Download 25 14. Let S be the set of even integers and T the set of odd integers

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14. Let S be the set of even integers and T the set of odd integers. Furthermore, let
P (a, b): a + b is odd.
(a) ∀a ∈ S, ∀b ∈ T , P (a, b).
(b) ∃a ∈ S, ∃b ∈ T , ∼ P (a, b).
(c) There exist an even integer a and an odd integer b such that a + b is even.
15. Let S be the set of even integers and T the set of odd integers. Furthermore, let
P (a, b): (a + 2)2 + (b + 3)2 = 0.
(a) ∃a ∈ S, ∃b ∈ T , P (a, b).
(b) ∀a ∈ S, ∀b ∈ T , ∼ P (a, b).
(c) For every even integer a and every odd integer b, (a + 2)2 + (b + 3)2 ̸= 0.
16. Let P (x, y): y < x2 .
(a) ∀x ∈ R, ∃y ∈ R+ , P (x, y).
(b) ∃x ∈ R, ∀y ∈ R+ , ∼ P (x, y).
(c) There exists a real number x such that for every positive real number y, y ≥ x2 .
17. Let P (a, b): ab ≥ 0.
(a) ∃a ∈ Z, ∀b ∈ Z, P (a, b).
(b) ∀a ∈ Z, ∃b ∈ Z, ∼ P (a, b).
(c) For every integer a, there exists some integer b such that ab < 0.
18. There exists an integer a such that for every integer b, | a+1
2 − b| > 1.
19. There exists an integer a such that for every integer b, | 2a+1
− b| ≥ 21 .
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Exercises for Section 3.2. Direct Proof
1. (a) R(2): 1 is even. (F)
R(4): 10 is even. (T)
R(6): 35 is even. (F)
n3 − n
is even. (F)
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n3 − n
(c) There exists n ∈ S such that
is even. (T)
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(b) For each n ∈ S,
2. (a) P (0): 0 is even. (T)
P (3): 14 is even. (T)
P (4): 30 is even. (T)
n(n + 1)(2n + 1)
(b) For each n ∈ S,
is even. (T)
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n(n + 1)(2n + 1)
is even. (T)
(c) There exists n ∈ S such that
6
3. Proof. Let x be a real number such that (x− 1)2 = 0. Thus x− 1 = 0 and so x = 1. Therefore,
x3 − 1 = 13 − 1 = 0.
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4. Proof. Let x be a real number such that x3 + x = x(x2 + 1) = 0. Hence x = 0 or x2 + 1 = 0.
Since x2 + 1 ≥ 1, it follows that x = 0. Thus x2 − x = 02 − 0 = 0 ≤ 0.
5. Proof. Assume that (x − 2)4 ≤ 0. Since (x − 2)4 ≥ 0, it follows that (x − 2)4 = 0 and so
x − 2 = 0. Hence x = 2. Therefore, 9 − x2 = 9 − 22 = 5 ≥ 0.
6. Proof. Let n be an integer such that 2n2 + n − 1 = 0. Since 2n2 + n − 1 = (2n − 1)(n + 1) = 0
and n is an integer, it follows that n + 1 = 0 and so n = −1. Hence n3 = (−1)3 = −1 < 0.
7. Proof. Let n be an even integer. So n = 2k for some integer k. Then 7n − 2 = 7(2k) − 2 =
14k − 2 = 2(7k − 1). Since 7k − 1 is an integer, 7n − 2 is even.
8. Proof. Let n be an odd integer. Therefore, n = 2k + 1 for some integer k. Thus 3n + 10 =
3(2k + 1) + 10 = 6k + 3 + 10 = 2(3k + 6) + 1. Since 3k + 6 is an integer, 3n + 10 is odd.
9. Proof. Let n be an odd integer. Therefore, n = 2k + 1 for some integer k. Then
7n2 − 2n + 15 =
=
=
7(2k + 1)2 − 2(2k + 1) + 15 = 7(4k 2 + 4k + 1) − (4k + 2) + 15
28k 2 + 28k + 7 − 4k − 2 + 15
28k 2 + 24k + 20 = 2(14k 2 + 12k + 10).
Since 14k 2 + 12k + 10 is an integer, 7n2 − 2n + 15 is even.
10. Proof. Let n be an odd integer in S. Then n = 3. Thus 3n + 1 = 3 · 3 + 1 = 10 is even.
11. Proof. Let n be a nonnegative integer in S. Then n = 0. Therefore, 2−n (n2 +3) = 20 (02 +3) =
3 is odd.
