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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

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

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 $a\in\mbox{{\bf R}}^+$

\begin{displaymath}(\vert x\vert<a)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(-a<x<a)\end{displaymath}

and
\begin{displaymath}
(\vert x\vert\leq a)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(-a\leq x\leq a).
\end{displaymath} (6.2)

We also have

\begin{displaymath}\vert x\vert\in\mbox{{\bf R}}_{\geq 0} \mbox{ for all } x\in\mbox{{\bf R}},\end{displaymath}

and

\begin{displaymath}\vert x\vert=0\hspace{1ex}\Longleftrightarrow\hspace{1ex}x=0.\end{displaymath}

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

\begin{displaymath}\vert x-a\vert<p\iff (a-p<x<a+p),\end{displaymath}

and

\begin{displaymath}\vert x-a\vert\leq p\iff (a-p\leq x\leq a+p).\end{displaymath}

Equivalently, we can say that

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

and

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

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

\begin{displaymath}\mbox{\rm dist}(x,y) = \vert x-y\vert. \end{displaymath}

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.


\begin{picture}(2.5,.25)(-.6,.25)
\put(-.5,.3){\line(1,0){2}}
\put(-.05,.25){\li...
....1}}
\put(-.23,.15){$a-p$}
\put(.43,.15){$a$}
\put(.78,.15){$a+p$}
\end{picture}
I remember the theorem by keeping the picture in mind.

6.5   Theorem (Triangle inequality.) For all real numbers $x$ and $y$
$\displaystyle \vert x + y\vert \leq \vert x\vert + \vert y\vert,$     (6.6)

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

\begin{displaymath}
-\vert x\vert \leq x \leq \vert x\vert \end{displaymath}

and

\begin{displaymath}
-\vert y\vert \leq y \leq \vert y\vert,
\end{displaymath}

so

\begin{displaymath}-(\vert x\vert + \vert y\vert) \leq x+y \leq (\vert x\vert + \vert y\vert).
\end{displaymath}

Hence (Cf. (6.2))

\begin{displaymath}\vert x + y\vert \leq \vert x\vert + \vert y\vert. \end{displaymath}

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 $a < d < b$.


\begin{picture}(2.5,.25)(-.6,.25)
\put(-.5,.3){\line(1,0){2}}
\put(-.05,.25){\li...
....15){$a$}
\put(.45,.15){$c$}
\put(.95,.15){$d$}
\put(1.3,.15){$b$}
\end{picture}
Then

\begin{displaymath}\vert c - d\vert < \vert b - a\vert = b - a. \end{displaymath}

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*}




\begin{picture}(2.5,.25)(-.6,.25)
\put(-.5,.3){\line(1,0){2}}
\put(-.05,.25){\li...
...1){.1}}
\put(-.23,.15){$-3$}
\put(.43,.15){$2$}
\put(0.9,.15){$7$}
\end{picture}
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

\begin{displaymath}A = (-3,7),\end{displaymath}

and

\begin{displaymath}B = (-\infty,-3) \cup (7,\infty).\end{displaymath}

Let

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

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 $-1$.


\begin{picture}(2.5,.25)(-.6,.25)
\put(-.5,.3){\line(1,0){2}}
\put(-.05,.25){\li...
...,1){.1}}
\put(-.1,.15){$-1$}
\put(.43,.15){$0$}
\put(.85,.15){$1$}
\end{picture}
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.



next up previous index
Next: 6.2 Approximation Up: 6. Limits of Sequences Previous: 6. Limits of Sequences   Index
Ray Mayer 2007-09-07