Chapter 1. Language, Logic, Sets, and Methods of Proof Section 1. Introduction We are about to embark on a wonderful mathematical journey. The two overarching goals of this course are to help you hone your mathematical critical thinking skills and to help you hone your mathematical communication skills. As we pointed out in Chapter 0 (Preliminaries), these goals are inextricably linked. Indeed, good writing helps us enhance our critical thinking skills by enabling us to uncover our understanding of the mathematics and, at times, to discover our misunderstanding of the mathematics. Mathematical critical thinking skills are not just about solving problems and writing proofs, but are also about the questioning of “facts” and ideas. For example, at this point in your mathematical career, if someone were to ask you if the square root of 2 is irrational, most likely you would not give it a second thought and would respond yes. But how do you know? Is it because you’ve heard that “fact” a million times and feel very comfortable with the notion that the square root of 2 is irrational? Or have you seen an argument that has convinced you that the square root of 2 is indeed irrational? The habit of mind of asking why “facts” are true will be important throughout your mathematical career and, indeed, throughout your life. Moreover if you have the honor of teaching a class, that habit of mind is what we hope you will develop in your own students. So what may you assume is true throughout the course? In Chapter 0 we provided you with a number of fundamental assumptions about the natural numbers, integers, rational numbers and real numbers that we will take as the basis for our journey.1 So, for example, we will assume that the sum of two integers is always an integer, but will not assume that the sum of two odd integers is an even integer. Indeed, that is something we will ask you to prove, and our starting point in that proof (and in many others) will be the formulation of good working definitions of our mathematical objects. We will, of course, need to know appropriate strategies for proving our statements and that is the topic of the next three sections of this chapter. Those strategies will be used repeatedly throughout this course and beyond. Then in the fifth section we will describe basic constructions involving sets and will apply our methods to prove some fundamental properties involving those constructions.

1 There are many books that develop the real numbers starting with the fundamental axioms that define the set of positive integers, and one of the best is The Structure of the Real Number System by Leon Cohen and Gertrude Ehrlich. The book was first published by Van Nostrand in 1963 as part of the University Series in Undergraduate Mathematics and updated versions are now available. We would encourage you to delve more deeply into the foundation of the real number system at the end of this course.

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Let’s now begin our mathematical journey by making some conjectures about the integers, rational numbers, and irrational numbers. We start with a few definitions that we will use over and over again. First, let’s recall from Chapter 0 that a real number a is a rational number if there exist integers c and d with

a = cd and d ≠ 0 . We proved in Chapter 0 that the sum of rational

numbers is rational and we asked you to prove that the product of rational numbers is rational. Now, suppose that m is an integer.

m is said to be even if there exists an integer n such that m = 2n. On the other hand, m is said to be odd if there exists an integer n such that m = 2n + 1.

Given the above definitions, two questions that quickly come to mind are the following:

1. Can an integer be both even and odd? 2. Is every integer either even or odd?

Let’s consider the first question. It’s easy to see that an integer m of the form 2n + 1 for some integer n cannot be of the form 2s for some integer s. Indeed, if 2s = m = 2n + 1 where s and n are integers, then 2(s− n) = 1 and so s− n = 1

2 . But that’s impossible since s− n is an integer. So, if m is an odd integer, then m cannot be even.

Remarks. Notice that in the above argument, we used two different letters n and s associated with our integer given m. Throughout our course we will need to “worry about” when we are allowed to use the same letter and when we must use different letters.

Now are we sure that for every integer m, there exists an integer n such m = 2n or there exists an integer n such that m = 2n + 1? It turns out that the answer is yes. In order to show that, we will need the following theorem from Chapter 2 (The Principle of Mathematical Induction):

Division Algorithm for Integers. If a and b are integers with b ≠ 0 , then there exist integers q and r such that a = bq + r and 0 ≤ r < | b | .

Let’s apply the Division Algorithm with b = 2. By the Division Algorithm, we know that every integer can be written in the form 2q + r where q and r are integers and 0 ≤ r < 2 . In other words, every integer can be written in the form 2q + 0 or 2q + 1 for some integer q. So every integer is indeed even or odd! In this course we would like you to try on new ideas and to make conjectures even if you are not completely sure that you are correct, and we will start practicing that skill in our first exercise below.

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Exercise 1. Write a conjecture to complete each of the following statements.

a) The sum of two even integers is ... b) The sum of two odd integers is … c) The sum of an odd integer with an even integer is … d) The quotient of two rational numbers is … e) The sum of two irrational numbers is … f) The product of two irrational numbers is … g) A rational number to a rational power is … h) An irrational number to an irrational power is … i) The product of an irrational number with a rational number is … j) An irrational number to a rational power is …

In this chapter, and throughout the course, we will learn techniques for proving our conjectures. So let’s get started! Section 2. Language and Logical Connectives Our basic building blocks throughout this course will be mathematical propositions. A proposition is a statement that is either true or false, but not both, and we will often denote propositions by capital letters such as P, Q, etc. So, for example, we’ll show in Section 4 that the statement “ 2 is irrational” is a true proposition. On the other hand, the statement “0 is neither even nor odd” is a false proposition. (Indeed, 0 is even since 0 is twice the integer 0.) Although we will usually limit ourselves to mathematical propositions, we will, on occasion, relax that restriction and consider common non-mathematical examples. When considering properties of the integers, we will find that it is not only important to know whether an integer is even or odd, but, more generally, whether an integer m is divisible by an integer n. Here’s the definition we will use:

If m and n are integers with n≠ 0, we say that n divides m (and m is a multiple of n) if there exists an integer c such that m = cn. In this case we say that n is a divisor of m.

Remarks. 1. It may surprise you to see that we stated our definition in terms of a product and not a quotient. Our definition can be generalized in other contexts, for example, when we are dealing with polynomials. 2. It is very important to understand the details of the definitions you will encounter throughout your mathematical journey. Some authors, for example, do not require n to be a nonzero integer in the above definition. In the case where n is 0, quotients cannot be used in the definition, whereas our “product” definition makes sense, even in that context. That’s another reason why most authors define

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the above property in terms of multiplication, even in the case where n is restricted to be nonzero.

3. Many authors use the notation n|m to denote the fact that n divides m.

In Chapter 2 (The Principle of Mathematical Induction) and Chapter 6 (Further Properties of the Integers) and in other sections of this course, we will often prove results concerning prime integers and we will need a good working definition of a prime. Here’s the one we will use:

An integer p is said to be prime if p > 1 and the only positive divisors of p are 1 and p itself.

Example 1. In this example, let’s consider several propositions concerning primes.

a) The statement “there is a prime greater than 100” is a true proposition since 101 is a prime and 101 > 100.

b) The statement “123,456,789 is a prime” is a false proposition since 3 is a divisor of 123,456,789. Indeed, 123,456,789 = 3(41,152,263). In general it is often hard to determine if a large positive integer is a prime, but we will develop a few strategies for doing so in Chapter 6 (Further Properties of the Integers).

c) The statement “there is a prime larger than 123,456,789” is a true proposition, although it is hard to provide an explicit example. But we will prove in Chapter 6 that there are infinitely many primes. Therefore, since there are only 123,456,789 positive integers between 1 and 123,456,789, we know that at least one prime must be larger than 123,456,789. (Actually, we know that infinitely many primes must be larger than 123,456,789.)

Example 2. Let’s consider the statement:

x3 − x −1= 0 . The above statement is neither true nor false but depends upon the value we assign to x. It is called an open sentence and there are many ways of making it into a proposition. One way of doing so is by choosing a specific x value. For example, the statement “ x3 − x −1= 0 when x = 1” is a false proposition since 1

3 −1−1= −1≠ 0 . There are other ways of modifying the open sentence so that it becomes a proposition. For example, what happens if we assert the existence of a real number x for which x3 − x −1= 0 ? In other words, let’s see what happens if we replace our original open sentence by the statement:

There exists a real number x for which x3 − x −1= 0 .

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The modified open sentence is a proposition, and is, in fact, a true proposition. How could we argue that? Well, it’s not clear what value we would choose for x. But can we assert the existence of x without finding a specific value for x? One way of proceeding is to notice that when we replace x by 10 and then by −10 , we have the following string of inequalities:

(−10)3 − (−10)−1< 0 <103 −10−1. Since the function f (x) = x3 − x −1 is a continuous function for all real numbers x, the Intermediate Value Theorem from first semester calculus tells us that there exists a real number x0 such that

−10 < x0 <10 and (x0 )3 − x0 −1= f (x0 ) = 0 . So we know that there exists a real number x0 for which (x0 )3 − x0 −1= 0 .

Notation. We will often use the notation P(x) or P(x, y) or P(x1,x2 ,...,xn) or … to denote open sentences depending upon the variables we wish to include in the sentence.

Exercise 2. Consider the open sentence x3 − x −1= 0 from Example 2. Can you find another way of making the sentence into a proposition? Exercise 3. Determine if the following statements are propositions. If the given statement is not a proposition, modify the statement so that it becomes a proposition.

a) sinπ4( ) > sin π

2( )

b) 2x2 + 3x + 4 = 0 c) 3+ n+ n2 d) The integers 2 and 3 have no common divisor greater than 1.

Logical Connectives and Symbols: ∧ , ∨ , and ~ Suppose that P and Q are propositions. There are a number of ways of constructing new propositions using P and Q and we will begin by discussing three basic constructions. Definition. Let P and Q be propositions.

1. The conjunction of P with Q, denoted by P∧Q, is the proposition “P and Q.” P∧Q is true in only one case, the case where both P and Q are true. 2. The disjunction of P with Q, denoted by P∨Q, is the proposition “P or Q.”

P∨Q is true whenever P is true or Q is true, or P and Q are both true. So P∨Q is false in only one case, the case where P and Q are both false.

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3. The negation of P, denoted by ~P, is the proposition “not P.” ~P is true precisely when P is false.

Remarks. In the above definitions, the propositional forms (P∧Q, P∨Q, and ~P) do not have truth values unless their basic components (P and Q) are assigned truth values.

In our definition, we noted that the proposition P∨Q is true when P and Q are both true. So the mathematical symbol ∨ is sometimes called the “inclusive” or. But in everyday common language, “or” often takes on an exclusive meaning. For example, suppose we said:

This semester I will take MA 225 at 11:45-12:35 on Mondays, Wednesdays, and Fridays or I will take MA 341 at 11:45-12:35 on Mondays, Wednesdays, and Fridays.

Here we would not be including the possibility that we would take both MA 225 and MA 341 at the same time. So we would not want to translate the statement above just by using our symbol ∨ for the disjunction. Example 3. Let’s define propositions P and Q as follows.

Let P be the proposition: 22 = 4 . Define Q to be the proposition: 2 is rational.

Then P is a true proposition, but Q is a false proposition. (As we mentioned previously, we’ll provide a proof that the square root of 2 is irrational in Section 4.) So P∧Q, the conjunction of P with Q, is false since Q is false, but the disjunction, P∨Q, is true since P is true. Furthermore, the negation, ~P, of P is false since P is true, and the negation, ~Q, of Q is true since Q is false. Example 4. Consider the following proposition: (−10)3 − (−10)−1< 0 <103 −10−1. Let’s use symbols and logical connectives to rewrite the string of inequalities using simpler components. There are two basic propositions: (i) the proposition (−10)3 − (−10)−1< 0 , and (ii) the proposition 0 <103 −10−1 . So let’s let P denote the proposition (−10)3 − (−10)−1< 0 and let Q denote the proposition 0 <103 −10−1 . In terms of P and Q, the proposition

(−10)3 − (−10)−1< 0 <103 −10−1 may be rewritten as P∧Q, the conjunction of P with Q. Exercise 4. In this exercise, we’ll define propositions P and Q as follows.

