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 $\alpha$ then:
 * $\norm {n_0} = n_0^\alpha$

By the definition of $n_0$ then:
 * $n_0^\alpha > 1$

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 then:
 * $n_0^{s+1} > n \ge n_0^s$.

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

Hence:

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

Hence:
 * $\norm {n} \ge C' n^\alpha$

Since $n \in \N$ was arbitrary then:
 * $\forall n \in N: \norm {n} \ge C' n^\alpha$

Let $n, N \in N$

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

Now:

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

By multiple rule for real sequences then:
 * $\sqrt[N]{C'} n^\alpha \to n^\alpha$ as $N \to \infty$

By Inequality Rule for Real Sequences, letting $N \to \infty$ for fixed $n$, then:
 * $\norm {n} \ge n^\alpha$

The result follows.