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Math 310/ 3–20–
Solutions to Problem Set 11
- Give a counterexample to each of the following statements.
(a) “|x + y| = |x| + |y| for all x, y ∈ R.”
Let x = 2 and y = −2. Then
|x + y| = |2 + (−2)| = | 0 | = 0, but |x| + |y| = | 2 | + | − 2 | = 4.
Therefore, |x + y| 6 = |x| + |y| for all x, y ∈ R.
(b) “For all x, y ∈ R, if cos x = cos y, then x = y.”
Let x = 0 and y = 2π. Then
cos 0 = 1 = cos 2π, but 0 6 = 2π.
Hence, it is not the case that for all x, y ∈ R, if cos x = cos y, then x = y.
(c) “For all x, y, z ∈ R, if
x y
≤ −1 and
y z
≤ −1, then
x z
Let x = 1, y = −1, and z = 1. Then x y
= − 1 and
y z
= − 1 , but
x z
Therefore, it is not the case that for all x, y, z ∈ R, if
x y
≤ −1 and
y z
≤ −1, then
x z
- Suppose that the universe is { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 }, and that
A = { 1 , 3 , 5 , 7 , 9 } and B = { 4 , 5 , 6 , 7 , 8 , 9 , 10 }.
List the elements of the following sets.
(a) A. A = { 2 , 4 , 6 , 8 , 10 }.
(b) A − B. A − B = { 1 , 3 }.
(c) A ∩ B. A ∩ B = { 5 , 7 , 9 }.
(d) A ∪ B. A ∪ B = { 1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 }.
[page 27]
Exercise 1.4. Write each of the following sets by listing its elements within braces.
(a) A = {n ∈ Z | − 4 < n ≤ 4 }. A = {− 3 , − 2 , − 1 , 0 , 1 , 2 , 3 , 4 }.
1
(b) B = {n ∈ Z | n^2 < 5 }. B = {− 2 , − 1 , 0 , 1 , 2 }.
(c) C = {n ∈ N | n^3 < 100 }. C = { 1 , 2 , 3 , 4 }.
(d) D = {x ∈ R | x^2 − x = 0}. D = { 0 , 1 }.
(e) E = {x ∈ R | x^2 + 1 = 0}. E = ∅.
Exercise 1.12. Find (^) P(A) and | P(A)| for:
(a) A = { 1 , 2 }.
P(A) =^ {∅,^ {^1 },^ {^2 },^ {^1 ,^2 }}^ and^ | P(A)|^ = 4.
(b) A = {∅, 1 , {a}}.
P(A) =^ {∅,^ {∅},^ {^1 },^ {{a}},^ {∅,^1 },^ {∅,^ {a}},^ {^1 ,^ {a}},^ {∅,^1 ,^ {a}}}.
| P(A)| = 8.
[page 103]
Exercise 4.36. Prove that A ∩ B = A ∪ B for every two sets A and B.
x ∈ A ∩ B ↔ ∼ x ∈ (A ∩ B) (Definition of complement) ↔ ∼ (x ∈ A ∧ x ∈ B) (Definition of intersection) ↔ ∼ x ∈ A∨ ∼ x ∈ B (DeMorgan) ↔ x ∈ A ∨ x ∈ B (Definition of complement) ↔ x ∈ (A ∪ B) (Definition of union)
This proves that A ∩ B = A ∪ B.
Exercise 4.38. Let A, B, and C be sets. Prove that (A − B) ∪ (A − C) = A − (B ∩ C).
x ∈ (A − B) ∪ (A − C) ↔ x ∈ (A − B) ∨ x ∈ (A − C) (Definition of union) ↔ (x ∈ A∧ ∼ x ∈ B) ∨ (x ∈ A∧ ∼ x ∈ C) (Definition of complement) ↔ x ∈ A ∧ (∼ x ∈ B∨ ∼ x ∈ C) (Distributivity) ↔ x ∈ A∧ ∼ (x ∈ B ∧ x ∈ C) (DeMorgan) ↔ x ∈ A∧ ∼ x ∈ (B ∩ C) (Definition of intersection) ↔ x ∈ [A − (B ∩ C)] (Definition of complement)
Therefore, (A − B) ∪ (A − C) = A − (B ∩ C).
[page 151]
Exercise 6.16. Prove that 1 +
n^2
n
for every positive integer n.
