Discrete Mathematics/Set theory/Answers

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) P ⊂ Q; R ⊂Q

(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) P ⊆ Q

(b) Q ⊆ P

4

(a)	(b)

(c)	(d)

(e)	(f)

5

(a) B ′

(b) A ∩ B ′

(c) (A ∪ B) ∩ (A ∩ B) ′ or (A ∩ B ′) ∪ (A ′ ∩ B)

(d) (A ∩ B) ∪ (A ′ ∩ B ′) or (A ∪ B) ′ ∪ (A ∩ B) or …?

6

(a) Region (b) represents A – B. So A – B = A ∩ B ′

(b) Region (c) represents A Δ B.

So A Δ B = (A ∩ B) ∪ (A ′ ∩ B ′) or (A ∪ B) ′ ∪ (A ∩ B)

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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

Back to Set Theory Exercise 4

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 × (L ∪ D) × (L ∪ D)}

4

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

Back to Set Theory Exercise 5

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