LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

RELATION & FUNCTION - PROBLEM & SOLUTION OF CARTESIAN PRODUCT

__CARTESIAN PRODUCT- PROBLEM & SOLUTION -__

**Example.1) Let A = {x ∈ N : x ≤ 5} and B = {y ∈ N : 2 < y ≤ 4}. Find out the followings –**

**(i) A X A (ii)
A X B (iii) B X A (iv) B X B**

**(v) n (A X A) (vi) n (A X B) (vii) n (B
X A) (viii) n (B X B)**

**Ans.) Here A = {1,
2, 3, 4, 5}, B = {3, 4}**

**(i) A X A = {(1, 1),
(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1),
(3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1),
(5, 2), (5, 3), (5, 4), (5, 5)} (Ans.)**

**(ii) A X B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4),
(4, 3), (4, 4), (5, 3), (5, 4)} (Ans.)**

**(iii) B X A = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4,
1), (4, 2), (4, 3), (4, 4), (4, 5)} (Ans.)**

**(iv) B X B = {(3, 3), (3, 4), (4, 3), (4, 4)} (Ans.)**

**(v) n (A X A) = 5 X 5 = 25**

**(Here it has been said that, the set A has ‘n’ = 5 elements
and the set B has ‘m’ = 2 elements, then the product set A X A has ‘n’ ‘n’
= 5 X 5 = 25 elements) (Ans.)**

**(vi) n (A X B) = 5 X 2 = 10**

**(Here it has been said that, the set A has ‘n’ = 5 elements and
the set B has ‘m’ = 2 elements, then the product set A X B has ‘n’ ‘m’ = 5 X 2 = 10 elements) (Ans.)**

**(vii) n (B X A) = 2 X
5 = 10**

**(Here it has been said that, the set A has ‘n’ = 5 elements
and the set B has ‘m’ = 2 elements, then the product set B X A has ‘m’ ‘n’
= 2 X 5 = 10 elements) (Ans.)**

**(viii) n (B X B) = 2 X
2 = 4**

**(Here it has been said that, the set A has ‘n’ = 5 elements
and the set B has ‘m’ = 2 elements, then the product set B X B has ‘m’ ‘m’
= 2 X 2 = 4 elements) (Ans.)**

**Example.2) If A = {1,
2, 3}, B = {4, 5}, C = {1, 2, 3, 4, 5} find (i) A X B, (ii) C X B, (iii) B X B,
hence prove that (C X B) – (A X B) = B X B**

**Ans.) Given, A = {1, 2, 3}, B = {4, 5}, C = {1, 2, 3, 4, 5}**

**(i) A X B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}**

**(ii) C X B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5),
(4, 4), (4, 5), (5, 4), (5, 5)}**

**(iii) B X B = {(4, 4), (4, 5), (5, 4), (5, 5)}**

**Now, (C X B) – (A X B) = {(1, 4), (1, 5), (2, 4), (2, 5), (3,
4), (3, 5), (4, 4), (4, 5), (5, 4), (5, 5)} - {(1, 4), (1, 5), (2, 4), (2, 5),
(3, 4), (3, 5)}**

** = {(4, 4), (4, 5), (5, 4), (5, 5)} = B X B (Proved)**

**Example.3) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6}, and
D = {5, 6, 7, 8}, verify that –**

**(i) A X (B ∩ C) = (A X B) ∩ (A X C)**

**(ii) (A X C) is a subset of (B X D)**

**Ans.) we have, A = {1, 2}, B = {1, 2, 3,
4}, C = {5, 6}, and D = {5, 6, 7, 8}**

**(i) (B ∩ C) = {1, 2, 3, 4} ∩ {5, 6} = φ**

**So, A X (B ∩ C) = {1, 2} X φ = φ …………………….(1)**

**Now, (A X B) = {1, 2} X {1, 2, 3, 4} =
{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}**

**And, (A X C) = {1, 2} X {5, 6} = {(1,
5), (1, 6), (2, 5), (2, 6)}**

**So, (A X B) ∩ (A X C) = {(1, 1), (1, 2), (1, 3), (1,
4), (2, 1), (2, 2), (2, 3), (2, 4)} ∩ {(1, 5), (1, 6), (2,
5), (2, 6)}**

** = φ …………………………..(2)**

**Hence, from (1) & (2), we can
conclude that A X (B ∩ C) = (A X B) ∩ (A X C) **

**(ii) (A X C) = {1, 2} X {5, 6} = {(1,
5), (1, 6), (2, 5), (2, 6)} ………………….(1)**

**And, (B X D) = {1, 2, 3, 4} X {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)} …………………….(2)**

**If I would like find the common element
between (A X C) & (B X D), then (A X C) ∩ (B X D) = {(1, 5), (1, 6), (2, 5), (2, 6)}, hence from (1) & (2) it
is clear that all the element of (A X C) are contained in (B X D). Hence, (A X
C) is subset of (B X D). (Ans.)**

**Example.4) Is A X φ
the empty set or not, where φ denotes the empty set and A is any set ?**

**Ans.)
φ is the empty set. It contains no element. Therefore, no ordered pair
is possible in A X φ. Hence, it is clear that A X φ contains no elements, i.e.,
A X φ is the empty set. (Ans.)**

**Example.5) A ⊆ B
and C ⊆ D,
prove that A X C ⊆ B X D**

**Ans.) Let (a, b) be any arbitrary
element of A X C. …………………..(1)**

**Then, (a, b) ∈ A X C **

**=> a ∈ A and b ∈ D [since A ⊆ B and C ⊆ D]**

**=> (a, b) ∈ B X D ……………………………..(2)**

**From (1) and (2) it follows that A X C ⊆ B X D (Proved)**