Answer :
Let f : N → N, be a function given by f(1) = f(2) = 1 and f(x) = x-1 for every x > 2.
Since, f(1) = 1 = f(2)
⇒ 1 and 2 does not have unique image.
Thus, f is not one-one.
Let f(x) = y
⇒ y = x-1
⇒ x = y+1
⇒ for each y ∈ R there exists x ∈ N such that f(x) = y.
Thus, f is onto.
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