AN INTRODUCTION TO LOGIC. and PROOF TECHNIQUES

Transcription

1 i AN INTRODUCTION TO LOGIC and PROOF TECHNIQUES Michael A. Henning School of Mathematical Sciences University of KwaZulu-Natal

2 ii

3 Contents 1 Logic Introduction Statements Negation Conjunction Disjunction The Implication The converse, inverse and contrapositive of an implication Compound Statements The Biconditional Tautologies Contradictions Logical Equivalence Negation of a Conditional Statement The Contrapositive of a Conditional Statement Quantified Statements The Universal Quantifier The Existential Quantifier Negation of Quantified Statements Universal Conditional Statements Review of Symbols iii

4 iv CONTENTS 2 Proof Techniques Introduction Trivial Proofs Vacuous Proofs Direct Proofs Proof by Contrapositive Proof by Cases and the Quotient-Remainder Theorem Proof by Contradiction Existence Proofs Disproof by Counterexample Proof by Mathematical Induction

5 Chapter 1 Logic 1.1 Introduction In this chapter we introduce the student to the principles of logic that are essential for problem solving in mathematics. The ability to reason using the principles of logic is key to seek the truth which is our goal in mathematics. Before we explore and study logic, let us start by spending some time motivating this topic. Mathematicians reduce problems to the manipulation of symbols using a set of rules. As an illustration, let us consider the following problem: Example 1.1 Joe is 7 years older than Themba. Five years from now Joe will be twice Themba s age. How old are Joe and Themba? Solution. To answer the above question, we reduce the problem using symbolic formulation. We let Themba s age be x. Then Joe s age is x + 7. We are given that five years from now Joe will be twice Themba s age. In symbols, (x + 7) + 5 = 2(x + 5). Solving for x yields x = 2. Therefore, Themba is 2 years old and Joe is 9. Our objective is to reduce the process of mathematical reasoning, i.e., logic, to the manipulation of symbols using a set of rules. The central concept of deductive logic is the concept of argument form. An argument is a sequence of statements aimed at demonstrating the truth of an assertion (a claim ). Consider the following two arguments. Argument 1. If x is a real number such that x < 2 or x > 2, then x 2 > 4. Therefore, if x 2 4, then x 2 and x 2. Argument 2. If it is raining or I am sick, then I stay at home. Therefore, if I do not stay at home, then it is not raining and I am not sick. Although the content of the above two arguments is very different, their logical form is the same. To illustrate the logical form of these arguments, we use letters of the alphabet (such as p, q and r) to represent the component sentences and the expression not p to refer to the sentence It is not the case that p. Then the common logical form of both the 1

6 2 CHAPTER 1. LOGIC arguments above is as follows: If p or q, then r. Therefore, if not r, then not p and not q. We start by identify and giving names to the building blocks which make up an argument. In Arguments 1 and 2, we identified the building blocks as follows: Argument 1. If x is a real number such that x } < {{ 2 } p Therefore, if x 2 4, then x 2 }{{} not r }{{} not p or x }{{ > 2 }, then x } 2 {{ > 4 }. q r. and x 2 }{{} not q Argument 2. If it is raining (p) or I am sick (q), then I stay at home (r). Therefore, if I do not stay at home (not r), then it is not raining (not p) and I am not sick (not q). 1.2 Statements In mathematics, we are constantly dealing with statements. By a statement we shall mean the following: Definition. A statement (or proposition) is a declarative sentence that is true or false, but not both. Example 1.2 In this example, we consider the following sentences. (i) One plus one equals two. (ii) One plus one equals three. (iii) He is a university student. Sentences (i) and (ii) are both statements (only the first of which is true). On the other hand, sentence (iii) is neither true nor false (the truth or falsity depends on the reference for the pronoun he. For some values of he the sentence is true; for others it is false), and so it is not a statement. Every statement has a truth value, namely true (denoted by T ) or false (denoted by F ). We often use p, q and r to denote statements, or perhaps p 1, p 2,..., p n if there are several statements involved. For example, we might write p : One plus one equals two. q : One plus one equals three.

7 1.3. NEGATION 3 We have seen that p and q are statements, where p has truth value T and q has truth value F. The possible truth values of a statement are often given in a table, called a truth table. The truth values for two statements p and q are given in Figure 1.1. Since there are two possible truth values for each of p and q, there are four possible combinations of truth values for p and q. It is customary to consider the four combinations of truth values in the order of TT, TF, FT, FF from top to bottom as shown in Figure 1.1. p T T F F q T F T F Figure 1.1 A truth table for two statements p and q. Exercises 1.1 Which of the following are statements? For those that are, indicate their truth value. (a) 2 is an integer. (b) Pietermaritzburg is the capital of KwaZulu-Natal. (c) Why should we study mathematics? (d) 3x + 1 is an odd integer. (e) Please be quiet. (f) Camels can fly. We now introduce four symbols, called logical connectives, used to build more complicated logical expressions out of simpler ones. We begin with the negation of a statement. 1.3 Negation Definition. Given a statement p, the negation of p is the statement not p or It is not the case that p and is denoted by p.

8 4 CHAPTER 1. LOGIC The negation of statement p has the opposite truth value from p: if p is true, then p is false; if p is false, then p is true. The truth table for p (in terms of the possible truth values of p) is given in Figure 1.2. p T F p F T Figure 1.2 A truth table for negation. For example, consider the statement p : The integer 2 is even. Then the negation of p is the statement p : It is not the case that the integer 2 is even. It would be better to write p : The integer 2 is not even. Or better yet to write p : The integer 2 is odd. Exercises 1.2 State the negation of each of the following statements. (a) 5 is an even integer. (b) 0 is not an odd integer. (c) It is cold. 1.3 Write the statement It is not the case that 0 is an odd integer in symbolic form and determine the truth value of the statement. 1.4 Conjunction Next we consider the conjunction of two statements.

