**Exercise 1.4 | Q 1. (i) | Page 22**

**If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (P ∪ Q) ∪ R**

#### Solution

P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}

P ∪ Q = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11}

= {1, 2, 3, 5, 7, 9, 11}

(P ∪ Q) ∪ R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9}

= {1, 2, 3, 4, 5, 7, 9, 11}

**Exercise 1.4 | Q 1. (ii) | Page 22**

**If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (P ∩ Q) ∩ S**

#### Solution

P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}

P ∩ Q = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11}

= {2, 5, 9}

(P ∩ Q) ∩ S = {2, 5, 9} ∩ {2, 3, 4, 5, 8}

= {2, 5}

**Exercise 1.4 | Q 1. (iii) | Page 22**

**If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (Q ∩ S) ∩ R**

#### Solution

P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}

Q ∩ S = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8}

= {2, 3, 5}

(Q ∩ S) ∩ R = {2, 3, 5} ∩ {3, 4, 5, 7, 9}

= {3, 5}

**Exercise 1.4 | Q 2 | Page 22**

**Test for the commutative property of union and intersection of the sets**

**P = {x : x is a real number between 2 and 7} and**

**Q = {x : x is an irrational number between 2 and 7}**

#### Solution

Commutative Property of union of sets

(A ∪ B) = (B ∪ A)

Here P = {3, 4, 5, 6}, Q =

P ∪ Q = {3, 4, 5, 6} ∪

=

Q ∪ P =

=

(1) = (2)

∴ P ∪ Q = Q ∪ P

∴ It is verified that union of sets is commutative.

Commutative Property of intersection of sets (P ∩ Q) = (Q ∩ P)

P ∩ Q =

Q ∩ P =

From (1) and (2)

P ∩ Q = Q ∩ P

∴ It is verified that intersection of sets is commutative.

**Exercise 1.4 | Q 3 | Page 22**

**If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets**

#### Solution

Associative Property of union of sets

A ∪ (B ∪ C) = (A ∪ B) ∪ C

(B ∪ C) = {m, n, q, s, t} ∪ {m, n, p, q, s}

= {m, n, p, q, s, t}

A ∪ (B ∪ C) = {p, q, r, s} ∪ {m, n, p, q, s, t}

= {m, n, p, q, r, s, t} ...(1)

(A ∪ B) = {p, q, r, s} ∪ {m, n, q, s, t}

= {m, n, p, q, r, s, t}

(A ∪ B) ∪ C = {m, n, p, q, r, s, t} ∪ {m, n, p, q, s}

= {m, n, p, q, r, s, t} ...(2)

From (1) and (2)

It is verified that A ∪ (B ∪ C) = (A ∪ B) ∪ C

**Exercise 1.4 | Q 4 | Page 22**

Verify the associative property of intersection of sets for A =

#### Solution

Associative Property of intersection of sets A ∩ (B ∩ C) = (A ∩ B) ∩ C

B ∩ C =

A ∩ (B ∩ C) =

A ∩ B =

(A ∩ B) ∩ C =

From (1) and (2),

It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C

**Exercise 1.4 | Q 5 | Page 22**

**If A = {x : x = 2 ^{n}, n ∈ W and n < 4}, B = {x : x = 2n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets**

#### Solution

A = {x : x = 2^{n}, n ∈ W, n < 4}

⇒ x = 2° = 1

x = 2^{1} = 2

x = 2^{2} = 4

x = 2^{3} = 8

∴ A = {1, 2, 4, 8}

B = {x : x = 2n, n ∈ N and n ≤ 4}

⇒ x = 2 × 1 = 2

x = 2 × 2 = 4

x = 2 × 3 = 6

x = 2 × 4 = 8

∴ B = {2, 4, 6, 8}

C = {0, 1, 2, 5, 6}

Associative property of intersection of sets

A ∩ (B ∩ C) = (A ∩ B) ∩ C

B ∩ C = {2, 6}

A ∩ (B ∩ C) = {1, 2, 4, 8} ∩ {2, 6}

= {2} ...(1)

A ∩ B = {1, 2, 4, 8} ∩ {2, 4, 6, 8}

= {2, 4, 8}

(A ∩ B) ∩ C = {2, 4, 8} ∩ {0, 1, 2, 5, 6}

= {2} ...(2)

From (1) and (2)

It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C