Q. 2 D3.9( 28 Votes )

# Check the injecti

Answer :

It is given that f : N → N given by f (x) = x^{3}

We can see that for x, y ϵ N,

f(x) = f(y)

⇒ x^{3} = y^{3}

⇒ 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^{3} = 2

⇒ f is not surjective.

Therefore, function f is injective but not surjective.

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