Sum of Absolute Values

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:

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

Corollary

For the number systems $\Z, \Q, \R$:

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

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:

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

$\blacksquare$

Proof of Corollary

Follows directly from:

• Integers form Ordered Integral Domain
• Rational Numbers form Ordered Integral Domain
• Real Numbers form Ordered Integral Domain

$\blacksquare$

Also see

• Triangle Inequality

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 2.7$: Theorem $11 \ \text{(ii)}$