Let N be the set of natural numbers and two functions f and g be defined as f , g : N → N such that f n = n + 1 2 , i f n i s o d d n 2 , i f n i s e v e n and g n = n - - 1 n . Then f o g is:

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We have, g n = n + 1 , i f n o d d n - 1 , i f n e v e n f g 1 = f ( 2 ) = 1 f g 2 = f ( 1 ) = 1 ∴ f g x is many one. f g 2 k = f ( 2 k - 1 ) = k f g 2 k + 1 = f 2 k + 2 = k + 1 ∴ f ( g ( x ) ) is onto.

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