MATH327 – HOMEWORK SOLUTIONS HOMEWORK #2

Section 1.1: Problems 1, 2, 4, 9 Section 1.2: Problems 1, 2, 5, 6, 12, 22

Krzysztof Galicki Problem 1.1.1.

a) \{−\sqrt{5}, \sqrt{5}\}.

b) {+1, −1, +3, −3, +5, −5, +15, −15}.

c) {0, −3/2}.

d) {−1, 0, 1, 2, 3}.

e) ∅.

Problem 1.1.2.

a) {−1, 1, 0, i, −i}.

b) \{1 − 2\sqrt{2}, 2 − 2\sqrt{2}, 3 − 2\sqrt{2}, 4 − 2\sqrt{2}, 5 − 2\sqrt{2}\}.

c) {3/4, −3/4, 4/3, −4/3, 0}.

d) {2, 3, 5, 6, 9}.

Problem 1.1.4.

a) TRUE

b) FALSE

c) TRUE

d) FALSE

e) FALSE

f) FALSE because 1 is odd but a + 2b is even if a is even.

g) FALSE because a + b\sqrt{2} = 0 and b ≠ 0 implies that \sqrt{2} = −a/b. As a, b ∈ ℚ this contradicts the fact that \sqrt{2} is irrational.

Problem 1.1.9.

a) 4 – A = {1, 2}, P(A) = {∅, {1}, {2}, {1, 2}}

b) 8 – A = {1, 2, 3}, P(A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

c) 2^{n}. This follows from the fact that one can choose k out of n elements in \binom{n}{k} = \frac{n!}{(n−k)!k!} ways and from the binomial expansion

2^{n} = (1 + 1)^{n} = \binom{n}{0} + \binom{n}{1} + ⋯ + \binom{n}{n − 1} + \binom{n}{n}.

Problem 1.2.1. a)

A = {1, 2, 3, 4, 5, 6}, B = {−1, 0, 1, 2, 3, 4, 5}, C = {−2, 0, 2}.

b)

A ∪ C = {−2, 0, 1, 2, 3, 4, 5, 6}, B ∩ C = {0, 2},

B \ C = {−1, 1, 3, 4, 5}, A ⊕ B = {−1, 0, 6},
C × (B ∩ C) = {(−2, 0), (−2, 2), (0, 0), (0, 2), (2, 0), (2, 2)},
(A \ B) \ C = {6}, A \ (B \ C) = {2, 6}, (B ∪ ∅) ∩ ∅ = ∅.

c)

T = {(1, 2), (2, 2)}.
S = {(1, −1), (2, 0), (3, 1), (4, 2), (5, 3), (6, 4)}.

Problem 1.2.2. If S = \{2, 5, \sqrt{2}, 25, π, 5/2\} and T = \{4, 25, \sqrt{2}, 6, 3/2\} then a)

S ∩ T = \{\sqrt{2}, 25\}, S ∪ T = \{2, 5, \sqrt{2}, 25, π, 5/2, 4, 6, 3/2\},
T × (S ∩ T) =
\{(4,\sqrt{2}),(4,25),(\sqrt{2},\sqrt{2}),(\sqrt{2},25),(25,\sqrt{2}),(25,25),(6,\sqrt{2}),(6,25),( \frac{3}{2} ,\sqrt{2}),(( \frac{3}{2} ,25)\}

b)

ℤ ∩ (S ∪ T) = {2, 5, 25, 4, 6}
(ℤ ∩ S) ∪ (ℤ ∩ T) = {2, 5, 25} ∪ {4, 25, 6} = {2, 4, 5, 6, 25}

We notice that

ℤ ∩ (S ∪ T) = (ℤ ∩ S) ∪ (ℤ ∩ T),

as they should be. c)

ℤ ∪ (S ∩ T) = ℤ ∪ \{\sqrt{2}, 25\} = \{\sqrt{2}, 0, ±1, ±2, ±3, ...\}

(ℤ ∪ S) ∩ (ℤ ∪ T) = \{\sqrt{2}, π, 5/2, 0, ±1, ±2, ±3, ...\} ∩ \{\sqrt{2}, 3/2, 0, ±1, ±2, ±3, ...\} =

= \{\sqrt{2}, 0, ±1, ±2, ±3, ...\}.

We notice that

ℤ ∪ (S ∩ T) = (ℤ ∪ S) ∩ (ℤ ∪ T),

as they should be. Problem 1.2.5. Let A = {a, b, c, {a, b}}. We have

a) A \ {a, b} = {c, {a, b}}.

b) {∅} \ P(A) = ∅.

c) A \ ∅ = A.

d) ∅ \ A = ∅.

e) {a, b, c} \ A = ∅.

f) ({a, b, c} ∪ {A}) \ A = {A}.

Problem 1.2.6. Let U = ℝ. We have

a) A = (1, ∞) ∪ (−∞, −2], A^{c} = (−2, 1].

b) A = (−3, ∞) ∩ (−∞, 4], A^{c} = (−∞, −3] ∪ (4, ∞).

c) A = {x ∈ ℝ | x^{2} ≤ −1} = ∅, A^{c} = ℝ. Problem 1.2.12. Let A_{n} = {a ∈ ℤ | a ≤ n}. We have

a) A_{3} ∪ A_{−3} = A_{3} as A_{−3} ⊂ A_{3}.

b) A_{3} ∩ A_{−3} = A_{−3} for the same reason.

c) A_{3} ∩ (A_{−3})^{c} = {−2, −1, 0, 1, 2, 3}.

d) ⋂_{i=0}^{4} A_{i} = A_{0} ∩ A_{1} ∩ A_{2} ∩ A_{3} ∩ A_{4} = A_{0} as A_{0} ⊂ A_{1} ⊂ A_{2} ⊂ A_{3} ⊂ A_{4}. Problem 1.2.22.

a) A ∪ B = A ∪ C does not imply B = C. This can be seen with A = {1, 2}, B = {1}, and C = {2}.

b) A ∩ B = A ∩ C does not imply B = C. This can be seen with A = {1}, B = {1, 2}, and C = {1, 3}.

c) A ⊕ B = A ⊕ C ⟹ B = C.

Proof. We prove it by contradiction. Suppose B ≠ C. Then there are four possible cases: Case 1. x ∈ B and x ∉ C and x ∈ A. In this case x ∉ A ⊕ B but x ∈ A ⊕ C which contradicts the hypothesis. Case 2. x ∈ B and x ∉ C and x ∉ A. In this case x ∈ A ⊕ B but x ∉ A ⊕ C which contradicts the hypothesis. Case 3. x ∉ B and x ∈ C and x ∈ A. In this case x ∈ A ⊕ B but x ∉ A ⊕ C which contradicts the hypothesis. Case 4. x ∉ B and x ∈ C and x ∉ A. In this case x ∉ A ⊕ B but x ∈ A ⊕ C which contradicts the hypothesis.

In each case we get the contradiction which shows that B = C.

d) A × B = A × C and A ≠ ∅ ⟹ B = C.

Proof. We prove it by contradiction. Suppose B ≠ C. Then there are two cases: Case 1. Either there is an element x ∈ B which is not in C or Case 2. There is an element y ∈ C which is not in B. In Case 1 choose any element a ∈ A (here we need the assumption that A is not empty). Then the pair (a, x) ∈ A × B but (a, x) ∉ A × C which contradict the hypothesis. In Case 2 choose any element a ∈ A. Then the pair (a, y) ∈ A × C but (a, y) ∉ A × B which contradict the hypothesis again.