Fill in the blanks to make the statement true ξ ' ∩ A = _ _ _ _

ξ is a universal set with a subset A . We know that, a universal set is a set which has elements of all the related sets. So, ξ ' is an empty set Therefore ξ ' & A are disjoint. Intersection of two disjoint sets is empty set. Hence, ξ ' ∩ A = ϕ

Show that A ∩ B = A ∩ C need not imply B = C

Let A = 1 , 2 , B = 1 , 3 and C = 1 , 4 . Intersection of two given sets is the largest set which contains all the elements that are common to both the sets. Then, A ∩ B = 1 , A ∩ C = 1 . We know that, equal sets are the sets having same elements in both the sets. Thus, A ∩ B = A ∩ C but B ≠ C .

Find the sets A , B and C such that A ∩ B , A ∩ C and B ∩ C are non-empty and A ∩ B ∩ C = ϕ .

Given A ∩ B , A ∩ C and B ∩ C are non-empty and A ∩ B ∩ C = ϕ . Let A = 0 , 1 , B = 1 , 2 and C = 2 , 0 . Intersection of two given sets is the largest set which contains all the elements that are common to both the sets. Here, A ∩ B = 1 B ∩ C = 2 A ∩ C = 0 ∴ A ∩ B , B ∩ C , and A ∩ C are non-empty.. However, A ∩ B ∩ C = ∅

Show that the following four conditions are equivalent : i A ⊂ B i i A - B = ∅ i i i A ∩ B = A i v A ∪ B = B

First, we have to show that ( i ) ⇔ ( i i ) ( i ) ⇒ ( i i ) : Let A ⊂ B To show: A − B = ϕ If possible, suppose A − B ≠ ϕ ⇒ There exists x ∈ A , x ∉ B , which is not possible as A ⊂ B . ∴ A − B = ϕ Thus, A ⊂ B ⇒ A − B = ϕ ( i i ) ⇒ ( i ) : Let A − B = ϕ To show: A ⊂ B Let x ∈ A Clearly, x ∈ B because if x ∉ B , then A − B ≠ ϕ which is wrong Thus, A − B = ϕ ⇒ A ⊂ B Hence, ( i i ) ⇔ ( i ) Next, we have to show that ( i ) ⇔ ( i i i ) ( i ) ⇒ ( i i i ) : Let A ⊂ B Clearly A ∩ B ⊂ A Let x ∈ A We have to show that As A ⊂ B , x ∈ B ∴ x ∈ A ∩ B ∴ A ⊂ A ∩ B Hence, A = A ∩ B ( i i i ) ⇒ ( i ) : We have A ∩ B = A Let x ∈ A ⇒ x ∈ A ∩ B ⇒ x ∈ A a n d x ∈ B ⇒ x ∈ B ∴ A ⊂ B Hence, ( i ) ⇔ ( i i i ) Next, we show that ( i ) ⇔ ( i v ) Let A ⊂ B To show: A ∪ B = B Clearly, B ⊂ A ∪ B Let x ∈ A ∪ B ⇒ x ∈ A o r x ∈ B Case I : x ∈ A ⇒ x ∈ B [ ∵ A ⊂ B ] ∴ A ∪ B ⊂ B ... ( i ) Case II : x ∈ B Then clearly A ∪ B ⊂ B ... ( i i ) So, from ( i ) and ( i i ) , we get A ∪ B = B ( i v ) ⇒ ( i ) : Let A ∪ B = B To show : A ⊂ B Let x ∈ A ⇒ x ∈ A ∪ B [ ∵ A ⊂ A ∪ B ] ⇒ x ∈ B [ ∵ A ∪ B = B ] ∴ A ⊂ B Hence, ( i v ) ⇔ ( i ) Hence, A ⊂ B , A - B = Φ , A ∩ B = A and A ∪ B = B are equivalent.

E M B I B E

Understanding ISC Mathematics Class 11 Volume 1>Chapter 1 - Sets>EXERCISE 1.5

M L Aggarwal Solutions for Chapter: Sets, Exercise 5: EXERCISE 1.5

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Sets, Exercise 5: EXERCISE 1.5

Attempt the practice questions on Chapter 1: Sets, Exercise 5: EXERCISE 1.5 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 11 Volume 1 solutions are prepared by Experienced Embibe Experts.