Ostrowski's Theorem/Non-Archimedean Norm

Theorem
Let $\norm {\, \cdot \,}$ be a non-trivial non-Archimedean norm on the rational numbers $\Q$.

Then $\norm {\, \cdot \,}$ is equivalent to the $p$-adic norm $\norm {\, \cdot \,}_p$ for some prime $p$.

Proof
From Characterisation of Non-Archimedean Division Ring Norms then:
 * $\forall n \in \N: \norm n \le 1$

Lemma 2.1
Let $n_0 = \min \set {n \in N : \norm n < 1}$.

Lemma 2.2
Let $p = n_0$.

Let $\alpha = - \dfrac {\log \norm p } {\log p}$ then:
 * $\norm p = p^{-\alpha} = \paren {p^{-1}}^\alpha = \norm p_p^\alpha$

Let $b \in N$

Case 1: $p \nmid b$
Let $p \nmid b$.

From Prime not Divisor implies Coprime:
 * $p$ and $b$ are coprime, that is, $p \perp b$

From Corollary 5 of Three Points in Ultrametric Space have Two Equal Distances:
 * $\norm b = 1$

By the definition of the $p$-adic norm:
 * $\norm b_p = 1$

Hence:
 * $\norm b = 1 = 1^\alpha = \norm b_p^\alpha$

Case 2: $p \divides b$
Let $p \divides b$.

Let $\nu = \map {\nu_p} b$ where $\nu_p$ is the $p$-adic valuation on $\Z$.

Then:
 * $b = p^\nu c$

where $p \nmid c$

From :
 * $\norm c = 1$

Hence:

In either case:
 * $\norm b = \norm b_p^\alpha$

Since $b$ was arbitrary, it has been shown:
 * $\forall b \in \N: \norm b = \norm b_p^\alpha$

From Equivalent Norms on Rational Numbers:
 * $\norm {\, \cdot \,}$ is equivalent to the $p$-adic norm $\norm {\, \cdot \,}_p$.