hw4_key - COT3100 Summer2001 Assignment#4(Solutions...

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COT3100 Summer’2001 Assignment #4. (Solutions). Assigned: 06/19, due: 07/03 on lecture. Total 100pts. 1. Which of the following are functions from the domain to the codomain given? Which functions are injective? Which functions are surjective? Find the inverse function for any bijective function. a) (5pts) f : Z N where f is defined by f ( x ) = x 2 +1. f is a function, since it returns a unique value for any x Z, but this function is not injective, because for instance f ( - 1)= f (1)=2. Neither it is surjective, because there is no- x Z such that f ( x ) = 0, 0 N. b) (5pts) g : N Q where g is defined by g ( x ) = 1/ x g is not a function, because there is no unique value assigned to x = 0, 0 N. c) (5pts) h : Z × N Q where h is defined by h ( z , n ) = z /( n +1) h is a function, since for any pair ( z , n ) Z × N, the given formula defines a unique value from Q. It is surjective and it is injective. d) (5pts) f : {1, 2, 3} { p , q , r } where f = {(1, q ), (2, r ), (3, p )} f defines a function, the function is injective and surjective. e) (5pts) g : N N where g is defined by g ( x ) = 2 x g is a function, it is injective, but not surjective Here N is the set of nonnegative integers, N={0, 1, 2, 3, …}, Z - the set of all integers, Z={…-3, -2, -1, 0, 1, 2, 3, …}, Q - the set of rational numbers, Q={ x / y | x , y Z and y 0 } 2 Let A = R - {2}, and let f be the function with domain A defined by the formula: 2 3 ) ( - = x x x f a) (6pts) Show that f is a bijective function from A to B for some set B R.
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