Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.3

Theorem
Let $R$ be a division ring.

Let $\norm{\,\cdot\,}_1: R \to \R_{\ge 0}$ and $\norm{\,\cdot\,}_2: R \to \R_{\ge 0}$ be norms on $R$.

Let $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ satisfy:
 * $\forall x \in R:\norm{x}_1 \lt 1 \iff \norm{x}_2 \lt 1$

Let $x_0, x \in R \setminus 0_R$ such that $\norm{x_0}_1, \norm {x}_1 \neq 1$.

Let $\alpha = \dfrac {\log \norm {x_0}_1 } {\log \norm {x_0}_2 }$ and $\beta = \dfrac {\log \norm {x}_1 } {\log \norm {x}_2 }$.

Then $\alpha = \beta$.

Proof
Let $r = n/m \in \Q$ be any rational number where $n, m \in \Z$ are integers and $m \neq 0$.

Then:

By Logarithm is Strictly Increasing then:
 * $\log \norm{x}_1^n \lt \log \norm{x_0}_1^m \iff \log \norm{x}_2^n \lt \log \norm{x_0}_2^m$

By Sum of Logarithms then:
 * $n\log \norm{x}_1 \lt m\log \norm{x_0}_1 \iff n\log \norm{x}_2 \lt m\log \norm{x_0}_2$

Since $\norm {x}_1 \neq 1$ by Lemma 2 then $\norm {x}_2 \neq 1$.

Hence:
 * $\log \norm{x}_1 \neq 0$
 * $\log \norm{x}_2 \neq 0$.

Since $m, \log \norm{x}_1, \log \norm{x}_2 \neq 0$ then:
 * $r = \dfrac n m \lt \dfrac {\log \norm{x_0}_1} {\log \norm{x}_1} \iff r = \dfrac n m \lt \dfrac {\log \norm{x_0}_2} {\log \norm{x}_2}$

By Between two Real Numbers exists Rational Number then:
 * $\dfrac {\log \norm{x_0}_1} {\log \norm{x}_1} = \dfrac {\log \norm{x_0}_2} {\log \norm{x}_2}$

Since $\norm {x_0}_1 \neq 1$ by Lemma 2 then $\norm {x_0}_2 \neq 1$.

Hence:
 * $\log \norm{x_0}_1 \neq 0$
 * $\log \norm{x_0}_2 \neq 0$.

Since $\log \norm{x_0}_2, \log \norm{x}_2 \neq 0$ then:
 * $\dfrac {\log \norm{x_0}_1} {\log \norm{x_0}_2} = \dfrac {\log \norm{x}_1} {\log \norm{x}_2}$

That is:
 * $\alpha = \beta$.