Discrete Mathematics. W. Ethan Duckworth. Fall 2017, Loyola University Maryland

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1 Discrete Mathematics W. Ethan Duckworth Fall 2017, Loyola University Maryland

2 Contents 1 Introduction Statements Constructing Direct Proofs Logical Reasoning Logical Combinations Boolean Algebra Introduction to Sets Quantifiers Constructing and Writing Proofs in Mathematics Divides and more direct proofs More methods of proof Proof by Contradiction Induction Ordinary Induction Variations on proof by induction Properties of Sets Operations Algebraic properties of set operations Functions Basic Definitions Injections, Surjections, Bijections Relations Relations Partial Orders and Equivalence Relations Cardinality Basic properties and Countable Sets Unions Rational Numbers

3 CONTENTS Uncountable Infinities

4 Chapter 1 Fundamentals, aka the language of mathematics 1.1 Statements, Exploring and Establishing Truth, Conditional Statements, Number Systems and Closure The topics in this section are sort of a grab-bag: different things that don t necessarily go together but that we ll combine anyway. It s kind of too bad that it s a grab-bag, I d rather have ideas that belong together and flow from one into the other. But, in a way, this is kind of inevitable in the first section, we have to introduce a few different, very basic things, before we start building up from there. The four basic things we introduce are: statements, exploring and establishing truth, conditional statements, number systems and closure Statements I can t exactly define what a mathematical statement is, but I know some of the properties it should have (from here on out, statement will be short for mathematical statement ). It should be a complete sentence in that it has a subject, a verb, and an object. It won t always be written in English (or some other natural language) as a complete sentence, but between the English and any mathematical symbols that appear it must be complete. It should make a logical assertion, i.e. the verb must imply something that is logical and is true or false. It must be unambiguously true or unambiguously false. In other words, it cannot be sometimes true and sometimes false. Example 1. For each of the following, is it a statement? (a) There exists a real number x such that x 2 = 50 (b) Let x = 3 ± 47 (c) Solve the equation 2x 2 3x 5 = Exploring and Establishing Truth You know, I can t define what truth is. Not even a little bit. I m not sure professional philosophers can either, but I know they spend a lot of time arguing about how to show something is true, and 4

5 CHAPTER 1. INTRODUCTION 5 also about which things are true. But in this class, we ll simply accept that some statements are true, and all other statements are false. Anything that could be both, or neither, or could change, is not a statement. How do we explore and establish truth? In math, we establish truth in exactly two ways: We prove a statement is true; or we provide a counter example that shows a statement is false. All other statements, for which we have no proof or no counter example, are unsolved, or unknown. How do we explore whether a statement is true? Examples. The more examples the better. It s also better if we look for ways we can vary our examples so they show all the relevant cases in a statement. Examples might be explored logically, or they might be explored using a computer, with graphs, calculations, etc. Similar statements and prior knowledge. Most statements are not made up in vacuum, they almost always relate to similar statements that were made before. If they are similar enough, we might have a good guess as to the truth value of the new statement. Example 2. Describe how to explore the following statements (a) If x and y are odd integers, then x + y is an even integer. (b) For all real numbers x, we have sin(2x) = 2 sin(x) Conditional Statements Of all the logical constructions in mathematics, the conditional statement, or implication, is the most important, and the one most often misunderstood. Here s what we mean by a conditional statement. Definition The statement If P, then Q. is a conditional statement. We assume that P and Q are statements as well. A conditional statement is a combination of P and Q, so in this sense we are making bigger statements out of smaller ones. People are often mistaken in thinking about the truth value of a conditional statement, however, thinking clearly about one example is all that is required to get it right. Example 3. Suppose I tell a student If you get 100% on the final exam, then you will get an A in the class. Now, consider the following situations, and decide if I told the truth or was a liar: The student got an 80% on the final, and I gave them an A. The student got an 80% on the final, and I gave them a B. The student got a 100% on the final, and I gave them an A. The student got a 100% on the final, and I gave them a B. I hope you ll agree with me that there would be only one situation in which I was a liar: the last. If we define two statements P : student gets an 100% on the final Q : I give the student an A in the class then If P, then Q is false only when P is true and Q is false. In all other situations If P, then Q is true.

6 CHAPTER 1. INTRODUCTION 6 Definition If P and Q are mathematical statements, then the new statement If P, then Q has truth value defined by the following: P Q If P, then Q T T T T F F F T T F F T We call P the hypothesis and Q the conclusion 1 As shown here, when P is false, we have that if P, then Q is true. We call this the vacuous case, or say that if P, then Q is vacuously true. This terminology is meant to suggest that there is no interesting property in this case, or content, or need to prove anything. Definition The following are synonyms for If P, then Q : P = Q, P implies Q, P is sufficient for Q, Q is necessary for P, P only if Q, Q if P, If P, Q. It s a good habit, that I will usually follow, to state things in the clearest way possible: If P, then Q. Use one of these synonyms only if you are really sure that it makes your communication clearer. This is where we ended on Friday, September 8 Example 4. The following are conditional statements (they happen to all be true as well). In all cases, we assume that x is an integer. (a) If x is even, then x 2 is even. (b) If x 2 is even, then x is even. (c) If x 2 4 = 0, then x = ± Number Systems: Defining Properties Mathematics is largely reductionist: we take a large, complicated thing and break it into smaller, simpler pieces. We do this for theorems: the Pythagorean Theorem is broken into statements about about parallel lines, and the sum of angles inside a triangle. These results are broken into yet simpler assertions, etc. We do this for definitions. The definition of derivative is given in terms of the definition of limit. This in turn is given in terms of logical statements about the real numbers. 1 The American Heritage Dictionary gives the following definition of hypothesis. 1. A tentative explanation that accounts for a set of facts and can be tested by further investigation; a theory. 2. Something taken to be true for the purpose of argument or investigation; an assumption. 3. The antecedent of a conditional statement. Definition (1) shows that in science and the world at large, a hypothesis is often something you re trying prove, and you ll be done when you prove it. This is contrary to the way it s used in mathematics. Here, a hypothesis is what you start with, what you get to assume, in order to prove something else.

7 CHAPTER 1. INTRODUCTION 7 In general, a mathematical text, and even all mathematical knowledge, has been produced after this reductionist goal has been pursued as far as possible. But there is a limit. If every definition had a prior definition then we would either need an infinite number of definitions, or our definitions would have to be arranged circularly. Mathematics is defined partly by its abhorrence of circular reasoning, and we wish to avoid an infinite list of definitions, so we (meaning mathematicians) agree to the following: we start each book or field with definitions. Within the definitions, we ll pursue the reductionist agenda as far as seems practical, and then perhaps one step further, and then agree to blankly assume those small, simple definitions. From there we will construct the simplest theorems, or the next simplest definitions, and so on. The simplest, smallest definitions then cannot be broken into other definitions. They are then, not really definitions in the mathematical sense at all. They are called primitive terms, and they are simply words which we assume describe something with certain agreed upon properties. In geometry, the primitive terms might be point and line which will not be defined, but will be assumed to satisfy certain simple properties that we think make sense. And then we prove things about them. For example, suppose we thought about defining even this way: An integer is even if it is divisible by 2. Could we break this large, complicated thing into smaller, simpler pieces? Could we define what some of the words in this definition mean? Well, what is an integer? What is 2? What does it mean to divide? It s pretty easy to define divide, but not as easy to define integer. In this class, we will start with the following undefined things, i.e. the following primitive terms. Pseudo-definition. We assume that the reader knows what the natural numbers are: 1, 2, 3, 4,... The number 1 is a natural number, and for any natural number n, we also have n + 1 is a natural number such that n+1 n, and ever natural number equals some finite combination of We abbreviate the set of all natural numbers with the symbol N. Pseudo-definition. We assume that the reader knows what the integers are: the natural numbers, the negatives of these numbers, and 0. Here are some of them, together with dots indicating the others:..., 4, 3, 2, 1, 0, 1, 2, 3, 4,.... We abbreviate the set of all integers with the symbol Z. Definition We assume that the reader knows what the rational numbers are: for each pair of integers, a, and b, when b 0, we define a rational number a. For each integer n we define b n = n and in this way view every integer as also being a rational number. 1 We abbreviate the set of all rational numbers with the symbol Q. Pseudo-definition. We assume that the reader feels like they know what the real numbers are. They can be modeled or visualized as all the points on a line, they can be modeled as decimal numbers (including both infinite and finite ones). We view every rational number as a real number. We abbreviate the set of all real numbers with the symbol R. Definition The set of irrational numbers is the set of those real numbers that are not rational.

8 CHAPTER 1. INTRODUCTION 8 Definition The set of real numbers R has operations + (addition) and (multiplication) defined on it, as well as relations = (equals) and < (less than) defined on it, as well as elements 0 and 1 such that the following hold: 1. Trichotomy: For all real numbers a and b either a < b or a = b or b > a holds, and not more than one of these relations holds. 2. Transitivity: For all real numbers a, b and c 3. Identities: For all real numbers a, we have and if a < b and b < c then a < c 0 + a = a (additive identity property) 1 a = a (multiplicative identity property). 4. Inverses: For each real number a there exists a number a such that a + ( a) = 0 (additive inverse property) and, if a 0, then there exists a number 1 a such that a 1 a = 1 (multiplicative inverse property). 5. Commutative, Associative, Distributive Laws: For all real numbers a, b and c we have (a) a + b = b + a (additive commutative property) (b) a b = b a (multiplicative commutative property) (c) a + (b + c) = (a + b) + c (additive associative property) (d) a (b c) = (a b) c (multiplicative associative property) (e) a (b + c) = a b + a c (distributive property) 6. Ordered Field: For all real numbers a, b, c (a) if a = b then a + c = b + c and a c = b c (b) if a < b then a + c < b + c (c) if a < b and c > 0 then a c < b c 7. Rationals are Dense: For every real number a and every real number ε > 0, there exists a rational number q such that a q < ε. We abbreviate some of the above notation: a b means a < b or a = b a > b means b < a ab and a b both mean a b a b means a + ( b) a 1 means 1 a a b means ab 1

9 CHAPTER 1. INTRODUCTION 9 Note: all of the properties in the previous definition hold for R, but they also hold for Q, and all of them except for the multiplicative inverse property hold in Z. Definition Let X is any of the number systems we described above (natural numbers, integers, rational numbers, real numbers, irrational numbers). Let be any of the standard four arithmetic operations defined above (+,,, ). If the following is true: for all a, b X we have a b X; then we say that X is closed under. If we apply this definition to the case where =, then we add the extra detail that b 0, so then the definition reads for all a, b X with b 0, we have a b X. In general in mathematics we use the word closed with slightly different meanings in different contexts. Here we ve said certain number systems are closed with certain operations. We also use the word closed to talk about intervals of the form [a, b], as opposed to open intervals (a, b). The word is used again in Analysis for certain kinds of sets. In all these contexts, though the specific meaning is a little different in each one, the word closed always means that a set contains things that are somehow produced by starting with elements of that same set. Example 5. (a) Z is closed under addition: if a and b are integers, then so is a + b. (b) Z is not closed under division: a = 5 and b = 2 are integers, but 5 2 is not. Example 6. Fill in the following table with C (for closed) or N (for not closed) N Z Q R irrationals Number Systems: Basic Propositions This section is a round up of basic algebraic properties of the real numbers. Everything here can be proven (and should be at some point) by the properties in Subsection Theorem (Algebraic Properties of R). The following properties hold for any a, b, c R: (a) If a + b = a then b = 0. (b) If a + b = a + c then b = c. (c) 0 a = a 0 = 0. (d) ( a) = a. (e) a = 1 a. (f) a ( b) = ( a) b = (a b). (g) ( a) ( b) = a b.

