3.1 Trivial and Vacuous Proofs

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1 CH3: DIRECT PROOF AND PROOF BY CONTRAPOSITIVE Lemma = is a mathematical result that is useful in verifying the truth of another result. Theorem/ Proposition = a true mathematical statement (that are especially significant) (Propositions are usually more important, and harder to prove than theorems.) Corollary = is a mathematical result that is a consequence of an earlier result. Most lemmas, theorems, propositions and corollaries are stated as an implication, or a characterization. 3.1 Trivial and Vacuous Proofs Most mathematical results are stated as an implication: P Q. Recall: P Q P Q T T T T F F F T T F F T Let S be a universal set, P (x) and Q(x) be two open sentences. For x S, it can be shown that Q(x) is true regardless of the truth value of P (x). Then P (x) Q(x) is true for all x S. In this case, P Q is true trivially. Such a proof is called a trivial proof. Result: Let x R. If x > 0, then x > 0. Proof: Since x 2 0, for all x R, it follows that x > x 2 0. Hence x Note that we didn t use the fact that x > 0, since it holds for all x R. Now, if it can be shown that P (x) is false for all x S (regardless of the truth value of Q(x)), then P (x) Q(x) is true for all values of x S. That is P Q is true vacuously. Such a proof is called a vacuous proof. Result: Let x R. If x < 0, then x 5 4. Proof Observe that x > x 2 0. Thus x < 0 is false for all x S, and so the implication is true Direct Proofs Direct Proof of P Q: Assume that P (x) is true for an arbitrary x S, and show that Q(x) is true for this x. In order to illustrate this type of proof we assume that we know: 1. The negative of an integer is an integer 2. The sum/difference/product of two integers is an integer 1

2 3. An integer n is even if n = 2k, k Z 4. An integer n is odd if n = 2k + 1, k Z. Result: 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 k Z. Thus 5n + 3 = 5(2k + 1) + 3 = 10k + 8 = 2(5k + 4). Since 5k + 4 is an integer, it follows that 5n + 3 is odd. Result: If n is an odd integer, then 4n 3 + 2n 1 is odd. Proof Assume that n is odd. So n = 2k + 1, for some integer k. Thus 4n 3 + 2n 1 = 4(2k + 1) 3 + 2(2k + 1) 1 = 4(4k k 2 + 6k + 1) + 4k = 16k k k + 5 = 2(8k k k + 2) + 1. Since 8k k k + 2 is an integer (as k Z), it follows that 4n 3 + 2n 1 is odd. Result : If n is an even integer, then 3n is odd. Proof Assume that n is an even integer. Then n = 2k, for some k Z. Thus 3n = 3(2k) = k = 2(2 4 3k ) + 1 = 2(48k ) + 1. Since 48k Z, it follows that 3n5 + 2 is odd. What s wrong with the proof? Be careful with what you claim to be an integer!! The result is false, but it can be fixed: If n is an even integer, then 3n is even. When writing a proof: 1. Write a proof so that somebody else can read it. 2. Write complete sentences, starting with Proof and ending with. 3. When introducing a new variable/symbol explain what the symbol is, and what set the variable belongs to. 4. If your equation wraps arund, then the equal sign goes at the end of the line, and not before the new line nor both within the text. Or you may choose to center it. 2

3 For example, within the text you should use it like this ab = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1. or ab = 4kl + 2k + 2l + 1 = 2(2kl + k + l) Proof by Contrapositive The contrapositive of an implication P Q is the implication Q P. Ex: Let P (x) : x = 2, and let Q(x) : x 2 = 4. P Q : If x = 2, then x 2 = 4. (T) Q P : If x 2 4, then x 2. (T) The implication P Q and Q P are logically equivalent: 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 A proof by contrapositive of P Q is a direct proof of Q P. Result: Let x Z. If 3x 15 is even, then x is odd. Proof Assume that x is even. Then x = 2a, for some integer a. Then 3x 15 = 3(2a) 15 = 6a 15 = 2(?) + 1 = 2(3a 8) + 1. Since 3a 8 Z, 3x 15 is odd. We can also prove it using a direct proof (use x = (3x 15) (2x 15)): Result: Let x Z. If 3x 15 is even, then x is odd. Proof Assume that 3x 15 is even. Then 3x 15 = 2a, for some integer a. Then x = (3x 15) (2x 15) = 2a 2x + 15 = 2(?) + 1 = 2(a x + 7)

