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%%%%%%%%%%MACRO FOR LONG DIVISION%%%%%%%%%%%%
\newdimen\digitwidth
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\def~{\hspace{\digitwidth}}
\def\divrule#1#2{%
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\vbox{\hrule width#2\digitwidth}}}
\newcount\gpten % power-of-ten, tells which
% digit we're doing
\newcount\rtot
% running total -- remainder so far
\newcount\scratch
\def\longdiv#1#2{%
% long division: #1/#2; integers only
\vtop{\offinterlineskip
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\the\scratch\strut}$\cr\the\rtot\strut\cr
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\global\rtot=#1\relax
\count0=\rtot\divide\count0by#2%
\edef\quotient{\the\count0}%
%
% make list macro out of digits in quotient:
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\edef\routine{\expandafter%
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%
% process list to give power-of-ten:
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% to display effect of one digit
% in quotient (zero ignored):
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by##1\relax \multiply\scratch by#2\relax
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% must hide \cr in a macro to skip it
}%
\tabskip=0pt
\halign{\hfil##\cr % \halign for entire
% division problem
$\quotient$\strut\cr
#2$\,\overline{\vphantom{\big)}%
\smash{\raise3.5\fontdimen8%
\textfont3\hbox{$\big)$}}%
\mkern2mu \the\rtot}$%
\cr\noalign{\kern-.2ex} \routine \cr
% do each digit in quotient
}}}
%%%%%%%%%ENDS MACRO FOR LONG DIVISION



%%%%%%%%%

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%%%%%%%




%%%%%%%%%%%%%%%%%%%%Two Column Table of Contents
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%%%%%%%%%%%%
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%%%%%%%%%

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%%%%%%%%%%%%%%%%%%
%%%%%%                  And voilà the document !!!

\begin{document}
\begin{frontmatter}
\pagenumbering{roman}
\begin{titlepage}
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\rput[lb]{90}(-0.3,-0.85\paperheight){\Huge\textcolor{lightgray}{Free
to photocopy and distribute}}
\rput[lb]{90}(-0.4,-0.3\paperheight){\red\Large\textbf{\today\quad
Version}}

\vspace*{3cm}
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%\centerline{\red\bighv{Andragogic Propaedeutic
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\vspace*{2cm} \centerline{\Large \textcolor{red}{David A. SANTOS}
\\ \href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu}}
\end{titlepage}

\begin{quote}
    Copyright \copyright{}  2007  David Anthony SANTOS.
    Permission is granted to copy, distribute and/or modify this document
    under the terms of the GNU Free Documentation License, Version 1.2
    or any later version published by the Free Software Foundation;
    with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts.
    A copy of the license is included in the section entitled ``GNU
    Free Documentation License''.
\end{quote}















 \clearpage {\small \twocoltoc{}}
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\lhead[\rm\thepage]{\it \rightmark} \rhead[\it
\leftmark]{\rm\thepage} \lfoot{\textcolor{lightgray}{Free to
photocopy and distribute}}\cfoot{}
\renewcommand{\sectionmark}{\markboth{\sectionname\ \thechapter}}
\renewcommand{\sectionmark}{\markright}

\chapter*{Preface}
%\markboth{}{}
%\addcontentsline{toc}{chapter}{Preface} \markright{Preface}
\QUOTEME{Mathematics knows no races or geographic boundaries; for
mathematics, the cultural world is one country. }{David~HILBERT}

I would like to thank Eleonora Chertok, Wimayra Luy, and Alain
Schremmer for their input.

\bigskip


I would like to thank the students who first tested these notes in
the Spring of 2008: Comfort Ademuwagun,  Takira Adkins, Roxan
Delisi,  Hai Do, Janie  Drayton, Jacquelin Fulton, Sean Horton,
Fatimah Johnson,  Wakeelah Lloyd, Ashley Mangum, Inmaculada
Martinez, Ruben Medacier, Leslie Pierre, Madeline Roman and Karina
Soriano.

\bigskip


\bigskip
I have not had the time to revise, hence errors will abound,
especially in the homework answers. I will be grateful to receive an
email pointing out corrections.


 \hfill \begin{tabular}{l}David A. SANTOS \\
\href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu} \end{tabular}




 \chapter*{To the Reader} \markboth{}{}
\addcontentsline{toc}{chapter}{To the Reader} \markright{To the
Reader} When writing these notes I set a certain number of goals and
presuppositions.
\begin{itemize}
\item I presume that you know basic addition, subtraction,
multiplication and division of whole numbers, but that you are not
very efficient at it.
\item My goal is to shew you how to perform these operations
efficiently and to think about the algorithms used to carry out
these operations.
\item At times it will seem that you already know what I am talking
about. This is natural, since I am not taking a linear approach to
the subject. Nevertheless, pay attention and do the homework.
Moreover, I would prefer that you use the methods I teach you
instead of some others you may remember from before. This is because
it is very likely that your memory as to why the correct methods
might be faulty.
\item I put  a very strong emphasis on story problems. The goal is
to see to teach you how to derive abstractions from concrete
situations. Therefore, I would prefer a verbose response to a curt
one.
\end{itemize}


These notes are provided for your benefit as an attempt to organise
the salient points of the course. They are a {\em very terse}
account of the main ideas of the course, and are to be used mostly
to refer to central definitions and theorems. The number of examples
is minimal. The {\em motivation} or informal ideas of looking at a
certain topic, the ideas linking a topic with another, the
worked-out examples, etc., are given in class. Hence  these notes
are not  a substitute to lectures: {\bf you must always attend to
lectures}. The order of the notes may not necessarily be the order
followed in the class.

\bigskip

Tutoring can sometimes help,  but bear in mind that whoever tutors
you may not be familiar with my conventions. Again, I am here to
help! On the same vein, other books may help, but the approach
presented here is at times unorthodox and finding alternative
sources might be difficult.

\bigskip

Here are more recommendations:
\begin{itemize}
\item Read a section before class discussion, in particular, read
the definitions.
\item Class provides the informal discussion, and you will profit
from the comments of your classmates, as well as gain confidence by
providing your insights and interpretations of a topic. {\bf Don't
be absent! }
\item I encourage you to form study groups and to discuss the
assignments. Discuss among yourselves and help each other but don't
be  {\em parasites!} Plagiarising your classmates' answers will only
lead you to disaster!
\item Once the lecture of a particular topic has been given, take a fresh look at the notes of the lecture topic.
\item Try to understand a single example well, rather than
ill-digest multiple examples.
\item Start working on the distributed homework ahead of time.
\item {\bf Ask questions during the lecture.} There are two main  types of questions
that you are likely to ask.
\begin{enumerate}
\item {\em Questions of Correction: Is that a minus sign there?} If you think that, for example, I have
missed out a minus sign or wrote $P$ where it should have been
$Q$,\footnote{My doctoral adviser used to say ``I said $A$, I wrote
$B$, I meant $C$ and it should have been $D$!} then by all means,
ask. No one likes to carry an error till line XLV because the
audience failed to point out an error on line I. Don't wait till the
end of the class to point out an error. Do it when  there is still
time to correct it!
\item {\em Questions of Understanding: I don't get it!} Admitting that you do not understand something is
an act requiring utmost courage. But if you don't, it is likely that
many others in the audience also don't. On the same vein, if you
feel you can explain a point to an inquiring classmate, I will allow
you time in the lecture to do so.  The best way to ask a question is
something like: ``How did you get from the second step to the third
step?'' or ``What does it mean to complete the square?''
Asseverations like ``I don't understand'' do not help me answer your
queries. If I consider that you are asking the same questions too
many times, it may be that you need extra help, in which case we
will settle what to do outside the lecture.
\end{enumerate}

\item Don't fall behind! The sequence of topics is closely
interrelated, with one topic leading to another.

\item Presentation is critical. Clearly outline your ideas. When writing solutions, outline major steps and write in
complete sentences. As a guide, you may try to emulate the style
presented in the scant examples furnished in these notes.
\end{itemize}






\end{frontmatter}
\clearpage


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\part{Sets}
\chapter{Sets and Inclusion-Exclusion}
\begin{df}\index{set}
By a {\em set} we will understand any well-defined collection of
objects. These objects are called the {\em elements} of the set.
 If $a$ belongs to the set $A$, then we write $a\in
A$, read ``$a$ is an element of $A$.'' If $a$ does not belong to the
set $A$, we write $a\not\in A$, read ``$a$ is  not an element of
$A$.'' \end{df}
\begin{exa}
Let $D=\{0,1,2,3,4,5,6,7,8,9\}$ be the set of the ten decimal
digits. Then $4\in D$ but $11\nin D$.
\end{exa}


\begin{rem} We will normally denote sets by capital letters, say $A, B, \Omega ,
\naturals$, etc. Elements will be denoted by lowercase letters, say
$a, b, \omega, r $, etc. The following sets will have the special
symbols below.
\begin{center}
\begin{tabular}{ll}
$\naturals= \{0,1,2, 3, \ldots\}$ & denotes the set of natural
numbers. \\
$\varnothing$ & denotes the empty set.\\
\end{tabular}
\end{center}
\end{rem}

\begin{exa}
Let $\Omega = \{1, 2, \ldots, 20\}$, that is, the set of integers
between $1$ and $20$ inclusive. A subset of $\Omega$ is $E = \{2, 4,
6, \ldots , 20\}$, the set of all even integers in $\Omega$. Another
subset of $\Omega $ is $P = \{2, 3, 5, 7, 11, 13, 17, 19\}$, the set
of primes in $\Omega$. Observe that, for example,  $4\in E$ but
$4\not\in P$.
\end{exa}
\begin{df}
 A
{\em subset}\index{set!subset} is a sub-collection of a set. We
denote that the set $B$ is a subset of $A$ by the notation $B
\subseteq A$.
\end{df}
\begin{exa}
The set $$ S=\{\mathrm{Comfort, Takira, Roxan, Hai, Janie,
Jacquelin, Sean, Fatimah, Wakeelah, Ashley, Inmaculada, Ruben,
Leslie, Madeline, Karina}\}$$is the set of students in a particular
section of Maths 016. This set can be split into two subsets: the
set $$ F=\{\mathrm{Comfort, Takira, Roxan, Hai, Janie, Jacquelin,
Fatimah, Wakeelah, Ashley, Inmaculada, Madeline, Karina}\}$$of
females in the class, and the set
$$ M=\{\mathrm{ Sean, Ruben,
Leslie}\}$$ of males in the class.  Thus we have $F\subseteq S$ and
$M \subseteq S$. Notice that it is {\em not true} that $F\subseteq
M$ or that $M \subseteq F$.
\end{exa}

\begin{exa}\label{exa:subsets3element}
Find all the subsets of $\{a, b, c\}$.
\end{exa}
\begin{solu} They are
\begin{multicols}{3}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{enumerate} \item[] $S_1  =
\varnothing$
\item[] $ S_2  =  \{a\} $
\item []$S_3  =  \{b\}$ \item[] $S_4  =  \{c\}$ \item[] $S_5  =  \{a,
b\}$
\item[]
$S_6  =  \{b, c\}$ \item[] $S_7  =  \{c, a\}$ \item[] $S_8  =  \{a,
b, c\} $
\end{enumerate}  \end{multicols}
\end{solu}

\begin{df}
The {\em union} of two sets  $A$ and $B$ is the set
$$A\cup B = \{x:x\in A \ {\mathrm{or}}\ x\in B\}.$$ \index{set!union}
Observe that this ``{\em or}'' is inclusive, that is, it allows the
possibility of $x$ being in $A$, or $B$, or possibly both $A$ and
$B$.


The {\em intersection} of two sets  $A$ and $B$,
is\index{set!intersection}
$$A\cap B = \{x:x\in A \ {\mathrm{and}}\ x\in B\}.$$



\index{set!difference} The difference of sets  $A$ {\em set-minus}
$B$, is
$$A\setminus B = \{x:x\in A \ {\mathrm{and}}\ x\not\in B\}.$$
\end{df}


Figures \ref{fig:a_union_b} through \ref{fig:a_minus_b} represent
these concepts pictorially, through the use of  {\em Venn Diagrams}.
\begin{df}\index{set!disjoint sets}\index{sets !disjoint}\index{sets !mutually exclusive}
Two sets  $A$ and $B$ are {\em disjoint} or {\em mutually exclusive}
if $A\cap B = \varnothing$.
\end{df}
\begin{df}\index{set!complement}
Let $A \subseteq \Omega$. The {\em complement} of  $A$ with respect
to $\Omega$ is $A^c = \{\omega \in \Omega: \omega \not\in A\} =
\Omega\setminus A$.
\end{df}
Observe that $A^c$ is all that which is outside $A$.  The complement
$A^c$ represents the event that $A$ does not occur. We represent
$A^c$ pictorially as in figure \ref{fig:a_complement}.




\vspace*{1cm}
\begin{figure}[h]
\begin{minipage}{3cm}
$$
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fillcolor=lightgray](-1,0){2}\pscircle[fillstyle=hlines,
fillcolor=lightgray](1,0){2} \uput[d](-1,-2){A}\uput[d](1,-2){B}
$$\meinecaption{1}{$A\cup B$} \label{fig:a_union_b}
\end{minipage}
\hfill
\begin{minipage}{3cm}
$$
\psset{unit=1pc} \pscircle(-1,0){2}\pscircle(1,0){2}
\uput[d](-1,-2){A}\uput[d](1,-2){B}
\pscustom[fillstyle=solid,fillcolor=green]{\psarc(1,0){2}{120}{240}\psarc(-1,0){2}{270}{60}}
$$\meinecaption{1}{$A\cap B$} \label{fig:a_intersection_b}
\end{minipage}
\hfill
\begin{minipage}{3cm}
$$
\psset{unit=1pc}
\pscircle[fillstyle=solid,fillcolor=green](-1,0){2}\pscircle(1,0){2}
\uput[d](-1,-2){A}\uput[d](1,-2){B}
\pscustom[fillstyle=solid,fillcolor=white]{\psarc(1,0){2}{120}{240}\psarc(-1,0){2}{270}{60}}
$$\meinecaption{1}{$A\setminus B$} \label{fig:a_minus_b}
\end{minipage}
\hfill
\begin{minipage}{3cm}
$$\psset{unit=1pc}
\pscustom[fillstyle=solid,fillcolor=white]{\psline(-3.5,2.5)(-3.5,-2.5)(3.5,-2.5)(3.5,2.5)(-3.5,2.5)}
\uput[r](1.5,1.5){A^c}
\pscircle[fillstyle=solid,fillcolor=green](0,0){2} \uput[u](0,0){A}
$$\meinecaption{1}{$A^c$} \label{fig:a_complement}
\end{minipage}
\end{figure}
\begin{exa}Let $\Omega = \{0,1,2,3,4,5,6,7,8,9\}$ be the universal set of the decimal digits and let
$A = \{0,2,4,6,8\}\subseteq \Omega$ be the set of even digits. Then
$A^c = \{1,3,5,7,9\}$ is the set of odd digits.
\end{exa}

\vspace*{4cm}
\begin{figure}[h]
\centering
\begin{minipage}{8cm}
$$
 \psset{unit=2.2pc}
 \pscustom[fillstyle=solid,fillcolor=blue]{\psarc(1,0){2}{120}{240}
\psarcn(-1,0){2}{300}{60}
 }
 %AB'
\pscustom[fillstyle=solid,fillcolor=yellow]{
\psarc(-1,0){2}{300}{60}\psarcn(1,0){2}{120}{240}}%A'B
\pscustom[fillstyle=solid,fillcolor=white]{
\psarc(1,0){2}{120}{240}\psarc(-1,0){2}{300}{60}
 }%AB
 {\tiny
 \uput[d](0,1){ A\cap B} \uput[dl](-1,0){A\cap
B^c}\uput[r](1,-.5){A^c\cap B}\uput[ur](2,2){(A\cup B)^c}}
\uput[d](-1,-2){A} \uput[d](1,-2){B}
$$\meinecaption{2}{Two
sets.}\label{fig:two_sets}
\end{minipage}\hfill
\begin{minipage}{8cm}
$$
 \psset{unit=2.2pc}
 \pscustom[fillstyle=solid,fillcolor=blue]{\psarc(1,0){2}{180}{240}
\psarcn(-1,0){2}{300}{120} \psarc(0,1.7320508076){2}{180}{240}
 }
 %AB'C'
\pscustom[fillstyle=solid,fillcolor=yellow]{\psarc(1,0){2}{240}{60}
\psarcn(0,1.7320508076){2}{0}{300} \psarc(-1,0){2}{300}{360}}%A'BC'
\pscustom[fillstyle=solid,fillcolor=green]{
\psarc(1,0){2}{180}{240}\psarc(-1,0){2}{300}{360}
 \psarcn(0,1.7320508076){2}{300}{240}}%ABC'
 \pscustom[fillstyle=solid,fillcolor=white]{\psarc(0,1.7320508076){2}{240}{300}\psarc(-1,0){2}{0}{60}
 \psarc(1,0){2}{120}{180}}%ABC
\pscustom[fillstyle=solid,fillcolor=violet]{\psarcn(1,0){2}{180}{120}
\psarc(-1,0){2}{60}{120} \psarc(0,1.7320508076){2}{180}{240}
 }%AB'C
 \pscustom[fillstyle=solid,fillcolor=red]{\psarc(-1,0){2}{60}{120}
 \psarcn(0,1.7320508076){2}{180}{0}\psarc(1,0){2}{60}{120}
 }%A'B'C
 \pscustom[fillstyle=solid,fillcolor=orange]{\psarcn(-1,0){2}{60}{0}
 \psarc(0,1.7320508076){2}{300}{360}\psarc(1,0){2}{60}{120}
 }
 %A'BC
{\tiny
 \uput[u](.1,2){
A^c\cap B^c\cap C} \uput[r](.5,1.5){ A^c\cap B\cap C} \uput[d](0,1){
A\cap B\cap C} \uput[dl](-1,0){A\cap B^c\cap C^c}
\uput[l](-.5,1.5){A\cap B^c\cap C} \uput[d](0,-.5){A\cap B \cap
C^c}\uput[r](1,-.5){A^c\cap B \cap C^c} \uput[ur](2,2){(A\cup B \cup
C)^c}} \uput[d](-1,-2){A} \uput[d](1,-2){B} \uput[u](0,4){C}
$$\meinecaption{2}{Three
sets.}\label{fig:three_sets}
\end{minipage}
\end{figure}





 The various intersecting regions for two and three sets can be
seen in figures \ref{fig:two_sets} and \ref{fig:three_sets}.


\begin{exa}
Let $A = \{1, 2, 3, 4, 5, 6\}$, and $B = \{1, 3, 5, 7, 9\}$. Then
$$A\cup B = \{1, 2, 3, 4, 5, 6, 7, 9\},\qquad A\cap B = \{1, 3,
5\},\qquad A\setminus B = \{2, 4, 6\},\qquad B\setminus A = \{7,
9\}.$$\end{exa}



\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small



\begin{pro} \label{exa:subsets4element}
Find all the subsets of $\{a, b, c, d\}$.
\begin{answer} The idea is the following. We use the result of example
\ref{exa:subsets3element}. Now, a subset of $\{a, b, c, d\}$ either
contains $d$ or it does not. This means that $\{a, b, c, d\}$ will
have $2 \cdot  8 = 16$ subsets. Since the subsets of $\{a, b, c\}$
do not contain $d$, we simply list all the subsets of $\{a, b, c\}$
and then to each one of them  we add $d$. This gives

\begin{multicols}{3} \columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small\begin{enumerate} \item[] $S_1  =
\varnothing$
\item[] $ S_2  =  \{a\} $
\item []$S_3  =  \{b\}$ \item[] $S_4  =  \{c\}$ \item[] $S_5  =  \{a,
b\}$
\item[]
$S_6  =  \{b, c\}$ \item[] $S_7  =  \{c, a\}$ \item[] $S_8  =  \{a,
b, c\} $

$S_9  = \{d\}$
\item[] $ S_{10}  =  \{a, d\} $
\item []$S_{11}  =  \{b, d\}$ \item[] $S_{12}  =  \{c, d\}$ \item[] $S_{13}  =  \{a,
b, d\}$
\item[]
$S_{14}  =  \{b, c, d\}$ \item[] $S_{15}  =  \{c, a, d\}$ \item[]
$S_{16} = \{a, b, c, d\} $
\end{enumerate}  \end{multicols}
\end{answer}
\end{pro}

\end{multicols}

\chapter{Inclusion-Exclusion}
\begin{df}
The {\em cardinality} \index{cardinality} of a set $A$, denoted by
$\card{A}$ is the number of elements that it has. If the set $X$ has
infinitely many elements, we write $\card{X} = \infty$.
\end{df}
\begin{exa}
If $A = \{0,1\}$ then $\card{A} = 2$. Also, $\card{\naturals} =
\infty$.
\end{exa}


We are now interested in finding the cardinality of a union of sets.
The following examples introduce the method of {\em
inclusion-exclusion}, which allows us to find the cardinality of a
union.


\begin{exa}
Of  $40$ people, $28$ smoke and  $16$ chew tobacco. It is also known
that  $10$ both smoke and chew. How many among the $40$ neither
smoke nor chew? \label{exa:incl_excl_1}\end{exa} \begin{solu}  We
fill up the Venn diagram in figure \ref{fig:incl_excl_1} as follows.
Since $\card{A \cap B} = 10$, we put a $10$ in the intersection.
Then we put a $28 - 10 = 18$ in the part that $A$ does not overlap
$B$ and a $16 - 10 = 6$ in the part of $B$ that does not overlap
$A$. We have accounted for $10 + 18 + 6 = 34$ people that are in at
least one of the set. The remaining $40 - 34 = 6$ are outside these
sets.
\end{solu}

 \vspace*{2cm}
\begin{figure}[h]
\centering
\begin{minipage}{7cm}
$$
\psset{unit=1.5pc} \pscircle(-1,0){2} \pscircle(1,0){2}
\rput(0,0){\tiny 10} \rput(-2, 0){\tiny 18} \rput(2, 0){\tiny 6}
\rput(2.5, 2.5){\tiny 6} \rput(-3, 1.5){\tiny A} \rput(3, 1.5){\tiny
B} $$\meinecaption{1}{Example
\ref{exa:incl_excl_1}.}\label{fig:incl_excl_1}
\end{minipage}
\begin{minipage}{7cm}
$$\psset{unit=1.5pc} \pscircle(-1,0){2} \pscircle(1,0){2}
\pscircle(0,1.41421356){2} \rput(2,0){\tiny 3} \rput(-2, 0){\tiny 1}
\rput(0,0.707106781){\tiny 3} \rput(0, 2.3){\tiny 1}
\rput(0,-0.8){\tiny 2} \rput(-1.43,1.43){\tiny 2}
\rput(1.43,1.43){\tiny 4} \rput(-3.8, 0){\tiny A} \rput(3.8,
0){\tiny B} \rput(0, 3.8){\tiny C}$$ \meinecaption{1}{Example
\ref{exa:3set_incl_excl}.} \label{fig:3set_incl_excl_exa}
\end{minipage}
\end{figure}
\begin{exa}
In a group of $30$ people, $8$ speak English, $12$ speak Spanish and
$10$ speak French. It is known that $5$ speak English and Spanish,
$5$ Spanish and French, and $7$ English and French. The number of
people speaking all three languages is $3$. How many do not speak
any of these languages?

\label{exa:3set_incl_excl}\end{exa} \begin{solu}  Let $A$ be the set
of all English speakers, $B$ the set of Spanish speakers and $C$ the
set of French speakers in our group. We fill-up the  Venn diagram in
figure \ref{fig:3set_incl_excl_exa} successively. In the
intersection of all three we put 8. In the region common to $A$ and
$B$ which is not filled up we put $5 - 2 = 3$. In the region common
to $A$ and $C$ which is not already filled up we put $5 - 3 = 2$. In
the region common to $B$ and $C$ which is not already filled up, we
put $7 - 3 = 4.$ In the remaining part of $A$ we put $8 - 2 - 3 - 2
= 1,$ in the remaining part of $B$ we put $12 - 4 - 3 - 2 = 3$, and
in the remaining part of $C$ we put $10 - 2 - 3 - 4 = 1$. Each of
the mutually disjoint regions comprise a total of $1 + 2 + 3 + 4 + 1
+ 2 + 3 = 16$ persons. Those outside these three sets are then $30 -
16 = 14.$ \end{solu}








\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

          \begin{pro}
A survey of a group's viewing habits over the last year revealed the
following information:
\begin{dingautolist}{202}
\item  $28\%$ watched gymnastics \item  $29\%$ watched baseball
\item $19\%$ watched soccer \item  $14\%$ watched gymnastics and
baseball \item $12\%$ watched baseball and soccer \item  $10\%$
watched gymnastics and soccer \item  $8\%$ watched all three sports.
\end{dingautolist}
Calculate the percentage of the group that watched none of the three
sports during the last year.
\begin{answer}
$52\%$
\end{answer}
    \end{pro}
\begin{pro}
In a group of $100$ camels, $46$ eat wheat, $57$ eat barley, and
$10$ eat neither. How many camels eat both wheat and
barley?
\begin{answer} Let $A$ be the set of camels eating wheat, and
$|A|$ its number, and $B$ be the set of camels eating barley, and
$|B|$ its number. Then
$$90 = 100 - 10 = |A\cup B| = |A| + |B| - |A\cap B| = 46 + 57 -
|A\cap B| = 103  - |A\cap B|,$$whence $|A\cap B| = 13.$
\end{answer}
\end{pro}
\begin{pro} At {\em Medieval High} there are forty students. Amongst
them, fourteen like Mathematics, sixteen like theology, and eleven
like alchemy. It is also known that seven like Mathematics and
theology, eight like theology and alchemy and five like Mathematics
and alchemy. All three subjects are favoured by four students. How
many students like neither Mathematics, nor theology, nor alchemy?
\begin{answer}  $15$
\end{answer}
\end{pro}
\begin{pro} Rusty has $20$ marbles of different colours: black, blue, green,  and yellow. Seventeen of the marbles are not
green, five are black, and $12$ are not yellow. How many blue
marbles does he have?
\end{pro}

\end{multicols}

\part{Symbols}

\chapter{The Necessity of Symbols\asterisk}
 \QUOTEME{If a man
who cannot count finds a four-leaf clover, is he entitled to
happiness?}{Stanislaw Jerzey}

\begin{center}
    \fcolorbox{blue}{yellow}{
    \begin{minipage}{.90\linewidth}
    \noindent\textcolor{red}{\textbf{Why bother?}} Symbols economise
    thought. For example, if we had \$12\ 345 in our bank account,
    one way to keep track of this money is to have 12345 pebbles,
    one for each dollar. But this is silly! We know of a perfectly
    efficient notation to represent this, namely with numerals.
\end{minipage}}
    \end{center}


\dropping{2}{T}he notions of {\em counting} and {\em number} are as
old as civilisation. Primitive men felt the need to count their
herds.\footnote{The word {\em men} is used here in its epicene
sense. The author has no patience for the politically correct humbug
of ``inclusive language.''} Once the idea of a {\em quantity} or
{\em number} was formed, symbols were introduced in order to
communicate these ideas. A {\em numeral} is a symbol used to
represent a number.

\bigskip

Early civilisations, like the Egyptians, used a variety of symbols
to represent the abstraction of number. A number was represented by
the aggregation of symbols, with no symbol appearing more than nine
times. Table \ref{tab:egyptian-numerals} gives the equivalence of
the symbols to numbers.


\begin{table}[h]
\centering
\begin{tabular}[t]{|c|c|c|c|c|c|c|}
\hline
\Huge{\egmil{1}}&\Huge{\eghuntho{1}}&\Huge{\egtentho{1}}&\Huge{\egtho{1}}&\Huge{\eghun{1}}&\Huge{\egten{1}}&\Huge{\egone{1}}\\
\hline
1,000,000&100,000&10,000&1,000&100&10&1\\
\hline
\end{tabular}
\meinecaption{1}{Egyptian Numerals.} \label{tab:egyptian-numerals}
\end{table}


The sign for {\em one} represents a vertical staff, for {\em one
thousand} a lotus flower, for {\em ten thousand} a pointing finger,
for a {\em hundred thousand} a burbot, and for a {\em million} a man
in astonishment. Egyptologists are not clear about the meaning of
the other symbols.
\begin{exa}In ancient Egyptian,  $276$ is
\mbox{\egyptify{0}{0}{0}{0}{2}{7}{6}} and $4622$ is
\mbox{\egyptify{0}{0}{0}{4}{6}{2}{2}}. The examples of $276$ and
$4622$ actually occur on a stone carving from Karnak, dating from
around $1500$ BC, now displayed in the Louvre in Paris.
\end{exa}





Perhaps more familiar to you is the system of Roman numerals. These
numbers were first developed by the Etruscans, passed to the Romans,
and then modified in the Middle Ages to produce the system we use
today. It is based on certain letters which are given values as
numerals. For large numbers (five thousand and above), a bar is
placed above a base numeral to indicate multiplication by $1000$.



\begin{table}[h]
\centering
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline I & V & X & L & C & D & M &   $\overline{V}$ & $\overline{X}$
& $\overline{L}$ & $\overline{C}$ & $\overline{D}$ & $\overline{M}$
 \\ \hline
 1 & 5 &  10 & 50 & 100 & 500 & 1000 &
 5,000 &  10,000 & 50,000 & 100,000 & 500,000 & 1,000,000   \\
 \hline
\end{tabular}
\end{table}



\bigskip


The rules for using Roman numerals have evolved across centuries.
For example, it was common to write IIII for $4$, rather than the IV
we use today. At Harvard Medical School's Library one reads
MDCCCCIIII for 1904, instead of \Romannum{1904}. The commonly used
modern rules are the following:

\begin{itemize}
\item Values are obtained by accretion of symbols.
\item A symbol may appear at most three consecutive times.
\item If a preexisting symbol exists for a number, do not aggregate
symbols of lower value to represent the number.
\item Preceding a higher-value symbol by a lower-value symbol
indicates subtraction.
\end{itemize}




\begin{exa}
Using the rule above, the first ten digits are
$$\begin{array}{|l|l|l|l|l|l|l|l|l|l|}
\hline 1=\Romannum{1} &   2=\Romannum{2} &
 3=\Romannum{3} &   4=\Romannum{4} &
 5=\Romannum{5} &   6=\Romannum{6} &
7=\Romannum{7} &   8=\Romannum{8} &
 9=\Romannum{9} &   10=\Romannum{10} \\
 \hline  \end{array}  $$
We start with the symbol of least value I, in order to produce I for
$1$. By accretion we produce II for  $2$. Again, by accretion we
produce III for $3$. Next is $4$. Since we cannot use the I more
than three consecutive times, we subtract from the next highest
valued symbol V, producing IV for $4$. We have a symbol for V. By
accretion we produce VI for $6$. Again by accretion, we produce VII
for $7$. We continue, producing VIII for $8$. Now, since we cannot
use three consecutive I's, we subtract from the next highest symbol,
producing IX for $9$. Finally, we have the symbol X for $10$.


Analogously we have

$$\begin{array}{|l|l|l|l|l|l|l|l|l|l|}
\hline 10=\Romannum{10} &   20=\Romannum{20} & 30=\Romannum{30} &
40=\Romannum{40} &
 50=\Romannum{50} &   60=\Romannum{60} &
 70=\Romannum{70} &   80=\Romannum{80} &
90=\Romannum{90} &   100=\Romannum{100} \\
 \hline  \end{array}  $$


and

$$\begin{array}{|l|l|l|l|l|l|l|l|l|l|}
\hline 100=\Romannum{100} &   200=\Romannum{200} &
 300=\Romannum{300} &   400=\Romannum{400} &
 500=\Romannum{500} &   60=\Romannum{600} &
  700=\Romannum{700} &   800=\Romannum{800} &
 900=\Romannum{900} &   1000=\Romannum{1000} \\
 \hline  \end{array}  $$
\end{exa}

\begin{exa}
Write, with explanation, $1989$ in Roman numerals.
\end{exa}
\begin{solu}
Write
$$1989 =1000 + 900 +80 +9.  $$
We first write $9$, which is IX. We then write $80$. This is LXXX.
We now write $900$, which is CM. Finally, we write $1000$, which is
M. We now concatenate these strings resulting in
$$1989 =1000 + 900 +80 +9=M + CM + LXXX + IX = \Romannum{1989}.  $$

\end{solu}


\begin{exa}
In Roman numerals, we have:
$$\begin{array}{|l|l|}
\hline 1215=\Romannum{1215} &   1492=\Romannum{1492} \\
\hline 1776=\Romannum{1776} &   1787=\Romannum{1787} \\
\hline 1863=\Romannum{1863} &   1966=\Romannum{1966} \\
 \hline 2000=\Romannum{2000} &   2001=\Romannum{2001} \\
\hline 2888=\Romannum{2888} &   2889=\Romannum{2889} \\
 \hline  \end{array}  $$
\end{exa}

\bigskip

Using letters of the alphabet as numbers can have amusing
consequences. During the Catholic-Protestant wars in Europe,
Protestants claimed that the Pope was the antichrist. To support
this claim, the Protestants came up with the hoax that  the Pope's
tiara bears the inscription {\bf VICARIVS FILII DEI}, and if one
counts the Roman numerals (giving the value of $0$ to the symbols
with no value), one obtains $666$. The Pope's tiara, of course, does
not bear that inscription, nor is this one of the Pope's titles.
Again, using numeric values in vogue for the Greek letters around
the time of the Apostle St. John, one obtains that the Emperor
Nero's name gave $666$.


\bigskip

The Egyptian and the Roman numeral are but two systems developed to
describe numbers. Other ancient cultures developed their own
systems. For example, the ancient Hebrews attached a numerical
values to the letters of the alphabet, having the following
equivalences. Numbers were represented by accretion of symbols. A
curiosity is that the symbols for $15$ and $16$, instead of being
written, as our decimal thinking dictates, $15=10+5=$\ \<yh> and
$16=10+6=$\ \<yw>, were written $9+6=$\<.tw> and $9+7=$ \<.tz>. This
is to avoid writing part of the name of G-d, which is not to be
uttered.

\bigskip

\begin{table}[h]
\centering \footnotesize
\begin{tabular}{|l||c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline  Hebrew Letter & Alef & Bet & Gimel & Dalet & Ha & Vav &
Zayin & Het & Tet & Yod  &  & \\
\hline  Hebrew Symbol &\<'> & \<b> & \<g> & \<d> & \<h>& \<w>& \<z>&
\<.h>&\<.t> &\<y>  & & \\
\hline  Value &1& 2 & 3 & 4 & 5& 6& 7& 8&9 &10  &  &\\ \hline Hebrew
Letter & Kaf & Lamed & Mem & Nun  & Samekh & Ayin  &
Pe & Tsadi & Qof & Resh & Shin  & Tav \\
\hline Hebrew Symbol &\<k> & \<l>& \<m>&\<n> & \<s>&\<`> & \<p>&
\<t>& \<q>& \<r>&\</s> & \<t>\\ \hline Value & 20 & 30& 40&50 & 60& 70 & 80& 90 & 100& 200 &  300 & 400\\
\hline
\end{tabular}
\meinecaption{1}{Hebrew Numerals.} \label{tab:hebrew-numerals}
\end{table}








\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Given that $\spadesuit=\diamondsuit\diamondsuit\diamondsuit$ and
that $\Gamma=\spadesuit\spadesuit$, how many $\diamondsuit$'s are
needed to write $\Gamma\Gamma\Gamma\spadesuit$ as a string of
$\diamondsuit$'s?
\begin{answer}
$21$
\end{answer}
\end{pro}
\begin{pro}
Use Egyptian numerals to write $1$, $12$, and $123$. Are there any
advantages gained by using Egyptian numerals? Any disadvantages?
Discuss.
\begin{answer}
$1=\egyptify{0}{0}{0}{0}{0}{0}{1}$,
\mbox{12=\egyptify{0}{0}{0}{0}{0}{1}{2}}, and
$123=\egyptify{0}{0}{0}{0}{1}{2}{3}$. Egyptian notation has the
advantage that you can write the symbols in any order and still get
the same number. With our notation, $321$ and $123$, for example,
have different meanings. On the other hand, performing arithmetic
operations like adding, subtracting, multiplying, and dividing,
seems more cumbersome. Moreover, for very large numbers, we would
have to use too many symbols. For example, if we say that the
current population of the USA is around 300 million, we would need
to write $\egyptify{1}{0}{0}{0}{0}{0}{0}$ three hundred times, which
is very impractical.
\end{answer}
\end{pro}
\begin{pro}
Use Roman numerals to write $1914$, $1917$,  $1939$,  $1963$, and
$1989$.
\begin{answer}
1914=\Romannum{1914},  1917=\Romannum{1917}, 1939=\Romannum{1939},
1963=\Romannum{1963},  1989=\Romannum{1989}.
\end{answer}
\end{pro}
\begin{pro}
Convert $\overline{V}\Romannum{321}$ to a  Hindu-Arabic numeral.
\begin{answer}
$5321$
\end{answer}
\end{pro}
\begin{pro}
Convert $\overline{V}MMM\Romannum{888}$ to a  Hindu-Arabic numeral.
\begin{answer}
$8888$
\end{answer}
\end{pro}
\begin{pro}
A question worth pondering is how did the ancient Egyptians and the
ancient Romans represent fractions. Do an internet or a library
search on this topic, consulting history of Mathematics books and
articles.
\end{pro}
\begin{pro}
The text tacitly points to a distinction between {\em numerals} and
{\em numbers}. Explain.
\end{pro}

\end{multicols}
\section*{References}
\cite{Eves} is an old, but reliable reference on ancient numerical
systems. \cite{Hof} contains interesting stories about the number of
the beast. \cite{Nic} is a very accessible Hebrew grammar for
beginners.

\begin{description}
\bibitem[Eves]{Eves} Howard EVES, ``An Introduction to the History of Mathematics''.
\bibitem[Hof]{Hof} Paul HOFFMAN, ``Archimedes Revenge,''
\bibitem[ManSte]{ManSte} Richard MANKIEWICZ and Ian STEWART , ``The Story of Mathematics''.
\bibitem[McL]{McL} John MCLEISH , ``The Story of Numbers: How Mathematics Has Shaped Civilization''.
\bibitem[Nic]{Nic} Sarah NICHOLSEN, ``Teach Yourself Biblical
Hebrew''.
\end{description}

\chapter{Positional Notation}
\QUOTEME { Numbers are the beginning and end of thinking. With
thoughts were numbers born. Beyond numbers thought does not
reach.}{Magnus Gustaf MITTAG-LEFFLER}
\begin{center}
    \fcolorbox{blue}{yellow}{
    \begin{minipage}{.90\linewidth}
    \noindent\textcolor{red}{\textbf{Why bother?}} In the preceding
    lecture you learned how different cultures represented the same
    idea of {\em number}. But try multiplying ,  say, $MCMLXI$ times
    $LXXVII$ in Roman numerals. Many ancients did! You will discover
    that it is painful and that our modern decimal notation is more
    suitable to carry out such computations.
\end{minipage}}
    \end{center}

\dropping{2}{O}ur system of writing numbers---the base $10$ or
decimal system of notation---comes from India, where it was
well-established by the VIII century. Around the same time, it was
independently discovered in China. It was introduced to Europe in
the XIII century through Arab merchants, and thus we call them {\em
Hindu-Arabic numbers.} Before their introduction to Europe, European
merchants kept their accounting records using Roman numerals.
Performing arithmetic in Roman numerals was a daunting task, and
this kept civilisation from advancing.



