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Ex 1.2, 5 Show that the Signum Function f: R → R, given by f(x) = {█(1 for 𝑥 >0@ 0 for 𝑥=0@−1 for 𝑥<0)┤ is neither one-one nor onto. f(x) = {█(1 for 𝑥 >0@ 0 for 𝑥=0@−1 for 𝑥<0)┤ For example: f(0) = 0 f(-1) = −1 f(1) = 1 f(2) = 1 f(3) = 1 Since, different elements 1,2,3 have the same image 1 , ∴ f is not one-one. Check onto f: R → R f(x) = {█(1 for 𝑥 >0@ 0 for 𝑥=0@−1 for 𝑥<0)┤ Value of f(x) is defined only if x is 1, 0, –1 For other real numbers(eg: y = 2, y = 100) there is no corresponding element x Hence f is not onto Thus, f is neither one-one nor onto

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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