Ostrowski's Theorem

Theorem
Every nontrivial norm on $$\mathbb{Q}$$ is Cauchy equivalent to either:
 * 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:
$$\forall n \in \mathbb{N}: \left\|{n}\right\| \le 1$$:

Let $$n_0 \ $$ be the least integer such that $$\left\|{n}\right\|<1 \ $$; such a number exists because we have assumed $$\left\|{*}\right\| \ $$ is non-trivial.

$$n_0 \ $$ must be prime, because if $$n_0 = n_1 n_2 \ $$ with $$n_1, n_2 < n_0$$, then $$\left\|{n_1}\right\| = \left\|{n_2}\right\| = 1 \ $$ and so $$\left\|{n_0}\right\| = \left\|{n_1}\right\| \left\|{n_2}\right\| =1 \ $$, a contradiction. So let $$n_o = p \ $$.

Claim: $$\left\|{q}\right\| = 1 \ $$ if $$q \ $$ is a prime not equal to $$p \ $$. Suppose not; then $$\left\|{q}\right\| < 1 \ $$, and for some large $$N \ $$ we have $$\left\|{q^N}\right\| = \left\|{q}\right\|^N < \tfrac{1}{2}$$.

Also, for some large $$M \ $$ we have $$\left\|{p}\right\|^M<\tfrac{1}{2}$$.

Since $$p^M, q^N \ $$ are relatively prime, by Bézout's Identity we can find integers $$n, m \ $$ such that $$mp^M + nq^N = 1 \ $$.

But then:


 * $$1= \left\|{1}\right\| = \left\|{mp^M+nq^n}\right\| \leq \left\|{mp^M}\right\|+\left\|{nq^N}\right\|=\left\|{m}\right\| \left\|{p^M}\right\|+\left\|{n}\right\| \left\|{q^N}\right\|,$$

by the definition of a norm. But $$\left\|{m}\right\|, \left\|{n}\right\| \leq 1 \ $$, so that


 * $$1 \leq \left\|{p^m}\right\|+\left\|{q^N}\right\| < \tfrac{1}{2}+\tfrac{1}{2} = 1$$,

a contradiction.

Hence $$\left\|{q}\right\|=1$$.

By the fundamental theorem of arithmetic, any positive integer can be factored into prime divisors: $$a = p_1^{b_1} p_2^{b_2} \dots p_r^{b_r} \ $$.

Then $$\left\|{a}\right\| = \left\|{p_1}\right\|^{b_1} \left\|{p_2}\right\|^{b_2} \dots \left\|{p_r}\right\|^{b_r}$$.

But the only $$\left\|{p_i}\right\| \ $$ which will not equal 1 will be $$\left\|{p}\right\| \ $$ if one of the $$p_i \ $$s is $$p \ $$. Its corresponding $$b_i \ $$ will be $$\text{ ord}_p (a) \ $$, where $$\text{ ord} \ $$ is defined as on the page Definition:P-adic Metric. Hence, if we let $$\rho = \left\|{p}\right\|<1 \ $$, we have


 * $$\left\|{a}\right\| = \rho^{\text{ord}_p (a) } \ $$

By the properties of norms, this same formula holds with any nonzero rational number in place of $$a \ $$.