Real Analysis Notes Suzanne Seager PDF Free Download

Real Analysis Notes Suzanne Seager 2015

Contents Introduction... 3 Chapter 1. Ordered Fields... 3 Section 1.1 Ordered Fields... 3 Field Properties... 3 Order Properties... 4 Standard Notation for Ordered Fields... 4 Section 1.2. Absolute Value... 9 Section 1.3. Multiple Quantifiers... 10 Section 1.4. More Multiple Quantifiers... 16 Chapter 2. Sequences...Error! Bookmark not defined. Section 2.1. Introduction to Sequences...Error! Bookmark not defined. Section 2.2. Sequences in Ordered Fields...Error! Bookmark not defined. Section 2.3. Limits of Sequences in Archimedean Ordered Fields...Error! Bookmark not defined. Section 2.4. Using Limits of Sequences...Error! Bookmark not defined. Section 2.5. Algebraic Properties of Converging Sequences...Error! Bookmark not defined. Section 2.6. Order Properties of Converging Sequences...Error! Bookmark not defined. Section 2.7. Cauchy Sequences...Error! Bookmark not defined. Chapter 3. The Real Numbers...Error! Bookmark not defined. Section 3.1. The Bisection Technique...Error! Bookmark not defined. Section 3.2 Upper and Lower Bounds...Error! Bookmark not defined. Section 3.4 The Real Numbers...Error! Bookmark not defined. Section 3.3. Properties of the Real Numbers...Error! Bookmark not defined. Section 3.5 Sequences of Real Numbers...Error! Bookmark not defined. Chapter 4. Functions on R...Error! Bookmark not defined. Section 4.1. Limits of Functions...Error! Bookmark not defined. Section 4.2 Continuity...Error! Bookmark not defined. Section 4.3 Continuous Functions on a Closed Interval...Error! Bookmark not defined. Section 4.4. Uniform Continuity...Error! Bookmark not defined. Section 4.5. Derivatives...Error! Bookmark not defined. Appendix A: Mathematical Proof...Error! Bookmark not defined. Index...Error! Bookmark not defined.

Introduction In introductory calculus you worked with an intuitive rather than a formal definition of limits, then used limits to define continuity and derivatives. For most practical purposes, such an intuitive definition is sufficient; historically, calculus developed from Isaac Newton and Gottfried Leibniz through the 17 th and 18 th centuries based on such definitions. However, the lack of a formal definition makes it difficult to answer more theoretical questions. For example, is it possible to define a function on the real numbers which is discontinuous everywhere? (What does your intuition suggest?) Is it possible to define a function which is continuous on the real numbers but not differentiable anywhere? By the 19 th century, mathematicians such as Bernard Bolzano and Augustin-Louis Cauchy were developing formal definitions so that such theoretical questions can be answered. Real Analysis is the theory that results from these formal definitions. Chapter 1. Ordered Fields The most basic concept required for calculus is that of the real numbers. The first numbers we learn about are the natural numbers N = {1, 2, 3, } used for counting. Augmenting these with zero and negatives results in the integers Z = {, -3, -2, -1, 0, 1, 2, 3, }, then division leads to the fractions that make up the rational numbers Q = { a/b a, b Z and b 0}. Extending these to the real numbers R is not quite so simple: what properties do the real numbers have that that the rational numbers do not? In order to answer this, we start with the most important properties that the real numbers, R, and the rational numbers, Q have in common. These properties define an ordered field. Section 1.1 Ordered Fields Definition 1.1.1. An ordered field is a set F with two binary operations, addition denoted by + and multiplication denoted by, and a less than order denoted by <, satisfying the following two sets of properties: Field Properties Commutative Property of Addition: For all a, b F, a + b = b + a. Associative Property of Addition: For all a, b, c F, (a + b) + c = a + (b + c). Existence of Zero: There exists an element 0 F such that for all a F, 0 + a = a and a + 0 = a. Existence of Negatives: For all a F there exists -a F such that a + (-a) = 0 and (-a) + a = 0. Commutative Property of Multiplication: For all a, b F, a b = b a. Associative Property of Multiplication: For all a, b, c F, (a b) c = a (b c). Existence of One: There exists 1 F with 1 0 such that for all a F, 1 a = a and a 1 = a. Existence of Reciprocals: For all a F\{0} there exists 1/a F such that a (1/a) = 1 and (1/a) a = 1. Distributive Property of Multiplication over Addition: For all a, b, c F, a (b + c) = a b + a c and (b + c) a = b a + c a.