12. Proof. Suppose that a and b are odd. Then a = 2x + 1 and b = 2y + 1, where x, y ∈ Z. Thus
ab + a + b
= (2x + 1)(2y + 1) + (2x + 1) + (2y + 1)
= 4xy + 2x + 2y + 1 + 2x + 1 + 2y + 1
= 4xy + 4x + 4y + 2 + 1 = 2(2xy + 2x + 2y + 1) + 1.
Since 2xy + 2x + 2y + 1 is an integer, ab + a + b is odd.
13. Proof. Let r and s be rational numbers. Then r = a/b and s = c/d, where a, b, c, d ∈ Z and
b, d ̸= 0. Therefore
5a 7c
5ad + 7bc
5r + 7s =
+
=
.
b
d
bd
Since 5ad + 7bc and bd are integers and bd ̸= 0, the number 5r + 7s is rational.
14. Proof. Let r and s be rational numbers. Then r = a/b and s = c/d, where a, b, c, d ∈ Z and
b, d ̸= 0. Therefore
a
c
ad − bc
r−s= − =
.
b
d
bd
Since ad − bc and bd are integers and bd ̸= 0, the number r − s is rational.
15. Proof. Let r and s be rational numbers. Then r = a/b and s = c/d, where a, b, c, d ∈ Z and
b, d ̸= 0. Therefore rs = ab · dc = ac
bd . Since ac and bd are integers and bd ̸= 0, the number rs is
rational.
16. Proof. Let a and b be positive integers. Observe that (a − b)2 ≥ 0. Thus
a2 − 2ab + b2 ≥ 0 and so a2 + b2 ≥ 2ab.
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Since a and b are positive, ab is positive. Dividing both sides of a2 + b2 ≥ 2ab by ab, we obtain
a2 + b 2
a
b
= + ≥ 2,
ab
b
a
as desired.
17. Proof. Let P and Q be statements such that P ∧ Q is true. Then P and Q are true and so
P ∨ Q is true.
18. (a) Proof. Since a and b are even, a = 2x and b = 2y for some integers x and y. Then
a + b = 2x + 2y = 2(x + y). Since x + y is an integer, a + b is even.
(b) Yes by (a).
(c) No.
(d) No.
19. (a) Proof. If T is an equilateral triangle, then its three sides have the same length and so at
least two of its sides have the same length. Therefore, T is isosceles.
(b) Yes by (a).
(c) No.
(d) No.
Exercises for Section 3.3. Proof by Contrapositive
1. Proof. Assume that n is an odd integer. Then n = 2k + 1 for some integer k. So
5n + 7 = 5(2k + 1) + 7 = 10k + 12 = 2(5k + 6).
Since 5k + 6 is an integer, 5n + 7 is even.
2. Proof. Assume that n is an even integer. Then n = 2k for some integer k. So
9n − 5 = 9(2k) − 5 = 18k − 5 = 18k − 6 + 1 = 2(9k − 3) + 1.
Since 9k − 3 is an integer, 9n − 5 is odd.
3. Proof. Assume that n is an even integer in S = {1, 2, 3}. Then n = 2. Thus 3n+4 = 3·2+4 =
10 is even.
4. Proof. First, we prove that if n is an even integer, then 3n − 11 is odd. Assume that n is
even. Then n = 2k for some integer k. So
3n − 11 = 3(2k) − 11 = 6k − 11 = 6k − 12 + 1 = 2(3k − 6) + 1.
Since 3k − 6 is an integer, 3n − 11 is odd.
Next, we verify the converse, namely if 3n − 11 is odd, then n is even. Assume that n is
odd. Then n = 2ℓ + 1 for some integer ℓ. Therefore,
3n − 11 = 3(2ℓ + 1) − 11 = 6ℓ − 8 = 2(3ℓ − 4).
Since 3ℓ − 4 is an integer, 3n − 11 is even.
5. Proof. First, we prove that if n is an even integer, then n3 is even. Assume that n is even.
Then n = 2k for some integer k. So n3 = (2k)3 = 8k 3 = 2(4k 3 ). Since 4k 3 is an integer, n3 is
even.
Next, we verify the converse, namely if n3 is even, then n is even. Let n be an odd integer.
Then n = 2ℓ + 1 for some integer ℓ. Thus
n3
=
(2ℓ + 1)3 = 8ℓ3 + 12ℓ2 + 6ℓ + 1
=
2(4ℓ3 + 6ℓ2 + 3ℓ) + 1.
Since 4ℓ3 + 6ℓ2 + 3ℓ is an integer, n3 is odd.