Let A = { x : x ∈ N , x is a multiple of 3 } and B = { x : x ∈ N and x is a multiple of 5 } . Write A ∩ B .

A = { x : x ∈ N and x is a multiple of 3 } = { 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , … } B = { x : x ∈ N and x is a multiple of 5 } = { 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , … } Therefore, it is clear that: A ∩ B = { 15 , 30 , 45 , … } = { x : x ∈ N where x is multiple of 15 }

If A = x ∈ C : x 2 = 1 and B = x ∈ C : x 4 = 1 , then write A - B and B - A .

As given, A = { x ∈ C : x 2 = 1 } x = 1 , - 1 ⇒ A = { - 1 , 1 } Then B = { x ∈ C : x 4 = 1 } ⇒ B = { x 4 - 1 = 0 } x 4 = 1 ⇒ B = i , - i , 1 , - 1 ⇒ A - B = ϕ The elements in set A that are not in B is ⇒ B - A = i , - i The elements in set B that are not in A is i , - i

If A and B are two sets such that A ⊂ B , then write B ' - A ' in terms of A and B .

If μ is universal set,then A ' = μ - A and B ' = μ - B ⇒ B ' - A ' = μ - A - μ - B = μ - A - μ + B ⇒ A - B = ∅ [given A ⊂ B ]

If A = ( x , y ) : y = 1 x , 0 ≠ x ∈ R and B = [ ( x , y ) : y = - x , x ∈ R ] then write A ∩ B

As we know, A = x , y : y = 1 x , x ∈ R - 0 = { 1 , 1 2 , 1 2 3 , 1 3 4 , 1 4 5 , 1 5 , … } Then B = x , y : y = - x , x ∈ R = 1 , - 1 2 , - 2 3 , - 3 4 , - 4 … … Therefore, we get ⇒ A ∩ B = 1 , 1 2 , 1 2 3 , 1 3 4 , 1 4 … … ∩ 0 , 0 1 , - 1 2 , - 2 3 , - 3 , … … There are no sets common between them. So, A ∩ B is or empty set. ∴ A ∩ B = ∅

If A = ( x , y ) : y = e x , x ∈ R and B = ( x , y ) : y = e - x , x ∈ R then write A ∩ B .

Given, A = ( x , y ) : y = e x , x ∈ R and B = ( x , y ) : y = e - x , x ∈ R ⇒ A = 0 , 1 , 1 , e , 2 , e 2 , 3 , e 3 , … ⇒ B = 0 , 1 1 , e - 1 2 , e - 2 3 , e - 3 … Therefore, it is clear that the common elements between two sets is A ∩ B ⇒ A ∩ B = 0 , 1