Order Properties Trichotomy Property: For all a, b F, exactly one of the following holds: a b. Transitivity Property: For all a, b, c F, if a < b and b < c then a < c. Adding Property: For all a, b, c F, if a < b then a + c < b + c. Multiplying by a Positive Property: For all a, b, c F, if a < b and 0 < c then a c < b c. We know that both R and Q satisfy these properties, so both the reals and the rationals are examples of ordered fields. Although the ordered fields are defined using only the operations and relations denoted by the symbols +,, =, and <, for convenience we define other standard notation (for all a, b F): Standard Notation for Ordered Fields Words Notation Represents Subtraction a - b a + (-b) Division a / b a 1/b, provided b 0 Multiplication ab a b Less than or equal to a b a b b < a Greater than or equal to a b b a Positive a 0 Negative a 0 Similarly, the only numbers in the definition are 0 and 1, but we define the standard symbols for the natural numbers: 2 is defined as 1 + 1, 3 as 2 + 1, and so on for all n N. Thus we can also define the nth power of an element a by a 1 = a, a 2 = a a, and so on for all n N. With this notation, the Order Properties can be rewritten in alternate forms to include the equality and/or greater than relations, for example the following: Trichotomy Property: For all a, b F, exactly one of the following holds: a b. Transitivity Property: For all a, b, c F, if a < b and b c then a < c. For all a, b, c F, if a b and b < c then a < c. For all a, b, c F, if a b and b c then a c. Adding Property: For all a, b, c F, if a b then a + c b + c. Multiplying by a Positive Property: For all a, b, c F, if a b and 0 c then ac bc. From the ordered field properties, all of the standard rules of algebra can be developed. We will take for granted the rules of algebra involving only the field properties, as these are studied in Abstract Algebra. However, as a review of both order and proofs we will prove the standard properties of order that apply to any ordered field F. Throughout this book, in order to demonstrate how proofs can be created, we give a detailed description of this process for several of the theorems. Comments in italics describe the thinking involved, and the Proof Analysis shows the step by step development (for more information on proofs and proof analyses

see Appendix A). The proof analysis is then rewritten into a more formal Proof using English paragraphs. Theorem 1.1.2 (a) For all a F, if a < 0 then -a > 0 (b) For all a F, if a > 0 then -a < 0. To prove part (a) of this theorem, we start from the logical structure of the statement: For all a F, if a < 0 then -a > 0. Since the statement begins with the universal quantifier "For all", we start the proof with a generic element a of F: a is generic. The remainder of the statement has the "if-then" structure, so we use the direct proof structure of assuming the and using it to reach the conclusion. Proof Analysis of (a) Let a F. Suppose a < 0. Reach -a > 0. generic element of the ordered field F assume the of the theorem reach the conclusion The heart of the proof is determining what steps can take us from the to the conclusion, and we do this by working both forwards from the and backwards from the conclusion. We will use the only tools we have at the moment: the ordered field properties and the definitions. Since all of the order properties use "<" rather than ">", we can start working backwards by using the definition of greater than to rewrite the conclusion as 0 < -a. For formal proofs it is essential to use exact definitions, so you should learn all definitions by heart as soon as possible. Proof Analysis of (a) Let a F generic element of the ordered field F Suppose a < 0 assume the of the theorem 0 < -a -a > 0 definition of greater than. We need to get from a < 0 to 0 < -a. We need inspiration, so we look through the ordered field properties for a way we can transform a to 0 while at the same time transforming 0 to -a. From the Existence of Negatives Property we have a + (-a) = 0 and from the Existence of Zero Property we have 0 + (-a) = -a, which suggests we try adding -a to both sides of the inequality a < 0. Can we justify this? Yes, by the Adding Property. Proof Analysis of (a) Let a F generic element of the ordered field F Suppose a < 0. assume the of the theorem a + (-a) < 0 + (-a) add (-a) to both sides by the Adding Property a + (-a) = 0 Existence of Negatives Property 0 + (-a) = -a Existence of Zero Property 0 < -a substitute 0 for a + (-a) and -a for 0 + (-a) in a + (-a) < 0 + (-a) -a > 0 definition of greater than This gives us a detailed proof analysis, which we now rewrite as a proof. We won't justify algebraic steps involving the Algebraic Properties in formal proofs. However, in this chapter we will include

detailed justification for every use of the Order Properties. It is particularly important to pay attention when multiplying both sides an inequality by an element to the sign of the element (positive, negative, or zero) as this is a common souce of careless errors: carefully justifying the properties involved helps to reduce such errors. As proofs are written in English paragraphs, we add connectives such as "and", "but", "then", "thus", "hence", "because", "therefore", etc. to indicate the flow of the proof and make it easier to read. Proof. (a) Let a F and suppose a < 0. Then -a + a < -a + 0 by adding -a to both sides. Simplifying both sides gives 0 < -a, and thus -a > 0. The proof of (b) is similar and is left as an exercise. Theorem 1.1.3. Multiplying by a Negative Property. For all a, b, c F, if a < 0 and b < c then ab > ac. Proof. Let a, b, c F. Suppose a < 0 and b < c. Then -a > 0 by Theorem 1.1.2, and so (-a)b < (-a)c by multiplying the inequality b < c by a positive, -a. Hence -ab < -ac and thus by adding ac to both sides we get -ab + ab + ac < -ac + ab + ac. This simplifies to ac < ab, and thus ab > ac. Recall from algebra the Zero Product Property: For all a, b F, ab = 0 if and only if a = 0 or b = 0 (can you prove this using only the Field Properties?). We can use the Zero Product Property to modify the order properties for multiplication to include equality: Multiplying by a Non-Negative Property: For all a, b, c F, if a < b and 0 c then ac bc. Multiplying by a Non-Positive Property. For all a, b, c F, if a 0 and b < c then ab ac. The next theorem extends the Zero Product Property to inequalities. Theorem 1.1.4. (a) For all a, b F, if either a > 0 and b > 0 or a < 0 and b < 0, then ab > 0. (b) For all a, b F, if either a < 0 and b > 0 or a > 0 and b < 0, then ab < 0. (c) For all a, b F, if ab > 0 then either a > 0 and b > 0, or a < 0 and b < 0. (d) For all a, b F, if ab < 0 then either a > 0 and b < 0, or a < 0 and b > 0. Proof Analysis of (a) We start with the logical structure. The is an "or" statement, which suggests using proof by cases (see Appendix AIII). Let a, b F. Suppose a > 0 and b > 0 or a < 0 and b < 0. Case 1. Suppose a > 0 and b > 0 Reach ab > 0 Case 2. Suppose a < 0 and b < 0 Reach ab > 0 generic a and b

Now we need to get from a > 0 and b > 0 to ab > 0 using only the Order Properties and previous theorems. Since ab is a multiplied by b, we multiply both sides of a > 0 by b to get ab > 0b. Can we justify this? Yes: since b > 0 we can use the Multiplying by a Positive Property. Case 1. Suppose a > 0 and b > 0 ab > 0b ab > 0 multiply a > 0 by a positive, b 0b = 0 by the Zero Product Property The second case is similar, resulting in the following proof. Proof of (a). Let a, b F. Suppose a > 0 and b > 0 or a < 0 and b < 0. Case 1. Suppose a > 0 and b > 0. Then ab > 0b by multiplying a > 0 by a positive, b, so ab > 0. Case 2. Suppose a < 0 and b < 0. Then ab > 0b by multiplying b < 0 by a negative, a, so ab > 0. The proof of (b) is similar and is left as an exercise. Proof of (c). Let a, b F. Suppose ab > 0. Then, by the Trichotomy Property, ab 0 and ab 0. Also by the Trichotomy Property, we have nine possibilities for a and b: (1) a > 0 and b > 0 (2) a > 0 and b = 0 (3) a > 0 and b < 0 (4) a = 0 and b > 0 (5) a = 0 and b = 0 (6) a = 0 and b < 0 (7) a < 0 and b > 0 (8) a < 0 and b = 0 (9) a < 0 and b < 0 Since ab 0, we have a 0 and b 0 so we can rule out (2), (4), (5), (6), and (8). Since not ab < 0, from the contrapositive of part (b) we have not a > 0 and b < 0 and not a < 0 and b > 0, which excludes (3) and (7). Thus the only possibilities left are (1) and (9): either a > 0 and b > 0, or a < 0 and b < 0. The proof of (d) is similar and is left as an exercise. Theorem 1.1.5. For all a F, if a 0 then a 2 > 0. In particular, 1 > 0. Proof. Let a F, a 0. Then a > 0 or a < 0 by the Trichotomy Property. So aa > 0 by Theorem 1.1.4, and thus a 2 > 0. In particular, since 1 0, by the Trichotomy Property 1 2 > 0 and thus 1 > 0. Theorem 1.1.6. The Reciprocal Property. For all a, b F, if 0 < a < b then 0 < 1/b < 1/a. Since 0 < 1/b < 1/a means 0 < 1/b and 1/b < 1/a, we first prove 0 < 1/b and then prove 1/b < 1/a. Proof. Let a, b F, with 0 < a < b. Then 0 0 by Theorem 1.1.5, and therefore 1/b > 0 by Theorem 1.1.4. Similarly, 1/a > 0, and thus (1/b)(1/a) > 0 by Theorem 1.1.4. But then (1/b)(1/a)a < (1/b)(1/a)b by multiplying a < b by a positive, (1/b)(1/a). Thus 1/b < 1/a.