Let P denote the proposition: sinπ4( ) > sin π

2( ) .

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Let Q be the proposition: f (x) = x2 is continuous at x = 2. Determine if each of the following statements is true or false.

a) P∧Q b) P∨Q c) P∨ (~ Q) d) (~P)∨ (~Q) e) ~ (P∧Q) f) ~ (P∨Q)

Exercise 5. In Exercise 3d we considered the following statement:

The integers 2 and 3 have no common divisor greater than 1. Suppose we define P to be “2 has no common divisor greater than 1,” Q to be “3 has no common divisor greater than 1,” and we then translate our original statement as P∧Q. Why are those that not appropriate choices for P and Q? As a tool, we will sometimes resort to looking at a truth table for a propositional form to remind ourselves when the form is true. For example, if we wish to construct the truth table for (P∧Q)∨ (~Q), we would set up a table that includes the case where P and Q are both true, the case where P and Q are both false, and the two cases where one is true and the other is false.

P Q P∧ Q ~Q (P∧ Q)∨ (~Q) T T T F T T F F T T F T F F F F F F T T

Remarks. In our example above, there are 4 combinations of T and F in our truth table having the two variables P and Q. If we wanted to set up a truth table for (P ∧ ~ Q )∨ S, there would be 8 combinations of T and F that we would need to examine. In general, there are 2n combinations for propositional forms with precisely n variables.

When constructing our proofs, it will be helpful to know when two propositional forms have the same truth tables. Such propositional forms are called equivalent forms and we can “substitute” one for the other in the construction of our proofs. For example, we’ll see in a moment that ~(P∧Q) is equivalent to (~P)∨ (~Q). So in the course of a proof, if we know that a certain statement is true and we know that our given statement can be symbolized as ~(P∧Q) (where P and Q are specific propositions), then we can also assert that the disjunction, (~P)∨ (~Q), is true.

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It will also be helpful to know when a propositional form is a tautology, that is, when the form is always true no matter what the truth values are of its basic components. We may always include a relevant tautology in the proof of a proposition. Example 5. The statement P∨ (~P) is always true no matter what proposition we substitute for P. Indeed, if P is true, then the disjunction of P with any other proposition is true. So, in particular, P∨ (~P) is true. On the other hand, if P is false, then we know that ~P is true and so the disjunction of ~P with P must be true. That argument shows us that P∨ (~P) is a tautology. For example, if x0 denotes a specific real number, we may assert that x0 > 0 or ~ (x0 > 0) . In other words, we may assert that x0 > 0 or x0 ≤ 0 . Exercise 6. Construct the truth table for the tautology P∨ (~P). Exercise 7. Let’s define ∨ to be the “exclusive” or. In other words, let’s define P∨Q to be true when either P is true or Q is true, but not both. Use our logical symbols and connectives (∧ , ∨ , and/or ~) to find a propositional form that is equivalent to P∨Q . In parts b-d of the theorem below, we provide some useful equivalent propositional forms. Theorem 1. Let P, Q, and R be propositions.

a) P∨ (~P) is a tautology. b) DeMorgan’s Laws:

(i) ~(P∨Q) is equivalent to (~P)∧ (~Q). (ii) ~(P∧Q) is equivalent to (~P)∨ (~Q).

c) ~(~P) is equivalent to P. d) Associative Properties:

(i) (P∧Q)∧R is equivalent to P∧ (Q∧R). (ii) (P∨Q)∨R is equivalent to P∨ (Q∨R).

In Example 5, we gave a justification for part a of Theorem 1. One way of proving parts b-e is to consider truth tables for each of the relevant propositional forms. Here’s the truth table we would use to argue that part b(i) of Theorem 1 is true:

P Q P∨ Q ~(P∨ Q) ~P ~Q (~P)∧ (~Q) T T T F F F F T F T F F T F F T T F T F F F F F T T T T

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Since Columns 4 and 7 are the same, the propositional forms ~(P∨Q) and (~P)∧ (~Q) are equivalent. Let’s come up with another argument to see that part b(i) is true. Proof of b(i). Let P and Q be propositions. We know that ~(P∨Q) is true precisely when the disjunction, P∨Q, of P with Q is false. But we also know that the disjunction P∨Q is false in only one case, the case where P is false and Q is false. Now P is false precisely when ~P is true and similarly Q is false precisely when ~Q is true. So ~(P∨Q) is true precisely when ~P and ~Q are both true, that is, precisely when (~P)∧ (~Q) is true. We therefore have that ~(P∨Q) is equivalent to (~P)∧ (~Q).

u Exercise 8. Use a truth table to show that part b(ii) of Theorem 1 is true. Now give a different argument to show that ~(P∧Q) is equivalent to (~P)∨ (~Q). Exercise 9. Suppose that P, Q, and R are propositions. Use a truth table to show that P∧ (Q∨R) is equivalent to (P∧Q)∨ (P∧R). Before defining our last two important logical connectives, we give one additional definition. Let P be a proposition. A denial of P is any proposition equivalent to ~P. So, for example, by Theorem 1(b), (~P)∧ (~Q) is a denial of P∨Q since (~P)∧ (~Q) is equivalent to ~(P∨Q). Example 6. Let x0 be a fixed integer. We’ll give a useful denial of the following:

x0 is an odd positive integer. To do so, let’s first notice that whenever x0 is a fixed integer, our statement above is actually the conjunction of P with Q where

P is the statement “ x0 is odd” and

Q is the statement “ x0 is positive.”

We know from Theorem 1b(ii) that ~(P∧Q) is equivalent to (~P)∨ (~Q). So one denial of the statement “ x0 is an odd positive integer” is the statement:

The integer x0 is not odd or x0 is not positive. Here’s a more “natural” way of stating our denial:

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The integer x0 is even or x0 ≤ 0 . Exercise 10. Suppose that x0 is a fixed real number. Give a useful denial of the following statement:

x0 < 3 or (x0 )2 = 16 .

Logical Connectives and Symbols: ⇒ and ⇔ We are now in position to define two of the most important logical connectives in our course: ⇒ and ⇔ . Definition. Let P and Q be propositions.

1. The conditional sentence P⇒Q is the proposition “If P, then Q.” (P⇒Q is also read as “P implies Q.”) P is called the antecedent or hypothesis and Q is called the consequent or conclusion. P⇒Q is false in only one case, the case where P is true but Q is false. (So the conditional P⇒Q is always true whenever the antecedent is false and the conditional is also true in the case where both P and Q are true.)

2. The converse of P⇒Q is the conditional statement Q⇒P. 3. The contrapositive of P⇒Q is the conditional statement (~Q)⇒ (~P). 4. The biconditional P⇔Q is the proposition “P if and only if Q” and is often

abbreviated as “P iff Q.” P⇔Q is true in exactly two cases, the case where P and Q are both true, and the case where P and Q are both false. In other words, P⇔Q is true precisely when P and Q are equivalent statements, that is, when P and Q have the same truth value.

Example 7. Let’s consider the conditional:

If 2 is irrational, then 3 is irrational. (*) “ 2 is irrational” is the antecedent of the above conditional and “3 is irrational” is the consequent. Notice that the antecedent is true, but the consequent is false (since we can write 3 as

31 ) and so our conditional statement is a false proposition. (We will discuss

rational numbers and irrational numbers in more depth later on in this chapter.) The converse of (*) is:

If 3 is irrational, then 2 is irrational.

The converse is actually a true proposition since the antecedent (“3 is irrational”) is false.

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The contrapositive of (*) is the conditional:

If 3 is not irrational, then 2 is not irrational. In other words, the contrapositive is the statement: If 3 is rational, then 2 is rational. Notice that the contrapositive is false since “3 is rational” is a true statement but “ 2 is rational” is a false statement. Exercise 11. Consider the proposition:

If 2 < 0 , then 2 + 3 = 5.

a) Is the conditional true or false? (Why?) b) What is the converse of the above conditional? Is the converse true or false? (Why?) c) What is the contrapositive of the original conditional? Is the contrapositive true or false? (Why?)

Example 8. Suppose we know that the conditional P⇒Q is true and we know that P is true. Then we may conclude that Q is true as well and that will often come into play in our proofs. (That is a rule of logic called modus ponens.) But suppose we know that P⇒Q is true and we know that P is false. In that case we can’t draw any conclusions about whether Q is true or Q is false. Indeed, when the antecedent, P, of the conditional P⇒Q is false, the conditional is true no matter what the truth value is for the consequent Q. Exercise 12. Suppose we know that the conditional P⇒Q is false. What can you conclude about the truth value for P and what can you conclude about the truth value for Q? Example 9. Suppose we are told that f is a fixed function and x0 is a fixed real number. In calculus we learned the following theorem:

If f is differentiable at x0 , then f is continuous at x0 . Here we are asserting that if we have a function f that is differentiable at x0 , then we may conclude that f is continuous at x0 . But what happens if f is not differentiable at x0 ? Is our theorem in conflict with our definition of when a conditional statement is true? For example, although we weren’t initially told what f and x0 are, suppose we find out that x0 = 0 and f is the function defined by

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f (x) = | x | if x ≠ 0

1 if x = 0.

⎧⎨⎪

⎩⎪

Here’s the graph of f:

. Our function f is neither differentiable nor continuous at x0 . Does that mean that our theorem from calculus is incorrect for this particular choice of f and x0 ? The answer is no. The only way our theorem would be incorrect for a given function f and a given real number x0 would be if f were differentiable at x0 , but not continuous at x0 .

Remark. In the next section we will see how we can use connectives and quantifiers to symbolize general statements such as “If f is differentiable at x0 , then f is continuous at x0 .” without first assuming that f is a fixed function and x0 is a fixed real number.

Exercise 13. As we mentioned previously, we will at times use non-mathematical examples and here is our first one! Suppose I tell you:

If I get paid on Friday, I will treat you to dinner. Under what circumstances have I lied to you? From our examples and exercises, there are a number of properties that we hope you have discovered (for example, that the contrapositive of a conditional is equivalent to the original conditional) and we capture some of those properties, as well as other properties, in the theorem below.

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Theorem 2. Let P and Q be propositions.

a) P⇒Q is equivalent to (~Q)⇒ (~P). b) P⇒Q is equivalent to (~P)∨Q. c) P⇔Q is equivalent to (P ⇒Q)∧ (Q ⇒ P) . d) ~(P⇒Q) is equivalent to P∧ (~Q).

Proof of a. We will prove part a by constructing a combined truth table for P⇒Q and (~Q)⇒ (~P). Here is the table:

P Q ~P ~Q P⇒Q (~Q)⇒ (~P) T T F F T T T F F T F F F T T F T T F F T T T T

We see that the last two columns are the same. So P⇒Q and (~Q)⇒ (~P) are equivalent.

u Below we’ll prove part d using some of our properties. Note that we will ask you to prove part b in Exercise 14 (by constructing a truth table) and we will use that result in the proof of part d. Proof of d. By part b, we know that P⇒Q is equivalent to (~P)∨Q and so ~(P⇒Q) is equivalent to ~((~P)∨Q). By DeMorgan’s Law (Theorem 1b(i)), ~((~P)∨Q) is equivalent to (~(~P))∧ (~Q). But by Theorem 1d, ~(~P) is equivalent to P. Consequently ~(P⇒Q) is equivalent to P∧ (~Q).

u Exercise 14. Show that part b of Theorem 2 is true by constructing a combined truth table for the two propositional forms.

