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)