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 $a_s \ne 0$

By Bounds for Integer Expressed in Base k:
 * ${n_0}^{s + 1} > n \ge {n_0}^s$

By Lemma 1.1:
 * $\norm { {n_0}^{s + 1} - n} \le \paren { {n_0}^{s + 1} - n}^\alpha$

Hence:

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:

By Limit of Root of Positive Real Number:
 * $\sqrt [N] {C'} \to 1$ as $N \to \infty$

By the Multiple Rule for Real Sequences:
 * $\sqrt [N] {C'} n^\alpha \to n^\alpha$ 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.