Remarks. Every definition is a biconditional. For example, let’s consider our “old” definition:

If m and n are integers with n nonzero, we say that n divides m if there exists an integer c such that m = cn.

By Theorem 2, our definition tells us two things: (i) if there exists an integer c such that m = cn, then n divides m, and (ii) if n divides m, then there exists an integer c such that m = cn.

Note that many authors drop the use of ( ) when using logical symbols and connectives, but we will usually include them for clarification purposes. Still, it is important to be

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able to understand the mathematical statements of other writers. The convention for the application of symbols is:

~, ∧ , ∨ , ⇒ , then ⇔ .

In other words, we first apply the symbol ~ to the proposition directly following it as we move from left to right. We then apply the symbol ∧ to the two propositions directly to the right and left of the symbol ∧ , etc. Example 10. Suppose that P, Q, and S are propositions. Let’s consider the proposition P⇒~Q∨ S. We first apply ~ to the proposition directly following it, that is to Q. So we now have P⇒ (~Q)∨ S. We then apply ∨ to the two propositions directly to the right and left of the symbol. That gives us the conditional P ⇒ (~ Q)∨ S( ) where the antecedent of our conditional is P and the consequent is (~Q)∨ S. Translations of English Phrases Using our Logical Symbols and Connectives In our definition of P⇒Q, we noted two common translations: “If P, then Q” and “P implies Q.” We also noted that P⇔Q is often translated as “P if and only if Q.” There are a number of other phrases that are symbolized as P⇒Q or P⇔Q.

P⇒Q P⇔Q

If P, then Q P if and only if Q P implies Q P is equivalent to Q

Q, if P P is necessary and sufficient for Q P only if Q

P is sufficient for Q Q is necessary for P

Q, whenever P Some of those phrases may be surprising to you. For example, why would we translate “P only if Q” as P⇒Q? Let’s look at an example. Suppose I told you:

You will graduate from NC State only if your cumulative GPA is 2.0 or greater. That, by the way, is a true statement! But have I guaranteed that if your GPA is 2.0 or greater, you will graduate from NC State? Or, have I said that if you did graduate from NC State, then your GPA must have been 2.0 or greater? The statement “If your GPA is 2.0 or greater, then you will graduate from NC State.” is false. Indeed, you must satisfy the other requirements in your major as well as your general education requirements in order to graduate. So having a GPA that is 2.0 or

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greater does not ensure your graduation. The conditional that captures the essence of the meaning of our original statement is:

If you graduate from NC State, then your GPA must be 2.0 or greater. We will try to make sure in our own writing that we are as clear as possible. So we will usually avoid statements such as “P only if Q.” But there will be a number of occasions when we will use the terminology “Q, whenever P.” Moreover, as we noted previously, it is important to be able to understand the writings of others. Example 11. Suppose that x and y are fixed real numbers. Let’s translate the following sentence using logical symbols and connectives:

If x and y are integers such that |x| = |y|, then x = ± y. There are five basic properties that we are asserting for x and y and we will define symbols for each of those properties.

1. Let P be the proposition: x is an integer. 2. Let Q be the proposition: y is an integer. 3. Let R be the proposition: |x| = |y|. 4. Let S be the proposition: x = + y. 5. Let T be the proposition: x = − y.

Our statement is symbolized as: (P∧Q)∧ R( )⇒ (S ∨T ) . Exercise 15. Suppose that a, b, and p are fixed integers with p≠ 0. Use logical symbols and connectives to symbolize:

p divides a or p divides b whenever p is a prime dividing ab. Section 3. Quantifiers In this section we will consider open sentences, that is, sentences that contain one or more variables that become propositions when specific values are substituted for those variables. As in Section 2, we will denote open sentences by P(x) or P(x,y) or

P(x1,x2 ,...,xn) or … depending upon the variables we wish to include in the sentence. In Example 2 of Section 2, we noted that there are other ways of modifying open sentences such as x2 − x −1= 0 so that they become propositions. In that example we asserted the existence of a real number x satisfying the condition x2 − x −1= 0 , and found that the modified sentence was a true proposition. Let’s consider one more example of that type before we examine open sentences more generally.

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Example 12. Consider the statement: x2 − x ≥ 0 . Let’s once again assert the existence of a real number x satisfying the condition, in other words, let’s replace the open sentence by:

There exists a real number x such that x2 − x ≥ 0 . It’s easy to see that the new proposition is true. Indeed, we need only assign the value 0 to x to show that there is, indeed, a real number x for which x2 − x ≥ 0 holds. Is x = 0 the only real number satisfying x2 − x ≥ 0 ? In other words, is 0 the unique real number for which x2 − x ≥ 0 ? Here again it is easy to see that the answer is no. For example, 1, 1.5, 2, π, … all satisfy the given inequality. We would have therefore created a false proposition if we had modified the sentence to read:

There exists a unique real number x such that x2 − x ≥ 0 . But do all real numbers satisfy the condition x2 − x ≥ 0 ? Again the answer is no. A quick look at the graph of the parabola f (x) = x2 − x might convince us we should look for x values between 0 and 1 for which the inequality fails. In fact, if we take x to be .5, then x

2 − x = (.5)2 − .5= .25− .5= −.25< 0 . So the proposition

x2 − x ≥ 0 for all real numbers x is also a false proposition. In Example 12 we described three ways of quantifying the open sentence x2 − x ≥ 0 to produce a proposition and that is the main focus of our present section. But notice in Example 12, we needed to know the universe from which to select our x values. Had we chosen a different universe, we may have found that our inequality was true for all x values in that universe. Exercise 16. Consider the open sentence x2 − x ≥ 0 where the universe for x is N, the set of natural numbers (i.e., N is the set of positive integers). Determine if the following propositions are true or false.

a) There exists a natural number x for which x2 − x ≥ 0 . b) There exists a unique natural number x for which x2 − x ≥ 0 . c) x2 − x ≥ 0 for all natural numbers x. (Hint: factor the expression x2 − x and then use the appropriate order properties from Chapter 0 (Preliminaries).)

Now let’s look more generally at the open sentence P(x1,x2 ,...,xn) . The truth set of

P(x1,x2 ,...,xn) is the set (x1,x2 ,...,xn ) : P(x1,x2 ,...,xn ) is true{ } . As in Example 12 and

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Exercise 16, we first need to identify the universe for each of our variables in order to determine the truth set of a given open sentence. Example 13. Consider the open sentence: x2 = 3 . If the universe is the set of real numbers, then the truth set has precisely two elements, namely the real numbers ± 3 . Now let’s take the set Z of integers as the universe for our open sentence. We know that if x is an integer satisfying x2 = 3 , then x is also a real number satisfying x2 = 3 . But

± 3 are the only real numbers satisfying x2 = 3 and neither of those values is an integer. So the truth set of x2 = 3 is empty if we assign the set of integers as the universe. Example 14. Let’s now consider the open sentence: x + y = 2. Let’s take as the universe for each of the two variables the set R of real numbers. Then the truth set of x + y = 2 consists of the points on the line y = 2− x . In other words, the truth set is:

(x, y)∈R × R: y = 2− x{ } .

Remarks. R × R denotes the set of all ordered pairs (x, y) where x and y are real numbers.

Exercise 17. Find the truth sets for each of the following open sentences.

a) x2 + 3x − 4 = 0 (where the universe is the set R of real numbers) b) x2 + 3x − 4 = 0 (where the universe is the set N of natural numbers) c) x is a prime ⇒ x is odd (where the universe is the set N of natural numbers)

(Hint: for which x values is the conditional statement false?) We are now in a great position to define our main quantifiers. As an illustration, we’ll do so for the open sentence P(x). Definition. Let P(x) be an open sentence in the variable x.

1. Universal Quantifier: (∀x)P(x) , read “for all x, P(x),” is true when P(x) is true for all values of x in the universe for the open sentence P(x).

2. Existential Quantifier: (∃x)P(x) , read “there exists an x such that P(x),” is true if the truth set for P(x) is nonempty.

3. Unique Existential Quantifier: (∃!x)P(x) , read “there exists a unique x such that P(x),” is true if the truth set for P(x) has precisely one element.

Remarks. 1. To say that the proposition (∃!x)P(x) is true means: (i) there exists an x0 in the universe for P(x) for which P( x0 ) is true and (ii) if y is any member of the universe for which P(y) is true, then y must equal x0 .

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2. To say that (∃!x)P(x) is false means: (i) P(x) is false for all x in the universe for P(x) or (ii) there are distinct elements y and z in the universe for P(x) such that P(y) and P(z) are both true.

Example 15. Let’s consider the open sentence x2 = 4 and let’s use as our universe the set of real numbers. The truth set has precisely two elements, namely the real numbers

±2 . So with our given universe, (∀x)(x2 = 4) is a false proposition, (∃x)(x2 = 4) is a true proposition, and (∃!x)(x2 = 4) is a false proposition. Now let’s change our universe and restrict our values to be integers. Then we still have that (∀x)(x2 = 4) is false, (∃x)(x2 = 4) is true, and (∃!x)(x2 = 4) is false. Let’s change our universe one more time and use the set N of natural numbers. This time the truth set of the open sentence x2 = 4 consists of exactly one natural number, namely the number 2. So with our new universe, (∀x)(x2 = 4) is false, (∃x)(x2 = 4) is true, and

(∃!x)(x2 = 4) is true as well. Exercise 18. Let’s consider P(x) where P(x) is the open sentence x2 > 0 .

a) Let’s take as our universe the set R of real numbers. Which of the following propositions are true: (∀x)P(x) , (∃x)P(x) , (∃!x)P(x) , (∃x)(~P(x)) ? (Make sure that you justify your answers!) b) Now let’s change our universe and use the set N of natural numbers. Determine whether each of the following propositions is true or false: (∀x)P(x) ,

(∃x)P(x) , (∃!x)P(x) , (∃x)(~P(x)) . (Again, you should make sure that you justify your answers!)

Example 16. Consider the open sentence: x + y = 2. Let’s take the set of real numbers as our universe for both x and y. We’ll first consider the proposition: (∀x)(∃y)(x + y = 2) . This proposition asserts:

For all real numbers x, there exists a real number y such that x + y = 2. Is the proposition true or false? In other words, if we start with a real number x, is there a real number y (possibly depending upon the x we initially chose) such that x + y = 2? The answer is certainly yes. Indeed if x is a real number, let’s choose y to be 2− x . Then we know that y is a real number and

x + y = x + (2− x) = 2 .

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But now let’s reverse the order of the quantifiers. So let’s consider the proposition:

(∃y)(∀x)(x + y = 2) . Our new proposition asserts:

There exists a real number y such that for all real numbers x, x + y = 2. So we are asserting that the same real number y “works” for all real numbers x, no matter which x we choose. That statement is false, but how might we try to convince a classmate that we are correct? Our strategy will be to start with an arbitrary real number y and to then produce a real number x for which x + y ≠ 2 . In order to find such an x, we’ll need to stay away from the real number 2− y since (2− y)+ y = 2 . Suppose y is a real number. Let x be the real number − y . Then

x + y = (− y)+ y = 0 ≠ 2 . So we cannot find a real number y having the property that x + y = 2 all real numbers x. In other words, the proposition (∃y)(∀x)(x + y = 2) is false!

