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.

The question is originally due to John H. 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 $\mathbb{N}[\frac 13]$.