9 1.4. CONJUNCTION 5 Definition. Given two statements p and q, the conjunction of p and q is the statement p and q and is denoted by p q. The conjunction p q is true only if both p and q are true; otherwise, p q is false. For example, consider the statements The conjunction of p and q, namely, p : The integer 2 is even. q : 4 is less than 3. p q : The integer 2 is even and 4 is less than 3. is a false statement since q is false (even though p is true). The truth table for the conjunction of two statements is shown in Figure 1.3. (As before, we write the truth values for p and q in the order of TT, TF, FT, FF from top to bottom in the table.) This truth table describes precisely when p q is true (or false). p q p q T T T T F F F T F F F F Figure 1.3 A truth table for conjunction. Exercises 1.4 Write each of the following statements in symbolic form and determine their truth value. (a) 1 < 2 and 2/3 is a rational number. (b) 3 is a prime number and 3 > Let p and q be the statements p : 1 is an odd integer. q : 1 < 2. Write each of the following sentences in terms of p, q and logical connectives, and find the truth values of the given statements. (a) 1 is an odd integer and 1 < 2. (b) 1 is not an odd integer and 1 < 2. (c) 1 is not an odd integer and 1 2.

10 6 CHAPTER 1. LOGIC 1.5 Disjunction Definition. Given two statements p and q, the disjunction of p and q is the statement p or q and is denoted by p q. The disjunction p q is true if at least one of p and q is true; otherwise, p q is false. Therefore, p q is true if exactly one of p and q is true or if both p and q are true. Thus for the statements p and q described earlier, the disjunction of p and q, namely, p q : The integer 2 is even or 4 is less than 3. is a true statement since at least one of p and q is true (in this case, p is true). The truth table for the disjunction of two statements is shown in Figure 1.3. p q p q T T T T F T F T T F F F Figure 1.4 A truth table for disjunction. Exercises 1.6 Write each of the following statements in symbolic form and determine their truth value. (a) 1 < 2 or 2/3 is a rational number. (b) 3 is a prime number or 3 > Let p and q be the statements p : 1 is an odd integer. q : 1 < 2. Write each of the following sentences in terms of p, q and logical connectives, and find the truth values of the given statements. (a) 1 is an odd integer or 1 < 2. (b) 1 is not an odd integer or 1 < 2. (c) 1 is not an odd integer or 1 2.

11 1.6. THE IMPLICATION The Implication Of special importance to us will be a connective called the implication (also called the conditional). Definition. Given two statements p and q, the implication of p and q is the statement If p, then q and is denoted by p q. We call p the hypothesis of the implication and q the conclusion. The implication p q can be expressed in words in several ways in addition to the wording If p, then q, namely: If p, then q. p implies q. q if p. p only if q. p is sufficient for q. q is necessary for p. The truth table for the implication p q is shown in Figure 1.5. p q p q T T T T F F F T T F F T Figure 1.5 A truth table for implication. Notice that the only situation for which the implication p q is false is when p is true and q is false. The truth table for p q is actually a definition, but let us convince ourselves with an example that the truth values in this truth table are indeed justified. Example 1.3 Suppose your boss makes you the following promise: If you meet the month-end deadline, then you will get a bonus. Under what circumstances are you justified in saying that your boss spoke falsely? Solution. The answer is: You do meet the month-end deadline and you do not get a bonus. Your boss s promise only says you will get a bonus if a certain condition (you meet

12 8 CHAPTER 1. LOGIC the month-end deadline) is met; it says nothing about what will happen if the condition is not met. So if the condition is not met, your boss did not lie (your boss promised nothing if you did not meet the month-end deadline); so your boss told the truth in this case. Are you convinced? Good! If not, let us then check the truth and falseness of the implication based on the various combinations of the truth values of the statements p : q : You meet the month-end deadline. You get a bonus. The given statement can be written as p q. Suppose first that p is true and q is true. That is, you meet the month-end deadline and you do get a bonus. Did your boss tell the truth? Yes, indeed. So if p and q are both true, then so too is p q, which agrees with the first row of the truth table of Figure 1.5. Second, suppose that p is true and q is false. That is, you meet the month-end deadline and you did not get a bonus. Then your boss did not do as he/she promised. What your boss said was false, which agrees with the second row of the truth table of Figure 1.5. Third, suppose that p is false and q is true. That is, you did not meet the month-end deadline and you did get a bonus. Your boss (who was most generous) did not lie (your boss promised nothing if you did not meet the month-end deadline); so he/she told the truth. This agrees with the third row of the truth table of Figure 1.5. Finally, suppose that p and q are both false. That is, you did not meet the month-end deadline and you did not get a bonus. Your boss did not lie here either. Your boss only promised you a bonus if you met the month-end deadline. So your boss told the truth. This agrees with the fourth row of the truth table of Figure 1.5. In summary, the implication p q is false only when p is true and q is false. A conditional (or implication) statement that is true by virtue of the fact that its hypothesis is false is said to be vacuously true or true by default. Thus the statement: If you meet the month-end deadline, then you will get a bonus is vacuously true if you do not meet the month-end deadline! Exercises 1.8 Write each of the following statements in symbolic form and determine their truth value. (a) If 0 is an odd integer, then South Africa will win the World Cup Soccer in (b) If Mtunzini is the capital of KwaZulu-Natal, then 2 is an even integer. 1.9 Let p and q be the statements p : 1 is an odd integer. q : 1 < 2.

13 1.7. THE CONVERSE, INVERSE AND CONTRAPOSITIVE OF AN IMPLICATION 9 Write each of the following sentences in terms of p, q and logical connectives, and find the truth values of the given statements. (a) If 1 is not an odd integer, then 1 2. (b) If 1 is an odd integer, then Suppose that p and q are statements so that p q is false. Find the truth values of each of the following: (a) p q (b) p q (c) q p. 1.7 The converse, inverse and contrapositive of an implication Definition. Let p and q be two statements. The statement q p is called the converse of the implication p q. The statement p q is called the inverse of the implication p q. The statement q p is called the contrapositive of the implication p q. Example 1.4 Write the converse, inverse and contrapositive of the statement in Example 1.3. Solution. Recall that the given statement can be written as p q where p and q are the statements p : You meet the month-end deadline. q : You get a bonus. The converse of this implication is q p, which is q p : If you get a bonus, then you have met the month-end deadline. The inverse of this implication is p q, which is p q : If you do not meet the month-end deadline, then you will not get a bonus. The contrapositive of this implication is q p, which is q p : If you do not get a bonus, then you will not have met the month-end deadline. Example 1.5 Write the converse, inverse and contrapositive of the following statements: If today is Saturday, then I will go for a 10km run. Solution. Let p and q be the following statements: p : q : Today is Saturday. I will go for a 10km run.

