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Class 11 RD Sharma Solutions – Chapter 2 Relations – Exercise 2.2
  • Last Updated : 28 Dec, 2020

Question 1: Given A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, find (A × B) ∩ (B × C).

Solution:

Given:

A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}

Let us find: (A × B) ∩ (B × C)

(A × B) = {1, 2, 3} × {3, 4}



= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(B × C) = {3, 4} × {4, 5, 6}

= {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

Therefore,

(A × B) ∩ (B × C) = {(3, 4)}

Question 2: If A = {2, 3}, B = {4, 5}, C = {5, 6} find A × (B ∪ C), (A × B) ∪ (A × C).

Solution:

Given: 

A = {2, 3}, B = {4, 5} and C = {5, 6}

Let us find: A x (B ∪ C) and (A x B) ∪ (A x C)

(B ∪ C) = {4, 5, 6}

A × (B ∪ C) = {2, 3} × {4, 5, 6}

= {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(A × B) = {2, 3} × {4, 5}

= {(2, 4), (2, 5), (3, 4), (3, 5)}

(A × C) = {2, 3} × {5, 6}

= {(2, 5), (2, 6), (3, 5), (3, 6)}

Therefore,

(A × B) ∪ (A × C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}



A × (B ∪ C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

Question 3: If A = {1, 2, 3}, B = {4}, C = {5}, then verify that:

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

(iii) A × (B – C) = (A × B) – (A × C)

Solution:

Given:

A = {1, 2, 3}, B = {4} and C = {5}

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

Let’s assume LHS: (B ∪ C)

(B ∪ C) = {4, 5}

A × (B ∪ C) = {1, 2, 3} × {4, 5}

= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

Now, RHS

(A × B) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

(A × C) = {1, 2, 3} × {5}

= {(1, 5), (2, 5), (3, 5)}

(A × B) ∪ (A × C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}

Therefore,

LHS = RHS

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Let’s assume LHS: (B ∩ C)

(B ∩ C) = ∅ (No common element)

A × (B ∩ C) = {1, 2, 3} × ∅

= ∅

Now, RHS

(A × B) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

(A × C) = {1, 2, 3} × {5}

= {(1, 5), (2, 5), (3, 5)}

(A × B) ∩ (A × C) = ∅

Therefore,

LHS = RHS

(iii) A × (B − C) = (A × B) − (A × C)

Let’s assume LHS: (B − C)

(B − C) = ∅

A × (B − C) = {1, 2, 3} × ∅

= ∅

Now, RHS

(A × B) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

(A × C) = {1, 2, 3} × {5}

= {(1, 5), (2, 5), (3, 5)}

(A × B) − (A × C) = ∅

Therefore,

LHS = RHS

Question 4: Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that:

(i) A × C ⊂ B × D

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Solution:

Given:

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) A x C ⊂ B x D

Let us consider LHS A x C

A × C = {1, 2} × {5, 6}

= {(1, 5), (1, 6), (2, 5), (2, 6)}

Now, RHS

B × D = {1, 2, 3, 4} × {5, 6, 7, 8}

= {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Since, all elements of A × C is in B × D.

Therefore,

We can say A × C ⊂ B × D

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Let’s assume LHS A × (B ∩ C)

(B ∩ C) = ∅

A × (B ∩ C) = {1, 2} × ∅

= ∅

Now, RHS

(A × B) = {1, 2} × {1, 2, 3, 4}

= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

(A × C) = {1, 2} × {5, 6}

= {(1, 5), (1, 6), (2, 5), (2, 6)}

Hence, there is no common element between A × B and A × C

(A × B) ∩ (A × C) = ∅

Therefore,

A × (B ∩ C) = (A × B) ∩ (A × C)

Question 5: If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find

(i) A × (B ∩ C)

(ii) (A × B) ∩ (A × C)

(iii) A × (B ∪ C)

(iv) (A × B) ∪ (A × C)

Solution:

Given:

