Partitioning $\mathbb Q^+_0$ in 2-element sets I'm having trouble solving the following exercise:

Find a partitioning of the set $\{x\in \mathbb Q | x \geq 0\}$ in disjoint subsets, each containing two elements.

Now at first I thought about pairs $(a,b) := \{\{a\}, \{a,b\}\}$ whith whom you could identify $\frac{a}{b}$, but they are rather subsets of $\mathcal P(\mathbb N)$, right? Maybe there is an easy solution, but neither I nor my friends couldn't come up with one yet.
 A: Note that if $\frac pq$ is a rational number such that $p$ and $q$ are positive integers and coprime, then $\frac qp\neq\frac pq$ as long as $p\neq q$.
This will leave you with exactly two numbers in your set which cannot be represented like that.
Formally, as requested by the OP in the comments, this would be, $$\{0,1\}\cup\left\{\left\{\frac pq,\frac qp\right\}\mathrel{}\middle|\mathrel{}\begin{array}{l}p,q\in\Bbb N\setminus\{0\},\\ \gcd(p,q)=1,\\ p\neq q\end{array}\right\}$$
Of course, if you define $\Bbb N$ as starting with $1$, you don't have to add that $\setminus\{0\}$ part.
A: You know that the (nonnegative) rationals are countable, so let $f$ be your favorite bijection to the natural numbers. It's easy to partition the range as required:   use the pairs consisting of an even number and the next odd one.
This is less constructive and less elegant than @AsafKaragila 's nice trick - but it works for any countable set.
A: For $0\ne x\ne 1$ let $F(x)=\{x,1/x\} .$ Let $F(0)=F(1)=\{0,1\}.$ Let $G=\{F(x): 0\leq x \in \mathbb Q\}.$
