Totally Ordered Ring Zero Precedes Element or its Inverse
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Theorem
Let $\struct {R, +, \circ, \preceq}$ be an ordered ring.
From the definition of ordered ring, $\preceq$ is compatible with $+$.
Let $0_R$ be the zero element of $R$.
Let $x \ne 0_R$ be a non-zero element of $R$.
Let $-x$ be the ring negative of $x$.
Then:
- $0_R \prec x \lor 0_R \prec -x$
but not both.
Proof
By the definition of total ordering, $\preceq$ is connected.
As $x \ne 0_R$, one of the following is true, but not both:
- $(1): \quad 0_R \prec x$
- $(2): \quad x \prec 0_R$
If $(2)$, because $\prec$ is compatible with $+$:
\(\ds x\) | \(\prec\) | \(\ds 0_R\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x + \paren {-x}\) | \(\prec\) | \(\ds 0_R + \paren {-x}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0_R\) | \(\prec\) | \(\ds -x\) | Definition of Ring Zero, Definition of Ring Negative |
$\blacksquare$