This theorem can also be proved indirectly. Recall the logical structure of a proof by contradiction: Proof Analysis by contradiction Suppose not. Reach a contradiction. We need the negation of the statement For all a, b F, if 0 < a < b then 0 < 1/b and 1/b < 1/a. Recall that the negation of "for all x P(x)" is "there exists x such that not P(x)", the negation of "if p then q" is "p and not q", and the negation of p and q is not p or not q. Proof Analysis by contradiction Suppose not. There exist a, b F such that 0 < a < b and either 0 1/b or 1/b /a negation 1/b 0 or 1/a 1/b Trichotomy Property Reach a contradiction. Or suggests proof by cases. In the first case, 1/b 0, so we can reach a contradiction by multiplying both sides by b. In the second case, 1/a 1/b, which suggests we multiply both sides by ab. Alternate Proof. Suppose not. Then there exist a, b F such that 0 < a < b, and either 0 1/b or 1/b 1/a. Thus either 1/b 0 or 1/a 1/b, by the Trichotomy Property. Case (i) Suppose 1/b 0. Since 0 < a < b, we have 0 0 by Theorem 1.1.5, a contradiction by the Trichotomy Property. Case (ii) Suppose 1/a 1/b. Since 0 < a < b, we have 0 0 by Theorem 1.1.4. Thus (1/a)ab (1/b)(ab) by multiplying 1/a 1/b by a positive, ab. Therefore b < a, a contradiction to a < b by the Trichotomy Property. We've now defined ordered fields and considered some important properties. As we saw earlier, every ordered field contains a copy of the natural numbers N, which can be identified with repeated sums of 1's: N = {1, 2, 3, } = {1, 1+1, 1+1+1, }. The Existence of Zero and the Existence of Negatives Properties give us 0 and -1, -2, -3, and thus the integers, Z = {, -3, -2, -1, 0, 1, 2, 3, }. The Existence of Reciprocals Property gives us 1/b for all non-zero integers b, so the products of integers by these reciprocals give us a copy of the rational numbers: Q = {a/b : a, b Z and b 0}. But since the rational numbers themselves form an ordered field, this is as far as we can go; we will need something more than the ordered field properties to describe the real numbers.

Section 1.2. Absolute Value Recall Theorem 1.1.2: For all a F, if a < 0 then -a > 0 and if a > 0 then -a < 0. This motivates the following definition: Definition 1.2.1. For all a F, the absolute value of a, denoted by a, is defined by a = -a if a 0 -a if a < 0. The following corollary gives some straightforward consequences of this definition; we leave the proof as an exercise. In future proofs, you may take this corollary for granted as part of the definition of absolute value. Corollary 1.2.2. For all a F, a 0, with a = 0 if and only if a = 0, and a a = -a. The next five theorems will be used extensively in the rest of this book. The proofs of the first two and the last are left as exercises. Theorem 1.2.3 The Product Property. For all a, b F, ab = a b. Theorem 1.2.4 The Quotient Property. For all a, b F, b 0, a/b = a / b. Theorem 1.2.5 The Unpacking Property. For all a, b F with b 0, a b if and only if -b a b. Recall that to prove an "if and only if" statement we must prove the "if" statement, if p then q, and also prove its converse, if q then p. Proof. Suppose a b. Then a a by the definition of absolute value (Corollary 1.2.2), so a b by transitivity. Also -a -a = a by the definition of absolute value, so -a b by transitivity. Thus a -b, by multiplying -a b by a negative, -1. Combining a b and a -b gives -b a b. Conversely, suppose -b a b. Then by trichotomy either a < 0 or a 0. Case (i) Suppose a < 0. Then a = -a by the definition of absolute value. Also (-1)(-b) (-1)a, by multiplying -b a by a negative, -1, so b -a. Thus b a and therefore a b. Case (ii) Suppose a > 0. Then a = a. Since a b, we get a b. Theorem 1.2.6 The Triangle Inequality. For all a, b F, a + b a + b. Proof. By the definition of absolute value, - a a a and - b b b. We can add these two inequalities to obtain - a + (- b ) a + b a + b (why?), and thus -( a + b ) a + b a + b. Hence a + b a + b by the Unpacking Property. We will find the Triangle Inequality particularly useful in the Subtract and Add Technique: a - b = a - c + c - b a - c + c - b by the Triangle Inequality. Corollary 1.2.7 The Reverse Triangle Inequality: For all a, b F, a - b a - b.

