Need to check if this function is bijective

I don't understand how $f : \mathbb N \to\mathbb N$ (where $0$ isn't included in the natural numbers set), $f(n) = n^2$ is not bijective. It seems both injective and surjective to me?

Thanks got it!

• For which $n$ is $f(n) = 3$? Apr 15 '14 at 23:20

It is surely not a surjection onto $\mathbb N$. For example, there is no n such that f(n) = 2.
It seems as though your notion of surjective is off. A function $f : X \rightarrow Y$ is surjective if and only if $\forall y \in Y$, $\exists x \in X$ such that $f(x) = y$. That is to say the set being mapped to gets completely "covered" by the map. In your case $\mathbb{N}$ does not get "covered" by the map.
We can only find $n$ such that $f(n)$ is a square number. Can we find an $n\in\mathbb{N}$ such that $f(n)$ is not a square number? Of course not! What does this imply about surjectivity? What does the definition of a bijection (injection+surjection) now tell us?