Trivial Norm on Division Ring is Norm

Theorem
Let $\left({R, +, \circ}\right)$ be a division ring, and denote its ring zero by $0_R$.

Then the trivial norm $\left \Vert{\cdot}\right \Vert: R \to \R_{\ge 0}$, which is given by:


 * $\left\Vert{x}\right\Vert = \begin{cases}

0 & \text{ if } x = 0_R\\ 1 & \text{ otherwise} \end{cases}$

defines a norm on $R$.

Proof
Proving each of the norm axioms one by one:

Proving $(N1) : \forall x \in R: \left\Vert {x} \right\Vert = 0 \iff x = 0_R$
This follows directly from the definition of the trivial norm.

Proving $(N2) : \forall x, y \in R: \left\Vert{x \circ y}\right\Vert = \left\Vert{x}\right\Vert \times \left\Vert{y}\right\Vert$
If $x=0_R$:

The reasoning is similar if $y = 0_R$.

If $x,y \ne 0_R$, then $x \circ y \ne 0_R$ by alternative definition $(3)$ of division rings. We get:

Proving $(N3) : \forall x, y \in R: \left\Vert {x + y}\right\Vert \leq \left\Vert{x}\right\Vert + \left\Vert{y}\right\Vert$
If $x=0_R$:

The reasoning is similar if $y = 0_R$.

If $x,y \ne 0_R$: