Consider a function f:R→Rf:R→R defined by f(x)=x2f(x)=x2. How does one show that ff is neither injective nor surjective?

Answer 1

To show that f is injective you must show that if f(a) = f(b), then a = b

So suppose that f(a)=f(b). Then a^2 = b^2, then a =b or a=-b. Hence f is NOT injective.

-9 is in R. Suppose f(c) = -9, then c^2 = -9. But there is no such c in R. That is -9 is not in the range of f. So f is not surjective.