Sum 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:
 * $\left \vert{a + b}\right \vert \le \left \vert{a}\right \vert + \left \vert{b}\right \vert$

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

Let $a \in D$.

If $P \left({a}\right)$ or $a = 0$ then $a \le \left \vert{a}\right \vert$.

If $N \left({a}\right)$ then by Properties of Negativity: $(1)$ and definition of absolute value:
 * $a < 0 < \left \vert{a}\right \vert$

and hence by transitivity $<$ we have:


 * $a < \left \vert{a}\right \vert$

By similar reasoning:


 * $-a < \left \vert{a}\right \vert$

Thus for all $a, b \in D$ we have:
 * $a \le \left \vert{a}\right \vert, b \le \left \vert{b}\right \vert$

As $<$ is compatible with $+$, we have:
 * $a + b \le \left \vert{a}\right \vert + \left \vert{b}\right \vert$

and:
 * $-\left({a + b}\right) = \left({-a}\right) + \left({-b}\right) \le \left \vert{a}\right \vert + \left \vert{b}\right \vert$

But either:
 * $\left \vert{a + b}\right \vert = a + b$

or:


 * $\left \vert{a + b}\right \vert = - \left({a + b}\right)$

Hence the result:


 * $\left \vert{a + b}\right \vert \le \left \vert{a}\right \vert + \left \vert{b}\right \vert$

Also see

 * Triangle Inequality