Sum of Absolute Values on Ordered Integral Domain

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:
 * $\size {a + b} \le \size a + \size b$

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 $D$.

Let $a \in D$.

If $\map P a$ or $a = 0$ then $a \le \size a$.

If $\map N a$ then by Properties of Negativity: $(1)$ and definition of absolute value:
 * $a < 0 < \size a$

and hence by transitivity $<$ we have:


 * $a < \size a$

By similar reasoning:


 * $-a < \size a$

Thus for all $a, b \in D$ we have:
 * $a \le \size a, b \le \size b$

As $<$ is compatible with $+$, we have:
 * $a + b \le \size a + \size b$

and:
 * $-\paren {a + b} = \paren {-a} + \paren {-b} \le \size a + \size b$

But either:
 * $\size {a + b} = a + b$

or:


 * $\size {a + b} = -\paren {a + b}$

Hence the result:


 * $\size {a + b} \le \size a + \size b$

Also see

 * Triangle Inequality