Question - 3 : -
If A = {1, 2, 3}, B = {4}, C = {5}, then verify that:
(i) A x (B ∪ C) = (A x B) ∪ (A x C)
(ii) A x (B ∩ C) = (A x B) ∩ (A x C)
(iii) A x (B – C) = (A x B) – (A x C)

Answer - 3 : -

Given:

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

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

Let us consider 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)}

∴ LHS = RHS

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Let us consider 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) = ∅

∴ LHS = RHS

(iii) A × (B − C) = (A × B) − (A × C)

Let us consider 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) = ∅

∴ 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 x C ⊂ B x D
(ii) A x (B ∩ C) = (A x B) ∩ (A x C)

Answer - 4 : -

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.

∴We can say A × C ⊂ B × D

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Let us consider 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)}

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

(A × B) ∩ (A × C) = ∅

∴ 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 x (B ∩ C)
(ii) (A x B) ∩ (A x C)
(iii) A x (B ∪ C)
(iv) (A x B) ∪ (A x C)

Answer - 5 : -

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) x C = (A x C) = (A x C) ∪ (B x C)
(ii) (A ∩ B) x C = (A x C) ∩ (B x C)

Answer - 6 : -

(i) (A ∪ B) x C = (A x C) = (A x C) ∪ (B x 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) x C = (A x C) ∩ (B x 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 x B ⊆ C x D and A ∩ B ∈ ∅, Prove that A ⊆ C and B ⊆ D.

Answer - 7 : -

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}

∴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.

Question - 8 : -

Answer - 8 : -

Question - 9 : -

Answer - 9 : -

Question - 10 : -
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and
h(z) = sin z for all x, y, z ϵ N.
show tht ho (gof) = (hog) of.

Answer - 10 : -

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RD Chapter 2 Relations Ex 2.2 Solutions

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