## Tamilnadu State Board Class 9 Maths Solutions Chapter 1 Set Language Ex 1.4

Question 1.

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

(i) (P ∪ Q) ∪ R

(ii) (P ∩ Q) ∩ S

(iii) (Q ∩ S) ∩ R

Solution:

(i) (P ∪ Q) ∪ R

(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}

(ii)(P ∩ Q) ∩ S

(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}

(iii) (Q ∩ S) ∩ R

(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}

Question 2.

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 = {√3, √5, √6}

P ∪ Q = {3, 4, 5, 6} ∪ {√3, √5, √6} = {3, 4, 5, 6, √3, √5, √6} … (1)

Q ∪ P = {√3, √5, √6} ∪ {3, 4, 5, 6}= {√3, √5, √6, 3, 4, 5, 6} …(2)

(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 = {3, 4, 5, 6} ∩ {√3, √5, √6} = { } … (1)

Q ∩ P = {√3, √5, √6} ∩ {3, 4, 5, 6} = { } …(2)

From (1) and (2)

P ∩ Q = Q ∩ P

∴ It is verified that intersection of sets is commutative.

Question 3.

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

(A ∪ B) ∪ C = {p, q, r, s, m, n, t} ∪ {m, n,p, q, s} = {p, q, r, s, m, n, t} … (2)

From (1) & (2)

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

Question 4.

Verify the associative property of intersection of sets for A = {-11, √2 ,√5 ,7}, B = {√3, √5, 6, 13} and C = {√2, √3,√5, 9}.

Solution:

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

B ∩ C = {√3, √5, 6, 13} ∩ {√2, √3, √5, 9} = {√3, √5}

A ∩ (B ∩ C) = {-11, √2, √5, 7} ∩ {√3, √5} = {√5} … (1)

A ∩ B = {-11, √2, √5, 7} ∩ {√3, √5, 6, 13} = {√5}

(A ∩ B) ∩ C = {√5} ∩ {√2, √3, √5 ,9} = {√5} …(2)

From (1) and (2), it is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C

Question 5.

If A = {x : x = 2^{n}, n ∈ W and n < 4}, B = {x: x = 2 n, 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 x 1 = 2

x = 2 x 2 = 4

x = 2 x 3 = 6

x = 2 x 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