14 10 CHAPTER 1. LOGIC Then the given statement can be written as p q. The converse of this implication is q p, which is q p : If I go for a 10km run, then today is Saturday. The inverse of this implication is p q, which is p q : If today is not Saturday, then I will not go for a 10km run. The contrapositive of this implication is q p, which is q p : If I do not go for a 10km run, then today is not Saturday. Exercises 1.11 Write the converse, inverse and contrapositive of each of the following statements. (a) If 0 is an odd integer, then South Africa will win the World Cup Soccer in (b) If Mtunzini is the capital of KwaZulu-Natal, then 2 is an even integer. 1.8 Compound Statements The symbols,, and are sometimes referred to as logical connectives. From given statements, we can use these logical connectives to form more intricate statements, called compound statements. Definition. A compound statement (or statement form or formula) is a statement made up of one or more statements with statement variables (such as p, q, and r) and at least one logical connective (such as,, and ). The truth table for a given statement form displays the truth values that correspond to the different combinations of truth values for the variables. For example, for given statements p and q, the conjunction p q is a compound statement. For a slightly more complex example, consider the compound statement given by (( p) (q r)) ( (s (q ( t)))). In compound statements, we avoid the use of many parentheses when no confusion arises. We often omit the outer pair of parentheses in a compound statement. For example, we write p rather than ( p). In expressions using the logical connectives,, and, we adopt the following order of operation:

15 1.8. COMPOUND STATEMENTS 11 performed first,, performed second, performed third. For example, p q = ( p) q. As in ordinary algebra, however, the order of operation can be overridden by the use of parentheses. Thus, (p q) represents the negation of the disjunction of p and q. Notice that the symbols and are coequal in order of operation. Therefore an expression such as p q r is considered ambiguous. It should be written as either (p q) r or p (q r). Example 1.6 Write each of the following sentences symbolically, letting p and q be the statements: p : It is hot. q : It is sunny. (a) (b) It is not hot and it is sunny. It is not hot and it is not sunny. Solution. (a) ( p) q. (b) ( p) ( q). Example 1.6 Construct the truth table for the compound statement (p q) (q p). Solution. Set up columns labelled p, q, p q, q p, and (p q) (q p). Complete the p and q columns with all four possible combinations of truth values for p and q (in the order of TT, TF, FT, FF from top to bottom). Then use the truth tables for (see Figure 1.5) to fill in the p q and q p columns. Finally, using the truth table for (see Figure 1.3), fill in the (p q) (q p) column. The resulting truth table is given in Figure 1.6. p q p q q p (p q) (q p) T T T T T T F F T F F T T F F F F T T T Figure 1.6 A truth table for (p q) (q q).

16 12 CHAPTER 1. LOGIC Example 1.7 Construct the truth table for the compound statement ( (p q)) (q p). Solution. Set up columns labelled p, q, p q, (p q), q p and ( (p q)) (q p). Complete the p and q columns with all four possible combinations of truth values for p and q. Then use the truth tables for (see Figure 1.4) and (see Figure 1.3) to write the truth values in the columns of p q and q p. Next, using the truth table for (see Figure 1.2), fill in the column of (p q). Finally, using the truth table for (see Figure 1.5), fill in the column for ( (p q)) (q p). The resulting truth table is given in Figure 1.7. p q p q (p q) q p ( (p q)) (q p) T T T F T T T F T F F T F T T F F T F F F T F F Figure 1.7 A truth table for ( (p q)) (q p). Exercises 1.12 Construct truth tables for the compound statement: (a) p (q p). (b) (p q) r. 1.9 The Biconditional For statements p and q, we are often interested in both the implication p q and its converse q p. Definition. Given two statements p and q, the conjunction (p q) (q p) of the implication p q and its converse q p is called the biconditional of p and q and is denoted by p q. The biconditional p q is often stated as p if and only if q. or p is equivalent to q.

17 1.10. TAUTOLOGIES 13 The truth table for the biconditional p q is shown in Figure 1.8. p q p q T T T T F F F T F F F T Figure 1.8 A truth table for a biconditional. Recall that p q represents the compound statement (p q) (q p) whose truth table we established earlier (see Figure 1.6). As mentioned earlier (in Section 1.6), the implication q p can be expressed in words as p is necessary for q, while p q can be expressed as p is sufficient for q. Therefore, p q can be expressed in words as p is necessary for q and p is sufficient for q. Another way to say this is p is necessary and sufficient for q. We remark that the phrase if and only if occurs often in mathematics. Many mathematicians abbreviate this phrase by writing iff. Although iff is informal and not a word, its use is common and you should be familiar with it Tautologies Definition. A compound statement S is called a tautology if the truth value of S is true for any assignment of truth values to the statement variables occurring in S. Example 1.8 Show that the compound statement p ( p) is a tautology. Solution. The truth table (see Figure 1.9) for p ( p) shows that this statement is always true regardless of the truth value of p. Hence, p ( p) is a tautology. p p p ( p) T F T F T T Figure 1.9 A truth table for p ( p).

18 14 CHAPTER 1. LOGIC Example 1.9 Show that the compound statement ( p) (q p) is a tautology. Solution. The truth table (see Figure 1.10) for ( p) (q p) shows that this statement is always true regardless of the truth values of p and q. Hence, this statement is a tautology. p q p q p ( p) (q p) T T F T T T F F T T F T T F T F F T T T Figure 1.10 A truth table for p ( p). Exercises 1.13 Show using truth tables that the following statements are tautologies: (a) ( p q) q. (b) ( p q) (p q) Contradictions Definition. A compound statement S is called a contradiction if the truth value of S is false for any assignment of truth values to the statement variables occurring in S. Example 1.10 Show that the compound statement p ( p) is a contradiction. Solution. The truth table (see Figure 1.11) for p ( p) shows that this statement is always false regardless of the truth value of p. Hence, this statement is a contradiction. p p p ( p) T F F F T F Figure 1.11 A truth table for p ( p).

