑧 < 𝑦 → 𝑧 ∈ 𝐼).

The expression above can be read “For all 𝑥, 𝑦 in 𝐼 and all 𝑧 ∈ ℝ, if 𝑥 is less than 𝑧 and 𝑧 is less than 𝑦, then 𝑧 is in 𝐼.”

Example 6.1:

1. The set 𝐴 = {0, 1} is not an interval. 𝐴 consists of just the two real numbers 0 and 1. There are

infinitely

many

real

numbers

between

0

and

1.

For

example,

the

real

number

1 2

satisfies

0 < 1 < 1, but 1 ∉ 𝐴.

2

2

2. ℝ is an interval. This follows trivially from the definition. If we replace 𝐼 by ℝ, we get ∀𝑥, 𝑦 ∈ ℝ ∀𝑧 ∈ ℝ (𝑥 < 𝑧 < 𝑦 → 𝑧 ∈ ℝ). In other words, if we start with two real numbers, and

take a real number between them, then that number is a real number (which we already said).

When we are thinking of ℝ as an interval, we sometimes use the notation (– ∞, ∞) and refer to this as the real line. The following picture gives the standard geometric interpretation of the real line.

In addition to the real line, there are 8 other types of intervals.

Open Interval:

(𝑎, 𝑏) = {𝑥 ∈ ℝ | 𝑎 < 𝑥 < 𝑏}

Closed Interval:

[𝑎, 𝑏] = {𝑥 ∈ ℝ | 𝑎 ≤ 𝑥 ≤ 𝑏}

Half-open Intervals:

(𝑎, 𝑏] = {𝑥 ∈ ℝ | 𝑎 < 𝑥 ≤ 𝑏} [𝑎, 𝑏) = {𝑥 ∈ ℝ | 𝑎 ≤ 𝑥 < 𝑏}

Infinite Open Intervals: (𝑎, ∞) = {𝑥 ∈ ℝ | 𝑥 > 𝑎}

(– ∞, 𝑏) = {𝑥 ∈ ℝ | 𝑥 < 𝑏}

Infinite Closed Intervals: [𝑎, ∞) = {𝑥 ∈ ℝ | 𝑥 ≥ 𝑎}

(– ∞, 𝑏] = {𝑥 ∈ ℝ | 𝑥 ≤ 𝑏}

It’s easy to check that each of these eight types of sets satisfies the definition of being an interval. Conversely, every interval has one of these nine forms. This will follow immediately from Theorem 6.1 and Problem 4 below.

Note that the first four intervals above (the open, closed, and two half-open intervals) are bounded. They are each bounded below by 𝑎 and bounded above by 𝑏. In fact, for each of these intervals, 𝑎 is the greatest lower bound and 𝑏 is the least upper bound. Using the notation from Lesson 5, we have for example, 𝑎 = inf(𝑎, 𝑏) and 𝑏 = sup(𝑎, 𝑏).

64

Example 6.2: 1. The half-open interval (– 2,1] = {𝑥 ∈ ℝ | – 2 < 𝑥 ≤ 1} has the following graph:
2. The infinite open interval (0, ∞) = {𝑥 ∈ ℝ | 𝑥 > 0} has the following graph:
Theorem 6.1: If an interval 𝐼 is bounded, then there are 𝑎, 𝑏 ∈ ℝ such that one of the following holds: 𝐼 = (𝑎, 𝑏), 𝐼 = [𝑎, 𝑏], 𝐼 = (𝑎, 𝑏], or 𝐼 = [𝑎, 𝑏).
Analysis: We will prove this by letting 𝑎 = inf 𝐼 and 𝑏 = sup 𝐼 (in other words, 𝑎 is the greatest lower bound of 𝐼 and 𝑏 is the least upper bound of 𝐼), and then doing each of the following:
(1) We will show 𝐼 ⊆ [𝑎, 𝑏]. (2) We will show (𝑎, 𝑏) ⊆ 𝐼. (3) We will then look at 4 different cases. As one sample case, if 𝑎, 𝑏 ∈ 𝐼, then we will have
𝐼 ⊆ [𝑎, 𝑏] and [𝑎, 𝑏] ⊆ 𝐼. It then follows from the “Axiom of Extensionality” that 𝐼 = [𝑎, 𝑏]. Recall: Given sets 𝑋 and 𝑌, the Axiom of Extensionality says that 𝑋 and 𝑌 are the same set if and only if 𝑋 and 𝑌 have precisely the same elements (See the technical note following Theorem 2.5 in Lesson 2). In symbols,
𝑋 = 𝑌 if and only if ∀𝑥(𝑥 ∈ 𝑋 ↔ 𝑥 ∈ 𝑌).
Since ∀𝑥(𝑥 ∈ 𝑋 ↔ 𝑥 ∈ 𝑌) is logically equivalent to ∀𝑥(𝑥 ∈ 𝑋 → 𝑥 ∈ 𝑌) ∧ ∀𝑥(𝑥 ∈ 𝑌 → 𝑥 ∈ 𝑋), we have
𝑋 = 𝑌 if and only if ∀𝑥(𝑥 ∈ 𝑋 → 𝑥 ∈ 𝑌) and ∀𝑥(𝑥 ∈ 𝑌 → 𝑥 ∈ 𝑋).
Therefore, to show that 𝑋 = 𝑌, we can instead show that 𝑋 ⊆ 𝑌 and 𝑌 ⊆ 𝑋. This is the approach we will take in the proof below.
Proof of Theorem 6.1: Let 𝐼 be a bounded interval. Since 𝐼 is bounded, by the Completeness of ℝ, 𝐼 has a least upper bound 𝑏. By Problem 10 in Lesson 5, 𝐼 has a greatest lower bound 𝑎. If 𝑥 ∈ 𝐼, then by the definitions of upper bound and lower bound, we have 𝑥 ∈ [𝑎, 𝑏]. Since 𝑥 was an arbitrary element of 𝐼, ∀𝑥(𝑥 ∈ 𝐼 → 𝑥 ∈ [𝑎, 𝑏]). So, 𝐼 ⊆ [𝑎, 𝑏].
Now, let 𝑧 ∈ (𝑎, 𝑏). It follows that 𝑎 < 𝑧 < 𝑏. Since 𝑏 is the least upper bound of 𝐼, 𝑧 is not an upper bound of 𝐼. So, there is 𝑦 ∈ 𝐼 with 𝑧 < 𝑦. Since 𝑎 is the greatest lower bound of 𝐼, 𝑧 is not a lower bound of 𝐼. So, there is 𝑥 ∈ 𝐼 with 𝑥 < 𝑧. Since 𝐼 is an interval, 𝑥, 𝑦 ∈ 𝐼, and 𝑥 < 𝑧 < 𝑦, it follows that 𝑧 ∈ 𝐼. Since 𝑧 was an arbitrary element of (𝑎, 𝑏), we have shown ∀𝑥(𝑥 ∈ (𝑎, 𝑏) → 𝑥 ∈ 𝐼). So, (𝑎, 𝑏) ⊆ 𝐼.
We have shown that (𝑎, 𝑏) ⊆ 𝐼 and 𝐼 ⊆ [𝑎, 𝑏]. There are now 4 cases to consider.
65

Case 1: If both the greatest lower bound of 𝐼 (namely, 𝑎) and the least upper bound of 𝐼 (namely, 𝑏) are elements of 𝐼, then we have [𝑎, 𝑏] ⊆ 𝐼 and 𝐼 ⊆ [𝑎, 𝑏]. So, 𝐼 = [𝑎, 𝑏].

Case 2: If 𝑎 ∈ 𝐼 and 𝑏 ∉ 𝐼, then we have [𝑎, 𝑏) ⊆ 𝐼 and 𝐼 ⊆ [𝑎, 𝑏). So, 𝐼 = [𝑎, 𝑏).

Case 3: If 𝑎 ∉ 𝐼 and 𝑏 ∈ 𝐼, then we have (𝑎, 𝑏] ⊆ 𝐼 and 𝐼 ⊆ (𝑎, 𝑏]. So, 𝐼 = (𝑎, 𝑏].

Case 4: If 𝑎 ∉ 𝐼 and 𝑏 ∉ 𝐼, then we have (𝑎, 𝑏) ⊆ 𝐼 and 𝐼 ⊆ (𝑎, 𝑏). So, 𝐼 = (𝑎, 𝑏).

□

Note: You will be asked to prove the analogous result for unbounded intervals in Problem 4 below.

Operations on Sets

In Lesson 2 we saw how to take the union and intersection of two sets. We now review the definitions from that lesson and introduce a few more.
The union of the sets 𝐴 and 𝐵, written 𝐴 ∪ 𝐵, is the set of elements that are in 𝐴 or 𝐵 (or both). 𝐴 ∪ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵}

The intersection of 𝐴 and 𝐵, written 𝐴 ∩ 𝐵, is the set of elements that are simultaneously in 𝐴 and 𝐵. 𝐴 ∩ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 and 𝑥 ∈ 𝐵}

The following Venn diagrams for the union and intersection of two sets can be useful for visualizing these operations. As usual, 𝑈 is some “universal” set that contains both 𝐴 and 𝐵.

𝑨∪𝑩

𝑨∩𝑩

The difference 𝐴 ∖ 𝐵 is the set of elements that are in 𝐴 and not in 𝐵. 𝐴 ∖ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 and 𝑥 ∉ 𝐵}

The symmetric difference between 𝐴 and 𝐵, written 𝐴 Δ 𝐵, is the set of elements that are in 𝐴 or 𝐵, but not both.
𝐴 Δ 𝐵 = (𝐴 ∖ 𝐵) ∪ (𝐵 ∖ 𝐴)

Let’s also look at Venn diagrams for the difference and symmetric difference of two sets.

66

𝑨∖𝑩 Example 6.3: Let 𝐴 = {0, 1, 2, 3, 4} and 𝐵 = {3, 4, 5, 6}. We have

𝑨 𝚫 𝑩

1. 𝐴 ∪ 𝐵 = {0, 1, 2, 3, 4, 5, 6} 2. 𝐴 ∩ 𝐵 = {3, 4} 3. 𝐴 ∖ 𝐵 = {0, 1, 2} 4. 𝐵 ∖ 𝐴 = {5, 6} 5. 𝐴 Δ 𝐵 = {0, 1, 2} ∪ {5,6} = {0, 1, 2, 5, 6}
Example 6.4: Let 𝐴 = (– 2,1] and 𝐵 = (0, ∞). We have 1. 𝐴 ∪ 𝐵 = (– 2, ∞) 2. 𝐴 ∩ 𝐵 = (0,1] 3. 𝐴 ∖ 𝐵 = (– 2,0] 4. 𝐵 ∖ 𝐴 = (1, ∞) 5. 𝐴 Δ 𝐵 = (– 2,0] ∪ (1, ∞)
Note: If you have trouble seeing how to compute these, it may be helpful to draw the graphs of 𝐴 and 𝐵 lined up vertically, and then draw vertical lines through the endpoints of each interval.
𝐴
𝐵

The results follow easily by combining these graphs into a single graph using the vertical lines as guides. For example, let’s look at 𝐴 ∩ 𝐵 in detail. We’re looking for all numbers that are in both 𝐴 and 𝐵. The two rightmost vertical lines drawn passing through the two graphs above isolate all those numbers nicely. We see that all numbers between 0 and 1 are in the intersection. We should then think about the two endpoints 0 and 1 separately. 0 ∉ 𝐵 and therefore, 0 cannot be in the intersection of 𝐴 and 𝐵. On the other hand, 1 ∈ 𝐴 and 1 ∈ 𝐵. Therefore, 1 ∈ 𝐴 ∩ 𝐵. So, we see that 𝐴 ∩ 𝐵 = (0,1].

67

Unions and intersections have many nice algebraic properties such as commutativity (𝐴 ∪ 𝐵 = 𝐵 ∪ 𝐴 and 𝐴 ∩ 𝐵 = 𝐵 ∩ 𝐴), associativity ((𝐴 ∪ 𝐵) ∪ 𝐶 = 𝐴 ∪ (𝐵 ∪ 𝐶) and (𝐴 ∩ 𝐵) ∩ 𝐶 = 𝐴 ∩ (𝐵 ∩ 𝐶)), and distributivity (𝐴 ∩ (𝐵 ∪ 𝐶) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶) and 𝐴 ∪ (𝐵 ∩ 𝐶) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)).
As an example, let’s prove that the operation of forming unions is associative. You will be asked to prove similar results in the problems below.
Theorem 6.2: The operation of forming unions is associative.
Note: Before beginning the proof, let’s draw Venn diagrams of the situation to convince ourselves that the theorem is true.

𝑨∪𝑩

𝑩∪𝑪

(𝑨 ∪ 𝑩) ∪ 𝑪 = 𝑨 ∪ (𝑩 ∪ 𝑪) Proof of Theorem 6.2: Let 𝐴, 𝐵, and 𝐶 be sets, and let 𝑥 ∈ (𝐴 ∪ 𝐵) ∪ 𝐶. Then 𝑥 ∈ 𝐴 ∪ 𝐵 or 𝑥 ∈ 𝐶. If 𝑥 ∈ 𝐶, then 𝑥 ∈ 𝐵 or 𝑥 ∈ 𝐶. So, 𝑥 ∈ 𝐵 ∪ 𝐶. Then 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵 ∪ 𝐶. So, 𝑥 ∈ 𝐴 ∪ (𝐵 ∪ 𝐶). If, on the other hand, 𝑥 ∈ 𝐴 ∪ 𝐵, then 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵. If 𝑥 ∈ 𝐴, then 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵 ∪ 𝐶. So, 𝑥 ∈ 𝐴 ∪ (𝐵 ∪ 𝐶). If 𝑥 ∈ 𝐵, then 𝑥 ∈ 𝐵 or 𝑥 ∈ 𝐶. So, 𝑥 ∈ 𝐵 ∪ 𝐶. Then 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵 ∪ 𝐶. So, 𝑥 ∈ 𝐴 ∪ (𝐵 ∪ 𝐶). Since 𝑥 was arbitrary, we have shown ∀𝑥(𝑥 ∈ (𝐴 ∪ 𝐵) ∪ 𝐶 → 𝑥 ∈ 𝐴 ∪ (𝐵 ∪ 𝐶)). Therefore, we have shown that (𝐴 ∪ 𝐵) ∪ 𝐶 ⊆ 𝐴 ∪ (𝐵 ∪ 𝐶).
A similar argument can be used to show 𝐴 ∪ (𝐵 ∪ 𝐶) ⊆ (𝐴 ∪ 𝐵) ∪ 𝐶 (the reader should write out the details).
68

Since (𝐴 ∪ 𝐵) ∪ 𝐶 ⊆ 𝐴 ∪ (𝐵 ∪ 𝐶) and 𝐴 ∪ (𝐵 ∪ 𝐶) ⊆ (𝐴 ∪ 𝐵) ∪ 𝐶, (𝐴 ∪ 𝐵) ∪ 𝐶 = 𝐴 ∪ (𝐵 ∪ 𝐶), and

therefore, the operation of forming unions is associative.

