Ostrowski's theorem for the rationals

Assume as the first case that $\mathrm{\mid }n\mathrm{\mid }\le 1$ for every integer $n$. Then [L1] makes $\mathrm{\mid }\cdot \mathrm{\mid }$ nonarchimedean. Because the absolute value is nontrivial, there is an integer $m$ with $\mathrm{\mid }m\mathrm{\mid }\mathrm{\ne }1$; the inequality $\mathrm{\mid }m\mathrm{\mid }\le 1$ forces $\mathrm{\mid }m\mathrm{\mid }<1$, so some prime divisor $p$ of $m$ has $\mathrm{\mid }p\mathrm{\mid }<1$ by [L4]. If also $\mathrm{\mid }q\mathrm{\mid }<1$ for a different prime $q$, then [L3] gives $up+vq=1$, and the nonarchimedean inequality yields $1=\mathrm{\mid }1\mathrm{\mid }\le \max \{\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }q\mathrm{\mid }\}<1$, impossible because integers all have absolute value at most $1$. Thus there is a unique prime $p$ with $\mathrm{\mid }p\mathrm{\mid }<1$.

Assume as the second case that $\mathrm{\mid }n\mathrm{\mid }>1$ for some integer $n$. Choosing a prime divisor of $n$ and using [L4], at least one prime $p$ satisfies $\mathrm{\mid }p\mathrm{\mid }>1$. Put $\alpha :=\log \mathrm{\mid }p\mathrm{\mid }\mathrm{/}\log p>0$ and, for each integer $t\in \{0,\dots ,p-1\}$, let $C_p:=\max \mathrm{\mid }t\mathrm{\mid }$. For every positive integer $m$, write the base-$p$ expansion $m=a_0+a_{1}p+\cdots +a_r{p}^r$ with $0\le a_i<p$. The triangle inequality gives $\mathrm{\mid }m\mathrm{\mid }\le {\sum }_{i=0}^r\mathrm{\mid }{a}_i\mathrm{\mid }\mathrm{\mid }p{\mathrm{\mid }}^i\le C_p(r+1)\mathrm{\mid }p{\mathrm{\mid }}^r\le C_p(1+{\log }_{p}m)m^{\alpha },$ because $p^r\le m<p^{r+1}$. Applying the same estimate to $m^k$ and taking $k$-th roots gives $\mathrm{\mid }m\mathrm{\mid }\le (C_p(1+k{\log }_{p}m){)}^{1\mathrm{/}k}{m}^{\alpha },$ so letting $k\to \mathrm{\infty }$ yields $\mathrm{\mid }m\mathrm{\mid }\le m^{\alpha }$. Now fix $m>1$ and put ${\alpha }_m:=\log \mathrm{\mid }m\mathrm{\mid }\mathrm{/}\log m$. Writing $p^k$ in base $m$ and repeating the same argument with $m$ in place of $p$ gives $\mathrm{\mid }p{\mathrm{\mid }}^k\le C_m(1+k{\log }_{m}p)p^{k{\alpha }_m}$ for some constant $C_m$, hence $\mathrm{\mid }p\mathrm{\mid }\le p^{{\alpha }_m}$ after taking $k$-th roots and letting $k\to \mathrm{\infty }$. Therefore $\alpha \le {\alpha }_m$. Since $\mathrm{\mid }m\mathrm{\mid }\le m^{\alpha }$ is exactly ${\alpha }_m\le \alpha$, we get ${\alpha }_m=\alpha$ for every $m>1$. Thus $\mathrm{\mid }m\mathrm{\mid }=m^{\alpha }$ for every positive integer $m$, and then $\mathrm{\mid }a\mathrm{/}b\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }\mathrm{/}\mathrm{\mid }b\mathrm{\mid }=\mathrm{\mid }a\mathrm{/}b{\mathrm{\mid }}_{\mathrm{\infty }}^{\alpha }$ for every nonzero rational.

For any prime $q\mathrm{\ne }p$, step 1.1 and [L3] applied to $p$ and $q$ give $\mathrm{\mid }q\mathrm{\mid }=1$. Hence if $x=\pm p^{k}a\mathrm{/}b$ with $p\nmid ab$, uniqueness of factorisation [L4] gives $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }p{\mathrm{\mid }}^k$. Writing $c:=-\log \mathrm{\mid }p\mathrm{\mid }\mathrm{/}\log p>0$, this becomes $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }x{\mathrm{\mid }}_p^c$, so $\mathrm{\mid }\cdot \mathrm{\mid }$ is equivalent to $\mathrm{\mid }\cdot {\mathrm{\mid }}_p$ by [L5].

Step 2.1 gives the nonarchimedean case and step 1.2 gives the archimedean case, and the two cases are disjoint because [L2] says every $p$-adic absolute value is nonarchimedean. Therefore every nontrivial absolute value on $\mathbb{Q}$ is equivalent either to the usual absolute value or to a unique $p$-adic one.