19 1.12. LOGICAL EQUIVALENCE 15 Example 1.11 Show that the compound statement (p q) (q p) is a contradiction. Solution. The truth table (see Figure 1.12) for (p q) (q p) shows that this statement is always false regardless of the truth values of p and q. Hence, this statement is a contradiction. p q p q p q p (p q) (q p) T T T F F F T F F F T F F T F T T F F F F T T F Figure 1.12 A truth table for (p q) (q p). Exercises 1.14 Show using truth tables that the following statements are contradictions: (a) ( p q) (p q). (b) (p q) (p q) Logical Equivalence Definition. Two statements R and S are logically equivalent if they have the same truth values for all combinations of truth values of the statement variables occurring in R and S. We denote the logical equivalence of R and S by R S. Example 1.12 Show that p q q p. Solution. The logical equivalence of p q and q p is verified in the truth table shown in Figure 1.13 since the corresponding columns of these two statements are identical. p q p q q p T T T T T F F F F T F F F F F F Figure 1.13 p q and q p are logically equivalent.

20 16 CHAPTER 1. LOGIC Example 1.13 Show that p q ( p) q. Solution. The logical equivalence of p q and ( p) q is verified in the truth table shown in Figure 1.14 since the corresponding columns of these two statements are identical. p q p p q ( p) q T T F T T T F F F F F T T T T F F T T T Figure 1.14 p q and ( p) q are logically equivalent. Example 1.14 Show that p q ( p q) ( q p). Solution. The logical equivalence of p q and ( p q) ( q p) follows directly from the definition of a biconditional and from Example p q (p q) (q p) (by definition of biconditional) ( p q) ( q p) (by Example 1.13). The logical equivalence of p q and ( p) q, and the logical equivalence of p q and ( p q) ( q p), is especially important. Strictly speaking this means that we can actually dispense with the logical connectives and. Remark. Any statement form containing or is logically equivalent to one containing only,, and. Example 1.15 Show that the statements (p q) and p q are not logically equivalent. Solution. The statements (p q) and p q have different truth values in rows 2 and 3 of in the truth table (shown in Figure 1.15), and are therefore not logically equivalent. p q p q p q (p q) p q T T F F T F F T F F T F T F F T T F F T F F F T T F T T Figure 1.15 (p q) and p q are not logically equivalent.

21 1.12. LOGICAL EQUIVALENCE 17 Example 1.16 Show that (p q) ( p) ( q). Solution. The logical equivalence of (p q) and ( p) ( q) is verified in the truth table shown in Figure 1.16 since the corresponding columns (columns 4 and 7) of these two statements are identical. p q p q (p q) p q ( p) ( q) T T T F F F F T F T F F T F F T T F T F F F F F T T T T Figure 1.16 (p q) ( p) ( q) are logically equivalent. Example 1.17 Show that p (q r) (p q) (p r). Solution. The logical equivalence of p (q r) and (p q) (p r) is verified in the truth table shown in Figure 1.17 since the corresponding columns (columns 5 and 8) of these two statements are identical. p q r q r p (q r) p q p r (p q) (p r) T T T T T T T T T T F F T T T T T F T F T T T T T F F F T T T T F T T T T T T T F T F F F T F F F F T F F F T F F F F F F F F F Figure 1.17 p q and ( p) q are logically equivalent. There are many fundamental logical equivalences that we often encounter. We summarize a number of these in Theorem 1 for future reference. Theorem 1 Let p, q and r be statements. Then the following logical equivalences hold. (1) Commutative Laws (i) p q q p (ii) p q q p (2) Associate Laws (i) (p q) r p (q r) (ii) (p q) r p (q r)

22 18 CHAPTER 1. LOGIC (3) Distributive Laws (i) p (q r) (p q) (p r) (ii) p (q r) (p q) (p r) (4) De Morgan s Laws (i) (p q) ( p) ( q) (ii) (p q) ( p) ( q) (5) Idempotent Laws (i) p p p (ii) p p p (6) Negation Laws (i) p ( p) T (ii) p ( p) F (7) Universal Bound Laws (i) p T T (ii) p F F (8) Identity Laws (i) p F p (ii) p T p (9) Double Negation Law ( p) p Proof. A proof of the Commutative Law 1(i) was given in Example 1.12, while proofs of the Distributive Law 3(i) and De Morgan s Law 4(i) were given in Examples 1.17 and 1.16, respectively. We leave the proofs of the other Laws as an exercise. We mention that De Morgan s Laws can be expressed in words as follows: De Morgan s Laws. The negation of an and statement is logically equivalent to the or statement in which each component is negated, while the negation of an or statement is logically equivalent to the and statement in which each component is negated. From Example 1.16 and the Double Negation Law, we have the logical equivalence Fact 1.1 p q ( p q).

23 1.12. LOGICAL EQUIVALENCE 19 Exercises 1.15 Determine which of the pairs of statement forms listed below are logically equivalent. Justify your answers using truth tables. (a) p (p q) and p. (b) p (p q) and p. (c) ( p) ( q) and (p q) Using Theorem 1, supply reasons for each step in the logical equivalence derived in (i) and (ii) below. (i) (p q) (p q) p ( q q) (by (a) ) p (q q) (by (b) ) p T (by (c) ) p (by (d) ). (ii) (p q) ( p q) ( q p) ( q p) (by (e) ) q (p p) (by (f) ) q F (by (g) ) q (by (h) ) Without using truth tables, verify the following logical equivalences using Theorem 1. (a) p (p q) and p. (b) p (p q) and p. (c) (p q) ( p q) and p. (d) ( p q) and p q Use the logical equivalences given in Example 1.13, Example 1.14, and Fact 1.1, to rewrite the given compound statements using only the connectives and. (a) (p q) r. (b) ( p q) (r q). (c) (p r) (q r).