 We start with two symbols, $0$ and $1$, and an
operation $+$, adjoining the elements
$$1+1,\quad  1+1+1,\quad 1+1+1+1,\quad 1+1+1+1+1,\quad \ldots .   $$
Observe that this set is {\em infinite} and {\em ordered}, that is,
you can compare any two elements and tell whether one is larger than
the other. We define the symbols
$$2=1+1,\quad  3=1+1+1,\quad 4=1+1+1+1,\quad 5=1+1+1+1+1,\quad  6=1+1+1+1+1+1,$$
$$7=1+1+1+1+1+1+1,\quad 8=1+1+1+1+1+1+1+1,\quad  9=1+1+1+1+1+1+1+1+1.
$$
Beyond $9$ we reuse these symbols by also attaching a meaning to
their place. Thus
$$ \begin{array}{lll}10 & = &  1+1+1+1+1+1+1+1+1+1, \\
 11 & = &  \psovalbox{1+1+1+1+1+1+1+1+1+1}+1, \\
 12 & = & \psovalbox{1+1+1+1+1+1+1+1+1+1}+1+1, \\
  & \vdots & \\
 21 & = & \psovalbox{1+1+1+1+1+1+1+1+1+1}+ \psovalbox{1+1+1+1+1+1+1+1+1+1}+1, \\
  & \vdots & \\
\end{array}$$
\begin{df}
A {\em positional notation} or {\em place-value notation system }is
a numeral system in which each position is related to the next by a
constant multiplier of that numeral system. Each position is
represented by a limited set of symbols. The resultant value of each
position is the value of its symbol or symbols multiplied by a power
of the base.
\end{df}


To form any number, we use the ten
digits:\index{digits}\footnote{From the Latin {\em digitum}, meaning
{\em finger}.} $0,1,2,3,4,5,6,7,8,9$. We can represent numbers as
large as we want by using their
 position to represent different values (place-value).\index{place-value} The first digit
right to left corresponds to the ones place,
 the next digit (again, right to left) corresponds to the tens place, the next digit to the hundreds
 place, and so on. We use  addition to explain the place-value.
Positional notation was such a great idea, and so greatly advanced
mathematical thought, that the great XIX century German
mathematician K. F. Gauss (1777-1855) wondered why the greatest
mathematician of antiquity,  Archimedes of Syracuse (ca 287 BC- ca
212 BC), had not come up with it. In fact, between the XII and the
XVI centuries  in Europe, there was a great divide between the {\em
algorists}, the ones who advocated the new system of positional
notation brought to them by the Arabs from India, doing computation
by writing the operations, and the {\em abacists}, who used the
abacus to perform the arithmetic operations.


 \begin{exa}Consider the number $45,321$. Observe that $ 45,321= 40,000 +5000 +300 +20 +1. $ The place-values are explained below. \vspace{1cm}
$$ \rput(-2,0){\psframebox{\scaletoheight{1cm}{4}}} \rput(-1,0){\psframebox{\scaletoheight{1cm}{5}}}\rput(0,-.25){\psframebox{\scaletoheight{.25cm}{,}}}
\rput(1,0){\psframebox{\scaletoheight{1cm}{3}}}\rput(2,0){\psframebox{\scaletoheight{1cm}{2}}}\rput(3,0){\psframebox{\scaletoheight{1cm}{1}}}
\uput[d]{-90}(-2,-.5){\mathrm{
ten-thousands}}\uput[d]{-90}(-1,-.5){\mathrm{thousands}}\uput[d]{-90}(1,-.5){\mathrm{hundreds}}
\uput[d]{-90}(2,-.5){\mathrm{tens}}\uput[d]{-90}(3,-.5){\mathrm{ones}}$$
\vspace{2cm}
\end{exa}



\bigskip

We now introduce some notation that will simplify our computations
with positional notation.



\begin{df}[Exponentiation]\index{exponentiation}
If $n$ is a  whole number greater than or equal to $1$ then the {\em
$n$-th power of $a$} is defined by
$$a^n = \underbrace{a\cdot a\cdots a}_{n  \ \mathrm{times}}  $$
Here, $a$ is the {\em base}, and $n$ is the {\em
exponent}.\index{base}\index{exponent}. If $a$ is any number
different from $0$ then we define
$$a^0 = 1.$$ We do not attach any meaning to $0^0$.\footnote{Much to the chagrin of
logicians and other spawns of Satan.}
\end{df}




Observe that it is very easy to compute powers of $10$. In fact
$$10^n = 1\underbrace{0 \ldots 0}_{n\ \mathrm{zeroes}},   $$that is,
$10^n$ is a $1$ trailed by $n$ zeroes. Thus $$10^0=1, \quad 10^1=10,
\quad 10^2 =100, \quad 10^3=1000,\ \ldots,
$$
 A {\em googol} is $10^{100}$, a term that was invented by the then nine-year-old  Milton
Sirotta in 1920. The name of the search engine {\em Google} is a
misspelling of googol.\index{googol}\index{Google}


\bigskip

Incidentally, here are the names of  powers of ten and million and
over:
\begin{table}[h]
\centering {\tiny
\begin{tabular}{|l|l|l|l|l|l|}
\hline $10^6$ & million & $10^9$ & billion & $10^{12}$ & trillion \\
\hline $10^{15}$ & quadrillion & $10^{18}$ & quintillion & $10^{21}$ & sextillion \\
\hline $10^{24}$ & septillion & $10^{27}$ & octillion & $10^{30}$ & nonillion \\
\hline $10^{33}$ & decillion & $10^{36}$ & undecillion & $10^{39}$ & duodecillion \\
\hline $10^{42}$ & tredecillion & $10^{45}$ & quattuordecillion & $10^{48}$ & quindecillion \\
\hline $10^{51}$ & sexdecillion & $10^{54}$ & septendecillion & $10^{57}$ & octodecillion \\
\hline $10^{60}$ & novendecillion & $10^{63}$ & vigintillion & $10^{100}$ & googol \\
\hline
\end{tabular}
} \meinecaption{1}{Names of some powers of ten.}
\label{tab:large-powers-10}

\end{table}









\begin{exa}
Powers of $2$ permeate computer culture. A {\em bit} is a binary
digit taking a value of either $0$ (electricity does not pass
through a  circuit) or $1$ (electricity passes through a circuit).
\index{bit} We have,
$$\begin{array}{lcl}2^1 = 2&\hspace{1cm} & 2^6=64 \\
2^2 = 4,&\hspace{1cm}  & 2^7=128 \\
2^3 = 8,&\hspace{1cm}  & 2^8=256 \\
2^4 = 16,&\hspace{1cm}  & 2^9=512 \\
2^5 = 32,&\hspace{1cm}  & 2^{10}=1024 \\
    \end{array}  $$
    Since $2^{10}\approx 1000$,\footnote{The symbol $\approx$ is read ``approximately.''} we call  $2^{10}$ a {\em
    kilobit}.\footnote{From the Greek {\em kilo}, meaning {\em thousand}.}
\end{exa}
\begin{rem}Notice that $2^3=8$ and $3^2=9$ are consecutive powers.
A 150 year old problem, called {\em Catalan's Conjecture} asserted
that these were the only strictly positive consecutive powers. This
conjecture was proved by the  number theorist Preda Mihailescu on 18
April 2002. This is one more example that not ``everything'' has
been discovered in Mathematics, that research still goes on today.
Again, notice that $a^b$ {\bf is not} $ab$. Thus $2^3 = (2)(2)(2) =
8$, and {\bf not} $(2)(3) = 6$.
\end{rem}


Returning to the topic of this lecture, we see that we can develop
any integer in decimal notation as the sum of powers of $10$. For
example,
$$45321=4\cdot 10^4 +5\cdot 10^3+3\cdot 10^2 +2\cdot 10 +1.  $$
We notice than in our system of notation, we have ten symbols and
every number can be expressed uniquely as sum of a multiple of one
of these symbols and a power of ten.



\bigskip

Suppose now that we only had two symbols: $0$ and $1$. Counting in
this system---called the {\em binary} system---takes the following
form:
$$ \begin{array}{|l|l|l|l|l|l|l|l|l|l|} \hline
1 = 1_2 & 2 = 10_2 & 3 = 11_2 & 4 = 100_2 & 5 = 101_2 & 6 = 110_2 &
7 = 111_2 & 8 = 1000_2 & 9 = 1001_2 & 10 = 1010_2 \\ \hline
 \end{array}   $$




If instead we use the the five symbols: $0$, $1$, $2$, $3$ and $4$,
counting now becomes.
$$ \begin{array}{|l|l|l|l|l|l|l|l|l|l|}
\hline 1 = 1_5 & 2 = 2_5 & 3 = 3_5 & 4 = 4_5 & 5 = 10_5 & 6 = 11_5 &
7 =
12_5 & 8 = 13_5 & 9 = 14_5 & 10 = 20_5 \\
\hline
 \end{array}   $$




Now that we know how to count in these systems, let us examine how
to convert a number represented in them to a decimal number.


\begin{exa}
Which (decimal) number is represented by the binary number
$11110100001001000000_2$?
\end{exa}
\begin{solu} The above counting method would not be very efficient
here. Notice that $11110100001001000000_2$ has twenty digits. We
have
$$\begin{array}{lll} 11110100001001000000_2 & = &
1\cdot 2^{19}+1\cdot 2^{18}+1\cdot 2^{17}+1\cdot 2^{16}+1\cdot
2^{14}+1\cdot 2^{9}+ 1\cdot 2^{6}\\
&  = &  524288+262144+131072 +65536+16384+512+64\\
& = & 1000000.
\end{array}$$
\end{solu}

\begin{exa}
Convert the octal (base $8$) number $777_8$ to a decimal.
\end{exa}
\begin{solu}
We have
$$777_8=7\cdot 8^2 + 7\cdot 8 + 7 =511.$$
\end{solu}

\begin{exa}
In the hexadecimal (base $16$) system, one uses the ten digits $0$,
$1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ and the letters $A$
($=10$ decimal), $B$ ($=11$ decimal), $C$ ($=12$ decimal), $D$
($=13$ decimal), $E$ ($=14$ decimal), and $F$ ($=15$ decimal). The
hexadecimal system is commonly used in the computer industry to
store the ``names'' of different shades of colour. For example, it
is used in HTML (the language in which most internet web pages are
written), where colour names are six-hexadecimal digits long.
Translate the hexadecimal numbers $AB21$ and $21AB$ into decimal
numbers.
\end{exa}
\begin{solu}
$$ AB21 = 10\cdot 16^3 + 11\cdot 16^2 + 2\cdot 16 + 1 =43809,  $$
and
$$ 21AB = 2\cdot 16^3 + 1\cdot 16^2 + 10\cdot 16 + 11 =8619. $$
\end{solu}

\bigskip

We now tackle the problem of converting a decimal into a number of a
different base. Consider the problem of converting $401$, say, to
base $3$. What is the largest power of $3$ that is smaller than
$401$? We quickly see that $$ 3^5=243 < 401<729=3^6.   $$By
division,
$$ 401=1\cdot 243+158=1\cdot 3^5+158. $$Now we do the same for the
remainder $158$, finding $$3^4=81<158<243=3^5.$$ As $150=1\cdot
81+77=1\cdot 3^4+77$, we deduce that
$$ 401=1\cdot 243+158=1\cdot 3^5+158= 1\cdot 3^5+1\cdot 3^4+77. $$
Now we do the same for the remainder $77$, finding
$$3^3=27<77<81=3^4.$$ As $77=2\cdot 27+23=2\cdot 3^3+23$, we
deduce that
$$ 401=1\cdot 243+158=1\cdot 3^5+158= 1\cdot 3^5+1\cdot 3^4+77= 1\cdot 3^5+1\cdot 3^4+2\cdot 3^3+23. $$
Now we do the same for the remainder $23$, finding
$$3^2=9<23<27=3^3.$$ As $23=2\cdot 9+5=2\cdot 3^2+5$, we
deduce that
$$ 401=1\cdot 243+158=1\cdot 3^5+158= 1\cdot 3^5+1\cdot 3^4+77= 1\cdot 3^5+1\cdot 3^4+2\cdot
3^3+23 = 1\cdot 3^5+1\cdot 3^4+2\cdot 3^3+2\cdot 3^2+5. $$ Finally,
we do the same for the remainder $5$, finding
$$3^1=3<5<9=3^2.$$ As $5=1\cdot 3+2$, we
deduce that
$$\begin{array}{lll} 401 & = & 1\cdot 243+158=1\cdot 3^5+158\\
& = &  1\cdot 3^5+1\cdot 3^4+77\\ & = &  1\cdot 3^5+1\cdot
3^4+2\cdot 3^3+23 \\ & = &  1\cdot 3^5+1\cdot 3^4+2\cdot 3^3+2\cdot
3^2+5\\
&  =  & 1\cdot 3^5+1\cdot 3^4+2\cdot 3^3+2\cdot 3^2+1\cdot
3+2\\
& = & 112212_3.
\end{array}$$


The above procedure may seem laborious, but we can actually display
it in a very convenient form that renders it trivial. The key is to
observe that finding the powers of $3$ that fit into $401$ is
equivalent to successively dividing $401$ by $3$.
 Thus we display
$$ \begin{array}{l|l|l}3 & 401 & 2\\
\hline 3 & 133 & 1 \\
\hline 3 & 44 & 2 \\
\hline 3 & 14 & 2 \\
\hline 3 & 4 & 1 \\
\hline 3 & 1 & 1
\end{array}\psline[linewidth=2pt,arrowscale=3]{->}(1,-2.5)(1,3)$$ Explanation:
Divide $401$ by $3$: we get a quotient of $133$ and remainder $1$.
Divide $133$ by $3$: we get a quotient of $44$ and a remainder of
$2$. Keep dividing until the quotient is smaller than $3$, then
stop. Read the successive remainders from the last to the first one
obtained.

\begin{exa}
Convert $1492$ to base $7$.
\end{exa}
\begin{solu}
We divide successively by $7$, noting each successive quotient in
the central column, and each successive remainder on the rightmost
column, obtaining
$$ \begin{array}{l|l|l}7 & 1492 & 1\\
\hline 7 & 213 & 3 \\
\hline 7 & 30 & 2 \\
\hline 7 & 4 & 4 \\
\end{array}\psline[linewidth=2pt,arrowscale=3]{->}(1,-2)(1,2)$$
We conclude that $1492=4231_7$. To check our result, we perform
$$4231_7 = 4\cdot 7^3 + 2\cdot 7^2 + 3\cdot 7 + 1 = 1372+98+21+1=1492.  $$
\end{solu}



Even though the decimal system dominates our arithmetic, this was
not always the case. The ancient Babylonians used a sexagesimal
(base $60$) system, the vestiges of which we can see in the fact
that a minute has $60$ seconds and an hour $60$ minutes. To
represent numbers is such a system one needs $60$ different symbols.







\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Give the value of the underlined digit.
\begin{multicols}{2}
\begin{enumerate}
\item $1\ 23\underline{4}$
\item $1\ 2\underline{3}4$
\item $1\ \underline{2}34$
\item $1\ 2\underline{3}4\ 056\ 789$
\end{enumerate}
\end{multicols}
\begin{answer}
\noindent
\begin{enumerate}
\item $4$ units
\item $3$ tens
\item $2$ hundreds
\item $3$ ten millions
\end{enumerate}

\end{answer}
\end{pro}

\begin{pro}
Light travels at approximately {\bf six hundred sixty nine million
six hundred thousand} miles per hour. Give a decimal numerical
representation for this quantity expressed in English.
\begin{answer}
$669600000$
\end{answer}
\end{pro}
\begin{pro}
Are $10$ and $010$ the same decimal number? Explain.
\begin{answer}
Yes. The leftmost zero in $010$ indicates that it has $0$
hundreds.\footnote{In Spanish, a derogatory term for an inferior,
insignificant person is {\em cero a la izquierda}, that is, a ``zero
to the left.''} Thus both numbers have $1$ ten and $0$ units and
nothing else, and are therefore equal.
\end{answer}
\end{pro}
\begin{pro}
Write in decimal numerals: ``twelve million twelve thousand twelve
hundred and twelve.''
\begin{answer}
This is
$$ 12,000,000+ 12,000+ 1,200+12= 12013212. $$
\end{answer}
\end{pro}
\begin{pro}
The names of the whole numbers from one to twelve are written down
in {\em lexicographical} order, that is, in the order in which they
appear in the dictionary. What is the fourth number on the list?
\begin{answer}
Four.
\end{answer}
\end{pro}
\begin{pro}
Convert $11111_2$ to decimal.
\begin{answer}
$11111_2 = 2^4+2^3+2^2+2+1 =31.$
\end{answer}
\end{pro}
\begin{pro}
Convert $11111_3$ to decimal.
\begin{answer}
$11111_3 = 3^4+3^3+3^2+3+1 =121.$
\end{answer}
\end{pro}
\begin{pro}
Convert $11111$ to binary.
\begin{answer}
$10101101100111_2.$
\end{answer}
\end{pro}
\begin{pro}
Convert $123_4$ to base $5$.
\begin{answer}
First we convert  $123_4$ to base $10$:
$$123=1\cdot 4^2+2\cdot 4+3=16+8+7=27.  $$
Now we convert $27$ to base $5$:
$$ \begin{array}{l|l|l}5 & 27 & 2\\
\hline 5 & 5 & 0 \\
\hline 5 & 1 & 1 \\
\end{array}$$
so $$123_4=27=102_5. $$
\end{answer}
\end{pro}
\begin{pro}
What integer follows $1266_7$ in base $7$?
\begin{answer}
$1300$
\end{answer}
\end{pro}
\begin{pro}
Convert $11111_3$ to decimal.
\begin{answer}
$11111_3 = 3^4+3^3+3^2+3+1 =121.$
\end{answer}
\end{pro}
\begin{pro}
In the Microsoft software {\em Excel}, columns are labelled by
letters. The first twenty six columns are labelled A through Z, the
twenty seventh is labelled AA, the twenty eighth AB, etc. What label
appears on the 2007th column? \vfill\hfill
\begin{answer}Writing in base $26$:
$2007= 2\cdot 26^2 + 25\cdot 26 + 5$. Thus the label is BYE.
\end{answer}

\end{pro}
\begin{pro}
How many (decimal) whole numbers are there between $1$ and $1000$ whose digits add up to $12$?
\begin{answer}
There is no such number with $1$ digit. With two digits there are $7$:
$$39, 48, 57, 66, 75, 84, 93.   $$With three digits there are $66$:
$$\begin{array}{lllllllllll}129 &  138 &  147 &  156 &  165 &  174 &  183 &  192 &
219 &  228 &  237\\   246 & 255 & 264 & 273 & 282 & 291 &
309 & 318 & 327 & 336 & 345\\   354 & 363 & 372 & 381 & 390 &
408 & 417 & 426 & 435 & 444 & 453\\ 462 & 471 & 480 &
507 & 516 & 525 & 534 & 543 & 552 & 561 & 570 \\
606 & 615 & 624 & 633 & 641  & 651 &  660 &
705 & 714 & 723 & 732 \\ 741 & 750 &
804 & 813 & 822 & 831 & 840 & 903 & 912 & 921 & 930. \end{array}$$
Thus there is a total of $7+66=73$ such numbers.
\end{answer}
\end{pro}
\begin{pro}
A {\em palindrome} \index{palindrome} is a natural number whose
decimal expansion is symmetric, that is, it reads the same backwards
as forwards. For example, $1$, $11$, $345543$, are all palindromes.
If the sequence of palindromes is written in ascending order
$$  1,2,3,4,5,6,7,8,9,11,22,\ldots , 99, 101, 111, 121, \ldots ,  $$
which palindrome follows $1991$?
\begin{answer}
$2002$
\end{answer}
\end{pro}
\begin{pro}A natural number  is called a {\em palindrome} if it is read
forwards as backwards, e.g., $1221$, $100010001$, etc., are palindromes.
The palindrome $10001$ is strictly between two other palindromes.
Which two?
\begin{answer}Between  $9999$ and $10101$.  \end{answer}
\end{pro}
\begin{pro}
Are there advantages and disadvantages of using the decimal system
over the binary system? Discuss.
\end{pro}
\begin{pro}
Without converting either number to decimal, decide which number is
larger: $1111_3$ or $1111_4$.
\begin{answer}
$1111_4$.
\end{answer}
\end{pro}
\begin{pro}
I am thinking of a whole number in the range from $0$ to $15$,
inclusive. You are allowed to ask me $4$ yes or no questions, to
which I will respond truthfully. Can you always guess my number in
at most four such questions?
\begin{answer}
Yes, you can always guess my number. Every whole number between $0$
and $15$  is one of the binary numbers
$$0000_2, \quad 0001_2, \quad 0010_2, \quad 0011_2, \quad \ldots , 1111_2.  $$
Thus you can ask me:
\begin{enumerate}
\item is the first binary digit (from left to right) of your number a zero? Yes or no.
\item is the second binary digit of your number a zero? Yes or no.
\item is the third binary digit of your number a zero? Yes or no.
\item is the fourth binary digit of your number a zero? Yes or no.
\end{enumerate}
In this way you completely identify my number. For example, if I am
thinking of $7=0111_2$, my answers to your questions are No, No, No,
and Yes, so you deduce my number.


Another equivalent way consists in successively halving the
intervals where my number may lie. So, again, if my number is $7$
you may ask:
\begin{enumerate}
\item Since halfway between $0$ and $15$ is $8$, you ask: Is your number smaller than $8$? I answer yes.
\item  Since halfway between $0$ and $7$ is $4$, you ask:  Is your number smaller than $4$? I answer no.
\item  Since halfway between $5$ and $7$ is $6$, you ask:  Is your number smaller than $6$? I answer no.
\item The number must be $7$.
\end{enumerate}


Again, with the method above, if my number were $13$:
\begin{enumerate}
\item  Since halfway between $0$ and $15$ is $8$, you ask: Is your number smaller than $8$? I answer
no.
\item  Since (almost) halfway between $8$ and $15$ is $11$, you ask: Is your number smaller than $11$? I answer
no.
\item  Since halfway between $11$ and $15$ is $13$, you ask: Is your number smaller than $13$? I answer no.
\item  Since halfway between $13$ and $15$ is $14$, you ask: Is your number smaller than $14$? I answer no.
\end{enumerate}
So you want a whole number that is smaller than $14$ but is not
smaller than $13$. The only option is $13$ and you have guessed my
number.
\end{answer}
\label{pro:guessing-numbers1}\end{pro}


\end{multicols}
\section*{References}
\cite{HaKn} is the book I learned Algebra from. Written in the XIX
century, I still think is unsurpassed in clarity of exposition,
careful selection of examples, and breadth of material. It covers
the material of this section more deeply than what was presented
here. Binary (base $2$), octal (base $8$) and hexadecimal (base
$16$) representations are essential in computer science, and
\cite{Lip} is a good reference for this.

\begin{description}
\bibitem[HaKn]{HaKn} H.S. HALL, and S.R. KNIGHT, ``Elementary Algebra for Schools ''.
\bibitem[Lip]{Lip} Seymour LIPSCHUTZ, ``Schaum's Outline of Essential Computer Mathematics''.

\end{description}





\chapter{Symbolical Expression}
\QUOTEME{Mathematicians are like Frenchmen: whatever you say to them
they translate into their own language and forthwith it is something
entirely different.}{GOETHE}\bigskip

\begin{center}
    \fcolorbox{blue}{yellow}{
    \begin{minipage}{.90\linewidth}
    \noindent\textcolor{red}{\textbf{Why bother?}} Part of the goal
    of this course is to prepare you for {\em Algebra}. In {\em
    Arithmetic} we deal with specific quantities, say, we compute
    $2+3$. The idea of Algebra is to generalise from Arithmetic, and
    in order to accomplish that level of abstraction, is necessary
    to have an efficient way of notation, so that we don't have to
    refer to specific instances.
\end{minipage}}
    \end{center}



\dropping{2}{I}n our arithmetical excursions it will be necessary to
introduce the abstraction of symbols. We will use {\em letters} to
denote arbitrary numbers. This will free us from a long enumeration
of cases. For example, suppose we notice that
$$ 1+0 = 1, \quad 2+0 = 2, \quad \dfrac{1}{2}+0 = \dfrac{1}{2}, \ldots
,$$etc. Since numbers are infinite, we could not possibly list all
cases. Here abstraction provides some economy of thought: we could
say that if $x$ is a number, then $$ x+0 = x, $$ with no necessity
of knowing what the arbitrary number $x$  is. Moreover, the
introduction of this symbolism will serve as prelude to Algebra.



\bigskip

We will normally associate the words {\em increase, increment,
augment,} etc., with addition. Thus if $x$ is an unknown number, the
expression ``a number increased by seven'' is translated into
symbols as $x+7$. We will later see that we could have written the
equivalent expression $7+x$.


\bigskip

We will normally associate the words {\em decrease, decrement, reduce, excess,
diminish, difference} etc., with subtraction. Thus if $x$ is an
unknown number, the expression ``a certain number decreased by
seven'' is translated into symbols as $x-7$. We will later see that
this differs from $7-x$, which is ``seven decreased by a certain
number.''



\bigskip

We will normally associate the word {\em product}  with
multiplication. Thus if $x$ is an unknown number, the expression
``the product of a certain number and seven'' is translated into
symbols as $7x$. Notice here that we use {\em juxtaposition} to
denote the multiplication of a letter and a number, that is, we do
not use the $ \times $ (times) symbol, or the $\cdot$ (central dot)
symbol. This will generally be the case, and hence the following are
all equivalent,
$$ 7x, \quad 7\times x, \quad 7(x), \quad (7)(x), \quad 7 \cdot  x.   $$
Notice again that a reason for {\em not} using $ \times $ when we use
letters is so that we do not confuse this symbol with the letter
$x$. We could have also have written $x7$, but this usage is just
plain weird.
\begin{rem}
We do need symbols in order to represent the product of two numbers.
Thus we write the product $5\cdot 6 = 5 \cdot  6 = (5)(6)= 30$ so
that we do not confuse this with the number $56$.
\end{rem}

\bigskip

A few other words are used for multiplication by a specific factor.
If the unknown quantity is $a$, then {\em twice} the unknown
quantity is represented by $2a$. {\em Thrice} the unknown quantity
is represented by $3a$. To {\em treble} a quantity is to triple it,
hence ``treble $a$'' is $3a$. The {\em square} of a quantity is that
quantity multiplied by itself, so for example, the square of $a$ is
$aa$, which is represented in short by $a^2$. Here $a$ is the {\em
base} and $2$ is the {\em exponent}. The {\em cube} of a quantity is
that quantity multiplied by its square, so for example, the cube of
$a$ is $aaa$, which is represented in short by $a^3$. Here $a$ is
the {\em base} and $3$ is the {\em exponent}.\index{exponent}

\bigskip

The word {\em quotient} will generally be used to denote division.
For example, the quotient of a number and $7$ is denoted by  $x\div
7$, or equivalently by $\dfrac{x}{7}$ or $x/7$.


\bigskip

Here are some more examples.
\begin{exa} If a number $x$ is trebled and if to this new number we add five, we obtain $3x+5$. \end{exa}
\begin{exa}
If $x$ is the larger between  $x$ and $y$, the difference between
$x$ and $y$ is $x - y$. However, if $y$ is the larger between  $x$
and $y$, the difference between $x$ and $y$ is $y - x$. \end{exa}
\begin{exa}If $a$ and $b$ are two numbers, then their product is $ab$, which we will later see that it is the same
as $ba$.
\end{exa}
\begin{exa}
The sum of the squares of $x$ and $y$ is $x^2 + y^2$. However, the
square of the sum of $x$ and $y$ is $(x + y)^2$.
\end{exa}

\begin{exa}
If $n$ is an integer, its predecessor is $n - 1$ and its successor
is $n + 1$.
\end{exa}
\begin{exa}
You begin the day with $E$  eggs. During the course of the day, you
fry $O$ omelettes, each requiring $A$ eggs. How many eggs are left?
\begin{solu}
$E-OA$, since $OA$ eggs are used in frying $O$ omelettes.
\end{solu}
\end{exa}


\begin{exa}An even natural number  has the form $2a$, where $a$ is a natural number.
  An odd natural number  has the form $2a+1$, where $a$ is a natural number.   \end{exa}
\begin{exa}A natural number  divisible by $3$ has the form $3a$, where $a$ is a natural number.
A natural number  leaving remainder $1$ upon division by $3$ has the
form $3a+1$, where $a$ is a natural number. A natural number leaving
remainder $2$ upon division by $3$ has the form $3a+2$, where $a$ is
a natural number.  \end{exa}




\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
If a person is currently $N$ years old, what was his age $20$ years
ago?
\begin{answer}
$N-20$.
\end{answer}
\end{pro}
\begin{pro}
If a person is currently $N$ years old, what will his age be in $20$
years?
\begin{answer}
$N+20$.
\end{answer}
\end{pro}


\begin{pro}
You start with   $x$ dollars. Then you treble this amount and
finally you increase what you now have by $10$ dollars. How many
dollars do you now have?
\begin{answer}
$3x+10$.
\end{answer}
\end{pro}
\begin{pro}
You start with   $x$ dollars. Then you add \$10 to this amount and
finally you treble what you now have. How many dollars do you now
have?
\begin{answer}
$3(x+10)$.
\end{answer}
\end{pro}
\begin{pro}
A knitted scarf uses three balls of wool. I start the day with $b$
balls of wool and knit $s$ scarves. How many balls of wool do I have
at the end of the day?
\begin{answer}
In knitting $s$ scarves, I need $3s$ balls of wool, and hence, at
the end of the day I have $b-3s$ balls of wool.
\end{answer}
\end{pro}

\begin{pro}
What is the general form for a natural number  divisible by $4$?
Leaving remainder $1$ upon division by $4$? Leaving remainder $2$
upon division by $4$? Leaving remainder $3$ upon division by $4$?
\begin{answer} $4n; 4n+1; 4n+2; 4n+3$, where $n$ is a natural number. \end{answer}
\end{pro}
\begin{pro}
If you have $a$ quarters and $b$ dimes only, how many coins do you
have?
\begin{answer}
$a+b$
\end{answer}
\end{pro}
\begin{pro}
If you have $a$ quarters and $b$ dimes only, how much money, in
cents, do you have?
\begin{answer}
$25a+10b$
\end{answer}
\end{pro}


\begin{pro}
If $d$ is a whole number, what is the next larger whole number?

\begin{answer}
$ d+1  $
\end{answer}
\end{pro}

\end{multicols}

\section*{References}
Again, \cite{HaKn} has much of this material, in very old-fashioned
language. So does \cite{Boy}, a book that is accessible online for
free, via Project Gutenberg.
\begin{description}
\bibitem[HaKn]{HaKn} H.S. HALL, and S.R. KNIGHT, ``Elementary Algebra for Schools ''.
\bibitem[Boy]{Boy} Wallace C. BOYDEN, ``A First Course in Algebra''.
\end{description}



\part{The Natural Numbers}

\chapter{Addition and Subtraction}
\begin{center}
    \fcolorbox{blue}{yellow}{
    \begin{minipage}{.90\linewidth}
    \noindent\textcolor{red}{\textbf{Why bother?}} Addition and
    subtraction of natural numbers are the most fundamental operations in Arithmetic.
    We will introduce some fancy vocabulary terms in this section,
    in order to facilitate our road to abstraction.
\end{minipage}}
    \end{center}

\begin{df}The  set of {\em natural numbers}  $\naturals$, is the set
$$ \naturals  = \{0,1,2,3,\ldots \}. $$
\end{df}

\vspace*{.5cm}

\begin{figure}[!h]
 \psset{unit=3pc}\psaxes[yAxis=false,labelFontSize=\tiny,
arrowscale=3,arrows={|->},labelsep=-.5](0,0)(0,0)(13.7,0)
\meinecaption{.5}{The Natural Numbers $\naturals$.}
\label{fig:natural-numbers}
\end{figure}

We will use the symbol $\in$, read {\em is in}, or {\em is an
element of}, to indicate that a certain element belongs to a certain
set. The negation of $\in$ is $\not\in$. For example, $1\in
\naturals$ because $1$ is a natural number, but
$\dfrac{1}{2}\not\in\naturals$.

\bigskip

Natural numbers are used for two main reasons:
\begin{enumerate}
\item  counting, as for example, ``there are $10$ sheep in the herd'',
\item or ordering, as for example, ``Los Angeles is the second largest city in the USA.''
\end{enumerate}




\bigskip
We can interpret the natural numbers as a linearly ordered set of
points, as in figure \ref{fig:natural-numbers}. This interpretation
of the natural numbers induces an order relation as defined below.
\begin{df}
Let $a$ and $b$ be two natural numbers. We say that {\em $a$ is
(strictly) less than $b$}, if $a$ is to the left of $b$ on the
natural number line. We denote this by $a<b$. Some other order
symbols follow:
\begin{center}\begin{enumerate} \item[]
$>$  {\em  is strictly  greater than}, \item[] $\leq$ {\em is less
than or equal to} and,
\item[]  $\geq$  {\em is   greater than or equal to}.
\end{enumerate}\end{center}
We also mention the symbols $=$, {\em equal to}, and $\neq$, {\em
not equal to}.
\end{df}

\begin{exa}
The farther to the right a number is on the number line, the larger
it is. Thus we have
$$1914<1939, \quad 666>100, \quad 1 \leq 1.  $$
\end{exa}

\bigskip

We now introduce the operation of $+$ or {\em addition} in
$\naturals$. We can think of $+$ as {\em joining}, {\em
concatenating}, {\em appending}, that is,  {\em and}. In fact, the
$+$ used today evolved from the fact that in Latin manuscripts
people used to write {\em et}---meaning {\em and}---for ``plus.''
This was shortened to just ``{\em t}'', and then this was slightly
altered to give the modern sign of plus. Let us introduce some fancy
terminology.
\begin{df}
In the expression $a+b=s$, $a$ is called the {\em augend}, $b$ is
called the {\em addend} and $s$ is called the {\em sum}. Both the
augend and the addend are called {\em summands}.
\end{df}
Thus in $1+4=5$, $1$ and $4$ are the summands and $5$ is their sum.
Now, joining  one marble to four marbles gives five marbles, that
is, $1$ marble  $+ 4$ marbles $= 5$ marbles. Concatenating a line
segment of length $b$ inches to one of length $a$ inches gives a
line segment of length $a+b$ inches. See figures \ref{fig:add-join}
and \ref{fig:add-conc}
 \vspace*{3cm}
\begin{figure}[h]
\begin{minipage}{7cm}
\centering \psset{unit=1pc} \pscircle*[linecolor=red](-5,0){.7}
\pscircle*[linecolor=red](-3,0){.7}
\pscircle*[linecolor=red](-1,0){.7}
\pscircle*[linecolor=red](1,0){.7}
\pscircle*[linecolor=red](3,0){.7}
\pspolygon(-4.15,1)(-4.15,-1)(3.9,-1)(3.9,1)
\meinecaption{1}{Joining.} \label{fig:add-join}
\end{minipage}
\hfill
\begin{minipage}{7cm}
\centering \psset{unit=1pc}
\psline[linecolor=gray,linewidth=2pt](-1,4)(1,4)
\pcline[offset=12pt]{|-|}(-1,4)(1,4)
\ncput*[nrot=:U]{$a$}\rput(0,3){$+$}
\psline[linecolor=blue,linewidth=2pt](1,2)(5,2)
\pcline[offset=-12pt]{|-|}(1,2)(5,2)
\ncput*[nrot=:U]{$b$}\rput(0,0){$=$}
 \psline[linecolor=green,linewidth=2pt](-1,-1)(5,-1)\pcline[offset=-12pt]{|-|}(-1,-1)(5,-1)
\ncput*[nrot=:U]{$a+b$}\meinecaption{1}{Concatenation.}
\label{fig:add-conc}
\end{minipage}
\end{figure}


Observe that in the examples above we added like quantities. Thus
marbles were added to marbles and inches to inches. We did not dare
to add marbles to inches! We call these similar entities the {\em
denominator} of the term. Hence in the sum $1$ marble  $+ 4$ marbles
$= 5$ marbles, marbles are the denominator. In the sum $1$ inch $+
4$ inches $= 5$ inches, inches are the denominator. It should not be
 a stretch of one's imagination to progressively reach the
following levels of abstraction:
\begin{itemize}
\item $1m + 4m=5m$
\item $1A + 4A=5A$
\item $\dfrac{1}{401} + \dfrac{4}{401}=\dfrac{5}{401}$
\item $1+4=5$
\end{itemize}
In the first sum, $m$ is the common denominator, we are adding two
like objects. Similarly, $A$ is the common denominator for the
second sum, and $\dfrac{\cdot}{401}$ is the one for the third sum.
The fourth sum is more abstract, in the sense that even though we do
not have a common denominator, we are asserting that if to a unit of
whatever we add four units of the same whatever, the result is five
of the whatever. It is important to realise that this last
abstraction came from our agreement that can only add terms with a
common denominator. Here are   more challenging examples.

\begin{exa}
Ariel walked $x$ miles and rode $10$ miles. How far did he go?
\end{exa}

\begin{solu}
The total of miles he went was $  x+10 $.
\end{solu}



\begin{exa}
Collect like terms: $$ a+2b+3a+4b. $$
\end{exa}
\begin{solu}
We combine the $a$'s with the $a$'s and the $b$'s with the $b$'s:
$$ a+2b+3a+4b = 4a+6b. $$
\end{solu}
\begin{rem}
In the preceding example there is the temptation to further combine
the $4a$ with the $6b$. But this temptation has no basis: we do not
know what the $a$'s are and what the $b$'s are. They are different
denominators and we cannot combine them without further information.
So, if the $a$'s were {\em apples} and the $b$'s {\em bananas}, we
cannot combine them because they are different fruit! But you may
object and say: ``Well, $4$ apples and $6$ bananas do give $10$
fruit! There! I combined them! But you have cheated! You have
created a common denominator, {\em fruit}, for apples and bananas,
and that was not asked in the problem. You cannot deviate from what
is given to you!
\end{rem}

Again, consider the following way of adding $731$ and $695$. Since
$$731= 7\cdot 10^2 + 3\cdot 10 + 1, \qquad 6\cdot 10^2 + 9\cdot 10 + 5,  $$
we could add in the following fashion, without worrying about
carrying:
$$(7\cdot 10^2 + 3\cdot 10 + 1)  +(6\cdot 10^2 + 9\cdot 10 + 5)= 13\cdot 10^2+12\cdot 10 + 6=1300+120+6=1426. $$
The above reasoning can be extended to other than powers of ten.
\begin{exa}
Collect like terms:
$$ (7x^2+3x+1)+(6x^2+8x+5).  $$
\end{exa}
\begin{solu}
We match the squares with the squares, the $x$ to the first power
with the $x$ to the first power, and the constant term with the
constant term to get
$$ (7x^2+3x+1)+(6x^2+8x+5)=13x^2+11x+6.  $$
\end{solu}





As we begin to see more examples of addition of natural numbers, the
following properties emerge.



\begin{axi}[Closure] The sum of any two natural numbers is a natural number, that is, $\naturals$ is {\em closed under addition}: if $a\in
\naturals$ and $b\in \naturals$ then also $a+b\in \naturals$.
\end{axi}
\begin{rem}
Thus if we add two natural numbers, we don't get eggs, we don't get
pastrami, we get a natural number.
\end{rem}
\begin{axi}[Additive Identity] If $0$ is added to any natural number, the result is unchanged, that is $0\in\naturals$ is the
{\em additive identity} of $\naturals$. It has the property that for
all $x\in\naturals$ it follows that $$x=0+x=x+0.  $$
\end{axi}
Again, it is easy to see that when two natural numbers are added,
the result does not depend on the order. This is encoded in the
following axiom.
\begin{axi}[Commutativity]Let $a\in\naturals$ and $b\in \naturals$ be arbitrary. Then
\mbox{$a+b=b+a$}.
\end{axi}The last axiom has to do with grouping.
\begin{axi}[Associativity]
Let $a, b, c$ be arbitrary natural numbers. Then the order of
parentheses when performing addition is irrelevant, that is,
$$ a+(b+c)=(a+b)+c = a+b+c.  $$
\end{axi}




How do we actually {\em add} in decimal notation? Let us first
present a geometric method to add integers along the number line.
\begin{exa}Use the number line to
perform the addition $2+3$. \end{exa}
\begin{solu}First locate $2$ by starting from $0$ and moving two
units right.

\vspace{.5cm}
$$\psset{unit=.5} \pstGeonode[PointName=none](0,0){O}(2,0){A}\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{O}{A}\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={->}](0,0)(0,0)(11.5,0)
\uput[d](0,0){0}\uput[d](2,0){2}
$$
\vspace{.5cm}

\bigskip Now, starting at $2$ move three units
right.


\vspace{.5cm}
$$\psset{unit=.5} \pstGeonode[PointName=none](0,0){O}(2,0){A}(5,0){B}\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{A}{B}\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={->}](0,0)(0,0)(11.5,0)
\uput[d](0,0){0}\uput[d](2,0){2}\uput[d](5,0){5}
$$\vspace{.5cm}


\bigskip
Since we landed at $5$, we conclude that $2+3=5$. \end{solu}



The above method of addition can, in principle, accomplish all the
sums of natural numbers that we think of. It suffers from the
limitation of us having to locate numbers on the number line and
counting at least twice. If the numbers are really large, then we
might miscount. Countless cultures that did not know our decimal
positional notation were able to carry out these computations. But
the cumbersomeness of these calculations meant that the operations
could be carried out by expert accountants. Today, one hopes, even a
child can perform these operations.




