Answers to Set Theory Exercise 1

1

(a) Yes; alphanumeric characters are A…Z, a…z and 0…9
(b) No; 'tall' is not well-defined
(c) Yes; the set is {12.5}
(d) Yes; the empty set
(e) No; 'good' is not well-defined

2

(a) T
(b) F
(c) T
(d) F; A is a subset of U (which we meet in the next section)
(e) F; {even numbers} means the set of all the even numbers, not just those between 2 and 10

3

(a) {4, 33, √9}
(b) {4, -5, 33, √9}
(c) {4, 2/3, -2.5, -5, 33, √9}
(d) {√2, π}

4

(a) F
(b) T
(c) F
(d) T

5 Examples might include:

(a) {London, Paris, Rome, …}
(b) {1, 3, 5, 7, …}, but not –3 or –1
(c) {5, -5}
(d) {3, 27, 243, …}

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Answers to Set Theory Exercise 2

1

In A? In B? In C? Region
Y Y Y vi
Y Y N iii
Y N Y v
Y N N ii
N Y Y vii
N Y N iv
N N Y viii
N N N i

2

They are all equal.

3

(a) True (b) False (c) True

4

(a)

(b) PQ; RQ
(c) False

5

(a)

(b)

(c)

(d)

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Answers to Set Theory Exercise 3

1

 (a) (b) A ∩ B = {6, 8} A ∪ C = {2, 3, 4, 6, 7, 8, 10} A ′ = {1, 3, 5, 7, 9} B ′ = {2, 4, 5, 9, 10} B ∩ A ′ = {1, 3, 7} B ∩ C ′ = {1, 6, 8} A – B = {2, 4, 10} A Δ B = {1, 2, 3, 4, 7, 10}

(c)
C - B = ø

2

(a) F
(b) F
(c) T

3

(a) PQ
(b) QP

4

5

(a) B
(b) AB
(c) (AB) ∩ (AB) ′ or (AB ′) ∪ (A ′ ∩ B)
(d) (AB) ∪ (A ′ ∩ B ′) or (AB) ′ ∪ (AB) or …?

6

(a) Region (b) represents AB. So AB = AB
(b) Region (c) represents A Δ B.
So A Δ B = (AB) ∪ (A ′ ∩ B ′) or (AB) ′ ∪ (AB)

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Answers to Set Theory Exercise 4

1

(a) P(A) = {ø, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2,4}, {3, 4}, {2, 3, 4}, {1, 3, 4}, {1, 2, 4}, {1, 2, 3}, {1, 2, 3, 4}}
| P(A) | = 16
(b) 32
(c) 210 = 1024

2

 Law Used (a) B ∪ (ø ∩ A) = B ∪ (A ∩ ø) Commutative = B ∪ ø Identity = B Identity (b) (A ' ∩ U) ' = (A' ) ' ∪ U' De Morgan = A ∪ U' Involution = A ∪ ø Complement = A Identity (c) (C ∪ A) ∩ (B ∪ A) = (A ∪ C) ∩ (B ∪ A) Commutative = (A ∪ C) ∩ (A ∪ B) Commutative = ((A ∪ C) ∩ B) ∪ ((A ∪ C) ∩ A) Distributive = ((A ∩ B) ∪ (C ∩ B)) ∪ (A ∪ (C ∩ A)) Distributive = ((B ∩ C) ∪ (A ∩ B)) ∪ (A ∪ (A ∩ C)) Commutative (2x) = ((B ∩ C) ∪ (A ∩ B)) ∪ ((A ∩ U) ∪ (A ∩ C)) Identity = ((B ∩ C) ∪ (A ∩ B)) ∪ (A ∩ (U ∪ C)) Distributive = ((B ∩ C) ∪ (A ∩ B)) ∪ (A ∩ U) Identity = ((B ∩ C) ∪ (A ∩ B)) ∪ A Identity = (B ∩ C) ∪ ((A ∩ B) ∪ A)) Associative = (B ∩ C) ∪ ((A ∩ B) ∪ (A ∩ U)) Identity = (B ∩ C) ∪ (A ∩ (B ∪ U)) Distributive = (B ∩ C) ∪ (A ∩ U) Identity = (B ∩ C) ∪ A Identity = A ∪ (B ∩ C) Commutative (d) (A ∩ B) ∪ (A ∩ B ' ) = A ∩ (B ∪ B ' ) Distributive = A ∩ U Complement = A Identity (e) (A ∩ B) ∪ (A ∪ B ' ) ' = (A ∩ B) ∪ (A ' ∩ (B ' ) ' ) De Morgan = (A ∩ B) ∪ (A ' ∩ B) Involution = (B ∩ A) ∪ (B ∩ A ' ) Commutative (× 2) = B ∩ (A ∪ A ' ) Distributive = B ∩ U Complement = B Identity (f) A ∩ (A ∪ B) = (A ∪ ø) ∩ (A ∪ B) Identity = A ∪ (ø ∩ B) Distributive = A ∩ (B ∩ ø) Commutative = A ∪ ø Identity = A Identity

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Answers to Set Theory Exercise 5

1

(a) X × Y = {(a, a), (a, b), (a, e), (a, f), (c, a), (c, b), (c, e), (c, f)}
(b) Y × X = {(a, a), (a, c), (b, a), (b, c), (e, a), (e, c), (f, a), (f, c)}
(c) X × X = {(a, a), (a, c), (c, a), (c, c)}
(d) They are equal: A = B

2

(a) (b, 2), (b, 4), (c, 1), (c, 5), (e, 1), (e, 5), (f, 2), (f, 4)
(b) P = C × R
(c) ((G × R) ∪ (C × T)) - (G × T)

3

V = {pqr | (p, q, r) ∈ L × (LD) × (LD)}

4

The shaded area is the same in each case, so it looks as though the proposition is true.

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