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:
 * $\forall x \in R:\norm x_1 < 1 \iff \norm x_2 < 1$

Let $x, y \in R \setminus 0_R$ such that $\norm x_1, \norm y_1 \ne 1$.

Let $\alpha = \dfrac {\log \norm x_1} {\log \norm x_2}$ and $\beta = \dfrac {\log \norm y_1} {\log \norm y_2}$.

Then:
 * $\alpha = \beta$

Proof
Because $x, y \in R \setminus 0_R$:
 * $\norm x_1, \norm y_1, \norm x_2 , \norm y_2 > 0$.

Because $\norm{x}_1, \norm {y}_1 \ne 1$, by Lemma 2:
 * $\norm x_2, \norm y_2 \ne 1$.

Hence:
 * $\log \norm x_1, \log \norm y_1, \log \norm x_2, \log \norm y_2 \ne 0$

and $\alpha, \beta$ are well-defined.

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

Then:

By Logarithm is Strictly Increasing:
 * $\log \norm y_1^n < \log \norm x_1^m \iff \log \norm y_2^n < \log \norm x_2^m$

By Sum of Logarithms:
 * $n \log \norm y_1 < m \log \norm x_1 \iff n \log \norm y_2 < m \log \norm x_2$

Because $m, \log \norm x_1, \log \norm x_2 \ne 0$:
 * $r = \dfrac n m < \dfrac {\log \norm x_1} {\log \norm y_1} \iff r = \dfrac n m < \dfrac {\log \norm x_2} {\log \norm y_2}$

By Between two Real Numbers exists Rational Number:
 * $\dfrac {\log \norm x_1} {\log \norm y_1} = \dfrac {\log \norm x_2} {\log \norm y_2}$

Because $\log \norm x_2, \log \norm y_2 \ne 0$:
 * $\dfrac {\log \norm x_1} {\log \norm x_2} = \dfrac {\log \norm y_1} {\log \norm y_2}$

That is:
 * $\alpha = \beta$