Cartesian Product of Sets and its Properties

Cartesian Product of Sets:

If A and B be the two non-empty sets, then the Cartesian product of set A with set B is denoted by A x B and is defined as the set of ordered pairs (x, y) such that x ∈ A and y ∈ B i.e.

A x B = {(x, y) : x ∈ A and y ∈ B}

Properties of Cartesian Product of Sets:

n (A x B) = n (A) . n (B)
A x B is an empty set if either A or B is empty.
A x B ≠ B x A (i.e. the cartesian product is not commutative).
A x B = B x A if A = B.

(1) If A = {1, 2} and B = {a, b, c}. Find A x B?

Solution: A = {1, 2}
B = {a, b, c}

Now, A x B = {(1, a) (1, b) (1, c) (2, a) (2, b ) (2, c)}

(2) If A = {1, 4}, B = {2, 3}, C = {-1, 2}. Verify that A x (B ∪ C) = (A x B) ∪ (A x C).

Solution: B ∪ C = {2, 3} ∪ {-1, 2} = {-1, 2, 3}

Also, A x (B ∪ C) = {(1, -1) (1, 2) (1, 3) (4, -1) (4, 2) (4, 3)} ……….(i)

Now A x B = {(1, 2) (1, 3) (4, 2) (4, 3)}
and A x C = {(1, -1) (1, 2) (4, -1) (4, 2)}

(A x B) ∪ (A x C) = {(1, 2) (1, 3) (4, 2) (4, 3) (1, -1) (4, -1)} ……….(ii)

From (i) and (ii)-
A x (B ∪ C) = (A x B) ∪ (A x C)

(3) If (P, 5) and (2q, 1) belong to the set {(x, y) : x + y = 3}. Find P and q?

Solution: {(x, y) : x + y = 3} and (P, 5) (2q, 1) ∈ set.
∴ P + 5 = 3 and 2q + 1 = 3
⇒ P = -2 and q = 1

(4) If x ∈ A = {2, 4, 6} and y ∈ B = {3, 4, 5, 7, 8}. Find the set R of all ordered pair (x, y) such that x < y and x is a factor of y?

Solution: R₁ = {(x, y) : x < y}
={(2, 3) (2, 4) (2, 5) (2, 7) (2, 8) (4, 5) (4, 7) (4, 8) (6, 7) (6, 8)}

R₂ = {(x, y) : x is a factor of y}
= {(2, 4) (2, 8) (4, 4) (4, 8)}

Now R = R₁ ∩ R₂ = {(2,4) (2, 8) (4, 8)}

(5) For any three sets A, B and C. Prove that-

(i) A x (B ∪ C) = (A x B) ∪ (A x C).

Solution: Let (x, y) ∈ A x (B ∪ C)
⇒ x ∈ A and y ∈ (B ∪ C)
⇒ x ∈ A and {y ∈ B or y ∈ C}
⇒ {x ∈ A and y ∈ B} or {x ∈ A and y ∈ C}
⇒ (x, y) ∈ (A x B) or (x, y) ∈ (A x C)
⇒ (x, y) ∈ (A x B) ∪ (A x C)

∴ A x (B ∪ C) ⊆ (A x B) ∪ (A x C) ……….(i)
Similarly, (A x B) ∪ (A x C) ⊆ A x (B ∪ C) ……….(ii)

From (i) and (ii)-
A x (B ∪ C) = (A x B) ∪ (A x C)

(ii) A x (B ∩ C) = (A x B) ∩ (A x C).

Solution: Let (x, y) ∈ A x (B ∩ C)
⇒ x ∈ A or y ∈ (B ∩ C)
⇒ x ∈ A or {y ∈ B and y ∈ C}
⇒ {x ∈ A or y ∈ B} and {x ∈ A or y ∈ C}
⇒ (x, y) ∈ (A x B) and (x, y) ∈ (A x C)
⇒ (x, y) ∈ (A x B) ∩ (A x C)

∴ A x (B ∩ C) ⊆ (A x B) ∩ (A x C) ……….(i)
Similarly, (A x B) ∩ (A x C) ⊆ A x (B ∩ C) ……….(ii)

From (i) and (ii)-
A x (B ∩ C) = (A x B) ∩ (A x C)

(iii) A x (B – C) = (A x B) – (A x C).

Solution: Let (x, y) ∈ A x (B – C)
⇒ x ∈ A and y ∈ (B – C)
⇒ x ∈ A and {y ∈ B and y ∉ C}
⇒ {x ∈ A and y ∈ B} and {x ∈ A and y ∉ C}
⇒ (x, y) ∈ (A x B) and (x, y) ∉ (A x C)
⇒ (x, y) ∈ (A x B) – (A x C)

∴ A x (B – C) ⊆ (A x B) – (A x C) ……….(i)
Similarly, (A x B) – (A x C) ⊆ A x (B – C) ……….(ii)

From (i) and (ii)-
A x (B – C) = (A x B) – (A x C)

(6) If A and B are any two non-empty sets then A x B = B x A ⇔ A = B?

Solution: Suppose A x B = B x A, then we have to prove that A = B.

Let x ∈ A
⇒ (x, a) ∈ A x B
⇒ (x, a) ∈ B x A [∵ A x B = B x A]
⇒ x ∈ B

∴ A ⊆ B
Similarly, B ⊆ A
∴ A = B

Conversely, suppose A = B, then we have to prove that A x B = B x A

Now, A x B = A x A [∵ B = A] ……….(i)
Also, B x A = A x A [∵ B = A] ……….(ii)

From (i) and (ii)-
A x B = B x A

(7) If A subset B and C subset D then prove that A x C is a subset of B x D or A x C ⊆ B x D?

Solution: Let (x, y) ∈ A x C
⇒ x ∈ A and y ∈ C
⇒ x ∈ B and y ∈ D [∵ A ⊆ B and C ⊆ D]
⇒ (x, y) ∈ (B x D)

∴ A x C ⊆ B x D

(8) For any three sets A, B and C. Prove that A x (B’ ∪ C’)’ = (A x B) ∩ (A x C).

Solution: A x (B’ ∪ C’)’
= A x [(B’)’ ∩ (C’)’] [Using De Morgan’s Law]
= A x (B ∩ C)
= (A x B) ∩ (A x C)

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