24 20 CHAPTER 1. LOGIC 1.13 Negation of a Conditional Statement Example 1.18 Show that (p q) p q Solution. We can prove this using truth tables. However, let us prove this without using truth tables and rather present a proof using the logical equivalences in Theorem 1. We have (p q) ( p q) (by Example 1.13) ( p) ( q) (by De Morgan s Laws) p q (by the Double Negation Law). The negation of a conditional statement can be expressed in words as follows: The negation of if p then q is logically equivalent to p and not q The Contrapositive of a Conditional Statement Recall that the contrapositive of a conditional statement p q is the statement q p. In this section, we show that: A conditional statement is logically equivalent to its contrapositive. Example 1.19 Show that p q q p Solution. We can prove this using truth tables. However, let us prove this without using truth tables and rather present a proof using the logical equivalences in Theorem 1. We have p q p q (by Example 1.13) q p (by the Commutative Laws) ( q) p (by the Double Negation Law) q p (by Example 1.13).

25 1.15. QUANTIFIED STATEMENTS Quantified Statements In the previous sections, we defined and discussed basic properties of compound statements. We were interested in whether a particular statement was true or false. This logic is called propositional logic or statement logic. However there are many arguments whose validity cannot be verified using propositional logic. Consider, for example, the sentence p : x is an even integer. This sentence is neither true nor false. The truth or falsity depends on the value of the variable x. For some values of x the sentence is true; for others it is false. Thus this sentence is not a statement. However, let us denote this sentence by P (x), i.e., P (x) : x is an even integer. Then, P (4) is true, while P (5) is false. To study the properties of such sentences, we need to extend the framework of propositional logic to what is called first-order logic. Definition. A predicate or propositional function is a sentence that contains a finite number of variables and becomes a statement when specific values are substituted for the variables. The domain of a predicate variable is the set of all values that may be substituted in place of the variables. In our earlier example, the sentence P (x) : x is an even integer is a predicate (or propositional function) with domain D the set of integers since for each x D, P (x) is a statement, i.e., for each x D, P (x) is true or false, but not both. Example 1.20 The following are examples of propositional functions: (a) The sentence P (x): x+3 is an odd integer with domain D the set of positive integers. (b) The sentence P (x): x + 3 is an odd integer with domain D the set of integers. (c) The sentence P (x, y, z): x 2 + y 2 = z 2 with domain D the set of positive integers. Some sets are encountered so often that they are given special names. These are summarized below: symbol N Z Q R for the set of natural number (positive integers) integers rational numbers real numbers

26 22 CHAPTER 1. LOGIC Thus, N = {1, 2, 3,...}, Z = {..., 2, 1, 0, 1, 2,...}, and Q = { a b a, b Z and b 0}. A real number that is not a rational number is called irrational The Universal Quantifier Example 1.21 Let P (x) be a predicate with domain D. Then the sentence Q(x) : for all x, P (x) is a statement. To see this, notice that either P (x) is true at each value x D (the notation x D indicates that x is in the set D, while x / D means that x is not in D) or P (x) is false for at least one value x D. If P (x) is true at each value x D, then Q(x) is true. However, if P (x) is false for at least one value x D, then Q(x) is false. Hence, Q(x) is a statement because it is either true or false (but not both). Definition. Each of the phrases every, for every, for each, and for all is referred to as the universal quantifier and is expressed by the symbol. Let P (x) be a predicate with domain D. A universal statement is a statement of the form x D, P (x). It is false if P (x) is false for at least one x D; otherwise, it is true. Example 1.22 Let P (x) be the predicate P (x) : x 2 x. Determine whether the following universal statements are true or false. (a) x R, P (x); (b) x Z, P (x). Solution. (a) Let x = 1 2 R. Then, ( 1 2 )2 = 1 4 < 1 2, and so P ( 1 2 ) is false. Therefore, x R, P (x) is false. (b) For all integers x, x 2 x is true, and so P (x) is true for all x Z. Hence, x Z, P (x) is true.

27 1.17. THE EXISTENTIAL QUANTIFIER 23 Exercises 1.19 Symbolize the following by using quantifiers, predicates, and logical connectives. (a) The square of any real number is nonnegative. (b) All fish can swim. (c) No integer has a square equal to Let P (x) be the predicate x is an odd integer, Q(x) the predicate x is a prime integer and R(x) the predicate x 2 is an odd integer. Write the following symbolic statements in English. (a) x Z, P (x) R(x). (b) x Z, P (x) Q(x) The Existential Quantifier Definition. Each of the phrases there exists, there is, for some, and for at least one is referred to as the existential quantifier and is denoted in symbols. Let P (x) be a predicate with domain D. An existential statement is a statement of the form x D such that P (x). It is true if P (x) is true for at least one x D; otherwise, it is false. Example 1.23 Let P (x) be the predicate P (x) : x 2 < x. Determine whether the following existential statements are true or false. (a) x R, P (x); (b) x Z, P (x). Solution. (a) Let x = 1 2 R. Then, ( 1 2 )2 = 1 4 < 1 2, and so P ( 1 2 ) is true. Therefore, x R, P (x) is true. (b) For all integers x, x 2 x is true, and so there is no integer x such that P (x) is true. Hence, x Z, P (x) is false.

28 24 CHAPTER 1. LOGIC Exercises 1.21 Symbolize the following by using quantifiers, predicates, and logical connectives. (a) Some rational numbers are integers. (b) Some integers are even. (c) There is an integer x such that x 2 = 4. (d) There exists an integer that is odd and prime. (e) Everybody trusts somebody. (f) Somebody trusts everybody. (g) Between every integer and its double there is a prime number Let P (x) be the predicate x is an odd integer and Q(x) the predicate x is a prime integer. Write the following symbolic statement in English. x Z, P (x) Q(x) Negation of Quantified Statements Fact 1.2. The negation of a universal statement of the form x D, P (x) is logically equivalent to an existential statement of the form x D such that P (x). Symbolically, ( x D, P (x)) x D such that P (x). Consider the universal statement x D, P (x). It is false if P (x) is false for at least one x D; otherwise, it is true. Hence it is false if and only if P (x) is false for at least one x D if and only if P (x) is true for at least one x D. Thus the negation of this statement is the statement x D such that P (x).

