Discrete Mathematics/Set theory/Answers

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Answers to Set Theory Exercise 1[edit]

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[edit]

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)
VennExercise2Answer4.jpg


(b) PQ; RQ
(c) False


5

(a)
VennExercise2Answer5a.jpg


(b)
VennExercise2Answer5b.jpg


(c)
VennExercise2Answer5c.jpg


(d)
VennExercise2Answer5d.jpg



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Answers to Set Theory Exercise 3[edit]

1

Answers to Set Theory Exercise 4[edit]

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 ∪ (øA) = B ∪ (Aø) Commutative
= Bø Identity
= B Identity
(b) (A ' ∩ U) ' = (A' ) ' ∪ U' De Morgan
= AU' Involution
= Aø Complement
= A Identity
(c) (CA) ∩ (BA) = (AC) ∩ (BA) Commutative
= (AC) ∩ (AB) Commutative
= A ∪ (CB) Distributive
= A ∪ (BC) Commutative
(d) (AB) ∪ (AB ' ) = A ∩ (BB ' ) Distributive
= AU Complement
= A Identity
(e) (AB) ∪ (AB ' ) ' = (AB) ∪ (A ' ∩ (B ' ) ' ) De Morgan
= (AB) ∪ (A ' ∩ B) Involution
= (BA) ∪ (BA ' ) Commutative (× 2)
= B ∩ (AA ' ) Distributive
= BU Complement
= B Identity
(f) A ∩ (AB) = (Aø) ∩ (AB) Identity
= A ∪ (øB) Distributive
= A ∩ (Bø) Commutative
= Aø Identity
= A Identity



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Answers to Set Theory Exercise 5[edit]

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

SetTheoryExercise5Answer4.jpg


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


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