Remarks. When considering the proposition (∃y)(∀x)(x + y = 2) in Example 16, once we were given y , our strategy was to define x in terms of y so that x would “work” no matter what our initial choice was for y . But x = − y was not our only option. For example, we could have chosen x to be 3− y or .25− y or 1− y or …

Exercise 19. Let’s once again consider the proposition (∃y)(∀x)(x + y = 2) from Example 16. Suppose we start with a real number y and we choose x to be y itself. With that choice of x, can we guarantee that x + y ≠ 2 ? Exercise 20. In this exercise we will see what happens when we try to distribute the universal quantifier across a disjunction. In parts a and b, let’s take as our universe the set Z of integers. Determine if each of the following propositions is true or false.

a) (∀x)(x is even∨ x is odd) b) (∀x)(x is even)∨ (∀x)(x is odd)

In several of the previous examples, we started with a universe and then restricted our open sentence to a subset of that universe. There are three related quantifiers that we could have used in each of those examples.

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Definition. Let P(x) be an open sentence and suppose that A is a nonempty subset of the universe for P(x).

1. (∀x ∈A)P(x) is the proposition “for all x in A, P(x).” (∀x ∈A)P(x) is true if P(x) is true for each element of A. In other words, (∀x ∈A)P(x) is true if and only if the proposition (∀x) (x ∈A)⇒ P(x)( ) is true.

2. (∃x ∈A)P(x) is the proposition “there exists an x in A such that P(x).”

(∃x ∈A)P(x) is true if and only if the proposition (∃x) (x ∈A)∧ P(x)( ) is true. 3. (∃!x ∈A)P(x) is the proposition “there exists a unique x in A such that P(x).”

(∃!x ∈A)P(x) is true if and only if the proposition

(∃x) (x ∈A)∧ P(x)( )∧ (∀y)(∀z) ( y ∈A)∧ (z ∈A)∧ P( y)∧ P(z)( )⇒ ( y = z)⎡⎣ ⎤⎦ is true.

Example 17. Let’s symbolize each of the following statements using quantifiers and logical connectives. In this problem, the universe for each variable is the set R of real numbers.

a) For all real numbers r, there exists a natural number n such that n > r.

We’ll present two different ways of symbolizing the above sentence, one where we use the set N of natural numbers as part of our quantifier and one where we do not. If we use N as part of our quantifier, the above proposition can be symbolized as:

(∀r)(∃n∈N)(n > r) . If we do not use N as part of our quantifier, our proposition becomes (in terms of quantifiers and logical connectives):

(∀r)(∃n)(n∈N ∧ n > r) .

b) All positive real numbers have square roots.

Again, let’s provide several different translations. Let R+ denote the set of positive real numbers. If we use R+ as part of our quantifier, the above proposition can be symbolized as:

(∀r ∈R+ )(r has a square root) .

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We know that the statement “r has a square root” means that there is a real number y satisfying r = y2 . So our proposition may also be symbolized as:

(∀r ∈R+ )(∃y)(r = y2 ) . Suppose we don’t use the set of positive real numbers in our quantifier. Then one way of symbolizing our proposition is:

(∀r)(r ∈R+ ⇒ r has a square root) . Exercise 21. Use as your universe the set R of real numbers, and, as in Example 17, let R+ denote the set of positive real numbers.

a) Symbolize

(∀r)(r ∈R+ ⇒ r has a square root) by replacing “r has a square root” with symbols and quantifiers as we did in our second translation in Example 17. b) Now symbolize

Some positive real numbers have a square root.

in two different ways. Example 18. Suppose that f(x) is a real-valued function whose domain is the set of real numbers R. Suppose as well that a and L are real numbers. In calculus we often define

limx→a

f (x) = L to mean: f(x) gets “arbitrarily close” to L as x gets “arbitrarily close” to a

(but does not equal a). But what does it mean for f(x) to get “arbitrarily close” to L or for x to get “arbitrarily close” to a? Mathematicians found that there were logical inconsistencies when using imprecise definitions such as the one above and so developed a formal definition of the notion of a limit. (You will study that definition in more depth in your advanced calculus and real analysis courses.) Here is the precise definition:

limx→a

f (x) = L if given any ε > 0 , there exists a δ > 0 such that

| f (x)− L |< ε whenever 0 < | x − a |< δ . One interpretation of the above definition is the following:

limx→a

f (x) = L means that no

matter how small our positive number ε is, we can force f(x) to be within ε units of L by taking x (where x ≠ a ) to be within δ units of a (where our positive number δ depends upon our original choice for ε ). (Note: mathematicians are fond of using lower case Greek letters to denote small positive real numbers!)

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Let’s see how we can use logical connectives and quantifiers to symbolize the proposition:

Given any ε > 0 , there exists a δ > 0 such that | f (x)− L |< ε whenever 0 < | x − a |< δ .

First of all, let’s take as our universe the set R of real numbers. Since we are imposing the condition that both ε and δ are greater than zero, it would be helpful to use the set R+ of positive real numbers in our formulation. So to say that ε > 0 means that ε ∈R+ . Similarly, we may symbolize δ > 0 as δ ∈R+ . In Section 2 we noticed that “Q, whenever P” is symbolized as P⇒Q. So the statement

| f (x)− L |< ε whenever 0 < | x − a |< δ should be symbolized as:

0 < | x − a |< δ ⇒ | f (x)− L |< ε . Finally, let’s notice that the quantifiers on ε and x should be universal quantifiers, and the quantifier that modifies the variable δ should be an existential quantifier. Now we’re ready to put all of the parts together. In terms of quantifiers and logical connectives, the statement:

Given any ε > 0 , there exists a δ > 0 such that | f (x)− L |< ε whenever 0 < | x − a |< δ .

may be symbolized as follows:

(∀ε ∈R+ )(∃δ ∈R+ )(∀x) 0 < | x − a |< δ ⇒| f (x)− L |< ε( ) . One of the pitfalls we may encounter is that common English usage and mathematical usage are not necessarily the same. We saw that with the usage of “or” and we’ll find that to be so in the next example as well. Example 19. Let’s examine the following Claim and Justification.

Claim. If x is any real number, then x2 ≥ 0 . Justification. Let x = 1. Then x is a real number such that

x2 = 12 ≥ 0 .

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So there is a real number x satisfying the inequality x2 ≥ 0 .

Does the justification convince you that the claim is correct? The argument the author used certainly shows us that we can find at least one real number satisfying the stated inequality. The problem is that although we used the word “any” in the statement of the claim, and often in English the word “any” means “some,” that is not the intent of our mathematical statement. In fact, the meaning of our mathematical statement is:

For all real numbers x, x2 ≥ 0 . So how could we convince a reader that the statement above is correct? One strategy would be to consider the graph of the parabola y = f (x) = x2 to see if the graph dips below the x-axis. But the problem with that strategy is that graphs often do not tell us the full story and, in fact, graphs may be misleading. So we need to think of how we would construct a formal justification. (We will do just that in Example 21 in Section 4.) Now let’s explore how we can give “useful” denials of our quantified statements. For example, suppose we want to find a denial of (∃x)P(x) that does not use ~ in front of the quantifier. We know that ~ (∃x)P(x) is true in precisely one case, the case where (∃x)P(x) is false. Moreover, to say that (∃x)P(x) is false means that P(x) is always false no matter what x we choose in the universe for P(x). In other words, to say that (∃x)P(x) is false means that ~P(x) is always true for all x in the universe for P(x). But that just tells us that (∀x)(~P(x)) is true. A similar argument shows that if (∀x)(~P(x)) is true, then

~ (∃x)P(x) must be true as well. More generally we have the following theorem. Theorem 3. Let P(x) be an open sentence and let A be a nonempty subset of the universe for P(x).

a) ~ (∀x)P(x) is equivalent to (∃x)(~P(x)) . b) ~ (∃x)P(x) is equivalent to (∀x)(~P(x)) . c) ~ (∃!x)P(x) is equivalent to (∀x)(~P(x))( )∨ (∃y)(∃z)( y ≠ z ∧ P( y)∧ P(z))( ) . d) ~ (∀x ∈A)P(x) is equivalent to (∃x ∈A)(~P(x)) . e) ~ (∃x ∈A)P(x) is equivalent to (∀x ∈A)(~P(x)) . f) ~ (∃!x ∈A)P(x) is equivalent to

(∀x ∈A)(~P(x))( )∨ (∃y ∈A)(∃z ∈A)( y ≠ z ∧ P( y)∧ P(z))( ) .

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As another illustration, we’ll consider part c of the above theorem. So why does part c make sense? To say that (∃!x)P(x) is false means that either P(x) is never true or P(x) is true for “too many” values. In other words, to say that (∃!x)P(x) is false means that either P(x) is never true or there exist at least two distinct elements y and z in the universe for which P(y) and P(z) are both true. That was the formulation that we tried to capture in part c of Theorem 3. Example 20. In Exercise 20, you were given the set Z of integers as your universe and you were asked to determine if each of the following propositions was true or false:

(∀x)(x is even∨ x is odd)

(∀x)(x is even)∨ (∀x)(x is odd) . The first proposition asserts that for all integers x, either x is even or x is odd. The second proposition asserts that every integer is even or every integer is odd. We hope that you came to the conclusion that the first proposition is true and the second is false. (One lesson we should take from those observations is that we can’t just distribute the universal quantifier across an “or” statement.) In this example, let’s give a useful denial of each proposition. In our final formulation of our denials we will make sure that we do not use ~ before any of the quantifiers. By Theorem 3, we know that ~ (∀x)(x is even∨ x is odd) is equivalent to

(∃x) ~ (x is even∨ x is odd)( ) . By DeMorgan’s Law (Theorem 1), we know that

~ (x is even∨ x is odd) is equivalent to ~ (x is even)( )∧ ~ (x is odd)( ) . We saw in Section 1 that if an integer x is not even, then x must be odd, and if the integer x is not odd, then it must be even. So a “more useful” way of expressing

~ (x is even)( )∧ ~ (x is odd)( ) is (x is odd)∧ (x is even) . Therefore,

~ (∀x)(x is even∨ x is odd) is equivalent to the statement (∃x)(x is odd ∧ x is even) . In other words, a denial of the statement:

For all integers x, either x is even or x is odd. is the statement:

There exists an integer x such that x is both odd and even. Now let’s consider the second proposition: (∀x)(x is even)∨ (∀x)(x is odd) . By DeMorgan’s Law and Theorem 3 we have that the following propositions are equivalent:

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~ (∀x)(x is even)∨ (∀x)(x is odd)( )~ (∀x)(x is even)( )∧ ~ (∀x)(x is odd)( )

(∃x)(~ x is even)∧ (∃x)(~ x is odd)(∃x)(x is odd)∧ (∃x)(x is even).

One way of translating the last statement is to assert that there exists an odd integer and there exists an even integer. (Note that in the last statement we are not asserting that the same x “works” in each of the two components of the conjunction.) Exercise 22. In each of the following, use quantifiers and logical connectives to symbolize each proposition. Then give a useful denial of each proposition. In your final formulation of the denial, make sure that you do not use ~ in front of any of your quantifiers.

a) If x is an even integer and y is an odd integer, then x + y is odd. (Take as your universe the set Z of integers.) b) If x and y are integers such that x < y, then there exists a real number r such that x < r < y. (Take as your universe the set of real numbers. But you may also use the set Z in your quantifiers.) c) If r is a positive real number, then there exists a positive integer n such that

1n < r . (Take as your universe the set of positive real numbers. You may also use the set N of positive integers in your quantifiers.)

