Question
f: Z → Z given by f(x) = x3
f(x) = x3
It is seen that for $\text{x},\text{y}\in\text{Z},$ f(x) = f(y) ⇒ x3 = y3 ⇒ x = y. $\therefore$ f is injective. Now, $2\in\text{N}.$ But, there does not exist any element x in domain Z such that f(x) = x3 = 2. $\therefore$ f is not surjective. Hence, function f injective but not surjective.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.