\chapter{Ordering -- absolute value}  % version définitive (!) 2000/09/10   
   
\Citation{Had I been present at the Creation, I would have given some useful hints for the better ordering of the 
  universe.\\[3pt] attributed to Alfonso `the Wise' \Detantatant{1221}{1284}--- King of Castile and Le\'on from 
  \oldstylenums{1252} --- on studying the Ptolemaic system.\footnotemark} \footnotetext{Ptolemaic 
  \pron{£bt£ol£i£hm£q£ik}: of or relating to Ptolemy \pron{£ht£ol£im£i}, (2nd century), Greek astronomer and 
  geographer.} 
   
\section{Ordering}   
   
Two real numbers can always been compared. Let $a$ and $b$ be two real numbers. Then or $a<b$ `$a$ is less than $b$' or 
$a = b$ `$a$ equals $b$' or $a>b$ `$a$ is greater than $b$'.\index{greater than}\index{less than}\index{$<$}\index{$>$} 
   
If $a<b$ is not true then $a\geq b$ \idest\  `$a$ is not less than $b$' or `$a$ is greater than or equal to $b$'.%
\index{not less than}\index{$\geq$} 
   
If $a>b$ is not true then $a\leq b$ \idest\  `$a$ is not greater than $b$' or `$a$ is less than or equal to $b$'.%
\index{not greater than}\index{$\leq$} 
   
\begin{prop}   
  Let $a$ and $b$ be two reals. $a > b$ if and only if $a-b$ is positive; $a \geq b$ if and only if $a-b$ is 
  non-negative; $a < b$ if and only if $a-b$ is negative; and $a \leq b$ if and only if $a-b$ is non-positive. 
\end{prop}   
   
\begin{rem}   
  From now on and for sake of convenience we shall replace the expression `if and only if' with its common abbreviation 
  `iff'.\index{iff} We can replace the expression `if and only if' with `means' or `is equivalent to'. 
     
  Let P and Q be propositions ---~statements, sentences that affirm or deny something and which are capable of being 
  true or false. Whenever we know that `If P then Q' and `If Q then P' we can state that `P iff Q'. 
\end{rem}   
   
As a consequence, for every real $a$: if $a>0$ then $a$ is positive; if $a<0$ then $a$ is negative. 
   
If $a=0$ then $a$ is said to be \nuvo{null} \pron{n£vl}. 
   
\subsection{Inequalities}   
   
One can add or subtract the same number from both sides of an \emph{inequality} \pron{£b£in£i£hkw£ol£et£i} and multiply or 
divide both sides by positive values without changing the inequality. But when one multiply or divide both sides of the 
statement by a negative value it is very important that one remembers that the inequality reverses.\index{inequality} We 
can sum this up with: 
   
\begin{prop}   
  Let $a$, $b$, and $c$ be reals. 
\begin{itemize}   
\item If $a\leq b$ then $a+c \leq b+c$. 
\item If $a\leq b$ and $c\geq0$ then $a\times c \leq b\times c$. 
\item If $a\leq b$ and $c\leq0$ then $a\times c \geq b\times c$. 
\end{itemize}   
\end{prop}   
   
As consequences of that proposition we can state the two following propositions. 
\begin{prop}   
  Let $a$, $b$, $c$, and $d$ be reals. If $a\leq b$ and $c \leq d$ then $a+c \leq b+d$. 
\end{prop}   
   
\begin{prop}   
  Let $a$, $b$, $c$, and $d$ be non-negative reals. If $a\leq b$ and $c \leq d$ then $ac \leq bd$. 
\end{prop}   
   
\subsection{Intervals}   
   
\subsubsection{Definitions}   
\begin{Def}   
  Let $a$ and $b$ be two reals such that $a<b$. The set of real numbers which are not less than $a$ and not greater than 
  $b$ is a \emph{closed interval} \pron{kl£wzd £h£int£evl} of \reel. It is denoted by \interfer{a}{b}. 
\end{Def}\index{interval}\index{$\interfer{a}{b}$}\index{interval!closed}   

\newpage
   
We can write $x \in \interfer{a}{b}$ iff $a \leq x \leq b$. The numbers $a$ and $b$ are the \emph{endpoints} of the 
interval \interfer{a}{b}.\index{endpoint} It can be graphically represented as below. NB: we hatch what does not belong 
to the interval. 
   