Section 1.3. Multiple Quantifiers Recall that to prove a statement of the form "For all x, P(x)" we normally start with a generic x and show that it has the property P. But to prove a statement of the form "There exists x, P(x)" we normally use construction: we find one specific value of x that has the property P. This means we work backwards from P(x) to find x. We start with a trivial example. Example 1.3.1 Prove that there exists x F such that 3x - 5 19. We don't know the value of x at the start, so we leave a blank for it and then work backwards to fill in the blank. Thus our initial proof structure, based on the logical structure of the statement, is as follows: Proof Analysis: Choose x = Reach 3x - 5 19. (we will fill this in later) conclusion Next we work backwards from 3x - 5 19 to find x. 3x - 5 19, so 3x 24, so x 8. To satisfy "there exists", we only need one value for x, we are not required to give all possible solutions. So we choose any x such that x 8. The easiest choice is x = 8, but we could also choose x = 5 or -½ or any other value that's at most 8 and the proof will be equally valid. Once we've made our choice for x we fill in the blank, then complete the proof by substitution: Proof. Let x = 8. Then 3x - 5 = 3(8) - 5 = 24-5 = 19 so 3x - 5 19. Note: In mathematics, the word "let" is commonly used to define a variable, whether the variable defined is arbitrary ("for all" statements) or specific ("there exists" statements). Because of the importance of the distinction between arbitrary and specific for the proofs in real analysis, in our proof analyses we will use "let" to indicate an arbitrary variable and "choose" to indicate a specific variable, replacing "choose" by "let" in the final proof. Statements involving multiple mixed quantifiers are the basis of much of real analysis, so in the rest of this section we investigate how to prove such statements. We have already seen examples of these statements: Existence of Zero: There exists an element 0 F such that for all a F, 0 + a = a and a + 0 = a. Existence of Negatives: For all a F there exists -a F such that a + (-a) = 0 and (-a) + a = 0. Both of these are existence properties, but one has quantifiers in the order "there exists - for all" and the other "for all - there exists". What is the difference? Think of the difference in meaning between "Everybody loves somebody" (different people may have different somebodies) and "Somebody loves everybody" (just one very loving somebody). Similarly, "There exists X for all Y" means there is just one X which works for all Y. This means X is independent of Y; to prove such a statement we must find X without knowing what Y is. We cannot

use Y to define X, which means we cannot express X in terms of Y. In the Existence of Zero Property, for example, 0 + 1 = 1, 0 + (-5) = -5, and 0 + 23 = 23: there is just one zero. On the other hand, "For all Y there exists X" means that each Y may have its own X; that is, the value of X may depend on the value of Y. In the Existence of Negatives Property, when a = 2 its negative is -2, but when a = 3 its negative is -3, not -2. When we want to emphasize that X depends on Y we may use the notation "X(Y)" in place of "X" and write "For all Y there exists X(Y)". Remember: when multiple quantifiers are involved in a proof we must pay particular attention to the order of the quantified variables. Example 1.3.2 Prove that there exists b > 0 such that for all x F, if x - 2 < b then -3x + 6 < 5. The logical form begins "There exists b for all x" so we need to find a value for b that satisfies the rest of the statement. Since "there exists b" comes before "for all x" in the statement, b must be independent of x; it would be invalid to write anything like "Let b = 2x + 3". However, since we don't know at the start what value we will need for b, we leave a blank for b and then work backwards to fill in the blank. Proof Analysis: Choose b =, so b > 0. (we will fill this in later) Let x F. generic element of F Suppose x - 2 < b -3x + 6 < 5. conclusion To prove -3x + 6 < 5 we start from -3x + 6 and try to reach something < 5. We expect to use the x - 2 0. (we will fill this in later) Let x F. generic element of F Suppose x - 2 < b -3x + 6 = -3(x - 2) factor = -3 x - 2 Product Property = 3 x - 2 definition of absolute value < 5 goal Next we find b. Our goal is now to reach 3 x - 2 < 5, so, working backwards, we want x - 2 < 5/3. From the x - 2 < b, which we can multiply by 3 to get 3 x - 2 < 3b. So we want 3b 5, which means b 5/3. We now have two conditions for b: b > 0 and b 5/3. Again, we only need one value for b which satisfies these, and any positive value up to 5/3 will work, so we just make the most straightforward choice, b = 5/3. We fill in the blank for b with 5/3 to complete the proof analysis.

Proof Analysis: Choose b = 5/3, so b > 0 (we now fill in the value 5/3 for b) Let x F generic element of F Suppose x - 2 < b -3x + 6 = -3(x - 2) factor = -3 x - 2 Product Property = 3 x - 2 definition of absolute value < 3b multiply the x - 2 0. Let x F and suppose x - 2 < b. Then -3x + 6 = -3(x - 2) = -3 x - 2 by the Product Property = 3 x - 2 by the definition of absolute value < 3b by multiplying x - 2 0 such that for all x F, if x - 1 0 Let x F Suppose x - 1 < b Reach x 2 + 4x - 5 < 3. (we will fill this in later) generic x conclusion We start with x 2 + 4x - 5 and want to reach something < 3. What can we do with x 2 + 4x - 5? One possibility is to use the Triangle Inequality. x 2 + 4x - 5 x 2 + 4x + 5 Triangle Inequality x 2 + 4 x + 5 Product Property < 3 goal This cannot work, since x 0, which means x 2 + 4 x + 5 5 > 3, contradicting what we want. So we abandon this approach and try something else. One possibility is to look back at how we proved the previous example and see if we can apply the same technique. We want to connect x 2 + 4x - 5 < 3 with the x - 1 < b, so we want a relationship between x 2 + 4x - 5 and x - 1. We can obtain this by factoring, just as we did in the previous example. x 2 + 4x - 5 = (x + 5)(x - 1) factor = x + 5 x - 1 Product Property

