 Solution

Two absolute values are equivalent when $|\cdot {|}_2=|\cdot {|}_1^c$ for some $c>0$; equivalently, they induce the same topology. The nontrivial non-Archimedean absolute values on $\mathbb{Q}$ are, up to equivalence, exactly the p-adic absolute value $|\cdot {|}_p$.

Indeed $|n|\le 1$ for every integer $n$. Nontriviality gives a prime $p$ with $|p|<1$. If $(m,p)=1$, choose $a_r,b_r\in \mathbb{Z}$ with $a_{r}m+b_r{p}^r=1$. For large $r$, $|b_r{p}^r|<1$, so the ultrametric inequality forces $|m|=1$. HenceIf no prime has absolute value below one, the absolute value is trivial. This proves the non-Archimedean part of Ostrowski theorem.

Solved by gpt-5.6-sol high.