A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}

(i) A × (B ∩ C)

(B ∩ C) = {4}

A × (B ∩ C) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

(ii) (A × B) ∩ (A × C)

(A × B) = {1, 2, 3} × {3, 4}

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(A × C) = {1, 2, 3} × {4, 5, 6}

= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}

(iii) A × (B ∪ C)

(B ∪ C) = {3, 4, 5, 6}

A × (B ∪ C) = {1, 2, 3} × {3, 4, 5, 6}

= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

(iv) (A × B) ∪ (A × C)

(A × B) = {1, 2, 3} × {3, 4}

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(A × C) = {1, 2, 3} × {4, 5, 6}

= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

Question 6: Prove that:

(i) (A ∪ B) × C = (A × C) = (A × C) ∪ (B × C)

(ii) (A ∩ B) × C = (A × C) ∩ (B × C)

Solution:

(i) (A ∪ B) × C = (A × C) = (A × C) ∪ (B × C)

Let (x, y) be an arbitrary element of (A ∪ B) × C

(x, y) ∈ (A ∪ B) C

Since, (x, y) are elements of Cartesian product of (A ∪ B) × C

x ∈ (A ∪ B) and y ∈ C

(x ∈ A or x ∈ B) and y ∈ C

(x ∈ A and y ∈ C) or (x ∈ Band y ∈ C)

(x, y) ∈ A × C or (x, y) ∈ B × C

(x, y) ∈ (A × C) ∪ (B × C) … (1)

Let (x, y) be an arbitrary element of (A × C) ∪ (B × C).

(x, y) ∈ (A × C) ∪ (B × C)

(x, y) ∈ (A × C) or (x, y) ∈ (B × C)

(x ∈ A and y ∈ C) or (x ∈ B and y ∈ C)

(x ∈ A or x ∈ B) and y ∈ C

x ∈ (A ∪ B) and y ∈ C

(x, y) ∈ (A ∪ B) × C … (2)

From 1 and 2, we get: (A ∪ B) × C = (A × C) ∪ (B × C)

(ii) (A ∩ B) × C = (A × C) ∩ (B × C)

Let (x, y) be an arbitrary element of (A ∩ B) × C.

(x, y) ∈ (A ∩ B) × C

Since, (x, y) are elements of Cartesian product of (A ∩ B) × C

x ∈ (A ∩ B) and y ∈ C

(x ∈ A and x ∈ B) and y ∈ C

(x ∈ A and y ∈ C) and (x ∈ Band y ∈ C)

(x, y) ∈ A × C and (x, y) ∈ B × C

(x, y) ∈ (A × C) ∩ (B × C) … (1)

Let (x, y) be an arbitrary element of (A × C) ∩ (B × C).

(x, y) ∈ (A × C) ∩ (B × C)

(x, y) ∈ (A × C) and (x, y) ∈ (B × C)

(x ∈A and y ∈ C) and (x ∈ Band y ∈ C)

(x ∈A and x ∈ B) and y ∈ C

x ∈ (A ∩ B) and y ∈ C

(x, y) ∈ (A ∩ B) × C … (2)

From 1 and 2, we get: (A ∩ B) × C = (A × C) ∩ (B × C)

Question 7: If A × B ⊆ C × D and A ∩ B ∈ ∅, Prove that A ⊆ C and B ⊆ D.

Solution:

Given:

A × B ⊆ C x D and A ∩ B ∈ ∅

A × B ⊆ C x D denotes A × B is subset of C × D that is every element A × B is in C × D.

And A ∩ B ∈ ∅ denotes A and B does not have any common element between them.

A × B = {(a, b): a ∈ A and b ∈ B}

Therefore,

We can say (a, b) ⊆ C × D [Since, A × B ⊆ C x D is given]

a ∈ C and b ∈ D

a ∈ A = a ∈ C

A ⊆ C

And

b ∈ B = b ∈ D

B ⊆ D

Hence proved.

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