Chapter 2: Relations – Exercise 2.2

Relations – Exercise 2.2 – Q.1

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}.

Relations – Exercise 2.2 – Q.2

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)}

Relations – Exercise 2.2 – Q.3

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.

Relations – Exercise 2.2 – Q.4

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.

Relations – Exercise 2.2 – Q.5

(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)}

Relations – Exercise 2.2 – Q.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)

Relations – Exercise 2.2 – Q.7

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