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1 Chapter 2 The Structure of R The purpose of this chapter is to explain to the reader why the set of real numbers is so special. By the end of this chapter, the reader should understand the difference between R and Q. We will see that R is complete while Q is not. We will also see that Q is countable while R is not. We will explain the consequences these differences have. 2.1 The set of Real Numbers Though it is possible to construct the set of real numbers from scratch (as did Cantor and Dedekind) and derive its properties from the fundamental axioms of set theory, this process is too big a task and beyond the scope of an introductory course in real analysis. Instead, we will make certain assumptions and derive all the important properties of R from these assumptions. Most of the properties we will derive are already known to you. The goal here is to see how they can be proven using only our assumptions and what has already been proven. More precisely, we assume we are given the following: 1. The set of real numbers R. 2. Two binary operations (addition, denoted + and multiplication, denoted.) 3. An order relation <. See section We will also assume a set of properties (see section 2.1.2). From these assumptions, we will prove most of the important results about real numbers, needed for this class. Finally, we will assume the axiom of completeness (see section 2.4). When we want to talk about the real numbers together with the two binary operations, we will use the notation (R, +,.). Before we do this though, let us give an intuitive presentation of the real numbers. 31

2 32 CHAPTER 2. THE STRUCTURE OF R Intuitive Definition of the Real Numbers The concept of number is fundamental to our way of life. The concept of natural numbers, or the numbers we use to count is grasped as an early age. This set, is denoted N. Thus, N = {1, 2, 3,...} Very soon one realizes that the natural numbers are not enough. If we think of numbers as used in counting to represent things one owns, then how does one represent what one owes? The answer is by putting a negative sign. If we combine the natural numbers, their negative and zero, we get the set of integers. This set is denoted by Z. Thus Z = {0, ±1, ±2, ±3,...} As one learns about multiplication and division, one realizes that Z is not enough. For example, if we try to divide 2 by 3, the answer is not an integer. Thus, we are quickly forced to introduce new numbers, the rational numbers. The set of rational numbers is denoted by Q. Its definition is { m } Q = n : m Z, n Z and n 0 Then, we ask the question: do we have all the numbers we ll ever need with the rational numbers? The answer is no, and it has been known for a long time. Around 500 BC, Pythagora knew that not every quantity could be expressed as a rational number. Consider for example a right triangle in which the length of the sides around the right angle is 1. Then, the length x of the hypotenuse satisfies = x 2. In other words x 2 = 2. What is the number x such that x 2 = 2? We know this number exists because we can construct such a triangle. We call this number 2. Pythagora proved that 2 is not a rational number. It is not the only one. π is not rational, neither is e. If p is a prime number, one can show that p is not rational. When a number is not rational, we say that it is irrational. There are more irrational numbers than rational ones. If we represent the rational numbers as dots on a line, there will be more empty spaces than dots. There are several categories of numbers which are irrational. We will not discuss those here. Before we look at the properties of the real numbers, we prove that 2 is irrational using Pythagora proof. We begin with a lemma which proof is left as an exercise. Lemma 79 Let n be an integer. If n 2 is even then n must also be even. Proof. See exercises. We are now ready for Pythagora proof. We state the result as a theorem and give its proof. Theorem 80 There does not exist a rational number r such that r 2 = 2.

3 2.1. THE SET OF REAL NUMBERS 33 Proof. We do a proof by contradiction. Suppose there exists integers a and b 0 such that r = a b, r2 = 2 and assume a is already in lowest terms that is b a and b do not have any common factors. Then, a2 b 2 = 2 or a2 = 2b 2. Thus a 2 is even. It follows that a is even from lemma 79. Thus there exists a number k such that a = 2k. It follows that 2b 2 = 4k 2 or b 2 = 2k 2. Hence, b 2 is even, thus b is even. Which contradicts the fact that a was in lowest terms. b Let us now look at the essential properties which define the set of real numbers Algebraic Properties of R (R, +,.) is a field that is the following properties are satisfied: Closure R is closed under both operations, that is if a R and b R then a+b R and ab R. (A1) a, b R, a + b = b + a. (+ is commutative) (A2) a, b, c R, (a + b) + c = a + (b + c). (+ is associative) (A3) There exists an element denoted 0 in R with the property: a R, a+0 = 0 + a = a. 0 is called the additive identity element. (A4) For each element a in R, there exists an element denoted a, called the additive inverse of a with the property: a + ( a) = ( a) + a = 0. (M1) a, b R, a.b = b.a. (. is commutative) (M2) a, b, c R, (a.b).c = a. (b.c). (. is associative) (M3) There exists an element denoted 1 0 in R with the property: a 0 R, a.1 = 1.a = a. 1 is called the multiplicative identity element. (M4) For each element a 0 in R, there exists an element denoted 1/a, called the multiplicative inverse of a with the property: a. (1/a) = (1/a).a = 1. (D) a, b, c R, a. (b + c) = (a.b) + (a.c) and (b + c).a = (b.a) + (c.a). (multiplication is distributive over addition) Remark Properties (A2) (A4) mean that (R, +) is a group. 2. Adding property (A1) means that (R, +) is a commutative group also called an abelian group.

4 34 CHAPTER 2. THE STRUCTURE OF R Important Basic Properties of Real Numbers The properties which follow do not depend on the fact that we are dealing with real numbers. They depend on the fact that R is a field. Thus, they do not really belong to a real analysis class. Students should have studied (or will study) in a modern algebra class these properties and how to prove them. They are given here because they are essential to the everyday manipulations we perform when we work with real numbers. However, we will not spend time on them. We take many of these properties for granted. Yet, without them, many of the things we do with real numbers would not be possible. First, we establish uniqueness of the identity element as well as uniqueness of the inverse under both operations. This fact is a direct consequence of the properties of each operation; it has nothing to do with the fact that we are working with real numbers. Theorem 82 The additive and multiplicative identity are unique. Proof. We only prove uniqueness of the additive identity. For this, we will prove that if x is an element of R such that x + a = a for every a R, then x = 0. x = x + 0 = x + (a + ( a)) = (x + a) + ( a) = a + ( a) = 0 Theorem 83 The additive and multiplicative inverses are unique. Proof. We only prove uniqueness of the additive inverse. For this, we will prove that if a and b are elements of R such that a + b = 0 then b = a. b = 0 + b = (( a) + a) + b = ( a) + (a + b) = ( a) + 0 = a Remark 84 Properties (A4) and (M4) guarantee the possibility of solving the equations a + x = 0 and a.x = 1. Theorem 83 guarantees the uniqueness of the solution. We now generalize this result. Many problems in mathematics involve solving one or more equations. The next two theorems play an important role in finding the solutions of equations.

5 2.1. THE SET OF REAL NUMBERS 35 Theorem Let a and b be arbitrary elements of R. then the equation a + x = b has the unique solution x = ( a) + b 2. Let a 0 and b be arbitrary elements of R. then the equation a.x = b has the unique solution x = b. (1/a) Proof. In both cases, we have to show that there is a solution. This can be done easily by verifying that the given solution is indeed a solution. Then, we have to show that it is unique. The details of the proof are left as an exercise. Theorem 86 (Cancellation Laws) Suppose that a, b and x are real numbers. Then the following is true: 1. a + x = b + x = a = b 2. If in addition x 0 then a.x = b.x = a = b Proof. We prove each part separately. (a) Choose y such that x + y = 0. Then, a + x = b + x = (a + x) + y = (b + x) + y = a + (x + y) = b + (x + y) = a + 0 = b + 0 = a = b (b) Since x 0, choose y such that xy = 1. Then a.x = b.x = (a.x).y = (b.x).y = a. (x.y) = b. (x.y) = a.1 = b.1 = a = b We now look at various miscellaneous properties of the set of real numbers. The order in which we list them is important in the sense that the proof of some of these properties depend on other properties. Theorem 87 If a and b are any elements of R then 1. a.0 = 0 2. a = ( 1).a 3. (a + b) = ( a) + ( b) 4. ( a) = a

