Trivial Norm on Division Ring is Norm

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

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


 * $\norm x = \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 $(N1) : \forall x \in R: \norm x = 0 \iff x = 0_R$
This follows directly from the definition of the trivial norm.

Proving $(N2) : \forall x, y \in R: \norm {x \circ y} = \norm x \times \norm y$
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 $(N3) : \forall x, y \in R: \norm {x + y} \le \norm x + \norm y$
If $x = 0_R$:

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

If $x, y \ne 0_R$: