Which functions are 1-1 and which are onto? Let $N$ denote the naturals, $Z$ the integers, $R$ the reals. Which of the following is 1-1? Onto?

$(a) f(x) = x^2$ on $N, Z, R$

Suppose $f(x_1), f(x_2) \in N$. Since $x_1, x_2 \in N : x_1 \neq x_2 \land f(x_1) \neq f(x_2)$, $f(x)$ is 1-1 on the domain $N$. However, since $\exists y [\forall x \in N : f(x) \neq y]$, e.g. $3$, $f(x)$ is not onto.
Suppose $f(x_1), f(x_2) \in Z$. Since $\exists x_1, x_2 \in Z : x_1 \neq x_2 \land f(x_1) = f(x_2)$ (namely, $[-2, 2]$), $f(x)$ is not 1-1 on the domain $Z$. Similarly, $f(x)$ is not onto.
Suppose $f(x_1), f(x_2) \in R$. Since $\exists x_1, x_2 \in R : x_1 \neq x_2 \land f(x_1) \neq f(x_2)$ (namely, $[-2, 2]$), $f(x)$ is not 1-1 on the domain $R$. Similarly, $R$ is not onto (there is no value mapping to some $a < 0 \in R$.
My question is, did I prove it adequately in the sense that I proved it by contradiction rather than using a constructive proof? I'd rather use a constructive proof and show it generally.

$(b) f(x) = x^3$ on $N, Z, R$

With similar reasoning, it can be shown that $f(x)$ is 1-1 on $N, Z, R$ and onto on $R$.
 A: First, a correction. You write e.g. 
$$x_1, x_2 \in N : x_1 \neq x_2 \land f(x_1) \neq f(x_2)$$
which is obviously false (suppose $x_1 = x_2 = 2)$. You meant, of course,
$$x_1, x_2 \in N : x_1 \neq x_2 \to f(x_1) \neq f(x_2)$$
and similarly in the other cases.
As to the query: your arguments are in fact fine by constructive standards. You aren't using the constructively unacceptable inference "$\neg C$ implies a contradiction, hence $C$". (That so-called "Classical Reductio" inference is, in the presence of other uncontentious assumptions, equivalent to the non-constructive rule "from $\neg\neg C$ infer $C$".)
Rather, you are e.g. disproving the generalized conditional (C) $\forall x\forall y(\varphi(x, y)\to \psi(x, y))$, e.g. the conditional that defines being "onto", by finding $a, b$ such that $\varphi(a, b)$ and $\neg\psi(a, b)$. That involves the inference "$C$ [plus side assumptions] implies a contradiction, hence $\neg C$". But that form of reductio -- often just called "negation introduction" -- is constructively acceptable.