6 36 CHAPTER 2. THE STRUCTURE OF R 5. ( 1). ( 1) = 1 Proof. 1. Since 0+0 = 0, we have (0 + 0).a = 0.a. Using the distributive law we can write it as 0.a+0.a = 0.a. This in turn can be written as 0.a+0.a = 0.a+0. Finally, using the cancellation law for addition, we obtain 0.a = Theorem 83 establishes the fact that every non-zero real number a has a unique additive inverse a. If we can show that ( 1).a has the properties of the additive inverse of a, the result will follow. 3. Left as an exercise a + ( 1).a = 1.a + ( 1).a = (1 + ( 1)).a = 0.a = 0 4. a + ( a) = 0 implies that a is the additive inverse of a. But the additive inverse of a is, by (A4) ( a). Since the additive inverse of a element is unique, it follows that ( a) = a. 5. The details are left as an exercise. Combine parts 2 and 4. Theorem If a R and a 0, then 1/a 0 and 1/ (1/a) = a 2. Suppose that a R, b R. Then, a.b = 0 a = 0 or b = Suppose that a R, b R. Then ( a). ( b) = a.b 4. If a R and a 0, then 1/ ( a) = (1/a) Proof. 1. Since a 0, 1/a exists by (M4). If we had 1/a = 0, we would have 1 = a. (1/a) = a.0 = 0 which is a contradiction. Thus, 1/a 0. Since a. (1/a) = 1, a is the multiplicative inverse of (1/a). But, by (M4), the multiplicative inverse of 1/a is 1/ (1/a). Since the multiplicative inverse of an element is unique, 1/ (1/a) = a. 2. Left as an exercise.

7 2.1. THE SET OF REAL NUMBERS Left as an exercise. 4. Since a 0, a 0, so 1/ ( a) exists. Furthermore, 1 = a. (1/a) = ( a) ( (1/a)) by part 3 of this theorem Thus, -(1/a) is the multiplicative inverse of a. But by (M4), the multiplicative inverse of a should be 1/ ( a). By uniqueness of the multiplicative inverse, it follows that 1/ ( a) = (1/a). From this point on, we will drop the use of the dot to denote multiplication. We will write: ab instead of a.b a 2 for aa, a 3 for ( a 2) a = aaa,... a n+1 for (a n ) a for n N. b a for b + ( a) or ( a) + b a 1 for 1/a and a n for 1/ (a n ) a b or a/b for a (1/b) Exercises As you do these problems, keep in mind that you already know all these results. The goal is to prove them using all the assumptions we have made as well as what you have already proven. When I ask you to prove a result stated in the notes, you can only use all the assumptions we have made as well as the results proven up to the result I am asking you to prove. If the question is a stand alone question, then you can use all the results stated in the notes as well as all the problems up to the question you are working on. 1. Finish proving theorem Finish proving theorem Finish proving theorem Finish proving theorem Finish proving theorem Prove that a + a = 0 = a = 0 for any arbitrary real number a. 7. Prove that if n is an integer and n is even, then n 2 is also even. 8. Prove lemma 79.

8 38 CHAPTER 2. THE STRUCTURE OF R 9. Prove that if n 2 is an odd integer, then n must also be odd. 10. Prove that 6 is not a rational number. 11. Prove that if x is irrational and r is rational, then x+r is irrational. Also, show that if r is a non-zero rational then xr is irrational. 12. Show by example that if x and y are irrational, then x + y and xy may be rational.

9 2.2. ORDER AXIOMS AND ORDER PROPERTIES OF R Order Axioms and Order Properties of R In this section, we introduce the order properties of R. These properties are very important and will be used heavily throughout the remainder of this course. Like in the previous section, we do not develop these properties from scratch. We assume a few of them, then we derive most of the properties from our assumptions. The properties we assume will be called axioms Order Axioms for R and Definitions On R there is a relation, denoted <, which satisfies the following axioms: Axiom 89 (O1) For all a and b in R, exactly one of the following holds: 1. a < b 2. a = b 3. b < a Axiom 90 (O2) For all a, b and c in R, if a < b and b < c then a < c. Axiom 91 (O3) For all a, b and c in R, if a < b then a + c < b + c. Axiom 92 (O4) For all a, b and c in R, if a < b and 0 < c then ac < bc. Remark 93 The first axiom, axiom 89, is known as the law of trichotomy. Remark 94 The second axiom, axiom 90, is simply the statement of the property called transitivity. Remark 95 If we represent the set of real numbers as a line stretching to infinity in both directions (the real line), then by convention if a < b, then a is positioned on the line to the left of b. Definition 96 We can now define some well known terms. Let a and b be in R. 1. If a < b, we say that a is less than b. 2. If a < b then we also write b > a and say b greater than a. 3. If a < b or a = b then we write a b and say a less than or equal to b 4. Similarly, we can define greater than or equal to. 5. If a > 0, we say that a is positive. If a < 0, we say that a is negative. 6. If a 0, we say that a is non-negative. If a 0, we say that a is non-positive.

