Trivial Norm on Division Ring is Norm
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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 Norm Axiom $\text N 1$: Positive Definiteness: $\forall x \in R: \norm x = 0 \iff x = 0_R$
This follows directly from the definition of the trivial norm.
$\Box$
Proving Norm Axiom $\text N 2$: Multiplicativity: $\forall x, y \in R: \norm {x \circ y} = \norm x \times \norm y$
If $x = 0_R$:
\(\ds \norm {x \circ y}\) | \(=\) | \(\ds \norm {0_R \circ y}\) | by substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {0_R}\) | Ring Product with Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Definition of Trivial Norm on Division Ring | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \times \norm y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm x \times \norm y\) | since $x = 0_R$ |
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:
\(\ds \norm {x \circ y}\) | \(=\) | \(\ds 1\) | since $x \circ y \ne 0_R$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm x \times \norm y\) | since $x, y \ne 0_R$ |
$\Box$
Proving Norm Axiom $\text N 3$: Triangle Inequality: $\forall x, y \in R: \norm {x + y} \le \norm x + \norm y$
If $x = 0_R$:
\(\ds \norm {x + y}\) | \(=\) | \(\ds \norm {0_R + y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 + \norm y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm x + \norm y\) | since $x = 0_R$ |
The reasoning is similar if $y = 0_R$.
If $x, y \ne 0_R$:
\(\ds \norm {x + y}\) | \(\le\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 1 + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm x + \norm y\) | since $x, y \ne 0_R$ |
$\blacksquare$