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 $\norm {\, \cdot \,}: V \to \R_{\ge 0}$ 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:
 * $\forall r, s \in R: \norm {r s}_R = \norm r_R \norm s_R$

Proof
Since $V$ is not a trivial vector space:


 * $\exists \mathbf v \in V: \mathbf v \ne 0$

By :
 * $\norm {\mathbf v} > 0$

Let $r, s \in R$:

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

The result follows.