Properties of Norm on Division Ring/Norm of Negative of Unity

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Theorem

Let $\struct {R, +, \circ}$ be a division ring with unity $1_R$.

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


Then:

$\norm {-1_R} = 1$


Proof

By Product of Ring Negatives:

$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$

So:

\(\ds \norm {-1_R}^2\) \(=\) \(\ds \norm {-1_R} \norm {-1_R}\)
\(\ds \) \(=\) \(\ds \norm {-1_R \circ -1_R}\) Norm Axiom $\text N 2$: Multiplicativity
\(\ds \) \(=\) \(\ds \norm {1_R}\) Product of Ring Negatives
\(\ds \) \(=\) \(\ds 1\) Norm of Unity of Division Ring

Thus:

$\norm {-1_R} = \pm 1$


By Norm Axiom $\text N 1$: Positive Definiteness:

$\norm {-1_R} \ge 0$

Hence:

$\norm {-1_R} = 1$

$\blacksquare$


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