Product of Absolute Values on Ordered Integral Domain

Theorem
Let $\left({D, +, \times}\right)$ be an ordered integral domain.

For all $a \in D$, let $\left \vert{a}\right \vert$ denote the absolute value of $a$.

Then:
 * $\left \vert{a}\right \vert \times \left \vert{b}\right \vert = \left \vert{a \times b}\right \vert$

Proof
Let $P$ be the positivity property on $D$, let $<$ be the ordering induced by it, and let $N$ be the negativity property on $D$.

We consider all possibilities in turn.

$(1): \quad a = 0$ or $b = 0$

In this case, both the LHS $\left \vert{a}\right \vert \times \left \vert{b}\right \vert$ and the RHS are equal to zero.

So:


 * $\left \vert{a}\right \vert \times \left \vert{b}\right \vert = \left \vert{a \times b}\right \vert$

$(2): \quad P \left({a}\right), P \left({b}\right)$

First:

Then:

So:
 * $\left \vert{a}\right \vert \times \left \vert{b}\right \vert = \left \vert{a \times b}\right \vert$

$(3): \quad P \left({a}\right), N \left({b}\right)$

First:

Then:

So:
 * $\left \vert{a}\right \vert \times \left \vert{b}\right \vert = \left \vert{a \times b}\right \vert$

Similarly $N \left({a}\right), P \left({b}\right)$.

$(4): \quad N \left({a}\right), N \left({b}\right)$

First:

Then:

So:
 * $\left \vert{a}\right \vert \times \left \vert{b}\right \vert = \left \vert{a \times b}\right \vert$

In all cases the result holds.