10 CHAPTER 1. INTRODUCTION 10 (h) a (b c) = a b a c. (i) If a b = a c, and a 0, then b = c. (j) If ab = 0 then a = 0 or b = 0. (k) 1 > 0. (l) If a > b > 0 then a 2 > b 2. (m) If a < b, and c < 0 then a c > b c. (n) If a < 0 and b > 0 then a b < 0. If a < 0 and b < 0 then a b > 0. (o) If a Z then there is no integer d Z such that a < d < a + 1. (p) There exists some n Z such that n b > a. Proof. Part (a). Suppose a, b R such that a + b = a. By the additive inverse property a exists. Add a to both sides to get a + (a + b) = a + a. By the associative property we can rewrite the left hand side to get ( a + a) + b = a + a. By definition of the additive inverse we have a + a = 0 and so this becomes 0 + b = 0. (In the future we will combine the last two steps by saying subtract a from both sides.) By definition of the additive identity this means that b = 0. Part (b). Suppose a, b, c R such that a + b = a + c. By the additive inverse property a exists. Adding a to both sides we get a+(a+b) = a+(a+c). Using the associative property we have ( a + a) + b = ( a + a) + c. By definition of the additive inverse we have 0 + b = 0 + c. (In the future we will combine the last two steps by saying subtract a from both sides.) By definition of the additive identity we have b = c. Part (c). Let a R. Note that 0 = and so 0a = (0 + 0)a. Apply the distributive law to get 0a = 0a + 0a. Subtract 0a from both sides to get 0a 0a = 0a + 0a 0a which comes 0 = 0a. Part (d). Let a R. By definition of the additive inverse we have a + ( a) = 0. Subtract ( a) from both sides to get a = ( a). Part (e). Let a R. Note that 0 = 1 + ( 1) and so 0a = (1 1)a. Apply the distributive law to get 0a = 1a + ( 1)a. Using the previous part, and the defining property of the multiplicative identity, we get 0 = a + ( 1)a. Subtracting a from both sides gives a = ( 1)a. Part (f). Let a, b R. By the previous part we have b = ( 1)b and a = ( 1)a and (ab) = ( 1)(ab). Then the following three expressions may be rewritten ( ) ( ) a ( b) = a ( 1)b, and ( a) b = ( 1)a b, and (a b) = ( 1)(ab). All the right hand sides are equal by the associative and commutative properties. Part (g). Let a, b R. Applying part (e) gives ( a) ( b) = ( 1)a ( 1)b Applying the commutative and associative properties gives ( ) RHS above = ( 1)( 1) ab Applying part (d) gives RHS above = 1ab

11 CHAPTER 1. INTRODUCTION 11 and applying the defining property of the multiplicative identity gives Part (h). Let a, b, c R. Then part (e) gives RHS above = ab. a(b c) = a(b + ( 1)c) Now we can apply the distributive property to get RHS above = ab + a( 1)c and the associative property and part (f) gives RHS above = ab ac. Part (m). Let a, b, c R with a < b, and c < 0. Subtract c from both sides of c < 0 to get 0 < c. Multiply both sides of a < b by c to get a( c) < b( c). Use part (f) to rewrite this as ab < bc. Add ab to both sides to get 0 < bc + ab and add bc to both sides to get bc < ab. Part (j). Let a, b R. We will prove if ab = 0 then a = 0 or b = 0 by contrapositive: We ll assume that a 0 and b 0 and prove that ab 0. There there are four cases: Case 1: a > 0 and b > 0, Case 2: a > 0 and b < 0, Case 3: a < 0 and b > 0, Case 4: a < 0 and b < 0. Of course, cases 2 and 3 are essentially the same, so we will just argue one of them. Case 1. Let a > 0 and b > 0. Apply Real Number Property 6 part 6c and multiply both sides of b > 0 by a to get ab > a0. By Proposition this becomes ab > 0, which shows that ab 0. Case 2. Let a > 0 and b < 0. Using the same results as before, multiply both sides of b < 0 by a to get ab < a0. This becomes ab < 0, which shows that ab 0. Case 3. This is the same proof as case 2, with a and b switched. Case 4. Let a < 0 and b < 0. Apply Theorem part (m) and multiply both sides of a < 0 by b to get ab < a0. This becomes ab < 0, which means that ab 0. We have shown that there are exactly 4 cases, and in every case we conclude that ab 0. Therefore, we have proven if a 0 and b 0, then ab 0. By contrapositive, this shows if ab = 0 then a = 0 or b = 0. Part (i). Let a, b, c R with a b = a c, and a 0. Subtract ac from both sides to get ab ac = 0. Factor (i.e. apply the distributive law) to get a(b c) = 0. By part (j) we see that either a = 0 or b c = 0. Since we assumed that a 0 we conclude that b c = 0. Then add c to both sides to get b = c. Part (k). (Probably we should have assumed this as part of our defining properties of R, but oh well.) Suppose this is not the case. Then by the trichotomy property we have either 1 = 0 or 1 < 0. If 1 = 0 then all natural numbers are 0, which is not the case since we assumed that n + 1 n for all natural numbers. If 1 < 0 then apply part (m) and multiply the inequality a < b by 1 to get 1a > 1b i.e. a > b. But this contradicts a < b, and so 1 < 0 is false. Part (l). Let a, b R and suppose that 0 < b < a. Then multiply both sides of b < a by b to get b 2 < ab. Multiply both sides of b < a by a to get ab < a 2. Combine these inequalities to get b 2 < a 2.

12 CHAPTER 1. INTRODUCTION 12 Part (n). Let a, b R. Suppose a < 0 and b > 0. Apply part Definition part 6c and multiply a < 0 by b to get ab < 0. Suppose a < 0 and b < 0. Apply part (m) and multiply a < b by b to get ab > 0. Part (o). Let a Z. Case 1: a > 0. Case 2: a 0. In case 1, we have that a is a natural number. By definition of the natural numbers, every natural number equals one of the following: 1, 2 = 1 + 1, 3 = , etc. Furthermore, 1 < 2 < 3 <.... As a consequence, 1 is the smallest natural number, i.e. 1 n for all n Z. Therefore, there is no natural number d such that 1 < d < 2. Because if there were, we would have d = m + 1 for some natural number m, and then m < 1. Similarly, there is no d between 2 and 3, or between 3 and 4, and by induction no d between a and a + 1. Case 2: if a 0 then apply the result just proven to a + 1. Part (p). Let a, b R. We first prove this the case where a, b N. Let n = a + 1 and note the following b 1 ab a ab + b > a (a + 1)b > a nb > a Now suppose that a, b are positive rational numbers. Then let a = c/d and b = e/f. Then nb > a n(e/f) > c/d n(ed) > cf and the list line follows from the case just proven for integers. Now suppose that a, b are positive real numbers. Let a and b be rational numbers such that b > b and a > b. Apply the case just proven for rational numbers to get nb > a. Then nb > nb > a > a. Finally, we leave the case where a or b is negative to the reader. Definition Let n be an integer. We say that n is even if there exists an integer k such that n = 2k. We say that n is odd if there exists an integer k such that n = 2k + 1. Theorem (Parity Results). 1. Every integer is either even or odd, and is not both. 2. The following hold: even + even = even even + odd = odd odd + odd = even even even = even even odd = even odd odd = odd We interpret the above formulas as follows: if we add two even numbers, the result is an even number; if we multiply two odd numbers, the result is an odd number, etc.

13 CHAPTER 1. INTRODUCTION 13 Proof. Part (1). We don t have the right proof technique at present to prove this part, but we ll borrow from the future the technique of proof by induction. Specifically, we ll prove the given statement for the natural numbers, and leave it to the reader to extend the proof to the set of integers. To be clear, we prove the following statement: for all n N we have that n is even or odd but not both. For k = 1 we have that k is odd since it equals Now we prove that it is not even. If it was even then 1 = 2m for some m Z. Then m = 1/2. Furthermore, if we divide the inequality 0 < 1 < 2 by 2 we get 0 < 1/2 < 1 meaning that 1/2 is between two successive integers. This contradicts part (o) of the previous theorem. Therefore k cannot be both even and odd. Now we assume our result holds for all natural numbers 1,..., k. Then k 1 is either even, or odd, but not both. Case 1: suppose k 1 is even. Then k 1 = 2m for some m Z. Then k = 2m + 1 which shows that k is odd. Now we show that k cannot also be even. If k was even, then k = 2m and k 1 = 2m 1 = 2(m 1) + 1 which would show that k 1 was odd. But k 1 cannot be both even and odd, and therefore k cannot be even. Case 2: suppose k 1 is odd. Then k 1 = 2m + 1 for some m Z. Then k = 2m + 2 = 2(m + 1) which shows that k is even. Now we show that k cannot also be odd. If k was odd, then k = 2m + 1 and k 1 = 2m which would show that k 1 was even. But k 1 cannot be both even and odd, and therefore k cannot be odd. Part (2). Let x, y Z. We have three cases: Case 1: both x and y are even; Case 2: both x and y are odd; Case 3: one of x and y is even and one is odd. Case 1: x = 2n and y = 2m for some n, m Z. Then x + y = 2(n + m) and xy = 2(2nm) which shows that x + y and xy are both even. Case 2: x = 2n + 1 and y = 2m + 1 for some n, m Z. Then x + y = 2(n + m + 1) and xy = 2(2nm + n + m) + 1 which shows that x + y is even and xy is odd. Case 3: x = 2n and y = 2m+1 for some n, m Z. Then x+y = 2(n+m)+1 and xy = 2(nm+n) which shows that x + y is odd and xy is even. 1.2 Constructing Direct Proofs Finally, we get to start proving things. Everything in this section that we prove can be interpreted as being in the form If P, then Q. Sometimes the if part will be implicit, as in the next example. Example 1. Translate each of the following statements into an if-then statement. (a) the product of two odd integers is odd. (b) 2 is irrational. Solution: (a) Here s one approach: If we have two odd numbers, their product is also odd. Here s a different way of writing it: If a and b are odd numbers, then the product ab is odd. Each of these has some advantage over the other. The first approach is very classical, it s the way Euclid would have written it, some 2000 years ago. Also, it s the way you would describe this result to someone else verbally. Finally, most excellent mathematical writers would advocate the following rule: If you don t have to name or give a symbol to something, don t. So the first approach has this going for it. On the other hand, the second version labels the numbers as a and b. If you were about to start proving this result, you would probably have to label these things, so you could start to work with them, form equations, etc. Also, perhaps, contrary to the advice in the previous paragraph, it feels more specific when we label them.

14 CHAPTER 1. INTRODUCTION 14 (b) Here s one approach: If you take the square root of 2, the result is irrational. This is a great start, but we need to go farther. It s not clear what kind of mathematical statement is the hypothesis here: you take the square root of 2. Is that a mathematical statement? Is it unambiguously true or false? Not quite. This is a good start, but we can do better. This is where we ended on Monday, September 11 A proof is an argument that something is logically true. Finding a proof may involve calculations, guessing, dead ends, applying definitions, applying other theorems, and following patterns of other proofs. But, the finished result for the proof should consist of the following: a discrete set of steps, each one being a clear, unambiguous statement, with each logical assertion given a justification. (Unless the justification is so basic that we assume the reader knows and remembers it without any mention being needed.) These steps should connect the starting statement with the conclusion. The starting statement might be a hypothesis, or a definition, or primitive terms. To prove an if-then statement we always start by assuming the hypothesis; in other words we take the hypothesis is a given truth, or more just given. The reason for this is that if the hypothesis is false, then there s nothing to prove, the statement is vacuously true. Here is a template for finding and writing a direct proof of an if-then statement: Template. Template for finding a direct proof. 0. Translate what you ve been given into an if-then format. (You do not have to write this step down, but it should help clarify things.) 1. Write the first sentence(s) of your proof by stating the hypothesis of your if-then, label any quantities you need and explicitly state what you re assuming about them. 2. Write the last sentence of your proof by stating the conclusion you re trying to reach. 3. Working either forwards or backwards, add sentences in the middle. Almost every sentence should be a mathematical statement, along with some justification for why that statement is true. Sentences that are not mathematical statements should be concise aids for the reader to help them understand the proof, or that define, introduce, or fix a term or symbol. 4. Keep going until you have a chain of links from the beginning to the end. steps 0, 1 and 2 should become automatic to you after you write a few proofs. They may still take a little practice, but there should be no question about what you re trying to do when you re working on them. It should also become automatic to apply or unravel definitions as needed anywhere in a proof. Unravel means that you replace some technical word with what it s definition is. Perhaps the biggest question beginning students have is about what needs to be justified. In practice here are some of the ways this question comes about: 1. I thought I justified a given step, but it turns out my justification needs to be broken down and justified. If I keep doing that, won t this go on forever? 2. C mon, some things don t have to be justified right? Like = 2? But the teacher took off points when I skipped something I thought was obvious? 3. There are always going to be some gaps between things, how small do those gaps have to be for me not to justify them or not to break them down further? I cannot answer these questions in exact terms, that s why they are what students struggle with! But I will say a few comments about each of the above questions. For #2: for now, you should justify every assertion. What counts as a justification? A definition, a previous result, or a previous line in your proof. For now, the only previous results we have are basic logic and what we ve assumed about number systems. So, if I ask you to justify everything, and if = 2 is a significant step in

