Ostrowski's Theorem

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

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

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

Let $n_0$ be the least such integer.

Since $\left \Vert {n_0}\right \Vert > 1$, it follows that $\exists \alpha \in \R_+$ 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 $n \ge n_0^s$.

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 integer $n$ and some large positive integer $N$, we can use this formula 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, this result extends to all $q \in \Q$.

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

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

Let $n_0$ be the least integer such that $\left \Vert {n}\right \Vert < 1$.

Such a number exists because we have assumed $\left \Vert {*}\right \Vert$ is non-trivial.

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

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

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

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

But then:


 * $1 = \left \Vert {1}\right \Vert = \left \Vert {mp^M+nq^n}\right \Vert \le \left \Vert {mp^M}\right \Vert + \left \Vert {nq^N}\right \Vert = \left \Vert {m}\right \Vert \left \Vert {p^M}\right \Vert + \left \Vert {n}\right \Vert \left \Vert {q^N}\right \Vert$

by the definition of a norm.

But $\left \Vert {m}\right \Vert, \left \Vert {n}\right \Vert \le 1$, so that


 * $1 \le \left \Vert {p^m}\right \Vert + \left \Vert {q^N}\right \Vert < \frac 1 2 + \frac 1 2 = 1$

which is a contradiction.

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

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

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


 * $\left \Vert {a}\right \Vert = \rho^{\operatorname{ord}_p \left({a}\right)}$

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