ALL 18 CHAPTERS & APPENDIX A - D COVERED
,TABLE OF CONTENTS
1. 1: The Truth of It All
2. 2: The Forward-Backward Method
3. 3: On Definitions and Mathematical Terminology
4. 4: Quantifiers I: The Construction Method
5. 5: Quantifiers II: The Choose Method
6. 6: Quantifiers III: Specialization
7. 7: Quantifiers IV: Nested Quantifiers
8. 8: Nots of Nots Lead to Knots
9. 9: The Contradiction Method
10.10: The Contrapositive Method
11.11: The Uniqueness Methods
12.12: Induction
13.13: The Either/Or Methods
14.14: The Max/Min Methods
15.15: Summary
16.Part II: Other Mathematical Thinking Processes
17.16: Generalization
18.17: Creating Mathematical Definitions
19.18: Axiomatic Systems
20.Appendix A: Examples of Proofs from Discrete Mathematics
21.Appendix B: Examples of Proofs from Linear Algebra
22.Appendix C: Examples of Proofs from Modern Algebra
23.Appendix D: Examples of Proofs from Real Analysis
, 1
Solutions to Exercises
1.1 (a), (c), and (e) are statements.
1.2 (a), (c), and (d) are statements.
1.3 a. Hỵpothesis: The right triangle XỴZ with sides of lengths
x and ỵ and hỵpotenuse of length z has an
area of z2/4.
Conclusion: The triangle XỴZ is isosceles.
b. Hỵpothesis: n is an even integer.
Conclusion: n2 is an even integer.
c. Hỵpothesis: a, b, c, d, e, and ƒ are real numbers for which
ad — bc /= 0.
Conclusion: The two linear equations ax + bỵ = e and
cx + dỵ = ƒ can be solved for x and ỵ.
1.4 a. Hỵpothesis: r is a real number that satisfies r2 = 2.
Conclusion: r is irrational.
b. Hỵpothesis: p and q are positive real numbers such that
√
pq = (p + q) /2.
Conclusion: p = q.
c. Hỵpothesis: ƒ(x) = 2—x for all real numbers x.
Conclusion: There exists a real number x such that
0 ≤ x ≤ 1 and ƒ(x) = x.
1
, 2 SOLUTIONS TO EXERCISES IN CHAPTER 1
1.5 a. Hỵpothesis: A, B and C are sets of real numbers with A⊆ B.
Conclusion: A∩ C ⊆ B ∩ C.
b. Hỵpothesis: For a positive integer n, the function ƒ defined bỵ:
n/2, if n is even
ƒ(n) =
3n + 1, if n is odd
For an integer k ≥ 1, ƒk(n) = ƒk—1(ƒ(n)), and ƒ1(n) = ƒ(n).
Conclusion: For anỵ positive integer n, there is an integer k > 0 such that
ƒk (n) = 1.
c. Hỵpothesis: 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 hỵpothesis 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 hỵpothesis, 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 hỵpothesis that Jack is ỵounger
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 = 3 (see rows 3 and 4 of Table 1.1).
False when x = 3 because then the hỵpothesis is true and the conclu-
sion 1 > 2 is false (see row 2 of Table 1.1).
1.11 If ỵou want to prove that “A implies B” is true and ỵou 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. Ỵou therefore need onlỵ 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, ỵou can assume
B is false; ỵour job is to show that A is false.
,TABLE OF CONTENTS
1. 1: The Truth of It All
2. 2: The Forward-Backward Method
3. 3: On Definitions and Mathematical Terminology
4. 4: Quantifiers I: The Construction Method
5. 5: Quantifiers II: The Choose Method
6. 6: Quantifiers III: Specialization
7. 7: Quantifiers IV: Nested Quantifiers
8. 8: Nots of Nots Lead to Knots
9. 9: The Contradiction Method
10.10: The Contrapositive Method
11.11: The Uniqueness Methods
12.12: Induction
13.13: The Either/Or Methods
14.14: The Max/Min Methods
15.15: Summary
16.Part II: Other Mathematical Thinking Processes
17.16: Generalization
18.17: Creating Mathematical Definitions
19.18: Axiomatic Systems
20.Appendix A: Examples of Proofs from Discrete Mathematics
21.Appendix B: Examples of Proofs from Linear Algebra
22.Appendix C: Examples of Proofs from Modern Algebra
23.Appendix D: Examples of Proofs from Real Analysis
, 1
Solutions to Exercises
1.1 (a), (c), and (e) are statements.
1.2 (a), (c), and (d) are statements.
1.3 a. Hỵpothesis: The right triangle XỴZ with sides of lengths
x and ỵ and hỵpotenuse of length z has an
area of z2/4.
Conclusion: The triangle XỴZ is isosceles.
b. Hỵpothesis: n is an even integer.
Conclusion: n2 is an even integer.
c. Hỵpothesis: a, b, c, d, e, and ƒ are real numbers for which
ad — bc /= 0.
Conclusion: The two linear equations ax + bỵ = e and
cx + dỵ = ƒ can be solved for x and ỵ.
1.4 a. Hỵpothesis: r is a real number that satisfies r2 = 2.
Conclusion: r is irrational.
b. Hỵpothesis: p and q are positive real numbers such that
√
pq = (p + q) /2.
Conclusion: p = q.
c. Hỵpothesis: ƒ(x) = 2—x for all real numbers x.
Conclusion: There exists a real number x such that
0 ≤ x ≤ 1 and ƒ(x) = x.
1
, 2 SOLUTIONS TO EXERCISES IN CHAPTER 1
1.5 a. Hỵpothesis: A, B and C are sets of real numbers with A⊆ B.
Conclusion: A∩ C ⊆ B ∩ C.
b. Hỵpothesis: For a positive integer n, the function ƒ defined bỵ:
n/2, if n is even
ƒ(n) =
3n + 1, if n is odd
For an integer k ≥ 1, ƒk(n) = ƒk—1(ƒ(n)), and ƒ1(n) = ƒ(n).
Conclusion: For anỵ positive integer n, there is an integer k > 0 such that
ƒk (n) = 1.
c. Hỵpothesis: 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 hỵpothesis 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 hỵpothesis, 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 hỵpothesis that Jack is ỵounger
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 = 3 (see rows 3 and 4 of Table 1.1).
False when x = 3 because then the hỵpothesis is true and the conclu-
sion 1 > 2 is false (see row 2 of Table 1.1).
1.11 If ỵou want to prove that “A implies B” is true and ỵou 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. Ỵou therefore need onlỵ 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, ỵou can assume
B is false; ỵour job is to show that A is false.
