Strict Negativity is equivalent to Strictly Preceding Zero

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 $a \in D$:

$\map N a \iff a < 0$

$\ds \map N a$

$\leadstoandfrom$

$\ds \map P {-a}$

Definition of Strict Negativity Property

$\ds $

$\leadstoandfrom$

$\ds \map P {-a + 0}$

$\ds $

$\leadstoandfrom$

$\ds a < 0$

$\blacksquare$

1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $10 \ \text {(i)}$