15 CHAPTER 1. INTRODUCTION 15 your proof (which is hard to believe, but you never know... ), you could say by basic properties of number systems. Later, of course, we won t justify everything. For #1 and #3, I ll say that no, it doesn t go on forever, and so the gaps aren t infinitely small. But, especially for the most elementary results, like the ones we ll practice at the beginning of the class, please think about the following. Always break the original statement to be proven into at least two simpler statements. Honestly I ve had students basically restate the original proposition into one equivalent statement and then think they were done. Also, always think about what material we were working with just before a given proposition. There will always be some definitions and/or theorems that come in the preceding pages or lectures. If these definitions or theorems are at all relevant to the proposition you ve been asked to prove, then you should break down your logical statements into small enough steps that you can show where the definitions and theorems are used. The textbook has something similar to the above template that it calls a know-show table. Such a table is pretty similar to something called a two-column proof, which is often used in basic Geometry classes. Template. Know-show table 1. State what you know in the first line of the table. The justification for this statement is that it s the hypothesis. 2. State your conclusion in the last line of the table. 3. Work from the beginning to the middle, and from the end to the middle, by applying (unraveling) definitions, simplifications, restatements, previous results, algebra, etc. Label each step and provide a justification for each step. 4. Keep going until you have a link from the beginning to the end. You may need to try more than one possibility at a time, until you find one statement to connect to the other. Step # Statement Justification Know P... Given P Q Show Q The labels P 1, P 2, etc. represent forward steps and the labels Q1, Q2, etc. represent backwards steps. Naming things Pseudo-definition. A theorem is a mathematical statement that has been proven true. Despite the pseudo-definition just given, in practice, we don t call every proved statement a theorem. We use different names to indicate how important a proved statement is, or how it fits into other results. Here is a list of some of these names. theorem a statement that has been proven true, and is of some importance.

16 CHAPTER 1. INTRODUCTION 16 lemma a statement that has been proven true, and whose main use is to prove other results. Usually this is stated, with a heading and a number, right before the theorem which it is used to prove. fact a statement that has been proven true, but is pretty minor. result a generic word for a statement that has been proven true. proposition a result that is less important than a theorem, but not entirely minor. corollary a result that follows from the previous theorem. claim similar to a lemma, but usually stated within a proof, not as a separate result. scholium a result that has been proven as part of a previous poof, but might be worth recording for later use A lengthy example We ll start with a simple proposition, and analyze the heck out of various proofs of it. To prove anything we need some primitives: we have those in our basic assumptions about number systems, and we can use those already to prove some stuff. Proposition x = 0. Proposition is the only number that acts like the additive identity. Proposition ( a) = a. Discussion. In class I wrote the preceding three results on the board and asked the students to try to prove them on their own. I did warn them that there was an implicit assumption in all three, so to some degree I was sloppy when I wrote them. But, on the other hand, that s the way many results will be written in various books, exercises, etc. So the first step is to figure out the implicit assumption. I asked the class what the hidden assumption was, and after only a few second someone said real numbers. So, I added that to the board, everything in these three propositions was a real number. The second step will be to rewrite these, or one of these, as an if-then statement. After a few minutes of silence, one student came up to the board and wrote the following: If x + y = x then y = 0 It turns out they saw the in the first result as a +. So the same student made the first correction: If x y = x then y = 0 I mentioned that this may be a good result, maybe a step we ll use, but what I want at first is just a restatement of the original Proposition, and this isn t it. A second student came up and offered the following If x is any real number then 0 x must equal 0. This is great. The hidden assumption, as mentioned above, about everything being a real number, is not explicit. And the conclusion is what we want, that 0 x = 0. Now I said they should know two lines of the proof: the first line and the last line. After about a minute, one student came up to the board and wrote the following

17 CHAPTER 1. INTRODUCTION x is a real number... 0 x = 0 I said this was great. Almost perfect. But, I asked if any one could add a word or two just to make it easier for another person to read. The statements here are good logical statements, but we tend to help the reader understand what s going on by adding things like Suppose... or Let... or We assume... or Given.... So, could anyone add something like that? A student made the following suggestion, which added to the board 1. Given x is a real number. I asked next if someone could add a word or phrase like that at the end. Someone suggested Therefore, and I added this, along with a little bit extra: 1. Given x is a real number Therefore 0 x = 0, which were trying to prove. Now I said we were at the hard part. All the steps in the middle. I let the students work on this for the rest of the period. In the first 5 minutes most people were just stuck, maybe one student had a pretty good idea of the proof, but even he needed to practice writing it in the forwards direction and pretty-ing it up. After about 5 minutes I started giving out hints: think about combining addition and subtraction and multiplication; at some point you ll need the distributive law, or, which amounts to the same thing, taking out a common factor. I explained why this kind of makes sense: the result we are proving uses 0, which is defined only by it s properties under addition, and which is multiplication. Out of all the things we ve assumed about numbers, the only property that has any connection at all between addition and multiplication is the distributive law. After another 5 minutes a lot of people started getting close, having results with x and x x and ab ac or something like that. This is where we ended on Wednesday, September 13 At the beginning of the next class I had another student come to the board and add some more lines, hopefully all the lines in the middle. They didn t have to be perfectly written, we could start with the logical steps. 1. Given x is a real number. 2. x x = 0 b/c Additive identity 3. (1 1)x = 0 b/c Factoring 4. 0 x = 0 b/c Additive identity 5. QED 0 x = Therefore 0 x = 0, which were trying to prove.

18 CHAPTER 1. INTRODUCTION 18 This is great. There are two things to do now: take each statement and wrap it in words to make an English language sentence that starts with a capitalized word and ends with a period; and triple check the logic of each step, is it a mathematical statement, is it true, does it follow only from our basic stated properties of the real numbers. This last one is the trickiest, because I can tell at a glance that all these statements are pretty basic. But I really mean is each justification exactly one of the properties I stated in the previous section about the real numbers. And I think that from step 2 to 3, there s more than one property being used. So I d add another step or two: 1. Given x is a real number. 2. x x = 0 b/c Additive identity 2.5 (1x) (1x) = 0 b/c multiplicative identity x + ( 1)x = 0 b/c...? 3. (1 1)x = 0 b/c Factoring 4. 0 x = 0 b/c Additive identity 5. QED 0 x = Therefore 0 x = 0, which were trying to prove. Hm, there appears to be a gap here: I really want to know that (1x) = ( 1)x, but that s certainly not one of our ultra-basic statements about the real numbers. I mean, I know it s true, but it s not listed in that list of properties in the last section. OK, so now I have a new goal Let s reason backwards prove: (1x) = ( 1)x (1x) = ( 1)x x = ( 1)x 0 = ( 1)x + x 0 = ( 1)x + 1x 0 = ( 1 + 1)x 0 = 0x Great, are we done? No. This was backwards reasoning: our intermediate step that we are trying to prove now, that (1x) = ( 1)x, can be used to prove that 0 = 0x. But we already knew that, and we still haven t proven (1x) = ( 1)x. OK, so this is still a gap, and maybe it s a dead end. Let s see if we can finish it off next time. This is where we ended on Friday, September 15 At the beginning of class today we had two more people come up and fill in their attempts at proofs. To the best of my recollection this is what they wrote 1. Given x is a real number = x + 0x = 0x mult. both sides 4. 0x + 0x 0x = 0x 0x additive identity 5. 0x + (0x 0x) = (0x 0x) associativity 7. 0x = 0, QED. This is very good. I d see a small gap or two here or there, but nothing that can t be filled in easily:

19 CHAPTER 1. INTRODUCTION Given x is a real number = (0 + 0)x = 0x mult. both sides 3. 0x + 0x = 0x distribute 4. 0x + 0x 0x = 0x 0x additive identity 5. 0x + (0x 0x) = (0x 0x) associativity 5.5 0x + 0 = 0 definition of additive identity 6. 0x + 0 = x = 0, QED. Of course, then we should wrap these in words (I ll indicated below what this might look like). The second student wrote something like this If I m being super-picky, I d add Given x is a real number. 2. x + ( x) = 0 additive inverse 3. 1x + ( x) = 0 multiplicative identity 4. (0 + 1)x + ( x) = 0 additive identity 5. 0x + 1x + ( x) = 0 distributive 6. 0x + 0 = 0 additive inverse 7. 0x = 0 additive identity 8. 0x = 0, QED. 1. Given x is a real number. 2. x + ( x) = 0 additive inverse 3. 1x + ( x) = 0 multiplicative identity 4. (0 + 1)x + ( x) = 0 additive identity 5. 0x + 1x + ( x) = 0 distributive 5.5 0x + x + ( x) = 0 mult identity 6. 0x + 0 = 0 additive inverse 7. 0x = 0 additive identity 8. 0x = 0, QED. I think in both proofs I could add a few steps with the associative identity. Honestly though, I find that wading through any steps with the associative identity a bit tedious, but on the other hand, if that s the task at hand, then I can do it. As far as writing this up with words, it should be easy. There are few standard constructions and a few standard phrases that can go in each one. E.g. By So From justification statement statement 1 conclude we see that because of we apply statement 2 etc. statement justification justification... and

20 CHAPTER 1. INTRODUCTION 20 If we apply this to the first proof we get something like the follow (at least as a start). Formal proof: Let x be any real number. Since 0 is the additive identity, we have 0+0 = 0. Multiplying both sides of this by x gives (0+0)x = 0x. Applying the distributive law gives 0x+0x = 0x. Now let s add 0x to both sides of this equation: (0x+0x) 0x = 0x 0x. Applying the associative property and the additive inverse property we get 0x + 0 = 0. From this we see that 0x = 0, QED. If we apply this to the second proof, and changing some words around a little just for a different style, we get something like the following: Formal proof: Let x R be given. Then the additive inverse property shows that x + ( x) = 0. By the multiplicative identity we have x = 1x and so we get 1x + ( x) = 0. Applying the additive identity we get 1 = and so (0 + 1)x + ( x) = 0. Now we distribute, ( 0x+1x ) +( x) = 0, apply the associative property and undo 1x = x to get 0x+0 = 0. Since the left hand side has the additive identity, this is equivalent to 0x = 0, QED. For the sake of contrast, here s a rather unpleasantly written, but clear and correct proof: Formal proof: Let x R. We will justify the following: 0 1 = x + ( x) 2 = 1x + ( x) 3 = (0 + 1)x + ( x) 4 = ( 0x + 1x ) + ( x) 5 = 0x + ( 1x + ( x) ) 6 = 0x + ( x + ( x) ) 7= 0x = 0x. Equality 1 is from the additive inverse property, equality 2 from multiplicative identity, equality 3 from additive identity, equality 4 from distribution, equality 5 from associativity, equality 6 from multiplicative identity, equality 7 from additive inverse, and equality 8 from additive identity. Why do I think this proof is unpleasant? Well, it s sort of the least friendly proof possible, a machine gun of logical statements and justifications with not even the slightest variation, emphasis, etc. Aside from that, it requires the reader to keep one eye on the equations and one eye on the justifications (or maybe one finger and one pencil, etc.) simultaneously. Does it have anything going for it? Sure: the chain of statements is crystal clear. I can imagine using such a proof, but only in the most severe of circumstances, when the chain of equalities I have is somewhat unclear and therefore could be improved by this sort of display, and the reasons are somewhat shallow and could therefore be stacked up all in a row. Long story short: avoid such proofs unless you are prepared to defend your choice to an unsympathetic reader (teacher, grader). This is where we ended on Monday, September 18

21 Chapter 2 Logical Reasoning 2.1 Logical Combinations And, Or, Not Definition If P and Q are mathematical statements, then new statement P and Q has truth value defined by the following: There s a symbolic way of writing and, We call and or conjunction Example 1. P Q P and Q T T T T F F F T F F F F P Q means P and Q P : rocks are hard J : fish have legs P and J : rocks are hard and fish have legs Definition If P is a mathematical statement, then the new statement not P is defined by the following: P not P T F F T There s a symbolic way of writing not, P means not P Finally, not and are both also called negation. 21

22 CHAPTER 2. LOGICAL REASONING 22 Example 2. P not P (suitably rewritten) rocks are hard rocks are not hard = Next we ll discuss the logical connector or. This requires some extra care because how we use it in math is a little different from how we use it in everyday English. A lot of times people use or in an exclusive sense, to compare two alternatives that cannot both happen: Would you like soup or salad? In everyday English: you can have soup, or salad, but not both. This is called an exclusive or. However, there are times in everyday English where we use or differently. In these situations, we know by context, that both alternatives can happen at the same time: People who get 100% on homework or 100% on tests will pass the class. This is an inclusive or, and people who do both will certainly pass the class. In mathematics, we need to be precise with our definitions, and we will pick one meaning of or and always use it: the inclusive definition. Definition If P and Q are mathematical statements, then the new statement P or Q has truth value defined by the following P Q P or Q T T T T F T F T T F F F There s a symbolic way of writing or, We call or or disjunction Example 3. P Q means P or Q P : Linnane is Loyola s President J : Duckworth is Superbowl MVP P or J : Linnane is Loyola s President or Duckworth is Superbowl MVP We now have four logical connectors and/or constructors: P = Q, P Q, P Q, and P. In some sense you can learn these by memorizing the truth tables, but in each case you should also learn to verbally describe what s going on. It might look like this: P = Q True except when P is true and Q is false P Q True when both are true P Q True when either are true P True only when P is false

23 CHAPTER 2. LOGICAL REASONING 23 Exclusive Or and Truth Tables Example 4. We have decided to let represent inclusive or. There is no universal standard symbol for exclusive or, but the following is used fairly often:. Fill in the following. P Q P Q T T F T F T F T T F F F Example 5. Come up with a symbolic formula for exclusive or, and verify that it gives the same truth table as the exclusive or in the previous example. The example we used for exclusive or was soup or salad. To make it extra clear, we added the phrase but not both. Thus, you can think of exclusive or as being the inclusive or, together with the phrase and not both. soup or salad soup or salad, but not both soup or salad, and not both soup or salad and not (soup and salad) P or L and not (P and L) (P L) (P L) To figure out the truth table for (P L) (P L) make a table starting with P and Q, then add a column for each piece of the whole formula (start with the smallest pieces, then the next smallest, etc., finishing with the largest). P L P L P L (P L) (P L) (P L) T T T T F F T F T F T T F T T F T T F F F F T F This is where we ended on Friday, September 15 If and only if Definition If P and Q are mathematical statements, then the new statement P if and only if Q means both of the following: if P, then Q and if Q, then P. Proposition The truth value of P if and only if Q is shown below: P Q P, if and only if Q T T T T F F F T F F F T

24 CHAPTER 2. LOGICAL REASONING 24 Proof. If P and Q are both true, then we can say P = Q and Q = P. If P and Q are both false, then we can say P = Q and Q = P by using the last row of Definition If P is true, and Q is false, then P = Q is false, from the second row in Definition Therefore P if and only if Q is also false. Similarly, if P is false and Q is true, then Q = P is false, and so P if and only if Q is false. One way to think about P if and only if Q is that for this to be true, P and Q need to have the same truth value. Definition The following are synonyms for P if and only if Q : 1. P Q 2. P iff Q 3. P is necessary and sufficient for Q 4. P is equivalent to Q 5. P is true exactly when Q is true. 2.2 Boolean Algebra Ordinary algebra uses letters to stand for numbers, studies what happens when you combine the letters using certain operations, and identifies the laws that apply to these combinations. Boolean algebra uses letters to stand for just two things: the values T (TRUE) and F (FALSE). We combine these letters using the logical connectives we ve seen before: and, not and or, if then, and if and only if. There s one further change of perspective in some of the discussion below. Before, when we had a statement like P = Q, we imagined P and Q standing for fixed values of Tand F. Now, at least some of the time, we ll imagine P and Q being variables. This is the same difference you had to come to terms with when you first learned the algebra of real numbers. When you first learned that you probably were told something like x represents a single unknown number. Therefore, in the statement 3x + 1 = 4, there was only one number that x could be. Later, you began to view x as a variable, so that it was allowed to take an infinite range of numbers. This makes a psychological difference when you see an expression like y = 3x+1. Something similar could happen here, e.g. I could say P T = F and then you could solve for the value of P. Or I could say (P = Q) T = F and then we can discus what range of values P and Q can have. For us, the purpose of Boolean algebra is to highlight some of the laws that describe logical combinations, and to make these laws easy to verify. We finish this section with two examples of such laws, and then a theorem that records most of the ones we ll use. Definition Two expressions in Boolean algebra are logically equivalent if they have the same truth value for all possible combinations of truth values of these variables. If X and Y are two logically equivalent statements then we can write this symbolically as (X Y ) = T. We typically will abbreviate this as X Y. Example 1. Show that X = Y and ( X) Y are logically equivalent.

25 CHAPTER 2. LOGICAL REASONING 25 Solution: X Y X Y ( X) ( X) Y T T T F T T F F F F F T T T T F F T T T Since columns 3 and 5 are the same, we see that X Y and ( X) Y are equivalent. Note that we can restate this property in familiar English: P = Q means the same thing as it is wrong to have P true and Q false, just as I would be a liar to have a student get 100% on the test, and give them a B in the class. Example 2. Go back and re-read the algebraic properties of the real numbers: identities, inverses, commutativity, associativity, distributive law. In each case see if you can imagine an analogous statement for statements made of truth values. For the moment, don t worry about whether the statement is true or false, just see if you can imagine what it would even look like. Solution: real numbers boolean algebra 0 + a = a? 1 a = a? a + ( a) = 0? a 0 = a 1 a = 1? a + b = b + a? a b = b a? a + (b + c) = (a + b) + c? a (b c) = (a b) c? a (b + c) = a b + a c? The real numbers, and the statements above, have two special numbers: 0 and 1. In boolean algebra we have only two values Tand F. But which of these should correspond to 0 and which to 1? A related question, actually in a deeper sense the same question, which of our operations =,,,, should correspond to + and which to? The operations and are the fundamental ones, so we ll use those. For most of these properties, it doesn t matter whether we think of being analogous to + or, and vice versa with. So there are four choices we can make: Real Boolean 0 F 1 T + or Real Boolean 0 F 1 T + or Real Boolean 0 T 1 F + or Real Boolean 0 T 1 F + All of the above choices will turn some of the algebraic properties of the real numbers into true algebraic properties of Boolean algebra. None of them will turn all the real number properties into true Boolean properties.

26 CHAPTER 2. LOGICAL REASONING 26 The standard choice is the second one. Using that one see if you can fill in the table above (warning: two of the lines will have to change). real numbers boolean algebra 0 + a = a F X X 1 a = a T X X a + ( a) = 0 X ( X) T a 0 = a 1 a = 1 X ( X) F a + b = b + a X Y Y X a b = b a X Y Y X a + (b + c) = (a + b) + c (X Y ) Z X (Y Z) a (b c) = (a b) c (X Y ) Z X (Y Z) X (Y Z) (X Y ) (X Z) a (b + c) = a b + a c and X (Y Z) (X Y ) (X Z) This is where we ended on Monday, September 18 Theorem If X, Y and Z are any Boolean variables, the following properties hold: X TRUE = X and X FALSE = X (Identity elements) X ( X) = FALSE and X ( X) = TRUE (Inverses) X Y = Y X and X Y = Y X (Commutative laws) X (Y Z) = (X Y ) Z and X (Y Z) = (X Y ) Z (Associative laws) X (Y Z) = (X Y ) (X Z) and X (Y Z) = (X Y ) (X Z) (Distributive laws) Proof. Some of these don t need to be proven, e.g. the commutative laws are built in to our definitions. The associative laws do need to be proven, but the proofs don t seem to offer any satisfaction. We ll prove one of the distributive laws. We ll do that using a truth table. Later, if we had more complicated expressions to prove, we might do that by combining simpler results like the ones we have in this theorem. X Y Z Y Z X (Y Z) X Y X Z (X Y ) (X Z) etc. etc. etc. etc. etc. etc. etc. There are a few algebraic properties for boolean algebra that are not analogues of the algebraic properties of the real numbers. We state the most important ones here. Theorem For all boolean variables X, Y and Z, we have ( X) = X and X X = X and X X = X (Idempotence) (X Y ) = ( X) ( Y ) and (X Y ) = ( X) ( Y ) (De Morgan s Laws) Converse and Contrapositive Definition Let P and Q be any statements, and form P = Q. The converse of P = Q is Q = P. The contrapositive of P = Q is Q = P.

27 CHAPTER 2. LOGICAL REASONING 27 Example 3. Consider the following statements X : If x = 3, then x 2 = 9 Y : If x 2 = 9, then x = 3 Z : If x 2 9, then x 3 W : If x 3, then x 2 9 (a) Find the truth value for each of X, Y, Z and W. (b) Which of Y, Z and W is the contrapositive of X? Which is the converse? Proposition Let P and Q be two logical statements. The following are equivalent: 1. P = Q 2. Q = P 3. ( P ) Q Proof. We give a truth table for the values of if P, then Q and if (not Q), then (not P ) : P Q P = Q Q P Q = P T T T F F T T F F T F F F T T F T T F F T T T T. Since the third and last columns are the same, we see that the third and last statements are logically equivalent. Definition Let P and Q be two logical statements. The contrapositive of the statement if P, then Q is the statement if (not Q), then (not P ). Template. Proof by contrapositive To prove if P, then Q by contrapositive, do the following. Clearly state We assume (not Q) (or some synonym). Fill in the rest of the steps of a direct proof of if (not Q), then (not P ). Label any quantities you need and explicitly state what you re assuming about them. Write the last sentence of your proof: Therefore we have (not P ) (or some synonym). Work from the beginning to the middle, and from the end to the middle, by unraveling definitions, simplifying, etc. Keep going until you have a link from the beginning to the end. When should you use a proof by contrapositive? Any time you feel like it. OK, I can be a little more helpful than that: if the condition (not Q) is simpler, or easier to work with or gives you some extra information than the condition P. Example 4. Discuss how one would prove the following statement directly, and by contrapositive Every student at Loyola University is under 8 feet tall. Solution: Let s start by putting this in if-then format: If a person is a student at Loyola University, then the person is under 8 feet tall.

28 CHAPTER 2. LOGICAL REASONING 28 Here s how a direct proof would go. We could go look at every single student at Loyola, all 4000 of them, measure their height, and see. This would be P = Q where P is a person is a student at Loyola University and Q is the person is under 8 feet tall. This would be very difficult, 4000 people is a lot. Here s how a contrapositive would work. Translate Q as the person is 8 feet tall or higher and translate P as the person is not a student at Loyola. Then we want to verify If a person is 8 feet tall or higher, then they are not a student at Loyola. What makes this easier is there aren t very many people over 8 feet tall in the whole world. We could go to Wikipedia and get a list and see that there are only 13 people listed from the last few hundred years who have been over 8 feet tall or higher, and none of them were Loyola Students. This is where we ended on Wednesday, September 20 Example 5. Prove the following statement: Let n be an integer. If n 2 is even, then n is even. (You can assume only our statements about numbers in Section 1.1.4, the definition of even and odd integers, the fact that every integer is either even or odd, and your homework problem about.) Solution: This has the form P = Q where P is n 2 is even and Q is n is even. Then Q is n is odd and P is n 2 is odd. So we want to prove If n is odd then n 2 is odd. Proving this is easier than a direct proof of P = Q because it s easier to work with n than to start with n 2. Proof. We prove this by contrapositive: suppose that n is odd. By definition, this means that we can write n = 2k + 1 for some k Z. Then n 2 = (2k + 1) 2 = 4k 2 + 2k + 1. Let K = k 2 + k and note that K Z. Then n 2 = 2(2k 2 + k) + 1 = 2K + 1. Therefore, n 2 is odd. The perceptive reader may have noticed that we never gave a proof template, or even a proof example, of an if-and-only-if statement. The reason for that is this: In many cases one direction of an if-and-only-if statement is easier proved using a contrapositive. Example 6. Prove the following: Let x and y be real numbers. Then xy = 0 if and only if x = 0 or y = 0. (You can assume only our statements about numbers in Section 1.1.4, and Theorem parts (c), (e) and (m)) Solution: Recall that we break an if and only if statement down into two if-thens. In this case we have: if xy = 0, then x = 0 or y = 0 and if x = 0 or y = 0, then xy = 0. We know how our template for if-then proofs goes too: we start with what we are given, and some number of steps later we will finish with what we were asked to prove. Sometimes we help the reader out by indicating which direction we re going to prove. We always help the reader out by explicitly giving the statement we are assuming and explicitly giving the statement we are going to prove. Filling all of this in, but without doing any of the actual steps in the middle of the proof, this is what we have so far: Let xy = 0. We will prove that x = 0 or y = 0... Therefore x = 0 or y = 0 is true, QED.

29 CHAPTER 2. LOGICAL REASONING 29 Let x = 0 or y = 0. We will prove that xy = 0... Therefore xy = 0, QED. Now we have to fill in some of the middle steps. As before, we can work forwards or backwards. We can also use things we have proven previously. Since we have proven Proposition in Section 1.2, we can use that for the direction. But which variable can we apply it to? In theory, we have two different cases: case 1, where x = 0, or case 2, where y = 0. Of course, the cases are basically exactly the same, but for now, let me pretend that I don t see that, and follow each one. Also, I should add that using cases is usually only needed when the statements that follow each case are complicated enough that it takes more than a single sentence. That may not be the case here, but let me act like it is, just to make the situation clear. Let x = 0 or y = 0. We will prove that xy = 0. We have two cases: case 1 is where x = 0, and case 2 is where y = 0. Case 1: suppose x = 0. Then xy is really 0 y. From Proposition we know that 0 y = 0. Therefore in this case xy = 0. Case 2: suppose y = 0. Then xy is really x 0. From Proposition we know that y 0 = 0. Therefore in this case xy = 0. To summarize: we have exactly two cases, and each of those cases concludes that xy = 0. Therefore, in all cases xy = 0, QED. Now we prove the other direction: every if-and-only-if proof will actually be two proofs. We start by stating the contrapositive. Why do we use the contrapositive? Because if we assume that x 0 and y 0, that gives us something explicit to work with. We will prove if xy = 0, then x = 0 or y = 0 by contrapositive. Assume, for contrapositive that x 0 and y 0. Then there are four cases: Case 1: x > 0 and y > 0, Case 2: x > 0 and y < 0, Case 3: x < 0 and y > 0, Case 4: x < 0 and y < 0. Of course, cases 2 and 3 are essentially the same, so we will just argue one of them. Case 1. Let x > 0 and y > 0. Apply Real Number Property 6 part 6c and multiply both sides of y > 0 by x to get xy > x0. By Proposition this becomes xy > 0, which shows that xy 0. Case 2. Let x > 0 and y < 0. Using the same results as before, multiply both sides of y < 0 by x to get xy < x0. This becomes xy < 0, which shows that xy 0. Case 3. This is the same proof as case 2, with x and y switched. Case 4. Let x < 0 and y < 0. Apply Theorem part (m) and multiply both sides of y < 0 by x to get xy < x0. This becomes xy < 0, which means that xy 0. We have shown that there are exactly 4 cases, and in every case we conclude that xy 0. Therefore, we have proven if x 0 and y 0, then xy 0. By contrapositive, this shows if xy = 0 then x = 0 or y = 0.

30 CHAPTER 2. LOGICAL REASONING 30 This is where we ended on Friday, September 22 If and only If Theorems We know already the logical definition of if and only if. We can adapt the proof template we had for direct if-then proofs to if-and-only-if theorems. Template. Direct proof of if-and-only-if To prove P if and only if Q (i.e. P Q) Prove if P, then Q (i.e. prove ) Prove if Q, then P (i.e. prove ) 2.3 Introduction to Sets Open Sentences Definition An open sentence is a sentence that involves variables x, y, z,... such that when values are assigned to those variables the result is a mathematical statement. Notation: P (x, y, z,... ). Synonyms: such a sentence is also called a predicate sentence or propositional function. Note: an open sentence does not have a truth value until we decide which values are being plugged in to the variables. Furthermore, by definition, we do not specify, or quantify, what values, or what range of values, are getting plugged in. If we do specify, then we have a statement, not an open sentence. Example 1. (a) 3x + 5 = 0 is an open sentence. If x is a real number, then 3x + 5 = 0 is statement (a false statement). There is some real number x such that 3x + 5 = 0 is also a statement (a true one). (b) x 2 0 is an open sentence. For all real numbers x, we have x 2 0 is a statement (a true one). Also There is some real number x such that x 2 0 is a true statement Sets The infinite, like no other problem, has always deeply moved the soul of men. The infinite, like no other idea, has had a stimulating and fertile influence upon the mind. But the infinite is also more than any other concept, in need of clarification. (Hilbert, Über das Unendliche) Pseudo-definition. A set is a collection of objects, with no sense of order or repetition (i.e. we do not assign any meaning to, or keep track of, what order the objects show up in, or whether or not they are repeated). Objects in a set are called elements or members of the set. Usual notation for sets: { } or {,,,..., }. We often represent a set with a capital letter such as A, B, C,....

31 CHAPTER 2. LOGICAL REASONING 31 Order and repetitions do not matter. The following are all the same set: {1, 2, 3}, {3, 2, 1}, {3, 2, 3, 1, 2, 1}. Membership in a set is denoted by. We read this symbol as is in or be in or something synonymous. Thus, 2 {1, 2, 3} reads as 2 is in {1,2,3} and let x A reads as let x be in A. A membership statement is a logical assertion: it is either true or false. For instance, 5 {1, 2, 3} is FALSE. Since this is false, it is true to say that 5 is not a member, which we write this way 5 {1, 2, 3}. Definition (Roster notation). Roster notation is when we list the elements of a set as explicitly as possible, sometimes relying upon a pattern for any that are not explicit. Example 2. (a) In roster notation we would describe N as N = {1, 2, 3, 4,... } (b) In roster notation, the set of prime numbers between 1 and 25 is {2, 3, 5, 7, 11, 13, 17, 19, 23}. Definition (Set-builder notation). Set-builder notation has the form {variable open sentence on the variable} to describe the set of all objects that satisfy the given condition. One condition is so common that we sometimes include it on the left hand side, namely what set the variable can be in. In this case, the notation looks like this {variable set condition on the variable} In all cases there needs to be an implicit or explicit statement about what the set is that the variable ranges over. It is called the universal set for that variable. If we make the notation a little more symbolic, it looks like this {x A P (x)} where A is some universal set, and P (x) is an open sentence defined on the variable x. (Note: half the books out there use : instead of for the divider in the middle between the variable and the open sentence. It doesn t matter which one you use, just be aware that you may see it both ways.) The author Sundstrom calls the set defined this way the truth set for the given open sentence. I.e. we are defining the set as all those elements that make the open sentence true. I m sure he didn t make that term up, but I haven t heard it before. My only point is this: most people just think of this as a set defined by a property, and they don t name the process, e.g. by calling it find the truth set. They just say, define the set this way, now find it explicitly or something like that. Example 3. Translate the following into/out of set-builder notation.

32 CHAPTER 2. LOGICAL REASONING 32 (a) {x x Z, x 2 9}. (b) {x R x 2 = 1}. (c) The set of all even numbers that can be written as the sum of two perfect squares. This is where we ended on Monday, September 25 Solution: (a) In words, this says that we want the set of all integers such that the square of the integer is 9. We can write these down explicitly, { 3, 2, 1, 0, 1, 2, 3}. (b) There is no x that satisfies this equation, so we have { }. (c) {x Z x = 2n for some n Z, and x = a 2 + b 2 for some a, b Z} Definition We say that two sets are equal if they have exactly the same elements. In other words, if A and B are two sets, then A = B means x A x B. We say that A is a subset of B if every element of A is also an element of B. In other words, The notation for A is a subset of B is A subset B means if x A then x B A B. Example 4. Let n Z. We say that 6 divides n if n = 6x for some x Z. Similarly we say that 18 divides n if n = 18x for some x Z. Define sets A and B as shown: Prove that B A. Template. Direct proof that a set is a subset To show A B: 1. Let x A Therefore x B. 5. Therefore A B. A = {x x Z and 6 divides x} B = {x x Z and 18 divides x} Example 5. Prove that the following two sets are equal: E = {x Z x is even}, and F = {x Z x = a + b for some a, b Z, with both a and b odd}. You can assume standard facts about even and odd numbers. Solution: Proving that E and F are equal means proving x E x F. As we know from proving if and only if statements this means proving If x E, then x F and If x F, then x E. We have templates for these if-then proofs, and so we have the beginning of a proof of this result:

33 CHAPTER 2. LOGICAL REASONING 33 Let x E Therefore x F. Let x F Therefore x E. This proves that x E x F, which means E = F. Now we fill in some of the missing steps. Let x E. Then x is even, by the definition of E. By definition of even, this means x = 2a for some a Z. Then 2a + 1 and 1 are both odd, by the definition of odd (note that 1 = 2( 1) + 1). So x = (2a + 1) + ( 1) is the sum of two odd numbers. Therefore x F. Let x F. Then x = a + b for some a, b Z, where a and b are odd. Applying the definition of odd we get a = 2c + 1 and b = 2d + 1 for some c, d Z. Combining these we have x = a + b = 2c d + 1 = 2(c + d + 1). This shows that x is even, which means x E. This proves that x E x F, which means E = F. Template. Direct proof that two sets are equal Let A and B be two sets. Two show that A = B, do the following: Suppose x A Therefore x B. Suppose x B Therefore x A. Therefore A = B. Example 6. Let x be any object and A any set. Prove that x A if and only if {x} A. Solution: Let x be any object, and A be any set. Let x A. We need to show that {x} A. In other words, we need to show that every element of {x} is also an element of A. But the only element of {x} is x. Since x A, we are done. Suppose {x} A. Since {x} is a subset of A, we have that every element of {x} is also an element of A. Since x is an element of {x}, we have that x is in A. In other words, x A. This is where we ended on Wednesday, September Quantifiers Often we have statements and propositions that include a variable: a symbol that represents an element of a set. As soon as we use variables, we need to describe two cases that can be applied: whether we are talking about all possibilities that the variable could be, or just about one possibility that the variable could be.

34 CHAPTER 2. LOGICAL REASONING 34 The notion of existence is one of the primitive concepts with which we must begin as given. It is the clearest concept we have. Gödel (quoted by Wang) Definition The symbol stands for the phrase there is or there exists. We use it in making logical statements as follows x A, assertions about x. A logical statement of this form is called an existential statement. The following are synonyms: x A, statement P, At least one value of x A makes P true, P is true for some x A, There exists x A such that P is true. Example 1. Translate the following into an English sentence using as few symbols as possible x R, such that x 2 = 10 Solution: There exists a real number such that when we square it we get 10. Template. Direct proof of existential statements To prove the statement x A, assertions about x, do the following: Let x equal... (give an explicit meaning or value: this is the one time where it s ok to prove by example ) Show that x satisfies the assertions. Therefore x A, assertions about x is true. Example 2. Prove the following: x R such that x 2 5x + 4 = 0. Solution: Let x = 4. Then 4 2 5(4) + 4 = = 0. Therefore x satisfies the equation x 2 5x + 4 = 0. Therefore x R such that x 2 5x + 4 = 0. Definition The symbol stands for the phrase for all or for every. making a logical statement as follows x A, assertions about x. A logical statement of this form is called an universal statement. The following are synonyms: x A, statement P, every value of x A makes P true, P holds for all x A without exception, for all x A, property P is true, if x X, then P is true. We use it in Example 3. Translate the following into an English sentence using as few symbols as possible n N, we have (n + 1) 2 4. Solution: For every natural number, if we add 1 to it and square the result, then we get a number that is at least 4.

35 CHAPTER 2. LOGICAL REASONING 35 The logical phrases for all and there exists are called quantifiers. Note that with the sort of statements we see in practice, it is generally harder for a for all statement to be true than it is for a there exists statement to be true. Example 4. Identify each of the following as true or false, and justify with one sentence. (a) x R such that x 2 x 1 = 0, (b) x R, we have x 2 x 1 = 0. Solution: We should be able to guess which of these is true and which is false without a lot of specific thought: universal statements are harder to make true. (a) is true since we can find at least one x that solves the equation: x = 1 ± 5 2 (b) is false, since there are values of x that do not satisfy the equation, such as x = 0. Template. Direct proof of universal quantifier statement To prove the statement x A, assertions about x do the following. Let x A. Show that x satisfies the assertions. Assume only that x satisfies the definition of elements of A. Make no other assumptions about x. Therefore x A, assertions about x is true. Example 5. Prove the following: n N, we have (n + 1) 2 4. Solution: We start by making this statement look more like an if-then: If n N, then (n + 1) 2 4. Now we should know exactly how to set up the proof of this, although we will still have work to do to fill in the middle steps. Let n N Therefore (n + 1) 2 4, QED. Filling in the middle steps always takes work and experience. In this case, my intuition is to start with an inequality, since that s where we re trying to get to. In fact, you could reason backwards: our goal is (n + 1) 2 4, if we take square roots then we should have n and so this should mean n 1. Now we write this forwards: Let n N. By definition of the natural numbers, we have n 1. By Definition we can add 1 to both sides of this inequality to get n By Definition we can multiply both sides of this inequality by n + 1 to get (n + 1) 2 (n + 1)2. We can also multiply both sides of n by 2 to get (n + 1)2 4. Combining the last two inequalities we get (n + 1) 2 (n + 1)2 4. Therefore, by transitivity, (n + 1) 2 4, QED. Scholium If a, b R with a b 0 then a 2 b 2.

36 CHAPTER 2. LOGICAL REASONING 36 This is where we ended on Friday, September 29 The two quantifiers we are discussing are related to each other by negation. To figure out the relationship, think about the following example. Example 6. (a) Suppose I say every student in this class will get an A. What does it take, to make me a liar? Can you state this in a way that uses a quantifier first, and then a negation? (b) Suppose I say there is a student in this class who will win the lottery. What does it take, to make me wrong? Can you state this in a way that uses a quantifier first, and then a negation? Solution: (a) I am a liar if not every student gets an A. To put the negation after the quantifier we could say that I m a liar if at least one student does not get an A. In other words, if there exists a student who does not get an A. (b) I am wrong if no one wins the lottery. To use a quantifier, and put the negation after it, we could say that I m wrong if every student fails to win the lottery. Fact The following rules describe the negation of existential and universal quantifier statements: ( x A, statement P ) = x A, P ( x A, statement P ) = x A, P In words: when you move a across a quantifier you switch the quantifier. Example 7. Translate into Math-English (avoid quantifier symbols), the negation of the following statements (a) x R, x = 0, (b) x Z, x Solution: (a) For every real number x, we have that x (b) There exist an integer x, such that x = 0. Logical statements that have only one quantifier (i.e. one of for all or there exists but not both), are usually simple to understand, at least at the logical level. Two quantifiers of the same type (i.e. both of the form for all or both of the form there exists ) are also relatively easy to understand as well: you just double up the quantifier and the variables, and order doesn t matter. Example 8. Translate the following into Math-English (avoid quantifier symbols). Which is true and which is false? (a) x R, y R, ln(x 2 + y 2 ) is defined. (b) x R, y R, ln(x 2 + y 2 ) is defined. Solution: (a) For all real numbers x and y we have that ln(x 2 + y 2 ) is defined. This is False. It suffices to prove this with an counter-example (in general, you can prove a statement false with a single counter example, but you cannot prove a statement true by using examples.) For instance, let x = 0, and y = 0. Then ln(x 2 + y 2 ) = ln(0) which is not defined. (b) There exist real numbers x and y such that ln(x 2 + y 2 ) is defined. This is True. It suffices to prove this with an example (in general, you can prove an statement true with an example, but you cannot prove an statement false by using examples). For instance, let x = 0 and y = 1. Then ln(x 2 + y 2 ) = ln(1) = 0.

37 CHAPTER 2. LOGICAL REASONING 37 Logical statements that have two quantifiers of different types are more interesting, and take more work to understand. x, y means that first we consider any x, and second we consider the existence of a y, which might depend upon what x is. In other words, if we start with a different x, then y might also change. x, y means that first we fix a single x, which might have special restrictions, and then see if that x will work for all possible values of y. Here, x cannot depend upon y, nor can y depend upon x. Example 9. An easy metaphoric example helps one understand the difference between x, y and y, y. Suppose we consider marriage, and to make it a symmetric relation, use is perfect marriage material. There exists a person P, who is perfect marriage material for each person Q in the world. For each person P, there exists a person Q who is perfect marriage material. The first statement says that one person P, could be a perfect match for everyone; the same person P could be a perfect match for me, and you, and President Obama! The second statement says that each person has a perfect match, but it presumably a different perfect match for each person. Example 10. Analyze the following two statements, and prove or disprove them. (a) x N, y N, x < y. (b) x N, y N, x < y. Solution: We start with English translations. For (a) we can start with For all x N, there exists y N, such that x < y. Here s a slightly less formal translation: Every natural number has a natural number that is bigger than it. For part (b) we can start with There exists x N so that for all y N, we have such that x y. Here s a slightly less formal translation: there is a natural number that is smaller than all natural numbers. This is where we ended on Monday, October 2 Now let s figure out true and false. Part (a) is true. To explore it, consider different values of x. If x = 11 we can take y = 12. If x = 1304 we can take y = We have to use different y s for different x s, since there is not a single natural number y that is larger than all other natural numbers. Now we prove that this statement is true (note: the real purpose of this proof is to model how we prove quantified statements like this: start with x, define y in terms of x, etc.). We prove that every natural number has a natural number that is bigger than it. Let x be any natural number. Set y = x + 1. Then x < y. Since x was arbitrary, we are done. Part (b) is false. For example, if we pick x = 10, then x y for y = 9. If we pick x = 1, then x y for y = 1. To prove a statement is false, we provide a counter-example. Actually, this is the same thing as proving the negation: ( x N, y N, x < y ) = x N, y N, x y. Thus, to prove that (b) is false, we can prove that the statement just given is true. We do this by following the template for universal quantifier statements. We prove that this statement is false by proving it s negation is true. It s negation may be stated as: For every x there exists a y such that x y. Let x be any natural number. Define y = x. Then x y.

38 CHAPTER 2. LOGICAL REASONING 38 Example 11. Definition: A function f(x) is continuous at the value x = a if the following is true ε > 0, δ > 0, x a < δ = f(x) f(a) < ε Prove that f(x) = 3x + 2 is continuous at x = 1. Solution: We start by translating this into Math-English: For every ε > 0 there exists δ > 0 such that if x 1 < δ then f(x) f(1) < ε. Now we know how to prove a for every statement, and we know how to prove and if-then statement. Let ε > 0 be given. Let δ = ε/3. Let x satisfy x 1 < δ. Then x 1 < ε 3. Multiply both sides by 3 to get Distribute the 3 and simplify to get 3 x 1 < ε 3x 3 < ε 3x < ε The last inequality is equivalent to f(x) f(1) < ε, which is what we were trying to prove. Therefore, f(x) is continuous at x = 1, QED. Example 12. (a) Negate the definition of continuous to define the following statement: The function g(x) is not continuous at x = 1. (b) Define a function g(x) as follows: { x 2 if x 1 g(x) = 0 if x = 1 Prove that g(x) is not continuous at x = 1. Solution: Part (a). We start by translating the definition of continuous into the present situation with g(x) and x = 1 ε > 0, δ > 0 such that x 1 < δ = g(x) g(1) < ε. Now we negate this ( ) ε > 0, δ > 0 such that x 1 < δ = g(x) g(1) < ε which becomes This is where we ended on Wednesday, October 4 ε > 0, δ > 0, x, such that x 1 < δ and g(x) g(1) ε.

39 CHAPTER 2. LOGICAL REASONING 39 The part with x may not be obvious, in fact the is implied. If I want to show that if x is real then x 2 0 is false I need to show the existence of an x that is real and with x 2 > 0. Similarly here: to show x 1 < δ = g(x) g(1) < ε is false I need to produce a counter-example, i.e. show the existence of an x such that x 1 < δ is true and g(x) g(1) < ε is false. So, in theory we know to to prove this, at least in outline form. The part with ε > 0 becomes Let ε =... where we need to fill in the dots. The part with δ > 0 becomes Let δ > 0 be given or Fix an arbitrary real number δ such that δ > 0. The part with x, x 1 < δ becomes Let x =... where we need to fill in the dots. So far, our proof looks like this: By definition of continuous, and the negation of that definition, it suffices to prove the following: that there exists an ε > 0, such that for all δ > 0, there exists an x such that x 1 < δ and g(x) g(1) ε. Let ε =.... Let δ be an arbitrary real number such that δ > 0. Let x =.... Then x 1 becomes x 1 =... Now we calculate g(x) g(1). This becomes =... < δ. g(x) g(1) = 0 =... So now we need to show that ε Therefore g(x) g(1) ε which shows that g is not continuous at x = 1. So to finish, there are a few things we have to fill in, but in fact only two real choices we have to make: what is ε and what is x. Once we make these choices, the other parts to fill in should be a little bit automatic. To pick ε, we have to understand what the definition of continuity means. The role that ε plays is the distance between y-values on the graph of g(x) and the fixed y-value g(1). In this case, the gap in the graph is 1, and so we can set ε = 1. Now we need to pick x such that x 1 < δ. In other words, we need the distance between x and 1 to be less than δ. How much less? In this problem, anything less than δ would do. How about, just to sort of show off, we make the distance between x and 1 come out to be 0.99δ, that is to say the distance is just 1% smaller than δ. That turns out to work. Now we finish the proof. By definition of continuous, and the negation of that definition, it suffices to prove the following: that there exists an ε > 0, such that for all δ > 0, there exists an x such that x 1 < δ and g(x) g(1) ε.

40 CHAPTER 2. LOGICAL REASONING 40 Let ε = 1. Let δ be an arbitrary real number such that δ > 0. Let x = δ. Then x 1 becomes x 1 = δ 1 = 0.99δ < δ. Now we calculate g(x) g(1). This becomes g(x) g(1) = g( δ) 0 = ( δ) 2 = ( δ) 2. So now we need to show that ( δ) 2 ε. We manipulate the given inequality δ > 0: δ > 0, 0.99δ > 0, δ > 1, ( δ) 2 > 1. Since ε = 1, this shows that g(x) g(1) ε which shows that g is not continuous at x = 1. That s a very good proof, but just to illustrate a different style, let s rewrite it with a few less words, and a little less vertical space. Just to be clear, I would prefer you write a proof more like the one I just gave, but if you want to see it distilled down closer to the bare minimum, here s another version. [More compact, but probably not as friendly of a proof.] We show that there exists an ε > 0, such that for all δ > 0, there exists an x such that x 1 < δ and g(x) g(1) ε. Let ε = 1, let δ > 0, and let x = δ. Then x 1 = δ 1 = 0.99δ < δ and g(x) g(1) = g( δ) 0 = ( δ) 2 = ( δ) 2. Starting with δ > 0 we get 0.99δ > 0 and δ > 1 and ( δ) 2 > 1 = ε. Therefore g(x) g(1) ε, QED. This is where we ended on Friday, October 6 Definition The notation!x A, statement about x means x A, statement about x and x is the only element in A with this property. In other words,!x means there exists a unique x.

41 CHAPTER 2. LOGICAL REASONING 41 To prove such a statement, requires proving two properties: existence, and uniqueness. Proving existence was described above. To prove uniqueness, the standard approach is as follows: Suppose there is another element y that satisfies the given property. Then prove that x = y. Sometimes you can prove uniqueness more directly: maybe you have calculated x or defined it in a way that is unique. Example 13. We will prove the following assertion: there exists a unique x Q such that 5x+7 = 3. Proof. We give essentially two separate, self-contained proofs: existence and uniqueness. Existence : Let x = 4/5. Then 5x + 7 = 5( 4/5) + 7 = = 3. This shows that there is a rational number x that satisfies the equation. Uniqueness : Suppose x is a rational number such that 5x + 7 = 3. Then 5x + 7 = 3 = 5x = 4 = x = 4/5. This shows that if there is a solution of the equation, then the solution must equal 4/5. In the previous example uniqueness followed from applying correct steps and getting a unique expression for x. In a sense you could say that x had no choice but to equal 1. (Note however that this proof did start with the assumption that x + 2 = 3 was valid; in a sense this means that we assumed existence out the outset.) The kind of uniqueness proof given in the previous example will only work if you have a sequence of steps each of which produces a unique result; usually this requires that you are able to get an explicit expression for x. The next example is in contrast. Example 14. We will prove the following assertion: there exists a unique differentiable function f satisfying f(0) = 1 and f (x) = 2 for all x. Proof. We start by proving existence of a function with the stated properties. Let f(x) = 2x + 1. Then f(0) = 1 and f (x) = 2 for all x. This shows that f satisfies the properties. Now we prove uniqueness of a function with the indicated properties. Suppose f and g are functions that satisfy the indicated properties, i.e. f(0) = 1 and f (x) = 2 for all x, and g(0) = 1 and g (x) = 2 for all x. Define a new function F (x) = f(x) g(x). Then F (0) = 0 and F (x) = f (x) g (x) = 2 2 = 0 for all x. By some theorem in Calculus (The Mean Value Theorem), this shows that F (x) = 0 for all x. Then f(x) g(x) = 0, so f(x) = g(x) for all x.

