MATH 225: Foundations of Higher Matheamatics Dr. Morton 3.4: Proof y Cases Chapter 3 handout page 12 prolem 21: Prove that for all real values of y, the following inequality holds: 7 2y + 2 2y 5 7. You often think of taking cases in dealing with asolute values. We have 2y + 2 if 2y + 2 0 2y + 2 = (2y + 2) if 2y + 2 < 0. Now 2y + 2 0 means 2y 2, so y 1. Likewise, and 2y + 2 < 0 means 2y < 2, i.e. y < 1. So 2y + 2 if y 1 2y + 2 = 2y 2 if y < 1. In the same way, 2y 5 if 2y 5 0 2y 5 = (2y 5) if 2y 5 < 0. Now 2y 5 0 means 2y 5, so y 5 2. Likewise, and 2y 5 < 0 means 2y < 5, i.e. y < 5 2. So 2y 5 if y 5 2y 5 = 2 2y + 5 if y < 5 2. Given the way the functions are roken apart, there are three cases: y < 1, 1 y < 5 2, and y 5 2. Notice that all real numers are in one of the three cases. (Continued on next page) 1
Case 1. y < 1. In this case, 2y + 2 2y 5 = (2y + 2) [ 2y + 5] = 2y 2 + 2y 5 = 7. Therefore, 7 2y + 2 2y 5 7 holds in this case. Case 2. 1 y < 5 2. In this case, 2y + 2 2y 5 = (2y + 2) [ 2y + 5] = 4y 3. We have to do some additional work to see whether the target inequality holds. We have 1 y < 5, so 4 < 4x 10, and 7 < 4y 3 7. 2 Therefore, 7 2y + 2 2y 5 7 holds in this case. Case 3. y 5 2. In this case, 2y + 2 2y 5 = (2y + 2) (2y 5) = 7. Therefore, 7 2y + 2 2y 5 7 holds in this case. Since 7 2y + 2 2y 5 7 holds all three cases, it is true for all y R. QED 2
Chapter 3 handout page 8 numer 3: Let n Z. If n 2 +n is odd then n 4 3n < 0. Proof: Let n Z. We have n 2 + n = n(n + 1). Now every integer n is either even or odd. We need two cases. n is even: Then n = 2k for some k Z, so n 2 +n = (2k) 2 +(2k) = 4k 2 +2k = 2(2k 2 +k). Let l = 2k 2 + k. Note that l Z ecause 2, k Z (y axioms). Thus n 2 + n = 2l where l is an integer, so n 2 + n is even. Thus in this case the hypothesis is false. The statement is therefore vacuously true in this case. n is odd: Then n = 2k + 1 for some k Z, so n 2 + n = (2k + 1) 2 + (2k + 1) = 4k 2 + 4k + 1 + 2k + 1 = 4k 2 + 6k + 2 = 2(2k 2 + 3k + 1). Let l = 2k 2 + 3k + 1. Note that l Z ecause 2, k, 3, 1 Z (y axioms). Thus n 2 + n = 2l where l is an integer, so n 2 + n is even. Thus in this case the hypothesis is false. The statement is therefore vacuously true in this case. In all cases the hypothesis is false, thus the theorem is true for all n Z. QED Chapter 3 handout page 8 prolem 4: Prove: If x is an integer, then x 2 x. Proof: By cases: Case 1: If x = 0, then x 2 = 0 2 = 0 = x, so x 2 x, in this case. Case 2: If x = 1, then x 2 = 1 2 = 1 = x, so x 2 x, in this case. Case 3: If x > 1, then x 2 = x x > x 1 = x (y axioms). Thus x 2 x, in this case. Case 4: If x < 0, then x 2 > 0 > x (again, y axioms). Thus x 2 x, in this case. This accounts for all integer values of x. 3
Chapter 3 handout page 8 prolem 7: Let a, Z. Prove that if a + and a are of the same parity, then a and are oth even. We prove the contrapositive: Let a, Z. If at least one of a, is odd then a + and a are of opposite parity. We need three cases. Case 1: a, are oth odd Case 2: a is even and is odd. Case 3: a is odd and is even. Note, though that cases 2 and 3 can e comined as oth a + and a are symmetric etween a and. So we will have two cases. Proof: (of Contrapositive): Case 1: a, are oth odd. Then there exist k, l Z such that a = 2k + 1, = 2l + 1. Then a + = (2k + 1) + (2l + 1) = 2k + 2k + 2 = 2(k + l + 1) and a = (2k + 1) (2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl +k +l) +1. Let m = k +l +1 and n = 2kl +k +l. Then m, n Z as k, l, 1, 2 Z (axioms). Thus a + = 2m, a = 2n + 1 where m, n Z, so a + is even, a is odd, and they have opposite parity in this case. Case 2: One of a, is even. Without loss of generality, assume a is even and is odd. Then there exist k, l Z such that a = 2k, = 2l + 1. Then a + = (2k) + (2l + 1) = 2k +2k +1 = 2(k +l)+1 and a = (2k) (2l+1) = 4kl+2k = 2(2kl+k). Let m = k +l and n = 2kl+k. Then m, n Z as k, l, 1, 2 Z (axioms). Thus a+ = 2m+1, a = 2n where m, n Z, so a + is odd, a is even, and they have opposite parity in this case. Thus in oth cases, a and a + have opposite parity. QED 4
