I'm reading the proof of Ostrowski's theorem in Gouvea's book on p-adic numbers and there is one step that I don't understand. Let $|\cdot|$ be a non-archimedean absolute value and $n=rp+s$ where $r,p,s$ are integers with $|rp|< 1$ and $|s|=1$. Then we conclude that $|n|=1$. I know that $|n|\leq 1$ because the absolute value is non-archimedean but I don't see why the other two values imply that it is not less than $1$.


$|n|\leq \operatorname{max}(|rp|,|s|)$ is $|n|\leq 1$ but we also have $s=n-rp$ which gives $|s|\leq \operatorname{max}(|n|,|rp|)$. Keeping in mind that $|rp|\lt 1$ we must have $|n|=|s|=1$. We could have seen that "geometrically". Every triangle is isosceles and the side $rp$ is smaller than $n$ so $|n|=|s|=1$

  • $\begingroup$ This is not very hard to see, I don't understand why I didn't got that myself ;) $\endgroup$ – Sebastian Bechtel May 2 '15 at 16:46
  • $\begingroup$ Non Archimedean distances are counterintuitive $\endgroup$ – marwalix May 2 '15 at 16:47
  • $\begingroup$ Indeed! But for the non-geometric argument this isn't really a point I would say, it's just about using the right inequalities. $\endgroup$ – Sebastian Bechtel May 2 '15 at 16:49
  • $\begingroup$ Whenever you have three points look for the equal sides in non Archimedean metric space. Even when you're not asked $\endgroup$ – marwalix May 2 '15 at 16:51

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