42 Chapter 3 Constructing and Writing Proofs in Mathematics 3.1 Divides and more direct proofs Definition Let a and b be integers 1. If a = bc for some integer c, then we say that b divides a. As synonyms we have: a is divisible by b, or a is a multiple of b, or b is a factor of a, or b is a divisor of a. The symbol 2 for this is b a. Example 1. (a) An integer n is even if and only if 2 n. (b) An integer n is a multiple of 3 if and only if 3 n. Theorem (Properties of Divides). Fix integers a, b, c. The following hold 1. a a (reflexive property) 2. a b a b (semi-symmetric property) 3. If a b and b c then a c (transitive property) 4. If a b and b a then a = ±b (semi-anti-symmetric property) 5. If ab ac and a 0 then b c (cancelation property) 6. If a b then a bc (multiplicative property) 7. If a b and a c then a (b ± c) (sum and difference property) 8. If a b and a c then a (ib + jc) for all i, j Z (linearity property) Proof. 1. Proof: Let a Z. Let c = 1. Then a = ac becomes a = 1a which is true. This shows that a a by definition of. This is where we ended on Monday, October 9 2. Proof: Let a, b Z such that a b. By definition of this means b = ac for some c Z. Then b = a( c) and so a b, by definition of. 1 By our discussion of let above this means that a and b could be any integers. We do not assume anything else about them yet, we do not get to choose what they are, etc. 2 This is not the same as b/a or b a or b : all of these are operations. The symbol b a is read more as an adjective; a it s not something you are meant to do, not an imperative verb, not like divide a by b. Rather it s giving a name to a relationship that a and b have together a and b are related by this property that we call divides. 42

43 CHAPTER 3. CONSTRUCTING AND WRITING PROOFS IN MATHEMATICS 43 Let a, b Z such that a b. By definition of this means b = ac for some c Z. Then b = a( c) and so a b, by definition of. 3. Proof: Let a, b, c Z such a b and b c. By definition of we have b = ax and c = by for some x, y Z. Substituting the first equation into the second gives c = (ax)y = a(xy). Since xy Z this shows that a c. 4. Proof: Let a, b Z with a b and b a. By definition of we have b = ax and a = by for some x, y Z. Substituting the first equation into the second gives a = (ax)y and so xy = 1. Since x and y are integers this implies that x = y = ±1. This implies a = ±b. 5. Proof: Let a, b, c Z such that ab ac and a 0. By definition of we have ac = ab(x) for some x Z. If a 0 then we can apply the multiplicative cancelation law (Theorem 1.1.8, part (i)) to get c = bx. By definition, this means b c. 6. Proof: Let a, b Z such that a b. By definition of we have b = ax for some x Z. Let c Z. Then bc = a(xc). This shows that a bc. 7. Proof: Let a, b, c Z such that a b and a c. By definition of we have b = ax and c = ay for some x, y Z. Then b ± c = ax ± ay = a(x ± y). This shows that a (b ± c). 8. Proof: Let a, b, c Z such that a b and a c. By definition of we have b = ax and c = ay for some x, y Z. Let i, j Z. Then ib + jc = iax + jay = a(ix + jy). This shows that a (ib + jc). This is where we ended on Wednesday, October More methods of proof We ve seen a few variations on proof techniques at this point: Direct proof (see Section 1.2.1), proof by contrapositive (see Section 2.2, Example 5), proof of biconditionals or if-and-only-ifs, (Section 2.2, Example 6 set equality (Section 2.3, Example 5), single quantifiers (Section 2.4, Example 2 and 5), combined quantifiers (Section 2.4, Examples 11 and 12), and a proof of existence by construction (Section 2.4, Examples 13 and 14 and. Example 1 (Using logical equivalence). Prove the following: If ab is even then a is even or b is even. Solution: As stated, we would probably need to consider four cases: a is even or odd, and b is even or odd. But, by using a logical equivalence, we can reduce this to two cases. Recall (or verify) the following: P = (Q R) (P Q) = R. The second formula is what we will prove. Proof. We will prove that if ab is even, and a is odd, then b is even. Suppose ab is even and a is odd. There are two cases for b: it can be even or odd. If b is odd, then ab would be odd, which it is not. Therefore b is not odd, and it is even. Example 2 (Nonconstructive existence). Prove that there is a solution of x + sin(x) = e 2. (You may use basic facts about calculus.)

44 CHAPTER 3. CONSTRUCTING AND WRITING PROOFS IN MATHEMATICS 44 Proof. Let f(x) = x + sin(x). Then f(0) = 0 + sin(0) = 0. Also, f(4π) = 4π + sin(4π) = 4π. Note that 0 < e < e 2 so f(0) < e 2. Similarly, e < 3 e 2 < 9 3 < π 12 < 4π e 2 < 4π so e 2 < f(4π). Therefore, by the Intermediate Value Theorem, there is some x between 0 and 4π such that f(x) = e Proof by Contradiction Reductio ad absurdum, which Euclid loved so much, is one of a mathematician s finest weapons. It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game. G.H. Hardy, A Mathematician s Apology Discussion. What we call a proof by contradiction is what Hardy calls reductio ad absurdum; it is also called an indirect proof. In the quote above, Hardy does not go on to explain what he means; perhaps he felt it was obvious, or perhaps he was a man of few words. I think he means this: in mathematics a logical contradiction is unacceptable. In fact, in a Platonic sense, if the field of mathematics contained a single, unmistaken, unremovable, contradiction, then the entire field would be destroyed. In offering a proof by contradiction, a mathematician is offering to sacrifice the entire game, all of mathematics. Example 1. Show that the following are equivalent: Solution: We produce construct a truth table: P = Q and (P Q) = F P Q P = Q P Q (P Q) = F T T T F T T F F T F F T T F T F F T F T Note that the most crucial line is the second: it is the only place that P = Q is false. Similarly, the only way (P Q) = F can be false is if P Q is true. This only happens if both P and Q are true, which means P is true and Q is false.

45 CHAPTER 3. CONSTRUCTING AND WRITING PROOFS IN MATHEMATICS 45 Example 2. Prove that 2 is irrational. You may assume the following: Prop 1 : If n is a natural number, and n 2 is even, then a is even. Prop 2 : Every rational number can be written in the form a where a and b have no common factors. b The first proposition follows from Section 3.2, Example 1. The second is presumably familiar, but we have not proven it yet in this course. Solution: We start by translating the statement into if-then form: If x 2 = 2, then x is irrational. Proof. We assume for contradiction that x 2 = 2 and that x is rational. By Prop 2 we can write x = a with a and b having no common factors. Squaring we get x 2 = 2 = a2, which we can b b2 rewrite as 2b 2 = a 2. This shows that a 2 is even, and so Prop 1 shows that a is even, and this means that a = 2g for some integer g. Now our equation becomes 2b 2 = (2g) 2 = 4g 2. If we cancel 2 from both sides we get b 2 = 2g 2. This shows that b 2 is even, and so Prop 1 shows that b is even. But if both a and b are even, then they have a common factor of 2, contradicting the second sentence of the proof. In the end this proof is both a classic, it goes back 2300 years to Euclid, and a little frustrating. At the end it doesn t feel like we ve shown anything. Proofs by contradiction are always a little like that, because in some sense they don t finish with what we want, they finish with a contradiction. But it s worth looking at in the big picture: If x is rational then we have a contradiction. The contradiction is not important, it s the fact that x Q leads to something false. Template. Proof by contradiction To prove if P, then Q Clearly state We assume for contradiction that P and Q are both true (or some synonym). Derive consequences of P and Q until you reach a contradiction (i.e. a statement that is known to be false), and clearly state this is a contradiction or some synonym. Conclude that Q must be false, i.e. Q is true. Discussion. Proof by contradiction can be justified by common sense If I assume that **** is true, then I can show that black is white. Therefore **** must be false. The difference between proof by contradiction and contrapositive is a little subtle, and you don t need to worry if you can t really see it. Practically speaking, the main difference is how P is used: In a true proof by contradiction, both P and Q will be used, from the beginning, and the eventual contradiction will not directly be about P or Q or Q, it will be about a somewhat distant mathematical fact. In a true proof by contrapositive only Q will be used, and the eventual contradiction will be only that it contradicts P. This is where we ended on Monday, October 16

46 Chapter 4 Induction The more you approach infinity, the deeper you penetrate terror. Gustave Flaubert 4.1 Ordinary Induction It s possible to combine the Well Ordered Property of the natural numbers with the proof technique of smallest counter example into a single proof technique that does not use contradiction. The result is the Principle of Mathematical Induction. Theorem (Mathematical Induction). Let P (n) be an open sentence defined on the natural numbers. Suppose the following statements are true: 1. P (1) is true and 2. n N if P (n) is true then P (n + 1) is true. Then n N we have P (n) is true. Discussion. Since this is a theorem, we should be able to prove it. However to prove it, we would have to give a more careful definition of the natural numbers. Basically, we would have to state that the natural numbers are defined by the following properties: (1) 1 N, (2) if n N then n + 1 N, (3) 1 n + 1 for any n N, (4) if n, m N and n + 1 = m + 1 then n = m, (5) if X N and X has the property that 1 X and if n X then n + 1 = X then X = N. But property (5), which we would assume by definition of N, is essentially the same property we are using in the theorem, so it seems to me like way too much work to define the natural numbers just to prove this theorem. Template. Proof by induction To prove that every natural number has some property, i.e. to prove n N, P (n) Prove that the result is true for n = 0 (the basis or base case ). Prove that if the result is true for k, then it is true for k + 1 (the inductive step ). I.e. 46

47 CHAPTER 4. INDUCTION 47 Suppose that n = k is some natural number that makes the result true (the inductive hypothesis ). Use the inductive hypothesis to show that the result is true for n = k + 1. State by induction, the result is true for all natural numbers (or some synonym). Discussion. In one step we switch to n instead of k. This is not necessary, but seems to be a common approach to writing proofs by induction. Maybe it makes more sense to think of n as variable and k as some fixed natural number that represents a known case, a case where P has already been shown to be true. For instance, suppose we verified directly that P (5) is true and then said Suppose that n = 5 is a natural number that makes the result true. We will prove that the result is true for n = 6. This is just what we mean by saying Suppose that n = k is a natural number that makes the result true. We will prove that the result is true for n = k + 1. And this is just what we mean by saying n N if P (n) is true then P (n + 1). Example 1. Let n be a natural number. Then n = n(n + 1). 2 Solution: A lot of this proof should be automatic template filling. For instance, we can write all of the following without having to do anything more than understand what the statements mean Proof. Base case. If k = 1 then the left hand side just has the term 0. The right hand side has 0(0+1) 2 which also equals 0. Since 0 = 0, the result is true for n = 0. Induction hypothesis. Assume that the result holds for n = k. Thus, k = k(k + 1). 2. So (k + 1)(k + 2) (k + 1) =. 2 This means that the result holds for n = k + 1. Therefore, the result hold for all natural numbers by induction. Now we need to fill in the missing middle steps. What we need in this case is to see how to turn the formula involving k into the formula involving k + 1. In general, we need a little insight into the kind of formula we are dealing with. In this case, the simplest part of the formula involves a sum. The left hand side is the sum of a bunch of numbers. For the inductive hypothesis the sum stops at k. We want to turn this sum into the formula involving k + 1. How can we turn a sum that stops at k into a sum that stops at k + 1? We can add k + 1. Can we do this just to the left hand side? No. We have an equation which we are assuming is true, the k equation. To get another new equation, we can do anything we want to both sides of the equation. So, we need to add k + 1 to both sides.