10 40 CHAPTER 2. THE STRUCTURE OF R Remark 97 If we have a < b and b < c, it is convenient to combine the two relations and write a < b < c. The same applies for the other inequalities. Remark 98 A field together with an order that is a relation satisfying axioms is called an ordered field. So, (R, +,., <) is an ordered field. Remark 99 The opposite of the statement a < b is a b. opposite of the statement a > b is a b. Similarly, the Properties of the Order We now establish the properties the relations we just defined satisfy. Most of these properties are already known to the reader. We will prove a few of them. The remaining proofs will be assigned as exercises. Theorem 100 For every xin R, the following is true: 1. If x < 0 then x > If x > 0 then x < 0. Proof. We only prove the first part. The second part is left as an exercise. If x < 0 then by O3 x + ( x) < 0 + ( x). It follows that 0 < x which is the same as x > 0. Theorem 101 Let a 0 R. 1. a 2 > > 0 Proof. We prove each part separately. 1. Since a 0, either a > 0 or a < 0. If a > 0, then a 2 = aa > 0.a = 0, hence a 2 > 0. If a < 0, then a > 0. Hence, ( a) ( a) > 0. ( a) = 0. But ( a) ( a) = aa = a 2. So, a 2 > Follows from part 1 and the fact that 1 = 1 2 Theorem 102 Let a, b, c be elements of R. 1. a < b a b < 0 2. a > b and c < 0 ac < bc 3. a > 0 1/a > 0 4. a < 0 1/a < 0

11 2.2. ORDER AXIOMS AND ORDER PROPERTIES OF R 41 Proof. left as an exercise. Theorem 103 If a and b are any real numbers, then a > b = a > 1 (a + b) > 2 b Proof. left as an exercise. Remark 104 Theorem 103 implies (by setting b = 0) that given any strictly positive number a, there exists another smaller strictly positive number ( 1 2 a). Thus there is no smallest strictly positive number. Theorem 105 Let a and b be arbitrary real numbers. If a > 0 and b > 0 then ab > 0 Proof. If a > 0 and b > 0 then by O4 ab > 0.b = 0. Therefore, ab > 0. Theorem 106 If ab > 0 then we either have a > 0 and b > 0 or we have a < 0 and b < 0. Proof. If ab > 0, then ab 0, hence neither a = 0 nor b ( = 0. By the law of trichotomy, either a > 0 or a < 0. If a > 0, then b = b a 1 ) = (ba) 1 a a = (ab) 1 a > 0 (why?) Hence, b > 0. The proof is similar if we assume that a < 0. Corollary 107 If ab < 0 then we either have a > 0 and b < 0 or we have a < 0 and b > 0. Proof. Left as an exercise Exercises 1. Finish proving theorem Finish proving theorem Finish proving theorem Finish proving theorem Finish proving corollary Prove that If a b and b a then a = b where a and b are arbitrary real numbers. 7. Prove the following where a, b, c and d denote arbitrary elements of R. (a) a < 0 and b > 0 = ab < 0 (b) a < 0 and b < 0 = ab > 0 (c) 0 < a < 1 = a 2 < a (d) 1 < a = a 2 > a

12 42 CHAPTER 2. THE STRUCTURE OF R (e) a < a + 1 (f) 0 < a < b = a 2 < b 2 (g) a b and c d = a + c b + d 8. If a, b R and a 2 + b 2 = 0, show that a = b = 0.

13 2.3. METRIC PROPERTIES OF R - ABSOLUTE VALUE Metric Properties of R - Absolute Value Definition and Properties The trichotomy property assures that if a 0 then either a > 0 or a < 0. Thus, we define: Definition 108 (Absolute value) The absolute value of a real number a, denoted a is defined by: { a if a 0 a = a if a < 0 The domain of this function is R, its range is the set of non-negative real numbers, and it maps the elements a and a into the same element. Geometrically, the absolute value of a number x can be interpreted as how far from 0 x is. Similarly, x a can be interpreted as the distance between x and a and therefore, one can think of the absolute value as a tool to measure distances between numbers. In particular, for a fixed number a, the set defined by {x R such that x a < δ} denotes the set of numbers within a distance δ of a. The inequality x a < δ is equivalent to δ < x a < δ or a δ < x < a + δ that is x (a δ, a + δ). This will be used a lot throughout this class. Remark 109 With a = 0 above, we see that δ < x < δ is equivalent to x < δ. The same is true if we replace < by. Students should review how to solve equations and inequalities involving absolute values as they too will be used a lot in this class. Theorem 110 The following properties hold: 1. a = 0 a = 0 2. a = a for all a in R 3. ab = a b for all a, b in R 4. 1 a = 1 for all a 0 in R a 5. a = a for all a, b 0 in R b b 6. If c 0, then a c c a c 7. a a a 8. a 2 = a Proof. See problems