□

Remember that associativity allows us to drop parentheses. So, we can now simply write 𝐴 ∪ 𝐵 ∪ 𝐶 when taking the union of the three sets 𝐴, 𝐵, and 𝐶.

Recall from Lesson 2 that sets 𝐴 and 𝐵 are called disjoint or mutually exclusive if 𝐴 ∩ 𝐵 = ∅. For example, the sets (−2, 0] and (1, ∞) are disjoint intervals. Here is a typical Venn diagram of disjoint sets 𝐴 and 𝐵.

𝑨∩𝑩=∅
In topology, we will often want to look at unions and intersections of more than two sets. Therefore, we make the following more general definitions.

Let 𝑿 be a nonempty set of sets. ⋃𝑿 = {𝑦 | there is 𝑌 ∈ 𝑋 with 𝑦 ∈ 𝑌} and

⋂𝑿 = {𝑦 | for all 𝑌 ∈ 𝑋, 𝑦 ∈ 𝑌}.

If you’re having trouble understanding what these definitions are saying, you’re not alone. The notation probably looks confusing, but the ideas behind these definitions are very simple. You have a whole bunch of sets (possibly infinitely many). To take the union of all these sets, you simply throw all the elements together into one big set. To take the intersection of all these sets, you take only the elements that are in every single one of those sets.

Example 6.5: 1. Let 𝐴 and 𝐵 be sets and let 𝑿 = {𝐴, 𝐵}. Then ⋃𝑿 = {𝑦 | there is 𝑌 ∈ 𝑋 with 𝑦 ∈ 𝑌} = {𝑦 | 𝑦 ∈ 𝐴 or 𝑦 ∈ 𝐵} = 𝐴 ∪ 𝐵. ⋂𝑿 = {𝑦 | for all 𝑌 ∈ 𝑋, 𝑦 ∈ 𝑌} = {𝑦 | 𝑦 ∈ 𝐴 and 𝑦 ∈ 𝐵} = 𝐴 ∩ 𝐵. 2. Let 𝐴, 𝐵, and 𝐶 be sets, and let 𝑿 = {𝐴, 𝐵, 𝐶}. Then ⋃𝑿 = {𝑦 | there is 𝑌 ∈ 𝑋 with 𝑦 ∈ 𝑌} = {𝑦 | 𝑦 ∈ 𝐴, 𝑦 ∈ 𝐵, or 𝑦 ∈ 𝐶} = 𝐴 ∪ 𝐵 ∪ 𝐶. ⋂𝑿 = {𝑦 | for all 𝑌 ∈ 𝑋, 𝑦 ∈ 𝑌} = {𝑦 | 𝑦 ∈ 𝐴, 𝑦 ∈ 𝐵, and 𝑦 ∈ 𝐶} = 𝐴 ∩ 𝐵 ∩ 𝐶. 3. Let 𝑿 = {[0, 𝑟) | 𝑟 ∈ ℝ+}. Then ⋃𝑿 = {𝑦 | there is 𝑌 ∈ 𝑋 with 𝑦 ∈ 𝑌} = {𝑦 | there is 𝑟 ∈ ℝ+ with 𝑦 ∈ [0, 𝑟)} = [0, ∞). ⋂𝑿 = {𝑦 | for all 𝑌 ∈ 𝑋, 𝑦 ∈ 𝑌} = {𝑦 | for all 𝑟 ∈ ℝ+, 𝑦 ∈ [0, 𝑟)} = {0}.

69

Notes: (1) Examples 1 and 2 give a good idea of what ⋃𝑿 and ⋂𝑿 look like when 𝑿 is finite. More generally, if 𝑿 = {𝐴1, 𝐴2, … , 𝐴𝑛}, then ⋃𝑿 = 𝐴1 ∪ 𝐴2 ∪ ⋯ ∪ 𝐴𝑛 and ⋂𝑿 = 𝐴1 ∩ 𝐴2 ∩ ⋯ ∩ 𝐴𝑛.
(2) As a specific example of Note 1, let 𝐴1 = (– ∞, 5], 𝐴2 = (0, 5), 𝐴3 = [2, 6), and 𝐴4 = (4, 99]. Let 𝑿 = {𝐴1, 𝐴2, 𝐴3, 𝐴4}. Then
⋃𝑿 = 𝐴1 ∪ 𝐴2 ∪ 𝐴3 ∪ 𝐴4 = (– ∞, 5] ∪ (0, 5) ∪ [2, 6) ∪ (4, 99] = (– ∞, 99]. ⋂𝑿 = 𝐴1 ∩ 𝐴2 ∩ 𝐴3 ∩ 𝐴4 = (– ∞, 5] ∩ (0,5) ∩ [2, 6) ∩ (4, 99] = (4, 5).
If you have trouble seeing how to compute the intersection, it may help to line up the graphs of the intervals, as was done in the Note following Example 6.4, and/or take the intersections two at a time: (– ∞, 5] ∩ (0, 5) = (0, 5) because (0, 5) ⊆ (– ∞, 5]. (0, 5) ∩ [2,6) = [2, 5) (draw the line graphs if you don’t see this). [2, 5) ∩ (4, 99] = (4,5) (again, draw the line graphs if you don’t see this).
(3) Let’s prove carefully that {𝑦 | there is 𝑟 ∈ ℝ+ with 𝑦 ∈ [0, 𝑟)} = [0, ∞). For convenience, let’s let 𝐴 = {𝑦 | there is 𝑟 ∈ ℝ+ with 𝑦 ∈ [0, 𝑟)}. If 𝑦 ∈ 𝐴, then there is 𝑟 ∈ ℝ+ with 𝑦 ∈ [0, 𝑟). So, 0 ≤ 𝑦 < 𝑟. In particular, 𝑦 ≥ 0. So, 𝑦 ∈ [0, ∞). Since 𝑦 ∈ 𝐴 was arbitrary, we have shown that 𝐴 ⊆ [0, ∞).
Let 𝑦 ∈ [0, ∞). Since (𝑦 + 1) − 𝑦 = 1 > 0, we have 𝑦 + 1 > 𝑦. So, 𝑦 ∈ [0, 𝑦 + 1). Since 𝑦 + 1 ∈ ℝ+, 𝑦 ∈ 𝐴. Since 𝑦 ∈ [0, ∞) was arbitrary, we have shown that [0, ∞) ⊆ 𝐴.
Since 𝐴 ⊆ [0, ∞) and [0, ∞) ⊆ 𝐴, it follows that 𝐴 = [0, ∞).
(4) Let’s also prove carefully that {𝑦 | for all 𝑟 ∈ ℝ+, 𝑦 ∈ [0, 𝑟)} = {0}. For convenience, let’s let 𝐵 = {𝑦 | for all 𝑟 ∈ ℝ+, 𝑦 ∈ [0, 𝑟)}. If 𝑦 ∈ 𝐵, then for all 𝑟 ∈ ℝ+, 𝑦 ∈ [0, 𝑟). So, for all 𝑦 ∈ ℝ+, 0 ≤ 𝑦 < 𝑟. So, 𝑦 is a nonnegative real number that is less than every positive real number. By Problem 8 in Problem Set 5, 𝑦 = 0. Therefore, 𝑦 ∈ {0}. Since 𝑦 ∈ 𝐵 was arbitrary, we have shown that 𝐵 ⊆ {0}.
Now, let 𝑦 ∈ {0}. Then 𝑦 = 0. For all 𝑟 ∈ ℝ+, 0 ∈ [0, 𝑟). So, 𝑦 ∈ 𝐵. It follows that {0} ⊆ 𝐵.
Since 𝐵 ⊆ {0} and {0} ⊆ 𝐵, it follows that 𝐵 = {0}.
(5) Note that the empty union is empty. Indeed, we have ⋃∅ = {𝑦 | there is 𝑌 ∈ ∅ with 𝑦 ∈ 𝑌} = ∅.
If 𝑿 is a nonempty set of sets, we say that 𝑿 is disjoint if ⋂𝑿 = ∅. We say that 𝑿 is pairwise disjoint if for all 𝐴, 𝐵 ∈ 𝑿 with 𝐴 ≠ 𝐵, 𝐴 and 𝐵 are disjoint. For example, if we let 𝑿 = {(𝑛, 𝑛 + 1) | 𝑛 ∈ ℤ}, then 𝑿 is both disjoint and pairwise disjoint.
Are the definitions of disjoint and pairwise disjoint equivalent? You will be asked to answer this question in Problem 5 below.
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Open and Closed Sets
A subset 𝑋 of ℝ is said to be open if for every real number 𝑥 ∈ 𝑋, there is an open interval (𝑎, 𝑏) with 𝑥 ∈ (𝑎, 𝑏) and (𝑎, 𝑏) ⊆ 𝑋.

In words, a set is open in ℝ if every number in the set has “some space” on both sides of that number inside the set. If you think of each point in the set as an animal, then each animal in the set should be able to move a little to the left and a little to the right without ever leaving the set. Another way to think of this is that no number is on “the edge” or “the boundary” of the set, about to fall out of it.

Example 6.6:

1. Every bounded open interval is open. To see this, let 𝑋 = (𝑎, 𝑏) and let 𝑥 ∈ 𝑋. Then 𝑋 = (𝑎, 𝑏)

itself

is

an

open

interval

with

𝑥

∈

(𝑎,

𝑏)

and

(𝑎,

𝑏)

⊆

𝑋.

For

example,

(0,

1)

and

(–

√2,

3)
5

are

open sets.

2. We will prove in the theorems below that all open intervals are open sets. For example, (– 2, ∞), (– ∞, 5), and (– ∞, ∞) are all open sets.

3. (0,1] is not an open set because the “boundary point” 1 is included in the set. If (𝑎, 𝑏) is any

open interval containing 1, then (𝑎, 𝑏) ⊈ (0,1] because there are numbers greater than 1 inside

(𝑎, 𝑏). For example, let 𝑥 = 1 (1 + 𝑏) (the average of 1 and 𝑏). Since 𝑏 > 1, we have that

2

𝑥

>

1 2

(1

+

1)

=

1 2

⋅

2

=

1.

So,

𝑥

>

1.

Also,

since

1

>

𝑎,

𝑥

>

𝑎.

Now,

since

1

<

𝑏,

we

have

that

𝑥

<

1 2

(𝑏

+

𝑏)

=

1 2

(2𝑏)

=

(1
2

⋅

2)

𝑏

=

1𝑏

=

𝑏.

So,

𝑥

∈

(𝑎,

𝑏).

4. We can use reasoning similar to that used in 3 to see that all half-open intervals and closed intervals are not open sets.

Theorem 6.3: Let 𝑎 ∈ ℝ. The infinite interval (𝑎, ∞) is an open set.

The idea behind the proof is quite simple. If 𝑥 ∈ (𝑎, ∞), then (𝑎, 𝑥 + 1) is an open interval with 𝑥 inside of it and with (𝑎, 𝑥 + 1) ⊆ (𝑎, ∞).

Proof of Theorem 6.3: Let 𝑥 ∈ (𝑎, ∞) and let 𝑏 = 𝑥 + 1.

Since 𝑥 ∈ (𝑎, ∞), 𝑥 > 𝑎. Since (𝑥 + 1) − 𝑥 = 1 > 0, we have 𝑏 = 𝑥 + 1 > 𝑥.

So, we have 𝑎 < 𝑥 < 𝑏. That is, 𝑥 ∈ (𝑎, 𝑏). Also, (𝑎, 𝑏) ⊆ (𝑎, ∞). Since 𝑥 ∈ (𝑎, ∞) was arbitrary, (𝑎, ∞)

is an open set.

□

In Problem 6 below (part (i)), you will be asked to show that an interval of the form (– ∞, 𝑏) is also an open set.

Theorem 6.4: ∅ and ℝ are both open sets.

Proof: The statement that ∅ is open is vacuously true (since ∅ has no elements, there is nothing to check).

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If 𝑥 ∈ ℝ, then 𝑥 ∈ (𝑥 − 1, 𝑥 + 1) and (𝑥 − 1, 𝑥 + 1) ⊆ ℝ. Since 𝑥 was an arbitrary element of ℝ, we

have shown that for every 𝑥 ∈ ℝ, there is an open interval (𝑎, 𝑏) with 𝑥 ∈ (𝑎, 𝑏) and (𝑎, 𝑏) ⊆ ℝ. So, ℝ

is open.

□

Many authors define “open” in a slightly different way from the definition we’ve been using. This next Theorem will show that the definition we have been using is equivalent to theirs.

Theorem 6.5: A subset 𝑋 of ℝ is open if and only if for every real number 𝑥 ∈ 𝑋, there is a positive real number 𝑐 such that (𝑥 − 𝑐, 𝑥 + 𝑐) ⊆ 𝑋.

Analysis: The harder direction of the proof is showing that if 𝑋 is open, then for every real number 𝑥 ∈ 𝑋, there is a positive real number 𝑐 such that (𝑥 − 𝑐, 𝑥 + 𝑐) ⊆ 𝑋.

To see this, suppose that 𝑋 is open and let 𝑥 ∈ 𝑋. Then there is an open interval (𝑎, 𝑏) with 𝑥 ∈ (𝑎, 𝑏) and (𝑎, 𝑏) ⊆ 𝑋. We want to replace the interval (𝑎, 𝑏) by an interval that has 𝑥 right in the center.

The following picture should help us to come up with an argument.