In order to explain an algorithm for addition, let us first think of
how we can give the decimal representation of an arbitrary number.
Let us agree that the notation
$$ \underline{d_nd_{n-1}\ldots d_2d_1d_0}, $$where the $d$'s are
digits, means
$$  \underline{d_nd_{n-1}\ldots d_1d_0} = d_n\cdot 10^n+d_{n-1}\cdot 10^{n-1}+\cdots +d_2\cdot 10^2 + d_2\cdot 10 + d_0.   $$
For example, if the numeral is $60637$, then this has $5$ digits,
and we have,\footnote{The arrow $\implies$ is read {\em implies}.}
$$\underline{d_4d_3d_2d_1d_0}=60637 \implies d_4=6, \quad d_3=0, \quad d_2 =6, \quad d_1 =3, \quad d_0 =7.  $$
If the numeral is  $1234567$, then this has $7$ digits, and we have
$$\underline{d_6d_5d_4d_3d_2d_1d_0}=1234567 \implies d_6=1, \quad d_5=2, \quad  d_4=3, \quad d_3=4, \quad d_2 =5, \quad d_1 =6, \quad d_0 =7.  $$


Suppose now that we have two arbitrary decimal integers, say
$$ \underline{d_nd_{n-1}\ldots d_2d_1d_0},\qquad \underline{e_pe_{p-1}\ldots e_2e_1e_0}. $$
Here it may be the case that $n=p$, that is, the integers have the
same number of digits, but we do not know that in advance. Let us
suppose for the sake of argument that $n<p$. To add the integers,
line them up at the units:
$$ \begin{array}{cccccccccc} & & & & d_n & d_{n-1} & \ldots  & d_2 & d_1 & d_0\\
+& e_p  & e_{p-1} & \ldots & e_n  & e_{n-1} & \ldots  & e_2 & e_1 & e_0\\
\hline
 \end{array}$$
The algorithm is then as follows:
\begin{enumerate}
\item Add $d_0+e_0$. If the sum  is smaller than $10$, put the
result below in this first column. If the result is $10$ or more,
put the units of the result below in the first column and carryover
the the tens of the result to the next column.
\item Add $d_1+e_1$ and any carryover from the units (if there is
any). If the sum  is smaller than $10$, put the result below in this
second column. If the result is $10$ or more, put the units of the
result below in the second  column and carryover the the tens of the
result to the next column. Continue a similar process until the
digits of the longest number have been exhausted.
\end{enumerate}
\begin{exa}
Use the above algorithm to perform the addition $17066+1492$.
\end{exa}
\begin{solu} We first line up the numbers at their last digit:
$$\opadd[voperator=bottom, resultstyle.1=\hole, resultstyle.2=\hole, resultstyle.3=\hole, resultstyle.4=\hole,  resultstyle.5=\hole, carryadd=false]{17066}{1492}  $$
Now we add the units: $6+2=8$ and there is no carrying since the
result is smaller than $10$:
$$\opadd[voperator=bottom, resultstyle.2=\hole, resultstyle.3=\hole, resultstyle.4=\hole,  resultstyle.5=\hole, carryadd=false]{17066}{1492}  $$
We add the tens: $6+9=15$. We put $5$ below on the column of tens and
we carry the $1$ to the column of hundreds:
$$\opadd[voperator=bottom,  resultstyle.3=\hole, resultstyle.4=\hole,  resultstyle.5=\hole]{17066}{1492}  $$
We add the hundreds, without forgetting the carry:  $1+0+4=5$. We
put $5$ below on the column of hundreds, and there is no carrying,
since the result is smaller than $10$:
$$\opadd[voperator=bottom,   resultstyle.4=\hole,  resultstyle.5=\hole]{17066}{1492}  $$
We add the thousands: $7+1=8$. We put $8$ below on the column of
thousands, and there is no carrying, since the result is smaller
than $10$:
$$\opadd[voperator=bottom,   resultstyle.5=\hole]{17066}{1492}  $$
Finally, we add the ten thousands: $1+0=1$ (remember that $1492$ is
the same as $01492$). We put $1$ below on the column of ten
thousands, and there is no carrying, since the result is smaller
than $10$:
$$\opadd[voperator=bottom]{17066}{1492}  $$
\end{solu}
\begin{exa}Add: $(x^4+7x^3+6x+6)+(x^3+4x^2+9x+2)$.
\end{exa}
\begin{solu}
We line up the terms that have the same exponents and add
vertically.
$$ \begin{array}{cccccccccccc} & x^4& + & 7x^3& & &  + & 6x & + & 6\\
+&  & &x^3& +& 4x^2& + & 9x& + & 2\\
\hline & x^4 & + & 8x^3 & + & 4x^2 & + & 15x & + & 8\\
 \end{array}$$
\end{solu}





Having now an idea of what it means to add  natural numbers, we
define subtraction  of natural numbers by means of addition. This is
often the case in Mathematics: we define a new procedure in terms of
old procedures.
\begin{df}[Definition of Subtraction]
Let $m, s, d$ be natural numbers, with $m\geq s$. Then the statement
$m-s=d$ means that $m=s+d$. In the expression $m-s=d$, $m$ is called
the {\em minuend}, $s$ the {\em subtrahend}, and $d$ the {\em
difference}.
\end{df}

\bigskip

Of course, our link to reality and subtraction is to think of
subtraction as {\em taking away}.

\bigskip


\begin{exa}
To compute $15-3$ we think of which number when added $3$ gives
$15$. Clearly then $15-3=12$ since $15=12+3$.
\end{exa}




Notice that subtraction of natural numbers  satisfies neither
closure, nor commutativity, nor associativity. For example,
$$ 3-5 \notin \naturals,  $$and so subtraction is not close in
$\naturals$. Since
$$ 5-3\neq 3-5, $$ subtraction is not commutative. Lastly, since
$$10-(7-4)\neq (10-7)-4 , $$subtraction is not associative.





Let us use a geometric method similar to the additive one in order
to perform subtraction.

 \begin{exa}Use the
number line to perform the subtraction $3-2$. \end{exa}
\begin{solu}First locate $3$ by starting from $0$ and moving three
units right.

\vspace{.5cm}
$$\psset{unit=.5} \pstGeonode[PointName=none](0,0){O}(3,0){A}\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{O}{A}\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={->}](0,0)(0,0)(11.5,0)
\uput[d](0,0){0}\uput[d](3,0){3}
$$
\vspace{.5cm}

\bigskip Now, starting at $3$ move two units
left.


\vspace{.5cm}
$$\psset{unit=.5} \pstGeonode[PointName=none](0,0){O}(3,0){A}(1,0){B}\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{A}{B}\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={->}](0,0)(0,0)(11.5,0)
\uput[d](0,0){0}\uput[d](3,0){3}\uput[d](1,0){1}
$$\vspace{.5cm}


\bigskip
Since we landed at $1$, we conclude that $3-2=1$. \end{solu}



We raise the same objections that we raised against the geometric
method for addition: it becomes cumbersome for large values and it
is prone to error if we miscount. Let us now discuss an algorithm
for subtraction.





Suppose again that  we have two arbitrary decimal integers, say
$$ \underline{d_nd_{n-1}\ldots d_2d_1d_0},\qquad \underline{e_pe_{p-1}\ldots e_2e_1e_0}, $$
with $n\leq p$ and  $\underline{d_nd_{n-1}\ldots d_2d_1d_0}\leq
\underline{e_pe_{p-1}\ldots e_2e_1e_0}$. To subtract the integers,
line them up at the units:
$$ \begin{array}{cccccccccc}
& e_p  & e_{p-1} & \ldots & e_n  & e_{n-1} & \ldots  & e_2 & e_1 & e_0\\
-& & & & d_n & d_{n-1} & \ldots  & d_2 & d_1 & d_0\\
\hline
 \end{array}$$
The algorithm is then as follows:
\begin{enumerate}
\item If  it follows that $d_k\leq e_k$ for all $k\leq n$, then there is no regrouping and we simply put
the result of $e_k-d_k$ in the $k$th column.
\item If the previous item is not true, consider a string $i \leq j \leq n$ for which $d_i>e_i$,
$d_{i+1}\geq e_{i+1}$, \ldots , $d_j\geq e_j$ and $d_{j+1}<
e_{j+1}$. Starting at the index $i+1$, successively regroup so that
now in the $i$-th column you have $10+e_i-d_i$, on the $i+1$th
column you have $10+e_{i+1}-1-d_{i+1}$, and so on, until the $j$th
column where you will have $10+e_j-1-d_j$ and on the $j+1$th column
you will have $e_{j+1}-d_{j+1}$. Now all these subtractions can be
accomplished. Put the result below in the respective column. Repeat
again after the $j+1$th column if the situation arises again.
\end{enumerate}

\begin{exa}
Use the above algorithm to perform the subtraction
$9876666789-1234554321$.
\end{exa}
\begin{solu}
Since $9876666789>1234554321$, the subtraction is permissible.
Notice that each digit of $1234554321$ is smaller than each
corresponding digit of $9876666789$, and so there is no carrying.
The result is displayed thus:
$$
\opsub[voperator=bottom,carrysub=true]{9876666789}{1234554321}
$$
\end{solu}


\begin{exa}
Use the above algorithm to perform the subtraction
$98723112389-72430014467$.
\end{exa}
\begin{solu}Since $98723112389>72430014467$, the subtraction is
permissible. The subtraction on the units and the tens is effected
without complications, since both digits in the subtrahend are
smaller than the corresponding digits in the minuend:
$$
\opsub[voperator=bottom,resultstyle.3=\hole,resultstyle.4=\hole,resultstyle.5=\hole,resultstyle.6=\hole,resultstyle.7=\hole,resultstyle.8=\hole,resultstyle.9=\hole,resultstyle.10=\hole,resultstyle.11=\hole,
resultstyle.12=\hole]{98723112389}{72430014467}
$$

When we turn to the hundreds, we find that $3<4$, and so, we must
regroup from the digits to the left. In the thousands we find $2<4$.
In the ten thousands we find $1\leq 1$. We stop at the hundred
thousands, because $1>0$. Thus we take $1$ from the hundred
thousands (where we now have $0$) to the ten thousands. In the ten
thousands now we have $10 + 1=11$, but we take $1$ to the thousands,
leaving $10$ in the ten thousands. We now have $10 + 2=12$ in the
thousands, but we take $1$ to the hundreds, leaving $11$ in the
thousands. Finally, we have $10+3=13$ in the hundreds. We subtract
until the hundred thousands getting,
$$
\opsub[voperator=bottom,resultstyle.7=\hole,resultstyle.8=\hole,resultstyle.9=\hole,resultstyle.10=\hole,resultstyle.11=\hole,
resultstyle.12=\hole]{98723112389}{72430014467}\oplput(-3,3){{\tiny
13}}\oplput(-4,3){{\tiny 11}} \oplput(-5,3){{\tiny 10}}
\oplput(-6,3){{\tiny 0}}
$$
We now turn to the millions. Since $3<0$, subtraction goes without
any complications:
$$
\opsub[voperator=bottom,resultstyle.8=\hole,resultstyle.9=\hole,resultstyle.10=\hole,resultstyle.11=\hole,
resultstyle.12=\hole]{98723112389}{72430014467}\oplput(-3,3){{\tiny
13}}\oplput(-4,3){{\tiny 11}} \oplput(-5,3){{\tiny 10}}
\oplput(-6,3){{\tiny 0}}
$$
In the ten millions we find that $2<3$, so we must regroup from the
digits to the left. In the hundred millions we see that $7>4$ so we
stop there. We take a $1$ from the hundred millions, and so we have
a $6$ in the hundred millions. In the ten millions we now have
$10+2=12$. We perform the subtractions:
$$
\opsub[voperator=bottom,resultstyle.10=\hole,resultstyle.11=\hole,
resultstyle.12=\hole]{98723112389}{72430014467}\oplput(-3,3){{\tiny
13}}\oplput(-4,3){{\tiny 11}} \oplput(-5,3){{\tiny 10}}
\oplput(-6,3){{\tiny 0}} \oplput(-8,3){{\tiny 12}}
\oplput(-9,3){{\tiny 6}}
$$

Finally, the subtraction in the billions and the ten billions occur
without major complications. We display the result:
$$
\opsub[voperator=bottom]{98723112389}{72430014467}\oplput(-3,3){{\tiny
13}}\oplput(-4,3){{\tiny 11}} \oplput(-5,3){{\tiny 10}}
\oplput(-6,3){{\tiny 0}} \oplput(-8,3){{\tiny 12}}
\oplput(-9,3){{\tiny 6}}
$$
\end{solu}


\begin{exa}
 A box of raisins was bought for $a$ dollars, and a firkin of
butter for $b$ dollars. If both were sold for $c$ dollars, how much
was gained?
\end{exa}
\begin{solu} The amount of money spent on raisins and butter gives
an original cost of $a+b$ dollars. Since we sold it for $c$, we must
take away our original cost, and hence we made a profit of  $c-(a+b)
$ dollars.
\end{solu}

\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

\begin{pro}
{\em Without availing from a calculator}, perform the addition
$123456789+987654321$.
\begin{answer}
$$\opadd[voperator=bottom]{123456789}{987654321}$$
\end{answer}
\end{pro}
\begin{pro}
{\em Without availing from a calculator}, perform the subtraction
$987654321-123456789$.
\begin{answer}
$$\opsub[voperator=bottom]{987654321}{123456789}$$
\end{answer}
\end{pro}


\begin{pro}\label{pro:number-cross}
Suppose we arrange the five numbers $1, 2, 3, 4, 5$ in each of the
five squares so that the horizontal and vertical lines both add to
$8$. Which number must go in the middle square?
\end{pro}
\vspace{1cm}
\begin{figurehere}
\centering \psset{unit=1pc}
\pspolygon[linewidth=2pt](-1,3)(-1,-3)(1,-3)(1,3)
\pspolygon[linewidth=2pt](3,-1)(-3,-1)(-3,1)(3,1)
\meinecaption{1}{Problem
\ref{pro:number-cross}.}\label{fig:number-cross}
\end{figurehere}
\begin{answer} $2$ \end{answer}
\begin{pro}
An oak tree is $42$ feet high. The oak tree is $18$ feet taller
than the fir tree. How tall is the fir tree?
\begin{answer}
$24 $ ft.
\end{answer}
\end{pro}
\begin{pro} Fill in the numbers from $1$ up to
$5$ in the five little circles, in such a way that in each triangle
the given number is the sum of the numbers on the vertices. Which
number appears in the shaded circle? \vspace*{2cm}
\begin{figurehere}
\rput(-8,0){
\begin{graph}(4,2)
\graphnodesize{.5} \roundnode{n1}(-2,0)[\graphnodecolour{1}]
\roundnode{n2}(0,0)[\graphnodecolour{1}]
\roundnode{n3}(2,0)[\graphnodecolour(1,.3,0)]
\roundnode{n4}(-1,-1)[\graphnodecolour{1}]\roundnode{n5}(1,-1)[\graphnodecolour{1}]
\edge{n1}{n2} \edge{n2}{n3} \edge{n1}{n4} \edge{n2}{n4}\edge{n2}{n5}
\edge{n3}{n5} \edge{n4}{n5} \rput(-1,-.5){6} \rput(0,-.6){10}
\rput(1,-.5){12}
\end{graph}}
\end{figurehere}
\begin{answer}
The figure below shows the correct arrangement. \vspace*{2cm}
\begin{figure}[h]
\begin{graph}(4,2)
\graphnodesize{.5} \roundnode{n1}(-2,0)[\graphnodecolour{1}]
\roundnode{n2}(0,0)[\graphnodecolour{1}]
\roundnode{n3}(2,0)[\graphnodecolour(1,.3,0)]
\roundnode{n4}(-1,-1)[\graphnodecolour{1}]\roundnode{n5}(1,-1)[\graphnodecolour{1}]
\edge{n1}{n2} \edge{n2}{n3} \edge{n1}{n4} \edge{n2}{n4}\edge{n2}{n5}
\edge{n3}{n5} \edge{n4}{n5} \rput(-1,-.5){6} \rput(0,-.6){10}
\rput(1,-.5){12} \nodetext{n1}{$1$} \nodetext{n2}{$3$}
\nodetext{n3}{$4$} \nodetext{n4}{$2$} \nodetext{n5}{$5$}
\end{graph}
\end{figure}

\end{answer}
\end{pro}
\begin{pro}
Can we find five even integers whose sum is $25$? \begin{answer} No.
When we add an even integer to another even integer the result is an
even integer. Thus the sum of five even integers is even, but $25$
is odd. \end{answer}
\end{pro}
\begin{pro}A bottle of wine and its cork cost $\$1$.
The bottle of wine costs $80$\textcent\ more than than the cork.
What is the price of the cork, in cents?
\begin{answer}
$10$\textcent. The cork costs  $10$\textcent\ and the bottle
$90$\textcent.
\end{answer}
\end{pro}
\begin{pro}
Ingrid made $686$ biscuits. She sold some of them. If
$298$ were left over, how many biscuits did she sell?
\begin{answer}
$686-298=388$
\end{answer}
\end{pro}
\begin{pro}
What is the value of $$1 + 2 + 3 + \cdots + 7 + 8 + 9 + 8 + 7 + \cdots + 3 + 2 + 1,$$ where all of the
integers from $1$ through $9$ and then back down to $1$ are added together?
\begin{answer}
$81$
\end{answer}
\end{pro}
\begin{pro}
How many numbers are there in the set
$$ \{2,4,6,\ldots , 2006\},  $$
that is, in the set of all strictly positive natural numbers between
$2$ and $2006$?
 What is
$$(2+4+\cdots + 2006) -(1+3+\cdots + 2005), $$
that is, what is the difference between the sum of all the odd
positive integers up to  $2005$ and the sum of all the even positive
integers up to (and including) $2006$?
\begin{answer}
Observe that
$$ 2=2\cdot 1, \quad 4=2\cdot 2, \quad 6 = 2\cdot 3, \ldots 2006 = 2\cdot 1003,  $$
and so there are $1003$ numbers.

\bigskip

The desired quantity is
$$(2-1) + (4-3) + \cdots + (2006-2005) = 1003.   $$
\end{answer}


\end{pro}
\begin{pro}
Luigi saved $\$184$. He saved $\$63$ more than Brian.
How much did Brian save?
\begin{answer}
$184-63=121$
\end{answer}
\end{pro}
\begin{pro}
Alice is $4$ inches taller than Betty. Caroline is $8$ inches
shorter than Alice. Betty is $69$ inches tall. How tall is Caroline,
in inches?
\begin{answer}
Alice is $69+4=73$ inches tall and Caroline is $73-8=65$ inches
tall.
\end{answer}
\end{pro}
\begin{pro}
A tree was planted $54$ years before $1961$. How old was that
tree in $2008$?
\begin{answer}
$ 101$ years.
\end{answer}
\end{pro}
\begin{pro}
Lilliam had $20$ pieces of candy. She gave two pieces to her
sister.
\begin{enumerate}\item How many did she have left?
\item  If she gave away 2 pieces each to 4 more people, how many
pieces would she have left?
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}\item $18$
\item $10$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
Buses need to be rented for $27$ children going on a field trip. Each bus
can take $12$ children in addition to the driver. How many buses must be
rented?
\begin{answer}
$3$
\end{answer}
\end{pro}
\begin{pro}
Collect like terms:
\begin{enumerate}
\item $(2a+8b)+(a+3b)$
\item $2a+8b-a-3b$
\item $(x^2+x+1)+(3x^2+2x+1)$
\end{enumerate}
\begin{answer}
\noindent
\begin{enumerate}
\item $3a+11b$
\item $a+5b$
\item $4x^2+3x+2$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
Bobby was playing on the elevator of a very tall building. Starting
from the floor where he was, he went five floors up, four down,
three up, four up, and two down. If he is now in the $30$th floor,
what was the original floor from where he started?
\begin{answer}
Start from $30$ and go back:
$$ 30+2-4-3+4-5=24, $$so he started on the $24$th floor.
\end{answer}
\end{pro}
\begin{pro}
 How many addition signs should be put between digits of the number $987654321$ and where should we put them to get a total of $99$?
\begin{answer}
Either $9+8+7+65+4+3+2+1 = 99 $ or $9+8+7+6+5+43+21 = 99  $.
\end{answer}
\end{pro}
\begin{pro}
You have a seven inch gold bar, that is already segmented into seven
equal pieces. You are allowed to make two cuts to it. How can you
pay an employee that demands to be paid one gold piece  daily for
the seven days that he works for you?
\begin{answer}
Cut the bar making pieces of $1$, $2$, and $4$ segments.
\end{answer}
\end{pro}
\begin{pro} Fill in the blank using either of the symbols '$=$' ,
'$\neq$', '$\in$', '$\notin$' as appropriate:
\begin{enumerate}
\item $3 \rule{1cm}{2pt} 3+0$ \item $5+1 \rule{1cm}{2pt} \naturals $ \item
$1+3 \rule{1cm}{2pt} 2+2$\item $64 \rule{1cm}{2pt} 604$
\item $\maltese \rule{1cm}{2pt} \naturals $
\item $ 5+6 \rule{1cm}{2pt} 7+ 8$
\end{enumerate}
\begin{answer}\noindent\begin{enumerate}
\item $3= 3+0$ \item $5+1\in \naturals $ \item
$1+3= 2+2$\item $64\neq 604$ \item $\maltese \notin \naturals $
\item $ 5+6 \neq 7+8$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
How many different sums can be made when two non-necessarily
distinct numbers from the set $\{1,3,4,5,7\}$ are taken?
\begin{answer}
From the addition table
$$ \begin{array}{|l||l|l|l|l|l|} \hline + & 1 & 3 & 4 & 5 & 7 \\
 \hline\hline 1 & 2 & 4 & 5 & 6 & 8 \\
 \hline 3 & 4 & 6 & 7 & 8 & 10 \\
 \hline 4 & 5 & 7 & 8 & 9 & 11 \\
 \hline 5 & 6 & 8 & 9 & 10 & 12 \\
 \hline 7 & 8 & 10 & 11 & 12 & 14 \\
 \hline
   \end{array}   $$the different sums belong to the set
   $\{2, 4, 5, 6, 7, 8,9,10,11,12,14\}$, and so there are eleven different sums.
\end{answer}
\end{pro}
\begin{pro}
A frog is in a $10$ ft well. At the beginning of each day, it leaps
$5$ ft up, but at the end of the day it slides   $4$ ft down. After how many days, if at all, will the
frog escape the well? \begin{answer} The frog will escape after
seven days. At the end of the sixth day, the frog has leaped $6$
feet. Then at the beginning of the seventh day, the frog leaps $5$
more feet and is out of the well.
 \end{answer}
 \end{pro}

\begin{pro}
Judith was imprisoned by a band of mathematicians and sent to
Guant\'{a}namo for crimes against Mathematics. Through the mercy of
the Brahmin  mathematician, she was given the choice of being
released after $10$ years or be  given freedom if she climbed the
$100$ steps of a $100$-step staircase subject to the following
rules:
\begin{enumerate}
\item She  climbs up or down only one step per day.
\item She climbs up on every day of  January, March, May, July,
September, and November.
\item She goes down on every day of February, April, June, August,
October, and December.
\end{enumerate}
Being adept at climbing, she chose this later option. If Judith
started on January 1 2001, when will she gain her freedom?
\begin{answer}
Let us see what happens in a typical non-leap year, and in a typical
leap year.

\bigskip

In a non-leap year\\

\begin{tabular}{|l|l|}\hline
by the end of & she has climbed \\
\hline 31 January & $31$ steps \\
\hline 28 February & $31-28=3$ steps \\
\hline 31 March & $3+31=34$ steps \\
\hline 30 April & $34-30=4$ steps \\
\hline 31 May & $31+4=35$ steps \\
\hline 30 June & $35-30=5$ steps \\
\hline 31 July & $31+5=36$ steps \\
\hline 31 August & $36-31=5$ steps \\
\hline 30 September & $30+5=35$ steps \\
\hline 31 October & $35-31=4$ steps \\
\hline 30 November  & $30+4=34$ steps \\
\hline 31 December & $34-31=3$ steps \\
\hline
\end{tabular}


\bigskip

In a leap year\\

\begin{tabular}{|l|l|}\hline
by the end of & she has climbed \\
\hline 31 January & $31$ steps \\
\hline 29 February & $31-29=2$ steps \\
\hline 31 March & $2+31=33$ steps \\
\hline 30 April & $33-30=3$ steps \\
\hline 31 May & $31+3=34$ steps \\
\hline 30 June & $34-30=4$ steps \\
\hline 31 July & $31+4=35$ steps \\
\hline 31 August & $35-31=4$ steps \\
\hline 30 September & $30+4=34$ steps \\
\hline 31 October & $34-31=3$ steps \\
\hline 30 November  & $30+3=33$ steps \\
\hline 31 December & $33-31=2$ steps \\
\hline
\end{tabular}

\bigskip

Now, $100-36=64$. Let us see how many years it takes her to climb
$64$ steps. By the end of the four-year range $2001-2004$, she
climbs $3+3+3+2=11$ steps. By the end of the four-year range
$2005-2008$, she has climbed $22$ steps. By the end of the four-year
range $2009-20012$, she has climbed $33$ steps. By the end of the
four-year range $2013-2016$, she has climbed $44$ steps. By the end
of the four-year range $2017-2020$, she has climbed $55$ steps. By
the end of the four-year range $2021-2024$, she has climbed $66$
steps. In fact, by 31 December 2023 she has climbed $64$ steps, and
by 31 July 2024 she has climbed $64+35=99$ steps. This means that
she needs to go into 2025. By 31 March 2025 she has climbed
$66+34=100$ steps and she is now free.

\end{answer}
\end{pro}
\begin{pro}
Using all the digits $\{0,1,2,3,4,5,6,7,8,9\}$, form two $5$-digit
numbers so that their difference is as large as possible.
\begin{answer}
$98765-10234=88531$. Why didn't we take $01234$ instead of $10234$?
\end{answer}
\end{pro}
\begin{pro}
Using all the digits $\{0,1,2,3,4,5,6,7,8,9\}$, form two $5$-digit
numbers so that their difference is as small as possible.
\begin{answer}
$50123-49876=247$.
\end{answer}
\end{pro}

\begin{pro}
You and I play the following game. I tell you to write down three
$2$-digit integers between $10$ and $89$. Then I write down three
$2$-digit integers of my choice. The answer comes to $297$, no
matter which three integers you choose (my choice always depends on
yours). For example, suppose you choose $12, 23, 48$. Then I choose
$87, 76, 51$. You add
$$12 + 23 + 48 + 87 + 76 + 51 = 297.$$ Again, suppose you chose
$33,  56, 89$. I then choose 66, 43, 10. Observe that
$$33 +  56 + 89 + 66 + 43 + 10 = 297.$$
Explain how I choose my numbers so that the answer always comes up
to be $297$ (!!!).  \begin{answer} Notice that I always choose my
number so that when I add it to your number I get $99$, therefore, I
end up adding $99$ three times and $3 \cdot  99 =
297$.\footnote{Using algebraic language, observe that if you choose
$x, y, z$, then I choose $(99 - x), (99 - y), (99 - z)$. This works
because
$$x + (99 - x) + y + (99 - y) + z + (99 - z) = 3(99) = 297.$$}
 \end{answer}
 \end{pro}
\begin{pro}
A merchant bought $a$ barrels of sugar and $p$ barrels of molasses.
How many barrels in all did he buy?

\begin{answer}
$ a+p  $
\end{answer}
\end{pro}
\begin{pro}
Jill bought a silk dress for $m$ dollars, a muff for $l$ dollars, a
shawl for $v$ dollars, and a pair of gloves for $c$ dollars. What
was the entire cost?
\begin{answer}
$m + l +v +c   $
\end{answer}
\end{pro}
\begin{pro}
You start the day with $q$ quarters and $d$ dimes. How much money do
you have? Answer in cents. If by the end of the day you have lost
$a$ quarters and  $b$ dimes, how much money do you now have? Answer
in cents. Write a one or two line explanation for your
answers.\begin{answer} You have $25q+10d$ cents at the beginning of
the day, and then you lose $25a+10b$, which leaves you with
$$(25q+10d)-(25a+10b)$$ cents.
\end{answer}
\end{pro}
\begin{pro}
Brian tore out several successive pages from a book. The first page that he tore up was page number $143$. The last page that he tore up
is also a three-digit number written with the same digits $\{1, 4, 3\}$ but in a different order. How many pages did he tear up?
\begin{answer}
The number of the last page torn must have opposite parity to the first one torn, thus it must be even, hence it must be  $314$ and hence
$314-143=171$ pages.
\end{answer}
\end{pro}
\begin{pro}
A man bought a hat for $h$ dollars. He then bought a jacket and a
pair of trousers. If the jacket is thrice as expensive as the hat
and the trousers are $8$ dollars cheaper than jacket, how much money
did he spend in total for the three items?
\begin{answer}
The jacket costs $3h$ and the trousers cost $3h-8$. Hence he paid in
total
$$ h + 3h + 3h-8 = 7h-8. $$
\end{answer}
\end{pro}
\begin{pro}
Peter is $x$ years old, Paul is $y$, and Mary is $z$ years. What is
the sum of their ages?
\begin{answer}
$ x+y+z  $
\end{answer}
\end{pro}
\begin{pro}
A boy bought a pound of butter for $y$ cents, a pound of meat for
$z$ cents, and a bunch of lettuce for $s$ cents. How much did they
all cost?
\begin{answer}
$y+z+s $\textcent
\end{answer}
\end{pro}
\begin{pro}
A man sells a carriage for $m$ dollars and loses $x$ dollars. What
was the cost of the carriage?
\begin{answer}
$ m+x  $
\end{answer}
\end{pro}
\begin{pro}
I paid $c$\textcent\ for a pound of butter, and $f$\textcent\ for a
lemon. How much more did the butter cost than the lemon?

\begin{answer}
$c-f$ \textcent
\end{answer}
\end{pro}
\begin{pro}
Sold a lot of wood for $b$ dollars, and received in payment a barrel
of flour worth $e$ dollars. How many dollars remain due?
\begin{answer}
$b-e$
\end{answer}
\end{pro}

\begin{pro}
Complete the following addition:
$$\opadd[voperator=bottom,operandstyle.1.1=\hole,
operandstyle.1.3=\hole, operandstyle.2.4=\hole, resultstyle.2=\hole,
carryadd=false] {3679}{5283}$$
\begin{answer}
$3679+5283=8962$
\end{answer}
\end{pro}
\begin{pro}
Find the missing digits:
$$\opadd[voperator=bottom,operandstyle.1.2=\hole,
operandstyle.2.3=\hole, operandstyle.2.1=\hole, resultstyle.3=\hole,
carryadd=false] {714}{189}$$
\begin{answer}
$714+189=903$.
\end{answer}
\end{pro}
\begin{pro}
Complete the following subtraction:
$$\opsub[voperator=bottom,operandstyle.1.3=\hole, operandstyle.2.1=\hole, resultstyle.2=\hole,
resultstyle.4=\hole,carrysub=false] {5671}{1920}$$
\begin{answer}
$5671-1920=3751$
\end{answer}
\end{pro}


\begin{pro}  By inspection (without any calculation) determine whether the following are true or false. Mention which property makes
the true statements true.
\begin{enumerate}
\item $1298+7459=7459+1298$ \item $1298+7459=7460+1298$
\item $(547+1250)+3=547+(1250+3)$ \item $(547+1250)+3=
547+(3+1250)$
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}
\item True by commutativity of addition.  \item False.
\item True by associativity of addition. \item True. First observe that  $(547+1250)+3= 547+(1250+3)$ by associativity and that
$(1250+3)=(3+1250)$ by commutativity. Then it follows that
$(547+1250)+3= 547+(3+1250)$ because equals replaced by equals are
equal.
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro} Each square represents a digit. Find the value of
each
missing digit. $$\begin{array}{lllllll}  & \blacksquare & 7 & 5 & \blacksquare & 6 \\
- & & \blacksquare & 5 & 6 & \blacksquare \\
\hline & 2 & 4 & \blacksquare & 7 & 5 \\
\end{array}
$$
\begin{answer}
$\psframebox{2}75\psframebox{3}6-\psframebox{2}56\psframebox{1} =
24\psframebox{9}75$.
\end{answer}
\end{pro}
\begin{pro}
Is it possible to replace the letter $A$  in the square below so
that every row has the same sum of every column?

\vspace*{1cm}

\PuzzleSolution
\begin{Puzzle}{3}{3}%
|{1}|{2}|{5}|.
|{3}|{3}|{2}|.
|{a}|{3} |{1}|.
\end{Puzzle}

\begin{answer}
Observe that the first and second rows, and the second and third
columns add up to $8$. Thus $A=4$ works.
\end{answer}
\end{pro}
\begin{pro}
Fill each square with exactly one number from
$$\{1,2,3,4,5,6,7,8,9\}$$ so that the square becomes a magic
square, that is, a square where every row has the same sum as every
column, and as every diagonal.

\vspace*{1cm}

\begin{Puzzle}{3}{3}%
|{\empty }|{\empty }|{\empty }|.
|{\empty }|{\empty }|{\empty }|.
|{\empty }|{\empty } |{\empty }|.
\end{Puzzle}
Is there more than one solution?
\begin{answer}
There are multiple solutions. They can be obtained by permuting the
entries of one another. Here are two:

\vspace*{1cm}

\begin{minipage}{5cm}
\PuzzleSolution
\begin{Puzzle}{3}{3}%
|{8}|{1}|{6}|.
|{3}|{5}|{7}|.
|{4}|{9} |{2}|.
\end{Puzzle}
\end{minipage}
\begin{minipage}{5cm}
\PuzzleSolution
\begin{Puzzle}{3}{3}%
|{4}|{9} |{2}|.
|{3}|{5}|{7}|.
|{8}|{1}|{6}|.
\end{Puzzle}
\end{minipage}

\end{answer}

\end{pro}

\begin{pro}
Vintik and Shpuntik agreed to go to the fifth car of a train.
However, Vintik went to the fifth car from the beginning, but
Shpuntik went to the fifth car from the end. How many cars has the
train if the two friends got to one and the same car?
\begin{answer}
Nine.
\end{answer}
\end{pro}
\begin{pro}
The desks in a classroom are arranged in straight rows. Jos\'{e} is in the third
row from the front and the fourth row from the back. He is also third from
the left end of a row and fifth from the right.
How many desks are in the classroom?
\begin{answer}
$42$.
\end{answer}
\end{pro}
\begin{pro}
In a contest to guess the number of balloons in a bunch, Adam guessed $25$,
Bob guessed $31$, Carlos guessed $29$, Daniel guessed $23$ and Edward
guessed $27$. Two guesses were wrong by $2$, and two guesses were wrong
by $4$. The other guess was correct. How many balloons are there?
\begin{answer}
$27$.
\end{answer}
\end{pro}
\begin{pro}
In the following array, some entries are missing. Find the value of
$x$. $$\begin{array}{|l|l|l|l|l||l|}
\hline & & & & &  {\rm Row\ Totals} \\
\hline
& ? & ? & 5 & 7 & 24    \\
\hline & 5 & ? & ? & 1 & 21  \\
\hline & 7 & 8 & ? & ? & 26 \\
\hline & 9 & 3 & 2 & x & ?  \\
\hline\hline {\rm Column\ Totals} & 29 & 21 & 22 & 19 & \\
\hline
\end{array}$$
\begin{answer}$6$\end{answer}
\end{pro}

\begin{pro}
 An urn contains $28$ blue marbles, $20$ red
marbles, $12$ white marbles, $10$ yellow marbles, and $8$ magenta
marbles. How many marbles must be drawn from the urn in order to
assure that there will be $15$ marbles of the same color?
\begin{answer} If all the magenta, all the
yellow, all the white, $14$ of the red and $14$ of the blue marbles
are drawn, then in among these $8 + 10 + 12 + 14 + 14 = 58$ there
are no $15$ marbles of the same color. Thus we need $59$ marbles in
order to insure that there will be $15$ marbles of the same color.
\end{answer}
\end{pro}

\begin{pro}
In a $4$ by $4$ magic square the sum of the four entries in each
row, in each column, and in each of the two main diagonals are all
the same. If the magic square shewn below is completed, find  $A+B$.


\vspace*{1cm}
\PuzzleSolution
\begin{Puzzle}{4}{4}%
|{\empty }|{\empty }|{7}|{12}|.
|{A}|{4}|{9}|{\empty}|.
|{10}|{5}|{\empty}|{3}|.
|{8}|{ 11 }|{\empty}|{B}|.\end{Puzzle}
\begin{answer}
$28$
\end{answer}
\end{pro}
\begin{pro}
In each of the following overlapping rings the sum of the entries is
$55$. What is $A-B$? \vspace*{2cm}
$$\psset{unit=3pc} \cnodeput(1;30){A}{11}  \cnodeput(1;90){B}{8}
 \cnodeput(1;150){C}{A}
\cnodeput(1;210){D}{9} \cnodeput(1;270){E}{9} \cnodeput(1;330){F}{B}
\ncline{A}{B}\ncline{B}{C}\ncline{C}{D}\ncline{D}{E}\ncline{E}{F}
\ncline{F}{A} \rput(1.75,0){\cnodeput(1;30){A}{2}
\cnodeput(1;90){B}{14}
 \cnodeput(1;150){C}{11}
\cnodeput(1;210){D}{B} \cnodeput(1;270){E}{7}
\cnodeput(1;330){F}{13}
\ncline{A}{B}\ncline{B}{C}\ncline{C}{D}\ncline{D}{E}\ncline{E}{F}
\ncline{F}{A}}      $$ \vspace*{1cm}
\begin{answer}
$2$
\end{answer}
\end{pro}
\begin{pro}
Let $\mathbb{E}=\{0,2,4,6, \ldots \}$ be the set of even natural
numbers. Is this set closed under addition? Is this set closed under
multiplication.
\begin{answer}
Yes. Yes.
\end{answer}


\end{pro}
\begin{pro}
Let $\mathbb{O}=\{1,3,5,7, \ldots \}$ be the set of odd natural
numbers. Is this set closed under addition? Is this set closed under
multiplication.
\begin{answer}
No. Yes.
\end{answer}
\end{pro}
\begin{pro}
Define the operation $\oplus$ of ``sooper-dooper addition'' by
$a\oplus b = a+b+10$, where $a$ and $b$ are natural numbers. For
example $2\oplus 3 = 2+3+10 = 15$ and $4\oplus 10 = 4+10+10 = 24$.
Is sooper-dooper addition a commutative operation? Is sooper-dooper
addition an associative operation?
\begin{answer}
It is commutative, because for arbitrary $a$ and $b$ we have
$a\oplus b = a+b+10$ and $b\oplus a = b+a+10$. Since $a+ b+10 =
b+a+10$ by the commutativity of the natural numbers, $\oplus$ is
commutative. It is also associative. For
$$ a\oplus (b\oplus c) = a\oplus (b+c+10) = a+(b+c+10)+10=a+b+c+20. $$Also
$$ (a\oplus b)\oplus c = (a+ b+10)\oplus c+ = (a+ b+10)+c+10=a+b+c+20. $$Since $ a\oplus (b\oplus c) =  (a\oplus b)\oplus c$, the operation is associative.
\end{answer}
\end{pro}

\begin{pro}\label{pro:domino}
Consider $2005$ hexagonal ``domino'' pieces with the numbers
$1,2,3,4,5,6$ written on the edges in clockwise fashion, as in
figure \ref{fig:domino}. A ``chain'' is formed so that domino rules
are observed. This means that two edges from different pieces are
joined only when the numbers on the edges agree. The numbers on the
edges joining two different domino pieces are now deleted and the
remaining numbers are now added. What is the maximum value of this
sum? \vspace*{1cm}
\begin{figurehere}
\rput(2,-.5){\begin{graph}(1,1) \roundnode{A}(1,0)
\roundnode{B}(.5,.866) \roundnode{C}(-.5,.866) \roundnode{D}(-1,0)
\roundnode{E}(-.5,-.866) \roundnode{F}(.5,-.866)  \edge{A}{B}
\edge{B}{C} \edge{C}{D} \edge{D}{E} \edge{E}{F} \edge{F}{A}
\edgetext{A}{B}{1} \edgetext{B}{C}{6}
 \edgetext{C}{D}{5}  \edgetext{D}{E}{4}  \edgetext{E}{F}{3}  \edgetext{F}{A}{2}
\end{graph}}
\meinecaption{1}{Problem \ref{pro:domino}.}\label{fig:domino}
\end{figurehere}
\begin{answer}In order to obtain the maximal sum, we must have a chain
where every piece is glued to the smallest possible of sides using
the smallest numbers. The chain then must have $2003$ pieces which
are joined on two sides and the two end pieces are joined at only
one side.  If the $1$ and the $2$ were shared, the first and second
piece must be joined at $1$ and the second and third at $2$. But
this is impossible because then the first and the third piece would
be joined at $3$ and $6$, violating domino rules. Hence only $3$'s
and $1$'s are shared. The middle $2003$ contribute $2+4+5+6=17$
each,  one of the end pieces contributes $2+3+4+5+6=20$, and the
other end piece contributes  $1+2+4+5+6=18$ hence the maximum sum is
$(2003)(17)+1(20) + 1(18)=34089.$
\end{answer}
\end{pro}
\end{multicols}

\chapter{Arithmetic Progressions}
\begin{df}
An arithmetic progression is a succession of numbers where you start
with a given number and form the next element by always adding the
same fix quantity, that is, there are numbers $a$ and $d$ such that
the succession of numbers takes the form:
$$ a, \quad a+d,\quad a+d+d=a+2d, \quad \quad a+d+d+d=a+3d, \ldots ,
$$Here $a$  is called the {\em first term},  $a+d$ is called the {\em second term}, etc., and $d$ is the {\em
common} difference.
\end{df}
\begin{exa}\label{exa:arit-pro-1}
The arithmetic progression $$ 3, 10, 17, 24, \ldots ,  $$has first
term $3$ and common difference $7$.
\end{exa}


\bigskip

Observe that in example \ref{exa:arit-pro-1}, the pattern of
additions is
$$3=3+7\cdot 0, \quad 10 = 3+7\cdot 1, \quad   17 = 3+7\cdot 2, \quad 24=3+7\cdot 3, \ldots,$$
Thus if we wanted to find the fifth term we would compute $3+7\cdot
4 =3+28=31$. If we wanted to compute the hundredth term we would
find $3+7\cdot 99 = 696$.

\begin{exa}
Consider the arithmetic progression
$$5, 11, 17, 23, \ldots  .  $$
\begin{enumerate}
\item Find its next term.
\item Find its hundredth term.
\item Find a formula for the term in position $n$.
\item Is $100$ a term in this progression? If so, which term is it?
\item Is $101$ a term in this progression? If so, which term is it?
\end{enumerate}
\end{exa}
\begin{solu}
The first term is $5$ and the common difference is $6$. The term
that follows $23$ is $23+6=29$. Notice that
$$5=5+6\cdot 0, \quad 11 = 5+6\cdot 1, \quad   17 = 5+6\cdot 2, \quad 23=5+6\cdot 3, \ldots,$$
and so the hundredth term is $5+6\cdot 99 =599$. From this pattern
we infer that the term in position $n$ is $5+6(n-1)$.



