= Solution
The non-Archimedean part of the <Ostrowski theorem> says that every nontrivial <Non-Archimedean absolute value> on $\mathbb Q$ is equivalent to the <p-adic absolute value> for a unique prime $p$.
Indeed, $|n|\leq1$ for every integer $n$. Nontriviality gives a prime $p$ with $|p|<1$, and there cannot be two such primes because the <Bezout identity> would make $1$ a sum of two terms of absolute value below one. If $a$ is coprime to $p$, another Bezout identity shows $|a|=1$. Consequently
$$
|x|=|p|^{v_p(x)}
$$
for every $x\in\mathbb Q^\times$, which is a positive real power of $|x|_p$.
More generally, the <Non-Archimedean absolute values on a number field> $K$ are indexed, up to equivalence, by the nonzero <prime ideals> $\mathfrak p\subset\mathcal O_K$. The value attached to $\mathfrak p$ is
$$
|x|_{\mathfrak p}=c^{-v_{\mathfrak p}(x)},
\qquad c>1.
$$
To prove completeness of the list, restrict an absolute value to $\mathbb Q$ and obtain a rational prime $p$. Its valuation ring contains $\mathcal O_K$ away from a unique prime above $p$; equivalently, its center
$$
\mathfrak p=\{a\in\mathcal O_K:|a|<1\}
$$
is a nonzero prime ideal. Since $\mathcal O_{K,\mathfrak p}$ is a <discrete valuation ring>, every $x\in K^\times$ is a unit times a power of a uniformizer, so the given value is equivalent to the displayed $\mathfrak p$-adic value. An absolute value trivial on $\mathbb Q$ is trivial on the algebraic extension $K$, so no further cases occur.
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