\begin{fgre}
  \IG{interfer}
\end{fgre}
   
There are nine types of intervals which are: {\setlength{\extrarowheight}{2pt} 
\begin{longtable}[c]{r @{ : } l}   
  \interfer{a}{b} & the set of real $x$ such that $a \leq x \leq b$\\[2pt] 
  \intferouv{a}{b} & the set of real $x$ such that $a \leq x < b$\\[2pt] 
  \intouvfer{a}{b} & the set of real $x$ such that $a < x \leq b$\\[2pt] 
  \interouv{a}{b} & the set of real $x$ such that $a < x < b$\\[2pt] 
  \intferouv{a}{\plusinf} & the set of real $x$ such that $a \leq x $\\[2pt] 
  \interouv{a}{\plusinf} & the set of real $x$ such that $a < x $\\[2pt] 
  \intouvfer{\moinsinf}{b} & the set of real $x$ such that $ x \leq b$\\[2pt] 
  \interouv{\moinsinf}{b} & the set of real $x$ such that $ x < b$\\[2pt] 
  \interouv{\moinsinf}{\plusinf} & the set of all reals \\[2pt] 
\end{longtable}}   
   
An interval such as \interouv{a}{b} is said to be \nuvo{open}.\index{interval!open} An interval such as \intferouv{a}{b} is 
said to be `closed in $a$ and open in $b$'.  The sign `\plusinf' is read `plus infinity' \pron{£in£hf£in£et£i}, it does 
\textbf{not} denote a number.  The sign `\moinsinf' is read `minus infinity', it does \textbf{not} denote a number 
either.\index{plus infinity}\index{minus infinity}\index{infinity}
   
\subsubsection{Intersection and union}   
   
\begin{Def}   
  The \emph{intersection} of two intervals $I$ and $J$ is the set of all reals which are in $I$ \textbf{and} in $J$.  It 
  is denoted by $I \cap J$ which is read `$I$ cap $J$'. 
\end{Def}\index{intersection}\index{$\cap$}\index{cap}
   
For example, let $I = \interfer{0}{10}$ and $J= \interfer{-5}{4}$, then $I\cap J = \interfer{0}{4}$ for $x$ belongs to 
$I\cap J$ if and only if $x \in I$ and $x \in J$ that is $0 \leq x \leq 10$ and $-5\leq x \leq 4$ which is equivalent to 
$0\leq x\leq4$. On the figure below the intersection is represented by the segment which is not hatched. 
   
\enlargethispage{-\baselineskip} 
   
\begin{fgre}
  \IG{intersect}
\end{fgre}
   
The intersection of $I = \interouv{0}{3}$ and $J = \interfer{5}{10}$ contains no real. One says that this intersection 
is equal to the \emph{null} \pron{n£vl} \emph{set} (or \emph{void} \pron{v£rd} \emph{set} or 
\emph{empty set}) which is denoted by $\vide$. On the figure below all the line is hatched.%   
\index{null!set}\index{void}\index{empty}\index{$\vide$}\index{set!void}\index{set!empty}\index{set!null} 
   
\begin{fgre}
  \IG{inters1}
\end{fgre}
   
\medskip   

The intersection of $I = \interfer{0}{3}$ and $J = \interfer{3}{10}$ contains only $3$.  One writes 
$I\cap J = \accol{3}$. % correction du 17/11/2000   
The intersection of $I = \intferouv{0}{3}$ and $J = \interfer{3}{10}$ does not contain any number for $3$ does not 
belong to $I$.  In the last two cases graphical representations are not really useful. 
   
\begin{Def}   
  The \emph{union} of two intervals $I$ and $J$ is the set of all reals which are in $I$ or in $J$.  It is denoted by $I 
  \cup J$ which is read `$I$ cup $J$'. 
\end{Def}\index{union}\index{$\cup$}\index{cap}   
   
For example, let $I = \interfer{0}{10}$ and $J= \interfer{-5}{4}$, then $I\cup J = \interfer{-5}{10}$ for $x$ belongs to 
$I\cup J$ if and only if $x \in I$ or $x \in J$ that is $0 \leq x \leq 10$ or $-5\leq x \leq 4$ which is equivalent to 
$-5\leq x\leq10$.

In the following figure each interval is represented with a bold line. The union is then represented by  the  whole bold
segment.
   
\begin{fgre}
  \IG{union1}
\end{fgre}

\medskip   

\begin{rem}   
  The union of two intervals is not always an interval for a subset of \reel{} is an interval iff it is represented by a 
  segment, a ray, or the whole line. For example $\interfer{-1}{1} \cup\interfer{3}{4}$ is not an interval. 

