Download 35 12. (a) The statement is true. Proof. Let n be a positive integer

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12. (a) The statement is true. Proof. Let n be a positive integer such that n2 − 4 = 0. Then
n2 − 4 = (n − 2)(n + 2) = 0 and so n − 2 = 0 or n + 2 = 0. Since n is a positive integer,
n + 2 ̸= 0 and so n − 2 = 0.
(b) The statement is false. Solution. Consider n = −2. Then n2 − 4 = (−2)2 − 4 = 0 but
n − 2 = −2 − 2 = −4 ̸= 0. Therefore, n = −2 is a counterexample.
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13. (a) The statement is true. Proof. Since there is no integer x such that 2x − 1 = 0, it follows
that 2x − 1 = 0 is false. Hence the implication is true.
(b) The statement is false. Solution. Consider x = 1/2. Then 2x − 1 = 0 but 2x2 − 3x − 2 =
−3 ̸= 0. So x = 1/2 is a counterexample.
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14. (a) The statement is true. Proof. Let n ∈ S. We have four cases.
Case 1. n = 1. Then (n2 − n)/2 = 0 is even.
Case 2. n = 4. Then (n2 − n)/2 = 6 is even.
Case 3. n = 5. Then (n2 − n)/2 = 10 is even.
Case 4. n = 8. Then (n2 − n)/2 = 28 is even.
(b) The statement is false. Solution. When n = 6, we have n ∈ S but (n2 − n)/2 = 15 is
not even. Therefore, n = 6 is a counterexample.
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15. The statement is false. Solution. Let n = 1. Then n2 + n + 2 = 4 is even but n = 1 is odd.
So n = 1 is a counterexample.
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16. The statement is false. Solution. Let a = b = 1. Then 2a + 3b = 5 is odd but it is not the
case that a is even and b is odd. . Thus a = 1 and b = 1 form a counterexample.
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17. The statement is false. Solution. Let a = 1 and b = 2. Then 3ab = 6 is even. Thus a = 1
and b = 2 form a counterexample.
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Exercises for Section 3.6. Existence Proofs
1. Proof. Let n = 11. Then 100 − n2 = 100 − (11)2 = 100 − 121 = −21 < 0.
2. Proof. Let m = 3 and n = 6. Thus m and n are of opposite parity. Then (m − 2)2 + (n − 6)2 =
1 ≤ 1.
√
√
3. Proof. Let x = 5. Then x2 = ( 5)2 = 5.
a+b
2 .
Then r is a rational number such that a < r < b.
√
√
5. Proof. Let a = 2 and b = 0. Then ab = ( 2)0 = 1 ∈ Q.
√
√
√
6. Proof. Let a = 2 and b = 1. Then ab = ( 2)1 = 2 is irrational.
√
√
√
7. Proof. Let x = 2. Then x4 − x2 − 2 = ( 2)4 − ( 2)2 − 2 = 4 − 2 − 2 = 0.
4. Proof. Let r =
8. Proof. Let n = 1. Then 4n2 − 8n + 3 = 4 − 8 + 3 = −1 < 0.
9. Proof. Let m = n = 3. Then
(m − 2)2 − (n − 3)3 = (3 − 2)2 − (3 − 3)3 = 12 − 02 = 1 − 0 = 1,
as desired.
10. Solution. Let x be a real number. Since
x4 − x2 + 2 = x4 − x2 + ( 12 )2 +
7
4
= (x2 − 21 )2 +
7
4
≥
7
4
> 0,
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it follows that x4 − x2 + 2 ̸= 0.
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√
√
11. Proof. Consider the number 3
these two possibilities separately.
Case 1.
2
. This number is either rational or irrational. We consider
√
√
√ √2
3 is rational. Letting a = 3 and b = 2 verifies that the statement is true.
√ √2
3 is irrational.
In this case, consider the number obtained by raising the√ (irra√
√ 2
√ √2
b
tional)
√ number 3 to the (irrational) power 2; that is, consider a , where a = 3 and
b = 2. Observe that
! √ "√2
√ √2·√2 √ 2
√ 2
3
= 3
= 3 = 3,
ab =
Case 2.
which is rational.
12. The statement is true. Proof. Let b ∈ Z. Now let a = |b| + 1. Thus a ∈ N and |a − |b|| =
|(|b| + 1) − |b|| = 1.
13. (a) The statement is true. Proof. Let a = 3 and b = 32 . Then
# $
(a − 1)(b − 1) = 2 12 = 1.
(b) The statement is true. Proof. Let a = 3 and b = 32 . Then
1
1
1 1
1 2
+ = + 3 = + = 1.
a b
3
3
3
2
Proof Analysis.
Observe that if a and b are two (distinct) rational numbers that
satisfy a1 + 1b = 1, then a+b
ab = 1 and so a + b = ab. Thus ab − a − b = 0, which is
equivalent to ab − a − b + 1 = 1 and so (a − 1)(b − 1) = 1. Therefore, two distinct rational
numbers a and b satisfy
(a − 1)(b − 1) = 1
if and only if a and b satisfy
1 1
+ =1
a b
if and only if a and b satisfy
a + b = ab. !
14. The statement is true. Proof.
Let a = 6 and b = 2. Then
a
b
+
3b
a
= 3 + 1 = 4.
15. The statement is false. Let A = {1}, which is nonempty, and let B be an arbitrary set. Since
1 ∈ A ∪ B, it follows that A ∪ B ̸= ∅ for every set B.
16. The statement is true. Proof. Let A be a proper subset of S and let B = S − A. Then
B ̸= ∅, A ∪ B = S and A ∩ B = ∅.
Exercises for Section 3.7. Proof by Contradiction
1. Proof. Assume, to the contrary, that there is a largest negative rational number r. Then
a
. Since s is a negative rational number and s > r,
r = − ab , where a, b ∈ N. Let s = 2r = − 2b
this is a contradiction.