We conclude our discussion of quantifiers with some observations that will be helpful in the next section. In Section 2 we considered the converse and contrapositive of the conditional P⇒Q where P and Q are propositions and we found that the contrapositive is equivalent to the original conditional. If we start instead with open sentences P(x) and Q(x) where we use the same universe for both sentences, we can also define the contrapositive and converse of the open sentence P(x)⇒Q(x) in a similar manner. So, for example, we define the contrapositive of the open sentence P(x)⇒Q(x) to be [~Q(x)]⇒ [~P(x)]. We know that when we take an arbitrary x0 in the universe for the open sentences P(x) and Q(x), the conditional P(x0 )⇒Q(x0 ) and the contrapositive [~Q(x0 )]⇒ [~P(x0 )] have the same truth value. So the truth sets for the two open sentences P(x)⇒Q(x) and

[~Q(x)]⇒ [~P(x)] are the same. That means that when we “quantify” both open sentences with the same quantifier, the resulting propositions are equivalent. For example,

(∀x) P(x)⇒Q(x)( ) and (∀x) [~Q(x)]⇒ [~P(x)]( ) are equivalent.

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The same argument works for the other pairs of equivalent propositions from Theorems 1 and 2 and we’ll examine just one more example.

From Theorem 1 we know that P⇒Q is equivalent to [~P]∨Q and therefore the truth sets of P(x)⇒Q(x) and [~P(x)]∨Q(x) are the same if we use the same universe for P(x) in both open sentences and the same universe for Q(x) in both open sentences. That tells us that if A is a subset of the common universe for P(x) and Q(x), then

(∃x ∈A) P(x)⇒Q(x)( ) and (∃x ∈A) [~P(x)]∨Q(x)( ) are equivalent propositions. Section 4. Methods of Proof In this section we will describe and apply two main strategies for proving mathematical propositions: direct proofs and proofs by contradiction. We will provide a number of examples of correct proofs but we will also give some examples of “proofs” that are incorrect. Critiquing “proofs” that contain errors can be quite helpful in learning how to construct correct proofs! Direct Proofs Our approach in using direct proofs will be to write down the assumptions in a particular problem and to then use those assumptions to create a sequence of logical steps from which the conclusion can be derived. At any point in the proof we are able to assert that a tautology is true (for example, if x is an integer, then x is even or x is odd). We will often use modus ponens to derive steps in the proof. (Recall that modus ponens tells us that if we know that R⇒S is true and we know that R is true, then we are able to assert that S is true as well.) Example 21. We’ll prove that if x is a real number, then x2 ≥ 0 . Discussion. Let’s first notice that the statement above is a “for all” statement. If we use as our universe the set R of real numbers, the above proposition may be symbolized as:

(∀x)(x2 ≥ 0) . So our strategy will be to start off by assuming that x is a real number and then we’ll use logical steps to deduce that x2 ≥ 0 . In our proof we’ll use our order properties from Chapter 0 (Preliminaries) to examine two cases: (i) the case where x ≥ 0 , and (ii) the case where x is not greater than or equal to 0, i.e., the case where x < 0. Proof. Let x be a real number. Then we know that either x ≥ 0 or x < 0.

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Case 1. Suppose that x ≥ 0 . By our order properties from Section 2 of Chapter 0 (Preliminaries), we have that x ⋅ x ≥ x ⋅0 = 0 . In other words, x2 ≥ 0 when x ≥ 0 . Case 2. Suppose that x < 0. Then −x > 0 . Indeed, by our order properties from Chapter 0, −x = (−1) ⋅(x) > (−1) ⋅0 = 0 . Now let’s apply our order properties to the inequality −x > 0 . We have that

x2 = (−x) ⋅(−x) > (−x) ⋅0 = 0 and so x2 ≥ 0 when x < 0.

In both cases we have shown that x2 ≥ 0 . Therefore, x2 ≥ 0 for all real numbers x.

u We will often find that the propositions we seek to prove are conditional statements of the form P⇒Q. We know that a conditional statement is false in only one case, the case where the antecedent is true and the consequent is false. So when using the direct proof approach, we will seek to eliminate that one case where the conditional is false by starting off with the assumption that the antecedent is true. We will then use logical steps to prove that the consequent must be true. We also know that the conditional statement P⇒Q is equivalent to its contrapositive [~Q]⇒ [~P], and that when we “quantify” the open sentences P(x)⇒Q(x) and [~Q(x)]⇒ [~P(x)] (using the same universe for P(x) and Q(x) and the same quantifier for both conditionals), the resulting propositions are equivalent as well. So we will have the option of providing a direct proof that the contrapositive is true. When we choose to do so, we will signal to the reader that we are going to use that approach. Example 22. We’ll prove that if n is an even integer, then n2 is even. Discussion. As in Example 21, let’s notice that the statement above is a “for all” statement. If we use as our universe the set Z of integers, the above proposition may be symbolized as: (∀n)(n is even ⇒ n2 is even) . So our strategy will be to start with an arbitrary integer n. We’ll assume that n is even and from there we will deduce that n2 must be even. In our proof we will use the definition of “even" to rewrite n in the form 2m for some integer m. Using that information, we will then show that n2 is twice an integer. Proof. Let n be an integer. Suppose that n is even. Then there exists an integer m such that n = 2m. So n

2 = (2m)2 = 4m2 = 2(2m2 ) . Since m is an integer, we know by our

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properties in Section 2 of Chapter 0 (Preliminaries) that 2m2 is an integer as well. Therefore, since n2 is twice an integer, n2 is even.

u Example 23. In this example let’s critique the proof that we give of the statement:

If n2 is even, then n is even. Before doing so, notice that the above statement is a universal statement. So what is the appropriate universe? The fact that “even” is used in the antecedent and consequent of the conditional statement tells us that an appropriate universe is once again the set of integers. So now let’s try to give a direct proof of our conditional. Proof. Let n be an integer. Suppose that n2 is even. Then n2 = 2k for some integer k.

So n = ± 2k = 2 ± 2k

2( ) and therefore n is twice the integer ±2k2 . Consequently n is

even. Discussion. The proof starts out fine. But it is not at all clear that either

2k2 or −

2k2

is an integer. We know that the product of integers is an integer and so 2k is an integer. But we certainly don’t know that the square root of an integer is an integer, and we don’t know that the quotient of integers is an integer. Is our conditional statement even correct? Suppose instead that we consider the contrapositive:

If n is odd, then n2 is odd. How would we proceed with a direct proof of the contrapositive? Well, we would start by assuming that n is odd and we would then write n as 2k + 1 for some integer k. Our next step would be to try to write n2 as twice an integer plus one. The analogous approach was a fruitful approach in Example 22 and we’ll mimic it in Example 24 below. Example 24. We will prove that the following statement is true:

If n2 is even, then n is even. Proof. We will prove that the contrapositive of our conditional is true. In other words, we will prove that if n is odd, then n2 is odd. Let n be an integer and assume that n is odd. Then n = 2k + 1 for some integer k. So

n2 = (2k +1)2 = 4k 2 + 4k +1= 2(2k 2 + 2k)+1 . Since k is an integer, so is 2k 2 + 2k .

Therefore, n2 is odd. u

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Example 25. Let’s suppose that a, b, and c are integers such that a is nonzero, a divides b, and a divides c. Let’s prove that a divides b + c. Discussion. We will start with our assumptions that a, b, and c are integers such that a is nonzero, a divides b, and a divides c. These assumptions tell us that there exist integers s and t such that b = as and c = at. In order to prove that a divides b + c, we hope to be able to find an integer u for which b + c = au. Proof. Let a, b, and c be integers such that a is nonzero, a divides b, and a divides c. Then there exist integers s and t such that b = as and c = at. So b + c = as + at = a(s + t). Since s and t are integers, so is s + t. Consequently, a divides b + c.

u Exercise 23. Let’s consider the following proposition:

If a, b, and c are integers such that a≠ 0, a divides b, and a divides c, then a divides bn + cm for all integers n and m.

a) Use quantifiers and logical connectives to symbolize the above proposition. (Take as your universe for each variable the set Z of integers.) b) Prove that the proposition is true.

Proofs by Contradiction Our proofs by contradiction will begin by asserting that our mathematical proposition P is false. From there we will derive a mathematical statement that we know must be false! If we are able to do so, then our initial assumption that P is false must have been incorrect, that is, P must be true! Note that when we use this approach, we will alert the reader that we are doing so. Example 26. Let’s now prove that 2 is irrational. Discussion. It is very hard to prove that a real number r is not rational by trying to prove that r cannot be written as a quotient of two integers with nonzero denominator. If we had started out by defining a real number r to be rational if r has a repeating decimal expansion, it would be even harder to give a direct proof of the fact that 2 is irrational! It is often much easier in these cases to use contradiction and that is the method we will use in our proof. So we will start by assuming that our statement is false, that is, we will start by assuming that 2 is rational. We will then write 2 as the quotient of two integers with nonzero denominator and will use that assumption to derive a statement we know must be false. Proof (by contradiction). Suppose not, that is, suppose that 2 is rational. Then there exist integers a and b with b nonzero such that

2 = a

b . By canceling common factors if

30

necessary, we may write 2 as cd where c and d are integers with d nonzero such that c

and d have no common factors greater than one. As 2 =

cd , d 2 = c and so 2d 2 = c2 . Therefore c2 is even (as d 2 is an integer) and so

c is even (by Example 24). Consequently, there exists an integer m such that c = 2m. We now have that 2d 2 = c2 = 4m2 and so d 2 = 2m2 . As m2 is an integer, d 2 is even and so d is even (by Example 24). We therefore know that d = 2n for some integer n. We have now shown that 2 is a common divisor of both c and d. But c and d have no common factor greater than one. So we have reached a contradiction. Consequently 2 is irrational.

u Suppose we wish to use the method of contradiction on a statement of the form (∀x)P(x) . We know that a denial of (∀x)P(x) is (∃x)(~P(x)) . So when using the strategy of proof by contradiction, we will start with the assumption that there exists an x (in the universe) for which ~P(x) is true, that is, for which P(x) is false. We will then use that assumption to derive a statement we know is false. Exercise 24. In each of the following, determine how you would begin a proof by contradiction.

a) (∃x)P(x) b) (∃!x)P(x)

How does our strategy of using contradiction work when our mathematical statement is a conditional statement of the form P⇒Q? Here we would be making two assumptions- that P is true and that Q is false. We would then use those assumptions to arrive at a contradiction. (Note that in our proofs, we need to determine from the start if our statement is a universal statement or existential statement.) Example 27. In Example 24 we gave a direct proof of the contrapositive of the following conditional:

If n2 is even, then n is even. Now let’s see how we would give a proof by contradiction of the above statement. In order to do so, let’s first notice that the statement we are trying to prove is a universal statement. So when we assume that the statement is false, we will actually be assuming that there exists an integer n0 for which the proposition:

If (n0 )2 is even, then n0 is even.

31

is false. In other words, we will start off by assuming that there exists an integer n0 for

which (n0 )2 is even and n0 is odd. Proof (by contradiction). Let’s assume that the statement:

If n2 is even, then n is even. is false. Then there exists an integer n0 such that (n0 )2 is even and n0 is odd. Since n0 is odd, there exists an integer m such that n0 = 2m+1 . Then

(n0 )2 = (2m+1)2 = 4m2 + 4m+1= 2(2m2 + 2m)+1. Since 2 and m are integers, 2m2 + 2m is an integer and therefore (n0 )2 is odd. By

assumption, (n0 )2 is even as well. But we know that an integer cannot be both even and odd. So we have now reached a contradiction and consequently our original proposition (If n2 is even, then n is even.) must be true.

u Additional Examples In the rest of this section, we will continue practicing our proof techniques. Exercise 25. Let a and b be integers. Consider the following proposition:

If ab is even, then a is even or b is even.

a) How would you begin a direct proof of the conditional? b) How would you begin a direct proof of the contrapositive? c) How would you begin a proof by contradiction of the original conditional? d) How would you begin a proof by contradiction of the contrapositive? e) Prove that the proposition is true.

