# Discrete Mathematics/Set theory/Answers

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## Contents

## 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, …}

Back to Set Theory Exercise 1

## 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)

- (b)
*P*⊂*Q*;*R*⊂*Q*

- (c) False

5

- (a)

- (b)

- (c)

- (d)

Back to Set Theory Exercise 2

## 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) 2
^{10}= 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) |
Distributive | ||

= 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[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*× (*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