= Solution
Every <Non-Archimedean absolute value> on $\mathbb Q$ is either the <trivial absolute value>, or $|\cdot|_p^c$ for a unique prime $p$ and some $c>0$. Here $|p|_p=p^{-1}$.
To prove this classification, the ultrametric inequality gives $|n|\leq1$ for all integers $n$. If every nonzero integer has absolute value one, multiplicativity makes the <absolute value on a field> trivial on all rational numbers. Otherwise some positive integer has absolute value less than one, and its prime factorization gives a prime $p$ with $0<|p|<1$. There cannot be two such primes: if $p\ne q$ and $ap+bq=1$ with integers $a,b$, then $1\leq\max(|p|,|q|)<1$, a contradiction. Every other prime consequently has absolute value one. Unique prime factorization now gives
$$
|x|=|p|^{v_p(x)}=p^{-c v_p(x)},\qquad c=-\frac{\log|p|}{\log p}>0.
$$
Conversely these are <Non-Archimedean absolute values>, by the valuation inequality $v_p(x+y)\geq\min(v_p(x),v_p(y))$. \b[Thus $\boxed{|\cdot|=|\cdot|_p^c\ (c>0)\text{ or the trivial value}}$.] This is the non-Archimedean part of <Ostrowski theorem>, including the trivial case.
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