Q1. Let A – {a, b, c, d}, B = {b, c, e} and C = {a, b, e}. Find:

1. A ∪ B = {a, b, c, d, e}
2. B ∪ C = {a, b, c, e}
3. A ∪ C = {a, b, c, d, e}
4. A ∩ B = {b, c}
5. B ∩ C = {b, e}
6. A ∩ C = {a, b}

Note:

The union of sets A and B, denoted by A ∪ B, is the set of all those elements, each one of which is either in A or in B or in both A and B

If there is a set A = {2, 3} and B = {a, b}, then A ∪ B = {2, 3, a, b}

• So if A ∪ B = {x | x ∈ A or x ∈ B} then
• x ∈ A ∪ B which means x ∈ A or x ∈ B
• And if x ∉ A ∪ B which means x ∉ A or x ∉ B
• The intersection of sets A and B is denoted by A ∩ B, and is a set of all elements that are common in sets A and B.
• If A = {1, 2, 3} and B = {2, 4, 5}, then A ∩ B = {2} as 2 is the only common element.
• Thus A ∩ B = {x: x ∈ A and x ∈ B}
• then x ∈ A ∩ B i.e. x ∈ A and x ∈ B
• And if x ∉ A ∩ B i.e. x ∉ A and x ∉ B

Q2. Let A = {2, 3, 4, 6}, B = {4, 6, 8, 9} and C = {2, 7, 8, 9}. Find:

1. A ∪ B = {2, 3, 4, 6, 8, 9}
2. B ∪ C = {2, 4, 6, 7, 8, 9}
3. A ∪ C = {2, 3, 4, 7, 8, 9}
4. A ∩ B = {4, 6}
5. B ∩ C = {8, 9}
6. A ∩ C = {2}

Q3. Let A = {1, 4, 7, 8} and B = {4, 6, 8, 9}. Find i) A – B and ii) B – A

1. A – B = {1, 4, 7, 8} – {4, 6, 8, 9} = {1, 7}
2. B – A = {4, 6, 8, 9} – {1, 4, 7, 8} = {6, 9}

Note:

• For any two sets A and B, the difference A – B is a set of all those elements of A which are not in B.
• i.e. if A = {1, 2, 3, 4, 5} and B = {4, 5, 6}
• Then A – B = {1, 2, 3} and B – A = {6}
• Therefore, A – B = {x | x ∈ A and x ∉ B}, then
• x ∈ A – B then x ∈ A but x ∉ B

Q4. Let ξ = {13, 14, 15, 16, 17, 18, 19, 20, 21}, A = {13, 17, 19} and B = {14, 16, 18, 20}. Find i) A′ and ii) B′

1. A′ = {14, 15, 16, 18, 20, 21}
2. B′ = {13, 15, 17, 19, 21}

Note:

• Let x be the universal set and let A⊆ x. Then the complement of A, denoted by A′ is the set of all those elements of x which are not in A.
• i.e. let ξ = {1, 2, 3, 4, 5,6 ,7 ,8} and A = {2, 3, 4}, then A′ = {1, 5, 6, 7, 8}
• Thus A′ = {x | x ∈ ξ and x ∉ A} clearly x ∈ A′ and x ∉ A
• ϕ′ = ξ and ϕ′= ξ
• A ∪ A′ = ξ and A ∩ A′ = ϕ

Q5. Let ξ = {x | x ∈ Z, -4 ≤ x ≤ 4}, A = {x | x ∈ W, x <4} and B = {x | x ∈ N, 2 <x ≤ 4}. Find i) A′ and ii) B′

Note: First find out the elements of the three given sets.

ξ = {-4, -3, -2, -1, 0, 1, 2, 3, 4}, A = {0, 1, 2, 3} and B = {3, 4}

A′ = {-4, -3, -2, -1, 4}

B′ = {-4, -3, -2, -1, 0, 1, 2}

Q6. Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36} and C = {x | x ∈ N, x is a factor of 48}. Find:

1. A′
2. B′
3. C′
4. A ∪ B
5. B ∪ C
6. A ∪ C
7. A ∩ B′
8. B ∩ C′
9. C – A
10. A – (B ∩ C)

Note: First calculate what the elements of each of these sets are. Therefore, we have the following:

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. This is the universal set.

