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 Strict 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:

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

$\blacksquare$

Also see

• Triangle Inequality

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $11 \ \text{(ii)}$