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if n(A) = p,
n(B) = q,

n(AxB) = pq

then 'HOW IS' the no. of relations = 2^pq

pls prove it..

$T a k e a n e x a p m l e, L e t A = {a, b} \Rightarrow n (A) = 2 A n d, B = {c, d} \Rightarrow n (B) = 2 T h e n, A \times B = {(a, c), (a, d), (b, d), (b, c)} S u b s e t s o f A \times B = {(a, c)}, {(a, d)}, {(b, d)}, {(b, c)}, {(a, c), (a, d)}, {(a, d), (b, d)}, {(b, d), (b, c)}, {(a, c), (b, d)}, {(a, c), (b, c)}, {(a, d), (b, c)} {(a, c), (a, d), (b, d)}, {(a, d), (b, d), (b, c)}, {(b, d), (b, c), (a, c)}, {(b, c), (a, c), (a, d)}, {(a, c), (a, d), (b, d), (b, c)}, φ \Rightarrow n u m b e r o f s u b s e t s A \times B = 16 \Rightarrow n u m b e r o f s u b s e t s A \times B = 2^{2 \times 2} S i n c e e a c h s u b s e t s o f A \times B d e f i n e s a r e l a t i o n, t h e r e f o r e, N u m b e r o f r e l a t i o n s = 2^{2 \times 2}$

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