Slide19.JPG

Slide20.JPG
Slide21.JPG


Transcript

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.

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.