Normed Vector Space Requires Multiplicative Norm on Division Ring

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

Let $V$ be a vector space that is not a trivial vector space.

Let $\norm {\,\cdot\,}:V \to \R_{\ge0}$ be a mapping from $V$ to the positive real numbers satisfying the vector space norm axioms.

Then $\norm {\,\cdot\,}_R$ is a multiplicative norm.

That is:
 * $\forall r,s \in R: \norm {rs}_R = \norm{r}_R \norm{s}_R$

Proof
Since $V$ is not a trivial vector space then:
 * $\exists v \in V: v \ne 0$

By Norm axiom (N1) then:
 * $\norm {v} > 0$

Let $r,s \in R$ then:

By dividing both sides of the equation by $\norm {v}$ then:
 * $\norm {rs}_R = \norm{r}_R \norm{s}_R$

The result follows.