Consider $f : \mathbb{N} → \mathbb{Z}$ defined as $f (n) = \frac{(−1)^n (2n−1)+1}{4}$. Find its inverse. I cannot find an inverse of this function for $f(n) = x$, where $x$ is an integer, that gives out a natural number. Some guidance would be very helpful... I already know the function is bijective so there has to be an inverse. 
Edited to add that I tried letting $n = 2x$ but since $x$ is an integer, there's still the possibility of having an input like $-2$ that gives a negative number, which is clearly not a natural number.
 A: The function does this:
$$1\to 0$$
$$2\to 1$$
$$3\to -1$$
$$4\to 2$$
$$5\to -2$$
$$\vdots$$
and so on.
The inverse would be $g\colon\mathbb{Z}\to\mathbb{N}$ defined by
$$g(n) = |2n|+\boldsymbol{1}_{\{n\leq 0\}},$$
where $\boldsymbol{1}_{\{n\leq 0\}}$ is the characteristic function on the event $\{n\leq 0\}$, i.e. it is $1$, when $n$ is non-positive, and is otherwise $0$.
You need to check that this is indeed the inverse.
A: \begin{align*}
f(n) &= \frac{(−1)^n (2n−1)+1}{4} \\
4f(n) &= (−1)^n (2n−1)+1  \\
4f(n)-1 &= (−1)^n (2n−1)  \\
(-1)^n(4f(n)-1) &= (2n−1)  \\
1+(-1)^n(4f(n)-1) &= 2n  \\
n &= \frac{1}{2}+\frac{(-1)^n}{2}(4f(n)-1) \\
n&= 1-2f(n) &\text{for n odd} \\
n&= 2f(n) &\text{for n even} \\
\end{align*}
Which is the inverse from $\mathbb{Z} \to \mathbb{N}$
For example, if $f(n) = 1$, you get $n=-1$ (odd) or $n=2$ (even).  Since the range of the inverse is $\mathbb{N}$, $n$ has to be 2.
This inverse can be compacted into a single expression, as has been done by others.
A: $$f^{-1}(z)=\begin{cases} 2|z| + 1 &\quad& z\leq 0\\ 2|z| &\quad& z>0 \end{cases}$$
We see that for $n$ odd, $f(n)=\frac{1-n}{2}$, so the inverse is as claimed.
Then, for $n$ even, $f(n)=\frac{n}{2}$, and the inverse is also as claimed.
