Split the set of real numbers into $2$-element sets How can we split $\mathbb{R}$ into disjoint sets, each consisting of $2$ elements?
I have found a similar (though much more general) question here.
But I am unable to deduce an answer to my specific question.

When taking the naive approach of $\{(x,-x)|x\in\mathbb{R^+}\}$ we exclude $0$.
Intuitively, I have concluded that $\mathbb{R}$ has an "odd" number of elements.
This leads me to think that splitting $\mathbb{R}$ into $2$-element sets is not feasible.
I do realize, however, that we cannot refer to an infinite number as odd or even.
 A: First consider the pairs $(2n,2n+1)$ for $n\in\Bbb Z$. Then consider the pairs $(-x,x)$ for $x\notin\Bbb Z$.
A: How about:
$\{(x,a-x)\mid x\in\mathbb{R^+}, a=1$ if $x\in\mathbb{Z}$ else $a=0\}$
A: Here's a solution that doesn't treat the integers as special:
Use sets of the form $\{x,x+1 \mid x\in [2n,2n+1), n\in\mathbb{Z}\}$.
The idea is to partition the real line by means of half open unit intervals, and pair corresponding elements of neighboring intervals.  So for instance elements of $[-2,-1)$ pair with elements of $[-1,0)$; elements of $[0,1)$ pair with elements of $[1,2)$, etc.
A: For negative numbers, just use the pairs $(x,\frac{1}{x})$. For nonnegative numbers, proceed as follows. To make binary representation of real numbers unique, always replace an infinite tail like $0.111\ldots$ by $1.0$, etc. Then pair up numbers that differ only in the first decimal place with respect to this unique representation.  Thus, $0.0$ will pair up with $0.1$, while $1.1011\ldots$ will pair up with $1.0011\ldots$, etc.