Each time we are adding a multiple of $6$ to $5$. Hence the numbers
in this progression leave remainder $5$ upon division by $6$. Since
$100=4+16\cdot 6$ leaves remainder $4$ upon division by $6$, the
number $100$ is not in this progression. However, $101=5+16\cdot 6$,
which means that $101$ is the $17$th term of this progression. We
have in fact,
$$5, 11, 17, 23,  29, 35, 41, 47, 53, 59, 65,  71, 77, 83, 89, 95, 101,\ldots  .  $$
so we explicitly see how to get to $101$.
\end{solu}


\begin{exa}
How many terms are in the arithmetic progression
$$ 7,\quad  15,\quad  23, \quad \ldots ,\quad  1207? $$
\end{exa}
\begin{solu}
Observe that the first term is $7$ and that the common difference is
$8$. The terms have the law of formation
$$7= 7 +8\cdot 0, \quad 15= 7 +8\cdot 1, \quad  23= 7 +8\cdot 2, \quad \ldots,  $$
so each time we are adding $7$ to a multiple of $8$. Now, simply
divide $1207$ by $8$ to find
$$1207 = 7 + 8\cdot 150.  $$This means that $1207$ is the $151$st
position, that is, that there are $151$ terms in this progression.
\end{solu}


The terms of an arithmetic progression display a symmetry that makes
summing consecutive terms easy. The following classic example is
reputed to have been solved in seconds by the then seven-year-old K.
F. Gauss (a XIX century mathematician).

\begin{exa}
Find the sum of all the natural numbers from $1$ to $100$,
inclusive.
\end{exa}
\begin{solu}
The trick is to consider the fifty pairs
$$ 100+1, \quad 99 + 2, \quad 98+3, \quad \ldots , \quad 51+50. $$
Each pair adds up to $101$ and there are $50$ of them, hence
$$ 1+2+3+\cdots + 100 = 101\cdot 50=5050. $$
\end{solu}
The series of multiples of a natural number $n$ forms an arithmetic
progression with common difference $n$. For example, the multiples
of $3$ are
$$0,3,6,9,12,15,\ldots ,  $$
an arithmetic progression with common difference $3$, and the
multiples of $4$ are
$$ 0,4,8,12,16, \ldots , $$
an arithmetic progression with common difference $4$. The {\em least
common multiple} $\lcm (3,4)$ is the smallest strictly positive
number that is common in both lists, in this case $\lcm (3,4) =12$.
\begin{exa}
Find $\lcm (12,45)$. \label{exa:lcm-12-45}
\end{exa}
\begin{solu}
We form the progressions of the multiples of both numbers and circle
the first nonzero repeat:
$$ 0,12,24,36,48,60,72,84,96,108,120, 132, 144,156,168, \psovalbox{180}, 192, \ldots, $$
$$ 0, 45, 90, 135, \pscirclebox{180}, 205,\ldots  $$ whence $\lcm
(12,45)=180$.
\end{solu}





The following sort of arithmetic progressions will be useful when we
study the division algorithm. Every natural number is either even or
odd, so we write
$$ \naturals = \{0,2,4,6,\ldots \}\cup \{1,3,5,7,\ldots\}, $$
where the symbol $\cup$ (read {\em union}) means that both sets are
joined. Notice that both sets are arithmetic progressions with
common difference $2$. Again, if the divisor is $3$, then the
possible remainders when we divide by $3$ are $0$, $1$, and $2$, and
we find
$$ \naturals = \{0,3,6,9,\ldots \}\cup \{1,4,7,10,\ldots\}\cup \{2,5,8,11,\ldots\}. $$
We say that these sets {\em partition} the natural numbers.
\begin{exa}
Partition the natural numbers into five infinite arithmetic
progressions.
\end{exa}
\begin{solu}Upon division by $5$, every natural number leaves
remainder $0$, $1$, $2$, $3$, or $4$. Hence we may take
$$ \naturals = \{0,5,10,15,\ldots \}\cup \{1,6,11,16,\ldots\}\cup \{2,7,12,17,\ldots\}\cup \{3,8,16,21,\ldots\}\cup \{4,9,13,18,\ldots\}. $$

\end{solu}


\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Find the $20$th number on the list
$$8, 11, 14, \ldots  $$ assuming that the pattern is preserved.
\begin{answer}
$8 + 3\cdot 19=65$.
\end{answer}
\end{pro}
\begin{pro}
Find the $1000$th number on the list
$$1, 10, 19, \ldots  $$ assuming that the pattern is preserved.
\begin{answer}
$1 + 9\cdot 999=8992$.
\end{answer}
\end{pro}
\begin{pro}
How many terms are there in the arithmetic progression
$$8, 11, 14, \ldots  3005? $$What is their sum?
\begin{answer}
Observe that to $8$ we must add multiples of $3$. Dividing $3005$ by
$3$ we obtain
$$ 3005=3\cdot 1001+2 = 3\cdot 999 + 3\cdot 2 +2 =3\cdot 999 +8, $$
and hence there are $999+1=1000$ terms.

Adding the $500$ pairs
$$ 8+3005=3013, \quad 11+3002=3013, \quad 14+2999=3013, \quad \ldots , \quad 1505+1508=3013, $$
we see that the sum is
$$8+ 11+14+ \cdots +  3005=500\cdot 3013 =1506500.  $$

\end{answer}
\end{pro}

\begin{pro}Find a formula for the $n$-th term of the arithmetic progression
$$ 2, 7, 12, 17, \ldots . $$
\begin{answer}
We start with $2$, and then keep adding $5$, thus
$$ 2= 2+5\cdot 0, \qquad  7= 2+5\cdot 1, \qquad  12= 2+5\cdot 2, \qquad  17= 2+5\cdot 3, \ldots. $$
The general $n$th term is therefore of the form $2+5(n-1)$, where
$n=1,2,3,\ldots$ is a natural number.
\end{answer}
 \end{pro}

\begin{pro}
Find a general formula for the $n$-th term of the arithmetic
progression $$1, 7, 13, 19, 25, \ldots .  $$
\begin{answer} $1+6(n-1)$ where $n=1,2,3,\ldots$.\end{answer}
\end{pro}

\begin{pro}
You start counting by $11$s, starting with $3$:\qquad\qquad  $$
3,14,25,36, \ldots .  $$ Which number appears in the $100$th
position? Give a formula for the number appearing in the $N$th
position.
\begin{answer}
 The pattern is as
follows,
$$ \begin{array}{|l|lll|}
\hline
\mathrm{Position} & \multicolumn{3}{|c|}{Number} \\
\hline
1 &  3 & = & 3 + 11\cdot 0 \\
2 &  14 & = & 3 + 11\cdot 1 \\
3 &  25 & = & 3 + 11\cdot 2 \\
4 &  36 & = & 3 + 11\cdot 3 \\
5 &  47 & = & 3 + 11\cdot 4 \\
6 &  59 & = & 3 + 11\cdot 5 \\
7 &  60 & = & 3 + 11\cdot 6 \\
\vdots &  \vdots & \vdots & \vdots \\
100 &  1092 & = & 3 + 11\cdot 99 \\

 \hline  \end{array}$$
A general formula is $3 + 11(N-1)$ for $N =1,2,3, \ldots$.
\end{answer}
\end{pro}

\begin{pro}
Find $\lcm (30,36)$. \begin{answer} $180$
\end{answer}
\end{pro}

\begin{pro}
Find $\lcm (24,36)$. \begin{answer} $72$
\end{answer}
\end{pro}
\begin{pro}
Find $\lcm (30,24)$. \begin{answer} $120$
\end{answer}
\end{pro}
\begin{pro}
Partition the natural numbers into the union of four infinite
arithmetic progressions.
\begin{answer}Upon division by $4$, every natural number leaves
remainder $0$, $1$, $2$, or $3$. Hence we may take
$$ \naturals = \{0,4,8,12,\ldots \}\cup \{1,5,9,13,\ldots\}\cup \{2,6,10,14,\ldots\}\cup \{3,7,11,15,\ldots\}. $$
\end{answer}
\end{pro}

\end{multicols}


\chapter{Multiplication}

\begin{df}
Given natural numbers $m$ and $n$, we define their {\em product}
$mn$ as
$$mn = \underbrace{n + n + \cdots + n}_{m\ {\rm times}}.  $$
The operation between $m$ and $n$  is called {\em multiplication}.
Here $m$ is called the {\em multiplier}, $n$ the {\em multiplicand}.
Both $m$ and $n$ are called {\em factors}.
\end{df}
\begin{rem} Ordinarily in Algebra, we use {\em juxtaposition}, that
is, putting one letter next to another, to indicate multiplication.
If we were multiplying two numbers, we would use Leibniz's raised
dot, as in $2\cdot 3$. We do not use the cross symbol $ \times $ so as
to avoid confusion with the letter $x$.
\end{rem}

A geometrical interpretation of multiplication is the following.
Consider an array with identical squares of area $1$ of width $m$
squares and height $n$, as in figure \ref{fig:mult-nat2}.
 The product $mn$ is the number of squares, which is also the area
 of the rectangle obtained once the squares are packed together and
 all the white space is eliminated.
\vspace{2cm}

\begin{figure}[h]
\centering \psset{unit=2pc}
\def\cuadradito{\pstGeonode[PointName=none](0,0){O}(.75,0){A}(.75,.75){B}(0,.75){C}\pscustom[fillcolor=gray,fillstyle=solid]{\pstLineAB{O}{A}\pstLineAB{A}{B}\pstLineAB{B}{C}\pstLineAB{C}{O}}}
\rput(-2.5,-2.5){\cuadradito} \rput(-1.5,-2.5){\cuadradito}
\rput(0,-2){$\bullet\bullet\bullet$}\rput(1.5,-2.5){\cuadradito}\rput(2.5,-2.5){\cuadradito}
\pcline[offset=-12pt]{|-|}(-2.5,-2.5)(3.5,-2.5)\ncput*{$m$ squares}
\rput(-2.5,-1.5){\cuadradito}\rput(2.5,-1.5){\cuadradito}
\rput(-2.5,-.5){\cuadradito}\rput(2.5,-.5){\cuadradito}
\rput(-2.5,1.5){\cuadradito}\rput(2.5,1.5){\cuadradito}
\rput(0,-1){$\bullet\bullet\bullet$}
\rput(0,0){$\bullet\bullet\bullet$} \rput(-2,1){$\bullet$}
\rput(-2,.75){$\bullet$}
\rput(2.75,1){$\bullet$}\rput(2.75,.5){$\bullet$}\rput(2.75,.75){$\bullet$}
\rput(-2,.5){$\bullet$}
\pcline[offset=-12pt]{|-|}(-3.5,-2.5)(-3.5,2.25)\ncput*[nrot=:U]{$n$
squares} \meinecaption{3}{Multiplication in $\naturals$.}
\label{fig:mult-nat2}
\end{figure}




Just like addition of natural numbers, multiplication satisfies the
following axioms.



\begin{axi}[Closure] The product of any two natural numbers is a natural number, that is, $\naturals$ is {\em closed under multiplication}: if $a\in
\naturals$ and $b\in \naturals$ then also $ab\in \naturals$.
\end{axi}
\begin{axi}[Multiplicative Identity] If $1$ is multiplied to any natural number, the result is unchanged, that is $1\in\naturals$ is the
{\em multiplicative identity} of $\naturals$. It has the property
that for all $x\in\naturals$ it follows that $$x=1\cdot x=x\cdot 1.
$$
\end{axi}
\begin{axi}[Commutativity]Let $a\in\naturals$ and $b\in \naturals$ be arbitrary. Then
\mbox{$ab=ba$}.
\end{axi}
\begin{axi}[Associativity]
Let $a, b, c$ be arbitrary natural numbers. Then the order of
parentheses when performing multiplication is irrelevant, that is,
$$ a(bc)=(ab)c = abc.  $$
\end{axi}

Since multiplication is stenography for addition, it is to be
understood that in a computation that involves sums and
multiplications, multiplications must be carried out first, unless
some other computation is   coerced by parentheses. For example, in
$$ 2+3\cdot 4 =2+12=14, $$
we perform the multiplication first, but in
$$ (2+3)\cdot 4=5\cdot 4=20, $$
the addition is coerced by the parentheses, and hence, we perform it
first.

\bigskip


The next property links multiplication and addition.

\begin{axi}[Distributive Law]
Let $a, b, c$ be natural numbers. Then
$$a(b+c) = ab+ac,  $$
and
$$(a+b)c=ac+bc.  $$
\end{axi}
For example,
$$3(4+3)=3\cdot 7=21, \qquad 3\cdot 4 + 3\cdot 3 = 12 +9 =21.  $$
In the case when the factors have each two summands, the distribute
law takes the form \begin{equation}\label{eq:mult-binom}
(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd.
\end{equation}
For example,
$$ (3+4)(5+6)=3\cdot 5+ 3\cdot 6+ 4\cdot 5+4\cdot 6=15+18+20+24=77.  $$
\begin{rem}
Needless to say, we could perform in a faster manner
$(3+4)(5+6)=7\cdot 11 =77,$ but our desire was to illustrate the
distributive law.
\end{rem}

\bigskip

Recall now the operation of exponentiation, which we have already
defined as, for $a\in \naturals$ and $n\in \naturals$, $n>0$,
$$a ^n =  \underbrace{a\cdot a\cdots a\cdot a}_{n\ a\mathrm{'s}}.  $$
We will have the convention that $a^0=1$ when $a\neq 0$, that is,
any non-zero number raised to the zeroth power is $1$ and that
$a^1=a$, that is, if the exponent is $1$ we do not write the
exponent.

\bigskip



We now analyse how to deal with products of exponential expressions
with the same base. For example, observe that
$$2^2=4, \quad 2^3=8,  \quad 2^5=32, \implies 2^2\cdot 2^3 =4\cdot 8 = 32 =2^5,  $$
and we deduce that $2^{2}\cdot 2^3=2^{2+3}=2^5$, that is, when
multiplying two exponential expressions with the same base, we added
the exponents. In general we have the following result.
\begin{thm}[First Law of Exponents]Let $a$ be a real number and $m, n$ natural numbers. Then
$$a^ma^n = a^{m+n}.  $$
\end{thm}
\begin{pf}
We have $$\begin{array}{lll}a^ma^n &  =  & \underbrace{a\cdot a
\cdots a}_{m\ a\mathrm{'s}}\cdot \underbrace{a\cdot a \cdots a}_{n\
a\mathrm{'s}} \\
& = & \underbrace{a\cdot a \cdots a}_{m+n\ a\mathrm{'s}}\\
 & = & a^{m+n}.
\end{array}$$
\end{pf}
\begin{exa}
We have: $(5x)(2x)=10x^{1+1}=10x^2$.
\end{exa}
\begin{exa}
We have: $(3x^2)(6x^4)=18x^{2+4}=18x^6$.
\end{exa}
Using the first law of exponents and the distributive law, we will
now give an algorithm for multiplication. Before explaining the
multiplication algorithm in general, let us consider an example.
Suppose we wanted to multiply $34$ to $56$.
 We go through
the digits of $56$, starting left to right. We  multiplied every
digit of $34$ by $6$, giving us
$$ 6\cdot 34=6(30+4)=6\cdot 30 + 6\cdot 4 = 180 + 24 =204.  $$
Now we move to the digit $5$ of $56$. But the value of this digit is
$50$, and so we multiply
$$ 50\cdot 34 = 50 (30+4)=50 \cdot 30 + 50\cdot 4 = 1500 +200 =1700. $$
We conclude that
$$ 34\cdot 56 = 34(50+6)=34\cdot 50+34\cdot 6 = 1700+204=1904.  $$
It is customary to display the result in the following manner:
$$ \opmul{34}{56} $$
Notice that here we shifted the multiplication by $5$ one unit left,
and pretended that we were in fact multiplying by $5$ and not $50$.
The two approaches have the same effect. If we wanted to see more
gory detail, and perhaps have a glimpse of how this algorithm
generalises to Algebra, we would probably proceed as follows. Since
$$ 34=3\cdot 10 +4, \quad 56=5\cdot 10+6, $$ using the distributive
law (\ref{eq:mult-binom}) we obtain,
$$ (3\cdot 10 +4)(5\cdot 10+6) = 3\cdot 10\cdot 5 \cdot 10  +3\cdot 10 \cdot 6 + 4\cdot 5 \cdot 10 + 4\cdot 6 =
1500+ 180+200+24= 1904. $$ The above ``vertical multiplication''
algorithm generalises to algebraic expressions. Suppose now that we
wanted to multiply $(3x+4)(5x+6)$. We first multiply every term of
$3x+4$ by $5x$:
$$\psmatrix &  & 3x & + & 4 \\
\cdot &  & 5x & + & 6 \\
\hline & \red{15x^2} & + & \red{20x} \\
 \endpsmatrix \ncline[arrows=->,linecolor=red]{2,3}{1,3}\ncline[arrows=->,linecolor=red]{2,3}{1,5}$$
We now multiply every term of $3x+4$ by $6$, lining up the terms of
the same exponent of $x$ (in our case, exponent $1$ and exponent
$0$) with those of the first partial product:
$$\psmatrix &  & 3x & + & 4 \\
\cdot &  & 5x & + & 6 \\
\hline & 15x^2 & + & 20x \\
& & + & \red{18x} & + & \red{24}  \\
 \endpsmatrix    \ncline[arrows=->,linecolor=red]{2,5}{1,3}\ncline[arrows=->,linecolor=red]{2,5}{1,5}$$

 \bigskip

 Finally, we add (collect) like terms:
 $$\begin{array}{llllll}  &  & 3x & + & 4 \\
\cdot &  & 5x & + & 6 \\
\hline & 15x^2 & + & 20x \\
& & + & 18x & + & 24  \\
\hline
& \red{15x^2} & + & \red{38x} & + & \red{24}  \\
 \end{array}$$
This multiplication can also be displayed horizontally, in a more
obvious display of (\ref{eq:mult-binom}):
$$ (3x+4)(5x+6)=(3x)(5x)+ (3x)(6)+4(5x)+4(6)=15x^2+18x+20x+24 =15x^2+38x+24. $$
Observe that if we substitute $x=10$ in the above we obtain,
$$15\cdot 10^2 +38\cdot 10 +24 =1500 +380 + 24=1904.  $$
\bigskip



 It is now easy to describe a multiplication algorithm akin to the
 one used for addition. Let
$$ \underline{d_nd_{n-1}\ldots d_2d_1d_0},\qquad \underline{e_pe_{p-1}\ldots e_2e_1e_0}, $$
be two arbitrary decimal integers. To multiply them:
\begin{enumerate}
\item Multiply every digit of $\underline{d_nd_{n-1}\ldots
d_2d_1d_0}$ by $e_0$, taking into account all the possible carries.
\item Multiply every digit of $\underline{d_nd_{n-1}\ldots
d_2d_1d_0}$ by $e_1$, taking into account all the possible carries,
and write the result below the partial result for $e_0$, but shift
this partial result one unit to the left.
\item Keep multiplying by the $e_k$, taking care of all carries, and
each time shifting to the left each partial products.
\item Add all these partial products.
\end{enumerate}

\begin{exa}
Multiply $123$ by $456$, and display your result using the
``vertical'' multiplication algorithm. Then multiply $x^2+2x+3$ by
$4x^2+5x+6$ ``horizontally.''
\end{exa}
\begin{solu}
We have, $$ \opmul{123}{456} $$

and $$\begin{array}{lll} (x^2+2x+3)(4x^2+5x+6) & = &
x^2(4x^2+5x+6)+2x(4x^2+5x+6)+3(4x^2+5x+6)\\
& = & 4x^4+5x^3+6x^2+8x^3+10x^2+12x+12x^2+15x+18 \\
& = & 4x^4+13x^3+28x^2+27x+18.
\end{array}$$
Observe that if we substitute $x$ for $10$ we obtain
$$ 4\cdot 10^4 + 13\cdot 10^3 +28\cdot 10^2 + 27\cdot 10 + 18 =40000+13000+2800+270+18 = 56088.$$
\end{solu}

To conclude this section we will introduce the symbol $!$.
\begin{df}If $n\in \naturals$ we define $n!$ ($n$ {\em factorial}) as follows:
$$ 0! =1 , \quad 1! = 1, \quad 2! = 1\cdot 2=2, \quad 3! = 1\cdot 2 \cdot 3 = 6, \quad 4! = 1\cdot 2 \cdot 3\cdot 4 = 24, \quad n! = 1 \cdot 2 \cdots n.   $$
\end{df}

\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

\begin{pro}
If the figure shewn is folded to form a cube, then three faces meet
at every vertex. If  for each vertex we take the product of the
numbers on the three faces that meet there, what is the largest
product we get?\vspace{1in}
$$
\psset{unit=2pc} \psline(-1, 0)(3,0)(3,1)(-1,1)(-1,0)
\psline(0,2)(0,-1)(1,-1)(1,2)(0,2) \psline(2,0)(2,1)
\rput(-.5,.5){4} \rput(.5, .5){2} \rput(1.5, .5){5} \rput(2.5,
.5){6} \rput(.5, 1.7){1} \rput(.5, -.7){3}
$$
\vspace{.25in}
\begin{answer}
Observe that $4$ and $5$ are on opposite sides, so they never meet.
The largest product is $3 \cdot  5  \cdot  6 = 90.$
\end{answer}
\end{pro}
\begin{pro}
Pencils are $8$ \textcent each. How much would $7$ pencils cost?
\begin{answer}
$56$ \textcent
\end{answer}
\end{pro}
\begin{pro}
Mr. Schremmer's class was doing a science experiment. There
were $7$ groups in the class. Each group got $4$ test tubes.
How many test tubes did the class use?
\begin{answer}
$28$
\end{answer}
\end{pro}
\begin{pro}
In some years an extra second is added on $31$ December to keep the
clocks in time with the rotation of the earth. How many seconds does
$31$ December have when this happens?
\begin{answer}
$24\cdot 60\cdot 60 +1=86401$
\end{answer}
\end{pro}
\begin{pro}
Sound travels at approximately $330$ meters per second. The sound of
an explosion took $28$ seconds to reach a person. Estimate how far
away was the person from the explosion.
\begin{answer}
$9240$ meters.
\end{answer}
\end{pro}
\begin{pro}
 According to experts the first $4$ moves in a chess game can be played in $197299$
totally different ways. If it takes $30$ seconds to make one move,
what time, in seconds, would it take one player to try every
possible set of $4$ moves?
\begin{answer}
$30 \cdot 197299 = 5918970$ seconds.
\end{answer}
\end{pro}
\begin{pro}
When writing all the natural numbers from $1$ to $999$ in a row, as
follows,
$$123456789101112\ldots 998999  $$how many digits have I used?
\begin{answer}
I have written $9$ $1$-digit integers ($1$ through $9$), $90$
$2$-digit integers (from $10$ to $99$) and $900$ $3$-digit integers
(from $100$ to $999$). Altogether I have used
$$1\cdot 9 + 2\cdot 90 + 3 \cdot 900=2889 $$ digits.
\end{answer}
\end{pro}
\begin{pro}
The integers from $1$ to $1000$ are written in succession. Find the
sum of all the digits.
\begin{answer}
Pair up the integers from $0$ to $999$ as
$$ (0, 999), \ (1, 998), \ (2, 997), \ (3, 996), \ \ldots , (499, 500).
$$Each pair has sum of digits $27$ and there are $500$ such pairs.
Adding $1$ for the sum of digits of $1000$, the required total is
$$27\cdot 500 + 1 = 13501.  $$
\end{answer}
\end{pro}
\begin{pro}
All the natural numbers---starting with $1$---are listed
consecutively
$$123456789101112131415161718192021\ldots$$Which digit occupies the $1002$nd place?
\begin{answer} When the number $99$ is written down, we have used
$$1\cdot 9 + 2\cdot 90 = 189$$ digits. If we were able to write
999, we would have used
$$1\cdot 9 + 2\cdot 90  + 3\cdot 900 =  2889$$digits, which is more than $1002$ digits. The $1002$nd
digit must be among the three-digit positive integers. We have $1002
- 189 = 813$ digits at our disposal, from which we can make
$\lfloor\dfrac{813}{3}\rfloor = 271$ three-digit integers, from 100
to $370$. When the $0$ in $370$ is written, we have used $189 +
3\cdot 271 = 1002$ digits. The $1002$nd digit is the $0$ in $370$.
\end{answer}
\end{pro}
\begin{pro}
Compute the following:
$$1^2, \quad 11^2, \quad 111^2, \quad 1111^2, \quad 11111^2, \quad 111111^2.   $$
Do you notice a pattern?
\begin{answer}We have,
$$1^2=1, \quad 11^2=121, \quad 111^2=12321, \quad 1111^2=1234321, \quad 11111^2=123454321, \quad 111111^2=12345654321.   $$
\end{answer}
\end{pro}
\begin{pro}
Multiply the numbers $12$ and $45$. Then multiply $x+2$ and $4x+5$.
\begin{answer}
We have,
$$ \opmul{12}{45} $$
Multiplying,
$$ \begin{array}{ccccc} &  &    x & + & 2 \\
\cdot  &  &    4x & + & 5 \\
\hline 4x^2 & + & 8x & &  \\
 &  & 5x & + & 10 \\
 \hline
 4x^2 & + & 13x & + & 10.
\end{array}$$
\end{answer}
\end{pro}
\begin{pro}
Compute $5! - 4!$.
\begin{answer}
$120-24 = 96$
\end{answer}
\end{pro}
\begin{pro}
Peter is $x$ years old, Paul is $y$, and Mary is $z$ years. What is
the product of their ages?
\begin{answer}
$ xyz  $
\end{answer}
\end{pro}
\begin{pro}
A farmer has $7$ ducks. He has $5$ times as many
chickens as ducks. How many more chickens
than ducks does he have?
\begin{answer}
$7\cdot 5 = 35 $
\end{answer}
\end{pro}
\begin{pro}
The sum of three consecutive integers is $2007$. Find their product.
\begin{answer}
$668\cdot 669\cdot 670=299417640$.
\end{answer}
\end{pro}
\begin{pro}
Multiply: $(2x^2)(4x^3)$.
\begin{answer}
$8x^5$.
\end{answer}
\end{pro}
\begin{pro}
Find the missing digits in the following product.
$$ \begin{array}{llllll}  &  &   &  & 2 & \blacksquare\\
\times &  &   &  &  \blacksquare & 9\\
\hline
 &  &   &  \blacksquare& \blacksquare & \blacksquare\\
 &  &   &  \blacksquare& 5 & \\
\hline
 &  &   &  4 & \blacksquare &  5\\
 \end{array}  $$
 \begin{answer}
$ 25\times 19 = 475.  $
 \end{answer}
\end{pro}

\begin{pro}
Find the numerical value of\quad  $3\cdot 4 + 4^2$.
\begin{answer}
$28$.
\end{answer}
\end{pro}
\begin{pro}
Find the numerical value of\quad  $1^12^23^3$.
\begin{answer}
$108$.
\end{answer}
\end{pro}
\begin{pro}
Prove that
$$ (3 + 2)^2 =25. $$

\end{pro}
\begin{pro}
 Find the
last two digits of the integer $$ 1! + 2! + 3! + \cdots + 100!.
$$
\begin{answer} For $k \geq 10$, $k!$ ends in two or more $0$'s. Thus the last
two digits of the desired sum are the last two digits of
$$1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 409113,
$$thus the desired sum ends in $13$.
\end{answer}
\end{pro}
\begin{pro}
Prove that
$$(2)(4) + 3^3  = 35.  $$
\end{pro}

\begin{pro}
Prove that
$$3^2 + 2^3 = 17. $$
\end{pro}
\begin{pro}
Prove that
$$(3^2)(4)(5) = 180.  $$
\end{pro}

\begin{pro}
Prove that
$$ (5 + (3 + 2(4))^2)^3 =   2000376.$$

\end{pro}
\begin{pro}
Calculate: $$(123^2\cdot 456^2 - 789  )^0 + 3\cdot 2^3.$$
\begin{answer}
 $(123^2\cdot 456^2 - 789  )^0 + 3\cdot 2^3=1+3\cdot 2^3=1+3\cdot 8 =
 1+24=25$.
\end{answer}
\end{pro}

\begin{pro}
 Dale should have divided a number by $4$, but instead he
subtracted $4$. He got the answer $48$. What should his answer have
been?
\begin{answer} We work backwards. He obtained $48$ from $48 + 4 = 52$. This means
that he should have performed $52 \div 4 = 13$.
 \end{answer}
 \end{pro}
\begin{pro} When a number is multiplied by $3$ and
then increased by $16$, the result obtained is $37$. What is the
original number?
\begin{answer} We work backwards as follows. We obtained $37$ by adding $16$ to
$37 - 16 = 21$. We obtained this $21$ by multiplying by $3$ the
number $21 \div 3 = 7$. Thus the original number was a $7$.
 \end{answer}
\end{pro}
\begin{pro}
A crossnumber is like a crossword except that the answers are
numbers with one digit in each square. Complete the following
crossnumber.

\vspace*{1cm}

\begin{minipage}{3cm}
\begin{Puzzle}{3}{3}%
|* |[1]{\empty}|[2]{\empty}|.
|[3]{\empty}|*|{\empty}|.
|[4]{\empty}|{\empty} |{\empty}|.
\end{Puzzle}
\end{minipage}
\begin{minipage}{5cm}
\begin{PuzzleClues}{\textbf{ACROSS}}\\
\Clue{1}{}{\mbox{See 3D.}}\\  \Clue{3}{}{Cube.}\\ \Clue{4}{}{\mbox{5 times 3D.}}\\
\end{PuzzleClues}
\begin{PuzzleClues}{\textbf{DOWN}}\\
\Clue{2}{}{Square} \\ \Clue{3}{}{\mbox{4 times 1A.}}\\
\end{PuzzleClues}
\end{minipage}

\end{pro}

\begin{pro}
Which of the digits $\{3,4,6,8,9\}$ cannot appear as one of the
missing digits in the product
$$\opmul[voperator=bottom,operandstyle.1.1=\hole, operandstyle.1.2=\hole,operandstyle.2.1=\hole, operandstyle.2.2=\hole, resultstyle.2=\hole,
resultstyle.4=\hole,carrysub=false] {5671}{1920}$$
\begin{answer}
$5671-1920=3751$
\end{answer}
\end{pro}
\begin{pro}
A quiz has $25$ questions with four points awarded for each correct
answer and one point deducted for each incorrect answer, with zero
for each question omitted. Anna scores $77$ points. How many
questions did she omit?
\begin{answer}
Anna answered $20$ questions correctly (She could not answer less
than $20$, because then her score would have been less than
$19 \cdot  4 = 76 < 77$; She could not answer more than $20$, because
her score would have been at least $21 \cdot  4 - 3=81>77$). To get
exactly $77$ points Anna had to answer exactly $3$ questions wrong,
which means she omitted $2$ questions.
\end{answer}
\end{pro}
\begin{pro}Doing only {\em one}  multiplication,
prove that
$$\begin{array}{lcl}(666)(222) +(1)(333)  +(333)(222) & & \\
  +(666)(333)  + (1)(445) + (333)(333) & & \\\qquad    + (666)(445) + (333)(445) +
(1)(222) & = & 1000000.  \end{array}$$
\begin{answer}
From the distributive law,
$$\begin{array}{lcl}
(666)(222) +(1)(333)  +(333)(222) & & \\
\quad  +(666)(333) + (1)(445) + (333)(333) & & \\
\quad + (666)(445) + (333)(445) + (1)(222) & = & (666 + 333 +
1)(445 + 333 + 222) \\
& = & (1000)(1000) \\
& = & 1000000.
\end{array}$$
\end{answer}
\end{pro}

\begin{pro}
A certain calculator  gives as the result of the product
$$ 987654  \cdot  745321 $$ the number $7.36119E11$, which means
$736, 119, 000, 000$. Explain how to find the last six missing
digits.
\begin{answer}
The trick is to use a technique analogous to the one for
multiplying, but this time three-digits at a time:
$$321 \cdot  654=209934,   $$
$$321 \cdot  987 = 316827,  $$
$$745 \cdot  654 = 487230,  $$
$$745  \cdot  987 = 735315.  $$
Thus
$$\begin{array}{llll}
  & & \psframebox{987} &\psframebox{654}\\
 \cdot  & & \psframebox{745} &\psframebox{321}\\
\hline  &  & 209 & 934 \\
& 316 & 827 & \\
& 487 & 230 & \\
735 & 315 & & \\
\hline
736& 119 & 266 & 934 \\
\end{array}   $$
\end{answer}
\end{pro}

\begin{pro}
 How many digits
does $4^{16}5^{25}$ have?
\begin{answer} There are $28$ digits, since
$$4^{16}5^{25} = 2^{32}5^{25} = 2^{7}2^{25}5^{25} = 128 \cdot
10^{25},
$$which is the $3$ digits of $128$ followed by $25$ zeroes.
\end{answer}
\end{pro}


\begin{pro}Each element of the set $$\{10, 11, 12, \ldots , 19, 20\}$$ is multiplied by each element of
the set $$\{21, 22, 23, \ldots , 29, 30\}.$$ If all these products
are added, what is the resulting sum? \begin{answer} This is asking
for the product $(10 + 11 + \cdots + 20)(21 + 22 + \cdots + 30)$
after all the terms are multiplied. But $10 + 11 + \cdots + 20 =
165$ and $21 + 22 + \cdots + 30 = 255.$ Therefore we want
$(165)(255) = 42075$.
\end{answer}
\end{pro}
\end{multicols}



\chapter{Primes and Factorisation}

We have already explored the multiplication of  natural numbers. We
will now consider the reverse procedure, that is, given a number, we
would like to obtain two (or more) numbers whose product is the
given number. \begin{df}To {\em factor} a natural number greater
than $1$ means to find two or more natural numbers whose product is
equal to the original number. The multiplication itself is called a
{\em factorisation} of the number. The numbers in the product are
called {\em factors}  of the original number.\end{df}
\begin{exa}One factorisation of $20$ is $20=4 \cdot  5$. Here $4$ and
$5$ are the factors. Other factorizations of $20$ are $20=1 \cdot
20$, or $20=2 \cdot  10$, or $20=2 \cdot  2  \cdot  5$.\end{exa}
\begin{df}A {\em prime number} is a natural number greater than $1$
with exactly two factors: $1$ and the number itself. If the natural
number $n>1$ is not prime, we say that it is {\em composite}.
Lastly, we say that $1$ is a {\em unit}.\end{df} Notice that the
list of the primes starts as $2,3,5,7,11,13,17,19,23,\ldots$.

\begin{rem}Every natural number is either a unit, prime, or composite.
\end{rem}

\begin{df}
We write $a|b$, if the natural number $a\neq 0$ divides the natural
number $b$. We say that $a$ is a {\em divisor} of $b$. Notice that
$1$ and $b$ (if $b\neq 0$) are always divisors of $b$. A divisor of
$b$ is {\em proper}, if it is smaller than $b$.
\end{df}


\bigskip

A procedure, called {\em Eratosthenes' Sieve} allows us to find out
the small primes. For example, let us find all the prime numbers
smaller than $25$. We first display these numbers:
$$\begin{array}{llllll} 2 & 3 & 4 & 5 & 6 & 7 \\
 8 & 9 & 10 & 11 & 12 & 13 \\
 14 & 15 & 16 & 17 & 18 & 19 \\
 20 & 21 & 22 & 23 & 24 & 25. \end{array} $$
We leave   $2$ uncrossed, and cross out every other  multiple of
$2$:
$$\begin{array}{llllll} 2 & 3 & \cancel{4} & 5 & \cancel{6} & 7 \\
 \cancel{8} & 9 & \cancel{10} & 11 & \cancel{12} & 13 \\
 \cancel{14} & 15 & \cancel{16} & 17 & \cancel{18} & 19 \\
 \cancel{20} & 21 & \cancel{22} & 23 & \cancel{24} & 25. \end{array} $$
The next uncrossed number after $2$ is $3$. Hence, we leave $3$
uncrossed and cross out every other multiple of $3$ (if a multiple
of $3$ is already crossed, it does not matter):
$$\begin{array}{llllll} 2 & 3 & \cancel{4} & 5 & \cancel{6} & 7 \\
 \cancel{8} & \cancel{9} & \cancel{10} & 11 & \cancel{12} & 13 \\
 \cancel{14} & \cancel{15} & \cancel{16} & 17 & \cancel{18} & 19 \\
 \cancel{20} & \cancel{21} & \cancel{22} & 23 & \cancel{24} & 25. \end{array} $$
The next uncrossed number after $3$ is $5$. Hence, we leave $3$
uncrossed and cross out every other multiple of $5$ (if a multiple
of $5$ is already crossed, it does not matter):
$$\begin{array}{llllll} 2 & 3 & \cancel{4} & 5 & \cancel{6} & 7 \\
 \cancel{8} & \cancel{9} & \cancel{10} & 11 & \cancel{12} & 13 \\
 \cancel{14} & \cancel{15} & \cancel{16} & 17 & \cancel{18} & 19 \\
 \cancel{20} & \cancel{21} & \cancel{22} & 23 & \cancel{24} & \cancel{25}. \end{array} $$
We continue this procedure with the next uncrossed numbers. We see
that after $13$ we are already out of range. The numbers that remain
uncrossed are the primes. Hence the primes smaller than $25$ are It
$$ 2, 3, 5, 7, 11, 13, 17, 19, 23, \ldots . $$







 Table \ref{tab:primes} lists all primes under $1014$.
\begin{table}[!h]
\centering {\tiny\begin{tabular}{|rrrrrrrrrr|} \hline
 2  &   3 &     5 &    7 &    11 &    13 &    17 &    19 &    23 &    29\\
   31&    37 &   41 &   43  &  47&    53 &   59 &   61&    67&    71\\
   73 &   79 &   83&    89  &  97  & 101 &  103  & 107  & 109  & 113\\
  127  & 131&   137 &  139 &  149   &151   &157&   163   &167   &173\\
  179  & 181&   191 &  193  & 197   &199&   211 &  223 &  227   &229\\
  233  & 239&   241 &  251  & 257   &263&   269 &  271 &  277   &281\\
  283  & 293&   307 &  311  & 313   &317&   331 &  337 &  347   &349\\
  353  & 359&   367  & 373  & 379   &383&   389 &  397 &  401   &409\\
  419  & 421&   431 &  433  & 439   &443&   449 &  457 &  461   &463\\
  467  & 479&   487  & 491  & 499   &503&   509 &  521 &  523   &541\\
  547  & 557&   563 &  569  & 571   &577&   587 &  593 &  599   &601\\
  607  & 613&   617  & 619  & 631   &641&   643 &  647 &  653   &659\\
  661  & 673&   677 &  683  & 691   &701&   709 &  719 &  727   &733\\
  739  & 743&   751  & 757  & 761   &769&   773 &  787 &  797   &809\\
  811  & 821&   823 &  827  & 829   &839&   853 &  857 &  859   &863\\
  877  & 881&   883  & 887  & 907   &911&   919 &  929 &  937   &941\\
  947  & 953&   967  & 971  & 977   &983&   991 &  997 & 1009  &1013\\
 \hline
\end{tabular}
} \meinecaption{1}{Some Prime Numbers}\label{tab:primes}
\end{table}
Euclid proved (around 300 BC) that there are infinitely many prime
numbers, hence we can find arbitrarily large prime numbers, which
means that the table \ref{tab:primes} can be made as long as we
wish. \begin{rem}All prime numbers greater than $2$ are odd. But not
every odd number is prime. For example, $9$ is not prime, since it
has as three factors, $1$, $3$, and $9$.\end{rem} \begin{df}The
factorisation of a natural number greater than $1$ in which every
factor greater than $1$ is prime, is called the {\em prime
factorisation} of the number.\end{df} It can be proved that the
factorisation of a natural number into primes is unique, up to the
order of the factors. This means that if you and I start with the
same number and we factor it into primes, then we will end up with
the same factorisations, except perhaps in the order in which
display the factors.



 \begin{exa}Find the prime
factorisations of $360$ and of $216$.\label{exa:fact-216}\end{exa}
\begin{solu}
A schematic procedure is shown in figures \ref{fig:factor-360}. The
idea is to start small and then substitute equals by equals. Then
choose all the prime factors in the circles. We conclude that
$$360=2 \cdot  2  \cdot  2  \cdot  3  \cdot  3  \cdot  5 = 2^3 \cdot  3^2 \cdot  5,  $$
and that
$$ 216=2^33^3. $$
\end{solu}

\begin{figure}[!h]
\begin{minipage}{7cm}\centering \psset{unit=.25pc}
\pstree{\Toval{360}}%
{\Toval{2}%
\pstree{\Toval{180}}%
{\Toval{2}%
\pstree{\Toval{90}} %
{\Toval{2}%
\pstree{\Toval{45}} %
{
\Toval{5}%
\pstree{\Toval{9}}  %
{
\Toval{3}%
\Toval{3}%
} } } } }\meinecaption{1}{Prime factorisation of
$360$.}\label{fig:factor-360}
\end{minipage}
\hfill
\begin{minipage}{7cm}\centering \psset{unit=.25pc}
\pstree{\Toval{216}}
{\Toval{2}%
\pstree{\Toval{108}}%
{\Toval{2}%
\pstree{\Toval{54}}%
{\Toval{2}%
\pstree{\Toval{27}}  %
{\Toval{3}%
\pstree{\Toval{9}}%
{ \Toval{3} \Toval{3} } } } } }  \meinecaption{1}{Prime
factorisation of $216$.}\label{fig:factor-216}
\end{minipage}
\end{figure}

\begin{exa}
How many  divisors does $100$ have?
\end{exa}
\begin{solu}Clearly
$100=2^25^2$. Thus every divisor will be of the form $2^a5^b$, with
$a=0, 1,$ or $2$, and $b=0, 1$ or $2$. Taking $a=0,b=0$ we get
$2^05^0=1$. Taking $a=1,b=0$ we get $2^15^0=2$. Taking $a=2,b=0$ we
get $2^25^0=4$. Taking $a=0,b=1$ we get $2^05^1=5$. Taking $a=1,b=1$
we get $2^15^1=10$. Taking $a=2,b=1$ we get $2^25^1=20$. Taking
$a=0,b=2$ we get $2^05^2=25$. Taking $a=1,b=2$ we get $2^15^2=50$.
Taking $a=2,b=2$ we get $2^25^2=100$. The divisors of $100$ are thus
$$1, 2, 4, 8, 10, 20, 25, 50, 100,  $$
$9$ in number.
\end{solu}
\begin{rem} In the preceding example there was no necessity of
listing all the divisors in order to know how many there are. Since
there are three choices for the exponent $a$, and three choices for
the exponent $b$, there is a total of $3\cdot 3 = 9$ divisors of a
$100.$
\end{rem}

\begin{df}
The {\em greatest common divisor} $\gcd (a, b)$ of two natural
numbers $a$ and $b$ is the largest natural number that divides both
$a$ and $b$.
\end{df}
For example, the divisors of $12$ are
$$ \{1,2,3,4,6,12\}, $$
and those of $30$ are
$$\{1,2,3,5,6,15,30\},  $$
and hence $\gcd (12,30)=6$, because $6$ is the largest number that
appears in both lists.


