Q2 of 107 Page 10

Check the injectivity and surjectivity of the following functions:

f : N → N given by f (x) = x3

It is given that f : N → N given by f (x) = x3

We can see that for x, y ϵ N,


f(x) = f(y)


⇒ x3 = y3


⇒ x = y


⇒ f is injective.


Now, let 2 ϵ N. But, we can see that there does not exists any x in N such that


f(x) = x3 = 2


⇒ f is not surjective.


Therefore, function f is injective but not surjective.


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