Is there a bijective function where $f: \mathbb Q \to \mathbb Q \setminus \{0\}$

Question

$f(x) = {1\over x}$ should be wrong, as the function isn't defined for $0$. Another could be: $f(x) = 2^x$, but is there anything else except functions of this type?
I was thinking of something with the injectivity to be provable by simply doing $f(x)=f(x')$, as I'm pretty much limited to doing it this way.
In case there's no such, how could I start on finding such a function?

Answer 1

Hint: $\mathbb Q$ is countable.

Answer 2

Use Robert's hint. Since both $\mathbb{Q}$ and $\mathbb{Q}\setminus\{0\}$ are both countably infinite, we may write $$ \begin{align} \mathbb{Q} &=\{a_n:n\in\mathbb{N}\} \\ \mathbb{Q\setminus\{0\}} &= \{b_n:n\in\mathbb{N}\} \end{align} $$ where $i\neq j$ implies $a_i\neq a_j$ and $b_i\neq b_j$. Now, the map $f:\mathbb{Q}\rightarrow\mathbb{Q}\setminus\{0\}$ given by $f(a_n)=b_n$ is invertible.

Answer 3

Recall that $\Bbb Q$ is countable therefore there is such a bijection. But there is actually more than just a bijection.

There is an order preserving bijection, that is, there is a bijection $f$ such that $f(x)<f(y)\iff x<y$. The reason is that every countable dense linear order without endpoints is isomorphic to the rational numbers. It is easy to check that $\Bbb Q\setminus\{0\}$ is such order (with the usual order restricted to exclude $0$, of course).

The proof is not very hard, but slightly technical, you can find the reasoning here, on Wikipedia.

Answer 4

Define a function $\varphi:\mathbb{Q}\to\mathbb{Q}\setminus\{0\}$ as follows. If $x$ is is a rational number which is not one of the numbers $0,1,2,3,\dots$, let $\varphi(x)=x$. It is left to you to decide on $\varphi(x)$ for the numbers where it has not yet been defined.

Remark: You may want to check into Hilbert's infinite hotel.