# Proof that a piecewise function is invertible.

Prove that the following function is invertible: $$g:\mathbb{Z}\rightarrow\mathbb{N}$$

$$g(x) = \left\{ \begin{array}{lr} -2x & : x\le0\\ 2x-1 & : x>0 \end{array} \right.$$

I'm having an issue proving that it is one-to-one in all cases. The cases where $x_{1},x_{2}\le0$ and $x_{1},x_{2}>0$ both pass proof by contrapostitive where $g(x_{1})=g(x_{2})$. I'm having a problem proving it is also one-to one if $x_{1}\le0$ but $x_{2}>0$ which would result in:

$$-2x_{1}=2x_{2}-1$$ $$x_{1}=-x_{2}+\frac{1}{2}$$

I know I'm going something wrong in my thinking and I have a sneaking suspicion it has something to do with the fact that I'm mapping to $\mathbb{N}$, but I'm completely unsure of where to go from here.

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Are you sure that $g:\mathbb{R} \rightarrow \mathbb{N}$? Because if $x=\sqrt{2}$, then surely $g(x)$ is not a natural number. – user39280 Oct 20 '12 at 18:19
Sorry, it's actually $\mathbb{Z}\rightarrow\mathbb{N}$. Let me edit my question. – kamikazekent Oct 20 '12 at 18:39

$-2x$ will always give you a nonnegative, even integer, when $x \leq 0$. And, $-2x$ is strictly decreasing. $2x - 1$ will always give a positive, odd integer, when $x > 0$. And, $2x - 1$ is strictly increasing. That's all you need.
Thanks, I think that makes sense. Normally I do have to prove that $g(x_{1})=g(x_{2})$, but what you're saying is in the third case I can apply a direct proof showing that it is impossible for an element to map to multiple elements in the co-domain since $x_{1}$ would always be even and $x_{2}$ would always be odd. Am I understanding that correctly? – kamikazekent Oct 20 '12 at 20:29
@kamikazekent Correct. Injective means that $g(x_1) = g(x_2)$ implies $x_1 = x_2$. Or, equivalently, the contrapositive is $x_1 \neq x_2$ implies $g(x_1) \neq g(x_2)$. So, pick two $x$ values that are both $x \leq 0$. The fact that the function is strictly decreasing implies $g(x_1) \neq g(x_2)$. Similarly if your two $x$ values are both $> 0$. And, if one is nonpositive and one is positive, then $g(x_1)$ is even and $g(x_2)$ is odd, so they're not equal then also. – Graphth Oct 20 '12 at 20:34