# CLASS-11RELATION & 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}

(iA 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)