In the picture, we have an open interval (𝑎, 𝑏), containing 𝑥. In this particular picture, 𝑥 is a bit closer to 𝑎 than it is to 𝑏. However, we should remember to be careful that our argument doesn’t assume this (as we have no control over where 𝑥 “sits” inside of (𝑎, 𝑏)).
In the picture, we see that 𝑥 − 𝑎 is the distance from 𝑎 to 𝑥, and 𝑏 − 𝑥 is the distance from 𝑥 to 𝑏. Since the distance from 𝑎 to 𝑥 is smaller, let’s let 𝑐 be that smaller distance. In other words, we let 𝑐 = 𝑥 − 𝑎. From the picture, it looks like the interval (𝑥 − 𝑐, 𝑥 + 𝑐) will be inside the interval (𝑎, 𝑏).
In general, if 𝑥 is closer to 𝑎, we would let 𝑐 = 𝑥 − 𝑎, and if 𝑥 is closer to 𝑏, we would let 𝑐 = 𝑏 − 𝑥. We can simply define 𝑐 to be the smaller of 𝑥 − 𝑎 and 𝑏 − 𝑥. That is, 𝑐 = min{𝑥 − 𝑎, 𝑏 − 𝑥}. From the picture, it seems like with this choice of 𝑐, the interval (𝑥 − 𝑐, 𝑥 + 𝑐) should give us what we want.
Proof of Theorem 6.5: Let 𝑋 be an open subset of ℝ and let 𝑥 ∈ 𝑋. Then there is an open interval (𝑎, 𝑏) with 𝑥 ∈ (𝑎, 𝑏) and (𝑎, 𝑏) ⊆ 𝑋. Let 𝑐 = min{𝑥 − 𝑎, 𝑏 − 𝑥}. We claim that (𝑥 − 𝑐, 𝑥 + 𝑐) is an open interval containing 𝑥 and contained in (𝑎, 𝑏). We need to show 𝑎 ≤ 𝑥 − 𝑐 < 𝑥 < 𝑥 + 𝑐 ≤ 𝑏.
Since 𝑐 = min{𝑥 − 𝑎, 𝑏 − 𝑥}, 𝑐 ≤ 𝑥 − 𝑎. So, – 𝑐 ≥ – (𝑥 − 𝑎). It follows that (𝑥 − 𝑐) − 𝑎 ≥ (𝑥 − (𝑥 − 𝑎)) − 𝑎 = (𝑥 − 𝑥 + 𝑎) − 𝑎 = 𝑎 − 𝑎 = 0.
So, 𝑥 − 𝑐 ≥ 𝑎.
Since 𝑐 = min{𝑥 − 𝑎, 𝑏 − 𝑥}, 𝑐 ≤ 𝑏 − 𝑥. So, – 𝑐 ≥ – (𝑏 − 𝑥). It follows that 𝑏 − (𝑥 + 𝑐) = 𝑏 − 𝑥 − 𝑐 ≥ 𝑏 − 𝑥 − (𝑏 − 𝑥) = 0.
So, 𝑏 ≥ 𝑥 + 𝑐, or equivalently, 𝑥 + 𝑐 ≤ 𝑏.
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Note that 𝑥 > 𝑎, so that 𝑥 − 𝑎 > 0, and 𝑥 < 𝑏, so that 𝑏 − 𝑥 > 0. It follows that 𝑐 > 0.

We have 𝑥 − (𝑥 − 𝑐) = 𝑐 > 0, so that 𝑥 > 𝑥 − 𝑐. We also have (𝑥 + 𝑐) − 𝑥 = 𝑐 > 0, so that 𝑥 + 𝑐 > 𝑥.

We have shown 𝑎 ≤ 𝑥 − 𝑐 < 𝑥 < 𝑥 + 𝑐 ≤ 𝑏, as desired.

Since (𝑥 − 𝑐, 𝑥 + 𝑐) ⊆ (𝑎, 𝑏) and (𝑎, 𝑏) ⊆ 𝑋, by the transitivity of ⊆ (Theorem 2.3 from Lesson 2), we have (𝑥 − 𝑐, 𝑥 + 𝑐) ⊆ 𝑋.

The converse is immediate since for 𝑥 ∈ 𝑋, (𝑥 − 𝑐, 𝑥 + 𝑐) is an open interval containing 𝑥.

□

The basic definition of a topological space involves open sets, unions, and intersections. We’re not going to talk about general topological spaces in this lesson (we will look at them in Lesson 14), but in the spirit of the subject, we will prove some results about unions and intersections of open sets in ℝ.

Theorem 6.6: The union of two open sets in ℝ is an open set in ℝ.

Proof: Let 𝐴 and 𝐵 be open sets in ℝ, and let 𝑥 ∈ 𝐴 ∪ 𝐵. Then 𝑥 ∈ 𝐴 or 𝑥 ∈ 𝐵. Without loss of

generality, we may assume that 𝑥 ∈ 𝐴 (see the Note below). Since 𝐴 is open in ℝ, there is an interval

(𝑎, 𝑏) with 𝑥 ∈ (𝑎, 𝑏) and (𝑎, 𝑏) ⊆ 𝐴. By Theorem 2.4, 𝐴 ⊆ 𝐴 ∪ 𝐵. Since ⊆ is transitive (Theorem 2.3),

(𝑎, 𝑏) ⊆ 𝐴 ∪ 𝐵. Therefore, 𝐴 ∪ 𝐵 is open.

□

Note: In the proof of Theorem 6.6, we used the expression “Without loss of generality.” This expression can be used when an argument can be split up into 2 or more cases, and the proof of each of the cases is nearly identical.

For Theorem 6.6, the two cases are (i) 𝑥 ∈ 𝐴 and (ii) 𝑥 ∈ 𝐵. The argument for case (ii) is the same as the argument for case (i), essentially word for word—only the roles of 𝐴 and 𝐵 are interchanged.

Example 6.7: (– 5, 2) is open by part 1 of Example 6.6 and (7, ∞) is open by Theorem 6.3. Therefore, by Theorem 6.6, (– 5, 2) ∪ (7, ∞) is also open.

If you look at the proof of Theorem 6.6 closely, you should notice that the proof would still work if we were taking a union of more than 2 sets. In fact, any union of open sets is open, as we now prove.

Theorem 6.7: Let 𝑿 be a set of open subsets of ℝ. Then ⋃𝑿 is open.

Proof: Let 𝑿 be a set of open subsets of ℝ and let 𝑥 ∈ ⋃𝑿. Then 𝑥 ∈ 𝐴 for some 𝐴 ∈ 𝑿. Since 𝐴 is open

in ℝ, there is an interval (𝑎, 𝑏) with 𝑥 ∈ (𝑎, 𝑏) and (𝑎, 𝑏) ⊆ 𝐴. By Problem 9 below (part (i)), we have

𝐴 ⊆ ⋃𝑿. Since ⊆ is transitive (Theorem 2.3), (𝑎, 𝑏) ⊆ ⋃𝑿. Therefore, ⋃𝑿 is open.

□

Example 6.8: 1. (1,2) ∪ (2,3) ∪ (3,4) ∪ (4, ∞) is open.

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2. ℝ ∖ ℤ is open because it is a union of open intervals. It looks like this: ⋯ (– 2, – 1) ∪ (– 1, 0) ∪ (0, 1) ∪ (1, 2) ∪ ⋯
ℝ ∖ ℤ can also be written as

⋃{(𝑛, 𝑛 + 1) | 𝑛 ∈ ℤ} or ⋃(𝑛, 𝑛 + 1)

𝑛∈ℤ

3.

If

we

take

the

union

of

all

intervals

of

the

form

(1
𝑛+1

,

1)
𝑛

for

positive

integers

𝑛,

we

get

an

open

set. We can visualize this open set as follows:

⋃

{(𝑛

1 +

1

,

𝑛1)

|

𝑛

∈

ℤ+}

=

⋯

∪

1 (5

,

1 4)

∪

1 (4

,

1 3)

∪

1 (3

,

1 2)

∪

1 (2

,

1)

Theorem 6.8: Every open set in ℝ can be expressed as a union of bounded open intervals.

The main idea of the argument will be the following. Every real number that is in an open set is inside an open interval that is a subset of the set. Just take the union of all these open intervals (one interval for each real number in the set).

Proof of Theorem 6.8: Let 𝑋 be an open set in ℝ. Since 𝑋 is open, for each 𝑥 ∈ 𝑋, there is an interval (𝑎𝑥, 𝑏𝑥) with 𝑥 ∈ (𝑎𝑥, 𝑏𝑥) and (𝑎𝑥, 𝑏𝑥) ⊆ 𝑋. We Let 𝒀 = {(𝑎𝑥, 𝑏𝑥) | 𝑥 ∈ 𝑋}. We will show that 𝑋 = ⋃𝒀.
First, let 𝑥 ∈ 𝑋. Then 𝑥 ∈ (𝑎𝑥, 𝑏𝑥). Since (𝑎𝑥, 𝑏𝑥) ∈ 𝒀, 𝑥 ∈ ⋃𝒀. Since 𝑥 was arbitrary, 𝑋 ⊆ ⋃𝒀.
Now, let 𝑥 ∈ ⋃𝒀. Then there is 𝑧 ∈ 𝑋 with 𝑥 ∈ (𝑎𝑧, 𝑏𝑧). Since (𝑎𝑧, 𝑏𝑧) ⊆ 𝑋, 𝑥 ∈ 𝑋. Since 𝑥 ∈ 𝑋 was arbitrary, ⋃𝒀 ⊆ 𝑋.

Since 𝑋 ⊆ ⋃𝒀 and ⋃𝒀 ⊆ 𝑋, it follows that 𝑋 = ⋃𝒀.

□

Theorem 6.9: The intersection of two open sets in ℝ is an open set in ℝ.

Proof: Let 𝐴 and 𝐵 be open sets in ℝ and let 𝑥 ∈ 𝐴 ∩ 𝐵. Then 𝑥 ∈ 𝐴 and 𝑥 ∈ 𝐵. Since 𝐴 is open, there

is an open interval (𝑎, 𝑏) with 𝑥 ∈ (𝑎, 𝑏) and (𝑎, 𝑏) ⊆ 𝐴. Since 𝐵 is open, there is an open interval (𝑐, 𝑑)

with 𝑥 ∈ (𝑐, 𝑑) and (𝑐, 𝑑) ⊆ 𝐵. Let 𝐶 = (𝑎, 𝑏) ∩ (𝑐, 𝑑). Since 𝑥 ∈ (𝑎, 𝑏) and 𝑥 ∈ (𝑐, 𝑑), 𝑥 ∈ 𝐶. By

Problem 6 below (part (ii)), 𝐶 is an open interval. By Problem 11 from Lesson 2 and part (ii) of Problem

3 below, 𝐶 ⊆ 𝐴 and 𝐶 ⊆ 𝐵. It follows that 𝐶 ⊆ 𝐴 ∩ 𝐵 (Prove this!). Since 𝑥 ∈ 𝐴 ∩ 𝐵 was arbitrary,

𝐴 ∩ 𝐵 is open.

□

In Problem 6 below (part (iii)), you will be asked to show that the intersection of finitely many open sets in ℝ is an open set in ℝ. In problem 8, you will be asked to show that an arbitrary intersection of open sets does not need to be open.

A subset 𝑋 of ℝ is said to be closed if ℝ ∖ 𝑋 is open.

ℝ ∖ 𝑋 is called the complement of 𝑋 in ℝ, or simply the complement of 𝑋. It consists of all real numbers not in 𝑋.

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Example 6.9:
1. Every closed interval is a closed set. For example, [0,1] is closed because its complement in ℝ is ℝ ∖ [0,1] = (– ∞, 0) ∪ (1, ∞). This is a union of open intervals, which is open.
Similarly, [3, ∞) is a closed set because ℝ ∖ [3, ∞) = (– ∞, 3), which is open.
2. Half-open intervals are neither open nor closed. For example, we saw in Example 6.6 that (0,1] is not an open set. We see that (0,1] is not closed by observing ℝ ∖ (0,1] = (– ∞, 0] ∪ (1, ∞), which is not open.
3. ∅ is closed because ℝ ∖ ∅ = ℝ is open. ℝ is closed because ℝ ∖ ℝ = ∅ is open. ∅ and ℝ are the only two sets of real numbers that are both open and closed.

Theorem 6.10: The intersection of two closed sets in ℝ is a closed set in ℝ.

Proof: Let 𝐴 and 𝐵 be closed sets in ℝ. Then ℝ ∖ 𝐴 and ℝ ∖ 𝐵 are open sets in ℝ. By Theorem 6.6 (or

6.7), (ℝ ∖ 𝐴) ∪ (ℝ ∖ 𝐵) is open in ℝ. Therefore, ℝ ∖ [(ℝ ∖ 𝐴) ∪ (ℝ ∖ 𝐵)] is closed in ℝ. So, it suffices

to show that 𝐴 ∩ 𝐵 = ℝ ∖ [(ℝ ∖ 𝐴) ∪ (ℝ ∖ 𝐵)]. Well, 𝑥 ∈ 𝐴 ∩ 𝐵 if and only if 𝑥 ∈ 𝐴 and 𝑥 ∈ 𝐵 if and

only if 𝑥 ∉ ℝ ∖ 𝐴 and 𝑥 ∉ ℝ ∖ 𝐵 if and only if 𝑥 ∉ (ℝ ∖ 𝐴) ∪ (ℝ ∖ 𝐵) if and only if

𝑥 ∈ ℝ ∖ [(ℝ ∖ 𝐴) ∪ (ℝ ∖ 𝐵)]. So, 𝐴 ∩ 𝐵 = ℝ ∖ [(ℝ ∖ 𝐴) ∪ (ℝ ∖ 𝐵)], completing the proof.

□

A similar argument can be used to show that the union of two closed sets in ℝ is a closed set in ℝ. This result can be extended to the union of finitely many closed sets in ℝ with the help of Problem 6 below (part (iii)). The dedicated reader should prove this. In Problem 10 below, you will be asked to show that an arbitrary intersection of closed sets in ℝ is closed. In problem 8, you will be asked to show that an arbitrary union of closed sets does not need to be closed.

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Problem Set 6

LEVEL 1

Full solutions to these problems are available for free download here:
www.SATPrepGet800.com/PMFBXSG

1. Draw Venn diagrams for (𝐴 ∖ 𝐵) ∖ 𝐶 and 𝐴 ∖ (𝐵 ∖ 𝐶). Are these two sets equal for all sets 𝐴, 𝐵, and 𝐶? If so, prove it. If not, provide a counterexample.

2. Let 𝐴 = {∅, {∅, {∅}}}, 𝐵 = {∅, {∅}}, 𝐶 = (−∞, 2], 𝐷 = (−1, 3]. Compute each of the following:
(i) 𝐴 ∪ 𝐵 (ii) 𝐴 ∩ 𝐵 (iii) 𝐴 ∖ 𝐵 (iv) 𝐵 ∖ 𝐴 (v) 𝐴 Δ 𝐵 (vi) 𝐶 ∪ 𝐷 (vii) 𝐶 ∩ 𝐷 (viii) 𝐶 ∖ 𝐷 (ix) 𝐷 ∖ 𝐶 (x) 𝐶 Δ 𝐷
LEVEL 2
3. Prove the following: (i) The operation of forming unions is commutative. (ii) The operation of forming intersections is commutative. (iii) The operation of forming intersections is associative.
4. Prove that if an interval 𝐼 is unbounded, then 𝐼 has one of the following five forms: (𝑎, ∞), (– ∞, 𝑏), [𝑎, ∞), (– ∞, 𝑏], (– ∞, ∞)
LEVEL 3
5. Prove or provide a counterexample: (i) Every pairwise disjoint set of sets is disjoint. (ii) Every disjoint set of sets is pairwise disjoint.

