Ostrowski's Theorem/Archimedean Norm/Lemma 1.1

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:
 * $\forall n \in N: \norm {n} \le n^\alpha$

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 $a_s \ne 0$

Since all of the $a_i < n_0$, by choice of $n_0$ then:
 * $\forall a_i: \norm {a_i} \le 1$

Then:

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

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

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

Let $n, N \in N$

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

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:
 * $\norm {n} \le n^\alpha$

The result follows.