Tamil Nadu Board Samacheer Kalvi solutions for Class 9th Mathematics Answers Guide Chapter 1 Set Language Exercise 1.4 [Page 22]

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 = ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6}\}}$

P ∪ Q = {3, 4, 5, 6} ∪ ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6}\}}$

= ${\displaystyle \{3,4,5,6,\sqrt{3},\sqrt{5},\sqrt{6}\}}$  ...(1)

Q ∪ P = ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6}\}}$ ∪ {3, 4, 5, 6}

= ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6},3,4,5,6\}}$ ...(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 = ${\displaystyle \{3,4,5,6\}\cap \{\sqrt{3},\sqrt{5},\sqrt{6}\}}$ = { } ...(1)

Q ∩ P = ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6}\}\cap \{3,4,5,6\}}$ = { } ...(2)

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 = ${\displaystyle \{-11,\sqrt{2},\sqrt{5},7\}}$, B = ${\displaystyle \{\sqrt{3},\sqrt{5},6,13\}}$ and C = ${\displaystyle \{\sqrt{2},\sqrt{3},\sqrt{5},9\}}$

Solution

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

B ∩ C = ${\displaystyle \{\sqrt{3},\sqrt{5},6,13\}\cap \{\sqrt{2},\sqrt{3},\sqrt{5},9\}=\{\sqrt{3},\sqrt{5}\}}$

 A ∩ (B ∩ C) = ${\displaystyle \{-11,\sqrt{2},\sqrt{5},7\}\cap \{\sqrt{3},\sqrt{5}\}=\left\{\sqrt{5}\right\}}$  ...(1)

A ∩ B  = ${\displaystyle \{-11,\sqrt{2},\sqrt{5},7\}\cap \{\sqrt{3},\sqrt{5},6,13\}=\left\{\sqrt{5}\right\}}$

(A ∩ B) ∩ C = ${\displaystyle \left\{\sqrt{5}\right\}\cap \{\sqrt{2},\sqrt{3},\sqrt{5},9\}=\left\{\sqrt{5}\right\}}$  ...(2)

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 = 2ⁿ, 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 = 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