29 1.18. NEGATION OF QUANTIFIED STATEMENTS 25 Example 1.24 What is the negation of the statement All mathematicians wear glasses? Solution. Let us write this statement symbolically. Let D be the set of all mathematicians and let P (x) be the predicate x wears glasses with domain D. The given statement can be written as x D, P (x). The negation is x D such that P (x). In words, the negation is There exists a mathematician who does not wear glasses or Some mathematicians do not wear glasses. Fact 1.3. The negation of an existential statement of the form x D such that P (x) is logically equivalent to a universal statement of the form x D, P (x). Symbolically, ( x D such that P (x)) x D, P (x). Consider the existential statement x D such that P (x) It is true if P (x) is true for at least one x D; otherwise, it is false. Hence it is false if and only if P (x) is false for all x D if and only if P (x) is true for all x D. Thus the negation of this statement is the statement x D, P (x). Example 1.25 What is the negation of the statement Some politicians are honest? Solution. Let us write this statement symbolically. Let D be the set of all politicians and let P (x) be the predicate x is honest with domain D. The given statement can be written as x D such that P (x). The negation is x D, P (x). In words, the negation is All politicians are not honest or No politician is honest.

30 26 CHAPTER 1. LOGIC Consider next the negation of a universal conditional statement. By Fact 1.2, we have that ( x D, (P (x) Q(x))) x D such that (P (x) Q(x)). But the negation of an if p then q statement is logically equivalent to an p and not q statement (see Example 1.18). Hence, (P (x) Q(x)) P (x) Q(x). Therefore we have the following fact: Fact 1.4. The negation of a universal conditional statement of the form x D, P (x) Q(x) is logically equivalent to a universal statement of the form x D such that (P (x) Q(x)). Symbolically, ( x D, P (x) Q(x)) x D such that (P (x) Q(x)). Written less symbolically, this becomes ( x D, if P (x) then Q(x)) x D such that P (x) and Q(x) Universal Conditional Statements Recall from Section 1.7 that a conditional statement has a contrapositive, a converse, and an inverse. These definitions can be extended to universal conditional statements.

31 1.19. UNIVERSAL CONDITIONAL STATEMENTS 27 Definition. Consider a universal conditional statement of the form x D, P (x) Q(x) 1. Its contrapositive is the statement x D, Q(x) P (x). 2. Its converse is the statement x D, Q(x) P (x). 3. Its inverse is the statement x D, P (x) Q(x). Example 1.26 Write the contrapositive, converse, and inverse of the statement: If a real number is greater than 3, then its square is greater than 9. Solution. Symbolically, the statement can be written as: x R, if x > 3 then x 2 > 9. (Here P (x) is the statement x > 3 and Q(x) the statement x 2 > 9.) 1. The contrapositive is: x R, if x 2 9 then x 3, or, equivalently, x R, if x 2 9 then x The converse is: x R, if x 2 > 9 then x > 3. (Note that the converse is false; take, for example, x = 4. Then, ( 4) 2 > 9 is true but 4 > 3 is false. Hence the statement if ( 4) 2 > 9 then 4 > 3 is false. Hence the universal statement x R, if x 2 > 9 then x > 3 is false.) 3. The inverse is: x R, if x 3 then x 2 9, or, equivalently, x R, if x 3 then x 2 9.

32 28 CHAPTER 1. LOGIC Exercises 1.23 Write the contrapositive, converse, and inverse of the following statements: (a) If the square of an integer is odd, then the integer is odd. (b) If an integer is divisible by 4, then it is even. (c) x R, if x(x + 1) > 0 then x > 0 or x < Review of Symbols We close this chapter with a review of the symbols that we have introduced. negation (not) disjunction (or) conjunction (and) implication biconditional universal quantifier (for every) existential quantifier (there exists)

33 Chapter 2 Proof Techniques 2.1 Introduction In this chapter, we discuss the topic of mathematical proofs. Our goal is to introduce the student to several important proof techniques for verifying mathematical statements. For a given true mathematical statement, how exactly can we verify that it is true? Finding answers to mathematical questions is all very well, but we must be certain that we are right and we must be able to able to convince others, not just ourselves, of this! A true mathematical statement is called a result. Interesting or significant mathematical results are called theorems (or propositions). For example, the mathematical statement 1+1 = 2 is true, but we would not call this a theorem, but rather a result. A corollary is a mathematical result that can be deduced from or is a consequence of some earlier result. A lemma is a mathematical result that is useful in proving another (more interesting) result. A lemma can be thought of as a helping result to prove some other result. Before presenting several proof techniques, we will need some elementary definitions in number theory. Definition. An integer n is even if and only if n = 2k for some integer k. An integer n is odd if and only if n = 2k + 1 for some integer k. Example 2.1 (a) The integer 8 is even since 8 = 2 4 (i.e., 8 = 2k where k = 4 Z). (b) The integer 5 is odd since 5 = 2( 3) + 1 (i.e., 5 = 2k + 1 where k = 3 Z). We shall show in Section 2.6 (using the so-called quotient-remainder theorem) that every integer is either even or odd. 29

