Verify why this is not a metric $d(x,y)=\|x-y\|_p$ ( $\|x\|_p=p^{-h}$ if $x=p^h\dfrac{m}{n}$).

Question

$d(x,y)=\|x-y\|_p$

$p$ is prime and $\|x\|_p=p^{-h}$ if $x=p^h\dfrac{m}{n}$, where $m, n$ are coprimes with $p$.

This is not a metric because if $x=y=p^k\dfrac{m}{n}$, then $x-y=0=p^0\dfrac0n$.

Hence $d(x,y)=p^{-0}=1 \ne0$.

Is this correct?

Answer 1

You didn't specify what $m,n$ are in the definition of $\|.\|_p$. Supposedly, they integers that are not divisible by $p$. However, $0$ is divisible by $p$, hence you cannot compute $\|0\|_p$ this way. Indeed, one needs to define explicitly that $\|0\|_p=0$; with this addon, $d$ is a metric.