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6. Prove the following: (i) For all 𝑏 ∈ ℝ, the infinite interval (– ∞, 𝑏) is an open set in ℝ. (ii) The intersection of two open intervals in ℝ is either empty or an open interval in ℝ. (iii) The intersection of finitely many open sets in ℝ is an open set in ℝ.
7. Let 𝐴, 𝐵, and 𝐶 be sets. Prove each of the following: (i) 𝐴 ∩ (𝐵 ∪ 𝐶) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶). (ii) 𝐴 ∪ (𝐵 ∩ 𝐶) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶). (iii) 𝐶 ∖ (𝐴 ∪ 𝐵) = (𝐶 ∖ 𝐴) ∩ (𝐶 ∖ 𝐵). (iv) 𝐶 ∖ (𝐴 ∩ 𝐵) = (𝐶 ∖ 𝐴) ∪ (𝐶 ∖ 𝐵).
LEVEL 4
8. Give an example of an infinite collection of open sets whose intersection is not open. Also, give an example of an infinite collection of closed sets whose union is not closed. Provide a proof for each example.
9. Let 𝑿 be a nonempty set of sets. Prove the following: (i) For all 𝐴 ∈ 𝑿, 𝐴 ⊆ ⋃𝑿. (ii) For all 𝐴 ∈ 𝑿, ⋂𝑿 ⊆ 𝐴.
LEVEL 5
10. Prove that if 𝑿 is a nonempty set of closed subsets of ℝ, then ⋂𝑿 is closed.
11. Let 𝐴 be a set and let 𝑿 be a nonempty collection of sets. Prove each of the following: (i) 𝐴 ∩ ⋃𝑿 = ⋃{𝐴 ∩ 𝐵 | 𝐵 ∈ 𝑿} (ii) 𝐴 ∪ ⋂𝑿 = ⋂{𝐴 ∪ 𝐵 | 𝐵 ∈ 𝑿} (iii) 𝐴 ∖ ⋃𝑿 = ⋂{𝐴 ∖ 𝐵 | 𝐵 ∈ 𝑿} (iv) 𝐴 ∖ ⋂𝑿 = ⋃{𝐴 ∖ 𝐵 | 𝐵 ∈ 𝑿}.
12. Prove that every closed set in ℝ can be written as an intersection ⋂𝑿, where each element of 𝑿 is a union of at most 2 closed intervals.
CHALLENGE PROBLEM
13. Prove that every nonempty open set of real numbers can be expressed as a union of pairwise disjoint open intervals.
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LESSON 7 – COMPLEX ANALYSIS THE FIELD OF COMPLEX NUMBERS
A Limitation of the Reals
In Lesson 5 we asked (and answered) the question “Why isn’t ℚ (the field of rational numbers) enough?” We now ask the same question about ℝ, the field of real numbers.
A linear equation has the form 𝑎𝑥 + 𝑏 = 0, where 𝑎 ≠ 0. If we are working inside a field, then this equation has the unique solution 𝑥 = – 𝑏𝑎−1 = – 𝑎𝑏. For example, the equation 2𝑥 − 1 = 0 has the unique solution 𝑥 = 2−1 = 12. Notice how important it is that we are working inside a field here. If we were allowed to use only the properties of a commutative ring, then we might not be able to solve this equation. For example, in ℤ (the ring of integers), the equation 2𝑥 − 1 = 0 has no solution.
A quadratic equation has the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎 ≠ 0. Is working inside a field enough to solve this equation? The answer is no! For example, a solution to the equation 𝑥2 − 2 = 0 must satisfy 𝑥2 = 2. In Lesson 5, we proved that this equation cannot be solved in ℚ. This was one of our main motivations for introducing ℝ. And, in fact, the equation 𝑥2 − 2 = 0 can be solved in ℝ. However, the equation 𝑥2 + 1 = 0 cannot be solved in ℝ. This follows immediately from Theorem 5.2, which says that if 𝑥 is an element of an ordered field, then 𝑥2 = 𝑥 ⋅ 𝑥 can never be negative.
Is there a field containing ℝ, where all quadratic equations can be solved? The answer is yes, and in fact, we can do much better than that. In this lesson we will define a field containing the field of real numbers such that every equation of the form 𝑎𝑛𝑥𝑛 + 𝑎𝑛−1𝑥𝑛−1 + ⋯ + 𝑎1𝑥 + 𝑎0 = 0 has a solution. Such an equation is called a polynomial equation, and a field in which every such polynomial equation has a solution is called an algebraically closed field.
The Complex Field
The standard form of a complex number is 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. So, the set of complex numbers is ℂ = {𝑎 + 𝑏𝑖 | 𝑎, 𝑏 ∈ ℝ}.
If we identify 1 = 1 + 0𝑖 with the ordered pair (1, 0), and we identify 𝑖 = 0 + 1𝑖 with the ordered pair (0, 1), then it is natural to write the complex number 𝑎 + 𝑏𝑖 as the point (𝑎, 𝑏). Here is a reasonable justification for this:
𝑎 + 𝑏𝑖 = 𝑎(1,0) + 𝑏(0,1) = (𝑎, 0) + (0, 𝑏) = (𝑎, 𝑏)
In this way, we can visualize a complex number as a point in The Complex Plane. A portion of the Complex Plane is shown to the right with several complex numbers displayed as points of the form (𝑥, 𝑦).
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The complex plane is formed by taking two copies of the real line and placing one horizontally and the other vertically. The horizontal copy of the real line is called the 𝑥-axis or the real axis (labeled 𝑥 in the above figure) and the vertical copy of the real line is called the 𝑦-axis or imaginary axis (labeled 𝑦 in the above figure). The two axes intersect at the point (0, 0). This point is called the origin. We can also visualize the complex number 𝑎 + 𝑏𝑖 as a directed line segment (or vector) starting at the origin and ending at the point (𝑎, 𝑏). Three examples are shown to the right. If 𝑧 = 𝑎 + 𝑏𝑖 is a complex number, we call 𝑎 the real part of 𝑧 and 𝑏 the imaginary part of 𝑧, and we write 𝑎 = Re 𝑧 and 𝑏 = Im 𝑧. Two complex numbers are equal if and only if they have the same real part and the same imaginary part. In other words,
𝑎 + 𝑏𝑖 = 𝑐 + 𝑑𝑖 if and only if 𝑎 = 𝑐 and 𝑏 = 𝑑. We add two complex numbers by simply adding their real parts and adding their imaginary parts. So,
(𝑎 + 𝑏𝑖) + (𝑐 + 𝑑𝑖) = (𝑎 + 𝑐) + (𝑏 + 𝑑)𝑖. As a point, this sum is (𝑎 + 𝑐, 𝑏 + 𝑑). We can visualize this sum as the vector starting at the origin that is the diagonal of the parallelogram formed from the vectors 𝑎 + 𝑏𝑖 and 𝑐 + 𝑑𝑖. Here is an example showing that (1 + 2𝑖) + (– 3 + 𝑖) = – 2 + 3𝑖.
The definition for multiplying two complex numbers is a bit more complicated: (𝑎 + 𝑏𝑖)(𝑐 + 𝑑𝑖) = (𝑎𝑐 − 𝑏𝑑) + (𝑎𝑑 + 𝑏𝑐)𝑖.
Notes: (1) If 𝑏 = 0, then we call 𝑎 + 𝑏𝑖 = 𝑎 + 0𝑖 = 𝑎 a real number. Note that when we add or multiply two real numbers, we always get another real number.
(𝑎 + 0𝑖) + (𝑏 + 0𝑖) = (𝑎 + 𝑏) + (0 + 0)𝑖 = (𝑎 + 𝑏) + 0𝑖 = 𝑎 + 𝑏. (𝑎 + 0𝑖)(𝑏 + 0𝑖) = (𝑎𝑏 − 0 ⋅ 0) + (𝑎 ⋅ 0 + 0𝑏)𝑖 = (𝑎𝑏 − 0) + (0 + 0)𝑖 = 𝑎𝑏 + 0𝑖 = 𝑎𝑏. (2) If 𝑎 = 0, then we call 𝑎 + 𝑏𝑖 = 0 + 𝑏𝑖 = 𝑏𝑖 a pure imaginary number. (3) 𝑖2 = – 1. To see this, note that 𝑖2 = 𝑖 ⋅ 𝑖 = (0 + 1𝑖)(0 + 1𝑖), and we have
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(0 + 1𝑖)(0 + 1𝑖) = (0 ⋅ 0 − 1 ⋅ 1) + (0 ⋅ 1 + 1 ⋅ 0)𝑖 = (0 − 1) + (0 + 0)𝑖 = – 1 + 0𝑖 = – 1.

(4) The definition of the product of two complex numbers is motivated by how multiplication should behave in a field, together with replacing 𝑖2 by – 1. If we were to naïvely multiply the two complex numbers, we would have
(𝑎 + 𝑏𝑖)(𝑐 + 𝑑𝑖) = (𝑎 + 𝑏𝑖)𝑐 + (𝑎 + 𝑏𝑖)(𝑑𝑖) = 𝑎𝑐 + 𝑏𝑐𝑖 + 𝑎𝑑𝑖 + 𝑏𝑑𝑖2 = 𝑎𝑐 + 𝑏𝑐𝑖 + 𝑎𝑑𝑖 + 𝑏𝑑(– 1) = 𝑎𝑐 + (𝑏𝑐 + 𝑎𝑑)𝑖 − 𝑏𝑑 = (𝑎𝑐 − 𝑏𝑑) + (𝑎𝑑 + 𝑏𝑐)𝑖.
The dedicated reader should make a note of which field properties were used during this computation.

Those familiar with the mnemonic FOIL may notice that “FOILing” will always work to produce the product of two complex numbers, provided we replace 𝑖2 by – 1 and simplify.

Example 7.1: Let 𝑧 = 2 − 3𝑖 and 𝑤 = – 1 + 5𝑖. Then 𝑧 + 𝑤 = (2 − 3𝑖) + (– 1 + 5𝑖) = (2 + (– 1)) + (– 3 + 5)𝑖 = 𝟏 + 𝟐𝒊. 𝑧𝑤 = (2 − 3𝑖)(– 1 + 5𝑖) = (2(– 1) − (– 3)(5)) + (2 ⋅ 5 + (– 3)(– 1))𝑖 = (– 2 + 15) + (10 + 3)𝑖 = 𝟏𝟑 + 𝟏𝟑𝒊.

With the definitions we just made for addition and multiplication, we get (ℂ, +, ⋅), the field of complex numbers. See Lesson 5 if you need to review the definition of a field.

Theorem 7.1: (ℂ, +, ⋅) is field.

The proof that (ℂ, +, ⋅) is a field is very straightforward and mostly uses the fact that (ℝ, +, ⋅) is a field. For example, to verify that addition is commutative in ℂ, we have
(𝑎 + 𝑏𝑖) + (𝑐 + 𝑑𝑖) = (𝑎 + 𝑐) + (𝑏 + 𝑑)𝑖 = (𝑐 + 𝑎) + (𝑑 + 𝑏)𝑖 = (𝑐 + 𝑑𝑖) + (𝑎 + 𝑏𝑖).

We have 𝑎 + 𝑐 = 𝑐 + 𝑎 because 𝑎, 𝑐 ∈ ℝ and addition is commutative in ℝ. For the same reason, we have 𝑏 + 𝑑 = 𝑑 + 𝑏.

We leave the full verification that (ℂ, +, ⋅) is a field as an exercise for the reader (Problem 2 below), and simply note a few things of importance here:

• The identity for addition is 0 = 0 + 0𝑖.

• The identity for multiplication is 1 = 1 + 0𝑖

• The additive inverse of 𝑧 = 𝑎 + 𝑏𝑖 is – 𝑧 = – (𝑎 + 𝑏𝑖) = – 𝑎 − 𝑏𝑖.

•

The

multiplicative

inverse

of

𝑧

=

𝑎

+

𝑏𝑖

is

𝑧−1

=

𝑎 𝑎2+𝑏2

−

𝑏 𝑎2+𝑏2

𝑖.

The reader is expected to verify all this in Problem 2.

Remark: By Note 1 above, we see that (ℝ, +, ⋅) is a subfield of (ℂ, +, ⋅). That is, ℝ ⊆ ℂ and (ℝ, +, ⋅) is a field with respect to the field operations of (ℂ, +, ⋅) (In other words, we don’t need to “change” the definition of addition or multiplication to get the appropriate operations in ℝ—the operations are already behaving correctly). Subfields will be covered in more detail in Lesson 11.

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Subtraction: If 𝑧, 𝑤 ∈ ℂ, with 𝑧 = 𝑎 + 𝑏𝑖 and 𝑤 = 𝑐 + 𝑑𝑖, then we define the difference 𝑧 − 𝑤 by
𝑧 − 𝑤 = 𝑧 + (– 𝑤) = (𝑎 + 𝑏𝑖) + (– 𝑐 − 𝑑𝑖) = (𝑎 − 𝑐) + (𝑏 − 𝑑)𝑖.
As a point, this difference is (𝑎 − 𝑐, 𝑏 − 𝑑). Here is an example illustrating how subtraction works using the computation (1 + 2𝑖) − (2 − 𝑖) = – 1 + 3𝑖.

Observe how we first replaced 2 − 𝑖 by – 2 + 𝑖 so that we could change the subtraction problem to the addition problem: (1 + 2𝑖) + (– 2 + 𝑖). We then formed a parallelogram using 1 + 2𝑖 and – 2 + 𝑖 as edges, and finally, drew the diagonal of that parallelogram to see the result.

Division: If 𝑧 ∈ ℂ and 𝑤 ∈ ℂ∗ with 𝑧 = 𝑎 + 𝑏𝑖 and 𝑤 = 𝑐 + 𝑑𝑖, then we define the quotient 𝑧 by
𝑤

𝑧 𝑤

=

𝑧𝑤−1

=

(𝑎

+

𝑏𝑖)

(𝑐2

𝑐 +

𝑑2

−

𝑐2

𝑑 +

𝑑2

𝑖)

=

𝑎𝑐 𝑐2

+ +

𝑏𝑑 𝑑2

+

𝑏𝑐 𝑐2

− +

𝑎𝑑 𝑑2

𝑖.

The definition of division in a field unfortunately led to a messy looking formula. However, when actually performing division, there is an easier way to think about it, as we will see below.

The conjugate of the complex number 𝑧 = 𝑎 + 𝑏𝑖 is the complex number 𝑧 = 𝑎 − 𝑏𝑖.

Notes: (1) To take the conjugate of a complex number, we simply negate the imaginary part of the number and leave the real part as it is.

(2) If 𝑧 = 𝑎 + 𝑏𝑖 ≠ 0, then at least one of 𝑎 or 𝑏 is not zero. It follows that 𝑧 = 𝑎 − 𝑏𝑖 is also not 0.

(3) The product of a complex number with its conjugate is always a nonnegative real number. Specifically, if 𝑧 = 𝑎 + 𝑏𝑖, then 𝑧𝑧 = (𝑎 + 𝑏𝑖)(𝑎 − 𝑏𝑖) = (𝑎2 + 𝑏2) + (– 𝑎𝑏 + 𝑎𝑏)𝑖 = 𝑎2 + 𝑏2.

