Ex 1.2 , 2 Check the injectivity and surjectivity of the following functions: (v) f: Z → Z given by f(x) = x3 f(x) = x3 Checking one-one (injective) f (x1) = (x1)3 f (x2) = (x2)3 Now, f (x1) = f (x2) (x1)3 = (x2)3 x1 = x2 Since if f (x1) = f (x2) , then x1 = x2 ∴ It is one-one (injective) Check onto (surjective) f(x) = x3 Let f(x) = y , such that y ∈ Z x3 = y x = 𝒚^(𝟏/𝟑) Here y is an integer i.e. y ∈ Z Let y = 2 x = 𝑦^(1/3) = 𝟐^(𝟏/𝟑) So, x is not an integer ∴ f is not onto (not surjective) Hence, function f is injective but not surjective.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.