Last updated at July 18, 2019 by Teachoo

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Ex 1.2, 1 Show that the function f: R* R* defined by f(x) = 1 x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*? Solving for f: R* R* f(x) = 1 x Checking one-one f (x1) = 1 x1 f (x2) = 1 x2 f (x1) = f (x2) 1 x1 = 1 x2 x2 = x1 x1 = x2 Hence, if f(x1) = f(x2) , x1 = x2 f is one-one Check onto f: R* R* f(x) = 1 x Let y = f(x) , such that y R* y = 1 x = 1 Since y not equal to 0, x is possible Thus we can say that if y R {0} , then x R {0} also Hence f is onto Show that the function f: R* R* defined by f(x) = 1 x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*? Now, domain R* is replaced by N , codomain remains R* Hence f : N R* f(x) = 1 x Checking one-one f (x1) = 1 x1 f (x2) = 1 x2 f (x1) = f (x2) 1 x1 = 1 x2 x2 = x1 x1 = x2 Hence, if f(x1) = f(x2) , x1 = x2 f f is one-one Check onto f: N R* f(x) = 1 x Let y = f(x) , , such that y R* y = 1 x = 1 Since y is real number except 0, x cannot always be a natural number Example: Let y = 2 x = 1 2 So, x is not a natural number Hence, f is not onto

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.