14 44 CHAPTER 2. THE STRUCTURE OF R The next result is often used in Mathematics. Theorem 111 (Triangle inequality) If a and b are any real numbers, then a b a + b a + b Proof. First, we prove the right side. Since a a a and b b b it follows that ( a + b ) a + b a + b. Thus, a + b a + b by remark 109 or part 6 of theorem 110. Next, we prove the left side. a = a + b b a + b + b = a + b + b Thus, a b a + b Similarly, b = a + b a a + b + a = a + b + a Thus b a a + b Combining the two inequalities, we get a b a + b a + b a b a + b It follows by part 6 of theorem 110 that a b a + b An application of the above theorem, often used when dealing with absolute values and inequalities is given by the corollary below. Corollary 112 If a, b and c are any given real numbers then a c a b + b c Proof. If we let a c = (a b) + (b c) and use the triangle inequality, we have a c = (a b) + (b c) a b + b c

15 2.3. METRIC PROPERTIES OF R - ABSOLUTE VALUE 45 Corollary 113 If a 1, a 2,...a n are any n real numbers, then Proof. See exercises. a 1 + a a n a 1 + a a n We finish with a theorem which appears simple in statement but whose proof can sometimes give a hard time to beginners. The theorem simply states that the only non-negative real number less that every positive number is 0. Theorem 114 Let x be a real number. If x < ɛ for each ɛ > 0 then x = 0. Proof. See exercises. Remark 115 This result is often used to show two numbers are equal. If a and b are two real numbers, to show a = b, it is enough to show that a b < ɛ for each ɛ > 0. By the theorem, it will follow that a b = 0 or a = b. Geometrically, we are simply saying that if the distance between a and b is less than any positive real number then a = b Equations and Inequalities Involving Absolute Values The purpose of this class is not to learn how to solve equations and inequalities involving absolute values. However, they will be used a lot throughout this class. Students should make sure they review them. We remind students of the basic principles used when solving equations and inequalities involving absolute values. For what follows, let a and b denote any expression. The equation a = b is equivalent to the two equations a = b and a = b. However, solutions must be checked, some may not be valid. The equation a = b is equivalent to the two equations a = b and a = b. However, solutions must be checked, some may not be valid. The inequality a < b is equivalent to b < a < b The inequality a b is equivalent to b a b The inequality a > b is equivalent to a > b or a < b The inequality a b is equivalent to a b or a b Notion of a Metric Space As we noted earlier, the absolute value can be used to measure distances between numbers. x y denotes the distance between x and y on the real line. For the rest of these notes, we will use the absolute value every time we need to measure distances between two real numbers. However, if the set in which our elements come from were not R, the absolute value may not be appropriate to measure

16 46 CHAPTER 2. THE STRUCTURE OF R distances. For example, you already know that in R 2, the distance between two points of coordinates (x 1, y 1 ) and (x 2, y 2 ) is given by (x 2 x 1 ) 2 + (y 2 y 1 ) 2. In more general settings, we replace the absolute value by a function we call a distance function or a metric. We give a formal definition for a metric. Definition 116 (metric) Let X be a set. A metric or a distance on X is a function d : X X R which satisfies the properties below. In the properties below, we let x, y and z be elements of X. 1. d (x, y) 0 (non-negativity) 2. d (x, y) = 0 x = y 3. d (x, y) = d (y, x) (symmetry) 4. d (x, z) d (x, y) + d (y, z) (triangle inequality) Definition 117 (metric space) A space X together with a metric d is called a metric space. We say that (X, d) is a metric space. Example 118 It is easy to see that the absolute value function is a metric for R, hence we say that (R,. ) is a metric space Exercises 1. Prove theorem Prove the second corollary of theorem Show that a + b = a + b ab Given real numbers a and b, prove that a b a b. 5. Given real numbers a and b, prove that a b a b. This is sometimes called the reverse triangle inequality. 6. Prove theorem Solve 2x 3 = x. 8. Solve 2x 3 = 3x. 9. Solve 2x + 5 > Solve 2x + 5 < 11.

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