Direct Proof and Proof by Contrapositive
Section 3.5: Proof Evaluations
5 37. Below is a proof o f a result.
Proof We consider two cases.
Case 1. a and b are even. Then a = 2r and b = 2s for integers r and s. Thus a 2 - b 2 = (2r )2 - (2s)2 = Ar2 - 4s 2 = 2(2r 2 - 2s2).
Since 2r 2 — 2s 2 is an integer, a 2 — b 2 is even.
Case 2. a and b are odd. Then a = 2r + 1 and b = 2s + 1 for integers r and s. Thus a 2 — b2 — (2r + l )2 - (2s + l )2 = (Ar2 + Ar + 1) - (As2 + As + 1)
= 4/-2 + Ar - As2 - 45 = 2 (2 r2 + 2r - 2s 2 - 2s).
Since 2r 2 + 2r — 2s 2 — 2s is an integer, a 2 — b 2 is even. ■
W hich o f the following is proved?
(1) Let a , b e Z. Then a and b are o f the same parity if and only if a 2 — b2 is even.
(2) Let a , b e Z. Then a 2 — b 2 is even.
(3) Let a, b e Z. If a and b are o f the same parity, then a 2 — b2 is even.
(4) Let a, b e Z. If a 2 — b 2 is even, then a and b are of the same parity.
5 58. Below is given a proof o f a result. W hat result is being proved?
P ro o f Assume that x is even. Then x = 2a for some integer a. So
3 x J - 4x - 5 = 3(2 a)2 - 4(2a) - 5 = 12a2 - 8a - 5 = 2 (6a2 - 4a - 3) + 1.
Since 6a 2 — 4a — 3 is an integer, 3.r2 — 4x — 5 is odd.
For the converse, assume that x is odd. So x = 2b + 1, where b e Z. Therefore,
3x2 - 4.v - 5 = 3(2/7 + l )2 - 4(2b + 1) - 5 = 3(4i>2 + 4b + 1) - &b - 4 - 5
= 12ft2 + 4b - 6 = 2(6b2 + 2b - 3).
Since 6b 2 + 2b — 3 is an integer, 3x2 — 4x — 5 is even.
3.39. Evaluate the proof of the following result.
R esult Let n e Z. If 3n — 8 is odd, then n is odd.
P ro o f Assume that n is odd. Then n = 2 k + \ for some integer k. Then 3n - 8 = 3(2* + 1) - 8 = 6k + 3 “ 8 = 6k — 5 = 2(3* — 3) + 1. Since 3k — 3 is an integer, 3n — 8 is odd.
3.40. Evaluate the proof o f the following result.
R esult Let a , b e Z. Then a — b i s even if and only if a and b are o f the same parity.
P ro o f We consider two cases.
Case 1. a and b are o f the same parity. We now consider two subcases.
Subcase 1.1. a and b are both even. Then a — 2 x and b = 2y, where x , y e Z. Then a — b = 2x — 2y = 2(x — y). Since x — y is an integer, a — b is even.
Subcase 1.2. a and b are both odd. Then a = 2x + 1 and b = 2y + 1, where x , y e Z. Then a — b = (2x + 1) — (2y + 1) = 2{.v — y). Since x — y is an integer, a — b is even.
Case 2. a and b are o f opposite parity. We again have two subcases.
Subcase 2.1. a is odd and b is even. Then a = 2x + 1 and b = 2 y, where x , y e Z. Then a — b = (2x + 1) — 2y = 2(x — y) + 1. Since x — y is an integer, a — b is odd.
Subcase 2.2. a is even and b is odd. Then a = 2x and b = 2 y + 1, where x, y e Z. Then
a — b — 2x — (2 y + 1) = 2x — 2_v — 1 = 2(x — y — 1) + 1. Since x — y — 1 is an integer, a — b i s odd.
3.41. The following is an attempted proof of a result. W hat is the result and is the attempted proof correct?
P ro o f Assume, w ithout loss of generality, that x is even. Then x = 2a for some integer a. Thus x y 2 = {2a)y2 = 2(ay 2).
Since a y 2 is an integer, x y 2 is even.
3.42. Given below is a proof of a result. W hat is the result?
P ro o f Assume, without loss o f generality, that x and y are even. Then x = 2a and y = 2b for integers a and b. Therefore,
x y + x z + y z = (2a)(2b) + (2 a)z + (2 b)z = 2(2 ab + az + bz).
Since 2ab + az + bz is an integer, xy + x z + y z is even.
3.43. W hat result is being proved below, and what procedure is being used to verify the result?
First, we present the following proof.
9 6 C h a p t e r 3 Direct Proof and Proof by Contrapositive
Additional Exercises for Chapter 3 9 7
Proof Assume that x is even. Then x = 2a for some integer a. Thus
7.v - 3 = 7(2a) - 3 = 14a - 3 = 14a - 4 + 1 = 2(7a - 2) + 1.
Since l a — 2 is an integer, I x — 3 is odd. •
We are now prepared to prove our main result.
Proof Assume that I x - 3 is even. From the result above, X is odd. So x = 2b + 1 for some integer b. Thus 3,v + 8 = 3(2 b + 1) + 8 = 6b + 11 = 2(3 b + 5) + 1.
Since 3b + 5 is an integer, 3x + 8 is odd. ■
5 .44. Consider the following statement.