34 30 CHAPTER 2. PROOF TECHNIQUES Definition. An integer n is prime if and only if n > 1 and for all positive integers r and s, if n = r s, then r = 1 or s = 1. An integer n is composite if and only if n = r s for some positive integers r and s with r 1 and s 1. Example 2.2 (a) The first six prime numbers are 2, 3, 5, 7, 11, 13. (b) The first six composite numbers are 4, 6, 8, 9, 10, 12. (c) Every integer greater than 1 is either prime or composite since the two definitions are negations of each other (see Section 1.18!). Definition. Two integers m and n are said to be of the same parity if m and n are both even or are both odd, while m and n are said to be of the opposite parity if one of m and n is even and the other is odd. Two integers are consecutive if one is one more than the other. So if one integer is n, the next consecutive integer is n + 1. Example 2.3 (a) The integers 2 and 18 are of the same parity (since they are both even). (b) The integers 7 and 12 are of the opposite parity (one is even, the other odd). Definition. Let n and d be integers with d 0. Then n is said to be divisible by d if n = d k for some integer k. Alternatively, we say that n is a multiple of d, or d is a factor of n, or d is a divisor of n, or d divides n. The notation d n is read d divides n. Example 2.4 (a) Is 16 divisible by 8? (b) Does 3 divide 21? (c) Is 21 a multiple of 7? (d) Does 5 30? (e) Is 7 a factor of 42? (f) Is 8 a factor of 8? Solution (a) Yes: 16 = 8 2. (b) Yes: 21 = 3 7. (c) Yes: 21 = 7 3. (d) Yes: 30 = 5 6. (e) Yes: 42 = 7 6. (f) Yes: 8 = 8 ( 1).

35 2.2. TRIVIAL PROOFS 31 Exercises 2.1 Let m and n be (fixed) integers. Justify your answers to each of the following questions: (a) Is 4mn + 10n even? (b) Is 8mn + 5 odd? (c) If m and n are positive, is m 2 + 2mn + n 2 composite? (d) If m n + 2 3, is m 2 n 2 composite? 2.2 (a) Is 45 divisible by 9? (b) Is 5k(3k + 6) divisible by 3? (c) Is 24 a multiple of 8? (d) Does 11 54? (e) Is 6 a factor of 42? (f) If n = 4k + 3, does 8 divide n 2 1? 2.2 Trivial Proofs Trivial proof. Let P (x) and Q(x) be statements with domain D. If Q(x) is true for all every x D, then the universal statement x D, P (x) Q(x) is true (regardless of the truth value of P (x)). Recall from the Truth Table 1.5, that if Q(x) is a true statement, then so too is the implication P (x) Q(x). Furthermore, the universal statement x D, P (x) Q(x) is a true statement provided P (x) Q(x) is true for every x D. In particular if Q(x) is true for all every x D, then immediately we can deduce that the statement x D, P (x) Q(x) is true. Such a proof we call a trivial proof. Result 2.1 For x R, if x > 5, then x > 0. Proof. Consider the statements P (x) : x > 5 and Q(x) : x > 0. Since x 2 0 for every x R, it follows that x > 0 for every x R. Hence, Q(x) is true for every x R. Thus, P (x) Q(x) is true for every x R, i.e., for x R, if x > 5, then x > 0.

36 32 CHAPTER 2. PROOF TECHNIQUES The symbol that occurs at the end of the proof of Result 2.1 indicates that the proof is complete. Note that the proof of Result 2.1 does not depend on x > 5. We could have, in fact, replaced x > 5 by any hypothesis and the result would still be true. It would have been better to replace the statement of Result 2.1 by If x R, then x > 0. We remark that the proof of Result 2.1 could be simplified if we do not introduce the statements P (x) and Q(x): Proof of Result 2.1. Since x 2 0 for every x R, it follows that x > 0 for every x R. Hence, x > 0. Result 2.2 If n is an odd integer, then 6n 3 + 4n + 3 is an odd integer. Proof. Since 6n 3 +4n + 3 = 2(3n 3 + 2n +1)+1 where 3n 3 +2n +1 Z (i.e., 6n 3 +4n +3 = 2k + 1 where k = 3n 3 + 2n + 1 Z), the integer 6n 3 + 4n + 3 is odd for every integer n. Note that in Result 2.2 the fact that 6n 3 + 4n + 3 is odd does not depend on n being odd. It would have been better to replace the statement of Result 2.2 by If n is an integer, then 6n 3 + 4n + 3 is odd. Exercises 2.3 For x R, prove that if x < 1, then x > Prove that if n is an odd integer, then 4n 3 + 6n is an even integer. 2.3 Vacuous Proofs Vacuous proof. Let P (x) and Q(x) be statements with domain D. If P (x) is false for all every x D, then the universal statement x D, P (x) Q(x) is true (regardless of the truth value of Q(x)). Recall from the Truth Table 1.5, that if P (x) is a false statement, then the implication P (x) Q(x) is true. Hence if P (x) is false for all every x D, then we can deduce that the statement x D, P (x) Q(x) is true. Such a proof we call a vacuous proof.

37 2.4. DIRECT PROOFS 33 Result 2.3 For x R, if x 2 2x + 1 < 0, then x > 4. Proof. Consider the statements P (x) : x 2 2x+1 < 0 and Q(x) : x > 4. Since x 2 2x+1 = (x 1) 2 0 for every x R, we have that x 2 2x + 1 < 0 is false for every x R. Hence, P (x) is false for every x R. Thus, P (x) Q(x) is true for every x R, i.e., for x R, if x 2 2x + 1 < 0, then x > 4. Note that in the proof of Result 2.3, the truth value of the statement x > 4 plays no role whatsoever. We could have, in fact, replaced x > 4 by any conclusion and the result would still be true. Exercises 2.5 Let x R. Prove that if 2x 2 4x + 4 < 0, then x Direct Proofs One of the most important proof techniques is the method of direct proof. Direct proof. Let P (x) and Q(x) be statements with domain D. If P (x) Q(x) is true for all x D for which P (x) is true, then the universal statement x D, P (x) Q(x) is true. Such a proof we call a direct proof. Thus to give a direct proof of the above universal statement, we assume P (x) is true for some particular but arbitrary element x D, and then show that Q(x) is true for this element x. Recall that the universal statement x D, P (x) Q(x) is true provided P (x) Q(x) is true for every x D. If P (x) is a false statement, then the implication P (x) Q(x) is true. Hence to show that the universal statement x D, P (x) Q(x) is true for every x D, we need only show that P (x) Q(x) is true for all x D for which P (x) is true. Such a proof we call a direct proof.