(4) We can change the quotient 𝑧 to standard form by multiplying the numerator and denominator by
𝑤
𝑤. So, if 𝑧 = 𝑎 + 𝑏𝑖 and 𝑤 = 𝑐 + 𝑑𝑖, then we have

𝑧 𝑧𝑤 (𝑎 + 𝑏𝑖)(𝑐 − 𝑑𝑖) (𝑎𝑐 + 𝑏𝑑) + (𝑏𝑐 − 𝑎𝑑)𝑖 𝑎𝑐 + 𝑏𝑑 𝑏𝑐 − 𝑎𝑑

𝑤 = 𝑤𝑤 = (𝑐 + 𝑑𝑖)(𝑐 − 𝑑𝑖) =

𝑐2 + 𝑑2

= 𝑐2 + 𝑑2 + 𝑐2 + 𝑑2 𝑖.

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Example 7.2: Let 𝑧 = 2 − 3𝑖 and 𝑤 = – 1 + 5𝑖. Then

𝑧 = 𝟐 + 𝟑𝒊.

𝑤 = – 𝟏 − 𝟓𝒊.

𝑧 𝑧𝑤 (2 − 3𝑖)(– 1 − 5𝑖) (– 2 − 15) + (– 10 + 3)𝑖 (– 17 − 7𝑖) 𝟏𝟕 𝟕

𝑤 = 𝑤𝑤 = (– 1 + 5𝑖)(– 1 − 5𝑖) =

(– 1)2 + 52

= 1 + 25 = – 𝟐𝟔 − 𝟐𝟔 𝒊.

Recall from Lesson 5 that in an ordered field, if 𝑎 > 0 and 𝑏 > 0, then 𝑎 + 𝑏 > 0 (Order Property 1) and 𝑎𝑏 > 0 (Order Property 2). Also, for every element 𝑎, exactly one of the following holds: 𝑎 > 0, 𝑎 = 0, or 𝑎 < 0 (Order Property 3).

Theorem 7.2: The field of complex numbers cannot be ordered.

Proof: Suppose toward contradiction that < is an ordering of (ℂ, +, ⋅). If 𝑖 > 0, then – 1 = 𝑖2 = 𝑖 ⋅ 𝑖 > 0 by Order Property 2.

If 𝑖 < 0, then – 𝑖 > 0, and therefore, – 1 = 𝑖2 = (– 1)(– 1)𝑖 ⋅ 𝑖 = (– 1𝑖)(– 1𝑖) = (– 𝑖)(– 𝑖) > 0, again by Order Property 2.

So, – 1 > 0 and it follows that 1 = (– 1)(– 1) > 0, again by order property 2. Therefore, we have

– 1 > 0 and 1 > 0, violating Order Property 3. So, (ℂ, +, ⋅) cannot be ordered.

□

Absolute Value and Distance
If 𝑥 and 𝑦 are real or complex numbers such that 𝑦 = 𝑥2, then we call 𝑥 a square root of 𝑦. If 𝑥 is a positive real number, then we say that 𝑥 is the positive square root of 𝑦 and we write 𝑥 = √𝑦.

For positive real numbers, we will use the square root symbol only for the positive square root of the number. For complex numbers, we will use the square root symbol for the principal square root of the number. The concept of principal square root will be explained in Lesson 15.

Example 7.3: 1. Since 22 = 4, 2 ∈ ℝ, and 2 > 0, we see that 2 is the positive square root of 4 and we write 2 = √4.
2. We have (– 2)2 = 4, but – 2 < 0, and so we do not write – 2 = √4. However, – 2 is still a square root of 4, and we can write – 2 = – √4.
3. Since 𝑖2 = – 1, we see that 𝑖 is a square root of – 1. 4. Since (– 𝑖)2 = (– 𝑖)(– 𝑖) = (– 1)(– 1)𝑖2 = 1(– 1) = – 1, we see that – 𝑖 is also a square root of
– 1. 5. (1 + 𝑖)2 = (1 + 𝑖)(1 + 𝑖) = (1 − 1) + (1 + 1)𝑖 = 0 + 2𝑖 = 2𝑖. So, 1 + 𝑖 is a square root of 2𝑖.

The absolute value or modulus of the complex number 𝑧 = 𝑎 + 𝑏𝑖 is the nonnegative real number |𝑧| = √𝑎2 + 𝑏2 = √(Re 𝑧)2 + (Im 𝑧)2

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Note: If 𝑧 = 𝑎 + 0𝑖 = 𝑎 is a real number, then |𝑎| = √𝑎2. This is equal to 𝑎 if 𝑎 ≥ 0 and – 𝑎 if 𝑎 < 0. For example, |4| = √42 = √16 = 4 and |– 4| = √(– 4)2 = √16 = 4 = – (– 4). The statement “|𝑎| = – 𝑎 for 𝑎 < 0” often confuses students. This confusion is understandable, as a minus sign is usually used to indicate that an expression is negative, whereas here we are negating a negative number to make it positive. Unfortunately, this is the simplest way to say, “delete the minus sign in front of the number” using basic notation. Geometrically, the absolute value of a complex number 𝑧 is the distance between the point 𝑧 and the origin. Example 7.4: Which of the following complex numbers is closest to the origin? 1 + 2𝑖, – 3 + 𝑖, or – 2 + 3𝑖?
|1 + 2𝑖| = √12 + 22 = √1 + 4 = √5 |– 3 + 𝑖| = √(– 3)2 + 12 = √9 + 1 = √10 |– 2 + 3𝑖| = √(– 2)2 + 32 = √4 + 9 = √13 Since √5 < √10 < √13, we see that 1 + 2𝑖 is closest to the origin. Notes: (1) Here we have used the following theorem: If 𝑎, 𝑏 ∈ ℝ+, then 𝑎 < 𝑏 if and only if 𝑎2 < 𝑏2. To see this, observe that 𝑎2 < 𝑏2 if and only if 𝑏2 − 𝑎2 > 0 if and only if (𝑏 + 𝑎)(𝑏 − 𝑎) > 0. Since 𝑎 > 0 and 𝑏 > 0, by Order Property 1, 𝑎 + 𝑏 > 0. It follows that 𝑎2 < 𝑏2 if and only if 𝑏 − 𝑎 > 0 if and only if 𝑏 > 𝑎 if and only if 𝑎 < 𝑏. Applying this theorem to 5 < 10 < 13, we get √5 < √10 < √13. (2) The definition of the absolute value of a complex number is motivated by the Pythagorean Theorem. As an example, look at – 3 + 𝑖 in the figure below. Observe that to get from the origin to the point (– 3,1), we move to the left 3 units and then up 1 unit. This gives us a right triangle with legs of lengths 3 and 1. By the Pythagorean Theorem, the hypotenuse has length √32 + 12 = √9 + 1 = √10.
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The distance between the complex numbers 𝑧 = 𝑎 + 𝑏𝑖 and 𝑤 = 𝑐 + 𝑑𝑖 is 𝑑(𝑧, 𝑤) = |𝑧 − 𝑤| = √(𝑐 − 𝑎)2 + (𝑑 − 𝑏)2.
Geometrically, we can translate the vector 𝑧 − 𝑤 so that the directed line segment begins at the terminal point of 𝑤 and ends at the terminal point of 𝑧. Let’s look one more time at the figure we drew for (1 + 2𝑖) − (2 − 𝑖) = – 1 + 3𝑖 and then translate the solution vector as we just suggested.
|– 𝟏 + 𝟑𝒊|
Notice that the expression for the distance between two complex numbers follows from a simple application of the Pythagorean Theorem. Let’s continue to use the same example to help us see this.
√𝟏𝟎 3
1
In the figure above, we can get the lengths of the legs of the triangle either by simply counting the units, or by subtracting the appropriate coordinates. For example, the length of the horizontal leg is 2 − 1 = 1 and the length of the vertical leg is 2 − (– 1) = 2 + 1 = 3. We can then use the Pythagorean Theorem to get the length of the hypotenuse of the triangle: 𝑐 = √12 + 32 = √1 + 9 = √10. Compare this geometric procedure to the formula for distance given above. While we’re on the subject of triangles, the next theorem involving arbitrary triangles is very useful. Theorem 7.3 (The Triangle Inequality): For all 𝑧, 𝑤 ∈ ℂ, |𝑧 + 𝑤| ≤ |𝑧| + |𝑤|.
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Geometrically, the Triangle Inequality says that the length of the third side of a triangle is less than or equal to the sum of the lengths of the other two sides of the triangle. We leave the proof as an exercise (see Problem 4 below). As an example, let’s look at the sum (1 + 2𝑖) + (– 3 + 𝑖) = – 2 + 3𝑖. In Example 7.4, we computed
|1 + 2𝑖| = √5, |– 3 + 𝑖| = √10, and |– 2 + 3𝑖| = √13. Note that √5 + √10 > √4 + √9 = 2 + 3 = 5, whereas √13 < √16 = 4. So, we see that
|(1 + 2𝑖) + (– 3 + 𝑖)| = |– 2 + 3𝑖| = √13 < 4 < 5 < √5 + √10 = |1 + 2𝑖| + |– 3 + 𝑖|. In the following picture, there are two triangles. We’ve put dark bold lines around the leftmost triangle and labeled the sides with their lengths.
|1 + 2𝑖|
Basic Topology of ℂ
A circle in the Complex Plane is the set of all points that are at a fixed distance from a fixed point. The fixed distance is called the radius of the circle and the fixed point is called the center of the circle. If a circle has radius 𝑟 > 0 and center 𝑐 = 𝑎 + 𝑏𝑖, then any point 𝑧 = 𝑥 + 𝑦𝑖 on the circle must satisfy |𝑧 − 𝑐| = 𝑟, or equivalently, (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟2. Note: The equation |𝑧 − 𝑐| = 𝑟 says “The distance between 𝑧 and 𝑐 is equal to 𝑟.” In other words, the distance between any point on the circle and the center of the circle is equal to the radius of the circle. Example 7.5: The circle with equation |𝑧 + 2 − 𝑖| = 2 has center 𝑐 = – (2 − 𝑖) = – 2 + 𝑖 and radius 𝑟 = 2. Note: |𝑧 + 2 − 𝑖| = |𝑧 − (– 2 + 𝑖)|. So, if we rewrite the equation as |𝑧 − (– 2 + 𝑖)| = 2, it is easy to pick out the center and radius of the circle. A picture of the circle is shown to the right. The center is labeled and a typical radius is drawn.
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An open disk in ℂ consists of all the points in the interior of a circle. If 𝑎 is the center of the open disk and 𝑟 is the radius of the open disk, then any point 𝑧 inside the disk satisfies |𝑧 − 𝑎| < 𝑟.
𝑁𝑟(𝑎) = {𝑧 ∈ ℂ | |𝑧 − 𝑎| < 𝑟} is also called the 𝒓-neighborhood of 𝒂.
Example 7.6: 𝑁2(– 2 + 𝑖) = {𝑧 ∈ ℂ | |𝑧 + 2 − 𝑖| < 2} is the 2 neighborhood of – 2 + 𝑖. It consists of all points inside the circle |𝑧 + 2 − 𝑖| = 2.
Notes: (1) A picture of the 2-neighborhood of – 2 + 𝑖 is shown to the right. The center is labeled and a typical radius is drawn. We drew the boundary of the disk with dashes to indicate that points on the circle are not in the neighborhood and we shaded the interior of the disk to indicate that every point inside the circle is in the neighborhood.
(2) The definitions of open disk and 𝑟-neighborhood of 𝑎 also make sense in ℝ, but the geometry looks a bit different. An open disk in ℝ is simply an open interval. If 𝑥 and 𝑎 are real numbers, then we have
𝑥 ∈ 𝑁𝑟(𝑎) ⇔ |𝑥 − 𝑎| < 𝑟 ⇔ √(𝑥 − 𝑎)2 < 𝑟 ⇔ 0 ≤ (𝑥 − 𝑎)2 < 𝑟2 ⇔ – 𝑟 < 𝑥 − 𝑎 < 𝑟 ⇔ 𝑎 − 𝑟 < 𝑥 < 𝑎 + 𝑟 ⇔ 𝑥 ∈ (𝑎 − 𝑟, 𝑎 + 𝑟).
So, in ℝ, an 𝑟-neighborhood of 𝑎 is the open interval 𝑁𝑟(𝑎) = (𝑎 − 𝑟, 𝑎 + 𝑟). Notice that the length (or diameter) of this interval is 2𝑟.
As an example, let’s draw a picture of 𝑁2(1) = (1 − 2, 1 + 2) = (– 1, 3). Observe that the center of this open disk (or open interval or neighborhood) in ℝ is the real number 1, the radius of the open disk is 2, and the diameter of the open disk (or length of the interval) is 4.
A closed disk is the interior of a circle together with the circle itself (the boundary is included). If 𝑎 is the center of the closed disk and 𝑟 is the radius of the closed disk, then any point 𝑧 inside the closed disk satisfies |𝑧 − 𝑎| ≤ 𝑟.
Notes: (1) In this case, the circle itself would be drawn solid to indicate that all points on the circle are included.
(2) Just like an open disk in ℝ is an open interval, a closed disk in ℝ is a closed interval.
(3) The reader is encouraged to draw a few open and closed disks in both ℂ and ℝ, and to write down the corresponding sets of points using set-builder notation and, in the case of ℝ, interval notation.
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A punctured open disk consists of all the points in the interior of a circle except for the center of the circle. If 𝑎 is the center of the punctured open disk and 𝑟 is the radius of the open disk, then any point 𝑧 inside the punctured disk satisfies |𝑧 − 𝑎| < 𝑟 and 𝑧 ≠ 𝑎.
Note that 𝑧 ≠ 𝑎 is equivalent to 𝑧 − 𝑎 ≠ 0. In turn, this is equivalent to |𝑧 − 𝑎| ≠ 0. Since |𝑧 − 𝑎| must be nonnegative, |𝑧 − 𝑎| ≠ 0 is equivalent to |𝑧 − 𝑎| > 0 or 0 < |𝑧 − 𝑎|.
Therefore, a punctured open disk with center 𝑎 and radius 𝑟 consists of all points 𝑧 that satisfy 𝟎 < |𝒛 − 𝒂| < 𝒓.
𝑁𝑟⨀(𝑎) = {𝑧 | 0 < |𝑧 − 𝑎| < 𝑟} is also called a deleted 𝒓-neighborhood of 𝑎. Example 7.7: 𝑁2⨀(– 2 + 𝑖) = {𝑧 ∈ ℂ | 0 < |𝑧 + 2 − 𝑖| < 2} is the deleted 2 neighborhood of – 2 + 𝑖. It consists of all points inside the circle |𝑧 + 2 − 𝑖| = 2, except for – 2 + 𝑖.
Notes: (1) A picture of the deleted 2-neighborhood of – 2 + 𝑖 is shown to the right. Notice that this time we excluded the center of the disk – 2 + 𝑖, as this point is not included in the set.
(2) In ℝ, we have 𝑁𝑟⨀(𝑎) = (𝑎 − 𝑟, 𝑎 + 𝑟) ∖ {𝑎} = (𝑎 − 𝑟, 𝑎) ∪ (𝑎, 𝑎 + 𝑟).
This is the open interval centered at 𝑎 of length (or diameter) 2𝑟 with 𝑎 removed.
Let’s draw a picture of 𝑁2⨀(1) = (– 1, 3) ∖ {1} = (– 1, 1) ∪ (1, 3).
(3) Notice how all the topological definitions we are presenting make sense in both ℝ and ℂ, but the geometry in each case looks different. You will continue to see this happen. In fact, these definitions make sense for many, many sets and structures, all with their own “look.” In general, topology allows us to make definitions and prove theorems that can be applied very broadly and used in many (if not all) branches of mathematics.
A subset 𝑋 of ℂ is said to be open if for every complex number 𝑧 ∈ 𝑋, there is an open disk 𝐷 with 𝑧 ∈ 𝐷 and 𝐷 ⊆ 𝑋.
In words, a set is open in ℂ if every point in the set has “space” all around it inside the set. If you think of each point in the set as an animal, then each animal in the set should be able to move a little in any direction it chooses without leaving the set. Another way to think of this is that no number is right on “the edge” or “the boundary” of the set, about to fall out of it.
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Example 7.8:
1. Every open disk 𝐷 is an open set. To see this, simply observe that if 𝑧 ∈ 𝐷, then 𝐷 itself is an open disk with 𝑧 ∈ 𝐷 and 𝐷 ⊆ 𝐷.
2. A closed disk is not an open set because it contains its “boundary.” As an example, let’s look at the closed unit disk 𝐷 = {𝑧 ∈ ℂ | |𝑧| ≤ 1}. Let’s focus on the point 𝑖. First note that 𝑖 ∈ 𝐷 because |𝑖| = √02 + 12 = √1 = 1 and 1 ≤ 1. Now, any open disk 𝑁 containing 𝑖 will contain points above 𝑖. Let’s say (1 + 𝜖)𝑖 ∈ 𝑁 for some positive real number 𝜖. Now, we have |(1 + 𝜖)𝑖| = √02 + (1 + 𝜖)2 = 1 + 𝜖, which is greater than 1. Therefore, (1 + 𝜖)𝑖 ∉ 𝐷. It follows that 𝑁 ⊈ 𝐷, and so, 𝐷 is not open.
3. We can use reasoning similar to that used in 2 to see that if we take any subset of a disk that contains any points on the bounding circle, then that set will not be open.
4. ∅ and ℂ are both open. You will be asked to prove this in Problem 7 below (parts (i) and (ii)).

