Product of Strictly Negative Element with Strictly Positive Element is Strictly Negative
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Theorem
Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $N$ be the (strict) negativity property on $D$:
- $\forall a \in D: \map N a \iff \map P {-a}$
Then for all $a \in D$:
- $\map N a, \map P b \implies \map N {a \times b}$
Proof
\(\ds \map N a, \map P b\) | \(\leadsto\) | \(\ds \map P {-a}, \map P b\) | Definition of Strict Negativity Property | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map P {\paren {-a} \times b}\) | Strict Positivity Property: $(P \, 2)$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map P {-\paren {a \times b} }\) | Product with Ring Negative | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map N {a \times b}\) | Definition of Strict Negativity Property |
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $10 \ \text {(v)}$