A = {1, 2, 3, 4, 6, 8, 12, 24}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Now we can easily calculate.

1. A′ = {9, 16, 18, 36, 48, 72, 144}
2. B′ = {8, 16, 24, 48, 72, 144}
3. C′ = {9, 18, 36, 72, 144}
4. A ∪ B = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}
5. B ∪ C = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 26, 48}
6. A ∩ C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
7. A ∩ B′ = {8, 24}
8. B ∩ C′ = {9, 18, 36}
9. C – A = {16, 48}
10. A – (B ∩ C)
• First calculate (B ∩ C) = {1, 2, 3, 4, 6, 12}
• Now Calculate A – (B ∩ C) = {8, 24}

Q7. Considering the sets given in Q.6 state whether the following statements are True or False.

1. A ∩ (B ∪ C) = A
2. A ⊂ C
3. B ⊆ C
4. A ∩ C′ = ϕ

Note: First calculate what the elements of each of these sets are. Therefore, we have the following:

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. This is the universal set.

A = {1, 2, 3, 4, 6, 8, 12, 24}

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

A ∩ (B ∪ C) = A

A = {1, 2, 3, 4, 6, 8, 12, 24}

B ∪ C = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

Therefore, A ∩ (B ∪ C) = {1, 2, 3, 4, 6, 8, 12, 24} = A

Hence Proved. TRUE

A ⊂ C

A = {1, 2, 3, 4, 6, 8, 12, 24}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} Every element of A is in C and there are two elements (16 and 18) which are in C but not in A. A is a proper subset of B.

Hence, TRUE.

B ⊆ C

B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

Here every element of B is not in C. Hence it is not a subset of C. Hence FALSE.

A ∩ C′ = ϕ

ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. This is the universal set.

A = {1, 2, 3, 4, 6, 8, 12, 24}

C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} Therefore C′ = {9, 18, 36, 72, 144}

If we look at the two sets A and C’, we see that there are no common elements and hence

A ∩ C′ = f or null set. Hence TRUE

Q8. Let A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Then verify the following identities:

B ∪ C = C ∪ B

B = {a, c, e, g} & C = {b, e, f, g}. Therefore, B ∪ C = {a, b, c, e, f, g}.

Similarly, C ∪ B = {a, b, c, e, f, g}.

Hence B ∪ C = C ∪ B

B ∩ C = C ∩ B

B = {a, c, e, g} & C = {b, e, f, g}. Therefore, B ∩ C = {a, b, c, e, f, g}.

Similarly, C ∪ B = {a, b, c, e, f, g}.

A ∪ (B ∪ C) = (A ∪ B) ∪ C

A = {a, b, c, d, e}, B = {a, c, e, g}, C = {b, e, f, g

Therefore:

LHS: A ∪ (B ∪ C) = {a, b, c, d, e} ∪ {a, b, c, e, f, g} = {a, b, c, d, e, f, g}

RHS: (A ∪ B) ∪ C = {a, b, c, d, e, g} ∪ {b, e, f, g} = {a, b, c, d, e, f, g}

Therefore, LHS = RHS. Hence Proved.

A ∩ (B ∩ C) = (A ∩ B) ∩ C

A = {a, b, c, d, e}, B = {a, c, e, g}, C = {b, e, f, g}

Therefore:

LHS: A ∩ (B ∩ C) = {a, b, c, d, e} ∩ {e, g} = {e}

RHS: (A ∩ B) ∩ C = {a, c, e} ∩ {b, e, f, g} = {e}

Therefore, LHS = RHS. Hence Proved.