\bigskip

Using prime factorisations, we may find the greatest common divisor
of two numbers without having to list all their divisors. We look at
which primes are common to both numbers and then select the least
power appearing of the primes.

\begin{exa}
Find $\gcd (12,30)$ by finding the prime factorisation of $12$ and
$30$ first.
\end{exa}
\begin{solu}
First observe that $12=2^2\cdot 3$ and that $30 =2^1\cdot 3^1\cdot
5^1 $. The primes $2$ and $3$ appear in both numbers. The $2$
appears with least exponent $1$ and so does the $3$, and so,
therefore, $\gcd (12,30)=2\cdot 3=6$
\end{solu}
\begin{exa}
Find $\gcd (300,360)$.
\end{exa}
\begin{solu}
First observe that $300=2^2\cdot 3\cdot 5^2$ and that $360 =2^3\cdot
3^2\cdot 5$. The primes $2$, $3$, and $5$ appear in both numbers.
The $2$ appears with least exponent $2$, the $3$ appears with least
exponent $1$, and the $5$ appears with least exponent $1$.
Therefore, $\gcd (300,360)=2^2\cdot 3 \cdot 5 =60$.
\end{solu}
\begin{exa}
Find $\gcd (132,504)$.
\end{exa}
\begin{solu}
First observe that $132=2^2\cdot 3\cdot 11$ and that $504 =2^3\cdot
3^2\cdot 7$. Only the primes $2$ and $3$ appear in both numbers. The
$2$ appears with least exponent $2$, the $3$ appears with least
exponent $1$. Therefore, $\gcd (132,504)=2^2\cdot 3 = 12$.
\end{solu}
\begin{exa}
Find $\gcd (100,343)$.
\end{exa}
\begin{solu}Observe that $100=2^25^2$ and $343=7^3$. They have no
prime factor in common, their only common divisor is $1$.
 Therefore, $\gcd (100,343)=1$.
\end{solu}

\begin{df}
Two natural numbers $a$ and $b$ are said to be {\em relatively
prime} if $\gcd (a,b)=1$.
\end{df}



A similar procedure for finding the least common multiple of two
numbers can now be outlined. We first find their prime
factorisations, take all the primes appearing in them, and choose
the powers with the maximum exponents for this primes.
\begin{exa}
Find $\lcm (12,45)$.
\end{exa}
\begin{solu}Notice that this is example  \ref{exa:lcm-12-45}.
The prime factorisations are $12=2^2\cdot 3$ and $45=3^2\cdot 5$.
All primes occurring at least once are $2$, $3$, and $5$. The prime
$2$ appears with maximum exponent $2$, the primes $3$ appears with
maximum exponent $2$ and the prime $5$ appears with maximum exponent
$1$. Hence $\lcm (12,45)=2^23^25 = 180$.\end{solu}
\begin{exa}
Find $\lcm (300,360)$.
\end{exa}
\begin{solu}
First observe that $300=2^2\cdot 3\cdot 5^2$ and that $360 =2^3\cdot
3^2\cdot 5$. The primes $2$, $3$, and $5$ appear at least once. The
$2$ appears with maximum exponent $3$, the $3$ appears with maximum
exponent $2$, and the $5$ appears with maximum exponent $2$.
Therefore, $\lcm (300,360)=2^3\cdot 3^2 \cdot 5^2 =1800$.
\end{solu}
\begin{exa}
Find $\lcm (132,504)$.
\end{exa}
\begin{solu}
First observe that $132=2^2\cdot 3\cdot 11$ and that $504 =2^3\cdot
3^2\cdot 7$. The primes $2$, $3$, $7$, and $11$ appear at least
once. The $2$ appears with maximum exponent $3$, the $3$ appears
with maximum exponent $2$, the $7$ appears with maximum exponent
$1$, and the $11$ appears with maximum exponent $1$. Therefore,
$$\lcm (132,504)=2^3\cdot 3^2 \cdot 7^1\cdot 11 ^1 =5544.$$
\end{solu}

\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Give the prime factorisation of $24$.
\begin{answer}
$24=2^33$
\end{answer}
\end{pro}
\begin{pro}
Give the prime factorisation of $36$.
\begin{answer}
$36=2^23^2$
\end{answer}
\end{pro}
\begin{pro}
How many prime numbers less than $10000$ have digits adding up to
$2$?
\begin{answer}
Three. They are $2$, $11$, $101$.
\end{answer}
\end{pro}
\begin{pro}
Demonstrate that $$\gcd (24,36)\cdot \lcm (24,36)=24\cdot 36.$$
\begin{answer}
Since $24=2^33$ and $36=2^23^2$, we have $\gcd (24,36) = 2^2\cdot
3=12$ and $\lcm (24,36) = 2^3\cdot 3^2=72$, from where $$\gcd
(24,36)\cdot \lcm (24,36)= 12\cdot 72 = 12\cdot 2 \cdot 36= 24\cdot
36.$$
\end{answer}
\end{pro}
\begin{pro}
How many  divisors does $36$ have?
\begin{answer} $36=2^23^2$, so it has $(2+1)(2+1)=9$ divisors.
\end{answer}
\end{pro}
\begin{pro}
How many  divisors does $38$ have?
\begin{answer} $38=2^1\cdot 19^1$, so it has $(1+1)(1+1)=4$ divisors.
\end{answer}
\end{pro}
\begin{pro}
How many  divisors does $40$ have?
\begin{answer} $40=2^35^1$, so it has $(3+1)(1+1)=8$ divisors.
\end{answer}
\end{pro}
\begin{pro}
An unsolved problem in Number Theory is the {\em Goldbach
Conjecture}, which asserts that every even natural number greater
than $6$ can be written as the sum of distinct primes. Verify
Goldbach's conjecture from every even integer between $8$ and $36$,
inclusive.
\begin{answer}In some cases there may be more than one answer. For
example, $20$ is $7+13$ and also $3+17$, etc.
$$ \begin{array}{lllll} 8 = 3 + 5 & 10 = 3 + 7 & 12 = 5 + 7 & 14 = 3 + 11 & 16 = 5 + 11 \\
18 = 7+11& 20 =3+17 & 22 = 3+19 & 24 = 7+17 & 26 =3+23 \\
28 = 11+17 & 30 =13+17 & 32 = 13+19 & 34 =11+23 & 36 =13+23 \\
 \end{array}  $$
\end{answer}
\end{pro}
\begin{pro} Given that there is a unique   digit
$d\in\{0,1,2,3,4,5,6,7,8,9\}$ so that the nine-digit number
$19700019d$ is  prime, find it.

\begin{answer}
$d$ cannot be even, as then the number would be divisible by $2$.
Now, $1+9+7+0+0+0+1+9+d=27+d$. If $d\in\{3,6,9\}$, then the sum of
the digital sum would be divisible by $3$, and so the number would
be divisible by $3$. If $d=5$, the number would be divisible by $5$.
If $d=7$, the number would be divisible by $197$. This leaves $d=1$
as the only possible digit, and so it must be this one.
\end{answer}
\end{pro}
\begin{pro}
A natural number is called {\em perfect} if it is the sum of all its
proper divisors. For example, the proper divisors of $6$ are $1, 2,
3$ and $6=1+2+3$. Also, the proper divisors of $28$ are $1, 2, 4, 7,
14$ and $28=1+2+4+7+14$, so $28$ is perfect. All perfect numbers
known to date are even, and it is a conjecture, still unsolved, that
there are no odd perfect numbers. Prove that $496$ is a perfect
number by finding all its proper divisors and adding them.
\begin{answer}
The proper divisors of $496$ are
$$ \{1,2,4,8,16,31,62,124,248\} $$ and we have
$$ 1+2+4+8+16+31+62+124+248=496. $$
\end{answer}
\end{pro}
\begin{pro}
How many  divisors does a  prime number have?
\begin{answer} If $p$ is a prime, then it has only $2$ divisors,
namely $1$ and $p$.
\end{answer}
\end{pro}
\begin{pro}
Factor $111111$ (six $1$'s) into primes.
\begin{answer}We have,$$\begin{array}{lll} 111111
& = & (111)(1001) \\
& = & (3)(37)(11)(91) \\
& = & (3)(37)(11)(7)(13) \\
& = & (3)(7)(11)(13)(37) \\
\end{array}
$$
\end{answer}
\end{pro}
\begin{pro}
Let $p$ be a prime number. List all the divisors of $p^2$.
\begin{answer}
$\{1,p,p^2\}$
\end{answer}
\end{pro}
\begin{pro}
Prove that among any $101$ integers taken from the set $\{1,2,\ldots
, 200\}$, there will always be $2$ which are relatively prime.
\begin{answer}
Notice that consecutive integers are always relatively primes. Any
$101$ integers is bound to have the pair of one the one hundred sets
$$ \{1,2\}, \{3,4\},\ldots , \{197,198\},\{199,200\}. $$

\end{answer}
\end{pro}

\begin{pro}
Which numbers have an odd number of divisors?
\begin{answer}
Only perfect squares.
\end{answer}
\end{pro}



 \begin{pro}
 Four comrades are racing side by side down a dusty staircase.
Frodo goes down two steps at a time, Gimli three, Legolas four, and
Aragorn five. If the only steps with all four's footprints are at
the top and the bottom, how many steps have just one footprint?
\begin{answer} Number the steps from $0$ to $N$,  where $N$ is the last step, and hence there are
$N+1$ steps. Notice that the top and the bottom of the stairs are
counted as steps. Let us determine first the number of steps. Frodo
steps on steps $0$, $2$, $4$, \ldots , $N$; Gimli steps on steps
$0$, $3$, $6$, \ldots , $N$; Legolas  steps on steps $0$, $4$, $8$,
\ldots , $N$; and Aragorn  steps on steps $0$, $5$, $10$, \ldots ,
$N$. For this last step to be common, it must be  a common multiple
of $2$, $3$, $4$, and $5$, and hence
 $$N = \lcm (2,3,4,5) = 60.  $$
Since every step that Legolas covers is also covered by Frodo, we
don't consider Legolas in our count. Frodo steps alone on the steps
which are multiples of $2$, but not multiples of $3$ or $5$. Hence
he steps on the $8$ steps
$$ 2, 14 , 22, 26, 34, 38,  46, 58. $$
Gimli steps alone on the $8$ steps that are multiples of $3$ but not
multiples of $2$ nor $5$:
$$ 3, 9, 21, 27, 33, 39, 51,
 57. $$
Finally, Aragorn steps alone on  the $4$ steps that are multiples of
$5$ but not multiples of $2$ or $3$:
$$5,25, 35, 55.$$
Our final count is thus $8+8+4=20$.


\end{answer}
\end{pro}

\begin{pro}[The Locker-room Problem] A locker room contains $100$ lockers, numbered $1$ through $100$.
Initially all doors are open. Person number $1$ enters and closes
all the doors. Person number $2$ enters and opens all the doors
whose numbers are multiples of $2$. Person number $3$ enters and if
a door whose number is a multiple of $3$ is open then he closes it;
otherwise he opens it. Person number $4$ enters and changes the
status (from open to closed and viceversa) of all doors whose
numbers are multiples of $4$, and so forth till person number $100$
enters and changes the status of door number $100$.  Which doors are
locked?
\begin{answer}
Notice that if  $d$ divides $n$ so does $\dfrac{n}{d}$. Thus we can
pair up every the different  divisors of $n$, and have an even
number of divisors as long as we do not have $d=\dfrac{n}{d}$. This
means that the integers with an even number of divisors will have
all doors open, and those with an odd number of divisors will all
all doors closed. This last event happens when
$d=\dfrac{n}{d}\implies n=d^2$, that is, when $n$ is a square. Hence
only the ten doors $$ 1, 4, 9, 16, 25, 36, 49, 64, 81, 100,  $$are
closed.
\end{answer}
\end{pro}
\begin{pro}
Find an infinite set of positive integers such that the sum of any
finite number of distinct elements of the set is not a square.
\begin{answer}
Consider the set $$ \{2^n: n\in \naturals, n\ \mathrm{odd}\}. $$
\end{answer}

\end{pro}
\end{multicols}
\section*{References}
Much of the material of this section belongs to a branch of
Mathematics formerly called ``the Higher Arithmetic'', nowadays
called Number Theory. Some  references follow.
\begin{description}
\bibitem[NiZuMo]{NiZuMo} I. Niven, H. Zuckermann, H. Montgomery ``Introduction to Number Theory''.
\end{description}



\chapter{Division}
\begin{df}[Definition of Division]
Let $m, n, x$ be natural numbers, with $n\neq 0$. Then the statement
$m\div n=x$ means that $m=xn$.
\end{df}
\begin{exa}
To compute $15\div 3$ we think of which number when multiplied $3$
gives $15$. Clearly then $15\div 3=5$ since $15=5\cdot 3$.
\end{exa}



\begin{rem}
Division is neither closed in $\naturals$, nor commutative, nor
associative. For example, $$4 \div 8 \notin \naturals, \quad 4\div 8
\neq 8 \div 4, \quad 8\div (4\div 2 )\neq (8\div 4)\div 2.
$$\end{rem}


Our links to division and reality are the ideas of {\em sharing} and
{\em containment}. The example above can be thought as the solution
to the following problem: $15$ customers are going to be handled
equally by  $3$ waitresses, how many people will each waitress
handle? The result $5$ means that $5+5+5=15$, that is, that each of
the $3$ waitresses will handle or receive an {\em equal share} of
$5$ customers. This corresponds to the idea of division as repeated
subtraction:
$$\underbrace{15-3 =12}_{\mathrm{first\ instance}}\to \underbrace{12-3 =9}_{\mathrm{second\ instance}} \to
\underbrace{9-3=6}_{\mathrm{third\ instance}} \to
\underbrace{6-3=3}_{\mathrm{fourth\ instance}} \to
\underbrace{3-3=0}_{\mathrm{fifth\ instance}}
$$
Since the subtraction was able to be evenly carried out five times,
$15\div 3=5$.






One may also think of the problem in the following way. If there are
$15$ customers to be served, and each waitress receives $5$, how
many waitresses are there? In this case we are asserting that
$3+3+3+3+3=15$.


\bigskip



The above  definition of division depends on the existence of a
natural number $x$ so that $m=xn$, which effectively means, that we
have division without remainder. What happens in general? Let us
first introduce some notation.

\begin{df}
By $\floor{x}$ (pronounced {\em floor of $x$}) we mean either $x$ if
$x$ is an integer, or the integer just to the left of $x$.
\end{df}
For example, $\floor{3}=3$, $\floor{3.1}=3$, $\floor{3.5}=3$,
$\floor{3.9}=3$, $\floor{4.1}=4$. Notice that this is not rounding.


\bigskip

Observe that the following inequality is satisfied:
\begin{equation}
\floor{x}\leq x < \floor{x}+1.
\end{equation}

Let us now consider the task of defining $a\div b$, by considering
first an example. Consider the problem of performing the division
$23\div 5$. We look for multiples of $5$ that straddle $23$. We
readily see that
$$ 20<23<25\implies 5\cdot 4 < 23 < 5\cdot 5. $$
We divide through by $5$ and assume that inequalities are preserved:
$$ 5\cdot 4 < 23 < 5\cdot 5
\implies (5\cdot 4)\div 5 < 23\div 5 < (5\cdot 5) \div 5 \implies 4
< 23\div 5 < 5.
$$This means that $$ \floor{23\div 5}=4. $$ In fact, we have
$$ 23=5\cdot 4 + 3 \implies 23=5\cdot\floor{23\div 5} +3.  $$
The excess $3$ over the multiple of $5$ just to the left of $23$ is
the {\em remainder}.


We are ready to state the following result, which we will admit
without proof.

\begin{thm}[Division Algorithm]
Given two natural numbers $a$ and $b$ with $b\neq 0$, we may find
two unique integers $\floor{a\div b}$ and $r$ with $0 \leq r <b$
such that
$$ a = b\floor{a\div b} +r. $$
Here $a$ is called the {\em dividend}, $b$ the {\em divisor},
$\floor{a\div b}$ the {\em quotient} and $r$ the {\em remainder}.
\end{thm}
Since $\floor{a\div b}=\dfrac{a-r}{b}$,
$$ \floor{a\div b}\leq a\div b < \floor{a\div b}+1\implies  \dfrac{a-r}{b}\leq a\div b <\dfrac{a-r}{b}+1. $$


The division algorithm gives us a way of classifying natural numbers
according to the remainder they leave. If $b>0$ is the divisor, then
the possible remainders are the $b$ numbers $0$, $1$, $2$, \ldots ,
$b-1$.
\begin{exa}
Take $b=2$ as the divisor. The possible remainders are $0$ and
$2-1=1$. Thus every natural number comes in one of the two flavours
$$ \{0,2,4,6,\ldots \}\quad \mathrm{or}\quad  \{1,3,5,7,\ldots\}, $$
that is, as a number that leaves remainder $0$ upon division by $2$
or as a number that leaves remainder $1$ upon division by $2$. In
fact, we see that
$$ \naturals = \{0,2,4,6,\ldots \}\cup \{1,3,5,7,\ldots\}, $$
and so the division algorithm serves to partition the natural
numbers into two parts. We can also put this in algebraic terms, by
saying that every natural number is of the form $2a$ or $2a+1$ for
some natural number $a$.
\end{exa}
\begin{exa}
Take $b=3$ as the divisor. The possible remainders are $0$, $1$, and
$3-1=2$. Thus every natural number comes in one of the three
flavours
$$  \{0,3,6,9,\ldots \}\quad \mathrm{or}\quad \{1,4,7,10,\ldots\}\quad \mathrm{or}\quad \{2,5,8,11,\ldots\}. $$
that is, as a number that leaves remainder $0$ upon division by $3$,
 as a number that leaves remainder $1$ upon division by $3$
or as a number that leaves remainder $2$ upon division by $2$. In
fact, we see that
$$ \naturals = \{0,3,6,9,\ldots \}\cup \{1,4,7,10,\ldots\}\cup \{2,5,8,11,\ldots\}. $$
and so the division algorithm serves to partition the natural
numbers into three parts. We can also put this in algebraic terms,
by saying that every natural number is of the form $3a$, $3a+1$ or
$3a+2$ for some natural number $a$.
\end{exa}

\begin{exa}
Prove that the sum of two even natural numbers is even. Can you find
five natural numbers whose sum is $25$?
\end{exa}
\begin{solu}
Let $2x$ and $2y$ be two even natural numbers. Then using the
distributive law in reverse,
$$ 2x+2y = 2(x+y). $$Since $x+y$ is a natural numbers, $2(x+y)$ is
twice a natural number and so even.



The answer to our second question is clearly NO, since $25$ is an
odd number and the sum of five even numbers would be even.
\end{solu}


\bigskip

We now discuss criteria for a given natural number to be (evenly)
divisible by another number. Granted, we will only consider a few
cases---just the criteria for divisibility by $2$, $3$, $5$, and
$9$---but these are important in their own merit.

\bigskip

The task we have at hand is the following. Suppose that we had a
very large number, let us say $1234509876.$  How do we know, {\em
without actually performing the division}, whether it is divisible
by $2$? By $3$? By $5$? It turns out that the decimal positional
notation makes answering these questions extremely easily.

\bigskip

For the case of divisibility by $2$, we want to investigate whether
$1234509876$ belongs to the set
$$  \{0,2,4,6,8,10,\ldots \} $$ of integers that leave no remainder
when divided by $2$. But we may write
$$ 1234509876 = 1234509870+6= 123450987\cdot 10+6=123450987\cdot 2\cdot 5 + 6.$$
Clearly, $123450987\cdot 2\cdot 5 $ is divisible by $2$. Hence the
divisibility of $1234509876$ hinges on whether its last digit, $6$
is divisible by $2$, which is clearly the case. We conclude that
thus $1234509876$ is divisible by $2$. Notice that we did not need
to perform any division. A more general argument can then establish
the following result.
\begin{thm}
A natural number is divisible by $2$ (even) if and only if its last
digit of its decimal expansion is even, that is, it ends in one of
$\{0,2,4,6,8\}$.
\end{thm}


For the case of divisibility by $5$, we want to investigate whether
$1234509876$ belongs to the set
$$  \{0,5,10,15,20,25,\ldots \} $$ of integers that leave no remainder
when divided by $5$. But we may write
$$ 1234509876 = 1234509870+6= 123450987\cdot 10+6=123450987\cdot 2\cdot 5 + 6.$$
Clearly, $123450987\cdot 2\cdot 5 $ is divisible by $5$. Hence the
divisibility of $1234509876$ hinges on whether its last digit, $6$
is divisible by $5$, which is clearly not the case. We conclude that
thus $1234509876$ is not divisible by $5$. Notice that we did not
need to perform any division. We state the general result.

\begin{thm}
A natural number is divisible by $5$  if and only if its last digit
of its decimal expansion ends in one of $\{0,5\}$.
\end{thm}
\begin{rem}
In fact, by writing $ 1234509876 = 1234509875+1$, since by the
preceding theorem $ 1234509875$ is evenly divisible by $5$, we see
that $ 1234509876$ leaves remainder $1$ upon division by $5$.
\end{rem}

Before we introduce a criterion for divisibility by $3$ or $9$, we
need the concept of {\em digital sum}. Consider any natural number,
say $1234509876$. Sum its digits successively, and then sum the
digits of each successive sum, until only a $1$-digit integer
remains:
$$1234509876\to 1+2+3+4+5+0+9+8+7+6=45\to 4+5=9,  $$
whence the digital sum is $9$.



\bigskip

Our criterion for divisibility by $3$ or $9$ will be linked to the
digital sum of the number in question. For the example at hand,
observe that
$$\begin{array}{lll}1234509876  & =  & 1\cdot 10^9 + 2\cdot 10^8 + 3\cdot 10^7+ 4\cdot 10^6+ 5\cdot 10^5
+ 0\cdot 10^4 + 9\cdot 10^3+ 8\cdot 10^2 + 7\cdot 10^1+6\\
& = & 1\cdot 999999999+2\cdot 99999999 + 3\cdot 9999999+4\cdot
999999 + 5\cdot 99999+ 0 \cdot 9999 + 9\cdot 999 + 8\cdot 99 + 7
\cdot 9\\
& & \qquad  + (1+2+3+4+5+0+9+8+7+6),
\end{array}
$$
Thus whether $1234509876$ is divisible by $3$ or $9$ depends
exclusively on whether its digital sum $1+2+3+4+5+0+9+8+7+6=45$ is.
In turn $45$ is divisible by $3$ or $9$ if $4+5=9$ is divisible by
$3$ or $9$. The string of sums make us conclude that $1234509876$ is
divisible by $9$. In general, we may state the following result.
\begin{thm}
A natural number is divisible by $3$ (or $9$) if and only if its
digital sum is divisible by $3$ (or $9$).
\end{thm}






\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
You put $54$ marbles into $6$ bags, ending up with the same number
of marbles in each bag. How many marbles would be in each bag
if there were $6$ bags?
\begin{answer}
$9$
\end{answer}
\end{pro}
\begin{pro}
Find the quotient and the remainder when $23$ is divided by $4$.
\begin{answer}
$23=4\cdot 5 + 3$ and so the quotient is $5$ and the remainder $3$.
\end{answer}
\end{pro}
\begin{pro}
Find the quotient and the remainder when $4$ is divided by $23$.
\begin{answer}
$4=0\cdot 23 + 4$ and so the quotient is $0$ and the remainder $4$.
\end{answer}
\end{pro}



\begin{pro}
If today is Thursday, what day will it be $100$ days from now?
\begin{answer}Since $100 = 7 \cdot  14 + 2$, $98$ days from today will be a
Thursday, and $100$ days from today will be a Saturday.
\end{answer}
\end{pro}

\begin{pro}
Bilbo and Frodo have just consumed a plateful of cherries. Each
repeats the rhyme `Tinker, tailor, soldier, sailor, \mbox{rich man,}
poor man, beggar man, thief' over and over again as he runs through
his own heap of cherry stones. Bilbo finishes on `sailor', whereas
Frodo finishes on `poor man'. What would they have finished on if
they had run through both heaps together?
\begin{answer}
This problem is from [Gard]. At `tailor.'
\end{answer}
\end{pro}
\begin{pro}
A runner ran $3000$ m in $8$ minutes. What is his average speed in
meters per second?
\begin{answer}
$6\tfrac{1}{4}$ meters per second.
\end{answer}
\end{pro}
\begin{pro}
Peter and Paul were selling magazines, of which Peter sold $60$ and
Paul $80$. Each magazine cost the same. If they received $\$700$
altogether for the magazines, how much money did Peter get?
\begin{answer}
$\$300$
\end{answer}
\end{pro}
\begin{pro}
Brian is drawing coloured camels in sequence. He draws a yellow
camel, a green camel, a blue camel, a red camel, a brown camel, an
orange camel, and again,  a yellow camel, a green camel, a blue
camel, etc. Which colour is the $2008$-th camel?
\begin{answer}
Red.
\end{answer}
\end{pro}
\begin{pro}
What is the greatest number of Mondays that can occur in $45$
consecutive days?
\begin{answer}
If one puts Monday at the very beginning of the period one obtains
$\ceil{\dfrac{45}{7}} = 7$ Mondays.
\end{answer}
\end{pro}

\begin{pro}
A club has $86$ members, and there are $14$ more girls than boys in
the club. How many are there of each?
\begin{answer}
There are $36$ boys and $50$ girls.
\end{answer}
\end{pro}

\begin{pro}
When a number is doubled and the result increased by nine, one obtains fifteen. What was the original number?
\begin{answer}
$3$
\end{answer}
\end{pro}

\begin{pro}
When a number is doubled and the result diminished by nine, one obtains fifteen. What was the original number?
\begin{answer}
$12$
\end{answer}
\end{pro}
\begin{pro}
When a number is halved and the result increased by nine, one obtains fifteen. What was the original number?
\begin{answer}
$12$
\end{answer}
\end{pro}

\begin{pro}
When a number is halved and the result diminished by nine, one obtains fifteen. What was the original number?
\begin{answer}
$48$
\end{answer}
\end{pro}
\begin{pro}
There are $5,064$ marbles that need to be packed in boxes. There are
$6$ boxes. We want to put the same number of marbles in each box.
How many marbles will fit into each box?
\begin{answer}
$844 $
\end{answer}
\end{pro}

\begin{pro}
The four hundred thirty seven students of a school were called to assembly for a drug raid. If the police line up as many as possible
into twenty five equal rows, how many pupils are left out?
\begin{answer}
$12$
\end{answer}
\end{pro}
\begin{pro}
A book publisher sent $140$ copies of a book to a bookstore. The
publisher packed the books in two types of boxes. One type held $8$
copies of the book and the other held $12$. The boxes were all full
and there were an equal number of each type of box. How many boxes
of each type were sent to the bookstores?
\begin{answer}
$7$
\end{answer}
\end{pro}
\begin{pro}
With what possible values can the digit  $a$ be replaced so that the
four-digit number $32a2$ be divisible by $9$?
\begin{answer}
The sum of the digits is $3+2+a+2=7+a$. We want $a$ to be one of
$\{0,1,2,3,4,5,6,7,8,9\}$ and also we want $7+a$ to be a multiple of
$9$. This happens only when $a=2$.
\end{answer}

\end{pro}
\begin{pro}
With what possible values can the digit  $a$ be replaced so that the
five-digit number $32aa2$ be divisible by $9$?
\begin{answer}
The sum of the digits is $3+2+a+a+2=7+2a$. We want $a$ to be one of
$\{0,1,2,3,4,5,6,7,8,9\}$ and also we want $7+2a$ to be a multiple
of $9$. This happens only when $a=1$.
\end{answer}

\end{pro}
\begin{pro}
A book publisher must bind $4500$ books. One contractor can bind all
these books in $30$ days and another contractor in $45$ days. How
many days would be needed if both contractors are working
simultaneously?
\begin{answer}
The first contractor binds $4500\div 30 = 150$ books per day, and
the second, $4500 \div 45 = 100$ books per day. Therefore, if they
worked simultaneously, they could bind $150+100 = 250$ books per
day, and it would take them $4500 \div 250 = 18$ days to bind all
books.
\end{answer}
\end{pro}
\begin{pro}
For commencement exercises, the students of a school are arranged in
nine rows of twenty eight students per row. How many rows could be
made with thirty six students per row?
\begin{answer}
Seven.
\end{answer}
\end{pro}

\begin{pro}
For which values of the natural number $n$ is $36 \div n$ a natural
number?
\begin{answer}
If  $36 \div n$ is a natural number, then $n$ must evenly divide
$36$, which means that $n$ is a divisor of $36$. Thus $n$ can be any
one of the values in  $\{1,2,3,4,6,9,12,18,36\}$.
\end{answer}
\end{pro}
\begin{pro}
A two-digit natural number is divided by the sum of its digits. What
is the largest possible remainder?
\begin{answer}
$15$
\end{answer}
\end{pro}

\begin{pro}
As a publicity stunt, a camel merchant has decided to pose the
following problem: ``If one gathers all of my camels into groups of
$4$, $5$ or $6$, there will be no remainder. But if one gathers them
into groups of $7$ camels, there will be $1$ camel left in one
group.'' The number of camels is the smallest positive integer
satisfying these properties. How many camels are there?
\begin{answer}The least common multiple of $4, 5$ and $6$ is $60$, hence
we want the smallest positive multiple of $60$ leaving remainder $1$
upon division by $7$. This is easily seen to be $120$.
\end{answer}
\end{pro}
\begin{pro}
What is the least and the largest number of Friday the $13$th that
there can be in a given year?
\begin{answer}
Least: $1$. Largest: $3$.
\end{answer}
\end{pro}
\begin{pro}
A number of bacteria are placed in a glass. One second later each bacterium splits in two, the next second each of the now existing bacteria
splits in two, et cetera. After one minute the glass is full. When was the glass half full?
\begin{answer}
At $59$ seconds.
\end{answer}
\end{pro}
\begin{pro}
A lotus flower doubles its surface each day, taking $32$ days to cover the whole surface of a pond.
How long will it take two lotus flowers to cover the same
pond?
\begin{answer}
Thirty one days. Explanation: consider just one lotus flower. On the second day it covers as much as two and thirty one days remain for it to
cover the pond.
\end{answer}
\end{pro}

\begin{pro}
A boy and a girl collected $24$ nuts. The boy collected twice as
many nuts as the girl. How many did each collect?
\begin{answer}
We can represent the boy's and girl's amounts by boxes, each box
having an equal amount of nuts, and there being two boxes for the
boy and one for the girl:
$$ \psframebox[fillstyle=solid,fillcolor=red]{B} \qquad \psframebox[fillstyle=solid,fillcolor=red]{B} \qquad \psframebox[fillstyle=solid,fillcolor=green]{G}  $$
Then clearly, each box must have $8$ nuts. The boy has $16$ and the
girl $8$. This problem was taken from [Toom].
\end{answer}
\end{pro}
\begin{pro}
Prove that the sum of two odd numbers is even.
\begin{answer}
If $2x+1$ and $2y+1$ are the odd numbers, their sum is
$$ 2x+1+2y+1=2x+2y+2 =2(x+y+1), $$twice an integer, and hence even.
\end{answer}
\end{pro}
\begin{pro}
Prove that the product of two odd numbers is odd.
\begin{answer}
If $2x+1$ and $2y+1$ are the odd numbers, their product is
$$ (2x+1)(2y+1)=(2x)(2y)+2x(1) + 1(2y)+1(1) = 2xy +2x+2y +1 =2(xy+x+y)+1, $$twice an integer plus one, and hence odd.
\end{answer}
\end{pro}
\begin{pro}
How many integers in the set $\{100,101,102, \ldots , 198, 199\}$ of
$100$ consecutive integers {\bf are not} the sum of four consecutive
integers?
\begin{answer}
If $n-1, n, n+1, n+2$ are four consecutive integers, then their sum
is $4n+2$, that is, the integers sought do not leave remainder $2$
upon division by $4$. Since three quarters of the integers leave
some other remainder, the answer is $\dfrac{3}{4}\cdot 100 = 75$.
\end{answer}
\end{pro}

\begin{pro} Prove that the square of any natural number either leaves
remainder $0$ when divided by $4$ or remainder $1$ when divided by
$4$. That is, prove that the square of any natural number is of the
form $4k$ or $4k + 1$.
\begin{answer} By the Division Algorithm, any integer comes in one of two
flavours: $2a$ or $2a + 1$. Squaring, $$ (2a)^2 = 4a^2,\,\,\, (2a +
1)^2 = 4(a^2 + a) + 1$$and so the assertion follows.
\end{answer}
\end{pro}


\begin{pro} Prove that no integer in the sequence $$ 11, 111, 1111, 11111, \ldots$$
is the square of an integer.
\begin{answer} The previous problem asserts that if a number leaves either
remainder $2$ or remainder $3$ when divided by $4$, then the number
cannot be a square.
 Observe that
$$ 11=8+3, 111= 100 + 8 +3, 1111=1100 +8 +3, 11111=11100+8+3, \ldots$$
so every integer in the sequence leaves remainder $3$ upon division
by $4$, and by the preceding problem, they cannot be squares.
\end{answer}
\end{pro}
\begin{pro}
{\em One Price Shoes} sell all their shoes for the same price, which
is an integral number of dollars. Anacleta and Sinforosa go shoe
shopping to {\em One Price Shoes}. Anacleta has \$200 and she buys
as many pairs of shoes as possible, and there remain \$32. Sinforosa
has \$150 and she buys as many pairs of shoes as possible, and there
remain \$24. What is the price of a pair of shoes?
\begin{answer}
Let $A$ be the number of pairs of shoes Anacleta buys and let $S$ be
the number of pairs of shoes Sinforosa buys. Let $P$ be the price,
in dollars, of each pair of shoes. Observe that $P > 32.$ Then $200
= AP + 32, 150 = SP + 24,$ whence
$$AP = 168 = 2^3\cdot 3\cdot 7, SP = 126 = 2\cdot 3^2\cdot 7.$$ The
price of each pair of shoes is a common divisor of 168 and 126, but
it must exceed $32$. The only such divisor is $42$. Hence the shoes
cost $\$42$.
\end{answer}
\end{pro}
\begin{pro}
Find the greatest multiple of $8$ using all ten digits and  with all
the digits different.
\begin{answer}
For a number to be divisible by $8$, the number formed by its last
three digits must the divisible by $8$. The largest number that can
be formed with all ten digits is $9876543210$, but this is not
divisible by $8$. A permutation of the last three digits to
$9876543120$ yields a number divisible by $8$, and the largest such.
\end{answer}
\end{pro}
\begin{pro}
Find the smallest strictly positive natural number $n$ such that
$\dis{\frac{n}{2}}$ is a (perfect) square, $\dis{\frac{n}{3}}$ is a
(perfect) cube, and $\dis{\frac{n}{5}}$ is  a (perfect) fifth power.
\begin{answer}
The integer must be of the form $n = 2^a3^b5^c,$ with $a, b, c$
positive integers. The conditions imply that $a - 1$ is even and
that $a$ is divisible by 15; that $b - 1$ is divisible by 3 and that
$b$ is divisible by 10; and that $c - 1$ is divisible by 5 and $c$
divisible by 6. Clearly,  the smallest positive integers satisfying
those conditions are $a = 15, b = 10, c = 6.$ Hence the integer
sought is $n = 2^{15}3^{10}5^6 = 30233088000000.$
\end{answer}
\end{pro}
\end{multicols}

\section*{References}
The excellent \cite{Aha} is a good guide for adults trying to teach
their children the basic of Arithmetic. I have borrowed some of his
ideas for these notes.
\begin{description}
\bibitem[Aha]{Aha} Ron AHARONI, ``Arithmetic for Parents''\end{description}


\chapter{Long Division}
Let us now discuss an algorithm for division. The way we display it
here is perhaps somewhat different from  what you learned in school,
but the result will be the same. We will start with simple examples
in the hopes of elucidating the general procedure. Before we begin,
let us introduce some notation.
\begin{df}[Fraction Notation] Let $a$, $b$ be  two natural numbers,
with $b\neq 0$. Then we define the {\em fraction} $\dfrac{a}{b}$ as
$$\dfrac{a}{b} = a\div b.  $$Here $a$ is the {\em numerator} of the
fraction and $b$ the {\em denominator}.
\end{df}

There it is! A fraction is a division that we are too lazy to carry
out!

\bigskip

Now, recall that from the distributive law, we have the equality
$$  a(b+c) = ab+ac,    $$
that is, we are able to ``multiply term by term.'' Suppose now that
instead of the factor $a$ we multiplied by $\dfrac{1}{a}$, with
$a\neq 0$. Then we would have
$$ \dfrac{1}{a}\left(b+c\right)=\dfrac{b+c}{a}=\dfrac{b}{a}+\dfrac{c}{a}, $$
which gives us a ``division term by term.''


\bigskip

For example, suppose we were considering the task of dividing $9639$
by $3$. Taking advantage of the decimal representation of the number
could proceed as follows:
$$9639\div 3 = \dfrac{9639}{3} =
\dfrac{9000+600+30+9}{3} =
\dfrac{9000}{3}+\dfrac{600}{3}+\dfrac{30}{3}+\dfrac{9}{3} =
3000+200+10+3=3213.
$$
Notice that we first isolated the thousands, divided them by $3$,
then the hundreds, divided them by $3$, etc. We could actually
display this ``vertically'' in the following manner.
$$ \opidiv[displayintermediary=all]{9639}{3}  $$
In the above display, we first begin by the thousands. Divide $9$ by
$3$, obtain $3$, write this in the space for the quotient. Multiply
$3$ by $3$, get $9$ and write this below the $9$ of the dividend.
Subtract and obtain $0$. Bring down the next digit from the
dividend, which is $6$. Divide $6$ by $3$, obtain $2$, write this in
the space for the quotient. Multiply $2$ by $3$, get $6$ and write
this below the $6$ of the dividend. Subtract and obtain $0$. Bring
down the next digit from the dividend, which is $3$. Divide $3$ by
$3$, obtain $1$, write this in the space for the quotient. Multiply
$1$ by $3$, get $3$ and write this below the $3$ of the dividend.
Subtract and obtain $0$. Bring down the next digit from the
dividend, which is $9$. Divide $9$ by $3$, obtain $3$, write this in
the space for the quotient. Multiply $3$ by $3$, get $9$ and write
this below the $9$ of the dividend. Subtract and obtain $0$. As a
check: $$9639 = 3\cdot 3213,  $$and so the quotient is $3213$ and
the remainder $0$.

\begin{rem}
You may prefer to display your division using a more Anglo Saxon
display: $$\longdiv{9639}{3}  $$
\end{rem}

The general algorithm is thus:
\begin{enumerate}
\item Find the number of times the divisor is contained into as many
of the dividend figures (as read from left to right) an place this
number in the place for the quotient.
\item Multiply the divisor by the number found above and place the
product under the figures used in the part above from the dividend.
\item Subtract the product found above from the said figures of the
dividend. Bring down the next figure from the dividend to the right
of the difference just obtained.
\item Rinse and repeat with the remainder just increased by the next
figure in the step above.
\end{enumerate}



\begin{exa} Perform the division: $21416\div 2$
\end{exa}
\begin{solu}
We write
$$\dfrac{21416}{2} =  \dfrac{20000}{2}+\dfrac{1000}{2}+ \dfrac{400}{2}+ \dfrac{10}{2}+\dfrac{6}{2} =
10000+500+200+5+3=10708. $$ Using the vertical algorithm for
division:
$$ \opidiv[displayintermediary=all]{21416}{2}  $$
This can also be displayed as
$$\longdiv{21416}{2}  $$
Verification: $21416=2\cdot 10708$, hence the quotient is $10708$
and the remainder $0$.
\end{solu}
\begin{exa}
Here we display $9634\div 3$:
$$ \opidiv[displayintermediary=all]{9634}{3}  $$
This can also be displayed as
$$\longdiv{9634}{3}  $$
Verification: $9634=3\cdot 3211+1$, hence the quotient is $3211$ and
the remainder $1$.
\end{exa}
\begin{exa}
Here we display $1326039\div 13$:
$$ \opidiv[displayintermediary=all]{1326039}{13}  $$
This can also be displayed as
$$\longdiv{1326039}{13}  $$
Verification: $1326039=13\cdot 102003$, hence the quotient is
$10708$ and the remainder $0$.
\end{exa}

\begin{exa}
Here we display $789\div 123$:
$$ \opidiv[displayintermediary=all]{789}{123}  $$
This can also be displayed as
$$\longdiv{789}{123}  $$
Verification: $789=123\cdot 6+51$, hence the quotient is $6$ and the
remainder $51$.
\end{exa}
\begin{exa}
Here we display $321123\div 123$:
$$ \opidiv[displayintermediary=all]{321123}{123}  $$
This can also be displayed as
$$\longdiv{321123}{123}  $$
Verification: $1326039=123\cdot 2610+93$, hence the quotient is
$2610$ and the remainder $93$.

