Product of Absolute Values on Ordered Integral Domain
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Theorem
Let $\struct {D, +, \times, \le}$ be an ordered integral domain whose zero is denoted by $0_D$.
For all $a \in D$, let $\size a$ denote the absolute value of $a$.
Then:
- $\size a \times \size b = \size {a \times b}$
Proof
Let $P$ be the (strict) positivity property on $D$.
Let $<$ be the (strict) total ordering defined on $D$ as:
- $a < b \iff a \le b \land a \ne b$
Let $N$ be the strict negativity property on $D$.
We consider all possibilities in turn.
$(1): \quad a = 0_D$ or $b = 0_D$
In this case, both the left hand side $\size a \times \size b$ and the right hand side are equal to zero.
So:
- $\size a \times \size b = \size {a \times b}$
$(2): \quad \map P a, \map P b$
First:
\(\ds \map P a, \map P b\) | \(\leadsto\) | \(\ds \size a = a, \size b = b\) | Definition of Absolute Value on Ordered Integral Domain | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \size a \times \size b = a \times b\) |
Then:
\(\ds \map P a, \map P b\) | \(\leadsto\) | \(\ds \map P {a \times b}\) | Strict Positivity Property: $(\text P 2)$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \size {a \times b} = a \times b\) | Definition of Absolute Value on Ordered Integral Domain |
So:
- $\size a \times \size b = \size {a \times b}$
$(3): \quad \map P a, \map N b$
First:
\(\ds \map P a, \map N b\) | \(\leadsto\) | \(\ds \size a = a, \size b = -b\) | Definition of Absolute Value on Ordered Integral Domain | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \size a \times \size b = -a \times b\) | Product with Ring Negative |
Then:
\(\ds \map P a, \map N b\) | \(\leadsto\) | \(\ds \map N {a \times b}\) | Properties of Strict Negativity: $(5)$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map P {-a \times b}\) | Definition of Strict Negativity Property | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \size {a \times b} = -a \times b\) | Definition of Absolute Value on Ordered Integral Domain |
So:
- $\size a \times \size b = \size {a \times b}$
Similarly $\map N a, \map P b$.
$(4): \quad \map N a, \map N b$
First:
\(\ds \map N a, \map N b\) | \(\leadsto\) | \(\ds \size a = -a, \size b = -b\) | Definition of Absolute Value on Ordered Integral Domain | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \size a \times \size b = a \times b\) | Product of Ring Negatives |
Then:
\(\ds \map N a, \map N b\) | \(\leadsto\) | \(\ds \map P {a \times b}\) | Properties of Strict Negativity: $(4)$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map P {a \times b}\) | Definition of Strict Negativity Property | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \size {a \times b} = a \times b\) | Definition of Absolute Value on Ordered Integral Domain |
So:
- $\size a \times \size b = \size {a \times b}$
In all cases the result holds.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $11 \ \text{(i)}$