Strict Negativity is equivalent to Strictly Preceding Zero
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Theorem
Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $\le$ be the total ordering induced by $P$, and let $<$ be its strict total ordering counterpart.
Let $N$ be the (strict) negativity property on $D$:
- $\forall a \in D: \map N a \iff \map P {-a}$
Then for all $a \in D$:
- $\map N a \iff a < 0$
Proof
\(\ds \map N a\) | \(\leadstoandfrom\) | \(\ds \map P {-a}\) | Definition of Strict Negativity Property | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \map P {-a + 0}\) | ||||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds a < 0\) | Strict Positivity Property induces Total Ordering |
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $10 \ \text {(i)}$