Ostrowski's Theorem/Archimedean Norm/Lemma 1.2

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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:

\(\displaystyle \norm n\) \(\ge\) \(\displaystyle \norm {n_0^{s + 1} } - \norm {n_0^{s + 1} - n}\) 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


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.

$\blacksquare$


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