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By taking suitable sets A, B, C, verify the following results: 

(i) A × (B ∩ C) = (A × B) ∩ (A × C). 

(ii) A × (B ∪ C) = (A × B) ∪ (A × C). 

(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A). 

(iv) C – (B – A) = (C ∩ A) ∪ (C ∩ B). 

(v) (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A).

(vi) (B – A) ∪ C = {1, 5, 8, 9, 10}

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To prove the following results let us take 

U = {1, 2, 5, 7, 8, 9, 10} 

A = {1, 2, 5, 7} 

B = {2, 7, 8, 9} 

C = {1, 5, 8, 7} 

(i) To prove: A × (B ∩ C) = (A × B) ∩ (A × C) 

B ∩ C = {8}; A = {1, 2, 5, 7} 

So A × (B ∩ C) = {1, 2, 5, 7} × {8} 

= {(1, 8), (2. 8), (5, 8), (7, 8)} 

Now A x B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)} …(1) 

A × C = {(1, 1), (1, 5),(1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)} 

(A × B) ∩ (A × C) = {(1, 8), (2, 8), (5, 8), (7, 8)} …(2) 

(1) = (2)

⇒ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To prove A × (B ∪ C) = (A × B) (A × C) 

B = {2, 7, 8, 9}, C = {1, 5, 8, 10) 

B ∪ C = {1, 2, 5, 7, 8, 9, 10} 

A = {1, 2, 5, 7} 

A × (B ∪ C) = {(1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)) … (1) 

A × B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)} 

A × C = {(1, 1), (1, 5), (1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)} 

(A × B) ∪ (A × C) = (1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1,9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)} … (2) 

(1) = (2) ⇒ A × (B ∪ C) = (A × B) ∪ (A × C) 

(iii) A × B = {(1, 2), (1, 7), (1, 8), (1, 9) (2, 2), (2, 7), (2, 8), (2, 9) (5, 2), (5, 7), (5, 8), (5, 9) (7, 2), (7, 7), (7, 8), (7, 9)} 

B × A = {(2, 1), (2, 2), (2, 5), (2, 7) (7, 1), (7, 2), (7, 5), (7, 7) (8, 1), (8, 2), (8, 5), (8, 7) (9,1), (9, 2), (9, 5), (9, 7)} 

L.H.S. (A × B) ∩ (B × A) = {(2, 2), (2, 7), (7, 2), (7, 7)} …(1) 

R.H.S. A ∩ B = {2, 7} 

B ∩ A = {2, 7} 

(A ∩ B) × (B ∩ A) = {2, 7} × {2, 7} 

= {(2, 2), (2, 7), (7, 2), (7, 7)} … (2) 

(1) = (2) ⇒ LHS = RHS

(iv) To prove C – (B – A) = (C ∩ A) ∪ (C ∩ B)

B – A = {8, 9} 

C = {1, 5, 8, 10}

∴ LHS = C – (B – A) = {1, 5, 10} … (1) 

C ∩ A = {1} 

U = {1, 2, 5, 7, 8, 9, 10} 

B = {2, 7, 8, 9} ∴ B’ = {1, 5, 10} 

C ∩ B = {1, 5, 10} 

R.H.S. (C ∩ A) ∪ (C ∩ B) = {1} ∪ {1, 5, 10} 

= {1, 5, 10} … (2) 

(1) = (2) ⇒ LHS = RHS

(v) To prove (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A) 

A = {1, 2, 5, 7}, B = {2, 7, 8, 9}, C = {1, 5, 8, 10} 

Now B – A = {8, 9} 

(B – A) ∩ C = {8} … (1) 

B ∩ C = {8} 

A = {1, 2, 5, 7} 

So (B ∩ C) – A = {8} … (2) 

C – A = {8, 10}

B = {2, 7, 8, 9} 

B ∩ (C – A) = {8} …. (3) 

(1) = (2) = (3)

(vi) To prove (B – A) ∪ C = {1, 5, 8, 9, 10} 

B – A = {8, 9}, 

C = {1, 5, 8, 10} 

(B – A) ∪ C = {1, 5, 8, 9, 10} ….. (1) 

B ∪ C = {1, 2, 5, 7, 8, 9, 10} 

A – C = {2, 7} 

(B ∪ C) – (A – C) = {1, 5, 8, 9, 10} ….. (2) 

(1) = (2) 

⇒ (B – A) ∪ C = (B ∪ C) – (A – C)

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