Example 28. Let’s try to prove the statement:

For each rational number x, there exists a unique rational number y such that xy = 1.

We will try to give a direct proof of our conditional. To do so we will start with a rational number x and then produce a real number y that has the following properties: (i) y is rational, (ii) xy = 1 and (iii) if z is a rational number such that xz = 1, then y = z. (Condition (iii) will tell us that our y is unique.)

32

So how shall we do that? From our assumption about x, we know we can write x in the form

ab where a and b are integers with b nonzero. So we’ll choose y to be

ba . We’ll

show that with this choice of y, y is rational, xy = 1, and if z is a rational number such that xz = 1, then y = z. Proof. Suppose that x is a rational number. Then there exist integers a and b with b ≠ 0 and

x = a

b . Define y to be the real number ba . Then y is quotient of two integers and

hence is a rational number. Moreover, y satisfies the equation xy = a b

ba = 1 .

Now suppose that z is a rational number satisfying xz = 1. We then have that xz = 1 = xy, that is, xz = xy. Therefore z = y by the Cancellation Law from Section 2 of Chapter 0 (Preliminaries). Discussion. There is a problem in the above proof. The conditional statement:

For each rational number x, there exists a unique rational number y such that xy = 1.

is false! Indeed 0 is a rational number and we know that if we choose x to be 0, then any y we take will have the property that xy = 0, not 1. So where did we go wrong in the above proof? There were several places. First when we wrote x as a quotient of two integers, we could not be sure that our numerator a was nonzero and we needed to know that in order to define y as

ba . So the two assertions that y is a real number and that y is a rational number were not justified. The last sentence of our proof was incorrect as well. Knowing that xz = xy does not allow us to conclude that z = y unless we have additional information such as x ≠ 0 . But our proof does suggest that there is a modification of the conditional statement that is true. Example 29. Let’s prove the statement:

For each nonzero rational number x, there exists a unique rational number y such that xy = 1.

Proof. Suppose that x is a nonzero rational number. Then there exist integers a and b such that

x = a

b , b ≠ 0 , and a ≠ 0 . Since a ≠ 0 , we may choose y to be ba . Since b and

a are integers with a ≠ 0 , y is a rational number. Moreover, y satisfies the equation

xy = a b

ba = 1 .

33

Now suppose that z is a rational number satisfying xz = 1. We then have that xz = 1 = xy, that is, xz = xy. Since x is a nonzero rational number, we may use our Cancellation Law from Section 2 of Chapter 0 (Preliminaries) to conclude that z = y.

u Exercise 26. Prove that if n is an integer, then there is a unique integer m such that n + m = 0. Example 30. In this example, let’s consider the following proposition:

x2 + x ≥ 0 for all real numbers x. Is our proposition true or false? One way of helping our intuition is to graph the parabola

y = x2 + x to see if the parabola dips below the x-axis. If we do so, we find that the parabola intersects the x-axis at the points (0, 0) and (−1, 0) and has a minimum at the point −

12 , − 1

4( ) . So that suggests that our inequality is false when x = −12 . Let’s just

make sure that is the case. (Note: there are certainly other values we could try as well.) Proof (that the above proposition is false). Let’s choose x to be −

12 . Then x is a real

number such that

x2 + x = − 1

2( )2+ − 1

2( ) = 14( ) + − 1

2( ) = − 14( ) < 0 .

Therefore our proposition:

x2 + x ≥ 0 for all real numbers x. is indeed false.

Remarks. In the above example, we showed that the proposition was false by producing a specific real number x, namely x = − 1

2 , for which the statement

x2 + x ≥ 0 does not hold. The number we used is called a counterexample and we will find that counterexamples are often quite useful in convincing our readers that a given universal proposition is incorrect.

Exercise 27. Prove or disprove the following proposition:

If n is an integer, then 2n ≤ n2 . Example 31. Let’s consider a (clever) proof of the following proposition:

There exists an irrational number r such that r 2 is rational.

34

Proof. We know that 2 is a real number and so

2( ) 2 is a real number that is either

rational or irrational.

Case 1.

2( ) 2 is rational.

In this case, let’s take r to be 2 . Example 26 tells us that r is irrational and therefore has the desired properties.

Case 2.

2( ) 2 is irrational.

Let r = 2( ) 2

. Then r is irrational and

r 2 = 2

2⎛⎝⎜

⎞⎠⎟

2

= 22⋅ 2( )

= 22= 2 = 2

1 .

Therefore r 2 is rational, and so in Case 2,

2( ) 2 has the desired properties.

In either case, there exists an irrational number r such that r 2 is rational.

u Is our proof correct? We did not produce a specific r, but found that either r = 2 or

r =

2( ) 2 “works.” So we have, indeed, constructed a legitimate argument.

Exercise 28. Is it possible to find rational numbers r and s such that r s is irrational? Why or why not? Example 32. In Exercise 25 we asked you to prove: If a and b are integers such that ab is even, then a is even or b is even. In this example, let’s consider the converse:

If a is even or b is even, then ab is even. (*) We’ll consider three different approaches to proving (*) and we will then decide which one to pursue. Direct Proof. In starting a direct proof, we would assume that a and b are integers such that a is even or b is even. So we would have two cases to consider: (i) the case where a is even, and (ii) the case where b is even. In each case we would seek to show that ab is even.

35

Direct Proof of the Contrapositive. The contrapositive of (*) is:

If ab is odd, then a is odd and b is odd. So we would start with the assumption that ab is odd and from there would seek to show that a is odd and b is odd. Proof by Contradiction. Here we would assume that there exist integers a and b for which a is even or b is even, but ab is odd. From that assumption we would try to reach a contradiction. The “hardest” approach would be to give a direct proof of the contrapositive. Indeed, we would end up writing ab as 2m + 1 for some integer m and would hopefully use that information to show that both a and b are odd. Let’s choose the first strategy instead. Proof. Assume that a and b are integers such that a is even or b is even.

Case 1. Suppose that a is even. Then there exists an integer n such that a = 2n. So ab = 2nb = 2(nb). Since n and b are integers, so is nb. Consequently ab is even.

Case 2. Suppose that b is even. Then there exists an integer m such that b = 2m. So ab = a(2m) = 2(am). Since m and a are integers, so is am. Therefore ab is even.

Since ab is even in each of the two cases above, we know that the proposition:

If a is even or b is even, then ab is even. is true.

u

Remarks. A natural question to ask is why we did not need to consider the case where a and b are both even. The reason is that Case 1 (as well as Case 2) already covers that situation. Indeed if a and b are both even, then a is even. So we know from Case 1 that ab is even (regardless of whether b is even or odd).

Exercise 29. Give a proof by contradiction of the following proposition from Example 32:

If a is even or b is even, then ab is even.

36

Section 5. Sets We warn the reader that this section will contain a number of definitions, many of which we hope are familiar to you. (In fact, we have already used some familiar notation in previous sections in this chapter and in Chapter 0.) In Section 5 we will begin with some basic definitions and we will then focus our attention on the construction of new sets. A set A is a collection of elements having the property that if x is any object, then either x is an element of A or x is not an element of A, but not both. We will use the notation x ∈A to mean that x is an element of the set A. Similarly, we will use the notation x ∉A to indicate that x is not an element of A. So if R denotes the set of real numbers and if N denotes the set of natural numbers, we have that 2 ∈R but 2 ∉N . If A and B are sets, we will say that A and B are equal if they have precisely the same elements. For example, since 2 ∈R but 2 ∉N , we know that R ≠ N . We will often use the notation {..} when describing sets. So {1, 2, 3} is the set consisting of the integers 1, 2, and 3, whereas {{1}, 2, 3} is the set consisting of the integers 2 and 3 and the set {1} containing the integer 1. (Notice that we are making a distinction between the integer 1 and the set {1} containing the integer 1.) It is very common to describe sets in terms of a given property P(x). We will denote the set of all objects x for which P(x) holds by {x: P(x)} . For example,

{x: x ∈R and x2 = 1} is the set of all real numbers x satisfying the equation x2 =1 . So

{x: x ∈R and x2 = 1}={±1}. On the other hand, {x: x ∈N and x2 = 1} is the set of all natural numbers having the property that x2 = 1 , that is, {x: x ∈N and x2 = 1}={1} . When P(x) is a conjunction of two properties as it is in the above examples, authors often put one of the conditions before the colon and the second condition after the colon. So, for example, {x: x ∈R and x2 = 1} is more commonly written as {x ∈R : x2 = 1}. Throughout these notes, we will use interval notation that we hope is familiar. For example, we will use (0,1) to denote the set of all real numbers strictly between 0 and 1, that is, (0,1) is the set {x ∈R: 0 < x <1} , [0,1] will denote the set {x ∈R: 0 ≤ x ≤1} ,

(2,∞) will denote the set {x ∈R: x > 2}, and (−∞,−2] will denote the set

{x ∈R: x ≤ −2} . When we want to include an endpoint in a set, we will use a squared off bracket ] or [. If we want to exclude an endpoint, we will use a rounded bracket. (We always use a rounded bracket with ∞ and with − ∞ .)

Remarks. The notation (a, b) can be used to denote an interval or an ordered pair. We will try to avoid confusion by telling the reader what we mean in a given situation if the notation is unclear.

37

The set having no elements is called the empty set or null set and will be denoted by ∅ . (Note that some authors denote the empty set by { }.) The empty set may be defined more formally as {x: x ≠ x} . If A and B are sets, we will say that B is a subset of A if A contains all of the elements of B. In that case we will write B ⊆ A or A ⊇ B . In terms of our quantifiers and logical symbols from Section 3, B ⊆ A means that the proposition

(∀x)(x ∈B ⇒ x ∈A) is true. For example, N⊆R. However, R is not a subset of N since our “old” friend,

2, is an element of R but not of N. We can reframe the notion of equality of sets in terms of subsets. Indeed, to say that two sets A and B are equal means that A ⊆ B and B ⊆ A . So in order to prove that A and B are equal, it suffices to prove that the following two conditional statements hold:

(i) (∀x)(x ∈A⇒ x ∈B) (ii) (∀x)(x ∈B ⇒ x ∈A) .

Example 33. Let A = {1,2,{1}} . Then 1∈A , 2∈A , and {1}∈A . {1}⊆ A since the only element of {1} is the integer 1 and we know that 1 is also an element in A. We also have that {{1}}⊆ A . Indeed, the only element of {{1}} is {1} and we know that {1} is an element of A. So in each of these two cases the relevant conditional statement ( (∀x)(x ∈B ⇒ x ∈A) ) is true.

Remark. Notice that since a conditional statement is always true when the antecedent is false, in our above discussion, we only examined the cases where the antecedent was true.

Example 33 continued. Let’s examine a few more potential subsets. The set {1,{1}} contains two elements, namely the elements 1 and {1}, both of which are elements of A. So {1,{1}}⊆ A . However, the set {{2}} is not a subset of A since {2} is an element of

{{2}} , but {2} is not an element of A. Finally, let’s notice that ∅⊆ A . Indeed, if x is any object then the proposition x ∈∅ is false and therefore the conditional (∀x)(x ∈∅⇒ x ∈A) is true! Exercise 30. Suppose that A is a set. Prove that ∅⊆ A . In the rest of this section we will construct new sets from “old” sets.

