# Ostrowski's Theorem/Archimedean Norm/Lemma 1.2

## Theorem

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

Let $n_0 = \min \set {n \in \N : \norm n > 1}$

Let $\alpha = \dfrac {\log \norm {n_0} } {\log n_0}$

Then:

$\forall n \in N: \norm n \ge n^\alpha$

## Proof

By the definition of $\alpha$:

$\norm {n_0} = n_0^\alpha$

By the definition of $n_0$:

$n_0^\alpha > 1$

Let $n \in \N$.

By Basis Representation Theorem then $n$ 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 $a_s \ne 0$

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

By Lemma 1.1:

$\norm {n_0^{s + 1} - n} \le \paren {n_0^{s + 1} - n}^\alpha$

Hence:

 $\displaystyle \norm n$ $\ge$ $\displaystyle \norm {n_0^{s + 1} } - \norm {n_0^{s + 1} - n}$ Reverse Triangle Inequality $\displaystyle$ $=$ $\displaystyle \norm {n_0}^{s + 1} - \norm {n_0^{s + 1} - n}$ Norm Axiom (N2) (Multiplicativity) $\displaystyle$ $=$ $\displaystyle n_0^{\alpha \paren {s + 1} } - \norm {n_0^{s + 1} - n}$ as $\norm {n_0} = n_0^\alpha$ $\displaystyle$ $\ge$ $\displaystyle n_0^{\alpha \paren {s + 1} } - \paren {n_0^{s + 1} - n}^\alpha$ as $\norm {n_0^{s + 1} - n} \le \paren {n_0^{s + 1} - n}^\alpha$ $\displaystyle$ $\ge$ $\displaystyle n_0^{\alpha \paren {s + 1} } - \paren {n_0^{s + 1} - n_0^s}^\alpha$ as $n \ge n_0^s$ $\displaystyle$ $=$ $\displaystyle n_0^{\alpha \paren {s + 1} } \paren {1 - \paren {1 - \frac 1 {n_0} }^\alpha}$ $\displaystyle$ $\ge$ $\displaystyle n^\alpha \paren {1 - \paren {1 - \frac 1 {n_0} }^\alpha}$ as $n_0^{s+1}>n$

Let $C' = \paren{1 - \paren{1 - \frac 1 {n_0} }^\alpha}$.

Then:

$\norm {n} \ge C' n^\alpha$

As $n \in \N$ was arbitrary:

$\forall n \in N: \norm n \ge C' n^\alpha$

Let $n, N \in N$.

Then:

$\norm {n^N} \ge C' \paren{n^N}^\alpha$

Now:

 $\displaystyle \norm {n^N} \ge C' \paren{n^N}^\alpha$ $\leadsto$ $\displaystyle \norm n^N \ge C' \paren {n^N}^\alpha$ Norm axiom (N2) (Multiplicativity) $\displaystyle$ $\leadsto$ $\displaystyle \norm n^N \ge C' \paren {n^\alpha}^N$ $\displaystyle$ $\leadsto$ $\displaystyle \norm n \ge \sqrt [N] {C'} n^\alpha$ taking $N$th roots
$\sqrt [N] {C'} \to 1$ as $N \to \infty$
$\sqrt [N] {C'} n^\alpha \to n^\alpha$ as $N \to \infty$

By Inequality Rule for Real Sequences, letting $N \to \infty$ for fixed $n$:

$\norm n \ge n^\alpha$

The result follows.

$\blacksquare$