Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition

Theorem
Let $\struct{R, \norm{\,\cdot\,}}$ be a normed division ring with unity $1_R$.

Then:
 * $\forall n \in \N_{>0}: \norm{n \cdot 1_R} \le 1 \implies \norm{\,\cdot\,}$ is non-Archimedean

where $n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{n \, times}$

Proof
Let:
 * $\forall n \in \N_{>0}: \norm{n \cdot 1_R} \le 1$

Let $x, y \in R$.

Let $y = 0_R$ where $0_R$ is the zero of $R$.

Then $\norm {x + y} = \norm {x} = \max \set{\norm {x}, 0} = \max \set{\norm {x}, \norm{y}}$

Lemma 1
Hence to complete the proof it is sufficient to prove:
 * $\forall x \in R: \norm{ x + 1_R} \le \max \set{\norm{x }, 1 }$

For $n \in \N$:

Lemma 2
Hence

Taking $n$th roots yields:
 * $\norm{x + 1_R} \le \paren {n+1}^{1/n} \max \set {\norm {x}, 1}$

Lemma 3
By multiple rule for real sequences then:
 * $\displaystyle \lim_{n \to \infty} \paren {n+1}^{1/n} \max \set {\norm {x}, 1} = \max \set {\norm {x}, 1}$

By Inequality Rule for Real Sequences then:
 * $\norm{x + 1_R} \le \max \set {\norm {x}, 1}$

The result follows.