Given an example of three sets A , B , C such that A ∩ C ≠ ϕ , B ∩ C ≠ ϕ , A ∩ C ≠ ϕ , and A ∩ B ∩ C = ϕ

Let A = { 1 , 2 } , B = { 2 , 3 , 4 } and C = { 1 , 3 , 5 } We know that, to find the intersection of two sets P and Q is a set which consists of all the elements which are common to both P and Q and it is denoted by P ∩ Q . Then, ( A ∩ B ) = { 1 , 2 } ∩ { 2 , 3 , 4 } = { 2 } ≠ ϕ . ( B ∩ C ) = { 2 , 3 , 4 } ∩ { 1 , 3 , 5 } = { 3 } ≠ ϕ . ( A ∩ C ) = { 1 , 2 , } ∩ { 1 , 3 , 5 } = { 1 } ≠ ϕ . ( A ∩ B ∩ C ) = { 1 , 2 } ∩ { 2 , 3 , 4 } ∩ { 1 , 3 , 5 } = ϕ . Thus, A ∩ B ≠ ϕ , B ∩ C ≠ ϕ , A ∩ C ≠ ϕ and ( A ∩ B ∩ C ) = ϕ .

For any sets A and B , prove that: ( A – B ) ∩ B = ϕ

We have to prove that ( A - B ) ∩ B = ϕ LHS = A - B ∩ B We know that, A ∩ B ' = A - B = A ∩ B ' ∩ B = A ∩ B ' ∩ B = A ∩ ϕ = ϕ ( A - B ) ∩ B = ϕ LHS = RHS Hence proved

For any sets A and B , prove that: A ∪ ( B – A ) = A ∪ B

We have to prove that: A ∪ ( B - A ) = A ∪ B LHS = A ∪ B - A Using the result B ∩ A ' = B - A = A ∪ B ∩ A ' Using the distributive property A ∪ B ∩ C = A ∪ B ∩ A ∪ C = A ∪ B ∩ A ∪ A ' = A ∪ B ∩ U Any set with intersection with universal set is set itself = A ∪ B A ∪ ( B - A ) = A ∪ B Hence proved.

For any sets A and B , prove that: ( A – B ) ∪ ( A ∩ B ) = A

We have to prove that: ( A - B ) ∪ ( A ∩ B ) = A LHS = A - B ∪ A ∩ B Using the result A ∩ B ' = A - B = A ∩ B ' ∪ A ∩ B Using the distributive property of intersection over union A ∩ B ∪ C = A ∩ B ∪ A ∩ C = A ∩ B ∪ B ' = A ∩ U Any set with the intersection with the universal set is set itself = A ( A - B ) ∪ ( A ∩ B ) = A Hence proved.

For any sets A and B , prove that: ( A ∪ B ) – B = A – B

We have to prove that: ( A ∪ B ) - B = A - B LHS = A ∪ B - B By De Morgan's law A ∩ B ' = A - B .....(i) = A ∪ B ∩ B ' Using the distributive property A ∩ B ∪ C = A ∩ B ∪ A ∩ C = A ∩ B ' ∪ B ∩ B ' = A ∩ B ' ∪ ϕ = A ∩ B ' From (i) equation, = A - B ( A ∪ B ) - B = A - B Hence proved

For any sets A and B , prove that: A – ( A ∩ B ) = A – B

We have to prove that: A - A ∩ B = A - B LHS = A ∪ B - B We know that, A ∩ B ' = A - B .....(i) = A ∩ A ∩ B ' = A ∩ A ' ∪ B ' Using the distributive property A ∩ B ∪ C = A ∩ B ∪ A ∩ C = A ∩ A ' ∪ A ∩ B ' = ϕ ∪ A ∩ B ' = A ∩ B ' From (i) equation, = A - B A - A ∩ B = A - B Hence proved

For any sets A and B , prove that: A ∩ B ’ = ϕ ⇒ A ⊂ B

Given A ∩ B ' = ϕ To show that: A ⊂ B Let x ∈ A ∵ A ∩ B ' = ϕ ∴ x ∉ B ' B ' is the complement set of B . This means that x ∈ B ∵ x ∈ A & x ∈ B This is true for all x ∈ A Hence, A ⊂ B

For any sets A and B , prove that: A ' ∪ B = U ⇒ A ⊂ B

To prove: A ⊂ B Given A ' ∪ B = U Let a ∈ A ⇒ a ∈ U ⇒ a ∈ A ' ∪ B ∵ U = A ' ∪ B ⇒ a ∈ B ∵ a ∉ A ' Thus, a ∈ A ⇒ a ∈ B Hence A ⊂ B

E M B I B E

Senior Secondary School Mathematics FOR CLASS 11>Set and Functions>Chapter 1 - Sets>EXERCISE 1E

R S Aggarwal Solutions for Exercise 5: EXERCISE 1E

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 5: EXERCISE 1E

Attempt the practice questions from Exercise 5: EXERCISE 1E with hints and solutions to strengthen your understanding. Senior Secondary School Mathematics FOR CLASS 11 solutions are prepared by Experienced Embibe Experts.