Recall that a function $f\colon X\times X \to \mathbb{R}_{\ge 0}$ is a *metric* if it satisfies:

* definiteness: $f(x,y) = 0$ iff $x=y$,
* symmetry: $f(x,y)=f(y,x)$, and
* the triangle inequality: $f(x,y) \le f(x,z) + f(z,y)$.

A function $f\colon X\times X \to X$ is *associative* if it satisfies:

* associativity: $f(x,f(y,z)) = f(f(x,y),z)$.

If $X=\mathbb{R}_{\ge 0}$, then it might be possible for the same function to be a metric and associative.   
 **Is there an associative  metric on the non-negative reals?**

Note that these demands actually make $X$ into a group.  The element $0$ is the identity because $f(f(0,x),x) = f(0,f(x,x)) = f(0,0) = 0$ by associativity and definiteness, so again by definiteness $f(0,x) = x$.  Every element is its own inverse because $f(x,x)=0$.  

In fact, the following question is equivalent.  **Is there an abelian group on the non-negative reals such that the group operation satisfies the triangle inequality?**

Note also that the answer is *yes* if $X=\mathbb{N}$, the non-negative numbers!  [Click here for a spoiler](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).  
"Also known as nim-sum."

The question is originally due to [John H. Conway](http://en.wikipedia.org/wiki/John_Horton_Conway).  To my knowledge, the question is unsolved even for $X = \mathbb{Q}_{\ge 0}$, but he does not seem to care about that case.  The spoiler above does extend to the non-negative dyadic rationals $\mathbb{N}[\frac 12]$, but apparently not to $\mathbb{N}[\frac 13]$.