This looks more promising. If we multiply both sides of our x - 1 < b by x + 5 we get an inequality involving x + 5 x - 1 : x - 1 < b so x + 5 x - 1 < x + 5 b But how do we justify this? We are multiplying both sides of an inequality by x + 5, but how do we know this doesn't change the inequality sign? We are multiplying by a positive if x + 5 > 0, but is this necessarily true? No, because x + 5 = 0 when x = -5, in which case x + 5 x - 1 = 0 and x + 5 b = 0 so we end up with x + 5 x - 1 = x + 5. Thus unless we can guarantee that x -5 we cannot conclude that x + 5 x - 1 < x + 5 b. The best we can conclude for now is that x + 5 x - 1 x + 5 b. x - 1 < b x + 5 x - 1 x + 5 b multiply x - 1 < b by a non-negative, x + 5 We now want to reach x + 5 b < 3, so if we knew what x was we could choose b < 3 / x + 5. But, since "There exists b" comes before "for all x" in the statement we are proving, we CANNOT choose an expression for b involving x, because x is not yet defined. Once we have chosen b, however, x must satisfy the condition x - 1 < b, so we can use x - 1 < b to ensure x + 5 b < 3. How can we relate x + 5 and x - 1. Can we rewrite x + 5 in terms of x - 1? Yes, with the Triangle Inequality, using our subtract and add technique to get the form we want. x + 5 = x - 1 + 1 + 5 = x - 1 + 6 subtract and add 1 x - 1 + 6 Triangle Inequality < b + 6 add 6 to both sides of x - 1 < b We want x + 5 b < 3, which would follow if we could get (b + 6)b < 3. We could try solving the quadratic inequality b 2 + 6b - 3 < 0 to get 0 < b < -3 + 2 3 0.464, and then choose, say, b = 0.4 to get (b + 6)b = (0.4 + 6)0.4 = 2.56 < 3. But in fact there is an easier way. Remember, we are free to choose b to be any positive number that will allow us to complete the proof. So we can restrict our choice of b to small positive numbers only by arbitrarily including an upper bound, say b 1. Then adding 6 to both sides gives us b + 6 1 + 6 = 7, and transitivity then gives us x + 5 < 7, which is much simpler to work with than x + 5 < b + 6. x + 5 = x - 1 + 1 + 5 = x 1 + 6 subtract and add 1 x - 1 + 6 Triangle Inequality 0, we can multiply both sides of x + 5 < 7 by b: x + 5 b < 7b multiply x + 5 < 7 by a positive, b Recall that we want x + 5 b < 3, so we want 7b < 3 which means b 3/7. Thus we now have three conditions on b: b > 0, b 1 and b 3/7. We can satisfy all three of these by choosing, say, b = 3/7:

x + 5 b < 7b = 7(3/7) = 3 substitute b = 3/7 Since this is a "there exists b" proof we only need one valid choice for b, and 3/7 0.428 is valid. But there are many other possible choices for b that we could have chosen instead. As we saw from the quadratic inequality, any choice of b between 0 and -3 + 2 3 would work. Since we did not have to find the "best" choice for b, we added the condition b 1 earlier and used it to choose b = 3/7. So now let's write up the whole proof: Proof. Let b = 3/7, so 0 < b 1. Let x F and suppose x - 1 < b. Then x 2 + 4x - 5 = (x + 5)(x - 1) = x + 5 x - 1 Product Property. Also, x + 5 x - 1 x + 5 b by multiplying x - 1 < b by a non-negative, x + 5. But x + 5 = x - 1 + 6 x - 1 + 6 by the Triangle Inequality < b + 6 by adding 6 to both sides of x - 1 < b 1 + 6 = 7 by adding 6 to both sides of b 1 so x + 5 7 by transitivity. Thus x + 5 b < 7b by multiplying x + 5 < 7 by a positive, b = 7(3/7) = 3 by substituting b = 3/7. Therefore x 2 + 4x - 5 < 3 by transitivity. Why did we pick 1 as the upper bound for b? This was arbitrary and just for convenience; we could have chose any other positive number instead. For example, suppose we added the upper bound b 2 instead. Then adding 6 to this would give us b + 6 < 2 + 6 = 8. So we would get x + 5 b < 8b, and thus want 8b < 3. Thus rather than choosing b = 3/7 we would have chosen b = 3/8, and the proof would have been equally valid with all the 7's replaced by 8's. The next example has one more "For all" quantifier at the start, so the statement has three quantifiers: "for all M", "there exists b", and "for all x": Example 1.3.4. Prove that for all M > 0 there exists b > 0 such that for all x F, if x - 1 0 generic M Choose b =, so b > 0. (we will fill this in later) Let x F. generic x Suppose x - 1 < b. x 2 + 4x - 5 < M. conclusion Apart from "For all M > 0" the only difference between this and the previous example is that the number 3 has been replaced by the variable M. So we can follow exactly the same steps as in the previous example (replacing the 3 with M) to obtain the three conditions b > 0, b 1 and b M/7 on b.