  The two bold segment do not touch or overlap.
\end{rem}   
   
\begin{fgre}
  \IG{union2}
\end{fgre}
   
\newpage

\section{Distance and absolute value}   
\subsection{Definitions}   
\begin{Def}   
  Let $x$ and $y$ be two reals and $M$ and $N$ the points on the real axis the abscissae of which are $x$ and $y$ 
  respectively. 
     
  The \emph{distance} \pron{£hd£ist£ens} between $x$ and $y$, denoted by $d\couple{x}{y}$ is defined by: 
  $$d\couple{x}{y}=MN.$$ 
\end{Def}\index{distance}\index{$d\couple{x}{y}$}   
   
For example $d\couple{7}{1} = 6$ for $6$ is the distance between the points with abscissae $1$ and $7$.  For analogous 
reason $d\couple{-4}{5}=9$. 
   
\begin{rem}   
  For any reals $x$ and $y$, $d\couple{x}{y}$ is non-negative for it equals a distance. Moreover, as $MN = NM$, then 
  $d\couple{x}{y} = d\couple{y}{x}$. 
\end{rem}   
   
\begin{Def}   
  Let $O$ be the origin on the real axis. Let $x$ be a real and $M$ the point on the real axis the abscissa of which is 
  $x$.  The \emph{absolute value} \pron{£h£abs£el£Ut £hv£alj£U} of $x$, denoted by \varabs{x}, is the distance from $x$ to 
  $0$.  It is also the distance from $M$ to $O$. We have:\index{value!absolute}\index{absolute}\index{absolute!value} 
  $$\varabs{x} = d\couple{0}{x} = OM.$$ 
\end{Def}\index{$\varabs{x}$}   
   
For example $\varabs{3}=3$ and $\varabs{-4}=4$. 
   
\begin{fgre}
  \IG{varabs}
\end{fgre}
   
\begin{rem}   
  For all real $x$, as $d\couple{0}{x}$ is non-negative, \varabs{x} is a non-negative real.\\ 
  For all real $x$, point $M'$, abscissa $-x$, is the reflection of point $M$, abscissa $x$, through $O$ therefore $OM = 
  OM'$ and so $\varabs{x}=\varabs{-x}$. 
\end{rem}   
   
\subsection{Properties}   
We now state useful properties some of which are proven. You can try to work out a proof for the remaining ones. 
   
The first one is easy to prove if you remember the geometrical interpretation of the absolute value as a distance on the 
real axis. (\textsl{Draw a figure}.) 
\begin{prop}   
  Let $x$ be a real. 
\begin{itemize}   
\item If $x$ is positive, then $\varabs{x} = x$. 
\item If $x$ is negative, then $\varabs{x} = -x$. 
\end{itemize}   
\end{prop}   
   
\begin{prop}   
  For all $x$ and $y$ real numbers. 
\begin{itemize}   
\item If $x \geq y $, then $\varabs{x - y } = x - y$. 
\item If $x \leq y $, then $\varabs{x - y } = y - x$. 
\item $d\couple{x}{y} = \varabs{x-y}$. 
\end{itemize}   
\end{prop}   
   
\begin{dem}   
  If $x \geq y $ then $x - y\geq 0 $ therefore $\varabs{x-y} = x-y$. 
   
  If $x \leq y $, then $x - y\leq 0 $ therefore $\varabs{x - y } = -\parent{x-y}=y - x$. 
\end{dem}   
   
\begin{prop}   
  Let §a be a non-negative real number and $a$ be any real. For all $x$ real: 
  $$ 
  \varabs{x-a} \leq §a \quad\text{iff}\quad a-§a \leq x \leq a+§a \quad\text{iff}\quad x \in \interfer{a-§a}{a+§a}. 
  $$ 
\end{prop}   
   
\begin{dem}   
  We have to distinguish two cases depending on the place of $x$ relatively to $a$. 
   
  \textbf{First case}: $x \leq a$. 
   
  Then $\varabs{x-a}\leq §a$ is equivalent to $a-x \leq §a$ \idest\  $a-§a\leq x$. Therefore in this case, $x$ fulfills 
  the inequality $\varabs{x-a}\leq §a$ iff $a-§a\leq x\leq a$. 
   
  \textbf{Second case}: $x \geq a$. 
   