As we mentioned in Lesson 6 right before Theorem 6.5, many authors define “open” in a slightly different way from the definition we’ve been using. Once again, let’s show that the definition we have been using is equivalent to theirs.

Theorem 7.4: A subset 𝑋 of ℂ is open if and only if for every complex number 𝑤 ∈ 𝑋, there is a positive real number 𝑑 such that 𝑁𝑑(𝑤) ⊆ 𝑋.

Analysis: The harder direction of the proof is showing that

if 𝑋 is open, then for every complex number 𝑤 ∈ 𝑋, there is a positive real number 𝑑 such that 𝑁𝑑(𝑤) ⊆ 𝑋.

𝐷

To see this, suppose that 𝑋 is open and let 𝑤 ∈ 𝑋. Then there is an open disk 𝐷 = {𝑧 ∈ ℂ | |𝑧 − 𝑎| < 𝑟} with 𝑤 ∈ 𝐷 and 𝐷 ⊆ 𝑋. We want to replace the disk 𝐷 with a disk that has 𝑤 right in the center.

To accomplish this, we let 𝑐 be the distance from 𝑤 to 𝑎. Then 𝑟 − 𝑐 is the distance from 𝑤 to the boundary of 𝐷. We will show that the disk with center 𝑤 and radius 𝑟 − 𝑐 is a subset of 𝐷.

The picture to the right illustrates this idea. Notice that 𝑐 + (𝑟 − 𝑐) = 𝑟, the radius of disk 𝐷.

Proof of Theorem 7.4: Let 𝑋 be an open subset of ℂ and let 𝑤 ∈ 𝑋. Then there is an open disk 𝐷 with 𝑤 ∈ 𝐷 and 𝐷 ⊆ 𝑋.

Suppose that 𝐷 has center 𝑎 and radius 𝑟. So, 𝐷 = {𝑧 ∈ ℂ | |𝑧 − 𝑎| < 𝑟}.

Let 𝑐 = |𝑤 − 𝑎| and let 𝑑 = 𝑟 − 𝑐. We will show that 𝑁𝑑(𝑤) ⊆ 𝐷. Let 𝑧 ∈ 𝑁𝑑(𝑤). Then |𝑧 − 𝑤| < 𝑑 = 𝑟 − 𝑐.

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By the Triangle Inequality (and SACT—see Note 2 below), |𝑧 − 𝑎| = |(𝑧 − 𝑤) + (𝑤 − 𝑎)| ≤ |𝑧 − 𝑤| + |𝑤 − 𝑎| < (𝑟 − 𝑐) + 𝑐 = 𝑟.

So, 𝑧 ∈ 𝐷. Since 𝑧 was an arbitrary element of 𝑁𝑑(𝑤), we showed that 𝑁𝑑(𝑤) ⊆ 𝐷.

So, we have 𝑁𝑑(𝑤) ⊆ 𝐷 and 𝐷 ⊆ 𝑋. By the transitivity of ⊆ (Theorem 2.3 from Lesson 2), we have 𝑁𝑑(𝑤) ⊆ 𝑋.

The converse is immediate since for 𝑤 ∈ 𝑋, 𝑁𝑑(𝑤) is an open disk containing 𝑤.

□

Notes: (1) The picture to the right shows how we used the Triangle Inequality. The three sides of the triangle have lengths |𝑧 − 𝑤|, |𝑤 − 𝑎|, and |𝑧 − 𝑎|.

(2) Notice how we used SACT (the Standard Advanced Calculus Trick) here. Starting with 𝑧 − 𝑎, we wanted to make 𝑧 − 𝑤 and 𝑤 − 𝑎 “appear.” We were able to do this simply by subtracting and then adding 𝑤 between 𝑧 and 𝑎. We often use this trick when applying the Triangle Inequality. SACT was introduced in Lesson 4 (Note 7 following Example 4.5).

(3) The same proof used here can be used to prove Theorem 6.5. The geometry looks different (disks and neighborhoods are open intervals instead of the interiors of circles, and points appear on the real line instead of in the complex plane), but the argument is identical. Compare this proof to the proof we used in Theorem 6.5.

A subset 𝑋 of ℂ is said to be closed if ℂ ∖ 𝑋 is open.

ℂ ∖ 𝑋 is called the complement of 𝑋 in ℂ, or simply the complement of 𝑋. It consists of all complex numbers not in 𝑋.

Example 7.9:
1. Every closed disk is a closed set. For example, 𝐷 = {𝑧 ∈ ℂ | |𝑧| ≤ 1} is closed because its complement in ℂ is ℂ ∖ 𝐷 = {𝑧 ∈ ℂ | |𝑧| > 1}. You will be asked to prove that this set 𝐷 is open in Problem 7 below (part (iii)).
2. If we take any subset of a closed disk that includes the interior of the disk, but is missing at least one point on the bounding circle, then that set will not be closed. You will be asked to prove this for the closed unit disk {𝑧 ∈ ℂ | |𝑧| ≤ 1} in Problem 10 below.
3. ∅ is closed because ℂ ∖ ∅ = ℂ is open. ℂ is closed because ℂ ∖ ℂ = ∅ is open. ∅ and ℂ are the only two sets of complex numbers that are both open and closed.

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Problem Set 7

Full solutions to these problems are available for free download here:
www.SATPrepGet800.com/PMFBXSG

LEVEL 1

1. Let 𝑧 = – 4 − 𝑖 and 𝑤 = 3 − 5𝑖. Compute each of the following:

(i) 𝑧 + 𝑤

(ii) 𝑧𝑤

(iii) Im 𝑤

(iv) 2𝑧 − 𝑤

(v) 𝑤

(vi)

𝑧 𝑤

(vii) |𝑧|

(viii) the distance between 𝑧 and 𝑤

LEVEL 2

2. Prove that (ℂ, +, ⋅) is field.

3. Let 𝑧 and 𝑤 be complex numbers. Prove the following:

(i)

Re 𝑧 = 𝑧+𝑧
2

(ii) Im 𝑧 = 𝑧−𝑧
2𝑖

(iii) 𝑧 + 𝑤 = 𝑧 + 𝑤

(iv) 𝑧𝑤 = 𝑧 ⋅ 𝑤

(v) ( 𝑧 ) = 𝑧
𝑤𝑤
(vi) 𝑧𝑧 = |𝑧|2

(vii) |𝑧𝑤| = |𝑧||𝑤|

(viii)

If

𝑤

≠

0,

then

|𝑧|
𝑤

=

|𝑧| |𝑤|

(ix) Re 𝑧 ≤ |𝑧|

(x) Im 𝑧 ≤ |𝑧|

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LEVEL 3
4. Prove the Triangle Inequality (Theorem 7.3).

5. Let 𝑧 and 𝑤 be complex numbers. Prove ||𝑧| − |𝑤|| ≤ |𝑧 ± 𝑤| ≤ |𝑧| + |𝑤|.

6. A point 𝑤 is an accumulation point of a set 𝑆 of complex numbers if each deleted neighborhood of 𝑤 contains at least one point in 𝑆. Determine the accumulation points of each of the following sets:

(i) {1 | 𝑛 ∈ ℤ+}
𝑛

(ii)

{ 𝑖 | 𝑛 ∈ ℤ+}
𝑛

(iii) {𝑖𝑛 | 𝑛 ∈ ℤ+}

(iv) {𝑖𝑛 | 𝑛 ∈ ℤ+}
𝑛
(v) {𝑧 | |𝑧| < 1}

(vi) {𝑧 |0 < |𝑧 − 2| ≤ 3}

LEVEL 4
7. Determine if each of the following subsets of ℂ is open, closed, both, or neither. Give a proof in each case. (i) ∅ (ii) ℂ (iii) {𝑧 ∈ ℂ | |𝑧| > 1} (iv) {𝑧 ∈ ℂ | Im 𝑧 ≤ −2} (v) {𝑖𝑛 | 𝑛 ∈ ℤ+} (vi) {𝑧 ∈ ℂ |2 < |𝑧 − 2| < 4}

8. Prove the following: (i) An arbitrary union of open sets in ℂ is an open set in ℂ. (ii) A finite intersection of open sets in ℂ is an open set in ℂ. (iii) An arbitrary intersection of closed sets in ℂ is a closed set in ℂ. (iv) A finite union of closed sets in ℂ is a closed set in ℂ. (v) Every open set in ℂ can be expressed as a union of open disks.

