Sum of Strictly Negative Elements is Strictly Negative

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 $a \in D$:

$\map N a, \map N b \implies \map N {a + b}$

Proof

\(\ds \map N a, \map N b\)

\(\leadsto\)

\(\ds \map P {-a}, \map P {-b}\)

Definition of Strict Negativity Property

\(\ds \)

\(\leadsto\)

\(\ds \map P {\paren {-a} + \paren {-b} }\)

Strict Positivity Property: $(P \, 1)$

\(\ds \)

\(\leadsto\)

\(\ds \map P {-\paren {a + b} }\)

\(\ds \)

\(\leadsto\)

\(\ds \map N {a + b}\)

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 $10 \ \text {(iii)}$