Q. 2 D3.8( 30 Votes )
Check the injecti
Answer :
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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