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1
Solutions to Exercises
1.1 (a), (c), and (e) are statements.
1.2 (a), (c), and (d) are statements.
1.3 a. Hypothesis: The right triangle XYZ with sides of lengths
x and y and hypotenuse of length z has an
area of z 2 /4.
Conclusion: The triangle XYZ is isosceles.
b. Hypothesis: n is an even integer.
Conclusion: n2 is an even integer.
c. Hypothesis: a, b, c, d, e, and f are real numbers for which
ad − bc 6= 0.
Conclusion: The two linear equations ax + by = e and
cx + dy = f can be solved for x and y.
1.4 a. Hypothesis: r is a real number that satisfies r 2 = 2.
Conclusion: r is irrational.
b. Hypothesis: p and q are positive real numbers such that
√
pq 6= (p + q) /2.
Conclusion: p 6= q.
c. Hypothesis: f(x) = 2−x for all real numbers x.
Conclusion: There exists a real number x such that
0 ≤ x ≤ 1 and f(x) = x.
1
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2 SOLUTIONS TO EXERCISES IN CHAPTER 1
1.5 a. Hypothesis: A, B and C are sets of real numbers with A ⊆ B.
Conclusion: A ∩ C ⊆ B ∩ C.
b. Hypothesis: For a positive integer n, the function f defined by:
n/2, if n is even
f(n) =
3n + 1, if n is odd
For an integer k ≥ 1, f k (n) = f k−1 (f(n)), and f 1 (n) = f(n).
Conclusion: For any positive integer n, there is an integer k > 0 such that
f k (n) = 1.
c. Hypothesis: x is a real number.
Conclusion: The minimum value of x(x − 1) ≥ −1/4.
1.6 Jack’s statement is true. This is because the hypothesis that Jack did
not get his car fixed is false. Therefore, according to rows 3 and 4 of Table
1.1, the if/then statement is true, regardless of the truth of the conclusion.
1.7 Jack’s statement is false. This is because the hypothesis, getting his
car fixed, is true while the conclusion, not missing the interview, is false.
Therefore, according to row 2 of the Table 1.1, the if/then statement is false.
1.8 Jack won the contest. This is because the hypothesis that Jack is younger
than his father is true, and, because the if/then statement is true, row 1 of
Table 1.1 is applicable. Therefore, the conclusion that Jack will not lose the
contest is also true.
1.9 a. True because A : 2 > 7 is false (see rows 3 and 4 of Table 1.1).
b. True because B : 1 < 2 is true (see rows 1 and 3 of Table 1.1).
1.10 a. True because 1 < 3 is true (see rows 1 and 3 of Table 1.1).
b. True if x 6= 3 (see rows 3 and 4 of Table 1.1).
False when x = 3 because then the hypothesis is true and the conclu-
sion 1 > 2 is false (see row 2 of Table 1.1).
1.11 If you want to prove that “A implies B” is true and you know that B
is false, then A should also be false. The reason is that, if A is false, then
it does not matter whether B is true or false because Table 1.1 ensures that
“A implies B” is true. On the other hand, if A is true and B is false, then
“A implies B” would be false.
1.12 When B is true, rows 1 and 3 of Table 1.1 indicate that the statement
“A implies B” is true. You therefore need only consider the case when B is
false. In this case, for “A implies B” to be true, it had better be that A is
false so that row 4 of Table 1.1 is applicable. In other words, you can assume
B is false; your job is to show that A is false.
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SOLUTIONS TO EXERCISES IN CHAPTER 1 3
1.13 (T = true, F = false)
A B C B⇒C A ⇒ (B ⇒ C)
T T T T T
T T F F F
T F T T T
T F F T T
F T T T T
F T F F T
F F T T T
F F F T T
1.14 (T = true, F = false)
A B C A⇒B (A ⇒ B) ⇒ C
T T T T T
T T F T F
T F T F T
T F F F T
F T T T T
F T F T F
F F T T T
F F F T F
1.15 (T = true, F = false)
A B B⇒A A⇒B
T T T T
T F T F
F T F T
F F T T
From this table, B ⇒ A is not always true at the same time A ⇒ B is true.
1.16 From row 2 of Table 1.1, you must show that A is true and B is false.
1.17 a. For A to be true and B to be false, it is necessary to find a real
number x > 0 such that log10 (x) ≤ 0. For example, x = 0.1 > 0, while
log10 (0.1) = −1 ≤ 0. Thus, x = 0.1 is a desired counterexample. (Any
value of x such that 0 < x ≤ 1 would provide a counterexample.)
b. For A to be true and B to be false, it is necessary to find an integer
n > 0 such that n3 < n!. For example, n = 6 > 0, while 63 = 216 <
720 = 6!. Thus, n = 6 is a desired counterexample. (Any integer
n ≥ 6 would provide a counterexample.)
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Solutions Manual for
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Solow (All Chapters)
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1
Solutions to Exercises
1.1 (a), (c), and (e) are statements.
1.2 (a), (c), and (d) are statements.
1.3 a. Hypothesis: The right triangle XYZ with sides of lengths
x and y and hypotenuse of length z has an
area of z 2 /4.
Conclusion: The triangle XYZ is isosceles.
b. Hypothesis: n is an even integer.
Conclusion: n2 is an even integer.
c. Hypothesis: a, b, c, d, e, and f are real numbers for which
ad − bc 6= 0.
Conclusion: The two linear equations ax + by = e and
cx + dy = f can be solved for x and y.
1.4 a. Hypothesis: r is a real number that satisfies r 2 = 2.
Conclusion: r is irrational.
b. Hypothesis: p and q are positive real numbers such that
√
pq 6= (p + q) /2.
Conclusion: p 6= q.
c. Hypothesis: f(x) = 2−x for all real numbers x.
Conclusion: There exists a real number x such that
0 ≤ x ≤ 1 and f(x) = x.
1
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2 SOLUTIONS TO EXERCISES IN CHAPTER 1
1.5 a. Hypothesis: A, B and C are sets of real numbers with A ⊆ B.
Conclusion: A ∩ C ⊆ B ∩ C.
b. Hypothesis: For a positive integer n, the function f defined by:
n/2, if n is even
f(n) =
3n + 1, if n is odd
For an integer k ≥ 1, f k (n) = f k−1 (f(n)), and f 1 (n) = f(n).
Conclusion: For any positive integer n, there is an integer k > 0 such that
f k (n) = 1.
c. Hypothesis: x is a real number.
Conclusion: The minimum value of x(x − 1) ≥ −1/4.
1.6 Jack’s statement is true. This is because the hypothesis that Jack did
not get his car fixed is false. Therefore, according to rows 3 and 4 of Table
1.1, the if/then statement is true, regardless of the truth of the conclusion.
1.7 Jack’s statement is false. This is because the hypothesis, getting his
car fixed, is true while the conclusion, not missing the interview, is false.
Therefore, according to row 2 of the Table 1.1, the if/then statement is false.
1.8 Jack won the contest. This is because the hypothesis that Jack is younger
than his father is true, and, because the if/then statement is true, row 1 of
Table 1.1 is applicable. Therefore, the conclusion that Jack will not lose the
contest is also true.
1.9 a. True because A : 2 > 7 is false (see rows 3 and 4 of Table 1.1).
b. True because B : 1 < 2 is true (see rows 1 and 3 of Table 1.1).
1.10 a. True because 1 < 3 is true (see rows 1 and 3 of Table 1.1).
b. True if x 6= 3 (see rows 3 and 4 of Table 1.1).
False when x = 3 because then the hypothesis is true and the conclu-
sion 1 > 2 is false (see row 2 of Table 1.1).
1.11 If you want to prove that “A implies B” is true and you know that B
is false, then A should also be false. The reason is that, if A is false, then
it does not matter whether B is true or false because Table 1.1 ensures that
“A implies B” is true. On the other hand, if A is true and B is false, then
“A implies B” would be false.
1.12 When B is true, rows 1 and 3 of Table 1.1 indicate that the statement
“A implies B” is true. You therefore need only consider the case when B is
false. In this case, for “A implies B” to be true, it had better be that A is
false so that row 4 of Table 1.1 is applicable. In other words, you can assume
B is false; your job is to show that A is false.
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SOLUTIONS TO EXERCISES IN CHAPTER 1 3
1.13 (T = true, F = false)
A B C B⇒C A ⇒ (B ⇒ C)
T T T T T
T T F F F
T F T T T
T F F T T
F T T T T
F T F F T
F F T T T
F F F T T
1.14 (T = true, F = false)
A B C A⇒B (A ⇒ B) ⇒ C
T T T T T
T T F T F
T F T F T
T F F F T
F T T T T
F T F T F
F F T T T
F F F T F
1.15 (T = true, F = false)
A B B⇒A A⇒B
T T T T
T F T F
F T F T
F F T T
From this table, B ⇒ A is not always true at the same time A ⇒ B is true.
1.16 From row 2 of Table 1.1, you must show that A is true and B is false.
1.17 a. For A to be true and B to be false, it is necessary to find a real
number x > 0 such that log10 (x) ≤ 0. For example, x = 0.1 > 0, while
log10 (0.1) = −1 ≤ 0. Thus, x = 0.1 is a desired counterexample. (Any
value of x such that 0 < x ≤ 1 would provide a counterexample.)
b. For A to be true and B to be false, it is necessary to find an integer
n > 0 such that n3 < n!. For example, n = 6 > 0, while 63 = 216 <
720 = 6!. Thus, n = 6 is a desired counterexample. (Any integer
n ≥ 6 would provide a counterexample.)
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