3.4 Absolute Value

Definition 3.3.

Let $x\in\mathbb{R}$ . Then the absolute value of $x$ is defined as

\lvert x\rvert=\begin{cases}x&\text{if}\ x\geq 0\\ -x&\text{if}\ x<0\end{cases}.

The absolute value has several useful properties.

Theorem 3.4 (Properties of Absolute Value)

Let $x,y\in\mathbb{R}$ , then

(i)

$\lvert x\rvert\geq 0$ ,
(ii)

$\lvert x\rvert=0\iff x=0\ \big{(}\lvert x-y\rvert=0\iff x=y\big{)}$ ,
(iii)

$\lvert-x\rvert=\lvert x\rvert=\max(x,-x)$ ,
(iv)

$x\leq\lvert x\rvert$ ,
(v)

$\lvert x\rvert\leq a\iff-a\leq x\leq a,\ \text{for}\ a\in\mathbb{R}$ ,
(vi)

Similarly, $\lvert x\rvert\geq a\iff x\leq-a$ or $x\geq a$ ,
(vii)

$\lvert xy\rvert=\lvert x\rvert\lvert y\rvert$ .

Proof.

(i), (ii), (iii), and (iv) all follow immediately from the definition.

(v)

Since $\lvert x\rvert\geq 0,a\geq 0$ . Therefore clearly $x\geq-a$ and $x\leq\lvert x\rvert\leq a$ .
(vi)

Follows similarly.
(vii)

The result follows immediately if $x$ or $y$ is 0, or if $x$ and $y$ are both either positive negative. Lets assume without loss of generality that $x<0$ , then, noting that $xy=-\lvert xy\rvert$ and $x=-\lvert x\rvert$ we find

$\lvert xy\rvert=-xy=(-x)(y)=\lvert x\rvert\lvert y\rvert.$

∎

These basic properties allow us to prove a basic theorem about inequalities with absolute values which will be absolutely indispensable in more advanced proofs.

Theorem 3.5 (Triangle Inequalities)

Let $x,y\in\mathbb{R}$ , then we have

(i)

(Triangle Inequality) $\lvert x+y\rvert\leq\lvert x\rvert+\lvert y\rvert$ ,
(ii)

(Reverse Triangle Inequality) $\big{\lvert}\lvert x\rvert-\lvert y\rvert\big{\rvert}\leq\lvert x-y\rvert$ .

Proof.

The proof of these statements follow directly from the basic properties of the absolute value, with the reverse triangle inequality following directly from the normal triangle inequality.

(i)

From (iv) we have $-\lvert x\rvert\leq x\leq\lvert x\rvert$ and similarly for $y$ . Adding these two inequalities we get

$\displaystyle-\lvert x\rvert-\lvert y\rvert$ $\displaystyle\leq x+y\leq\lvert x\rvert+\lvert y\rvert$ (3.5)

$\displaystyle-\big{(}\lvert x\rvert+\lvert y\rvert\big{)}$ $\displaystyle\leq x+y\leq\lvert x\rvert+\lvert y\rvert,$ (3.6)

and thus from (v) we have $\lvert x+y\rvert\leq\lvert y\rvert$ .
(ii)

Following from the triangle equality, note that

$\displaystyle\lvert x\rvert$ $\displaystyle=\lvert(x-y)+y\rvert\leq\lvert x-y\rvert+\lvert y\rvert$ (3.8)

$\displaystyle\lvert y\rvert$ $\displaystyle=\lvert(y-x)+x\rvert\leq\lvert y-x\rvert+\lvert x\rvert.$ (3.9)

Rearranging each equation in turn, we get

$\displaystyle\lvert x\rvert-\lvert y\rvert$ $\displaystyle\leq\lvert x-y\rvert$ (3.10)

$\displaystyle\lvert y\rvert-\lvert x\rvert=-\big{(}\lvert x\rvert-\lvert y% \rvert\big{)}$ $\displaystyle\leq\lvert y-x\rvert=\lvert x-y\rvert,$ (3.11)

and so using (vi) we obtain $\big{\lvert}\lvert x\rvert-\lvert y\rvert\big{\rvert}\leq\lvert x-y\rvert$ .

∎

	$\displaystyle-\lvert x\rvert-\lvert y\rvert$	$\displaystyle\leq x+y\leq\lvert x\rvert+\lvert y\rvert$		(3.5)
	$\displaystyle-\big{(}\lvert x\rvert+\lvert y\rvert\big{)}$	$\displaystyle\leq x+y\leq\lvert x\rvert+\lvert y\rvert,$		(3.6)

	$\displaystyle\lvert x\rvert$	$\displaystyle=\lvert(x-y)+y\rvert\leq\lvert x-y\rvert+\lvert y\rvert$		(3.8)
	$\displaystyle\lvert y\rvert$	$\displaystyle=\lvert(y-x)+x\rvert\leq\lvert y-x\rvert+\lvert x\rvert.$		(3.9)

	$\displaystyle\lvert x\rvert-\lvert y\rvert$	$\displaystyle\leq\lvert x-y\rvert$		(3.10)
	$\displaystyle\lvert y\rvert-\lvert x\rvert=-\big{(}\lvert x\rvert-\lvert y% \rvert\big{)}$	$\displaystyle\leq\lvert y-x\rvert=\lvert x-y\rvert,$		(3.11)