Metrics on Q, Ostrowski's theorem Can you help med find an example of a metric on $\mathbb Q$ and $x,y\in \mathbb Q$ such that the product $d(x,0)\cdot d(y,0)\neq d(xy,0)$?
Reason: I am looking into Ostrowski's theorem (which appears to be given for valuations), and I was wondering whether the result could be stated for metrics as well. Clearly, a valuation gives rise to a metric, but does every metric induce a valuation?
Thank you
 A: What about $d(x,y)=2\lvert x-y\rvert$?
A: $$d(x,y)=\|x-y\|, \qquad \|x\| = \sup_{v\in S} |x|_v$$ where $S$ is any set of (inequivalent) absolute values: $|.|_\infty$ is the euclidean absolute value and $|.|_p$ is the $p$-adic absolute value and $|0|_{tr}=0,|x|_{tr}=1$ is the trivial absolute value.
A: As you mentioned, in the case of concepts relating to Ostrowski theorem, we deal with valuations or equivalently, norms on fields. Specially, we define a norm on a field $K$ to be a function $\|.\|:K\to[0,\infty)$ satisfying


*

*$\|0\|=0$

*(Triangle inequality): for any $x,y\in K$: $\|x+y\|\leq \|x\|+\|y\|$ (and strong triangle inequality $\|x+y\|\leq \max(\|x\|,\|y\|)$ in the case of non-Archimedean norms)

*$\|xy\|\leq \|x\| \|y\|$, for all $x,y\in K$ and

*$\|1\|\leq 1$
where from 3 and 4 we find out $\|1\|=1$.
The condition $d(x,0)d(y,0)\ne d(xy,0)$ therefore, becomes $\|x\|\cdot\|y\|\ne \|xy\|$ which by above definition, is equivalent to $$\|xy\|<\|x\|\cdot \|y\|\tag{*}$$
i.e. you look for norms which satisfy $(*)$.
You may note that Ostrowki theorem is about special norms called "absolute values"
A: If you really just want a metric on $\mathbb Q$ with the property that $d(x,0)\cdot d(y,0) \neq d(xy,0)$ for some $x,y$, then there are plenty of options; because you almost view $\mathbb Q$ just as a countable set. In particular, take any bijection $f: \mathbb Q \rightarrow S$ to your favourite other countable set which comes with a metric $(S, d_S)$, and then take $d(x,y) := d_S(f(x),f(y))$. Like
$f:\mathbb Q \rightarrow \mathbb N$
or $f:\mathbb Q \rightarrow \mathbb Z$
or $f:\mathbb Q \rightarrow \mathbb N^{17}$
or $f:\mathbb Q \rightarrow \mathbb Q(\sqrt{-35}, \sqrt[71]{12})$
or $f:\mathbb Q \rightarrow \lbrace x \in5\mathbb N: x \ge 11000 \rbrace$
or $f:\mathbb Q \rightarrow \mathbb Z[1/3] \cap (0,1)$
or even $f:\mathbb Q \rightarrow \mathbb Q$, $f(x) := \dfrac{-19}{13}x-7.829964573$,
where in each case you just take the restriction of the Euclidean metric (embed the fourth one into $\mathbb C$, say) of the set on the right. Now it's highly unlikely that for any $x,y$ you have $d(x,0)d(y,0)=d(xy,0)$. Actually, this is guaranteed to fail for $x=y=1$ for the last three, and for the others, just choose them so $f(0)$ and $f(1)$ do not have distance $1$.
Note that the other answers presuppose that your metric still has plenty to do with the arithmetic of $\mathbb Q$, which is of course reasonable, because as you should see here, otherwise all hell breaks loose.