4 Since a x + 7 Z, x is odd. Result: Let x Z. Then 5x + 3 is even if and only if x is odd. Proof Assume that x is even. Then x = 2k, k Z. Thus 5x + 3 = 5(2k) + 3 = 10k + 3 = 2(5k + 1) + 1. Since 5k + 1 Z, it follows that 5x + 3 is even. For the converse, let x be odd. So x = 2k + 1, k Z. Hence, 5x + 3 = 5(2k + 1) + 3 = 10k + 8 = 2(5k + 4). There are two implications to prove here: (1) If 5x + 3 is even, then x is odd. (2) If x is odd, then 5x + 3. The next result is useful to use, so we refer to it as a theorem. Theorem: Let x Z. Then x 2 is even if and only x is even. Proof: Assume that x is odd. Thus x = 2k + 1, for some k Z. Therefore x 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1. Since 2k 2 + 2k is an integer, x 2 is odd. For the converse, assume that x is even. Then x = 2k, for some k an integer. Thus x 2 = (2k) 2 = 4k 2 = 2(2k 2 ). Recall that the contrapositive of P Q is Q P. The contrapositive of the Theorem above gives us another theorem: Theorem: Let x Z. Then x 2 is odd if and only x is odd. 3.4 Proof by Cases Result: Let n Z. Then n 2 + 3n + 5 is an odd integer. Proof We proceed by cases, according to whether n is even or odd. 1. Case 1. n is even. Then n = 2x, x Z. So n 2 + 3n + 5 = 4x 2 + 6x + 5 = 2(2x 2 + 3x + 2) Case 2.n is odd. Then n = 2x + 1, x Z. So n 2 + 3n + 5 = 4x 2 + 4x x = 2(2x 2 + 5x + 4)

5 Thus n 2 + 3n + 5 is an odd integer. Note: The result must be true for all the cases. If it fails for one case, then the result is false. What are the typical cases? Ex1: Case 1. n is even Case 2. n is odd Ex2: Case 1. n > 0 Case 2. n = 0 Case 3. n < 0 Ex3: We can have a combination of x and y Case 1. xy < 0 Subcase 1.1. x > 0 and y > 0 Subcase 1.2. x < 0 and y < 0 Case 2. xy = 0 Then x = 0 or y = 0. Case 3. xy > 0 Subcase 3.1. x > 0 and y < 0 Subcase 3.2. x < 0 and y > 0 Two integers x and y are of the same parity if x and y are both even or both odd. Two integers x and y are of opposite parity if x is even and y is odd, or x is odd and y is even. Result: Let x, y Z. Then 3x + 3y is an odd integer if and only iff x and y are of opposite parity. Proof Assume that x and y are of the same parity. We consider two cases. 1. Case 1. x and y are both even. Then x = 2k, y = 2l, k, l Z. So 3x + 3y = 6k + 6l = 2(3k + 3l). Since 3k + 3l is an integer, 3x + 3y is even. 2. Case 2. x and y are both odd. Then x = 2k + 1, y = 2l + 1, k, l Z. So 3x + 3y = 6k l + 3 = 2(3k + 3l + 3). Since 3k + 3l + 3 is an integer, 3x + 3y is even. Thus 3x + 3y is an even integer. We now verify the converse. Assume that x and y are of opposite parity. We consider two cases. 1. Case 1. x is even and y is odd. Then x = 2k, y = 2l + 1, k, l Z. So 3x + 3y = 6k + 6l + 3 = 2(3k + 3l + 1) + 1. Since 3k + 3l + 1 is an integer, 3x + 3y is odd. 2. Case 2. x is odd and y is even. Then x = 2k + 1, y = 2l, k, l Z. So 3x + 3y = 6k l = 2(3k l)

6 Since 3k + 3l + 1 is an integer, 3x + 3y is odd. Thus 3x + 3y is an odd integer. Note: In this particular theorem, we could have said: Without loss of generality (WLOG) assume that x is even and y is odd (and this would also cover the case of x is odd and y is even.) However, this is not always possible: if you would have to show that 3x + y is either even or odd, you cannot use WLOG since this s not a symmetric expression. Result: Let a and b be integers. Then ab is even iff a is even or b is even. Proof: Assume that a is odd and b is odd. Then a = 2k + 1, b = 2l + 1, k, l Z. So ab = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1. Since 2kl + k + l is an integer, ab is odd. For the converse, assume that a is even or b is even. Without loss of generality say that a is even. Then a = 2k, k Z. So Since kb is an integer, ab is even. ab = (2k)b = 2(kb). Result: Let x, y Z. Prove that if xy and x + y are even, then both x and y are even. Proof: Assume that x is odd or y is odd, WLOG say x is odd. We wish to show that xy or x + y is odd. Then x = 2k + 1. So y could be even or y could be odd. We consider two cases. 1. Case 1. y is odd. Then y = 2l + 1, l Z. So xy = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1. Since 2kl + k + l is an integer, xy is odd. 2. Case 2. y is even. Then y = 2l, l Z. So Since k + l is an integer, x + y is odd. x + y = (2k + 1) + 2l = 2(k + l) + 1. Read Proof evaluations: gives you good hints on what not to do in terms of mistakes. 6

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