38 34 CHAPTER 2. PROOF TECHNIQUES Result 2.4 If n is an even integer, then 3n + 5 is an odd integer. Proof. If we let P (n): n is even and Q(n): 3n + 5 is odd, then we need to show that the universal statement n Z, P (n) Q(n) is true. To do this, we assume P (n) is true for some particular but arbitrary element n Z and show that Q(n) is true for this element n. Since P (n) is true, n = 2k for some integer k. Hence, 3n + 5 = 3(2k) + 5 = 6k + 5 = 2(3k + 2) + 1 = 2m + 1, where m = 3k + 2. Since k Z, we must have m Z (since the product of two integers is an integer, and the sum and difference of two integers is an integer). Hence, 3n + 5 = 2m + 1 for some integer m, whence Q(n) is true. Thus by the method of direct proof, we have proven our desired result. We remark that the proof of Result 2.4 could be simplified if we do not introduce the statements P (x) and Q(x). Proof of Result 2.4. Assume that n is an even integer. Then, n = 2k for some integer k. Hence, 3n + 5 = 3(2k) + 5 = 6k + 5 = 2(3k + 2) + 1 = 2m + 1, where m = 3k + 2. Since k Z, we must have m Z. Hence, 3n + 5 = 2m + 1 for some integer m, whence 3n + 5 is an odd integer. Result 2.5 If n is an odd integer, then 5n + 3 is an even integer. Proof. Assume that n is an odd integer. Then, n = 2k + 1 for some integer k. Hence, 5n + 3 = 5(2k + 1) + 3 = 10k + 8 = 2(5k + 4) = 2m, where m = 5k + 4. Since k Z, we must have m Z. Hence, 5n + 3 = 2m for some integer m, whence 5n + 3 is an even integer. Result 2.6 If n is an odd integer, then n 2 + n is even. Proof. Assume that n is an odd integer. Then, n = 2k + 1 for some integer k. Hence n 2 + n = (2k + 1) 2 + (2k + 1) = 4k 2 + 6k + 2 = 2m, where m = 2k 2 + 3k + 1. Since k Z, we must have m Z. Hence, n 2 + n is even. Result 2.7 If the sum of any two integers is even, then so is their difference. Proof. Assume that m and n are (particular but arbitrarily chosen) integers such that m + n is even. (We show that m n is even.) Then, m + n = 2k for some integer k. Thus, m = 2k n. Hence, m n = (2k n) n = 2k 2n = 2(k n) = 2l,

39 2.5. PROOF BY CONTRAPOSITIVE 35 where l = k n. Since k, n Z and the difference between two integers is an integer, l Z. Hence, m n = 2l where l Z. Thus, m n is even. Exercises 2.6 Use the method of direct proof to prove that: (a) The product of two odd integers is an odd integer. (b) The sum of two even integers is an even integer. (c) If n is an odd integer, then 5n is an even integer. (d) If n is an even integer, then 3n 2 4n 5 is an odd integer. (e) If n is an even integer, then 5n 3 is an even integer. (f) If n Z and 7n 3 is odd, then n is even. 2.7 Find the mistake in the proof of the following result, and provide a correct proof. Result. If m is an even integer and n is an odd integer, then 2m + 3n is an odd integer. Proof. Since m is an even integer and n is an odd integer, m = 2k and n = 2k + 1 for some integer k. Therefore, 2m + 3n = 2(2k) + 3(2k + 1) = 10k + 3 = 2(5k + 1) + 1 = 2l + 1, where l = 5k + 1. Since k Z, l Z. Hence, 2m + 3n = 2l + 1 for some integer l, whence 2m + 3n is an odd integer. 2.8 Find the mistake in the proof of the following result, and provide a correct proof. Result. For all integers n 1, n 2 + 2n + 1 is composite. Proof. Let n = 4. Then, n 2 + 2n + 1 = (4) + 1 = 25 and 25 is composite. 2.5 Proof by Contrapositive Recall that the contrapositive of a conditional statement P (x) Q(x) is the statement Q(x) P (x). In Section 1.14, we showed that a conditional statement P (x) Q(x) is logically equivalent to its contrapositive Q(x) P (x).

40 36 CHAPTER 2. PROOF TECHNIQUES Let P (x) and Q(x) be statements with domain D. A proof by contrapositive of the statement x D, P (x) Q(x) is a direct proof of its contrapositive x D, Q(x) P (x); that is, we assume that Q(x) is true for some particular but arbitrary element x D, and then show that P (x) is true for this element x. Result 2.8 Let n Z. If n is odd, then n is even. Proof. Let P (n) be the statement n is odd and let Q(n) be the statement n is even. Then we need to show that the universal statement n Z, P (n) Q(n) is true. To do this, we use a proof by contrapositive. We give a direct proof to show that Q(n) P (n). Hence we assume that Q(n) is true for some particular but arbitrary element n Z and show that P (n) is true for this element n. Since Q(n) is true, n is not even. Thus, n is odd, and so n = 2k + 1 for some integer k. Hence, n = (2k + 1) = 4k 2 + 4k + 6 = 2(k 2 + 2k + 3) = 2m, where m = k 2 + 2k + 3. Since k Z, we must have m Z. Hence, n = 2m for some integer m, and so n is even, i.e., n is not an odd integer. Thus, P (n) is true. Therefore by the method of direct proof, we have proven that Q(n) P (n) is true. Hence, P (n) Q(n) is true. Therefore, n Z, P (n) Q(n). We remark that the proof of Result 2.8 could be simplified if we do not introduce the statements P (x) and Q(x). Proof of Result 2.8. Assume that n is odd. Then, n = 2k + 1 for some integer k. Hence, n = (2k + 1) = 4k 2 + 4k + 6 = 2(k 2 + 2k + 3) = 2m, where m = k 2 + 2k + 3. Since k Z, we must have m Z. Hence, n = 2m for some integer m, and so n is an even integer. Result 2.9 Let n Z. If n 2 is even, then n is even. Proof. We use a proof by contrapositive. Assume that n is odd. Then, n = 2k + 1 for some integer k. Hence, n 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 2(k 2 + 2k) + 1 = 2m + 1, where m = k 2 + 2k. Since k Z, we must have m Z. Hence, n 2 = 2m + 1 for some integer m, and so n 2 is an odd integer.