Absolute Value

Here is an example where it is natural to use a two part definition of a function. We write

| x| = $$\left\{\vphantom{\begin{array}{ll} x&\mbox{if $x \ge 0$;}\\ -x&\mbox{if $x < 0$.} \end{array}}\right.$$$$\begin{array}{ll} x&\mbox{if $x \ge 0$;}\\ -x&\mbox{if $x < 0$.} \end{array}$$

An equivalent definition is | x| = $\sqrt{{x^2}}$. This is the absolute value or modulus of x. It's particular use is in describing distances; we interpret | x - y| as the distance between x and y. Thus

(a - $$\delta$$, a + $$\delta$$) = {X $$\in$$ $$\mathbb {R}$$ : | x - a| < $$\delta$$},

so a $\delta$ - neighbourhood of a consists of all points which are closer to a than $\delta$.

Note that we can always ``expand out'' the inequality using this idea. So if | x - y| < k, we can rewrite this without a modulus sign as the pair of inequalities - k < x - y < k. We sometimes call this ``unwrapping'' the modulus; conversely, in order to establish an inequality involving the modulus, it is simply necessary to show the corresponding pair of inequalities.

Proposition 1.11 (The Triangle Inequality.)   For any x, y $\in$ $\mathbb {R}$,

| x + y|$$\le$$| x| + | y|.

Proof. Since - | x|$\le$x$\le$| x|, and the same holds for y, combining these we have

- | x| - | y|$$\le$$x + y$$\le$$| x| + | y|

and this is the same as the required result. $\qedsymbol$

Exercise 1.12   Show that for any x, y, z $\in$ $\mathbb {R}$, | x - z|$\le$| x - y| + | y - z|.

Proposition 1.13   For any x, y $\in$ $\mathbb {R}$,

| x - y|$$\ge$$$$\Big\vert$$| x| - | y|$$\Big\vert$$.

Proof. Using 1.12 we have

| x| = | x - y + y|$$\le$$| x - y| + | y|

and so | x| - | y|$\le$| x - y|. Interchanging the rôles of x and y, and noting that | x| = | - x|, gives | y| - | x|$\le$| x - y|. Multiplying this inequality by -1 and combining these we have

- | x - y|$$\le$$| x| - | y|$$\le$$| x - y|

and this is the required result. $\qedsymbol$

Example 1.14   Describe {x $\in$ $\mathbb {R}$ : | 5x - 3| > 4}.

Proof. Unwrapping the modulus, we have either 5x - 3 < - 4, or 5x - 3 > 4. From one inequality we get 5x < - 4 + 3, or x < - 1/5; the other inequality gives 5x > 4 + 3, or x > 7/5. Thus

{x $$\in$$ $$\mathbb {R}$$ : | 5x - 3| > 4} = (- $$\infty$$, - 1/5) $$\cup$$ (7/5,$$\infty$$).

$\qedsymbol$

Exercise 1.15   Describe {x $\in$ $\mathbb {R}$ : | x + 3| < 1}.

Exercise 1.16   Describe the set {x $\in$ $\mathbb {R}$ : 1$\le$x$\le$3} using the absolute value function.