Ostrowski's Theorem/Archimedean Norm

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

Then $\norm {\, \cdot \,}$ is equivalent to the absolute value, $\size {\, \cdot \,}$.

Proof
$\exists n \in \N$ such that $\norm n > 1$:

Let $n_0$ be the least such number.

Since $\left \Vert {n_0}\right \Vert > 1$, it follows that $\exists \alpha \in \R_{>0}$ such that $\left \Vert {n_0}\right \Vert = n_0^\alpha$.

From the Basis Representation Theorem, any positive integer $n$ can be written:


 * $n = a_0 + a_1 n_0 + a_2 n_0^2 + \cdots + a_s n_0^s$, where $0 \le a_i < n_0$ and $a_s \ne 0$

Then:

Since all of the $a_i < n_0$, we have $\left \Vert {a_i}\right \Vert \le 1$.

Hence:

because $n \ge n_0^s$.

By Sum of Infinite Geometric Progression (and since $n_0^\alpha > 1$), the expression in brackets is a finite constant; call it $C$.

Hence $\left \Vert {n}\right \Vert \le C n^\alpha$ for all positive integers.

For any positive integers $n$ and $N$, we can use this inequality on $n^N$ and take $N$th roots to obtain:

$\left \Vert {n}\right \Vert \le \sqrt[N]{C} n^\alpha$.

Letting $N \to \infty$ for fixed $n$ gives $\left \Vert {n}\right \Vert \le n^\alpha$.

Now consider again the formulation of $n$ in base $n_0$.

We have $n_0^{s+1} > n \ge n_0^s$.

Since:


 * $\left \Vert {n_0^{s+1}}\right \Vert = \left \Vert {n+n_0^{s+1} - n}\right \Vert \le \left \Vert {n}\right \Vert + \left \Vert {n_0^{s+1} - n}\right \Vert$

we have:

since $\left \Vert {n_0^{s+1}}\right \Vert = \left \Vert {n_0}\right \Vert^{s+1}$, and by the first inequality ($\left \Vert {n}\right \Vert \le n^\alpha$) on the term being subtracted.

Thus:

for some constant $C'$ which may depend on $n_0$ and $\alpha$, but not on $n$.

As before, for very large $N$, use this inequality on $n^N$, take $N$th roots and let $N \to \infty$, to get:


 * $\left \Vert {n}\right \Vert \ge n^\alpha$

These two results imply $\left \Vert {n}\right \Vert = n^\alpha$.

By the second property of norms (namely multiplicativity), this result extends to all $q \in \Q$.

Suppose a series $\left\{{x_1, x_2, \ldots}\right\}$ is Cauchy on the [Definition:Euclidean Metric on Real Number Line|Euclidean metric]].

We have $\left \Vert {x_j - x_i}\right \Vert \le \left \vert {x_j - x_i}\right \vert$, and so the series is Cauchy on $\left \Vert {*}\right \Vert $.

Now suppose a series is Cauchy on $\left \Vert {*}\right \Vert$.

Then for any $N$ such that $\forall i, j > N: \log_\alpha \left \vert{x_j - x_i}\right \vert < \epsilon, \left \Vert {x_j - x_i}\right \Vert < \epsilon$, so the series is Cauchy on the Euclidean metric.

Thus, $\left \Vert {*}\right \Vert$ is Cauchy equivalent to the Euclidean metric.