We now need to choose b (which may be in terms of M since "for all M" occurs before "there exists b") to satisfy these three conditions. But we don't know the value of M, so we don't know whether M/7 1 or M /7 > 1. If M/7 1 then as above we can choose b = M/7. However, if M /7 > 1, to satisfy b 1 we cannot choose b = M/7 but we can choose b = 1 instead. Thus we choose b to be M/7 if M /7 1, and 1 if M /7 > 1. This is the minimum of 1 and M/7, which we write as min(1, M/7). Let M > 0 generic M M/7 > 0 multiply by a positive, 1/7 Choose b = min(1, M/7) (we fill in our formula for b in terms of M) b > 0 since 1 > 0 and M/7 > 0 b 1 and b M/7 definition of minimum For example, if M = 3 we would choose b = min(1, 3/7) = 3/7 as above; if M = ½ we would choose b = min(1, (½)/7) = 1/14; and if M = 50 we would choose b = min(1, 50/7) = 1. The rest follows as in the previous example, so we put this together into a proof. Proof. Let M > 0, and let b = min(1, M/7). Then b > 0 since M > 0, and also b 1 and b M/7. Let x F and suppose x - 1 < b. Then x 2 + 4x - 5 = (x + 5)(x - 1) = x + 5 x - 1 Product Property. Also x + 5 x - 1 x + 5 b by multiplying x - 1 < b by a non-negative, x + 5. But x + 5 = x - 1 + 6 x - 1 + 6 by the Triangle Inequality < b + 6 by adding 6 to x - 1 < b 1 + 6 = 7 by adding 6 to b 1 so x + 5 7 by transitivity. Thus x + 5 b < 7b by multiplying x + 5 < 7 by a positive, b 7(M/7) = M by multiplying b M/7 by a positive, 7. Therefore x 2 + 4x - 5 < M. by transitivity. Here's a similar example. Try it for yourself before reading the proof below. Example 1.3.5. Prove that for all M > 0 there exists b > 0 such that for all x F, if x - 2 0. Choose b = min(1, M/5). Then b > 0, b 1, and b M/5. Let x F and suppose x - 2 < b. Then x 2-4 = (x - 2)(x + 2) = x - 2 x + 2 by the Product Property b x + 2 by multiplying x - 2 < b by a non-negative, x + 2. But x + 2 = (x - 2) + 4 x - 2 + 4 by the Triangle Inequality < b + 4 by adding 4 to x - 2 < b 1 + 4 = 5 by adding 4 to b 1. So x + 2 < 5 by transitivity. Thus x 2-4 b x - 2 from above < 5b by multiplying x + 2 < 5 by a positive, b 5(M/5) = M by multiplying b M/5 by a positive, 5 Therefore x 2-4 M. by transitivity

Section 1.4. More Multiple Quantifiers We conclude this chapter with a multiple quantifier example involving division. Example 1.4.1. Prove that for all M > 0 there exists b > 0 such that for all x F, x 2/3, if x + 4 < b then x + 4 / 3x - 2 < M. Hint: Use the Reciprocal Property. We start with the logical structure then work backwards from the conclusion. Wherever we can we will reuse techniques from the previous examples. Proof Analysis. Let M > 0 Choose b =, so b > 0 Let x F, x 2/3 Suppose x + 4 < b Reach x + 4 / 3x - 2 < M generic M (we will fill this in later) generic x conclusion There is no polynomial to factor but we can rewrite x + 4 / 3x - 2 as x + 4 ( 1 / 3x - 2 ). So we can relate the to the conclusion just by multiplying by 1 / 3x - 2, which is positive: Suppose x + 4 1 / M and thus as 3x - 2 > b / M. Since we need an upper bound on 1 / 3x - 2, we want a lower bound on its reciprocal 3x - 2. As in the previous examples, to make the x + 4 < b easier to work with we restrict b by adding the condition b 1 x + 4 < 1 from x + 4 < b by transitivity, PROVIDED b 1. In order to use the we need to rewrite 3x - 2 in terms of the x + 4. As before we use the Triangle Inequality, first factoring out the 3 then using our subtract and add technique. 3x - 2 = 3(x - 2/3) = 3 x - 2/3 Product Property = 3 x + 4-4 - 2/3 = 3 x + 4-14/3 subtract and add 4 3( x + 4 + -14/3 ) Triangle Inequality = 3( x + 4 + 14/3) = 3 x + 4 + 14 definition of absolute value. 3 x + 4 + 14 < 3 + 14 = 17 multiply x + 4 < 1 by a positive, 3, and add 14 3x - 2 < 3b + 17 transitivity Unfortunately, the inequality sign is the wrong way around: we need 3x - 2 > b / M rather than 3x - 2 > b / M. Sometimes we just have abandon an approach and all of the work involved and try again another way. Since we can't get what we want from Triangle Inequality, we try the Unpacking Property instead, starting from the x + 4 < 1. -1 < x + 4 < 1 Unpacking Property -3 < 3x + 12 < 3 multiply by a positive, 3