\end{exa}


We now discuss algebraic division. Let us start with what is perhaps
the easiest case: powers of $10$. Observe that
$$ \dfrac{10^5}{10^2} = \dfrac{100000}{100}= \dfrac{1000\cancel{0}\cancel{0}}{1\cancel{0}\cancel{0}} = 1000 =10^3, $$
that is to perform the division, we cancelled as many zeroes as the
smaller power had. But notice what happened to the exponents. We in
fact had
$$\dfrac{10^5}{10^2} =10^{5-2}=10^3, $$
that is, to perform division of powers with the same base, we
subtracted the exponents. To further our evidence we again observe
that
$$  \dfrac{2^6}{2^2} = \dfrac{64}{4}=16=2^4 \implies  \dfrac{2^6}{2^2} = 2^{6-2} =2^4.  $$
Again,
$$\dfrac{a^5}{a^2} = \dfrac{aaaaa}{aa} = \dfrac{aaa\cancel{a}\cancel{a}}{\cancel{a}\cancel{a}} = a^3,  $$
and, of course, $a^3=a^{5-2}$.  This generalises as follows.
\begin{thm}[Second Law of Exponents]Let $a\neq 0$ be a real number and $m, n$ natural numbers, such that $m \geq n$. Then
$$\dfrac{a^m}{a^n} = a^{m-n}.  $$
\end{thm}
\begin{pf}
We have $$\begin{array}{lll}\dfrac{a^m}{a^n} &  =  &
\dfrac{\underbrace{a\cdot a \cdots a}_{m\ a\mathrm{'s}}}{
\underbrace{a\cdot a \cdots a}_{n\
a\mathrm{'s}}} \\
& = & \underbrace{a\cdot a \cdots a}_{m-n\ a\mathrm{'s}}\\
 & = & a^{m-n}.
\end{array}$$
\end{pf}

\begin{exa}
$$ \dfrac{2^9}{2^4} = 2^{9 - 4} = 2^5 = 32. $$
\end{exa}
\begin{exa}$\dfrac{15x^5y^7z^4}{5x^2y^2z^2} = 3x^3y^5z^2.$\end{exa}

\bigskip

In the case when the numerator possesses more than one term, we use
the distributive law. \begin{exa} $\dfrac{x^3+x^4}{x^2}=
\dfrac{x^3}{x^2}+\dfrac{x^4}{x^2}=x+x^2$.
\end{exa}

We will postpone the study when the denominator possesses more than
one term till an Algebra course.


\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Perform the division without a calculator: $100200300400\div 25$.
\begin{answer}
$\opidiv[displayintermediary=all]{100200300400}{25}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $123123\div 123$.
\begin{answer}
$\longdiv{123123}{123}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $123123\div 1001$.
\begin{answer}
$\longdiv{123123}{1001}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $1234567890\div 90$.
\begin{answer}
$\longdiv{1234567890}{90}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $10111213\div 9$.
\begin{answer}
$\longdiv{10111213}{9}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $10111213\div 14$.
\begin{answer}
$\longdiv{10111213}{14}$
\end{answer}
\end{pro}
\begin{pro}
Divide: $\dfrac{a^{12}}{a^6}$
\begin{answer} $a^6$
\end{answer}
\end{pro}
\begin{pro}
Divide: $\dfrac{a^{8}x^6}{a^6x}$
\begin{answer} $a^2x^5$
\end{answer}
\end{pro}
\begin{pro}
Divide term by term: $\dfrac{30x^6+9x^9}{3x^3}$
\begin{answer} $10x^3+3x^6$
\end{answer}
\end{pro}
\begin{pro}
Divide term by term: $\dfrac{x^2+x^3}{x^2}$
\begin{answer} $1+x$
\end{answer}
\end{pro}


 \begin{pro}Find the exact numerical value of: \quad $\dfrac{20^6}{25^5}$.
\begin{answer} $\dfrac{4096}{625}$ \end{answer}

 \end{pro}
\begin{pro}Calculate:  $$\dfrac{100^3+100^3+100^3+100^3+100^3+100^3}{100^2+100^2+100^2}.$$
\begin{answer}  $\dfrac{100^3+100^3+100^3+100^3+100^3+100^3}{100^2+100^2+100^2}=\dfrac{6\cdot 100^3}{3\cdot 100^2}=2\cdot 100=200$.  \end{answer}
\end{pro}

  \begin{pro}Find the exact numerical value of:  $$\dfrac{25^2+25^2+25^2+25^2 }{25^4}.$$
\begin{answer} $\dfrac{4}{625}$ \end{answer}

   \end{pro}

\begin{pro}
If
$$\left(\dfrac{4^5 + 4^5 + 4^5 + 4^5}{3^5 + 3^5 + 3^5}\right)\left(\dfrac{6^5 + 6^5 + 6^5 + 6^5 + 6^5 + 6^5}{2^5 + 2^5}\right) = 2^n,$$find $n$.
\begin{answer}
The product equals
$$\frac{4(4^5)}{3(3^5)}\cdot\frac{6(6^5)}{2(2^5)} = 4(4^5) =
2^{12},$$so $n=12$.
\end{answer}
\end{pro}
\begin{pro}
If
$$ 3^{2001} + 3^{2002} + 3^{2003} = a3^{2001},$$find $a$.
\begin{answer}
 We have  $$ 3^{2001} + 3^{2002} + 3^{2003} = 3^{2001}(1 +
3 + 3^2) = (13)3^{2001},$$whence $a = 13$.
\end{answer}
\end{pro}



\end{multicols}



















\chapter{Roots and Geometric Progressions}

We introduce now the operation of extracting roots. Notice that we
will introduce this ``new'' operation by resorting to the
``reverse'' of an ``old'' operation. This is often the case in
Mathematics.\index{root}\index{root extraction}
\begin{df}[Roots] Let $m$ be a natural number greater than or equal
to $2$, and let $a$ and $b$ be any natural numbers. We write that
$\sqrt[m]{a}=b$ if $a=b^m$. In this case we say that $b$ is the {\em
$m$-th root} of $a$. The number $m$ is called the {\em index} of the
root.
\end{df}
\begin{rem}
In the special case when $m=2$, we do not write the index. Thus we
will write $\sqrt{a}$ rather than $\sqrt[2]{a}$. The number
$\sqrt{a}$ is called the {\em square root of $a$}. The number
$\sqrt[3]{a}$ is called the {\em cubic root of $a$}.
\end{rem}

\begin{exa}
We have\\

\centerline{\begin{tabular}{lll} $\sqrt{1}=1$ & because &
$1^2=1$,\\
$\sqrt{4}=2$ & because &
$2^2=4$,\\
$\sqrt{9}=3$ & because &
$3^2=9$,\\
$\sqrt{16}=4$ & because &
$4^2=16$,\\
$\sqrt{25}=5$ & because &
$5^2=25$,\\
$\sqrt{36}=6$ & because &
$6^2=36$.\\
\end{tabular}}
\end{exa}

\begin{exa}
We have\\

\centerline{\begin{tabular}{lll} $\sqrt[10]{1}=1$ & because &
$1^{10}=1$,\\
$\sqrt[5]{32}=2$ & because &
$2^5=32$,\\
$\sqrt[3]{27}=3$ & because &
$3^3=27$,\\
$\sqrt[3]{64}=4$ & because &
$4^3=64$,\\
$\sqrt[3]{125}=5$ & because &
$5^3=125$.\\
$\sqrt[10]{1024}=2$ & because &
$2^{10}=1024$.\\
\end{tabular}}
\end{exa}


\begin{df}
Let $a, r$ be non-zero natural numbers. A {\em geometric
progression} is a sequence of the form
$$ a, ar, ar^2, ar^3, \ldots, $$that is, a sequence where we
successively multiply by the same number. Here $a$ is the {\em first
term} of the progression, and $r$ is its {\em common ratio}.
\end{df}

\begin{exa}
What is the next sequence of the geometric progression
$$  3, 6, 12, \ldots? $$
\end{exa}
\begin{solu}
Observe that each time we are multiplying by $2$, hence the next
term is $12\cdot 2 = 24$.
\end{solu}


\begin{exa}Find a formula for the $n$-th term of the geometric progression
$$ 6, 12, 24, 48, \ldots . $$  \end{exa}
\begin{solu}
We start with $6$, and then keep multiplying by $2$, thus
$$ 6= 6\cdot 2^0, \qquad  12=6\cdot 2^1, \qquad  24= 6\cdot 2^2, \qquad  48= 6\cdot 2^3, \ldots. $$
The general term is therefore of the form $3\cdot 2^{n-1}$, where
$n=1,2,3,\ldots$ is a natural number.
\end{solu}

\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

\begin{pro}
Find $\sqrt{49}$. \begin{answer} $7$
\end{answer}
\end{pro}
\begin{pro}
Find $\sqrt{64}$. \begin{answer} $8$
\end{answer}
\end{pro}
\begin{pro}
Find $\sqrt[5]{243}$. \begin{answer} $3$
\end{answer}
\end{pro}
\begin{pro}
Find $\sqrt{81}+\sqrt{100}$. \begin{answer} $9+10=19$
\end{answer}
\end{pro}
\begin{pro}
Find $\sqrt[3]{343}$. \begin{answer} $7$
\end{answer}
\end{pro}
\begin{pro}Find
$$ \sqrt{1+2000\sqrt{1+2001\sqrt{1+2002\sqrt{1+2003\cdot 2005}}}}.  $$
\begin{answer}
$2001$
\end{answer}
\end{pro}
\begin{pro}
What is the closest (integer) perfect square to $2003$?
\begin{answer}  Observe that $\sqrt{2003}
> 44$. We see that $44^2 = 1936$ and $45^2 = 2025$. Thus the closest square is $2025$.
\end{answer}
\end{pro}
\begin{pro}
Find the next term of the geometric progression $$ 2, 6, 18, \ldots
.
$$
\begin{answer} $54$.
\end{answer}
\end{pro}
\begin{pro}  The positive non-squares are ordered
in increasing order: $$  2,\quad 3, \quad 5, \quad 6, \quad 7, \quad
8, \quad 10, \ldots .
$$Find the $100$-th term.
\begin{answer}
Since there are $10$ squares between $1$ and $110$ inclusive $110$
is the $100$-th term.
\end{answer}
\end{pro}

\begin{pro}
A crossnumber is like a crossword except that the answers are
numbers with one digit in each square. Complete the following
crossnumber. \vspace*{2cm}
\begin{minipage}{3cm}
\begin{Puzzle}{3}{3}%
|* |[1]{\empty}|[2]{\empty}|.
|[3]{\empty}|{\empty}|{\empty}|.
|{\empty}|* |{\empty}|.
\end{Puzzle}
\end{minipage}
\begin{minipage}{5.5cm}
\begin{PuzzleClues}{\textbf{ACROSS}}\\
\Clue{1}{}{\mbox{Prime Number}}\\  \Clue{3}{}{\mbox{Square of 3D.}}\\
\end{PuzzleClues}
\begin{PuzzleClues}{\textbf{DOWN}}\\
\Clue{1}{}{\mbox{Prime Number.}} \\ \Clue{2}{}{\mbox{Square of 1D.}}\\
\Clue{3}{}{\mbox{ Sqr. root of 3A.}} \\
\end{PuzzleClues}
\end{minipage}
\end{pro}

\begin{pro}
The product of three consecutive even integers is $87*****8$, where
the asterisks represent five missing digits. What is the sum of the
integers?
\begin{answer}Let the integers be $n-2, n, n+2$. Forcedly, they must end in $2, 4, 6$, since their product ends in
$8$. The product of the integers is $n^3-4n \approx n^3$. Now $n^3
\approx 87000008 \implies n \approx 443$. Inspection gives $n=444$,
and the sum is $442+444+446=1332$.
\end{answer}

\end{pro}

\end{multicols}



\part{Fractions}
\chapter{Fractions}
\QUOTEME{I continued to do arithmetic with my father, passing
proudly through fractions to decimals. I eventually arrived at the
point where so many cows ate so much grass, and tanks filled with
water in so many hours I found it quite enthralling.}{\mbox{Agatha
CHRISTIE}}
\bigskip
\dropping{2}{I}n this section we review some of the arithmetic
pertaining fractions. We have already defined fractions in the
lecture on long division, but here is the definition again.
\begin{df}
A {\em (positive numerical) fraction} is a number of the form $m\div
n=\dfrac{m}{n}$ where $m$ and $n$ are natural numbers and $n\neq 0$.
Here $m$ is the {\em numerator} or the fraction and $n$ is the {\em
denominator} of the
fraction.\index{fraction}\index{numerator}\index{denominator}
\end{df}



Given a natural number $n\neq 0$, we divide the interval between
consecutive natural numbers $k$ and $k+1$ into $n$ equal pieces.
Figures \ref{fig:den-2-frac} \ref{fig:den-3-frac}, and
\ref{fig:den-4-frac}, shew examples with $n=2$,  $n=3$, and $n=4$,
respectively. Notice that the larger $n$ is, the finer the
partition. \vspace*{1cm}
\begin{figure}[h]
\psset{unit=6pc} \psaxes[yAxis=false,
subticks=2]{|->}(0,0)(0,0)(7.2,0) \multido{\nx=0+.5,
 \n =0+1}{15}{\uput[d](\nx, 0){$\frac{\n}{2}$}}
 \meinecaption{1}{Fractions with denominator $2$.}
 \label{fig:den-2-frac}
\end{figure}
\vspace*{1cm}
\begin{figure}[h]
\psset{unit=6pc} \psaxes[yAxis=false,
subticks=3]{|->}(0,0)(0,0)(7.2,0)
 \multido{\nx=0+.33333333,
 \n =0+1}{21}{\uput[d](\nx, 0){$\frac{\n}{3}$}}
 \meinecaption{1}{Fractions with denominator $3$.}
 \label{fig:den-3-frac}
\end{figure}
 \vspace*{1cm}
\begin{figure}[h]
\psset{unit=6pc} \psaxes[yAxis=false,
subticks=4]{|->}(0,0)(0,0)(7.2,0)
 \multido{\nx=0+.25,
 \n =0+1}{28}{\uput[d](\nx, 0){$\frac{\n}{4}$}}
 \meinecaption{1}{Fractions with denominator $4$.}
 \label{fig:den-4-frac}
\end{figure}
\begin{rem}You will notice that in many of the fractions written
above, the numerator is larger than the denominator. In primary
school vocabulary these fractions are commonly called {\em
improper}, and perhaps, you were taught to convert them to a {\em
mixed number}. For example, when we write $\dfrac{9}{2}$, your
primary school teacher might have preferred to write $4\frac{1}{2}$.
We will, however, prefer to use improper fractions as this is the
common usage in Algebra and in modern computer algebra
programmes.\index{mixed number}\index{fraction!improper}
\end{rem}




Note that some divisions of natural numbers are effectively natural
numbers. Any fraction whose denominator is a factor of the numerator
is a natural number. In particular, if the denominator is $1$, or if
the numerator is $0$, the fraction is indeed a natural number. Let
us see some examples. \begin{exa}Since the divisors of $6$ are $1$,
$2$, and $3$, all the fractions
$$\dfrac{6}{1}=6, \quad \dfrac{6}{2}=3, \quad \dfrac{6}{6}=1,    $$are natural numbers.\end{exa}
\begin{exa} $\dfrac{6}{3}=2$ is a natural number.\end{exa} \begin{exa}$7=\dfrac{7}{1}$  is a
natural number.\end{exa} \begin{exa}$\dfrac{0}{3}=0$ is a natural
number.\end{exa}
\begin{exa}$\dfrac{4}{0}$ is  undefined.\end{exa}

\par
We now try to understand the meaning of $\dfrac{m}{n}$ when it is
not a natural number. We first consider the following two cases when
$m$ and $n$ are non-zero natural numbers with: (i) $m<n$, (ii)
$m>n$.

\bigskip

Let us consider the first case.


\begin{exa}A picture representation of the concept of $\dfrac{3}{4}$ can
be obtained as follows. Call a square a {\em unit}, that is a $1$,
and divide it into four equal pieces. If we shade three of these
four equal squares, we have shaded $\dfrac{3}{4}$ of the area. There
is, of course, nothing special about squares. We may have used pies,
line segments, etc. It is clear that $0<\dfrac{3}{4}<1$. See figures
\ref{fig:3/4-square}, \ref{fig:3/4-pie}, and
\ref{fig:3/4-segment}.\end{exa} \vspace{1cm}
\begin{figure}[!h]
\begin{minipage}{5cm}
\centering \psset{unit=.8pc} \pspolygon(-4,-4)(-4,4)(4,4)(4,-4)
\pscustom[fillstyle=solid,fillcolor=blue]{\pspolygon(0,0)(4,0)(4,4)(0,4)}
\pscustom[fillstyle=solid,fillcolor=blue]{\pspolygon(0,0)(-4,0)(-4,-4)(0,-4)}
\pscustom[fillstyle=solid,fillcolor=blue]{\pspolygon(0,0)(-4,0)(-4,4)(0,4)}
\pscustom[fillstyle=solid,fillcolor=gray]{\pspolygon(0,0)(4,0)(4,-4)(0,-4)}
\meinecaption{2}{$\dfrac{3}{4}$ of a square.}\label{fig:3/4-square}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=1pc} \SpecialCoor \degrees[100]
%%the circle is divided into "100" degrees
%%one start the wedges in intervals: here [0;10][]
\pswedge[fillstyle=solid,fillcolor=blue]{2}{ 0 }{75}
\pswedge[fillstyle=solid,fillcolor=gray]{2}{75}{100}
\meinecaption{2}{$\dfrac{3}{4}$ of a pie.}\label{fig:3/4-pie}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=.8pc}
\psline[linecolor=blue,linewidth=2pt]{|-|}(-1,-1)(2,-1)\psline[linecolor=gray,linewidth=2pt]{|-|}(2,-1)(3,-1)
\meinecaption{2}{$\dfrac{3}{4}$ of a
segment.}\label{fig:3/4-segment}
\end{minipage}
\end{figure}


\clearpage

Let us consider now the second case.



 \begin{exa}A picture representation of the concept of $\dfrac{5}{2}$ can
be obtained as follows. Call a square a {\em unit}, that is a $1$,
and divide it into two equal pieces. Since we only have two pieces,
it is clear that we need three squares. It follows that
$2<\dfrac{5}{2}<3$. If we shade five of these  six equal pieces, we
have shaded $\dfrac{5}{2}$ of the area. See figures
\ref{fig:5/2-square}, \ref{fig:5/2-pie}, and
\ref{fig:5/2-segment}.\end{exa} \vspace{2cm}
\begin{figure}[!h]
\begin{minipage}{5cm}
\centering \psset{unit=.25pc}
\rput(-10,8){\pspolygon[fillstyle=solid,fillcolor=blue](-4,-4)(0,-4)(0,4)(-4,4)
\pspolygon[fillstyle=solid,fillcolor=blue](0,-4)(4,-4)(4,4)(0,4)}
\rput(0,0){\pspolygon[fillstyle=solid,fillcolor=blue](-4,-4)(0,-4)(0,4)(-4,4)
\pspolygon[fillstyle=solid,fillcolor=blue](0,-4)(4,-4)(4,4)(0,4)}
\rput(10,-8){\pspolygon[fillstyle=solid,fillcolor=blue](-4,-4)(0,-4)(0,4)(-4,4)
\pspolygon[fillstyle=solid,fillcolor=gray](0,-4)(4,-4)(4,4)(0,4)}
\meinecaption{2}{$\dfrac{5}{2}$ squares.}\label{fig:5/2-square}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=.4pc} \SpecialCoor \degrees[100]
%%the circle is divided into "100" degrees
%%one start the wedges in intervals: here [0;10][]
\rput(-10,8){\pswedge[fillstyle=solid,fillcolor=blue]{2}{ 0 }{50}
\pswedge[fillstyle=solid,fillcolor=blue]{2}{50}{100}}
\rput(0,0){\pswedge[fillstyle=solid,fillcolor=blue]{2}{ 0 }{50}
\pswedge[fillstyle=solid,fillcolor=blue]{2}{50}{100}}
\rput(10,-8){\pswedge[fillstyle=solid,fillcolor=blue]{2}{ 0 }{50}
\pswedge[fillstyle=solid,fillcolor=gray]{2}{50}{100}}
\meinecaption{2}{$\dfrac{5}{2}$ pies.}\label{fig:5/2-pie}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=1pc}
\rput(0,3){\psline[linecolor=blue,linewidth=2pt]{|-|}(-1,-1)(1,-1)\psline[linecolor=blue,linewidth=2pt]{|-|}(1,-1)(3,-1)}
\rput(0,0){\psline[linecolor=blue,linewidth=2pt]{|-|}(-1,-1)(1,-1)\psline[linecolor=blue,linewidth=2pt]{|-|}(1,-1)(3,-1)}
\rput(0,-3){\psline[linecolor=blue,linewidth=2pt]{|-|}(-1,-1)(1,-1)\psline[linecolor=gray,linewidth=2pt]{|-|}(1,-1)(3,-1)}
\meinecaption{2}{$\dfrac{5}{2}$ segments.}\label{fig:5/2-segment}
\end{minipage}



\end{figure}









\bigskip




By doing more of the line diagrams above you may have noticed that
there are multiple names for, say, the natural number $2$. For
example
$$ 2 = \dfrac{2}{1} =  \dfrac{4}{2} = \dfrac{6}{3} = \dfrac{8}{4} = \dfrac{10}{5} = \dfrac{12}{6}, $$
etc. This observation is a particular case of the following result.
\begin{thm}[Cancellation Law]
Let $m,n, k$ be natural numbers with $n\neq 0$ and $k \neq 0$. Then
$$\dfrac{mk}{nk} = \dfrac{m}{n}.  $$
\label{thm:cancellation-law} \index{cancellation law}
\end{thm}
\begin{pf} We will prove this for $m\leq n$. For $m>n$ the argument
is similar. Divide the interval $[0;1]$ into  $nk$ pieces. Consider
the $k$-th, $2k$-th, $3k$-th, \ldots , $nk$-th markers. Since
$\dfrac{nk}{nk}=1$, the $nk$-th marker has to be $1$. Thus the $n$
markers $k$-th, $2k$-th, $3k$-th, \ldots , $nk$-th, form a division
of $[0;1]$ into $n$ equal spaces. It follows that the $k$-th marker
is $\dfrac{1}{n}$, that is, $\dfrac{k}{nk}=\dfrac{1}{n}$, the
$2k$-th marker is $\dfrac{2}{n}$, that is,
$\dfrac{2k}{nk}=\dfrac{2}{n}$, etc., and so the  $mk$-th marker is
$\dfrac{m}{n}$, that is, $\dfrac{mk}{nk}=\dfrac{m}{n}$, as we wanted
to prove.\end{pf} Thus given a fraction, if the numerator and the
denominator have any common factors greater than $1$, that is, any
non-trivial factors, we may reduce the fraction and get an equal
fraction. \begin{df}Two fractions such that $\dfrac{a}{b} =
\dfrac{x}{y}$ are said to be {\em equivalent}. If $b<y$, then
$\dfrac{a}{b}$ is said to be a {\em reduced form} of $\dfrac{x}{y}$.
\end{df}\index{fraction!equivalent} \index{fraction!reduced form}
\begin{rem}
It is possible to prove that any fraction has a  unique reduced form
with minimal denominator, the so called {\em equivalent fraction in
lowest terms}. This depends on the fact that the natural numbers can
be factored uniquely into primes, and hence, though we will accept
this result, we will not prove it here.
\end{rem}

\begin{exa}
 A geometric interpretation of  why  $\dfrac{1}{2}=\dfrac{2}{4}$ is given in figures
\ref{fig:equiv-half-quarter1} and \ref{fig:equiv-half-quarter2}.
\end{exa} \vspace{1cm}
\begin{figure}[!h]
\begin{minipage}{7cm}\centering\psset{unit=.7pc}
\pspolygon(-4,-4)(-4,4)(4,4)(4,-4)
\pscustom[fillstyle=solid,fillcolor=white]{\pspolygon(0,-4)(4,-4)(4,4)(0,4)}
\pscustom[fillstyle=solid,fillcolor=red]{\pspolygon(0,-4)(-4,-4)(-4,4)(0,4)}
\meinecaption{1}{$\dfrac{1}{2}$} \label{fig:equiv-half-quarter1}
\end{minipage}
\hfill
\begin{minipage}{7cm}\centering\psset{unit=.7pc}
\pspolygon(-4,-4)(-4,4)(4,4)(4,-4)
\pscustom[fillstyle=solid,fillcolor=white]{\pspolygon(0,0)(4,0)(4,4)(0,4)}
\pscustom[fillstyle=solid,fillcolor=red]{\pspolygon(0,0)(-4,0)(-4,-4)(0,-4)}
\pscustom[fillstyle=solid,fillcolor=red]{\pspolygon(0,0)(-4,0)(-4,4)(0,4)}
\pscustom[fillstyle=solid,fillcolor=white]{\pspolygon(0,0)(4,0)(4,-4)(0,-4)}
\meinecaption{1}{$\dfrac{2}{4}$} \label{fig:equiv-half-quarter2}
\end{minipage}
\end{figure}

\begin{exa}
To reduce $\dfrac{104}{120}$ to lowest terms, observe that
$$ \dfrac{104}{120} = \dfrac{104\div 4}{120 \div 4} = \dfrac{26\div 2}{30\div 2} = \dfrac{13}{15}.
$$Since $13$ and $15$ do not have a non-trivial factor in common,
$\dfrac{13}{15}$ is the desired reduction.
\end{exa}
\begin{rem}
The reduction steps above are not unique. For example, if we had
seen right away that $8$ was a common factor, we would have obtained
$ \dfrac{104}{120} = \dfrac{104\div 8}{120 \div 8} =\dfrac{13}{15}$,
obtaining the same result. Hence, no matter how many steps you take,
as long as there is valid cancellation to do and you perform them
all, you will always obtain the right result.
\end{rem}
\begin{exa}
Find a fraction with denominator $120$ equivalent to
$\dfrac{11}{24}$.
\end{exa}
\begin{solu}
Observe that $120\div 24 =5$. Thus
$$\dfrac{11}{24} =\dfrac{11\cdot 5}{24\cdot 5} =\dfrac{55}{120}.   $$
\end{solu}

As we mentioned before, we will prefer to use improper fraction to
mixed numbers. Here is an example of how to convert from one to the
other and viceversa.

\begin{exa}
Convert the mixed number $2\frac{3}{4}$ into an improper fraction.
\end{exa}
\begin{solu}
We have $$ 2\tfrac{3}{4} = 2 + \dfrac{3}{4}
=\dfrac{8}{4}+\dfrac{3}{4} = \dfrac{11}{4}.$$ Some people prefer to
write this explanation as $ 2\tfrac{3}{4} = \dfrac{2\cdot 4 + 3}{4}
=\dfrac{11}{4}$.
\end{solu}
\begin{exa}
Convert the improper fraction $\dfrac{19}{4}$ to a mixed number.
\end{exa}
\begin{solu}
We perform the long division of $\dfrac{19}{4}=19\div 4 $: $$
\opidiv[displayintermediary=all]{19}{4}$$ whence
$\dfrac{19}{4}=4\tfrac{3}{4}$.
\end{solu}

\section*{Homework}
\begin{pro}
Locate the following fractions on the number line: $\dfrac{3}{4}$,
$\dfrac{7}{4}$, and $\dfrac{11}{4}$. \vspace{1cm}
\begin{figure}[h]
\centering \psset{unit=6pc}\rput(-1.5,0){\psaxes[yAxis=false,
subticks=4,arrowscale=3,]{|->}(0,0)(0,0)(3.2,0)}
 \meinecaption{1}{Problem \ref{pro:deno-4}.}
 \label{fig:deno-4}
\end{figure} \label{pro:deno-4}

\end{pro}
\begin{pro}
Find two fractions equivalent to ${\dfrac{5}{3}}$. Show all the
steps.
\end{pro}
\begin{pro}
Find two fractions equivalent to ${\dfrac{3}{5}}$. Show all the
steps.
\end{pro}
\begin{pro}
Find a fraction equivalent to ${\dfrac{3}{10}}$ with denominator
$50$. Show all the steps.
\begin{answer}
$\dfrac{3}{5}=\dfrac{3\cdot 5}{10\cdot 5}=\dfrac{15}{50}$
\end{answer}
\end{pro}
\begin{pro}
Circle {\em Yes} or {\em No} depending on whether the pair of
fractions are equivalent. Justify your answer.
\begin{enumerate}
\item ${\dfrac{3}{7}}$,\quad  ${\dfrac{12}{28}}$ \quad \pscirclebox{Yes} \quad \pscirclebox{No}
\item ${\dfrac{2}{3}}$,\quad   ${\dfrac{5}{6}}$ \quad \pscirclebox{Yes} \quad \pscirclebox{No}
\item ${\dfrac{0}{5}}$,\quad   ${\dfrac{0}{3}}$ \quad \pscirclebox{Yes} \quad \pscirclebox{No}
\item ${\dfrac{7}{7}}$,\quad
${\dfrac{345}{345}}$ \quad \pscirclebox{Yes} \quad
\pscirclebox{No}\item ${\dfrac{14}{2}}$, ${\dfrac{21}{3}}$\quad
\pscirclebox{Yes} \quad \pscirclebox{No}
\item ${\dfrac{1}{16}}$,\quad   ${\dfrac{1}{32}}$\quad \pscirclebox{Yes} \quad \pscirclebox{No}
\item ${\dfrac{1}{4}}$,\quad   ${\dfrac{2}{8}}$\quad \pscirclebox{Yes} \quad \pscirclebox{No}
\end{enumerate}
\end{pro}
 \begin{pro}
What fraction of an hour is $36$ minutes?
\begin{answer}
$\dfrac{36}{60} = \dfrac{3}{5}$ of an hour.
\end{answer}
 \end{pro}
\begin{pro}
In making a garden fertiliser, a gardener mixes 2 lbs. of  nitrate,
3 lbs. of phosphate, and 6 lbs. of potash. What is the ratio of
nitrate to the total amount of fertiliser?
\begin{answer}
$\dfrac{2}{2+3+6}=\dfrac{2}{11}$.
\end{answer}
\end{pro}
\begin{pro}
Express  $\dfrac{102}{210}$ in least terms.
\begin{answer}$\dfrac{17}{35} $. \end{answer}
\end{pro}
\begin{pro}
Find an equivalent fraction to  $\dfrac{102}{210}$ with denominator
$3990$.
\begin{answer}Observe that $3990=210\cdot 19$. Thus $\dfrac{102}{210}= \dfrac{102\cdot 19}{210\cdot 19}=\dfrac{1938}{3990}$. \end{answer}
\end{pro}
\begin{pro}
Arrange in increasing order: \quad $\dfrac{2}{3}$, $\dfrac{3}{5}$,
$\dfrac{8}{13}$. \begin{answer} Observe that $3\cdot 5 \cdot 13 =
195$ is a common denominator. Now
$$\dfrac{2}{3}= \dfrac{130}{195}; \quad \dfrac{3}{5}= \dfrac{117}{195}; \quad \dfrac{8}{13}= \dfrac{120}{195},  $$
whence $\dfrac{3}{5}$ is the smallest, $\dfrac{8}{13}$ is in the
middle, and $\dfrac{2}{3}$ is the largest.
\end{answer}
\end{pro}
\begin{pro}
Reduce to lowest terms: $\dfrac{116690151}{427863887}$.
\begin{answer}$\dfrac{116690151}{427863887}=\dfrac{3\cdot 38896717}{11\cdot 38896717} = \dfrac{3}{11}.$
\end{answer}
\end{pro}
\begin{pro}
Convert the improper fraction to a mixed number: $\dfrac{13}{7}$
\begin{answer}
$1\tfrac{6}{7}$
\end{answer}
\end{pro}
\begin{pro}
Convert the improper fraction to a mixed number: $\dfrac{23}{5}$
\begin{answer}
$4\tfrac{3}{5}$
\end{answer}
\end{pro}
\begin{pro}
Convert the mixed number to an improper fraction: $3\tfrac{1}{3}$
\begin{answer}
$\dfrac{10}{3}$
\end{answer}
\end{pro}
\begin{pro}
Convert the mixed number to an improper fraction: $2\tfrac{4}{7}$
\begin{answer}
$\dfrac{18}{7}$
\end{answer}
\end{pro}

\begin{pro}
Four singers take part in a musical round of $4$ equal lines, each
finishing after singing the round through $3$ times. The second
singer begins when the first singer begins the second line, the
third singer begins when the first singer begins the third line, the
fourth singer begins when the first singer begins the fourth line.
What is the fraction of the total singing time when all the singers
are singing simultaneously?
\begin{answer}
Three lines have been sung before the fourth singer starts, and
after that he sings 12 more lines. So the total span of lines is 15.
They start singing simultaneously from line 4, and the first singer
is the first to end, in line 12. Thus $12 - 4 + 1 = 9$ lines out of
15 are sung simultaneously and the fraction sought is $\frac{9}{15}
= \frac{3}{5}.$
\end{answer}

\end{pro}


\chapter{Addition and Subtraction of Fractions}
\QUOTEME{A man is like a fraction whose numerator is what he is and
whose denominator is what he thinks of himself. The larger the
denominator the smaller the fraction.}{Count Lev Nikolgevich
TOLSTOY}
\bigskip

\dropping{2}{W}e now define addition of fractions. We would like
this definition to agree with our definition of addition of natural
numbers. Recall that we defined addition of natural numbers $x$ and
$y$ as the concatenation of two segments of length $x$ and $y$. We
thus define the addition of two fractions $\dfrac{a}{b}$ and
$\dfrac{c}{d}$ as the concatenation of two segments of length
$\dfrac{a}{b}$ and $\dfrac{c}{d}$.  The properties of associativity,
commutativity, etc., stem from this definition. The problem we now
have is how to concretely apply this definition to find the desired
sum? From this concatenation definition it follows that natural
numbers $a$ and $b\neq 0$,
\begin{equation}  \dfrac{a}{b} =\underbrace{\dfrac{1}{b}+\cdots +\dfrac{1}{b}}_{a\ \mathrm{times}}.\end{equation}
It follows from the above property that
\begin{equation}  \dfrac{x}{b}+\dfrac{y}{b} = \dfrac{x+y}{b}.\label{eq:frac-same-den}\end{equation}
\begin{exa}
We have $$ \dfrac{2}{13} + \dfrac{5}{13} = \dfrac{7}{13}. $$
\end{exa}


We now  determine a general formula for adding fractions of
different denominators.
\begin{thm}[Sum of Fractions]
Let $a, b, c, d$ be natural numbers with $b\neq 0$ and $d\neq 0$.
Then $$\dfrac{a}{b}+\dfrac{c}{d} = \dfrac{ad+bc}{bd}.$$
\label{thm:sum-of-fractions}
\end{thm}
\begin{pf}
From the Cancellation Law (Theorem \ref{thm:cancellation-law}),
$$\dfrac{a}{b}+\dfrac{c}{d} = \dfrac{ad}{bd}+\dfrac{bc}{bd} =\dfrac{ad+bc}{bd},
$$proving the theorem.
\end{pf}
The formula obtained in the preceding theorem agrees with that of
(\ref{eq:frac-same-den}) when the denominators are equal. For, using
the theorem,
$$  \dfrac{x}{b}+\dfrac{y}{b} =  \dfrac{xb}{b\cdot b}+\dfrac{yb}{b\cdot b} = \dfrac{xb+by}{b\cdot b} =
\dfrac{b(x+y)}{b\cdot b} = \dfrac{x+y}{b},  $$ where we have used
the distributive law.