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LEVEL 5
9. A complex number 𝑧 is an interior point of a set 𝑆 of complex numbers if there is a neighborhood of 𝑧 that contains only points in 𝑆, whereas 𝑤 is a boundary point of 𝑆 if each neighborhood of 𝑤 contains at least one point in 𝑆 and one point not in 𝑆. Prove the following: (i) A set of complex numbers is open if and only if each point in 𝑆 is an interior point of 𝑆. (ii) A set of complex numbers is open if and only if it contains none of its boundary points. (iii) A set of complex numbers is closed if and only if it contains all its boundary points.
10. Let 𝐷 = {𝑧 ∈ ℂ | |𝑧| ≤ 1} be the closed unit disk and let 𝑆 be a subset of 𝐷 that includes the interior of the disk but is missing at least one point on the bounding circle of the disk. Show that 𝑆 is not a closed set.
11. Prove that a set of complex numbers is closed if and only if it contains all its accumulation points. (See Problem 6 for the definition of an accumulation point.)
12. Prove that a set consisting of finitely many complex numbers is a closed set in ℂ. (Hint: Show that a finite set has no accumulation points.)
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LESSON 8 – LINEAR ALGEBRA VECTOR SPACES
Vector Spaces Over Fields
Recall the following: 1. In previous lessons, we looked at three structures called fields: ℚ (the field of rational numbers), ℝ (the field of real numbers), and ℂ (the field of complex numbers). Each of these fields come with two operations called addition and multiplication. Also, ℚ is a subfield of ℝ and ℝ is a subfield of ℂ. This means that every rational number is a real number, every real number is a complex number, and addition and multiplication in ℚ, ℝ, and ℂ all work the same way.
2. Fields have a particularly nice structure. When working in a field, we can perform all the arithmetic and algebra that we remember from elementary and middle school. In particular, we have closure, associativity, commutativity, identity elements, and inverse properties for both addition and multiplication (with the exception that 0 has no multiplicative inverse), and multiplication is distributive over addition.
3. The standard form of a complex number is 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. We add two complex numbers using the rule (𝑎 + 𝑏𝑖) + (𝑐 + 𝑑𝑖) = (𝑎 + 𝑐) + (𝑏 + 𝑑)𝑖.
To give some motivation for the definition of a vector space, let’s begin with an example.
Example 8.1: Consider the set ℂ of complex numbers together with the usual definition of addition. Let’s also consider another operation, which we will call scalar multiplication. For each 𝑘 ∈ ℝ and 𝑧 = 𝑎 + 𝑏𝑖 ∈ ℂ, we define 𝑘𝑧 to be 𝑘𝑎 + 𝑘𝑏𝑖.
The operation of scalar multiplication is a little different from other types of operations we have looked at previously because instead of multiplying two elements from ℂ together, we are multiplying an element of ℝ with an element of ℂ. In this case, we will call the elements of ℝ scalars.
Let’s observe that we have the following properties: 1. (ℂ, +) is a commutative group. In other words, for addition in ℂ, we have closure, associativity, commutativity, an identity element (called 0), and the inverse property (the inverse of 𝑎 + 𝑏𝑖 is – 𝑎 − 𝑏𝑖). This follows immediately from the fact that (ℂ, +, ⋅) is a field. When we choose to think of ℂ as a vector space, we will “forget about” the multiplication in ℂ, and just consider ℂ together with addition. In doing so, we lose much of the field structure of the complex numbers, but we retain the group structure of (ℂ, +). 2. ℂ is closed under scalar multiplication. That is, for all 𝒌 ∈ ℝ and 𝒛 ∈ ℂ, we have 𝒌𝒛 ∈ ℂ. To see this, let 𝑧 = 𝑎 + 𝑏𝑖 ∈ ℂ and let 𝑘 ∈ ℝ. Then, by definition, 𝑘𝑧 = 𝑘𝑎 + 𝑘𝑏𝑖. Since 𝑎, 𝑏 ∈ ℝ, and ℝ is closed under multiplication, 𝑘𝑎 ∈ ℝ and 𝑘𝑏 ∈ ℝ. It follows that 𝑘𝑎 + 𝑘𝑏𝑖 ∈ ℂ. 3. 𝟏𝒛 = 𝒛. To see this, consider 1 ∈ ℝ and let 𝑧 = 𝑎 + 𝑏𝑖 ∈ ℂ. Then, since 1 is the multiplicative identity for ℝ, we have 1𝑧 = 1𝑎 + 1𝑏𝑖 = 𝑎 + 𝑏𝑖 = 𝑧.
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4. For all 𝒋, 𝒌 ∈ ℝ and 𝒛 ∈ ℂ, (𝒋𝒌)𝒛 = 𝒋(𝒌𝒛) (Associativity of scalar multiplication). To see this, let 𝑗, 𝑘 ∈ ℝ and 𝑧 = 𝑎 + 𝑏𝑖 ∈ ℂ. Then since multiplication is associative in ℝ, we have (𝑗𝑘)𝑧 = (𝑗𝑘)(𝑎 + 𝑏𝑖) = (𝑗𝑘)𝑎 + (𝑗𝑘)𝑏𝑖 = 𝑗(𝑘𝑎) + 𝑗(𝑘𝑏)𝑖 = 𝑗(𝑘𝑎 + 𝑘𝑏𝑖) = 𝑗(𝑘𝑧).
5. For all 𝒌 ∈ ℝ and 𝒛, 𝒘 ∈ ℂ, 𝒌(𝒛 + 𝒘) = 𝒌𝒛 + 𝒌𝒘 (Distributivity of 1 scalar over 2 vectors). To see this, let 𝑘 ∈ ℝ and 𝑧 = 𝑎 + 𝑏𝑖, 𝑤 = 𝑐 + 𝑑𝑖 ∈ ℂ. Then since multiplication distributes over addition in ℝ, we have
𝑘(𝑧 + 𝑤) = 𝑘((𝑎 + 𝑏𝑖) + (𝑐 + 𝑑𝑖)) = 𝑘((𝑎 + 𝑐) + (𝑏 + 𝑑)𝑖) = 𝑘(𝑎 + 𝑐) + 𝑘(𝑏 + 𝑑)𝑖 = (𝑘𝑎 + 𝑘𝑐) + (𝑘𝑏 + 𝑘𝑑)𝑖 = (𝑘𝑎 + 𝑘𝑏𝑖) + (𝑘𝑐 + 𝑘𝑑𝑖) = 𝑘(𝑎 + 𝑏𝑖) + 𝑘(𝑐 + 𝑑𝑖) = 𝑘𝑧 + 𝑘𝑤. 6. For all 𝒋, 𝒌 ∈ ℝ and 𝒛 ∈ ℂ, (𝒋 + 𝒌)𝒛 = 𝒋𝒛 + 𝒌𝒛 (Distributivity of 2 scalars over 1 vector). To see
this, let 𝑗, 𝑘 ∈ ℝ and 𝑧 = 𝑎 + 𝑏𝑖 ∈ ℂ. Then since multiplication distributes over addition in ℝ, we have
(𝑗 + 𝑘)𝑧 = (𝑗 + 𝑘)(𝑎 + 𝑏𝑖) = (𝑗 + 𝑘)𝑎 + (𝑗 + 𝑘)𝑏𝑖 = (𝑗𝑎 + 𝑘𝑎) + (𝑗𝑏 + 𝑘𝑏)𝑖 = (𝑗𝑎 + 𝑗𝑏𝑖) + (𝑘𝑎 + 𝑘𝑏𝑖) = 𝑗(𝑎 + 𝑏𝑖) + 𝑘(𝑎 + 𝑏𝑖) = 𝑗𝑧 + 𝑘𝑧.
Notes: (1) Since the properties listed in 1 through 6 above are satisfied, we say that ℂ is a vector space over ℝ. We will give the formal definition of a vector space below.
(2) Note that a vector space consists of (i) a set of vectors (in this case ℂ), (ii) a field (in this case ℝ), and (iii) two operations called addition and scalar multiplication.
(3) The operation of addition is a binary operation on the set of vectors, and the set of vectors together with this binary operation forms a commutative group. In the previous example (Example 8.1), we have that (ℂ, +) is a commutative group.
(4) Scalar multiplication is not a binary operation on the set of vectors. It takes pairs of the form (𝑘, 𝑣), where 𝑘 is in the field and 𝑣 is a vector to a vector 𝑘𝑣. Formally speaking, scalar multiplication is a function 𝑓: 𝔽 × 𝑉 → 𝑉, where 𝔽 is the field of scalars and 𝑉 is the set of vectors (see the beginning of Lesson 3 for a brief explanation of this notation).
(5) We started with the example of ℂ as a vector space over ℝ because it has a geometric interpretation where we can draw simple pictures to visualize what the vector space looks like. Recall from Lesson 7 that we can think of the complex number 𝑎 + 𝑏𝑖 as a directed line segment (which from now on we will call a vector) in the complex plane that begins at the origin and terminates at the point (𝑎, 𝑏).
For example, pictured to the right, we can see the vectors 𝑖 = 0 + 1𝑖, 1 + 2𝑖, and 2 = 2 + 0𝑖 in the complex plane.
We can visualize the sum of two vectors as the vector starting at the origin that is the diagonal of the parallelogram formed from the original vectors. We see this in the first figure on the left below. In this figure, we have removed the complex plane and focused on the vectors 1 + 2𝑖 and 2, together with their sum (1 + 2𝑖) + (2 + 0𝑖) = (1 + 2) + (2 + 0)𝑖 = 3 + 2𝑖.
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A second way to visualize the sum of two vectors is to translate one of the vectors so that its initial point coincides with the terminal point of the other vector. The sum of the two vectors is then the vector whose initial point coincides with the initial point of the “unmoved” vector and whose terminal point coincides with the terminal point of the “moved” vector. We see two ways to do this in the center and rightmost figures below.
Technically speaking, the center figure shows the sum (1 + 2𝑖) + 2 and the rightmost figure shows the sum 2 + (1 + 2𝑖). If we superimpose one figure on top of the other, we can see strong evidence that commutativity holds for addition.
2

1 + 2𝑖

3 + 2𝑖

1 + 2𝑖

3 + 2𝑖

3 + 2𝑖

1 + 2𝑖

(0, 0) 2

(0, 0)

(0, 0) 2

We can visualize a scalar multiple of a vector as follows: (i) if 𝑘 is a positive real number and 𝑧 ∈ ℂ, then the vector 𝑘𝑧 points in the same direction as 𝑧 and has a length that is 𝑘 times the length of 𝑧; (ii) if 𝑘 is a negative real number and 𝑧 ∈ ℂ, then the vector 𝑘𝑧 points in the direction opposite of 𝑧 and has a length that is |𝑘| times the length of 𝑧; (iii) if 𝑘 = 0 and 𝑧 ∈ ℂ, then 𝑘𝑧 is a point.
In the figures below, we have a vector 𝑧 ∈ ℂ, together with several scalar multiples of 𝑧.

2𝑧

– 2𝑧

𝑧

1𝑧

– 𝑧 = (– 1)𝑧

2

–

1 2

𝑧

We are now ready for the general definition of a vector space.
A vector space over a field 𝔽 is a set 𝑉 together with a binary operation + on 𝑉 (called addition) and an operation called scalar multiplication satisfying:
(1) (𝑉, +) is a commutative group. (2) (Closure under scalar multiplication) For all 𝑘 ∈ 𝔽 and 𝑣 ∈ 𝑉, 𝑘𝑣 ∈ 𝑉. (3) (Scalar multiplication identity) If 1 is the multiplicative identity of 𝔽 and 𝑣 ∈ 𝑉, then 1𝑣 = 𝑣. (4) (Associativity of scalar multiplication) For all 𝑗, 𝑘 ∈ 𝔽 and 𝑣 ∈ 𝑉, (𝑗𝑘)𝑣 = 𝑗(𝑘𝑣). (5) (Distributivity of 1 scalar over 2 vectors) For all 𝑘 ∈ 𝔽 and 𝑣, 𝑤 ∈ 𝑉, 𝑘(𝑣 + 𝑤) = 𝑘𝑣 + 𝑘𝑤. (6) (Distributivity of 2 scalars over 1 vector) For all 𝑗, 𝑘 ∈ 𝔽 and 𝑣 ∈ 𝑉, (𝑗 + 𝑘)𝑣 = 𝑗𝑣 + 𝑘𝑣.
Notes: (1) Recall from Lesson 3 that (𝑉, +) a commutative group means the following: • (Closure) For all 𝑣, 𝑤 ∈ 𝑉, 𝑣 + 𝑤 ∈ 𝑉.
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• (Associativity) For all 𝑣, 𝑤, 𝑢 ∈ 𝑉, (𝑣 + 𝑤) + 𝑢 = 𝑣 + (𝑤 + 𝑢).
• (Commutativity) For all 𝑣, 𝑤 ∈ 𝑉, 𝑣 + 𝑤 = 𝑤 + 𝑣.
• (Identity) There exists an element 0 ∈ 𝑉 such that for all 𝑣 ∈ 𝑉, 0 + 𝑣 = 𝑣 + 0 = 𝑣.
• (Inverse) For each 𝑣 ∈ 𝑉, there is – 𝑣 ∈ 𝑉 such that 𝑣 + (– 𝑣) = (– 𝑣) + 𝑣 = 0.
(2) The fields that we are familiar with are ℚ (the field of rational numbers), ℝ (the field of real numbers), and ℂ (the field of complex numbers). For our purposes here, we can always assume that 𝔽 is one of these three fields.
Let’s look at some basic examples of vector spaces.
Example 8.2: 1. Let ℝ2 be the set of all ordered pairs of real numbers. That is, ℝ2 = {(𝑎, 𝑏 ) | 𝑎, 𝑏 ∈ ℝ} We define addition by (𝑎, 𝑏) + (𝑐, 𝑑) = (𝑎 + 𝑐, 𝑏 + 𝑑). We define scalar multiplication by 𝑘(𝑎, 𝑏) = (𝑘𝑎, 𝑘𝑏) for each 𝑘 ∈ ℝ. With these definitions, ℝ2 is a vector space over ℝ. Notice that ℝ2 looks just like ℂ. In fact, (𝑎, 𝑏) is sometimes used as another notation for 𝑎 + 𝑏𝑖. Therefore, the verification that ℝ2 is a vector space over ℝ is nearly identical to what we did in Example 8.1 above. We can visualize elements of ℝ2 as points or vectors in a plane in exactly the same way that we visualize complex numbers as points or vectors in the complex plane. 2. ℝ3 = {(𝑎, 𝑏, 𝑐) | 𝑎, 𝑏, 𝑐 ∈ ℝ} is a vector space over ℝ, where we define addition and scalar multiplication by (𝑎, 𝑏, 𝑐) + (𝑑, 𝑒, 𝑓) = (𝑎 + 𝑑, 𝑏 + 𝑒, 𝑐 + 𝑓) and 𝑘(𝑎, 𝑏, 𝑐) = (𝑘𝑎, 𝑘𝑏, 𝑘𝑐), respectively. We can visualize elements of ℝ3 as points in space in a way similar to visualizing elements of ℝ2 and ℂ as points in a plane. 3. More generally, we can let ℝ𝑛 = {(𝑎1, 𝑎2, … , 𝑎𝑛) | 𝑎𝑖 ∈ ℝ for each 𝑖 = 1, 2, … , 𝑛}. Then ℝ𝑛 is a vector space over ℝ, where we define addition and scalar multiplication by (𝑎1, 𝑎2, … , 𝑎𝑛) + (𝑏1, 𝑏2, … , 𝑏𝑛) = (𝑎1 + 𝑏1, 𝑎2 + 𝑏2, … , 𝑎𝑛 + 𝑏𝑛). 𝑘(𝑎1, 𝑎2, … , 𝑎𝑛) = (𝑘𝑎1, 𝑘𝑎2, … , 𝑘𝑎𝑛). 4. More generally still, if 𝔽 is any field (for our purposes, we can think of 𝔽 as ℚ, ℝ, or ℂ), we let 𝔽𝑛 = {(𝑎1, 𝑎2, … , 𝑎𝑛) | 𝑎𝑖 ∈ 𝔽 for each 𝑖 = 1, 2, … , 𝑛}. Then 𝔽𝑛 is a vector space over 𝔽, where we define addition and scalar multiplication by (𝑎1, 𝑎2, … , 𝑎𝑛) + (𝑏1, 𝑏2, … , 𝑏𝑛) = (𝑎1 + 𝑏1, 𝑎2 + 𝑏2, … , 𝑎𝑛 + 𝑏𝑛). 𝑘(𝑎1, 𝑎2, … , 𝑎𝑛) = (𝑘𝑎1, 𝑘𝑎2, … , 𝑘𝑎𝑛).
Notes: (1) Ordered pairs have the property that (𝑎, 𝑏) = (𝑐, 𝑑) if and only if 𝑎 = 𝑐 and 𝑏 = 𝑑. So, for example, (1,2) ≠ (2,1). Compare this to the unordered pair (or set) {1, 2}. Recall that a set is determined by its elements and not the order in which the elements are listed. So, {1, 2} = {2, 1}.
We will learn more about ordered pairs in Lesson 10.
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(2) (𝑎1, 𝑎2, … , 𝑎𝑛) is called an 𝒏-tuple. So, ℝ𝑛 consists of all 𝑛-tuples of elements from ℝ, and more generally, 𝔽𝑛 consists of all 𝑛-tuples of elements from the field 𝔽.

For example, (3, 2 − 𝑖, √2 + √3𝑖, – 3𝑖) ∈

ℂ4

and

(1,

1 2

,

1 3

,

1 4

,

1 5

,

1 6

,

1 7

,

1)
8

∈

ℚ8

(and

since

ℚ8

⊆ ℝ8

⊆ ℂ8,

we can also say that this 8-tuple is in ℝ8 or ℂ8).

(3) Similar to what we said in Note 1, we have (𝑎1, 𝑎2, … , 𝑎𝑛) = (𝑏1, 𝑏2, … , 𝑏𝑛) if and only if 𝑎𝑖 = 𝑏𝑖 for all 𝑖 = 1, 2, … , 𝑛. So, for example, (2, 5, √2, √2) and (2, √2, 5, √2) are distinct elements from ℝ4.

(4) You will be asked to verify that 𝔽𝑛 is a vector space over the field 𝔽 in Problem 3 below. Unless stated otherwise, from now on we will always consider the vector space 𝔽𝑛 to be over the field 𝔽.

Let’s look at a few other examples of vector spaces.