-3-14 < 3x + 12-14 < 3-14 add -14-17 < 3x - 2 < -11 simplify Remember, we want an expression involving 3x - 2, not 3x - 2. By transititivy, 3x - 2 is negative here, so 3x - 2 = -(3x - 2), and thus 3x - 2 = - 3x - 2. So we can substitute - 3x - 2 for 3x - 2. We will then need to multiply all sides of our inequality by -1, which will reverse all inequality signs. 3x - 2 < 0 transitivity, since -11 < 0 3x - 2 = -(3x - 2) definition of absolute value 3x - 2 = - 3x - 2 multiply by -1-17 < - 3x - 2 < -11 substitute into -17 < 3x - 2 < -11 (-1)(-17) > (-1)(- 3x - 2 ) > (-1)(-11) multiply by a negative, -1 (reversing inequality signs) 17 > 3x - 2 > 11 simplify We want 3x - 2 > b / M and we have 3x - 2 > 11, so by transitivity all we need now is 11 b / M, which means b 11M since M > 0. Thus, PROVIDED b 1 and PROVIDED b 11M, we can reach 3x - 2 > b / M. To satisfy both of these conditions, we choose b to be the minimum of 1 and 11M. Choose b = min(1, 11M), so b > 0, b 1, and b 11M Suppose x + 4 3x - 2 > 11 as above Thus 3x - 2 / b > 11 / b multiply by a positive 1/b so b / 3x - 2 0 and let b = min(1, 11M), so b > 0, b 1, and b 11M. Let x F, x 2/3, and suppose x + 4 < b. Then x + 4 / 3x - 2 < b / 3x - 2 by multiplying by a positive, 1/ 3x - 2. But x + 4 < 1 by transitivity, since b 1. Then -1 < x + 4 < 1 by the Unpacking Property, so -3 < 3x + 12 < 3 by multiplying by a positive, 3, so -3-14 < 3x + 12-14 < 3-14 by adding -14 to all sides, and thus -17 < 3x - 2 < -11. Since -11 < 0 we have 3x - 2 < 0 by transitivity, and thus by the definition of absolute value 3x - 2 = -(3x - 2) and 3x - 2 = - 3x - 2. Hence -17 < - 3x - 2 < -11 by substitution. Then (-1)(-17) > (-1) 3x - 2 > (-1)(-11) by multiplying by a negative, -1, so 17 > 3x - 2 > 11, so 17 / b > 3x - 2 / b > 11 / b by multiplying by a positive, 1/b. so b / 3x - 2 < b / 11 by the Reciprocal Property. But b / 11 11M / 11 = M by multiplying b 11M by a positive, 1/11. Thus x + 4 / 3x - 2 < M by transitivity. Finally, we summarize the main techniques from this section, which we will use repeatedly throughout the rest of this book.

Summary of Useful Techniques for Proofs with Multiple Quantifiers 1. Logical Structure: Use "Let" for variables from "for all" and "Choose = " for variables from "there exists", preserving the order of the quantifiers. 2. Technique: To simplify absolute values of polynomials try factoring then using the Product Property. 3. Technique: BOUND FACTORS. To reach p(x) q(x) < M when you can choose b so that p(x) < b, find a bound c so that q(x) < c and choose b = c / M. 4. Technique: SUBTRACT AND ADD. To convert an inequality involving x - r into one involving x - s, rewrite x - r as x - s + s - r then use either the Triangle Inequality or the Unpacking Property. 5. Technique: To simplify chosing a value of b > 0 to satisfy an inequality, add an upper bound restriction on b (say b 1). Practice Exercises on Proofs with Multiple Quantifiers Prove each of the following: 1. For all M > 0 there exists b > 0 such that for all x F, if x + 3 0 there exists b > 0 such that for all x F, if x + 1 0 there exists b > 0 such that for all x F, x 2, if x - 2 < b then x - 2 / x + 2 < M.