\bigskip

Observe that the trick for adding the fractions in the preceding
theorem was to convert them to fractions of the same denominator.
\begin{df}To express two fractions in a {\em common denominator} is
to write them in the same denominator. The smallest possible common
denominator is called the {\em least common denominator}.
\end{df}\index{fraction!least common denominator}\index{fraction!common denominator}

\begin{exa}
Add: $\dfrac{3}{5}+\dfrac{4}{7}$.
\end{exa}
\begin{solu}
A common denominator is $5\cdot 7=35$. We thus find
$$\dfrac{3}{5}+\dfrac{4}{7}= \dfrac{3\cdot 7}{5\cdot 7}+\dfrac{4\cdot 5}{7\cdot 5} = \dfrac{21}{35}+\dfrac{20}{35} = \dfrac{41}{35}.  $$
\end{solu}

\begin{rem}In the preceding example, $35$ is not the only denominator
that we may have used. Observe that $\dfrac{3}{5} = \dfrac{42}{70}$
and $\dfrac{4}{7}=\dfrac{40}{70}$. Adding,
$$\dfrac{3}{5}+\dfrac{4}{7} = \dfrac{3\cdot 14}{5\cdot 14}+\dfrac{4\cdot 10}{7\cdot 10} = \dfrac{42}{70}+\dfrac{40}{70} =\dfrac{82}{70}=\dfrac{82\div 2}{70\div 2} =\dfrac{41}{35}.$$
This shews that it is not necessary to find the {\em least common
denominator} in order to add fractions, simply a common
denominator.\end{rem} In fact,  let us list  the multiples of $5$
and of $7$ and let us circle the common multiples on these lists:
\begin{center}\begin{tabular}{l}
The multiples of $5$ are \quad $5,10,15,20,25,30,\pscirclebox{35},40,45,50,55,60,65,\pscirclebox{70},75,\ldots , $\\
The multiples of $7$ are \quad $7,14,21,28,\pscirclebox{35},42,49,56,63,\pscirclebox{70},77,\ldots .  $\\
\end{tabular}
\end{center}
The sequence $$ 35, 70, 105, 140, \ldots , $$is a sequence of common
denominators for $5$ and $7$.

\begin{exa}
To perform the addition $\dfrac{2}{7}+\dfrac{1}{5}+\dfrac{3}{2}$,
observe that $7\cdot 5 \cdot 2 = 70$ is a common denominator. Thus
$$
\begin{array}{lll}\dfrac{2}{7}+\dfrac{1}{5}+\dfrac{3}{2} & = &
\dfrac{2\cdot 10}{7\cdot 10}+\dfrac{1\cdot 14}{5\cdot
14}+\dfrac{3\cdot 35}{2\cdot 35}\\
& = & \dfrac{20}{70}+\dfrac{14}{70}+\dfrac{105}{70}\\
& = & \dfrac{20+14+105}{70}\\
& = & \dfrac{139}{70}.\\

\end{array}
$$
\end{exa}

A few words about subtraction of fractions. Suppose
$\dfrac{a}{b}\geq \dfrac{c}{d}$. Observe that this means that
$ad\geq bc$, which in turns means that $ad-bc\geq 0$. Then from a
segment of length $\dfrac{a}{b}$ we subtract one of $\dfrac{b}{d}$.
In much the same manner of addition of fractions then
$$ \dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad}{bd}-\dfrac{bc}{bd} = \dfrac{ad-bc}{bd}. $$
Now, since $ad-bc\geq 0$, $ad-bc$ is a natural number and hence
$\dfrac{ad-bc}{bd}$ is a fraction.


\begin{exa}
We have
$$ \dfrac{3}{4} - \dfrac{2}{5} = \dfrac{15}{20} - \dfrac{8}{20} = \dfrac{7}{20}. $$
\end{exa}

\bigskip

If one of the terms is a mixed number, it should first be converted
to an improper fraction.
\begin{exa}
We have $$3\frac{3}{4}-2\dfrac{3}{5} = \dfrac{15}{4} -
\dfrac{13}{10} =\dfrac{75}{20}-\dfrac{26}{20} = \dfrac{49}{20}.
$$
\end{exa}
\begin{exa}
We have $$2\frac{1}{4}+\dfrac{1}{3} = \dfrac{9}{4} + \dfrac{1}{3}
=\dfrac{27}{12}+\dfrac{4}{12} = \dfrac{31}{12}.
$$
\end{exa}


\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Jack and Jill were eating cherries. Jack grabbed $\dfrac{1}{3}$ of
the initial amount of cherries and Jill grabbed $\dfrac{1}{6}$ of
the initial amount of cherries. What fraction of the original amount
of cherries remains?
\begin{answer}
$1-\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{2}$.
\end{answer}
\end{pro}
\begin{pro}
Add the fractions and reduce to least terms: $\dfrac{1}{7} +
\dfrac{1}{5}$. If your result is an improper fraction, leave it in
this form
\begin{answer} $\dfrac{12}{35}$  \end{answer}
\end{pro}
\begin{pro}
Perform the operations  and reduce to least terms: $\dfrac{1}{7} +
\dfrac{1}{5} - \dfrac{2}{35}$. If your result is an improper
fraction, leave it in this form
\begin{answer} $\dfrac{2}{7}$  \end{answer}
\end{pro}
\begin{pro}
Perform the operations  and reduce to least terms:
$2\frac{1}{2}-\tfrac{1}{4}$. If your result is an improper fraction,
leave it in this form
\begin{answer} $\dfrac{9}{4}$  \end{answer}
\end{pro}
\begin{pro}
Perform the operations  and reduce to least terms:
$3\frac{5}{9}-\tfrac{2}{9}$. If your result is an improper fraction,
leave it in this form
\begin{answer} $\dfrac{10}{3}$  \end{answer}
\end{pro}
\begin{pro}
Perform the operations  and reduce to least terms:
$3\frac{5}{9}-\tfrac{8}{9}$. If your result is an improper fraction,
leave it in this form
\begin{answer} $\dfrac{8}{3}$  \end{answer}
\end{pro}
\begin{pro}
Perform the operations  and reduce to least terms:
$3\frac{5}{9}+2\tfrac{8}{9}$. If your result is an improper
fraction, leave it in this form
\begin{answer} $\dfrac{58}{9}$  \end{answer}
\end{pro}
\begin{pro}
Perform the operations  and reduce to least terms:
$3\frac{5}{9}-2\tfrac{8}{9}$. If your result is an improper
fraction, leave it in this form
\begin{answer} $\dfrac{2}{3}$  \end{answer}
\end{pro}

\end{multicols}



\chapter{Multiplication and Division of Fractions}

We now approach multiplication of fractions. We saw that a possible
interpretation for the product $xy$ of two natural numbers $x$ and
$y$ is the area of a rectangle with sides of length $x$ and $y$. We
would like to extend this interpretation in the case when $x$ and
$y$ are fractions. That is, given a rectangle of sides of length
$x=\dfrac{a}{b}$ and $y=\dfrac{c}{d}$, we would like to deduce that
$\dfrac{ac}{bd}$ is the area of this rectangle.
\begin{thm}[Multiplication of Fractions]
Let $a, b, c, d$ be natural numbers with $b\neq 0$ and $d\neq 0$.
Then $$\dfrac{a}{b}\cdot \dfrac{c}{d} = \dfrac{ac}{bd}.$$
\label{thm:prod-of-fractions}
\end{thm}
\begin{pf}First consider the case when $a=c=1.$ Start with a unit
square and cut it horizontally into $b$ equal segments. Then cut it
vertically into $d$ equal segments. We have now $bd$ equal pieces,
each one having an area of $\dfrac{1}{bd}$. Since each piece is in
dimension $\dfrac{1}{b}$ by $\dfrac{1}{d}$, we have shewn that
$$ \dfrac{1}{b}\cdot \dfrac{1}{d} = \dfrac{1}{bd}. $$

\bigskip
Construct now a rectangle of length $\dfrac{a}{b}$ and width
$\dfrac{c}{d}$. Such a rectangle is obtained by concatenating along
its length $a$ segments of length $\dfrac{1}{b}$ and along  its
width $c$ segments of length $\dfrac{1}{d}$. This partitions the
large rectangle into $ac$ sub-rectangles, each of area
$\dfrac{1}{bd}$. Hence the area of the $\dfrac{a}{b}$ by
$\dfrac{c}{d}$ rectangle is $ab\left(\dfrac{1}{bd}\right)$, from
where
$$ \dfrac{a}{b}\cdot \dfrac{c}{d} = ac\left(\dfrac{1}{bd}\right)=\dfrac{ac}{bd}, $$
proving the theorem.
\end{pf}


\begin{exa}
We have
$$ \dfrac{2}{3}\cdot \dfrac{3}{7} = \dfrac{6}{21}=\dfrac{2}{7}. $$
Alternatively, we could have cancelled the common factors, as
follows,
$$ \dfrac{2}{\cancel{3}}\cdot \dfrac{\cancel{3}}{7} =\dfrac{2}{7}. $$
\end{exa}

\bigskip

Again, if one of the terms  is a mixed number, it must first be
converted to an improper fraction.
\begin{exa}
We have
$$ \left(2\tfrac{3}{4}\right)\cdot \tfrac{2}{11} = \dfrac{11}{4}\cdot \dfrac{2}{11} = \dfrac{2}{4}=\dfrac{1}{2}. $$
\end{exa}

\begin{exa}
Find the exact value of the product
$$\left(1 - \dfrac{2}{5}\right)\left(1 - \dfrac{2}{7}\right)\left(1 - \dfrac{2}{9}\right) \cdots \left(1 - \dfrac{2}{99}\right)\left(1 - \dfrac{2}{101}\right) .$$
\end{exa}\begin{solu} We have,
$$\begin{array}{lll}\left(1 - \dfrac{2}{5}\right)\left(1 - \dfrac{2}{7}\right)\left(1 - \dfrac{2}{9}\right) \cdots \left(1 - \dfrac{2}{99}\right)\left(1 - \dfrac{2}{101}\right)
& = &
\dfrac{3}{5}\cdot\dfrac{5}{7}\cdot\dfrac{7}{9}\cdot\dfrac{9}{11}\cdots
\dfrac{97}{99}\cdot\dfrac{99}{101}\\
& = & \dfrac{3}{101}.
\end{array}$$
\end{solu}
We now tackle division of fractions. Recall that we defined division
of natural numbers as follows. If $n\neq 0$ and $m, x$ are natural
numbers then $m\div n = x$ means that $m=xn$. We would like a
definition of fraction division compatible with this definition of
natural number division. Hence we give the following definition.
\begin{df}
Let $a, b, c, d$ be natural numbers with $b\neq 0$, $c\neq 0$,
$d\neq 0$. We define the fraction division
$$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{x}{y} \iff \dfrac{a}{b} = \dfrac{x}{y}\cdot \dfrac{c}{d}.  $$
\label{df:div-fraction}\end{df} We would like to know what
$\dfrac{x}{y}$ above is in terms of $a, b, c, d$. For this purpose
we have the following theorem.
\begin{thm}[Division of Fractions]\label{thm:div-of-fractions}
Let $a, b, c, d$ be natural numbers with $b\neq 0$, $c\neq 0$,
$d\neq 0$. Then
$$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c} = \dfrac{ad}{bc},
$$ that is, $\dfrac{x}{y}$ in definition \ref{df:div-fraction} is
$\dfrac{x}{y}=\dfrac{ad}{bc}$.
\end{thm}
\begin{pf}
Let us  prove that $\dfrac{x}{y}=\dfrac{ad}{bc}$ satisfies the
definition of division of fractions. Observe that
$$ \dfrac{x}{y}\cdot \dfrac{c}{d} = \dfrac{ad}{bc}\cdot \dfrac{c}{d} =
\dfrac{adc}{bcd}=\dfrac{a}{b},$$ and hence $\dfrac{x}{y}$ so chosen
is the right result for the division of fractions. Could there be
another fraction, say $\dfrac{x'}{y'}$ that satisfies definition
\ref{df:div-fraction}? Suppose
$$\dfrac{a}{b} = \dfrac{x'}{y'}\cdot \dfrac{c}{d}.  $$Then
$$\dfrac{x}{y}= \dfrac{a}{b} \cdot \dfrac{d}{c} =  \dfrac{x'}{y'}\cdot \dfrac{c}{d}\cdot \dfrac{d}{c}= \dfrac{x'}{y'}.$$
Hence $\dfrac{x}{y}=\dfrac{a}{b} \cdot \dfrac{d}{c}$ is the only
fraction that satisfies the definition of fraction division.
\end{pf}
\begin{df}
Let $c\neq 0$, $d\neq 0$ be natural numbers. The {\em reciprocal} of
the fraction $\dfrac{c}{d}$ is the fraction $\dfrac{d}{c}$.
\end{df}
Theorem \ref{thm:div-of-fractions} says that in order to divide two
fractions we must simply multiply the first one by the reciprocal of
the other.
\begin{exa}
The reciprocal of the fraction $\dfrac{3}{4}$ is $\dfrac{4}{3}$.
\end{exa}
\begin{exa}
The reciprocal of the fraction $\dfrac{1}{2}$ is $\dfrac{2}{1}$,
that is $2$.
\end{exa}
\begin{exa}
The reciprocal of the mixed number $1\tfrac{3}{4}$ is the same as
the reciprocal of the improper fraction $\dfrac{7}{4}$, which is  is
$\dfrac{4}{7}$.
\end{exa}


\begin{exa}
We have,
$$\dfrac{24}{35}\div \dfrac{20}{7} = \dfrac{24}{35}\cdot \dfrac{7}{20} = \dfrac{4\cdot 6}{7\cdot 5}\cdot \dfrac{7\cdot 1}{4\cdot 5}
=\dfrac{6\cdot 1}{5\cdot 5}=\dfrac{6}{25}.$$
\end{exa}
\begin{exa}
We have,
$$2\tfrac{1}{3}\div 3\tfrac{2}{3}= \dfrac{7}{3}\div \dfrac{11}{3} = \dfrac{7}{3}\cdot \dfrac{3}{11} =
\dfrac{7}{11}.$$
\end{exa}
\begin{exa}
Find the value of $$ \dfrac{1}{1+ \dfrac{1}{1+ \dfrac{1}{1+
\dfrac{1}{1}}}}. $$
\end{exa}
\begin{solu}
Plainly,
$$ \dfrac{1}{1+ \dfrac{1}{1+ \dfrac{1}{1+
\dfrac{1}{1}}}}= \dfrac{1}{1+ \dfrac{1}{1+ \dfrac{1}{2}}}=
\dfrac{1}{1+ \dfrac{1}{\dfrac{3}{2}}}=\dfrac{1}{1+
\dfrac{2}{3}}=\dfrac{1}{\dfrac{5}{3}}=\dfrac{3}{5}.
$$
\end{solu}

\begin{exa}
 Calculate: $\dfrac{4}{5}\cdot \dfrac{5}{6}+\dfrac{4}{5}\div \dfrac{5}{6}$
\end{exa}
\begin{solu}
We have
 $$\dfrac{4}{5}\cdot \dfrac{5}{6}+\dfrac{4}{5}\div
 \dfrac{5}{6}=\dfrac{2}{3}+\dfrac{24}{25}=\dfrac{122}{75}.$$
\end{solu}
\begin{exa}
Calculate:
$\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{\dfrac{1}{2}-\dfrac{1}{3}}$.
\end{exa}
\begin{solu} We have
$$\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{\dfrac{1}{2}-\dfrac{1}{3}}=\dfrac{\dfrac{5}{6}}{\dfrac{1}{6}}=5.$$
\end{solu}


\section*{Homework}  \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
 What is one third of a half of a fifth of a quarter of $99000$?
\begin{answer}
This is $\dfrac{1}{3}\cdot \dfrac{1}{2}\cdot \dfrac{1}{5}\cdot
\dfrac{1}{4}\cdot 99000=825$.
\end{answer}
\end{pro}
\begin{pro}
If $4$ times a number is $48$, what is $\dfrac{1}{3}$ of the number?
\begin{answer}
$4$
\end{answer}
\end{pro}
\begin{pro}
John had $\$240$ of which he lost $\dfrac{5}{8}$ betting at the
horses. How much money does he have left?
\begin{answer}
$\$150$
\end{answer}
\end{pro}


\begin{pro}Complete the ``fraction puzzle'' below. \vspace*{1cm}
 \PuzzleSolution
\begin{Puzzle}{5}{5}%
|{$\frac{1}{3}$}|{$+$}|{\empty}|{$=$}|{2}|. |{$+$}|*|{$-$}|*|*|.
|{$\frac{3}{4}$}|{$ \cdot $}|{\empty}|{$=$}|{\empty}|.
|{$=$}|*|{$=$}|*|*|.
|{\empty}|{$\div$}|{$\frac{2}{3}$}|{$=$}|{\empty}|.
\end{Puzzle}
\begin{answer}
Here are the solutions. \\

\PuzzleSolution
\begin{Puzzle}{5}{5}%
|{$\frac{1}{3}$}|{$+$}|{$\frac{5}{3}$}|{$=$}|{2}|.
|{$+$}|*|{$-$}|*|*|.
|{$\frac{3}{4}$}|{$ \cdot $}|{$1$}|{$=$}|{$\frac{3}{4}$}|.
|{$=$}|*|{$=$}|*|*|.
|{$\frac{13}{12}$}|{$\div$}|{$\frac{2}{3}$}|{$=$}|{$\frac{13}{8}$}|.
\end{Puzzle}

\end{answer}
\end{pro}
\begin{pro}
Tito was pigging out on cookies, and in the course of three days, he
ate $420$ cookies. On the first day, he ate $\dfrac{2}{5}$ of the
cookies. On the second day, he ate $\dfrac{5}{6}$ of the remaining
cookies. How many cookies did he eat on the third day?
\begin{answer}
On the first day he ate $\dfrac{2}{5}\cdot 420 = 168$ cookies, so
there remain $420-168=252$ cookies after the first day. On the
second day he ate $\dfrac{5}{6}\cdot 252 = 210$ cookies, so there
remain $252-210=42$ cookies to be eaten on the third day. Answer:
$42$ cookies.
\end{answer}
\end{pro}
\begin{pro}
David spent $\dfrac{2}{5}$ of his money on a storybook. The
storybook cost $\$20$. How much money did he have
at first?
\begin{answer}
$\dfrac{5}{2}\cdot 20 = 50$ dollars.
\end{answer}
\end{pro}
\begin{pro}
Anna and Tina agree to wash $15$ windows for \$90. If Anna  washed $9$ windows and Tina $6$, and each agrees to receive the same amount per window,
how much will each get?
\begin{answer}
Anna gets $\tfrac{9}{15}\cdot 90=54$ dollars and Tina $\tfrac{6}{15}\cdot 90 36$ dollars.
\end{answer}
\end{pro}
\begin{pro}
When the cathedral clock strikes four, it takes $8$ seconds between the first and the last strokes. How many seconds does it take to
strike twelve?
\begin{answer}
$29\tfrac{1}{3}$ seconds.
\end{answer}
\end{pro}


\begin{pro}
Find the exact numerical value of  $$\left(\dfrac{2}{51}\right)\div
\left(\dfrac{3}{17}\right) \cdot\left(\dfrac{7}{10}\right) .$$
\begin{answer}
$\dfrac{7}{45}$.
\end{answer}
\end{pro}
\begin{pro}
Find the exact numerical value of
$$\dfrac{\tfrac{4}{7}-\tfrac{2}{5}}{\tfrac{4}{7}+\tfrac{2}{5}}.$$
\begin{answer}
$\dfrac{3}{17}$.
\end{answer}
\end{pro}
\begin{pro}
Find the value of $$\dfrac{10 + 10^2}{\dfrac{1}{10} +
\dfrac{1}{100}}. $$ \begin{answer} We have,  $$  \dfrac{10 +
10^2}{\dfrac{1}{10} + \dfrac{1}{100}} = \dfrac{10^3 + 10^4}{10 + 1}
= \dfrac{11000}{11} = 1000.$$

\end{answer}
\end{pro}

\begin{pro}
Find the exact numerical value of \quad
$$\dfrac{1^3+2^3+3^3-3(1)(2)(3)}{(1+2+3)(1^2+2^2+3^2-1\cdot 2-2\cdot
3-3\cdot 1)}.$$
\begin{answer}
$1$.
\end{answer}
\end{pro}
\begin{pro}
 If
$$ \cfrac{1}{1 + \cfrac{1}{5}} = \frac{a}{b},$$ where the fraction
$\frac{a}{b}$ is in least terms, find $a^2 + b^2$.
\begin{answer}We have, $$ \cfrac{1}{1 + \cfrac{1}{5}} =  \dfrac{1}{
\dfrac{6}{5}} = \frac{5}{6} = \frac{a}{b}, $$whence $a^2 + b^2 = 5^2
+ 6^2 =  61$.

\end{answer}
\end{pro}

\begin{pro}
Find the exact numerical value of \quad
$$\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdots  \left(1-\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right). $$
\begin{answer}
$\dfrac{1}{100}$.
\end{answer}
\end{pro}
\begin{pro}
A painter had $25$ gallons of paint. He used $2\tfrac{1}{2}$ gallons
of paint every hour. He finished the job in $5\tfrac{1}{2}$ hours.
How much paint did he have left?
\begin{answer}
$25-(2\tfrac{1}{2})\cdot
(5\tfrac{1}{2})=\dfrac{45}{4}=11\tfrac{1}{4}$ gallons.
\end{answer}
\end{pro}
\begin{pro}
Naomi has $16$ yards of gift-wrap in order to wrap the gifts for the
Festival of Lights at the community centre. Each gift requires
$1\tfrac{7}{8}$ yards of paper. How many gifts can she wrap?
\begin{answer} We have $1\tfrac{7}{8}=\dfrac{15}{8}$ and
$$16 \div \dfrac{15}{8}=\dfrac{16}{1}\cdot \dfrac{8}{15}
=\dfrac{128}{15} =8\tfrac{8}{15},$$so she is able to wrap $8$ gifts.
\end{answer}
\end{pro}
\begin{pro}
Evaluate $$\cfrac{1}{2 - \cfrac{1}{2 - \cfrac{1}{2 - \cfrac{1}{2}}}}
.$$

\begin{answer}
 Proceeding from the innermost fraction one easily sees that
$$\dfrac{1}{2 - \dfrac{1}{2 - \dfrac{1}{2 - \dfrac{1}{2}}}} =
\dfrac{1}{2 - \dfrac{1}{2 - \dfrac{2}{3}}} = \dfrac{1}{2 -
\dfrac{3}{4}} = \dfrac{4}{5}.$$
\end{answer}
\end{pro}
\begin{pro}
Find the value of $$1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4}}}. $$
\begin{answer} $\dfrac{43}{30}$ \end{answer}
\end{pro}
\begin{pro}
What is the angle between the hands of a clock at a quarter to five?
\begin{answer}
Measure angles clockwise, with origin ($0^\circ$), at 12:00. Each
minute ran by the minute hands accounts for $\frac{360^\circ}{60} =
6^\circ$, and so, each, when jumping from hour to hour, the hands
travel $5(6^\circ) = 30^\circ$. When the minute hand is on the $8$,
it has  travelled $\frac{40}{60} = \frac{2}{3}$ of a circumference,
that is
$$\frac{2}{3}360^\circ = 240^\circ$$and the hour hand has moved
$\frac{2}{3}$ of the way from $4$ to $5$, travelling
$$\left(4 + \frac{2}{3}\right)(30^\circ) = 140^\circ. $$Hence, angle between
the hands is $240^\circ - 140^\circ = 100^\circ.$
\end{answer}
\end{pro}


\end{multicols}


\chapter{Ratio and Proportion}
\begin{df}
Two fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are said to be {\em proportional} or {\em in proportion} if
$$ \dfrac{a}{b}  = \dfrac{c}{d}.$$
\end{df}
Proportions occur in many practical problems and our interest is to be able to provide a framework that helps us solve a variety
of proportion problems. For example, suppose a certain recipe demands three eggs for two pints of milk. It does not require much thought to deduce that
doubling the recipe requires six eggs for four pints of milk and that trebling the recipe would require nine eggs for six pints of milk. The problem is now to
figure out how many eggs are required if we are only given information about the milk, or viceversa, how much milk is required if we are only given information
about the eggs. To do this we use a procedure that used to be called the {\em rule of three}. Suppose that we would like to find out how many eggs are necessary
if we had $9$ pints of milk. The trick is to form products of fractions that have only eggs as numerator, cancelling all references to milk. The original proportion is
$$ \dfrac{3\ \mathrm{eggs}}{2\ \mathrm{pints}}.  $$In order to cancel the pints from the denominator we must multiply by pints. Since we are interested in how many eggs
per $9$ pints, we multiply by $9$ pints, obtaining,
$$ \dfrac{3\ \mathrm{eggs}}{2\ \cancel{\mathrm{pints}}}\cdot (9\ \cancel{\mathrm{pints}}) = \dfrac{27}{2} \ \mathrm{eggs}, $$which means we need
$13\tfrac{1}{2}$ eggs. Again, if we had available $20$ eggs, we would need
$$ \dfrac{2\ \mathrm{pints}}{3\ \cancel{\mathrm{eggs}}}\cdot (20\ \cancel{\mathrm{eggs}}) = \dfrac{40}{3} \ \mathrm{pints}, $$that is we would need
$13\tfrac{1}{3}$ pints of milk.

\bigskip

We will now give more examples.
\begin{exa}Yosida can eat  $18$ hotdogs in $4$ minutes. At this rate, how many hotdogs can he eat in $3$ minutes?
\end{exa}
\begin{solu}
The ``fundamental ratios'' here are
$$  \dfrac{18 \ \mathrm{hotdogs}}{4 \ \mathrm{minutes}} \qquad \mathrm{or}  \qquad  \dfrac{4 \ \mathrm{minutes}}{18 \ \mathrm{hotdogs}}.  $$
Since we are interested in the number of hotdogs, we use the one that has hotdogs on the numerator, and cancel the minutes on the denominator. Thus he can eat
$$  \dfrac{18 \ \mathrm{hotdogs}}{4 \ \cancel{\mathrm{minutes}}} \cdot (3\ \cancel{\mathrm{minutes}}) = \dfrac{54}{4} \ \mathrm{hotdogs} = \dfrac{27}{2}  \ \mathrm{hotdogs}, $$
that is, he can eat $13\tfrac{1}{2}$ hotdogs in $3$ minutes.
\end{solu}
\begin{exa}Under normal conditions, a car gives $27$ miles per gallon of gasoline. Under the same conditions, how many gallons of gasoline are required in order
to cover $900$ miles?
\end{exa}
\begin{solu}
The ``fundamental ratios'' here are
$$  \dfrac{27 \ \mathrm{miles}}{1 \ \mathrm{gallon}} \qquad \mathrm{or}  \qquad  \dfrac{1 \ \mathrm{gallon}}{27 \ \mathrm{miles}}.  $$
Since we are interested in the number of gallons, we use the ratio that has gallons on the numerator, and cancel the miles on the denominator.
Thus we need
$$  \dfrac{1 \ \mathrm{gallon}}{27 \ \cancel{\mathrm{miles}}} \cdot (900\ \cancel{\mathrm{miles}}) = \dfrac{900}{27} \ \mathrm{gallons} = \dfrac{100}{3}  \ \mathrm{gallons}, $$
that is, we need  $33\tfrac{1}{3}$ gallons of gasoline in order to ride  $900$ miles.
\end{solu}
The method we used for solving proportion problems can be used in unit conversion problems. Here is an example.
\begin{exa}
The current rate of exchange is $1$ dollar per $40$ Thai Bahts, and 1 euro for 65 Thai baht. How many dollars is $20$ euros?
\end{exa}
\begin{solu}
We want dollars on the numerator. We proceed as follows,
$$  \dfrac{1 \ \mathrm{dollar}}{40 \ \cancel{\mathrm{baht}}} \cdot  \dfrac{65 \ \cancel{\mathrm{baht}}}{1 \ \cancel{\mathrm{euro}}} \cdot (20 \ \cancel{\mathrm{euros}})
= \dfrac{1300}{40}\ \mathrm{dollars}= \dfrac{65}{2}\ \mathrm{dollars},$$
which is \$32.50.
\end{solu}
\begin{exa}
Milk, dark, and white chocolate miniature bars are mixed together in
a ratio of $5:4:2$ to form $275$ lbs of a {\em Chocoholic-Enticer}
mixture. How much of each type of bars was used for this mixture?
\end{exa}
\begin{solu}
The denominator is $5+4+2=11$. One wants $\dfrac{5}{11}\cdot
275=125$ lbs of milk chocolate, $\dfrac{4}{11}\cdot 275=100$ lbs of
dark chocolate, and $\dfrac{2}{11}\cdot 275=50$ lbs of white
chocolate.
\end{solu}


\begin{exa}
John takes $2$ hours to paint a room, whereas Bill takes $3$ hours
to paint the same room. How long would it take if both of them start
and are working simultaneously?
\end{exa}
\begin{solu}
In one hour John does $\dfrac{1}{2}$ of the job. In one hour Bill
does $\dfrac{1}{3}$. Thus in one hour they together accomplish
$\dfrac{1}{2}+\dfrac{1}{3} = \dfrac{5}{6}$ of the job. Thus in
$\dfrac{6}{5}$ of an hour they accomplish the job. Since
$\dfrac{6}{5}\cdot 60= 72$, it takes them $72$ minutes for them to
finish the job.
\end{solu}


\section*{Homework}  \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Rahnnya takes 20 minutes to solve 3 maths problems. At this rate,
how many minutes will it take her to solve 2880 problems?
\begin{answer}
She will take
$19200$ minutes.
\end{answer}
\end{pro}
\begin{pro}
Oscar rides his bike, being able to cover $6$ miles in $54$ minutes.
At that speed, how long does it take him to cover a mile?
\begin{answer}
$9$ minutes.
\end{answer}
\end{pro}
\begin{pro}
If there are $300$ calories in $100$ g of a certain type of food,
how many calories are there in a $30$ g portion of this food?
\begin{answer}
$90$ calories.
\end{answer}
\end{pro}
\begin{pro}
If the ratio of $7$ to $13$ is the same as the ratio of $x$ to $52$,
what is $x$?
\begin{answer} $28$\end{answer}
\end{pro}

\begin{pro}
Brazilian, Colombian, and Kenyan coffee beans are mixed together in
ratio $4:7:2$ to form $195$ lbs of a {\em Ya Betta Drink Me} blend.
How much of each type of beans was used for this mixture?
\begin{answer}
The denominator is $4+7+2=13$. One uses $\dfrac{4}{13}\cdot 195=60$
lbs of Brazilian,  $\dfrac{7}{13}\cdot 195=105$ lbs of Colombian,
and
 $\dfrac{2}{13}\cdot 195=30$
lbs of Kenyan.
\end{answer}
\end{pro}

\begin{pro}
On a certain map City A is $4$ inches apart from City B. If the
scale is such that $1\tfrac{1}{5}$ inches represent $30$ miles, find
the actual distance, in miles, between City A and City B.
\begin{answer}
We have $\left(\dfrac{30 \ \mathrm{miles}}{1\tfrac{1}{5} \
\mathrm{inches}}\right)(4 \ \mathrm{inches})= 100$. City A is $100$
miles apart from City B.
\end{answer}
\end{pro}
\begin{pro}
Nanette is going to make {\em cr\^{e}pes}. Her recipe requires $3$
fl oz of beer for every $2$ eggs. If she is going to utilise $8$
eggs, how many fluid ounces of beer does she need?
\begin{answer}
She will need $12$ fluid ounces of beer.
\end{answer}
\end{pro}

\begin{pro}
A car with five tyres (four road tyres and a spare tyre) travelled
$30, 000$ miles. If all five tyres were used equally, how many
miles' wear did each tyre receive?
\begin{answer}
Travelling $30,000$ miles with $4$ tyres is as travelling $120,000$
miles on one tyre. The average wear of each of the $5$ tyres is thus
$120000 \div 5 = 24000$ miles.
\end{answer}
\end{pro}

\begin{pro}
Andy and Brandy run in opposite directions on a circular track,
starting at diametrically opposite points. Each one runs at a
constant speed. They first meet after Andy has run $100$ meters.
They next meet after Brandy has run $150$ meters past their first
meeting point. What is the length of the track, in meters?
\begin{answer}
$350$ meters.
\end{answer}
\end{pro}

\begin{pro}
Anton, Banton, and Canton are mowing their lawns. Anton's lawn is
twice as large as Banton's, and thrice as large as Canton's.
Canton's lawn mower cuts half as fast as Banton's, and one third as
fast as Anton's. If they start mowing their lawns at the same time,
who will finish first.
\begin{answer}
Banton finishes first.  Anton and Canton take the same amount of
time.
\end{answer}
\end{pro}


\end{multicols}







\part{Decimals}
\chapter{Decimals}
We now consider a set of special fractions.
\begin{df}
A {\em decimal fraction} is a fraction whose denominator is a power
of $10$.
\end{df}
For example, all of the following are decimal fractions:
$$ \dfrac{3}{10}, \quad \dfrac{7}{1000}, \qquad \dfrac{213}{100}. $$
Decimal fractions are ubiquitous in calculator computations. Since
they are so common, we use a shortcut notation for them, which we
will call {\em decimal notation}. The idea is to separate a number
from its floor and use a period, called the {\em decimal point} to
separate the integral part from its fractional part. For example,
instead of writing
$$123\tfrac{789}{1000},  $$we write---in decimal notation---$123.789$.
Since
$$\dfrac{789}{1000}=\dfrac{700}{1000}+\dfrac{80}{1000}+\dfrac{9}{1000} = \dfrac{7}{10}+\dfrac{8}{100}+\dfrac{9}{1000},   $$
the convention we are making is that on the first place to the right
after the decimal point we put the tenths, on the second place we
put the hundredths, on the third place we put the thousandths.


\bigskip

Again, suppose we were given the fraction $$ \dfrac{7813}{100}. $$We
first compute
$$\floor{\dfrac{7813}{100}}=78.  $$ This means that
$$\dfrac{7813}{100}=\dfrac{7800}{100}+\dfrac{13}{100} = 78 + \dfrac{13}{100}.  $$ Hence, the
equivalent decimal notation is
$$ \dfrac{7813}{100} = 78+ \dfrac{13}{100}=78.13. $$
Again, we might interpret $\dfrac{7813}{100}$ as $7813\div 100 $.
Hence, another way of proceeding would be to perform the long
division
$$  \opdiv[displayintermediary=all]{7813}{100}$$



\bigskip

The procedure exhibited above allows us to convert any decimal
fraction to decimal notation, and viceversa. We will see in the
examples below that it also allows to convert certain other
fractions to decimals, and viceversa.
\begin{exa}
Write in decimal notation: $\dfrac{78}{100}$.
\end{exa}
\begin{solu}
The denominator is $100=10^2$, and so we use at most two decimal
places. We have
$$ \dfrac{78}{100} =0.78.$$
Another way of proceeding would be to observe that
$$ \dfrac{78}{100}  = 78\div 100 = \opdiv[displayintermediary=all]{78}{100}$$



Notice that we have written a $0$ as the floor of $\dfrac{78}{100}$.
It is really not necessary to do this and you will often see the
following written:
$$ \dfrac{78}{100} =.78,$$even by computer programs such as Maple.
We will stick to the former way of writing for easy reading.
\end{solu}
\begin{exa}
Write in decimal notation: $\dfrac{78}{10000}$.
\end{exa}
\begin{solu}
The denominator is $10000=10^4$, and so we use at most four decimal
places. We have
$$ \dfrac{78}{10000} =0.0078.$$Another way of proceeding would be to observe that
$$ \dfrac{78}{10000}  = 78\div 10000 = \opdiv[displayintermediary=all]{78}{10000}$$

\end{solu}
\begin{exa}
Write in decimal notation: $\dfrac{780}{10000}$.
\end{exa}
\begin{solu}
The denominator is $10000=10^4$, and so we use at most four decimal
places. We have
$$ \dfrac{780}{10000} =0.0780.$$Now, the last zero in $0.0780$ does
not contribute to the magnitude of $0.0780$ and thus we may drop it.
We will write then
$$ \dfrac{780}{10000} =0.078,$$and so adopt the convention that we
do not write trailing zeroes. Another way of proceeding would be to
observe that
$$ \dfrac{780}{10000}  = 780\div 10000 = \opdiv[displayintermediary=all]{780}{10000}$$

\end{solu}

So far we have converted decimal fractions to decimal notation. Can
we also convert non-decimal fractions? The answer is yes, but we
have some caveats. Since $10^n = 2^n5^n$, it turns out that the only
fractions whose denominators we can multiply by a natural number and
obtain a power of $10$ are those fractions whose denominators are
only divisible by $2$ or $5$. All other fractions cannot be
converted in this manner. But we can always rely on the fact that
$\dfrac{a}{b}=a\div b$, and hence use division in order to convert
fractions to decimal.


\begin{exa}\label{exa:3/5-into-decimal}
Write in decimal notation: $\dfrac{3}{5}$.
\end{exa}
\begin{solu}Here we are given a fraction whose denominator is not a
power of ten. Our first task is to write this fraction in an
equivalent form as a fraction whose denominator is a power of ten.
In this case we easily find
$$ \dfrac{3}{5} =\dfrac{6}{10}.$$Now, we convert  $\dfrac{6}{10}$ to a decimal, finding
$$ \dfrac{3}{5}=\dfrac{6}{10} =0.6.$$
Another way of proceeding would be to observe that
$$ \dfrac{3}{5}  = 3\div 5 = \opdiv[displayintermediary=all]{3}{5}$$

\end{solu}
\begin{rem}
In the above example we said that we need to convert the fraction to
an equivalent one whose denominator is a power of ten. Of course, we
know that there are infinitely many ways of doing that. Which one to
take, then? The beauty of the method is that it works regardless of
your choice. For example, you could have written
$$ \dfrac{3}{5}=\dfrac{600}{1000} =0.600=0.6,$$following our
convention of deleting trailing zeroes.

