Ostrowski's Theorem

Theorem
Every nontrivial norm on $$\mathbb{Q}$$ is Cauchy equivalent to the P-adic Metric $\left|{*}\right|_p$ for some prime $$p \ $$ or the Euclidean metric.

Proof
Let $$\left\|{*}\right\|$$ be a norm.

Case 1:
$$\exists n \in \mathbb{N}$$ such that $$\left\|{n}\right\| > 1$$:

Let $$n_0 \ $$ be the least such integer.

Since $$\left\|{n_0}\right\| > 1$$, it follows that $$\exists \alpha \in \mathbb{R}_+$$ such that $$\left\|{n_0}\right\| = n_0^\alpha$$.

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 $$\left\|{a_i}\right\| \leq 1$$.

Hence:

$$ $$ $$

because $$n \ge n_0^s$$.

The expression in brackets is a finite constant; call it $$C$$.

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

For any positive integer $$n$$ and some large positive integer $$N$$, we can use this formula to obtain:

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

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

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

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

Since $$\left\|{n_0^{s+1}}\right\| = \left\|{n+n_0^{s+1} - n}\right\| \leq \left\|{n}\right\| + \left\|{n_0^{s+1} - n}\right\|$$, we have:

$$ $$

since $$\left\|{n_0^{s+1}}\right\| = \left\|{n_0}\right\|^{s+1}$$, and by the first inequality ($$\left\|{n}\right\| \le n^\alpha$$) 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 $$\left\|{n}\right\| \ge n^\alpha$$.

These two results imply $$\left\|{n}\right\| = n^\alpha \ $$. By the second property of norms, this result extends to all $$q \in \mathbb{Q}$$.

Suppose a series $$\left\{{x_1, x_2, \ldots}\right\}$$ is Cauchy on the Euclidean metric.

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

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

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

Case 2:
$$\forall n \in \mathbb{N}: \left\|{n}\right\| \le 1$$: