What is an example of function $f: \Bbb{N} \to \Bbb{Z}$ that is a bijection? Could you give me an example of function $ f \colon \mathbb N \to \mathbb Z$ that is both one-to-one and onto? Does this work: $f(n) := n \times (-1)^n$?
N starts with zero.
 A: $$f\colon \Bbb N \to \Bbb Z,
\qquad
f(n)=
\begin{cases}
\quad\dfrac n2\qquad \text{if $n$ is even}\\
-\dfrac {n+1}{2}\quad \text{if $n$ is odd}
\end{cases}
$$
A: Here are lots of answers . . . well not really, they are actually all in effect the same.  In each of them we take $f(0)=0$ and I give instructions for the rest.


*

*$f(n)=k$ if $n$ is the $k$th prime, $f(n)=-k$ if $n$ is the $k$th non-prime.

*$f(n)=k$ if $n$ is the $k$th (non-zero) square, $f(n)=-k$ if $n$ is the $k$th non-square.

*$f(n)=k$ if $n$ is the $k$th power of $2$, $f(n)=-k$ if $n$ is the $k$th non-power of $2$.


. . . and so on . . . not forgetting


*$f(n)=k$ if $n$ is the $k$th even number, $f(n)=-k$ if $n$ is the $k$th odd number,


which is actually the example already given by many people.
A: Here is a slight modification of OP's suggestion: The function $f(n):=\sum_{k=0}^n (-1)^k k$ is a bijection $f:\mathbb{N}_0\to \mathbb{Z}$.  
A: $$f(n)=\frac{1-(2n+1)\cos n\pi}4$$
A: $$f(n)=(-1)^n \left \lceil \frac{n}{2} \right \rceil$$
where $ \lceil x \rceil $ rounds up a real value (see ceiling function).
A: $$f(n) = \left\{\begin{array}{cc}  
\frac{n}{2} & n \textrm{ is even,}\\  
\frac{-(n+1)}{2} & n \textrm{ is odd.}
\end{array}  
\right.$$
Hence, $$f(0) = 0,$$ $$f(1) = \frac{-(1+1)}{2} = -1,$$ $$f(2) = \frac{2}{2} = 1,$$ $$f(3) = \frac{-(3+1)}{2} = -2 \cdots$$
A: First note that $\Bbb{Z}$ contains all negative and positive integers. As such, we can  think of $\Bbb{Z}$ as (more or less) two pieces. Next, we know that every natural number is either odd or even (or zero for some people) so again we can think of $\Bbb{N}$ as being in two pieces. lastly, let's try to make a map that takes advantage of the "two pieces" observation . That is, let's make a function from evens/odds to positives/negatives. Let $f: \Bbb{N} \to \Bbb{Z}$ where 
$$f(n) = \begin{cases}
    \frac{n}{2} & n\text{ is even} \\
    -\frac{n + 1}{2} & \text{else}
\end{cases}$$
This map is a bijection, although I will leave the proof of that up to you.
A: $$f(n) = n\text{th number in the sequence }0, 1, -1, 2, -2, \ldots$$
