Equivalent Norms are both Non-Archimedean or both Archimedean

Theorem
Let $R$ be a division ring with [Definition:Unity of Ring|unity]] $1_R$.

Let $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ be equivalent norms on $R$.

Then $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ are either both non-Archimedean or both Archimedean.

Proof
By Characterisation of Non-Archimedean Division Ring Norms then:
 * $\norm{\,\cdot\,}_1$ is non-Archimedean $\iff \forall n \in \N_{>0}: \norm{n \cdot 1_R}_1 \le 1$.

By the definition of norm equivalence then:
 * $\forall n \in \N: \norm {n \cdot 1_R}_1 \le 1 \iff \norm {n \cdot 1_R}_2 \le 1$

Similarly, by Characterisation of Non-Archimedean Division Ring Norms then:
 * $\forall n \in \N_{>0}: \norm{n \cdot 1_R}_2 \le 1 \iff \norm{\,\cdot\,}_2$ is non-Archimedean.

The result follows.