Properties of Norm on Division Ring/Norm of Negative

Theorem
Let $\struct {R, +, \circ}$ be a division ring.

Let $0_R$ be the zero and $1_R$ the unity of $R$.

Let $\norm {\, \cdot \,}$ be a norm on $R$.

Let $x \in R$

Then:
 * $\left\Vert{-x}\right\Vert = \left\Vert{x}\right\Vert$

Proof
First, we prove that $\left\Vert{-1_R}\right\Vert = 1$.

By Product of Ring Negatives, $-1_R \circ -1_R = 1_R \circ 1_R=1_R$.

So:

Thus:


 * $\left\Vert{-1_R}\right\Vert = \pm 1$

But $\left\Vert{-1_R}\right\Vert \ge 0$ by definition of norm, and it follows that:


 * $\left\Vert{-1_R}\right\Vert = 1$

Now:

as desired.