Example 8.3:

1. Let 𝑀 = {[𝑎𝑐 𝑑𝑏] | 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℝ} be the set of all 2 × 2 matrices of real numbers. We add two

matrices using the rule [𝑎𝑐

𝑑𝑏] + [𝑔𝑒

𝑓ℎ]

=

[𝑐𝑎

+𝑒 +𝑔

𝑏 𝑑

+ +

𝑓ℎ],

and

we

multiply

a

matrix

by

a

real

number using the rule 𝑘 [𝑎𝑐 𝑑𝑏] = [𝑘𝑘𝑎𝑐 𝑘𝑘𝑑𝑏]. It is straightforward to check that 𝑀 is a vector

space over ℝ.

2. For 𝑚, 𝑛 ∈ ℤ+, an 𝑚 × 𝑛 matrix over a field 𝔽 is a rectangular array with 𝑚 rows and 𝑛 columns,

and entries in 𝔽. For example, the matrix 𝐴 = [ 5

2

1
5 ] is a 2 × 3 matrix over ℝ. We

– 3 √3 7

will generally use a capital letter to represent a matrix, and the corresponding lowercase letter

with double subscripts to represent the entries of the matrix. We use the first subscript for the row and the second subscript for the column. Using the matrix 𝐴 above as an example, we see that 𝑎21 = – 3 because the entry in row 2 and column 1 is – 3. Similarly, we have 𝑎11 = 5, 𝑎12 = 2, 𝑎13 = 15, 𝑎22 = √3, and 𝑎23 = 7.

Let 𝑀𝑚𝔽 𝑛 be the set of all 𝑚 × 𝑛 matrices over the field 𝔽. We add two matrices 𝐴, 𝐵 ∈ 𝑀𝑚𝔽 𝑛 to

get 𝐴 + 𝐵 ∈ 𝑀𝑚𝔽 𝑛 using the rule (𝑎 + 𝑏)𝑖𝑗 = 𝑎𝑖𝑗 + 𝑏𝑖𝑗. We multiply a matrix 𝐴 ∈ 𝑀𝑚𝔽 𝑛 by a

scalar 𝑘 ∈ 𝔽 using the rule (𝑘𝑎)𝑖𝑗 = 𝑘𝑎𝑖𝑗.

2 –5 4

For example, if we let 𝐴 be the matrix above and 𝐵 = [

5 ], then we have

– 1 – √3 1

𝐴+𝐵 =[

7 –4

–3 0

1 8

]

and

2𝐴 = [ 10 –6

4 2√3

2
5 ]. 14

Notice that we get the entry in the first row and first column of 𝐴 + 𝐵 as follows:

(𝑎 + 𝑏)11 = 𝑎11 + 𝑏11 = 5 + 2 = 7

Similarly, we get the other two entries in the first row like this:

(𝑎 + 𝑏)12 = 𝑎12 + 𝑏12 = 2 + (– 5) = – 3

14 5 (𝑎 + 𝑏)13 = 𝑎13 + 𝑏13 = 5 + 5 = 5 = 1

97

I leave it to the reader to write out the details for computing the entries in the second row of 𝐴 + 𝐵.

We get the entries in the first row of 2𝐴 as follows: (2𝑎)11 = 2𝑎11 = 2 ⋅ 5 = 10 (2𝑎)12 = 2𝑎12 = 2 ⋅ 2 = 4

12 (2𝑎)13 = 2𝑎13 = 2 ⋅ 5 = 5

I leave it to the reader to write out the details for computing the entries in the second row of 2𝐴.

With the operations of addition and scalar multiplication defined as we have above, it is not too hard to show that 𝑀𝑚𝔽 𝑛 is a vector space over 𝔽. 3. Let 𝑃 = {𝑎𝑥2 + 𝑏𝑥 + 𝑐 | 𝑎, 𝑏, 𝑐 ∈ ℝ} be the set of polynomials of degree 𝟐 with real coefficients. We define addition and scalar multiplication (with scalars in ℝ) on this set of polynomials as follows:
(𝑎𝑥2 + 𝑏𝑥 + 𝑐) + (𝑑𝑥2 + 𝑒𝑥 + 𝑓) = (𝑎 + 𝑑)𝑥2 + (𝑏 + 𝑒)𝑥 + (𝑐 + 𝑓). 𝑘(𝑎𝑥2 + 𝑏𝑥 + 𝑐) = (𝑘𝑎)𝑥2 + (𝑘𝑏)𝑥 + (𝑘𝑐).
For example, if 𝑝(𝑥) = 2𝑥2 + 3𝑥 − 5 and 𝑞(𝑥) = – 5𝑥 + 4, then 𝑝(𝑥), 𝑞(𝑥) ∈ 𝑃 and we have
𝑝(𝑥) + 𝑞(𝑥) = (2𝑥2 + 3𝑥 − 5) + (– 5𝑥 + 4) = 2𝑥2 − 2𝑥 − 1.
3𝑝(𝑥) = 3(2𝑥2 + 3𝑥 − 5) = 6𝑥2 + 9𝑥 − 15.

It is straightforward to check that 𝑃 is a vector space over ℝ.

Subspaces

Let 𝑉 be a vector space over a field 𝔽. A subset 𝑈 of 𝑉 is called a subspace of 𝑉, written 𝑈 ≤ 𝑉, if it is also a vector space with respect to the same operations of addition and scalar multiplication as they were defined in 𝑉.

Notes: (1) Recall from Note 2 following Example 3.3 that a universal statement is a statement that describes a property that is true for all elements without mentioning the existence of any new elements. A universal statement begins with the quantifier ∀ (“For all”) and never includes the quantifier ∃ (“There exists” or “There is”).

Properties defined by universal statements are closed downwards. This means that if a property defined by a universal statement is true in 𝑉 and 𝑈 is a subset of 𝑉, then the property is true in 𝑈 as well.

For example, the statement for commutativity is ∀𝑣, 𝑤(𝑣 + 𝑤 = 𝑤 + 𝑣). This is read “For all 𝑣 and 𝑤, 𝑣 + 𝑤 = 𝑤 + 𝑣.” The quantifier ∀ is referring to whichever set we are considering. If we are thinking about the set 𝑉, then we mean “For all 𝑣 and 𝑤 in 𝑉, 𝑣 + 𝑤 = 𝑤 + 𝑣.” If we are thinking about the set 𝑈, then we mean “For all 𝑣 and 𝑤 in 𝑈, 𝑣 + 𝑤 = 𝑤 + 𝑣.”

If we assume that + is commutative in 𝑉 and 𝑈 ⊆ 𝑉, we can easily show that + is also commutative in 𝑈. To see this, let 𝑣, 𝑤 ∈ 𝑈. Since 𝑈 ⊆ 𝑉, we have 𝑣, 𝑤 ∈ 𝑉. Since + is commutative in 𝑉, we have 𝑣 + 𝑤 = 𝑤 + 𝑣. Since 𝑣 and 𝑤 were arbitrary elements in 𝑈, we see that + is commutative in 𝑈.

98

(2) Associativity, commutativity, and distributivity are all defined by universal statements, and therefore, when checking if 𝑈 is a subspace of 𝑉, we do not need to check any of these properties— they will always be satisfied in the subset 𝑈.
(3) The identity property for addition is not defined by a universal statement. It begins with the existential quantifier ∃ “There is.” Therefore, we do need to check that the identity 0 is in a subset 𝑈 of 𝑉 when determining if 𝑈 is a subspace of 𝑉. However, once we have checked that 0 is there, we do not need to check that it satisfies the property of being an identity. As long as 0 ∈ 𝑈 (the same 0 from 𝑉), then it will behave as an identity because the defining property of 0 contains only the quantifier ∀.
(4) The inverse property for addition will always be true in a subset 𝑈 of a vector space 𝑉 that is closed under scalar multiplication. To see this, we use the fact that – 1𝑣 = – 𝑣 for all 𝑣 in a vector space (see Problem 4 (iv) below).
(5) Since the multiplicative identity 1 comes from the field 𝔽 and not the vector space 𝑉, and we are using the same field for the subset 𝑈, we do not need to check the scalar multiplication identity when verifying that 𝑈 is a subspace of 𝑉.
(6) The main issue when checking if a subset 𝑈 of 𝑉 is a subspace of 𝑉 is closure. For example, we need to make sure that whenever we add 2 vectors in 𝑈, we get a vector that is also in 𝑈. If we were to take an arbitrary subset of 𝑉, then there is no reason this should happen. For example, let’s consider the vector space ℂ over the field ℝ. Let 𝐴 = {2 + 𝑏𝑖 | 𝑏 ∈ ℝ}. 𝐴 is a subset of ℂ, but 𝐴 is not a subspace of ℂ. To see this, we just need a single counterexample. 2 + 𝑖 ∈ 𝐴, but (2 + 𝑖) + (2 + 𝑖) = 4 + 2𝑖 ∉ 𝐴 (because the real part is 4 and not 2).
(7) Notes 1 through 6 above tell us that to determine if a subset 𝑈 of a vector space 𝑉 is a subspace of 𝑉, we need only check that 0 ∈ 𝑈, and 𝑈 is closed under addition and scalar multiplication.
(8) The statements for closure, as we have written them do look a lot like universal statements. For example, the statement for closure under addition is “For all 𝑣, 𝑤 ∈ 𝑉, 𝑣 + 𝑤 ∈ 𝑉.” The issue here is that the set 𝑉 is not allowed to be explicitly mentioned in the formula. It needs to be understood.
For example, we saw in Note 1 that the statement for commutativity can be written as “∀𝑣, 𝑤(𝑣 + 𝑤 = 𝑤 + 𝑣).” The quantifier ∀ (for all) can be applied to any set for which there is a notion of addition defined. We also saw that if the statement is true in 𝑉, and 𝑈 is a subset of 𝑉, then the statement will be true in 𝑈.
With the statement of closure, to eliminate the set 𝑉 from the formula, we would need to say something like, “For all 𝑥 and 𝑦, 𝑥 + 𝑦 exists.” However, there is no way to say “exists” using just logical notation without talking about the set we wish to exist inside of.
We summarize these notes in the following theorem.
Theorem 8.1: Let 𝑉 be a vector space over a field 𝔽 and let 𝑈 ⊆ 𝑉. Then 𝑈 ≤ 𝑉 if and only if (i) 0 ∈ 𝑈, (ii) for all 𝑣, 𝑤 ∈ 𝑈, 𝑣 + 𝑤 ∈ 𝑈, and (iii) for all 𝑣 ∈ 𝑈 and 𝑘 ∈ 𝔽, 𝑘𝑣 ∈ 𝑈.
Proof: Let 𝑉 be a vector space over a field 𝔽, and 𝑈 ⊆ 𝑉.
99

If 𝑈 is a subspace of 𝑉, then by definition of 𝑈 being a vector space, (i), (ii), and (iii) hold.

Now suppose that (i), (ii), and (iii) hold.

By (ii), + is a binary operation on 𝑈.

Associativity and commutativity of + are defined by universal statements, and therefore, since they hold in 𝑉 and 𝑈 ⊆ 𝑉, they hold in 𝑈.

We are given that 0 ∈ 𝑈. If 𝑣 ∈ 𝑈, then since 𝑈 ⊆ 𝑉, 𝑣 ∈ 𝑉. Since 0 is the additive identity for 𝑉, 0 + 𝑣 = 𝑣 + 0 = 𝑣. Since 𝑣 ∈ 𝑈 was arbitrary, the additive identity property holds in 𝑈.
Let 𝑣 ∈ 𝑈. Since 𝑈 ⊆ 𝑉, 𝑣 ∈ 𝑉. Therefore, there is – 𝑣 ∈ 𝑉 such that 𝑣 + (– 𝑣) = (– 𝑣) + 𝑣 = 0. By (iii), – 1𝑣 ∈ 𝑈 and by Problem 4 (part (iv)), – 1𝑣 = – 𝑣. Since 𝑣 ∈ 𝑈 was arbitrary, the additive inverse property holds in 𝑈.
So, (𝑈, +) is a commutative group.

By (iii), 𝑈 is closed under scalar multiplication.

Associativity of scalar multiplication and both types of distributivity are defined by universal statements, and therefore, since they hold in 𝑉 and 𝑈 ⊆ 𝑉, they hold in 𝑈.

Finally, if 𝑣 ∈ 𝑈, then since 𝑈 ⊆ 𝑉, 𝑣 ∈ 𝑉. So, 1𝑣 = 𝑣, and the scalar multiplication identity property holds in 𝑈.

Therefore, 𝑈 ≤ 𝑉.

□

Example 8.4:

1. Let 𝑉 = ℝ2 = {(𝑎, 𝑏) | 𝑎, 𝑏 ∈ ℝ} be the vector space over ℝ with the usual definitions of addition and scalar multiplication, and let 𝑈 = {(𝑎, 0) | 𝑎 ∈ ℝ}. If (𝑎, 0) ∈ 𝑈, then 𝑎, 0 ∈ ℝ, and so (𝑎, 0) ∈ 𝑉. Thus, 𝑈 ⊆ 𝑉. The 0 vector of 𝑉 is (0, 0) which is in 𝑈. If (𝑎, 0), (𝑏, 0) ∈ 𝑈 and 𝑘 ∈ ℝ, then (𝑎, 0) + (𝑏, 0) = (𝑎 + 𝑏, 0) ∈ 𝑈 and 𝑘(𝑎, 0) = (𝑘𝑎, 0) ∈ 𝑈. It follows from Theorem 8.1 that 𝑈 ≤ 𝑉.

This subspace 𝑈 of ℝ2 looks and behaves just like ℝ, the set of real numbers. More specifically, we say that 𝑈 is isomorphic to ℝ. Most mathematicians identify this subspace 𝑈 of ℝ2 with ℝ,
and just call it ℝ. See Lesson 11 for a precise definition of “isomorphic.”

In general, it is common practice for mathematicians to call various isomorphic copies of certain

structures by the same name. As a generalization of this example, if 𝑚 < 𝑛, then we can say

ℝ𝑚 ≤ ℝ𝑛 by identifying (𝑎1, 𝑎2, … , 𝑎𝑚) ∈ ℝ𝑚 with the vector (𝑎1, 𝑎2, … , 𝑎𝑚, 0, 0, … ,0) ∈ ℝ𝑛

that

has

a

tail

end

of

𝑛

−

𝑚

zeros.

For

example,

we

may

say

that

(2,

√2,

7,

–

1 2

,

0,

0,

0)

is

in

ℝ4,

even though it is technically in ℝ7. With this type of identification, we have ℝ4 ≤ ℝ7.

2. Let 𝑉 = ℚ3 = {(𝑎, 𝑏, 𝑐) | 𝑎, 𝑏, 𝑐 ∈ ℚ} be the vector space over ℚ with the usual definitions of addition and scalar multiplication and let 𝑈 = {(𝑎, 𝑏, 𝑐) ∈ ℚ3 | 𝑐 = 𝑎 + 2𝑏}. Let’s check that
𝑈 ≤ 𝑉.

100