Let n G Z. Then (n - 5)(n + 7)(« + 13) is odd if and only if n is even.
Which of the following would be an appropriate way to begin a proof of this statement?
(a) Assume that (n - 5)(n + l ) (n + 13) is odd.
(b) Assume that (n — 5 )(n + l ) (n + 13) is even.
(c) Assume that n is even.
(d) Assume that n is odd.
(e) We consider two cases, according to whether n is even or n is odd.
A D D IT IO N A L
e x e r c i s e sFO R C H A P T E R 3
5 45. Let x G Z. Prove that if I x - 8 is even, then x is even.
: 46. Let x e Z. Prove that x 3 is even if and only if x is even.
: 47. Let x e Z. Use one or two lemmas to prove that 3x3 is even if and only if 5x2 is even.
5 48. Give a direct proof of the following: Let x G Z. If 1 lx - 5 is odd, then x is even.
- 49. Let x , y G Z. Prove that if x + y is odd, then x and y are of opposite parity.
- 50. Let x, y e Z. Prove that if 3x + 5y is even, then x and y are of the same parity.
5 51. Let i j g Z . Prove that (x + l) y 2 is even if and only if x is odd or y is even.
52. Let x, y G Z. Prove that if xy and x + y are even, then both x and y are even.
5 53. Prove, for every integer x, that the integers 3x + 1 and 5x + 2 are of opposite parity.
: 54. Prove the following two results:
(a) Result A: Let k g Z. If rc3 is even, then n is even.
(b) Result B: If n is an odd integer, then 5n9 + 13 is even.
5 55. Prove for every two distinct real numbers a and b, either > a or > b.
56. Let X , y G Z. Prove that if a and b are even integers, then a x + by is even.
5 57. Evaluate the proof of the following result.
R esult Let x , y G Z and let a and b be odd integers. If a x + by is even, then x and y are of the same parity.
P ro o f Assume that x and y are of opposite parity. Then X = 2 p and y = 2q + 1 for some
integers p and q. Since a and b are odd integers, a = 2r + 1 and b = 2s + 1 for integers r and s. Hence a x + by = (2 r + 1)(2 p) + {2s + 1)(2 q + 1)
= Apr + 2 p + Aqs + 2s + 2q + 1
= 2(2 p r + p + 2qs + s + q) + 1.
Since 2 p r + p + 2qs + s + q is an integer, a x + by is odd. a
9 8 Chapter 3 Direct Proof and Proof by Contrapositive
3.58. Let S = {a, b, c, d } be a set o f four distinct integers. Prove that if either (I) for each .v e S. the integer x and the sum of any two of the remaining three integers of S are of the same parity or (2) for each x G S, the integer x and the sum of any two of the remaining three integers of S are of opposite parity, then every pair o f integers of S is of the same parity.
3.59. Prove that if a and b are two positive integers, then a 2(b + 1) + b 2(a + 1) > 4ab, 3.60. Let a, b e Z. Prove that if ab = 4, then (a — b )3 — 9 (a — b) = 0.
3.61. Let a . b and c be the lengths of the sides o f a triangle T where a < b < c. Prove that if T is a right triangle, then
3.62. Consider the following statement.
Let n G Z. Then 3/z3 + 4n 2 + 5 is even if and only if n is even.
Which of the following would be an appropriate way to begin a proof o f this statement?
(a) Assume that 3« 3 + An2 + 5 is odd.
(b) Assume that 3//3 + 4n 2 + 5 is even.
(c) Assume that n is even.
(d) Assume that n is odd.
(e) We consider two cases, according to whether n is even or n is odd.
3.63. Let V = {A, B , C } be a partition o f a set S of integers, where A - { n € S: n is odd and n > 0}, B = {n e S : n is odd and n < 0} and C = {n € S : n is even and n > 0}. Prove that if .v and v are elements of S belonging to distinct subsets in V. then xy is either odd, even and greater than 1, or even and less than — 1.
3.64. Let /? G N. Prove that if n 3 — 5n — 10 > 0, then n > 3.
3.65. Prove for every odd integer a that (a 2 + 3)(a2 + 7) = 32b for some integer b.
3.66. Prove for every two positive integers a and b that
(a + b) ( '• | ' ) > 4.
\ a b J
3.67. Which result is being proved below, and what procedure is being used to verify the result?
We begin with the following proof.
P ro o f First, assume that x is even. Then x = 2a, where a G Z. Thus 3x - 2 = 3(2a) - 2 = 6a - 2 = 2(3a - 1).
Since 3a — 1 is an integer, 3x — 2 is even.
Next, suppose that x is odd. Then x = 2b + I for some integer b. So 3* - 2 = 3(2b + 1) - 2 = 6b + 1 = 2(3,/) + 1.
Since 3a is an integer, 3x — 2 is odd.
We can now give the following proof.
P ro o f First, assume that 3x - 2 is even. From the preceding result, x is even and so x = 2a, where a g Z. Thus
5.\ + 1 = 5(2 a) + 1 = 2(5a) + 1.
Since 5a is an integer, 5x + 1 is odd.
Next, assume that 3x - 2 is odd. Again, by the preceding result, x is odd. Hence x = 2b + 1 for some integer b. Therefore,
5x + 1 = 5(2 b + 1) + 1 = W b + 6 = 2(5 b + 3).
Since 5b + 3 is an integer, 5x + 1 is even.