Absolute Value on Ordered Integral Domain is Strictly Positive except when Zero

Theorem

Let $\struct {D, +, \times, \le}$ be an ordered integral domain.

For all $a \in D$, let $\size a$ denote the absolute value of $a$.

Then $\size a$ is (strictly) positive except when $a = 0$.

Let $<$ be the (strict) total ordering defined on $D$ as:

$a < b \iff a \le b \land a \ne b$

From the trichotomy law, exactly one of three possibilities holds for any $ a \in D$:

$(1): \quad \map P a$:

In this case $0 < a$ and so $\size a = a$.

So:

$\map P a \implies \map P {\size a}$

$(2): \quad \map P {-a}$:

In this case $a < 0$ and so $\size a = -a$.

So:

$\map P {-a} \implies \map P {\size a}$

$(3): \quad a = 0$:

In this case $\size a = a$.

So:

$a = 0 \implies \size a = 0$.

Hence the result.

$\blacksquare$

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