48 CHAPTER 4. INDUCTION 48 Proof. Base case. If n = 0 then the left hand side just has the term 0. The right hand side has 0(0+1) 2 which also equals 0. Since 0 = 0, the result is true for n = 0. Induction hypothesis. Assume that the result holds for n = k. Thus, k(k + 1) k =. 2 Inductive step. Now we add k + 1 to both sides of this equation, k + (k + 1) = We get a common denominator on the right k + (k + 1) = k(k + 1) 2 + 2(k + 1) 2 k(k + 1) 2 = + (k + 1). k(k + 1) + 2(k + 1) 2 = (k + 1)(k + 2). 2 So (k + 1)(k + 2) (k + 1) =. 2 This means that the result holds for n = k + 1. Therefore, the result hold for all natural numbers by induction. Example 2. For each natural number n, we have n 2 (2n + 1)(n + 1)(n) =. 6 Proof. Base step. If n = 0 then the equation becomes 0 = 0, which is true. This is where we ended on Monday, October 25 Inductive hypothesis. Assume that the result holds for n = k. In other words, we assume that the following equation is true: k 2 = (2k + 1)(k + 1)(k). 6 Inductive step. Now we add (k + 1) 2 to both sides of the equation: k 2 + (k + 1) 2 (2k + 1)(k + 1)(k) = 6 The right hand side of this last equation can be rewritten as (2k + 1)(k + 1)(k) + 6 6(k + 1)2 6 ( (2k + 1)(k) + 6(k + 1) (2k + 1)(k + 1)(k) + 6(k + 1)2 = ) 6 (k + 1) = (2k2 + 7k + 6)(k + 1) 6 + (k + 1) 2. (2k + 3)(k + 2)(k + 1). 6 = = 6 So we have ( )( ) 2(k + 1) + 1 (k + 1) + 1 (k + 1) k 2 + (k + 1) 2 =. 6 This means that the given statement is true for n = k + 1. Therefore, by induction, the result holds for all natural numbers.

49 CHAPTER 4. INDUCTION Variations on proof by induction Theorem (Principle of Strong Mathematical Induction). Let P (n) be an open sentence defined on the natural numbers. Suppose the following statements are true: 1. P (1) is true and 2. n N if P (1), P (2),...,P (n) are true then P (n + 1) is true. Then n N we have P (n) is true. Discussion. The difference between this theorem and Theorem is this: in condition (2) we use a stronger hypothesis in the if-then, namely we assume not just that P (n) is true but P (1),...,P (n) are all true. Template. Proof by strong induction To prove that every natural number has some property: Prove that the result is true for n = 0 (the basis or base case ). Prove that if the result is true for 0, 1,..., k, then it is true for k + 1 (the inductive step ). I.e. Suppose that the result is true for n = 0, 1,... k (the strong inductive hypothesis ). Use the inductive hypothesis to show that the result is true for n = k + 1. State by induction, the result is true for all natural numbers (or some synonym). Example 1. (Scheinermann, #22.18) Recall the Fibonacci numbers. They are defined as follows: F 0 = 1, F 1 = 1, F 2 = = 2, F 3 = = 3, F 4 = = 5,..., F n = F n 1 + F n 2. The first 11 Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. Show that every positive integer is equal to the sum of distinct Fibonacci numbers. Solution: To help understand the statement of the result, and perhaps to see how to prove it, we write out its conclusions for the first few positive integers: N Fibonacci numbers 1 = 1 2 = 2 = = 3 = 4 = = = 5 6 = = = The ones we have crossed out show how we can write a number as a sum of Fibonacci numbers, but the Fibonacci numbers used are not distinct. So those sums don t show the conclusion we are trying to prove. Let s skip a few integers. How can we write 22 as the sum of Fibonacci numbers? One way is 22 =

50 CHAPTER 4. INDUCTION 50 but this does not use distinct Fibonacci numbers. There are a few other ways to write 22 as the sum of Fibonacci numbers, 22 = = = How would you go about starting with an integer like n = 75? The crucial issue is, what is the best way to get the first Fibonacci number to use? To be efficient, we should start with 55 since that s the biggest Fibonacci number we can use. This leaves 75 = , and then we can think about how to write 20 as a sum of Fibonacci numbers. Proof. Base case. If n = 1 then n is the sum of one integer, 1. Strong inductive hypothesis. Suppose that for all n = 1, 2,..., k we can write n as the sum of distinct Fibonacci numbers. Now, let n = k + 1 be fixed. We need to show that n is the sum of distinct Fibonacci numbers. Let F x be the largest Fibonacci number less than or equal to n and let This is where we ended on Wednesday, October 25 r = n F x so that n = F x + r. If r = 0, then n = F x and we are done. Otherwise, note that r < n. Then the strong inductive hypothesis shows that r = F i1 + F i2 + + F it where F i1,..., F it are distinct Fibonacci numbers. This means that n = F i1 + F i2 + + F it + F x, and all we have left to do is show that F x does not equal F i1, F i2,..., F i2. We will show that r < F x. Suppose this is not the case, so that r F x. Then n F x F x and so n 2F x = F x + F x F x + F x 1 = F x+1. This contradicts the assumption that F x was the largest Fibonacci number less than or equal to n. Therefore r < F x and so F x > F i1, F i2,..., F it, and so F x is distinct from F i1, F i2,..., F it. This proof makes me a little sad. Why? Because the original idea, was so simple: take a number like 75 and use the Fibonacci number 55 to say that 75 = Now the question becomes can we split 20 up further, and that s not too hard. But the simplicity of this example covers up at at least two important things. First, I can tell at a glance that 20 is a lot smaller than 55 and so I know that whatever Fibonacci numbers I use to split up 20 can t possibly equal 55. But that s not obvious when I have F x and some other unknown numbers. Second, everything seems very safe and obvious when we have small numbers, like 75. Maybe it would help, a tiny bit, to consider much larger numbers. Suppose n = Then the largest Fibonacci number that is less than or equal to n is F 62 = and using this we get r equal to r = n F 62 = So, now it s not so obvious that everything will work out fine with r. Discussion. The previous example is really cool. It s quite rare to be able to write any natural number as a sum of distinct numbers of from some given sequence. The interested reader should ask themselves a few questions about this example. What makes it work? What s the really important property that the Fibonacci numbers have that we use in the proof? Could the same result be true for some other sequence of numbers? Can you find another sequence that has the same property? Can you classify all sequences for which the same result holds?

51 CHAPTER 4. INDUCTION 51 Cauchy Induction Theorem Let P (n) be an open sentence. If P (2) is true, and if P (n) = P (2n), and if P (n) = P (n 1) then P (n) is true for all n 2. Example 2. The Arithmetic Mean Geometric Mean Inequality.

52 CHAPTER 4. INDUCTION 52 Exercises 1. For all of the statements below, state what the inductive hypothesis would be, and what the conclusion of the inductive step would be. Identify for which natural numbers the following statements are true, and then prove them by induction. 2. 2n + 1 n 2 3. n < 2 n 4. n 2 2 n 5. n 3 2 n 6. n! n n 7. nm n 8. If n and m are relatively prime, then an + bm = 1 for some integers a, b Z n n 10. cos(nx) equals a polynomial in cos(x) 11. A polynomial of degree n has at most n roots. 12. The sum of the first k nth powers is a polynomial in k (induct on n). 13. A 2 n 2 n grid of unit squares with one square removed can be covered by trionimos. (A trionimo is the shape formed by three unit squares making an L-shape, i.e. what you get when you start with the 2 2 grid and remove one square.) 14. If p is prime and p ab then p a or p b 15. If p is prime and p a 1 a 2 a n then p a i for some i with 1 i n. 16. If p is prime and p a n then p a 17. Let f 1,..., f n be differentiable functions. Then (f 1 f n ) = 18. Let f(x) = e x2. Then x n 19. lim x e x = 0 n f 1 f i f n i=1 f (n) (x) = some polynomial e x x n e x dx = n!. 21. Let T be a finite, directed graph that has no cycles, and let n be the number of vertices and e be the number of edges of T. Then n = 1 + e. 22. We can write n = p 1... p k where each p i is prime. 23. If F n is the nth Fibonacci number then F n = 1 5 (ϕ n (1 ϕ) n ) where ϕ is the golden ratio (2n 1) = n (7 n 4 n ) 26. 3n 2 + 3n + 1 n n 1 = 2 n (9 n 1) n = n 2 + n na 2 < (1 + a) n 31. (a + b) n = 32. n k=0 ( n k ) a n k b k 1 + x + x x n = 1 xn+1 1 x 33. n 2 5n! (10 n 1) 35. The sum of the interior angles of a convex n-gon is 180(n 2). 36. If we write 3 n in usual base 10 notation, then the ( second ) from last digit ( is ) even If A = then A n 1 0 =. 1 1 n Strengthen the result we proved earlier about Fibonacci numbers to the following: every natural number can be written as the sum of distinct, nonconsecutive Fibonacci numbers. (It turns out that the Fibonacci numbers that appear in this way are unique. To find out more about this you should read about the Zeckendorf Theorem.)

53 Chapter 5 Properties of Sets If the doors of perception were cleansed every thing would appear to man as it is, Infinite. William Blake, The Marriage of Heaven and Hell 5.1 Operations In this section we define union, intersection, complement, Venn diagrams, proper subsets, the empty set, the power set, cardinality of a finite set. In the next section we establish some of the algebraic properties of these operations. Definition Let A and B be any sets. The union of A and B is the set of all elements that are in A or B or both. We write the union as A B. In other words, A B = {x x A or x B}. The intersection of A and B is the set of all elements that are in both A and B. We write the intersection as A B. In other words, A B = {x x A and x B}. Example 1. Let A be the set of integers that are multiples of 6 and B be the set of multiples of 10: A = {x Z : 6 x}, B = {x Z : 10 x}. (We ve used the alternative set-builder notation : for such that since visually it s nice not to mix this up with divides.) Describe the union A B and intersection A B. Solution: We have that A B equals all the integers that are multiples of 6 or 10: A B = {0, ±6, ±10, ±12, ±18, ±20, ±24,... }, = {x Z : 6 x or 10 x}. 53

54 CHAPTER 5. PROPERTIES OF SETS 54 We have that A B equals all the integers that are multiples of both 6 and 10: A B = {0, ±30, ±60, ±120,... } = {x Z : 30 x} This is where we ended on Friday, October 27 Definition Let A and B be any sets. The relative complement or set difference is A B = {x A x B} Suppose both A and B are subsets of some fixed, universal set U. If A U then we call U A the complement of A. We use the notation A c for the complement: Example 2. Define the sets A and B as A c = U A. A = {x Z 3 x}, B = {x Z 5 x}. The most plausible definition of the universal set here is U = Z, which we will assume. A = 3Z and B = 5Z. Describe A B. (a) Define the sets (b) Let U = Z. Describe A c. Solution: There is a useful way of picturing the combinations of sets that we ve been talking about: Venn diagrams. In these pictures we use a circle, or oval to represent a set. Given more than one set we draw overlapping circles and shade in the appropriate parts of the picture. A B A B A B A B Like with unions and intersections, we can picture the set difference using Venn Diagrams: A B A B We can picture the complement in a similar fashion to Venn diagrams. However, people usually don t draw the universal set U as a circle, but as a rectangle:

55 CHAPTER 5. PROPERTIES OF SETS 55 A c A U Example 3. Let A = 3Z and B = 5Z and U = Z. Draw the Venn diagram for A B, A B, A B, B A, A c, A c B c, etc. Example 4. Make a Venn diagram showing the the Latin, Greek and Cyrillic letters. Solution: The Power Set From ive-always-found-this-cool-venn-diagram-of-shared Definition For any set A, the power set of A is the collection (or set) of all subsets of A. The notation for the power set is P(A). Example 5. Find all the subsets of {4, 5, 6}. Solution: We can try to write all of these subsets in no particular order: {4, 5} {4, 5, 6} {5} {4} {6} but this makes it hard to see if we got all the subsets. In any case, we want a more systematic approach that we can generalize for our next theorem! Let s use lists. We ll look at all the possible lists we can make, where each entry keeps track of whether or not an element of {4, 5, 6} should be included in a subset. Thus each list should have three entries, and we can put yes or no in each list entry to indicated whether the corresponding