38

Power Sets The set of all subsets of a set A is called the power set of A and is denoted by P(A). So

P(A) = {X: X ⊆ A}. Example 34. Let A = ∅ . The only subset of A is ∅ itself. So P(A) = {∅} . Example 35. Let A = {1}. Then A has two subsets, namely, {1} and ∅ , and therefore

P( A) ={∅,{1}} . Exercise 31. In Example 33, we listed four subsets of the set {1,2,{1}} , namely, {1} ,

{{1}} , {1,{1}} , and ∅ , and each is therefore an element of the power set of {1,2,{1}} . Find all of the other elements of P({1,2,{1}}) . Exercise 32. Let A = {1, 2}. Find P(A). Unions and Intersections The union of A with B, denoted by A∪ B , is the set of all elements that appear in at least one of the two sets A and B. In other words,

A∪ B ={x : x ∈A or x ∈B}.

So if A = 1, 2, 3{ } and if B ={1, 4, 2}, then A∪ B ={1, 2, 3, 4, 2}.

Remark. In the above example, it was not necessary to list the integer 1 twice in the set A∪ B .

Unions have a natural counterpart. The intersection of A with B, denoted by A∩ B , is the set of all elements that A and B have in common. In our example where A ={1, 2, 3}

and B ={1, 4, 2}, A∩ B ={1}. We will say that A and B are disjoint if A∩ B =∅ . So for example, the sets A ={1,2,3} and B ={π ,4} are disjoint. Example 36. Let C = [0, 2) and let D = [1, 3). If you graph C and D on the real line, you should come up with the conjecture that C ∪ D = [0, 3) and C ∩ D = [1, 2) . Let’s prove that C ∪ D = [0, 3) . In order to do so, we’ll prove that C ∪ D ⊆ [0, 3) and that

[0, 3)⊆ C ∪ D . Proof (that C ∪ D = [0, 3) ). We will first prove that C ∪ D ⊆ [0, 3) . Let x ∈C ∪ D . Then x ∈C or x ∈D .

39

Case 1. Suppose that x ∈C . Then 0 ≤ x < 2 and so 0 ≤ x < 3 as well. Consequently, x ∈[0, 3) . Case 2. Suppose that x ∈D . Then 1≤ x < 3 and so 0 ≤ x < 3 . Therefore in this case, x ∈[0, 3) as well.

We have now shown that C ∪ D ⊆ [0, 3) . Next we’ll show that [0, 3)⊆ C ∪ D . Let x ∈[0, 3) . Then 0 ≤ x < 3 . We will consider two cases, the case where 0 ≤ x < 2 and the case where 2 ≤ x < 3.

Case 1. Suppose that 0 ≤ x < 2 . Then x ∈[0, 2) = C and so x ∈C ∪ D . Case 2. Suppose that 2 ≤ x < 3. Then 1≤ x < 3 and so x ∈[1, 3) = D . Therefore, we also have that x ∈C ∪ D .

So in either case, x ∈C ∪ D and consequently [0, 3)⊆ C ∪ D . We have shown that C ∪ D ⊆ [0, 3) and [0, 3)⊆ C ∪ D . Hence, C ∪ D = [0, 3) .

u Exercise 33. Define C and D as in Example 36. Prove that C ∩ D = [1, 2) . We will now generalize the notions of unions and intersections to any collection of sets. (We will include the possibility that our collections have infinitely many sets.) More generally, we may define the union

A

A∈A∪ of any family of sets {A : A∈A} to be

the set of all objects x that appear in at least one set in our family. More formally,

A

A∈A∪ = x : x ∈A for some A∈A{ } .

We note that if the collection A is indexed by the natural numbers, that is, if

A = An: n∈N{ } , then

AA∈A∪ is usually denoted by

An

n∈N∪ or by

An

n=1

∞

∪ . In general if

n0 is a fixed integer and A = An: n∈Z and n ≥ n0{ } ,

AA∈A∪ is often denoted by

An

n=n0

∞

∪ .

As with unions, we may define the intersection of any collection of sets as follows:

40

A

A∈A∩ = x : x ∈A for all A∈A{ } .

We use comparable notation with intersections as we did with our unions. For example,

when A = An: n∈N{ } , we will often denote

AA∈A∩ by

An

n∈N∩ or by

An

n=1

∞

∩ . A similar

convention exists when A = An: n∈Z and n ≥ n0{ } for some fixed integer n0 . Example 37. Let’s consider an example of an intersection of an infinite family of sets. For each positive integer n, let

An = 0,1+ 1

n⎡⎣ ) . So, A1 = 0,1+ 1

1⎡⎣ ) = 0, 2⎡⎣ ) ,

A2 = 0,1+ 1

2⎡⎣ ) = 0,112⎡⎣ ) ,

A100 = 0,1+ 1

100⎡⎣ ) = 0,1 1100⎡⎣ ) , etc. Here’s a graphical snapshot

of the sets we have defined:

Now, let’s look at

Ann=1

∞

∩ . Our sets have the property that A1 ⊇ A2 ⊇ A3 ⊇!A100 ⊇! .

We know from calculus that limn→∞

1n( ) = 0 . So the right hand endpoints of our intervals are

converging to 1 and our left hand endpoints remain constant at 0. We also know that for each natural number n, 1+

1n >1 . Therefore, 1 is an element of each set we’ve defined.

It seems reasonable to conjecture that

Ann=1

∞

∩ = 0,1[ ] and in this example we will give a

proof of that conjecture using our knowledge of calculus. In our proof we will use a result from calculus known as the Squeezing Lemma for Sequences. There are many formulations of the Squeezing Lemma and the one we will use states that if

an n=1

∞,

bn n=1

∞, and

cn n=1

∞ are sequences such that an ≤ bn ≤ cn for all

n, then limn→∞

an ≤ limn→∞

bn ≤ limn→∞

cn as long as each of the limits exists.

41

Proof. As before we will show that two set inclusions hold: 0,1[ ]⊆ An

n=1

∞

∩ and

An

n=1

∞

∩ ⊆ 0,1[ ] . Suppose that x is an element of 0,1[ ] . Then for each positive integer n,

1n > 0 and so

0 ≤ x ≤1<1+ 1n . Consequently, x ∈An for all n in N and we have that

x ∈ An

n=1

∞

∩ . We

have therefore shown that 0,1[ ]⊆ An

n=1

∞

∩ .

Now suppose that x ∈ An

n=1

∞

∩ . Then for all n in N, x ∈An and so 0 ≤ x <1+ 1n for all n.

Let’s apply the Squeezing Lemma to the three sequences

an n=1

∞,

bn n=1

∞, and

cn n=1

∞

where an = 0 for all n, bn = x for all n, and cn = 1+ 1n for all n. Notice that

limn→∞

(0) = 0 ,

limn→∞

(x) = x , and limn→∞

1+ 1n( ) = 1 . So by the Squeezing Lemma we have that 0 ≤ x ≤1, that

is, x ∈[0,1] . Therefore

Ann=1

∞

∩ ⊆ 0,1[ ] and we may now conclude that

Ann=1

∞

∩ = 0,1[ ] .

u

Remarks. Some of you may not have been familiar with the Squeezing Lemma and that’s okay. We will make a point of letting you know what calculus results we plan to use in our examples, particularly those results that are not covered in a first semester calculus course.

Exercise 34. For each positive integer n, define An as in the previous example, that is,

let An = 0,1+ 1n⎡⎣ ) .

a) Find

Ann=1

∞

∪ .

b) Prove your result from part a. Complements Suppose a universe U is specified for a given problem and all of our objects are contained in that universe. Let A be a subset of U. The complement of A, denoted by Ac , is the set of all elements in U not contained in A.

42

It is very important that we specify the universe for a given problem. For example, suppose that A is the set of natural numbers N and U is the set of integers. Then

Ac ={0, −1, − 2, − 3,… ,} . But if A is the set of natural numbers N and U is the set of real

numbers, then Ac = x ∈R : x ∉N{ } . So 2 ∈Ac when the universe is R, but not when

the universe is Z. Exercise 35. If A is the set of natural numbers N and U is the set of real numbers, write Ac as a union of open intervals. It is possible to generalize the above definition as follows. Suppose that A and B are sets. The complement of A in B, denoted by B − A or by B \ A , is the set consisting of all elements in B not contained in A, that is, B − A ={x ∈B : x ∉A} . So if A = (0, 1] and

B = 12 , 2[ ] , then B − A = (1, 2] .

Exercise 36. Let A = (0, 1] and let B = 1

2 , 2[ ] .

a) Find A− B . b) In general, what does part a and our computation above tell you about B − A and A− B ?

In trying to understand the properties of our newly constructed sets, it is often helpful to be able to picture our constructions using a Venn diagram or some other representation. For example, A∪ B may be represented as:

. We note that although it is often said that “a picture is worth a thousand words,” when we give formal proofs, we will use the words. Fortunately, we will never find it necessary to use one thousand words when proving our propositions! (We also note that the “red” we used in the above Venn diagram was the closest we could find in our toolbox to Wolfpack red!) There are many useful set properties and we list a few of those properties in Theorem 4 below. We’ll prove just one part of Theorem 4. But before providing the formal proof, we will first examine the Venn diagram to help convince us that the property makes sense.

43

Theorem 4. Suppose that A and B are sets. Then

(i) A∪ Ac = U

(ii) ( Ac )c = A (iii) B − A = B∩ ( Ac ) (iv) ( A∪ B)c = ( Ac )∩ (Bc ) , and (v) ( A∩ B)c = ( Ac )∪ (Bc ) .

More generally, if {A : A∈A} is a family of sets, then

(vi)

AA∈A∪

⎛⎝⎜

⎞⎠⎟

c

= Ac( )A∈A∩ , and

(vii)

AA∈A∩

⎛⎝⎜

⎞⎠⎟

c

= Ac( )A∈A∪ .

Example 38. Let’s consider property (iv) above. Here are the Venn diagrams for the sets Ac and Bc :

. If we compute the intersection of those sets, we see that we exclude precisely the elements in A∪ B . In other words, the intersection of the sets Ac and Bc may also be represented as:

and that is what property (iv) tells us. (Note that to be fair to our neighbors to the west of us, we used Carolina blue in the above Venn diagram!) Now let’s give a formal proof.

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Proof of Theorem 4(iv). We will first prove that ( Ac )∩ (Bc )⊆ ( A∪ B)c . Suppose that

x ∈( Ac )∩ (Bc ) . Then x ∈Ac and x ∈Bc . Therefore x ∉A and x ∉B . So x ∉( A∪ B) , that is, x ∈( A∪ B)c . Consequently, ( Ac )∩ (Bc )⊆ ( A∪ B)c . We now prove that ( A∪ B)c ⊆ ( Ac )∩ (Bc ) . Suppose that x ∈( A∪ B)c . Then

x ∉( A∪ B) , and hence x ∉A and x ∉B . Therefore x ∈Ac and x ∈Bc . So

x ∈( Ac )∩ (Bc ) . Consequently, ( A∪ B)c ⊆ ( Ac )∩ (Bc ) and we may now conclude that

( Ac )∩ (Bc ) = ( A∪ B)c . u

Exercise 37. Let A and B be sets.

a) Prove that B − A = B∩ ( Ac ) . b) Prove that ( A∩ B)c = ( Ac )∪ (Bc ) .

Cartesian Products We have one last definition to present in this section. Let A and B be sets. The Cartesian product of A with B, denoted by A × B , is the set of all ordered pairs (a, b) where the first coordinate comes from the set A and the second coordinate comes from the set B. In other words,

A× B = (a,b): a ∈A and b∈B{ } . For example, [4,6]× [2,3] is the set (x, y)∈R × R: 4 ≤ x ≤ 6 and 2 ≤ y ≤ 3{ } . We may represent [4,6]× [2,3] in the plane as a solid rectangle with vertices (4, 2), (4, 3), (6, 2) and (6, 3) as follows:

. The set R ×R is represented graphically by the entire plane. Exercise 38. Graph the set (−∞,6]× [2,3] in the plane.

