6.1 Absolute Value

6.1   Definition (Absolute values.) Recall that if $x$ is a real number, then the absolute value of $x$, denoted by $\vert x\vert$, is defined by

$$\vert x\vert = \cases{x &if $x > 0$,\cr 0 &if $x=0$,\cr -x &if $x<0$.\cr}$$

We will assume the following properties of absolute value, that follow easily from the definition:

For all real numbers $x,y,z$ with $z\neq 0$

$$\begin{eqnarray*} \vert x\vert &=& \vert-x\vert \cr \vert xy\vert &=& \vert x\ve... ... {\vert z\vert}}\cr \cr -\vert x\vert &\leq& x\leq \vert x\vert. \end{eqnarray*}$$

For all real numbers $x$, and all $a\in\mbox{{\bf R}}^+$

$$(\vert x\vert<a)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(-a<x<a)$$

and

$$(\vert x\vert\leq a)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(-a\leq x\leq a).$$ (6.2)

We also have

$$\vert x\vert\in\mbox{{\bf R}}_{\geq 0} \mbox{ for all } x\in\mbox{{\bf R}},$$

and

$$\vert x\vert=0\hspace{1ex}\Longleftrightarrow\hspace{1ex}x=0.$$

6.3   Theorem. Let $a\in\mbox{{\bf R}}$ and let $p\in\mbox{${\mbox{{\bf R}}}^{+}$}$. Then for all $x \in \mbox{{\bf R}}$ we have

$$\vert x-a\vert<p\iff (a-p<x<a+p),$$

and

$$\vert x-a\vert\leq p\iff (a-p\leq x\leq a+p).$$

Equivalently, we can say that

$$\{x\in \mbox{{\bf R}}: \vert x-a\vert < p\} = (a-p,a+p)$$

and

$$\{x\in \mbox{{\bf R}}: \vert x-a\vert \leq p \} = [a-p,a+p].$$

Proof:    I will prove only the first statement. I have

$$\begin{eqnarray*} \vert x-a\vert<p & \iff & -p<x-a<p \\ & \iff & a-p<a+(x-a)<a+p \\ & \iff & a-p<x<a+p. \mbox{ $\diamondsuit$} \end{eqnarray*}$$

6.4   Definition (Distance.) The distance between two real numbers $x$ and $y$ is defined by

$$\mbox{\rm dist}(x,y) = \vert x-y\vert.$$

Theorem 6.3 says that the set of numbers whose distance from $a$ is smaller than $p$ is the interval $(a-p,a+p)$. Geometrically this is clear from the picture.

I remember the theorem by keeping the picture in mind.

6.5   Theorem (Triangle inequality.) For all real numbers $x$ and $y$

$$\vert x + y\vert \leq \vert x\vert + \vert y\vert,$$ (6.6)

Proof      For all $x$ and $y$ in R we have

$$-\vert x\vert \leq x \leq \vert x\vert$$

and

$$-\vert y\vert \leq y \leq \vert y\vert,$$

so

$$-(\vert x\vert + \vert y\vert) \leq x+y \leq (\vert x\vert + \vert y\vert).$$

Hence (Cf. (6.2))

$$\vert x + y\vert \leq \vert x\vert + \vert y\vert.$$

6.7   Exercise. Can you prove that for all $(x,y)\in\mbox{{\bf R}}^2\Big( \vert x-y\vert\leq \vert x\vert-\vert y\vert\Big)$?

Can you prove that for all $(x,y)\in \mbox{{\bf R}}^2\Big(\vert x-y\vert \leq \vert x\vert + \vert y\vert\Big)$?

Remark: Let $a, b, c, d$ be real numbers with $a < c < b$ and $a < d < b$.

Then

$$\vert c - d\vert < \vert b - a\vert = b - a.$$

This result should be clear from the picture. We can give an analytic proof as follows.

$$\begin{eqnarray*} (a < c < b \mbox{ and }a < d < b) &\mbox{$\Longrightarrow$}& (... ...\\ & \mbox{$\Longrightarrow$}& \vert c-d\vert < \vert b-a\vert. \end{eqnarray*}$$

6.8   Examples. Let

$$\begin{eqnarray*} A &=& \{ x \in \mbox{{\bf R}}: \vert x-2\vert < 5 \},\\ B&=& \{ x\in\mbox{{\bf R}}: \vert x-2\vert > 5\}. \end{eqnarray*}$$

Then a number $x$ is in $A$ if and only if the distance from $x$ to $2$ is smaller than $5$, and $x$ is in $B$ if and only if the distance from $x$ to $2$ is greater than $5$. I can see by inspection that

$$A = (-3,7),$$

and

$$B = (-\infty,-3) \cup (7,\infty).$$

Let

$$C = \{ x \in \mbox{{\bf R}}: \left\vert {x-1 \over x+1} \right\vert < 1\}.$$

If $x \in \mbox{{\bf R}}\setminus \{-1\}$, then $x$ is in $C$ if and only if $\vert x-1\vert < \vert x+1\vert$, i.e. if and only if $x$ is closer to $1$ than to $-1$.

I can see by inspection that the point equidistant from $-1$ and $1$ is $0$, and that the numbers that are closer to $1$ than to $-1$ are the positive numbers, so $C = (0,\infty)$. I can also do this analytically, (but in practice I wouldn't) as follows. Since the alsolute values are all non-negative

$$\begin{eqnarray*} \vert x-1\vert < \vert x+1\vert &\mbox{$\Longleftrightarrow$}&... ...e{1ex}& 0 < 4x \hspace{1ex}\Longleftrightarrow\hspace{1ex}0 < x. \end{eqnarray*}$$

6.9   Exercise. Express each of the four sets below as an interval or a union of intervals. (You can do this problem by inspection.)

$$\begin{eqnarray*} A_1 & = & \{x\in\mbox{{\bf R}}\colon \vert x-{1\over 2}\vert< ... ...\in\mbox{{\bf R}}\colon \vert{3\over 2}+x\vert\geq {3\over 2}\}. \end{eqnarray*}$$

6.10   Exercise. Sketch the graphs of the functions from $\mbox{{\bf R}}$ to $\mbox{{\bf R}}$ defined by the following equations:

$$\begin{eqnarray*} f_1(x)&=&\vert x\vert, \\ f_2(x)&=&\vert x-2\vert, \\ f_3(x)... ... \\ f_6(x)&=&\vert x^2-1\vert, \\ f_7(x)&=&\vert x^2-1\vert^2. \end{eqnarray*}$$

(No explanations are expected for this problem.)

6.11   Exercise. Let $f_1,\cdots ,f_7$ be the functions described in the previous exercise. By looking at the graphs, express each of the following six sets in terms of intervals.

$$\begin{eqnarray*} S_1&=&\{x\in\mbox{{\bf R}}\colon f_1(x)<1\} \\ S_2&=&\{x\in\m... ...olon f_5(x)<3\} \\ S_6&=&\{x\in\mbox{{\bf R}}\colon f_6(x)<3\}. \end{eqnarray*}$$

Let $S_7=\{x\in\mbox{{\bf R}}\colon f_7(x)<{1\over 2}\}$. Represent $S_7$ graphically on a number line.

Remark: The notation $\vert x\vert$ for absolute value of $x$ was introduced by Weierstrass in 1841 [15][Vol 2,page 123]. It was first introduced in connection with complex numbers. It is surprising that analysis advanced so far without introducing a special notation for this very important function.