we have,
A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
∴ A × B = {1, 2, 3} × {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
and, B × C = {3, 4} × {4, 5, 6} = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
∴ (A × 8) ∩ {B × C} = {3, 4}.
We have,
A = {2, 3} , B = {4, 5} and C = {5,6}
∴ B ∪ C = {4,5} ∪ {5,6} = {4,5,6}
∴ A × {B ∪ C} = {2,3} × {4,5,6}
= {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Now,
B ∩ C = {4,5} ∩ {5,6} = {5}
∴ A × (B ∩ C) = {2,3} × {5} = {(2, 5), (3, 5)}
Now,
A × B = {2, 3} × {4, 5}
= {(2,4), (2,5), (3, 4), (3, 5)}
and, A x C = {2,3} × {5,6}
= {(2, 5), (2, 6), (3, 5), (3, 6)}
∴ (A × B) ∪ (A × C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
We have,
A = {1, 2, 3} × {4}
∴ B ∪ C = {4} ∪ {5} = {4, 5}
∴ A × (B ∪ C) = {1, 2, 3} × {4, 5}
⟹ A × (B ∪ C) = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} ....(i)
Now,
A × B = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
and, 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)}
⟹ (A × B) ∪ (A × C) = {(1,4), (1,5), (2, 4), (2, 5), (3, 4), (3, 5)} ......(ii)
From equation (i) and (ii), we get
A × (B ∪ C) = (A × B) ∪ (A × C)
Hence verified.
We have,
A = {1, 2, 3}, B = {4} and C = {5}
∴ B ∩ C = {4} ∩ {5} = Φ
∴ A × (B ∩ C) = {1,2,3} × Φ
⟹ A × (8 ∩ C) = Φ ....(i)
Now,
A × B = {1 ,2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
and, 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)}
⟹ (A × B) ∩ (A × C) = Φ ...(ii)
From equation (i) and equation (ii), we get
A × (B ∩ C) = (A × B) ∩ (A × C)
Hence verified.
We have,
A = {1, 2, 3}, B = {4} and C = {5}
∴ B - C = {4}
∴ A × (B - C) = {1, 2, 3} × {4}
⟹ A × (B - C) = {(1, 4), (2, 4), (3, 4)} ... (i)
Now,
A × B = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
and, A × C = {1, 2, 3} × {5} = {(1, 5), (2, 5), (3, 5)}
∴ (A × B) - (A × C) = {(1, 4), (2, 4), (3, 4)} ...(ii)
From equation (i) and equation (ii), we get
A × (B - C) = (A × B) - (A × C)
Hence verified.
We have,
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
∴ 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)} …..(i)
and, A × C = (1, 2) × (5, 6)
= {(1, 5), (1, 6), (2, 5), (2, 6)} ---(ii)
Clearly from equation (i) and equation (ii), we get
A × C ⊂ B × D
Hence verified.
We have,
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
∴ B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
A × (B ∩ C) = {1, 2} × Φ = Φ ...(i)
Now,
A × B = {1, 2} × {1, 2, 3, 4}
= {(1,1), (1, 2), (1,3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
and, A × C = {1, 2} × {5, 6} = {(1, 5), (1, 6), (2, 5), (2, 6)}
∴ (A × B) ∩ (A × C) = θ ---(ii)
From equation (i) and equation (ii), we get
A × (B ∩ C) = (A × B) ∩ (A × C)
Hence verified.
(i) we have,
A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
∴ B ∩ C = {3, 4} ∩ {4, 5, 6} = {4}
∴ A × (B ∩ C) = {1, 2, 3,} × {4}
= {(1, 4), (2, 4), (3, 4)}
⟹ A × (8 ∩ C) = {(1, 4) , (2, 4), (3, 4)}
(ii) We have,
A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
∴ A × B = {1, 2, 3,} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
And
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) we have,
A = {1, 2, 3,}, B = (3, 4) and C = (4, 5, 6)
∴ B ∪ C = {3, 4} ∪ {4, 5, 6} = (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)}
Let (a, b) bean arbitrary element of (A ∪ B) × C. Then,
(a, b) ϵ (A ∪ B) × C
⟹ a ϵ A ∪ B and b ϵ C [By defination]
⟹ (a ϵ A or a ϵ B) and b ϵ C [By defination]
⟹ (a ϵ A and b ϵ C) or (a ϵ B and b ϵ C)
⟹ (a, b) ϵ A × C or (a, b) ϵ B × C
⟹ (a, b) ϵ (A × C) ∪ (B × C)
⟹ (a, b) ϵ (A ∪ B) × C
⟹ (a, b) ϵ (A × C) ∪ (B × C)
⟹ (A ∪ B) × C ⊆ (A × C) ∪ (B × C) ... (i)
Again, let (x, y) be an abitrary element of (A × C) ∪ (B × C). Then,
(x, y) ϵ (A × C) ∪ (B × C)
⟹ (x, y) ϵ A × C or (x, y) ϵ B × C
⟹ x ϵ A and y ϵ C or x ϵ B and y ϵ C
⟹ (x ϵ A or x ϵ B) and y ϵ C
⟹ x ϵ A ∪ B and y ϵ C
⟹ (x, y) ϵ (A ∪ B) × C
⟹ (x, y) ϵ (A × C) ∪ (B × C)
⟹ (x, y) ϵ (A ∪ B) × C
⟹ (A × C) ∪ (B × C) ⊆ (A ∪ B) × C ...(ii)
Using equation (i) and equation (ii), we get
(A ∪ B) × C = (A × C) ∪ (B × C)
Hence proved.
Let (a, b) be an arbitrary element of (A ∩ 8) × C. Then,
(a, b) ϵ (A ∩ B) × C
⟹ a ϵ A ∩ B and b ϵ C
⟹ (a ϵ A and a ϵ B) and b ϵ C [by defination]
⟹ (a ϵ A and b ϵ C) and (a ϵ B and b ϵ C)
⟹ (a, b) ϵ A × C and (a, b) ϵ B × C
⟹ (a, b) ϵ (A × C) ∩ (B × C)
⟹ (a, b) ϵ (A ∩ B) × C
⟹ (a, b) ϵ (A × C) ∩ (8 × C)
⟹ (A ∩ 8) × C ⊆ (A × (B × C) ---(i)
Let (x, y) be an arbitrary element of (A × C) ∩ (B × C). Then,
(x, y) ϵ (A × C) ∩ (B × C)
⟹ (x, y) × A × C and (x, y) ϵ B × C [By defination]
⟹ (x ϵ A and y ϵ C) and (x ϵ B and y ϵ C)
⟹ (x ϵ A and x ϵ B) and y ϵ C
⟹ x ϵ A ∩ B and y ϵ C
⟹ (x, y) ϵ (A ∩ B) × C
⟹ (x, y) ϵ (A × C) ∩ (B × C)
⟹ (x, y) ϵ (A ∩ B) × C
⟹ (A × C) ∩ (B × C) ⊆ (A ∩ B) × C ---(ii)
Using equation (i) and equation (ii), we get
(A ∩ B) × C = (A × C) ∩ (B × C)
Let (a, b) be an arbitrary element of A × B. then,
(a, b) ϵ A × B
⟹ a ϵ A and b ϵ B ...(i)
NOW
(a, b) ϵ A × B
⟹ (a, b) ϵ C × D [∵ A × B ⊆ C × D]
⟹ a ϵ C and b ϵ D ---(ii)
∴ a ϵ A ⟹ a ϵ C [Using (i) and (ii)]
⟹ A ⊆ C
and,
b ϵ B ⟹ b ϵ D
⟹ B ⊆ D
Hence, proved