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Remarks. Cartesian products are important in calculus and in many other fields of mathematics. We will use Cartesian products over and over again in Chapter 3 (Functions) and in Chapter 4 (Equivalence Relations).

Concluding Example We conclude this section with a caution that some collections are not sets even if we use {..} notation! Example 39. We’ll prove that A = A: A is a set and A∉A{ } is not a set. Discussion. We will use the strategy of proof by contradiction to prove our result. So we will start off by assuming that A is a set. We will then come up with an object X that is neither in A nor not in A . At each stage in our proof we will need to consider the condition that defines A , that is, we will need to consider the condition:

X is a set and X ∉X . Proof (by contradiction). Let

A = A: A is a set and A∉A{ } and suppose that A is a set. Since A itself is an object as well as a set, we know that either A ∈A or A ∉A , but not both.

Case 1. A ∈A By definition, A = A: A is a set and A∉A{ } and since A ∈A , A must satisfy the two conditions: A is a set and A ∉A . But we are in Case 1 where A ∈A . So in this case we have that A ∈A and A ∉A , which is impossible. Therefore Case 1 cannot hold.

Case 2. A ∉A Since A ∉A , A cannot satisfy the following: A is a set and A ∉A . So at least one of the two conditions must fail. We assumed in the beginning of our proof that A is a set. So the second condition must be false, that is we must have that A ∈A . But since we are in Case 2, we also have that A ∉A , which is again impossible. Therefore Case 2 is does not hold and we may therefore conclude that our original assumption was false. So A is, in fact, not a set!

u

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End of Chapter Problems I. Quantifiers and Logical Connectives 1. In this problem, the universe is the set of all integers, P(x) is "x is a positive," S(x) is "x is a perfect square," and E(x) is "x is even". Use quantifiers and logical connectives to symbolize each of the following:

a) There is an even integer that is also a perfect square. b) Some perfect squares are positive but not all perfect squares are positive. c) No positive integer is even.

2. In this problem, use as your universe the set of all real numbers. Use quantifiers and logical connectives to symbolize:

If a is a positive real number, then there exists a negative real number b such that a + b = 0.

3. Consider the following proposition:

(∃x) [P(x)∧Q(x)]⇒ [~R(x)]( ) . Give a denial of the proposition that does NOT use ~ in your final answer. 4. Determine if each of the following statements is TRUE or FALSE. The universe for each variable is the set of real numbers. (Be sure to justify your answer.)

a) (∀x)(∃y)(∀z)(xy = zy) b) (∀x)(x2 is irrational⇒ x is irrational) c) (∀x)(∀y) [x is irrational ∧ y is irrational]⇒ x + y is irrational( ) d) (∃!y)( y > 3∧ y <10) e) (∃x)(0 < x ⇒ x2 = −1)

5. Find the converse and contrapositive for each of the following conditionals:

a) If x2 − x −12 > 0 , then x < −3 and x > 4 . (Here x is a real number.) b) If f has a maximum at a, then ′f (a) = 0 or f is not differentiable at a. (Here, f is a function whose domain is the set of real numbers and a is a real number.) c) If a > 0 and b > 0, then ab > 0. (Here, a and b are real numbers.)

6. Give a useful denial for each of the statements below.

a) For every real number x, there exists a real number y such that y > x. b) There exists a real number x such that x7 + 2x +1= 0 . c) There exists a unique real number x such that x7 + 2x +1= 0 .

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7. Suppose that y = f(x) is a function. f is said to be strictly increasing if for all real numbers a and b, f (a) < f (b) whenever a < b . f is said to be strictly decreasing if for all real numbers a and b, f (a) > f (b) whenever a < b .

a) Give an example of a strictly increasing function. b) Give an example of a strictly decreasing function. c) Use quantifiers and logical connectives to symbolize: for all real numbers a and b, f (a) < f (b) whenever a < b . d) Use quantifiers and logical connectives to symbolize: for all real numbers a and b, f (a) > f (b) whenever a < b . e) Formulate a statement that explains what is meant by the sentence: f is not strictly increasing. f) Formulate a statement that explains what is meant by the sentence: f is not strictly decreasing. g) If f is not strictly increasing, is f necessarily strictly decreasing? (Why or why not?)

II. Proofs and Counterexamples 1. Prove each of the following:

a) The sum of two even integers is even. b) The sum of two odd integers is even. c) The product of two even integers is even. d) The product of two odd integers is odd. e) The product of an even integer with an odd integer is even. f) If a and b are integers such that ab is even, then a is even or b is even. g) If n2 is an odd integer, then n is odd.

2. Prove each of the following:

a) If s and t are rational numbers, then st2 is rational. b) If t2 is irrational, then t is irrational. c) There exists a natural number n such that n2 = 400. d) If n is a nonzero integer such that n divides a and n divides b, then n divides ac− bd for all integers c and d. e) For all real numbers x greater than 2, there exists a negative real number y such that

x = 2 y

1+y .

f) For all real numbers x greater than 2, there exists a unique negative real number y such that

x = 2 y

1+y .

g) There is no largest natural number. 3. Prove or give a counterexample in each of the following:

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a) If a and b are irrational, then a + b is irrational. b) If a and b are irrational, then ab is irrational. c) If a and b are rational, then ab is rational. d) If x is rational, then there exists a rational number y such that x + y is an integer. e) If x is rational, then there exists a unique rational number y such that x + y is an integer. f) If a and b are rational and b ≠ 0 , then

ab is rational.

g) If f is continuous at x0 and f is continuous at x1 , then f is continuous at

x0 + x1 .

h) Let f(x) = x2 + x +1 . If a < b, then f(a) < f(b).

III. Construction of Sets 1. Let A = (−∞,1)∪ (2, 3) and let B = (−2,∞) . Find each of the following:

a) A∪ B b) ( A∪ B)c (Use as your universe the set R.) c) A∩ B d) ( A∩ B)c (Use as your universe the set R.) e) Ac (Use as your universe the set R.) f) Bc (Use as your universe the set R.) g) ( Ac )∪ (Bc ) (Use as your universe the set R.) h) ( Ac )∩ (Bc ) (Use as your universe the set R.) i) B − A j) A− B

2. For each positive integer n, let An = − n

2 , 1+ 1n⎡⎣ ) . Find

An

n=1

∞

∩ . (Express your answer

using interval notation.) 3. For each positive integer n, let An =

1n , 3− 1

n⎡⎣ ) . Find each of the following and express your answers using interval notation:

a)

Ann=1

∞

∩

b)

Ann=1

∞

∪

c)

( Anc )

n=1

∞

∪ (In part c, the universe is the set of real numbers.)

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4. For each positive integer n, let An = {7,n}. Find each of the following:

a)

Ann=1

∞

∩

b)

Ann=1

∞

∪

c)

( Anc )

n=1

∞

∪ (In part c, the universe is the set of natural numbers.)

5. Determine if each of the following statements is true or false:

a) 1,2,{3}{ }∈ 1,2,{3}{ } .

b) 1,2,{3}{ }⊆ 1,2,{3}{ } . c) A⊆ P( A) . (Here, A is an arbitrary set.) d) A∈P( A) . (Here, A is an arbitrary set.) e) {A}⊆ P( A) . (Here, A is an arbitrary set.) f) ∅⊆ P( A) . (Here, A is an arbitrary set.) g) ∅∈P( A) . (Here, A is an arbitrary set.) h) P( A∩ B) = P( A)∩ P(B) . (Here, A and B are arbitrary sets.) i) P( A∪ B) = P( A)∪ P(B) . (Here, A and B are arbitrary sets.) j) If A, B, and C are sets such that A∩ B ≠ ∅ and A∩C ≠ ∅ , then B∩C ≠ ∅ .

6. Give a counterexample for each of the following conditional statements:

a) If P( A) ≠ ∅ , then A ≠ ∅ . b) If A⊆ B∪C and B∩C =∅ , then A⊆ B or A⊆ C . c) If A∩ B∩C =∅ , then A∩ B =∅ or A∩C =∅ or B∩C =∅ .

7. Give a useful denial for each of the statements below.

a) For all x in A, (x,x)∈R . (Here, A and R are sets.) b) For all x and y in A, if (x, y)∈R , then ( y,x)∈R . (Here, A and R are sets.) c) For all x, y, and z in A, if (x, y)∈R and ( y, z)∈R , then (x, z)∈R . (Here, A and R are sets.)

IV. Incorrect Arguments In each of the following, the given "proof" is incorrect. Find all the errors in the proof. Determine if the "claim" is correct or incorrect and justify your answer. 1. Let A, B, and C be sets.

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Claim: If A ⊆ B and B ⊆ C , then A ⊆ C .

Proof. Let x ∈A . Since A ⊆ B , x ∈B . Since B ⊆ C and x ∈B , x ∈C . Therefore, x ∈A and x ∈C . So A ⊆ C .

2. Let A and B be sets.

Claim: If A ⊆ B , then P( A)⊆ P(B) .

Proof. Let x ∈A . Since A ⊆ B , x ∈B . Therefore, if {x}∈P( A) , then

{x}∈P(B) . So P( A)⊆ P(B) . 3. Let A and B be sets.

Claim. ( A∩ B)c ⊆ ( Ac )∩ (Bc ) .

Proof. Let x ∈( A∩ B)c . Then x ∉A∩ B . So x ∉A and x ∉B . Therefore

x ∈Ac and x ∈Bc . So x ∈( Ac )∩ (Bc ) . We have proven that if x ∈( A∩ B)c , then x ∈( Ac )∩ (Bc ) . Consequently, ( A∩ B)c ⊆ ( Ac )∩ (Bc ) .

V. Proofs without Claims In each problem below, we provide a “proof” (where the last line is incomplete). Your job is to determine the last line of the proof and an appropriate claim! 1. Claim:

Proof. Suppose that r and s are rational numbers. Then there exist integers a, b, c, and d such that b ≠ 0 , d ≠ 0 ,

r = ab , and

s = cd . So,

r + s = ab +

cd

= ad+bcbd .

Since a, b, c, and d are integers with b ≠ 0 and d ≠ 0 , ad + bc and bd are integers with bd ≠ 0 . Consequently,

2. Claim:

Proof. Let r = 2 and let s = 2 . Then r and s are irrational. But

rs = 2 2 = 2 = 21 ,

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a rational number. Therefore,

3. Claim:

Proof. Suppose that A, B, and C are sets such that A⊆ B∩C . Let a ∈A . Since a ∈A and A⊆ B∩C , a ∈B∩C . Therefore, a ∈B and a ∈C . We have proven the following two statements:

(i) whenever a ∈A , a ∈B , and

(ii) whenever a ∈A , a ∈C .

Consequently, VI. Investigations 1. In Exercise 1 of Section 1, you were asked to make a number of conjectures. Let’s

revisit that exercise by making conjectures to complete each of the following statements in parts a-j. (Be sure to prove that your statements are correct.)

a) The sum of two even integers is ... b) The sum of two odd integers is … c) The sum of an odd integer with an even integer is … d) The quotient of two rational numbers is … e) The sum of two irrational numbers is … f) The product of two irrational numbers is … g) A rational number to a rational power is … h) An irrational number to an irrational power is … i) The product of an irrational number with a rational number is … j) An irrational number to a rational power is … k) Are there other conjectures that you have?

2. a) Find the power set of each of the following sets: {1, 2}, {3, 4,5}, and {6, 7,8,9} . b) Find the number of elements in the power set for each set in part a. c) Let’s suppose that n is a positive integer and A is a set having precisely n elements. Make a conjecture about the number of elements in the power set of A.