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

A = {a, b, c, d, e}, B = {a, c, e, g}, C = {b, e, f, g}  Therefore:

LHS: A ∪ (B ∩ C) = {a, b, c, d, e} ∪ {e, g} = {a, b, c, d, e, g}

RHS: (A ∪ B) ∩ (A ∪ C) = {a, b, c, d, e, g} ∩ {a, b, c, d, e, f, g} = {a, b, c, d, e, g}

Therefore, LHS = RHS. Hence Proved.

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

A = {a, b, c, d, e}, B = {a, c, e, g}, C = {b, e, f, gTherefore:

LHS: A ∩ (B ∪ C) = {a, b, c, d, e} ∩ {a, b, c, e, f, g} = {a, b, c, e}

RHS: (A ∩ B) ∪ (A ∩ C) = {a, c, e} ∩ {b, e} = {a, b, c, e}

Q9. Let A = {b, c, d, e} and B = {d, e, f, g} be the two sub sets of the universal set ξ = {b, c, d, e, f, g}. Then verify the following:

(A ∪ B)′ = (A′∩ B′)

LHS: A ∪ B = {b, c, d, e} ∪ {d, e, f, g} = {b, c, d, e, f, g}

Therefore, (A ∪ B)′ = ϕ

RHS: A′ = {f, g} and B′ = {b, c}

Therefore, A′∩ B′ = ϕ

Therefore, LHS = RHS. Hence proved.

(A ∩ B)′ = (A′∪ B′)

x = {b, c, d, e, f, g}

LHS: A ∩ B = {b, c, d, e} ∩ {d, e, f, g} = {d, e}

Therefore, (A ∩ B)′ = {b, c, f, g}

RHS: A′ = {f, g} and B′ = {b, c}

Therefore, A′∪ B′ =  {b, c, f, g}

Therefore, LHS = RHS. Hence proved.

Q10. Fill in the blanks

1. A ∪ A = A
2. A ∩ A = A
3. A ∪ f = A
4. A ∩ f = A
5. (A ∪ B)′ = (A′∩ B′)
6. (A ∩ B)′ = (A′∪ B′)

Q11. Let ξ ={x | x ∈ N 4 ≤ x < 18} and A, B, and C are subsets of ξ. Given A = {x | x is a multiple of 2}, B = {x | x is a multiple of 3}, C = {x | x ∈ N x < 11}. Then verify, the following:

Note: First identify the elements of all the 4 sets in this question. That would make the solution easier.

ξ= {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

A = {4, 6, 8, 10, 12, 14, 16}

B = {6, 9, 12, 15}

C = {4, 5, 6, 7, 8, 9, 10}.

(A ∪ B)′ = (A′∩ B′)

LHS: (A ∪ B) = {4, 6, 8, 9, 10, 12, 14, 15, 16}

(A ∪ B)′ = {5, 7, 11, 13, 17}

RHS: (A′∩ B′) = {5, 7, 9, 11, 13, 15, 17} ∩ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}

(A′∩ B′) = {5, 7, 11, 13, 17}

Therefore, LHS = RHS. Hence proved.

(A ∩ B)′ = (A′∪ B′)

LHS: (A ∩ B)′ = {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}

RHS: (A′∪ B′) = {5, 7, 9, 11, 13, 15, 17} ∪ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}

= {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}

Therefore, LHS = RHS. Hence proved.

A – B = A ∩ B′

A – B = {4, 6, 8, 10, 12, 14, 16} – {6, 9, 12, 15} = {4, 8, 10, 14, 16}

A ∩ B′ = {4, 6, 8, 10, 12, 14, 16} ∩ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17} = {4, 8, 10, 14, 16}

Therefore, LHS = RHS. Hence proved.

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

A = {4, 6, 8, 10, 12, 14, 16}

B = {6, 9, 12, 15}

C = {4, 5, 6, 7, 8, 9, 10}

A ∪ (B ∩ C) = {4, 6, 8, 9, 10, 12, 14, 16}

(A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 9, 10, 12, 14, 15, 16} ∩ {4, 5, 6, 7, 8, 9, 10, 12, 14, 16} = {4, 6, 8, 9, 10, 12, 14, 16}

Therefore, LHS = RHS. Hence proved