  Then $\varabs{x-a}\leq §a$ is equivalent to $x-a \leq §a$ \idest\  $x\leq a+§a$. Therefore in this case, $x$ fulfils the 
  inequality $\varabs{x-a}\leq §a$ iff $a\leq x\leq a+§a$. 
   
  For all $x$ real, $x \leq a$ or $x \geq a$. So $\varabs{x-a}\leq §a$ is equivalent to $a-§a\leq x\leq a$ or $a\leq 
  x\leq a+§a$ \idest\  $\varabs{x-a}\leq §a$ is equivalent to $a-§a\leq x\leq a+§a$.  \Qed\,\footnote{QED stands for latin 
    `quod erat demonstrandum' and means `which was to be proven'.} 
\end{dem}   
   
\begin{prop}   
  Let $x$ and $y$ be two real numbers. 
\begin{itemize}   
\item $\dps\varabs{xy} = \varabs{x}\,\varabs{y}$; 
\item If $y \neq 0$ then $\dps\varabs{\dfrac{x}{y}} = \dfrac{\varabs{x}}{\varabs{y}}$; 
\item $\dps \varabs{x+y} \leq \varabs{x} + \varabs{y}$ (triangle inequality). 
\end{itemize}   
\end{prop}   
   
\section{Approximation}   
   
There are basically two types of errors which we need to consider in mathematics: \emph{measurement errors} 
\pron{£hme£j£em£ent £her£e£Z}\index{error}\index{error!measurement} and \emph{round-off} \pron{£hr£Wnd£of} \emph{errors}. 
\index{error!round-off} A measurement error is always present whenever a physical \pron{£hf£iz£ikl} quantity 
\pron{£hkw£ont£et£i} is measured \pron{£hme£j£ed}. A round-off error is present whenever a number is rounded to a 
specified \pron{£hspes£if£pd} number of decimal places (d.p.) or significant \pron{s£ig£hn£if£ik£ent} figures (s.f.). 
Round-off errors are also present when a computer \emph{chops off} ---~or \emph{truncate} \pron{tr£v£n£hk£qt}~--- 
numbers which it is working with or displaying. \textsl{By \emph{errors} we do not mean \emph{mistakes}.} 
   
Measuring a value to the nearest unit means deciding that it is nearer one mark than another, in other words it is 
within half a unit of that mark. For example, if David's age is 14 to the nearest year that means he is actually between 
$13\demi$ and $14\demi$ and, more precisely, that the age is not less than $13\demi$ and \textbf{less than} $14\demi$. 
   
If $x$ is an approximation to an exact number $X$, then the \emph{absolute error} in $x$ is $e = x-X$.%
\index{absolute!error}\index{error}\index{error!absolute}\index{absolute!error bound} The \emph{absolute error bounds} 
are the limits between which the absolute error lies. So, if $\varabs{e} \leq §e$, where $§e >0$, then $§e$ is called 
the maximum absolute error bound of $x$.\index{error!bound} 
   
Knowing that, we can write $X = x \pm §e$. That means that the true value of number $X$ lies between $x-§e$ and $x+§e$, 
\idest\  $x-§e \leq X < x+§e$. 
   
In general, the maximum absolute error bound when a number has been rounded to $n$ decimal places is 
$\demi\times10^{-n}$.  For example, `The Sun is 93 million miles from the Earth' means that the Sun is $9.3\times 10^7$ 
miles from Earth, to the nearest million miles. Here, the absolute error bound is $0.05\times10^7$ miles \idest\  if $d$ 
is the distance, in miles, from the Earth to the Sun, 
$$\parent{9.3 -0.05}\times10^7\leq d < \parent{9.3 +0.05}\times10^7.$$ 
   
To \nuvo{round down} a number is to approximate it to a certain number of significant digits or to a whole number or 
number of tens, hundreds, etc{.} by replacing the remaining digits by zero. For example, $432.25$ can be rounded down to 
$432.2$, $432$, $430$, or $400$ according to circumstances. 
   
To \nuvo{round up} a number is to approximate it to a certain number of significant digits or to a whole number or 
number of tens, hundreds, etc{.} by increasing the relevant digit by one and replacing the remainder by zeros.  For 
example, $486.75$ can be rounded up to $486.8$, $487$, $490$, or $500$ according to requirements.  \endinput 
   
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% correction du 17/11/2000 
% correction pour v1.1 2001-08-02 
% correction pour v1.2 2002-03-11
% correction pour v1.3 2003-06-15
% correction pour v1.4 2003-07-04
