## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.4

**Exercise 1.4 Class 9 Maths Samacheer 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}

**Exercise 1.4 Class 9 Maths Solution Samacheer 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 = {

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 = {3, 4, 5, 6} ∩ {

Q ∩ P = {

From (1) and (2)

P ∩ Q = Q ∩ P

∴ It is verified that intersection of sets is commutative.

**Samacheer Kalvi 9th Maths Exercise 1.4 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

**9th Standard Maths Exercise 1.4 Question 4.**

Verify the associative property of intersection of sets for A = {-11,

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

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

**9th Maths Exercise 1.4 Question 5.**

If A = {x : x = 2n, 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 = 2n, n ∈ W, n < 4}

⇒ x = 2°= 1

x = 21 = 2

x = 22 = 4

x = 23 = 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

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