Q. 2 A3.9( 79 Votes )

# Check the injecti

Answer :

It is given that f : N → N given by f (x) = x^{2}

We can see that for x, y ϵ N,

f(x) = f(y)

⇒ x^{2} = y^{2}

⇒ 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) = x^{2} = 2

⇒ f is not surjective.

Therefore, function f is injective but not surjective.

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