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Ex 1.2, 2 Check the injectivity and surjectivity of the following functions: (ii) f: Z → Z given by f(x) = x2 f(x) = x2 Checking one-one (injective) f (x1) = (x1)2 f (x2) = (x2)2 Putting f (x1) = f (x2) (x1)2 = (x2)2 ∴ x1 = x2 or x1 = –x2 Since x1 does not have unique image, It is not one-one (not injective) Example f(–1) = (–1)2 = 1 f(1) = (1)2 = 1 Here, f(–1) = f(1) , but –1 ≠ 1 Hence, it is not one-one Check onto (surjective) f(x) = x2 Let f(x) = y , such that y ∈ Z x2 = y x = ±√𝒚 Note that y is an integer, it can be negative also Putting y = −3 x = ±√((−𝟑)) Which is not possible as root of negative number is not an integer Hence, x is not an integer So, f is not onto (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 and Computer Science at Teachoo