Chapter 3 handout page 8 prolem 9: Prove that if a and are real numers, where 0, then a = a. Proof: Assume that a and are real numers and 0. We will consider the following cases. CASE 1 : Assume that a 0 and > 0, so that a 0 (y axioms). Then a = a, =, and a = a = a. CASE 2 : Assume that a 0 and < 0, so that a 0 (y axioms). Then a = a, =, and a = a = a = a. CASE 3 : Assume that a < 0 and > 0, so that a < 0 (y axioms). Then a = a, =, and a = a = a = a. CASE 4 : Assume that a < 0 and < 0, so that a > 0 (y axioms). Then a = a, =, and a = a = a = a. Thus, for all possile cases, it has een proven that a = a. QED Chapter 3 handout page 8 prolem 10: Prove that for all real values of a and, a = a. Proof: Assume that a and are real numers. We will consider the following cases. CASE 1 : Assume that a 0 and 0, so that a 0 (y axioms). Then a = a, =, and a = a = a. CASE 2 : One of a, 0 and one of a, < 0. Without loss of generality, assume that a 0 and < 0, so that a 0 (y axioms). Then a = a, =, and a = a = a = a. CASE 3 : Assume that a < 0 and < 0, so that a > 0 (y axioms). Then a = a, =, and a = a = a = a. Thus, for all possile cases, it has een proven that a = a. QED 5
Page 10 Chapter 3 handout, prolem 2: Discuss the following proof: The difference etween any odd integer and any even integer is odd. Proof: Suppose n is any odd integer and m is any even integer. By definition of odd, n = 2k + 1 where k is an integer, and y definition of even m = 2k where k is an integer. Then n m = 2k + 1 2k = 1. But 1 is odd. Therefore, the difference etween any odd integer and any even integer is odd. Solution: This proof uses k to define oth m and n, which relates them. But they were supposed to e general integers. They should have used n = 2k + 1, m = 2l where k, l Z. Page 10 Chapter 3 handout, prolem 3: Discuss the following proof: The product of an even integer and an odd integer is even. Proof: Suppose m is an even integer and n is an odd integer. If m n is even, then y definition of even, there exists an integer r such that m n = 2r. Also, since m is even, there exists an integer p such that m = 2p, and since n is odd, there exists an integer q such that n = 2q + 1. Thus m n = (2p) (2q + 1) = 2r where r is an integer. Thus, y definition of even, m n is even, as was to e shown. Solution: This proof starts out y assuming that the product is even, which is what was supposed to e proved, not assumed! Page 10 Chapter 3 handout, prolem 4: Discuss the following proof: The sum of any two even integers equals 4k for some integer k. Proof: Suppose m, n are any two even integers. By definition of even, m = 2k for some integer k and n = 2k for some integer k. By sustitution, m + n = 2k + 2k = 4k. This is what was to e shown. Solution: Again, this proof uses the same variale to define m and n. (Note: This theorem is false: 2+4=6, which is not a multiple of 4.) Page 10 Chapter 3 handout, prolem 5: Discuss the following proof: If m + n is even, then m and n are even. Proof: Suppose m, n are even. Then m = 2k, n = 2l where k, l Z. Then m + n = 2k + 2l = 2(k + l). Let p = k + l. Note that p Z ecause k, l Z. Thus m + n = 2p where p Z, so m + n is even. Solution: This proof proved the converse of the statement, which is not logically equivalent to the original statement. Whereas the converse is true, the original statement is false (for example, 1+3=4 is even, ut neither 1 nor 3 is even.) 6