\end{rem}
\begin{exa}\label{exa:13/8-into-decimal}
Write in decimal notation: $\dfrac{13}{8}$.
\end{exa}
\begin{solu}
First observe that $8=2^3$. The smallest number we need to  multiply
$8$ by in order to get a power of ten is $5^3=125$ and then
$2^35^3=1000$. Thus
$$ \dfrac{13}{8} = \dfrac{13\cdot 125}{8\cdot 125} = \dfrac{1625}{1000} = 1.625. $$
Another way of proceeding would be to observe that
$$\dfrac{13}{8}\qquad =\qquad 13\div 8 \qquad \rightsquigarrow\qquad  8\, \begin{tabular}[b]{@{}r@{}}
1 \\ \hline \big)\begin{tabular}[t]{@{}l@{}}
13 \\
~8 \\ \divrule{0}{2}
~5 \\
\end{tabular}
\end{tabular}
\qquad \rightsquigarrow\qquad  8\, \begin{tabular}[b]{@{}r@{}} 1.625
\\ \hline \big)\begin{tabular}[t]{@{}l@{}}
13.000 \\
~8 \\ \divrule{0}{5}
~50 \\
~48\\ \divrule{1}{2}
~~20\\
~~16\\ \divrule{2}{2}
~~~40\\
~~~40\\ \divrule{3}{2} ~~~~0
\end{tabular}
\end{tabular}
$$
\end{solu}
\begin{exa}
Write in decimal notation: $\dfrac{7}{25}$.
\end{exa}
\begin{solu}
First observe that $25=5^2$. The smallest number we need to multiply
$25$ by in order to get a power of ten is $2^2=4$ and then
$2^25^2=100$. Thus
$$ \dfrac{7}{25} = \dfrac{7\cdot 4}{25\cdot 4} = \dfrac{28}{100} = 0.28. $$
Another way of proceeding would be to observe that
$$ \dfrac{7}{25}  = 7\div 25 = \opdiv[displayintermediary=all]{7}{25}$$
\end{solu}
\begin{exa}\label{exa:1/3-into-decimal}
Write in decimal notation: $\dfrac{1}{3}$.
\end{exa}
\begin{solu}
Since the denominator is neither divisible by $2$ nor $5$ only,
there is no way we can find an equivalent fraction with power of ten
on the denominator equivalent to this fraction. We proceed by long
division, Another way of proceeding would be to observe thatWe have,
$$\dfrac{1}{3}\qquad =\qquad 1\div 3 \qquad \rightsquigarrow\qquad  3\, \begin{tabular}[b]{@{}r@{}}
0 \\ \hline \big)\begin{tabular}[t]{@{}l@{}}
1 \\
0 \\ \divrule{0}{1}
1 \\
\end{tabular}
\end{tabular}
\qquad \rightsquigarrow\qquad   3\, \begin{tabular}[b]{@{}r@{}} 0.3
\\ \hline \big)\begin{tabular}[t]{@{}l@{}}
1.0 \\
0 \\ \divrule{0}{2}
10 \\
~9\\\divrule{1}{2} ~1
\end{tabular}
\end{tabular}
\qquad \rightsquigarrow\qquad   3\, \begin{tabular}[b]{@{}r@{}} 0.33
\\ \hline \big)\begin{tabular}[t]{@{}l@{}}
1.00 \\
0 \\ \divrule{0}{3}
10 \\
~9\\\divrule{1}{2}
~10\\
~~9\\\divrule{2}{2} ~~1
\end{tabular}
\end{tabular}
$$
It is not going to take long to arrive at the conclusion that the
remainder will always be $1$, meaning that there is no way to end
the division with a remainder of zero! The meaning of this is that
the decimal notation that we are looking for will \textbf{not} be
\textbf{ terminating}. In other words, the digits after the decimal
point will never stop. However, there is a pattern that repeats
infinitely (in this case the digit $3$). For that reason, we say
that the decimal expression is \textbf{repeating}. In order to
indicate the pattern, we will use a horizontal bar on top of it. So,
in this case we write, $\dfrac{1}{3}=0.\overline{3}$. As always, the division may also be displayed as
$$ \opdiv[displayintermediary=all]{1}{3}$$
\end{solu}
\begin{rem}
Examples \ref{exa:3/5-into-decimal} and \ref{exa:13/8-into-decimal}
shew two fractions having terminating expansions. Example
\ref{exa:1/3-into-decimal} shews an example of a decimal with a
repeating expansion. It is possible (but somewhat hard to prove),
that a fraction in reduced form with denominator of the form
$2^m5^n$ where $m, n$ are natural numbers, will always terminate,
while a fraction in reduced form whose denominator does not have
this form will always repeat.
\end{rem}
\begin{exa}
Convert ${\dfrac{15}{16}}$ to decimal.
\end{exa}
\begin{solu}
Since the denominator is $16=2^4$, a power of $2$, the decimal will
terminate. In fact, we have
$$ \dfrac{15}{16}= \dfrac{15\cdot 5^4}{16\cdot 5^4} = \dfrac{15\cdot 625}{16\cdot 625} = \dfrac{9375}{10000} =0.9375.$$
Or, performing the long division,
$$16\, \begin{tabular}[b]{@{}r@{}}
0.9375 \\ \hline \big)\begin{tabular}[t]{@{}l@{}}
15.0000 \\
0 \\ \divrule{0}{6}
150 \\
144\\ \divrule{0}{3}
~~60\\
~~48\\ \divrule{2}{2}
~~120\\
~~112\\ \divrule{2}{3}
~~~80\\
~~~80\\ \divrule{3}{2} ~~~~0
\end{tabular}
\end{tabular}
$$
\end{solu}
\begin{exa}Convert ${\dfrac{101}{165}}$ into a decimal.\end{exa}
\begin{solu}
Again, since the denominator $165$ is not of the form $2^m5^n$, the
decimal will  repeat. Performing the long division,
$$ 165\, \begin{tabular}[b]{@{}r@{}}
0.61212 \\ \hline \big)\begin{tabular}[t]{@{}l@{}}
101.00000 \\
0 \\ \divrule{0}{8}
1010\\
~990\\ \divrule{0}{4}
~~200\\
~~165\\ \divrule{2}{3}
~~~350\\
~~~330\\ \divrule{3}{3}
~~~~200\\
~~~~165\\ \divrule{4}{3}
~~~~~350\\
~~~~~330\\ \divrule{5}{3}
~~~~~~20\\

\end{tabular}
\end{tabular} $$
Since we notice that the remainder will be alternatively $20$ and
$35$, we conclude that this decimal notation is not terminating, it
is
 repeating. The pattern that repeats is $12$. This means that
$\dfrac{101}{165}=0.6\overline{12}.$
\end{solu}

The procedure of converting a decimal to a fraction is even easier.
\begin{exa}
Convert to a fraction and write the result in lowest terms: $0.125$.
\end{exa}
\begin{solu}
Notice that there are three decimal places. Hence we start with a
denominator of $10^3=1000$. We have
$$0.125 = \dfrac{125}{1000} = \dfrac{1}{8}.  $$
\end{solu}
\begin{exa}
Convert to a fraction and write the result in lowest terms:
$0.0028$.
\end{exa}
\begin{solu}
Notice that there are four decimal places. Hence we start with a
denominator of $10^4=10000$. We have
$$0.0028 = \dfrac{28}{10000} = \dfrac{7}{2500}.  $$
\end{solu}
\begin{exa}
Convert to a fraction and write the result in lowest terms: $3.24$.
\end{exa}
\begin{solu}
Notice that there are two decimal places. Hence we start with a
denominator of $10^2=100$. We have
$$3.24 = \dfrac{324}{100} = \dfrac{324}{100} = 3.24.  $$
\end{solu}


A special kind of decimal fraction is the {\em percent}. Basically,
the notation for percent \% means $\div 100$. Thus $$ 11\% =
\dfrac{11}{100} = 0.11,
$$for example.

\begin{exa}
Convert the percent to a decimal: $55\%$.
\end{exa}
\begin{solu}
We have
$$ 55\% = \dfrac{55}{100} = 0.55. $$
\end{solu}
\begin{exa}
Convert the decimal to a percent: $1.2$.
\end{exa}
\begin{solu}
We have
$$ 1.2 = 1+\dfrac{2}{10} = \dfrac{100}{100} + \dfrac{20}{100} = \dfrac{120}{100} = 120\%. $$
\end{solu}

\begin{exa}
From a class of $36$, $6$ students are absent. Write the percent of
students absent in a mixed fraction/percent form.
\end{exa}
\begin{solu}
This is $\dfrac{6}{36}=\dfrac{1}{6} = \dfrac{100}{6}\cdot
\dfrac{1}{100}=16\tfrac{2}{3}\%$
\end{solu}











We now tackle the problem of comparing decimals. We compare positive
decimals {\em lexicographically}, that is, in much the same manner
that we look up words in a dictionary. Let $a_1, a_2, a_3, \ldots$
and $b_1, b_2, b_3, \ldots$ be digits and let $P_1$ and $P_2$ be
natural numbers. To compare the numbers
$$P_1.a_1a_2a_3\ldots, \qquad P_2.b_1b_2b_3\ldots,  $$ we first compare $P_1$ and $P_2$. If $P_1>P_2$, then
$$P_1.a_1a_2a_3\ldots > P_2.b_1b_2b_3\ldots.  $$If $P_1=P_2$, we now compare $a_1$ and $b_1$. If $a_1>b_1$, then
$$P_1.a_1a_2a_3\ldots > P_2.b_1b_2b_3\ldots.  $$If $P_1=P_2$ and $a_1=b_1$, we now compare $a_2$ and $b_2$. If $a_2>b_2$, then
$$P_1.a_1a_2a_3\ldots > P_2.b_1b_2b_3\ldots.  $$If $P_1=P_2$, $a_1=b_1$, and $a_2=b_2$, we now compare $a_3$ and $b_3$. If $a_3>b_3$, then
$$P_1.a_1a_2a_3\ldots > P_2.b_1b_2b_3\ldots,  $$and so on. Thus the procedure
compares digits from left to right until we find the first spot
where the digits differ. \begin{exa}$5.1>2.99998877$ since from
their integral parts, $5>2$.\end{exa} \begin{exa}$5.222>5.199999$.
Their integral parts are both $5$, so now we compare the first
decimal place. Since $2>1$, we obtain the conclusion.\end{exa}
\begin{exa}$5.212>5.209$. Their integer parts are both $5$, and
their first decimal place are both $2$, so now we compare the second
decimal place. Since $1>0$, we obtain the conclusion.\end{exa}
\begin{exa}$5.02=5.02000$. This is so because $5.02=5.02+0.00000 =
5.02000$.\end{exa} \begin{rem}The preceding example shows that the
{\em trailing zeroes} after the right-most non-zero digit after the
decimal point are immaterial.\end{rem} \begin{exa}$0.5 <
0.\overline{5}$.\end{exa}
\begin{exa}$0.\overline{5}<0.56$.\end{exa}
\begin{exa}Since $\dfrac{1}{3} = 0.33333\ldots = 0.\overline{3}$, it
is verified that $0.3 < \dfrac{1}{3}<0.34$.\end{exa}


\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Convert into a fraction and express in lowest terms: $0.204$
\begin{answer}
$\dfrac{51}{250}$
\end{answer}
\end{pro}
\begin{pro}
Convert into a decimal:\quad  $\dfrac{9}{11}$.
\begin{answer}
$0.\overline{81}$
\end{answer}
\end{pro}
\begin{pro}
Convert to a decimal: $\dfrac{3}{20}$
\begin{answer}
$ 0.15$
\end{answer}
\end{pro}
\begin{pro}
Convert to a decimal: $\dfrac{3}{200}$
\begin{answer}
$ 0.015$
\end{answer}
\end{pro}
\begin{pro}
Convert to a decimal: $\dfrac{7}{15}$
\begin{answer}
$ 0.4\overline{6} $
\end{answer}
\end{pro}
\begin{pro}
Convert to a decimal: $\dfrac{1}{7}$
\begin{answer}
$ 0.\overline{142857} $
\end{answer}
\end{pro}
\begin{pro}
Convert to a fraction: $24.24$
\begin{answer}
$ 24\tfrac{6}{25} $
\end{answer}
\end{pro}
\begin{pro}
Convert to a percent: $0.12$
\begin{answer}
$ 12\% $
\end{answer}
\end{pro}
\begin{pro}
Convert to a percent: $0.012$
\begin{answer}
$ 1.2\% $
\end{answer}
\end{pro}
\begin{pro}
Arrange in increasing order: $$ 0.45,\quad  0.445,\quad
0.4\overline{4}, \quad  0.4\overline{45}, \quad 0.446.
$$
\begin{answer}
$0.4\overline{4}<0.445<0.4\overline{45}<0.446 <0.45  $
\end{answer}
\end{pro}
\begin{pro}
Arrange in increasing order: $$ \dfrac{2}{3},  \quad \dfrac{3}{4},
\quad 0.67, \qquad 0.666.
$$
\begin{answer}
$ 0.666< \dfrac{2}{3}< 0.67<\dfrac{3}{4}$
\end{answer}
\end{pro}
\begin{pro}
In the infinite repeating decimal $$ 0.567895678956789\ldots =
0.\overline{56789},
$$what is the 2008th digit after the decimal point?
\begin{answer} $7$ \end{answer}
\end{pro}

\begin{pro}
Add and express your result in fraction form: $0.2+ \dfrac{1}{4}$
\begin{answer}
$ 0.2+\dfrac{1}{4} =\dfrac{2}{10}+\dfrac{1}{4} =
\dfrac{4}{20}+\dfrac{5}{20}=\dfrac{9}{20}$
\end{answer}
\end{pro}
\begin{pro}
Multiply and express your result in fraction form. If your answer is
an improper fraction, leave the result in this form:
$\left(\dfrac{2}{5}\right)(3.3)$
\begin{answer}
$ \left(\dfrac{2}{5}\right)(3.3)=
\left(\dfrac{2}{5}\right)\left(3\dfrac{3}{10}\right) =
\left(\dfrac{2}{5}\right)\left(\dfrac{33}{10}\right)=\dfrac{66}{50}=\dfrac{33}{25}
$
\end{answer}
\end{pro}
\begin{pro}
You start with $\$100$. You give $20\%$ to your friend. But it turns
out that you need the $\$100$ after all in order to pay a debt. By
what percent  should you increase your current amount in order to
restore the $\$100$? The answer is {\em not}  $20\%$!
\begin{answer}
You need to increase $\$20$, and $\$20$ is $25\%$ of $\$80$, and so
you need to increase by $25\%$.
\end{answer}
\end{pro}
\end{multicols}

\chapter{Addition, Subtraction, and Multiplication}
The addition and subtraction algorithms explained in the lesson on
integers extend to addition and subtraction of decimals. That is, in
order to add or subtract decimals, we line the numbers at the
decimal point and proceed as if they were integers.
\begin{exa} Add:
$123.456789 + 98.7654321$.
\end{exa}
\begin{solu}
We line up the numbers at the decimal point and add:
$$ \opadd{123.456789}{98.7654321} $$
\end{solu}
\begin{exa} Subtract:
$123.456789 - 98.7654321$.
\end{exa}
\begin{solu}
We line up the numbers at the decimal point and subtract:
$$ \opsub{123.456789}{98.7654321} $$
Notice that we added a zero to the end of $123.456789$ in order to
line up with all the decimals of $98.7654321$.
\end{solu}
Multiplication of decimals is only slightly more complicated.
Suppose we wanted to perform the product $1.1\cdot 1.23$. Since
$$ 1.1\cdot 1.23=\dfrac{11}{10} \cdot \dfrac{123}{100} = \dfrac{1353}{1000} =1.353,$$
that is, we simply multiplied the numbers, with no regard of the
decimal point, and then, counting from the rightmost digit of the
product, we moved the decimal point to the left as many places as
there appear in the factors, in our case $1+2=3$ places.
\begin{exa}
Multiply: $12.121 \cdot 34.34$.
\end{exa}
\begin{solu}There are $3+2=5$ decimal places in the factors. We
multiply as we would ordinary integers and then from the rightmost
digit of the product we move the decimal point five units to the
left. We have
$$\opmul{12.121}{34.34}
$$
\end{solu}

We now present some examples involving percents.

\begin{exa}
Find $24\%$ of $80$.
\end{exa}
\begin{solu}
Essentially, we interpret ``of'' as multiplication:
$$ (24\%)(80) = (0.24)(80)=19.2. $$
\end{solu}




\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

\begin{pro}
How much would it cost to fill a $12$-gallon tank of gasoline at
\$3.18 per gallon?
\begin{answer}
$\$38.16$
\end{answer}
\end{pro}

\begin{pro}
Compute: $11\% + 2.1$.
\begin{answer}
$2.21$
\end{answer}

\end{pro}
\begin{pro}
Compute: $\left(\dfrac{3}{4}\right)(1.24) $.
\begin{answer}
$0.93 $
\end{answer}

\end{pro}
\begin{pro}
Compute: $(2.1\%)(0.12\%) $.
\begin{answer}
$0.0000252 $
\end{answer}

\end{pro}
\begin{pro}
Compute: $(0.1)(0.02)(0.003)(0.0004)$.
\begin{answer}
$0.0000000024$
\end{answer}

\end{pro}
\begin{pro}
Compute: $0.1^2+0.3\cdot 0.4$.
\begin{answer}
$0.13$
\end{answer}

\end{pro}

\begin{pro}
You have $\$8.00$. You buy $2$ oranges and $3$ juices. Each orange
costs $\$0.35$ and each juice costs $\$0.90$. How much do
you have left?
\begin{answer}
$\$4.60$
\end{answer}
\end{pro}
\begin{pro}
 Find $30\%$ of $810$.
 \begin{answer}
$243$
 \end{answer}
\end{pro}

\begin{pro}
$840$ is $28\%$ of what number?
\begin{answer}
$3000$
\end{answer}
\end{pro}

\begin{pro}
What percent of $408$ is $34$?
\begin{answer}
$\dfrac{25}{3}\%$
\end{answer}
\end{pro}

\begin{pro}
Hillary trades goats for a $40\%$ profit. If the original price of a
goat is $\$90$, find the new selling price.
\begin{answer}
\$126.
\end{answer}
\end{pro}
\begin{pro}
After a night of beer and pizza, the bill comes to a total of
\$34.20, including sales tax. If you are leaving a 15\% tip, how
much will you end up paying?
\begin{answer}
 \$39.33.
\end{answer}
\end{pro}
\begin{pro}
The area of a living room in a one-bedroom apartment is 60\% of the
total area (living-room, bedroom, and kitchen). The area of a
bedroom is 40\% of the living-room area. Find the area of a kitchen,
if the living-room area is 468 sq. feet.
\begin{answer}
The area  of the apartment  is $468\div 0.6=780$ sq feet. The area
of the bedroom is $0.4\cdot 468=187.2$ sq feet. The area of the
kitchen is thus $780-468-187.2=124.8$ sq feet.
\end{answer}
\end{pro}
\begin{pro}
The mass of $200$ kg cucumbers consists of $99$\% of water. The
cucumbers are drying out due to the sun, till the mass consists of
$98$\% of water. Determine the weight of the cucumbers now.
\end{pro}
\begin{pro}
 At a certain college $99\%$ of the $100$ students are female,
but only $98\%$ of the students living on campus are female. If some
females live on campus, how many students live off campus?
\begin{answer} Of the $100$ students, only one is male. He is $2\%$ of the
on-campus population. Thus the whole on-campus population consists
of $50$ students, so there are $100-50=50$ off-campus students.
\end{answer}
\end{pro}
\begin{pro}
Abdullah had $200$ camels. Eighty died, and all but $25\%$ of those remaining ran away. How many were left?
\begin{answer}
$30$
\end{answer}
\end{pro}
\begin{pro}
You mix in a recipient $3$ litres of water and $1$ litre of juice. The juice is composed of $20\%$ pulp and $80\%$ water.
In this new $4$-litre mixture, what percent is water? What percent is pulp?
\begin{answer}
$95\%$ water, $5\%$ pulp.
\end{answer}
\end{pro}

\begin{pro}
What would be the price of a $5\frac{1}{2}$-mile trip with the
following taxi-cab company? \\ \vspace*{2mm}
\begin{flushright}\begin{tabular}{|lcr|}
\hline
& {\sc Smiling Camel Taxi Services}  & \\
First $\frac{1}{4}$ mi & & \$ .85 \\
Additional $\frac{1}{4}$ mi & &  \$ .40 \\ \hline \end{tabular}
\end{flushright}
\begin{answer}
Observe that $5\tfrac{1}{2}=\dfrac{11}{2}=\dfrac{22}{4}.$ Hence
$\dfrac{21}{4}$ miles are at the rate of $40$\textcent. The trip
costs $\$0.85 + \$.40\cdot 21 = \$.85+\$8.40=\$9.25$.
\end{answer}
\end{pro}
\begin{pro}If $a$ is 50 \% larger than $c$, and $b$ is 25 \% larger
than $c$, then $a$ is what percent larger than $b$? \begin{answer}
One has $a = 1.5c$ and $b = 1.25c$ Thus $a = \dis{\frac{1.5b}{1.25}
= 1.2b}$. Therefore $a$ is $20\%$ larger than $b$.
\end{answer}
\end{pro}
\begin{pro}
Consider the two infinite repeating decimals
$$a = 0.\overline{12345} = 0.1234512345\ldots ; \qquad b = 0.\overline{98765} = 0.9876598765\ldots .  $$
If the sum $a+b$   is written as an improper fraction in lowest
terms $\dfrac{p}{q}$, find $p^2+q^2$.
\end{pro}

\end{multicols}
\chapter{Division}We now consider division of decimals. Let us start with the case where the divisor is a natural number. For example:
$$3.99\div3=\dfrac{399}{100}\div \dfrac{3}{1}=
\dfrac{399}{100} \cdot
\dfrac{1}{3}=\dfrac{399}{100 \cdot 3}=\dfrac{3 \cdot 133}{100 \cdot 3}=\dfrac{133}{100}=1.33.$$

In many cases, however, the numbers may not be ``nice'' enough to be
handled in the manner of the preceding example. We would like to
know, therefore, a version of the the long division algorithm
allowing us to perform more intricate calculations.

The first thing to do is to divide the natural number before the
decimal point by the divisor (in the usual way). This basically
means that in the first step of the division we can not go beyond
the point. We go on with the usual process. When bringing down the
first digit after the decimal point, before dividing, we need to put
a decimal point in the quotient and then proceed with the division.

Keep in mind that if necessary, we can always add zeros at the end
of the decimal number (it would not change its value), so we can go
on with the division until either the remainder is zero, or we
determine that the quotient will not be terminating but a repeating
decimal. \begin{exa}Shew that \opdiv[style=text]{1.21}{2} by means
of the long division algorithm.\end{exa}
\begin{solu} Notice the sequence of divisions:
$$2\, \begin{tabular}[b]{@{}l@{}}
\ 0 \\ \hline \big)\begin{tabular}[t]{@{}l@{}}
1.21 \\
0 \\ \divrule{0}{3} 1
\end{tabular}
\end{tabular}  \qquad  \rightsquigarrow \qquad
2\, \begin{tabular}[b]{@{}l@{}} \ 0.6 \\ \hline
\big)\begin{tabular}[t]{@{}l@{}}
1.21 \\
0 \\ \divrule{0}{3}
12\\
12 \\ \divrule{0}{2}
~0 \\
\end{tabular}
\end{tabular}  \qquad  \rightsquigarrow \qquad  2\, \begin{tabular}[b]{@{}r@{}}
 0.60 \\ \hline
\big)\begin{tabular}[t]{@{}l@{}}
1.21 \\
0 \\ \divrule{0}{3}
12\\
12 \\ \divrule{0}{2}
~01\\
~~0\\ \divrule{2}{2} ~~1
\end{tabular}
\end{tabular}  \qquad  \rightsquigarrow \qquad
 2\, \begin{tabular}[b]{@{}r@{}}
 0.605 \\ \hline
\big)\begin{tabular}[t]{@{}l@{}}
1.210 \\
0 \\ \divrule{0}{3}
12\\
12 \\ \divrule{0}{2}
~01\\
~~0\\ \divrule{2}{2}
~~10\\
~~10\\ \divrule{2}{2} ~~~0
\end{tabular}
\end{tabular}      $$

We explain the steps as follows.
\begin{itemize}
\item We divide the natural number before the decimal point by the
divisor.
\item Now we bring down the first digit after the decimal point ($2$),
put a point at the quotient and then divide.
\item Continue with the division bringing down the next digit ($1$).
\item
Since the remainder is not zero and there are no more digits to
bring down, we may add a zero at the end of the dividend and
continue with the division.

\end{itemize}
\end{solu}
We will now provide more examples, but without narrating the
explanation. \begin{exa}Prove that
\opdiv[style=text,period,hrulewidth=0.2pt,
vruleperiod=0.7,afterperiodsymbol=]{24.382}{12}
\end{exa}
\begin{solu}We perform the long division
$$12\, \begin{tabular}[b]{@{}r@{}}
 2.031833 \\ \hline
\big)\begin{tabular}[t]{@{}l@{}}
24.382000 \\
24 \\ \divrule{0}{8}
~03\\
~~0 \\ \divrule{1}{2}
~~38\\
~~36\\ \divrule{2}{2}
~~~22\\
~~~12\\ \divrule{3}{2}
~~~100\\
~~~~96 \\ \divrule{4}{3}
~~~~~40 \\
~~~~~36 \\ \divrule{5}{2}
~~~~~~40 \\
~~~~~~36 \\ \divrule{6}{2} ~~~~~~~4
\end{tabular}
\end{tabular}    $$
We can see that from this point forward, the remainder will always
be $4$. That means that the decimal expression of the quotient is
not terminating and repeating. In this case the pattern that repeats
is the digit $3$. \end{solu} \begin{exa}Prove that
\opdiv[style=text,period,hrulewidth=0.2pt,
vruleperiod=0.7,afterperiodsymbol=]{135.36}{3}
\end{exa}

\begin{solu}We perform the long division
$$3\, \begin{tabular}[b]{@{}r@{}}
 45.12\\ \hline
\big)\begin{tabular}[t]{@{}l@{}}
135.36 \\
12 \\ \divrule{0}{5}
~15\\
~15 \\ \divrule{1}{2}
~~03\\
~~~3\\ \divrule{2}{2}
~~~06\\
~~~~6\\ \divrule{3}{2}
~~~~0\\
\end{tabular}
\end{tabular}    $$
\end{solu}






Let us combine a few operations with fractions and decimals.

\begin{exa}Perform the operation and write the result in decimal
form:
$$ \dfrac{3}{4}+1.2\% + 1.2. $$
\end{exa}
\begin{solu}
We have
$$ \dfrac{3}{4}+1.2\% + 1.2=0.75+0.012+1.2=1.962. $$

\end{solu}

\begin{exa}
Write as a fraction: $\dfrac{1}{0.1+\dfrac{1}{0.2}}$.
\end{exa}
\begin{solu}We have
$$\begin{array}{lll}\dfrac{1}{0.1+\dfrac{1}{0.2}} & = &
\dfrac{1}{0.1+5}\\
 & = & \dfrac{1}{\tfrac{1}{10}+5}\\
& = & \dfrac{1}{\dfrac{51}{10}}\\
& = & \dfrac{10}{51}.
\end{array}$$
\end{solu}


Here are some percent problems.

\begin{exa}What percent of $72$ is equal to $45$?\end{exa}
\begin{solu}
We know that the multiplication of the decimal corresponding to
the percent times $72$ should be equal to $45$. But this is the
same as saying that $45$ divided by $72$ is equal to the decimal
corresponding to the percent. Since $$45\div72=0.625,$$
we  conclude that $62.5$\% of $72$ is $45$.
\end{solu}
\begin{exa}
$45$ is $75\%$ of which number?
\end{exa}
\begin{solu}
If we multiply $0.75$ by the given number then we obtain $45$. Hence, the given number is
$$ 45\div 0.75 = 4500\div 75 = 60. $$
\end{solu}

\begin{exa}If a store offers a $35$\% discount on a item, and the price
after the discount is \$52, what was the original price of the
item?\end{exa}
\begin{solu}
First we need to understand that if $35$\% was discounted, then
the final price is actually $65$\% of the original price. Therefore, we
know that $65$\% of the original price is equal to
$52$. This means that $0.65$ times the original price will be
equal to $52$. Then the original price is the result of dividing
$52$ by $0.65$:
$$52\div0.65=5200\div65=80,$$
from where it follows that  the original price of the item was \$80.
\end{solu}
\begin{exa}
A company reports that  $23\%$ of its employees are smokers. If
there are $92$ smokers in the company, how many employees in total
work there?
\end{exa}
\begin{solu}
If $x$ is the number of employees, we want to solve the proportion
$$  \dfrac{23x}{100}=92 \implies x = 92\cdot \dfrac{100}{23}=400,$$
and so $400$ employees work at the company.
\end{solu}





\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Calculate:\quad $1020.4016\div 637.751$.
\begin{answer}
$1.6$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $1.23\div 0.006$.
\begin{answer}
$205$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $1.023\div 0.15$.
\begin{answer}
$6.82$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $1.023\div 0.9$.
\begin{answer}
$1.13\overline{6}$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $8\div 7$.
\begin{answer}
$1.\overline{142857}$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $12\div 11$.
\begin{answer}
$1.\overline{09}$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $\dfrac{1-0.1}{2-0.2}$.
\begin{answer}
$0.5$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $\dfrac{1-0.01}{2-0.02}$.
\begin{answer}
$0.5$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $\dfrac{123}{12.3-0.123}$.
\begin{answer}
$10.\overline{10}$
\end{answer}
\end{pro}
\begin{pro}
Convert to a fraction:\quad
$\dfrac{1}{0.1+\dfrac{1}{0.2+\dfrac{1}{0.3}}}$.
\begin{answer}
$\dfrac{530}{203}$
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad $\dfrac{0.1+0.3+0.5+0.7+0.9}{0.2+0.4+0.6+0.8}$.
\begin{answer}
$1.25$
\end{answer}
\end{pro}

\begin{pro}
Jill bought $6$ pounds of apples for $\$1.38$. How much did each pound
cost?
\begin{answer}
$23$\textcent
\end{answer}
\end{pro}
\begin{pro}
You bought some furniture for \$424.53, price which included a 6\%
sales tax. What was the price of piece, before sales tax?
\begin{answer}
\$400.50
\end{answer}
\end{pro}


\begin{pro}
The total weight of a pile of $500$ of  salt crystals is $6.5$ g.
What is the average weight of a salt crystal?
\begin{answer}
$\dfrac{6.5}{500}$ grams, that is, $0.013$ grams.
\end{answer}
\end{pro}


\begin{pro}
The only buttons that work on a calculator are\psframebox{$4$},
\psframebox{$\div$}, \psframebox{$ \times $}, \psframebox{$+$},
\psframebox{$-$}, \psframebox{$($}, \psframebox{$)$},
\psframebox{!}, \psframebox{.}, and \psframebox{$=$}. Produce every
natural number from $1$ to $20$ on the screen of this calculator
with the extra constraint that you must press the \psframebox{$4$}
{\bf exactly} four times! For example, to get $5$  on the screen,
you could type
$$ \psframebox{(}\psframebox{4} \psframebox{ \times } \psframebox{4}\psframebox{+} \psframebox{4} \psframebox{)} \psframebox{\div}  \psframebox{4} \psframebox{=}. $$
\begin{answer}
There is more than one possible answer in some cases.

\begin{center}

\begin{tabular}{|l|||l|}
\hline
$1$ is  \psframebox{$4$}\psframebox{$4$} \psframebox{$\div$} \psframebox{$4$}\psframebox{$4$} \psframebox{$=$} & $11$ is  \psframebox{$4$} \psframebox{$\div$} \psframebox{.}  \psframebox{$4$} \psframebox{$+$} \psframebox{$4$} \psframebox{$\div$} \psframebox{$4$}   \psframebox{$=$}\\
$2$ is  \psframebox{$4$}\psframebox{$\div$} \psframebox{$4$} \psframebox{$+$}  \psframebox{$4$} \psframebox{$\div$} \psframebox{$4$} \psframebox{$=$}  & $12$ is \psframebox{$($} \psframebox{$4$} \psframebox{$4$} \psframebox{$+$} \psframebox{$4$}  \psframebox{$)$} \psframebox{$\div$} \psframebox{$4$} \psframebox{$=$}\\

$3$ is  \psframebox{$($}\psframebox{$4$} \psframebox{$+$} \psframebox{$4$}  \psframebox{$+$} \psframebox{$4$} \psframebox{$)$} \psframebox{$\div$} \psframebox{$4$}\psframebox{$=$} & $13$ is \psframebox{$4$} \psframebox{$!$} \psframebox{$-$} \psframebox{$4$} \psframebox{$4$}  \psframebox{$\div$} \psframebox{$4$} \psframebox{$=$}\\
$4$ is  \psframebox{$4$}\psframebox{$+$} \psframebox{$4$} \psframebox{$ \times $}  \psframebox{$($} \psframebox{$4$} \psframebox{$-$} \psframebox{$4$} \psframebox{$)$} \psframebox{$=$} & $14$ is \psframebox{$4$} \psframebox{$ \times $} \psframebox{$($} \psframebox{$4$} \psframebox{$-$}  \psframebox{$.$} \psframebox{$4$} \psframebox{$)$}\psframebox{$-$}  \psframebox{$.$} \psframebox{$4$} \psframebox{$=$} \\

$5$ is \psframebox{$($}\psframebox{$4$} \psframebox{$ \times $} \psframebox{$4$}\psframebox{$+$} \psframebox{$4$} \psframebox{$)$} \psframebox{$\div$}  \psframebox{$4$} \psframebox{$=$} &  $15$ is \psframebox{$4$}\psframebox{$4$} \psframebox{$\div$} \psframebox{$4$}\psframebox{$+$} \psframebox{$4$} \psframebox{$=$}  \\
$6$ is \psframebox{$($}\psframebox{$4$} \psframebox{$+$}\psframebox{$4$} \psframebox{$)$}\psframebox{$\div$} \psframebox{$4$} \psframebox{$+$}   \psframebox{$4$} \psframebox{$=$} &  $16$ is \psframebox{$4$}\psframebox{$ \times $} \psframebox{$4$} \psframebox{$+$}\psframebox{$4$} \psframebox{$-$}\psframebox{$4$} \psframebox{$=$}  \\
$7$ is \psframebox{$4$}\psframebox{$4$} \psframebox{$\div$}\psframebox{$4$} \psframebox{$-$}\psframebox{$4$}  \psframebox{$=$} &  $17$ is \psframebox{$4$}\psframebox{$ \times $} \psframebox{$4$} \psframebox{$+$}\psframebox{$4$} \psframebox{$\div$}\psframebox{$4$} \psframebox{$=$}  \\
$8$ is \psframebox{$4$}\psframebox{$ \times $} \psframebox{$4$}\psframebox{$-$} \psframebox{$4$}\psframebox{$-$} \psframebox{$4$}  \psframebox{$=$} &  $18$ is \psframebox{$4$}\psframebox{$\div $} \psframebox{$.$} \psframebox{$4$}\psframebox{$+$} \psframebox{$4$}\psframebox{$+$}\psframebox{$4$} \psframebox{$=$}  \\
$9$ is \psframebox{$4$}\psframebox{$\div$} \psframebox{$.$}\psframebox{$4$}\psframebox{$-$} \psframebox{$4$}\psframebox{$\div$} \psframebox{$4$}  \psframebox{$=$} &  $19$ is \psframebox{$4$}\psframebox{$!$} \psframebox{$-$} \psframebox{$4$}\psframebox{$-$} \psframebox{$4$}\psframebox{$\div$}\psframebox{$4$} \psframebox{$=$}  \\
$10$ is \psframebox{$($}\psframebox{$4$} \psframebox{$4$}\psframebox{$-$}\psframebox{$4$} \psframebox{$)$}\psframebox{$\div$} \psframebox{$4$}  \psframebox{$=$} &  $20$ is \psframebox{$4$}\psframebox{$\div $} \psframebox{$.$} \psframebox{$4$}\psframebox{$+$} \psframebox{$4$}\psframebox{$\div$}\psframebox{$.$}\psframebox{$4$} \psframebox{$=$}  \\
\hline
\end{tabular}
\end{center}
\end{answer}
\end{pro}
\end{multicols}


\part{Real Numbers}

\chapter{Integers: Addition and Subtraction}
\QUOTEME { God created the integers. Everything else is the work of
Man.}{\mbox{Leopold KROENECKER}}
\bigskip
\dropping{2}{T}he introduction of fractions in previous chapters
helped solve the problem that the natural numbers are not closed
under division. We now solve the problem that the natural numbers
are not closed under subtraction.
\begin{df}
A natural number not equal to $0$ is said to be {\em positive}. The
set
$$ \{1,2,3,4,5,\ldots\} $$ is called the set of {\em positive
integers}.
\end{df}
\begin{df}
Given a natural number $n$, we define its opposite $-n$ as the
unique number $-n$ such that
$$n + (-n)=(-n)+n=0.  $$
\begin{rem}
Observe that $-0=0$  is the only integer that equals its opposite.
\end{rem}



The collection
$$ \{-1,-2,-3,-4,-5,\ldots\}  $$ of all the opposites of the natural
numbers is called the set of {\em negative integers}. The collection
of natural numbers together with the negative integers is the set of
{\em integers}, which we denote by the symbol\footnote{From the
German word for number: {\em Zählen}.}
 $\integers$.
\end{df}
A graphical representation of the integers is given in figure
\ref{fig:integers}.



 \vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering\psaxes[yAxis=false,labelFontSize=\tiny,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0) \meinecaption{.5}{The
Integers $\integers$.} \label{fig:integers}
\end{figure}








\begin{exa}\label{exa:plottingpoints}If each of the markings in figure
\ref{fig:plottingpoints} represents a unit, then we plot and label
the numbers $-5$, $-1$, and $4$. \end{exa}


\vspace{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)
\uput[d](0,0){$0$}\uput[d](-5,0){$-5$}\uput[d](-1,0){$-1$}
\uput[d](4,0){$4$}\psdots[dotscale=1.5,dotstyle=*,linecolor=red](-5,0)\psdots[dotscale=1.5,dotstyle=*,linecolor=blue](-1,0)
\psdots[dotscale=1.5,dotstyle=*,linecolor=green](4,0)
\meinecaption{.5}{Example \ref{exa:plottingpoints}.}
\label{fig:plottingpoints}
\end{figure}



There seems to be no evidence of usage of negative numbers by the
Babylonians, Pharaonic Egyptians, or the ancient Greeks. It seems
that the earliest usage of them came from China and India. In the
7th Century, negative numbers were used for bookkeeping in India.
The Hindu astronomer Brahmagupta, writing around A.D. 630, shews a
clear understanding of the usage of negative numbers.

\bigskip

Thus it took humans a few millennia to develop the idea of negative
numbers. Since, perhaps, our lives are more complex now, it is not
so difficult for us to accept their existence and understand the
concept of negative numbers.

\bigskip


Let $a\in \integers$ and $b\in\integers$. If $a>0$, then $-a<0$. If
$b<0$, then $-b>0$. Thus either the number, or its ``mirror
reflexion'' about $0$ is positive, and in particular, for any
$a\in\integers$, $-(-a)=a$. This leads to the following definition.
\begin{df}
Let $a\in\integers$. The {\em absolute value} of $a$ is defined and
denoted by
$$|a| = \left\{\begin{array}{ll}a & \mathrm{if}\ a \geq 0, \\ -a & \mathrm{if}\ a < 0, \\ \end{array}\right.  $$
\end{df}
\begin{exa}
$|5| = 5$ since $5>0$. $|-5|=-(-5)=5$, since $-5<0$.
\end{exa}
\begin{rem}
Letters have no idea of the sign of the numbers they represent. Thus
it is a {\bf mistake} to think, say, that $+x$ is always positive
and $-x$ is always negative.
\end{rem}


We would like to define addition, subtraction, multiplication and
division in  the integers in such a way that these operations are
consistent with those operations over the natural numbers and so
that they  enjoy  closure, commutativity, associativity, and
distributivity under addition and multiplication.

\bigskip

We start with addition. Recall that we defined addition of two
natural numbers and of two fractions as the concatenation of two
segments. We would like this definition to extend to the integers,
but we are confronted with the need to define what a ``negative
segment'' is. This we will do as follows. If $a<0$, then $-a>0$. We
associate with $a$ a segment of length $|-a|$, but to the left of
$0$ on the line, as in figure \ref{fig:negative-segment}.


 \vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)
\pstGeonode[PosAngle={-90}](-3,0){a}(0,0){O}
\pstGeonode[PointName=none](3,0){A}\uput[d](3,0){$-a$}
\ncline[offset=12pt]{|-|}{a}{O}\ncput*[nrot=:U]{$a$}\meinecaption{.5}{A
negative segment. Here $a<0$.} \label{fig:negative-segment}
\end{figure}

Hence we define the addition of integers $a$, $b$, as the
concatenation of segments. Depending on the sign of $a$ and $b$, we
have four cases. (We exclude the cases when at least one of $a$ or
$b$ is zero, these cases being trivial.)

\vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering
\pstGeonode[PosAngle={-90}](3,0){a}(0,0){O}(2,0){b}
\pstGeonode[PointName=none](5,0){c}
\ncline[offset=-20pt]{|-|}{a}{O}\ncput*[nrot=:D]{$a$}
\ncline[offset=15pt]{|-|}{b}{O}\ncput*[nrot=:D]{$b$}
\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{a}{c}\ncput*[nrot=:U]{$a+b$}
\uput[d](5,0){$a+b$}\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)\meinecaption{.5}{$a+b$
with $a>0$, $b>0$.} \label{fig:a>0b>0}
\end{figure}



\begin{exa}[Case $a>0, b>0$] To add $b$ to $a$, we first locate $a$
on the line. From there, we move $b$ units right (since $b>0$),
landing at $a+b$. Notice that this case reduces to addition of
natural numbers, and hence, we should obtain the same result as for
addition of natural numbers. This example is illustrated in figure
\ref{fig:a>0b>0}. For a numerical example (with $a=3$, $b=2$), see
figure \ref{fig:3+2}.


\label{exa:a+b,both-positive}\end{exa}


\vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering
\pstGeonode[PosAngle={-90}](0,0){O}
\pstGeonode[PointName=none](5,0){c}(3,0){a}(2,0){b}
\ncline[offset=-20pt]{|-|}{a}{O}\ncput*[nrot=:D]{$3$}
\ncline[offset=15pt]{|-|}{b}{O}\ncput*[nrot=:D]{$2$}
\uput[d](2,0){$2$}\uput[d](3,0){$3$}\uput[d](5,0){$5$}
\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{a}{c}\ncput*[nrot=:U]{$3+2$}
\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)\meinecaption{.5}{$3+2$.}
\label{fig:3+2}
\end{figure}






\begin{exa}[Case $a<0, b<0$] To add $b$ to $a$, we first locate $a$
on the line. From there, we move $b$ units left (since $b<0$),
landing at $a+b$. This example is illustrated in figure
\ref{fig:a<0b<0}. For a numerical example (with $a=-3$, $b=-2$), see
figure \ref{fig:-3+(-2)}. \label{exa:a+b,both-negative}\end{exa}


\begin{rem}
Examples   \ref{exa:a+b,both-positive} and
\ref{exa:a+b,both-negative} conform to the following intuitive idea.
If we associate positive numbers to ``gains'' and negative numbers
to ``losses'' then a ``gain'' plus a ``gain'' is a ``larger gain''
and a ``loss'' plus a ``loss'' is a ``larger loss.''
\end{rem}


\vspace*{2cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering
\pstGeonode[PosAngle={-90}](-3,0){a}(0,0){O}(-2,0){b}
\pstGeonode[PointName=none](-5,0){c}
\ncline[offset=-20pt]{|-|}{a}{O}\ncput*[nrot=:U]{$a$}
\ncline[offset=15pt]{|-|}{b}{O}\ncput*[nrot=:U]{$b$}
\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{a}{c}\ncput*[nrot=:D]{$a+b$}
\uput[d](-5,0){$a+b$}\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)\meinecaption{.5}{$a+b$
with $a<0$, $b<0$.} \label{fig:a<0b<0}
\end{figure}


\vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering
\pstGeonode[PosAngle={-90}](0,0){O}
\pstGeonode[PointName=none](-5,0){c}(-3,0){a}(-2,0){b}
\ncline[offset=-20pt]{|-|}{a}{O}\ncput*[nrot=:U]{$-3$}
\ncline[offset=15pt]{|-|}{b}{O}\ncput*[nrot=:U]{$-2$}
\uput[d](-2,0){$-2$}\uput[d](-3,0){$-3$}\uput[d](-5,0){$-5$}
\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{a}{c}\ncput*[nrot=:D]{$-3+(-2)$}
\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)\meinecaption{.5}{$-3+(-2)$.}
\label{fig:-3+(-2)}
\end{figure}




\begin{exa} We have,$$ (+1) + (+3) + (+5) = +9,   $$ since we are adding three gains, and
we thus obtain a larger gain.
\end{exa}
\begin{exa}We have,
$$ (-11) + (-13) + (-15) = -39,   $$
since we are adding three losses, and we thus obtain a larger loss.
\end{exa}


We now tackle the cases when the summands have opposite signs. In
this case, borrowing from the preceding remark, we have a ``gain''
plus a ``loss.'' In such a case it is impossible to know before hand
whether the result is a gain or a loss. The only conclusion we could
gather, again, intuitively, is that the result will be in a sense
``smaller'', that is, we will have a ``smaller gain'' or a smaller
loss.''  Some more thinking will make us see that if the ``gain'' is
larger than the loss, then the result will be a ``smaller gain,''
and if the ``loss'' is larger than the ``gain'' then the result will
be a ``smaller loss.''

\begin{exa}[Case $a<0, b>0$] To add $b$ to $a$, we first locate $a$
on the line. Since $a<0$, it is located to the left of $O$. From
there, we move $b$ units right (since $b>0$), landing at $a+b$. This
example is illustrated in figure \ref{fig:a<0b>0}. For a numerical
example (with $a=-3$, $b=2$), see figure \ref{fig:-3+2}. Again, we
emphasise, in the sum $(-3)+(+2)$, the ``loss'' is larger than the
``gain.'' Hence when adding, we expect a ``smaller loss'', fixing
the sign of the result to be minus.
\label{exa:a+b,a-neg,b-pos}\end{exa}



\vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering
\pstGeonode[PosAngle={-90}](-3,0){a}(0,0){O}(2,0){b}
\pstGeonode[PointName=none](-1,0){c}
\ncline[offset=-20pt]{|-|}{a}{O}\ncput*[nrot=:U]{$a$}
\ncline[offset=15pt]{|-|}{b}{O}\ncput*[nrot=:D]{$b$}
\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{a}{c}\ncput*[nrot=:U]{$a+b$}
\uput[d](-1,0){$a+b$}\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)\meinecaption{.5}{$a+b$
with $a<0$, $b>0$.} \label{fig:a<0b>0}
\end{figure}

\vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering
\pstGeonode[PosAngle={-90}](0,0){O}
\pstGeonode[PointName=none](-1,0){c}(-3,0){a}(2,0){b}
\ncline[offset=-20pt]{|-|}{a}{O}\ncput*[nrot=:U]{$-3$}
\ncline[offset=15pt]{|-|}{b}{O}\ncput*[nrot=:D]{$2$}
\uput[d](2,0){$2$}\uput[d](-3,0){$-3$}\uput[d](-1,0){$-1$}
\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{a}{c}\ncput*[nrot=:U]{$-3+2$}
\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)\meinecaption{.5}{$-3+2$.}
\label{fig:-3+2}
\end{figure}



\bigskip




\begin{exa}[Case $a>0, b<0$] To add $b$ to $a$, we first locate $a$
on the line. From there, we move $b$ units left (since $b<0$),
landing at $a+b$.  This example is illustrated in figure
\ref{fig:a>0b<0}. For a numerical example (with $a=3$, $b=-2$), see
figure \ref{fig:3+(-2)}. \label{exa:a+b,a-pos,a-neg}\end{exa}

\vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering
\pstGeonode[PosAngle={-90}](3,0){a}(0,0){O}(-2,0){b}
\pstGeonode[PointName=none](1,0){c}
\ncline[offset=20pt]{|-|}{a}{O}\ncput*[nrot=:D]{$a$}
\ncline[offset=-15pt]{|-|}{b}{O}\ncput*[nrot=:U]{$b$}
\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{a}{c}\ncput*[nrot=:D]{$a+b$}
\uput[d](1,0){$a+b$}\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)\meinecaption{.5}{$a+b$
with $a>0$, $b<0$.} \label{fig:a>0b<0}
\end{figure}

\vspace*{1cm}
\begin{figure}[!h]
 \psset{unit=3pc}\centering
\pstGeonode[PosAngle={-90}](0,0){O}
\pstGeonode[PointName=none](1,0){c}(3,0){a}(-2,0){b}
\ncline[offset=20pt]{|-|}{a}{O}\ncput*[nrot=:D]{$3$}
\ncline[offset=-15pt]{|-|}{b}{O}\ncput*[nrot=:U]{$-2$}
\uput[d](-2,0){$-2$}\uput[d](3,0){$3$}\uput[d](1,0){$1$}
\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{a}{c}\ncput*[nrot=:D]{$3+(-2)$}
\psaxes[yAxis=false,labels=none,
 ticks=none,
arrowscale=3,arrows={<->}](0,0)(-6.5,0)(6.5,0)\meinecaption{.5}{$3+(-2)$.}
\label{fig:3+(-2)}
\end{figure}

\bigskip

\begin{exa}We have,
$$ (+19) + (-21) = -2,   $$since the loss of $21$ is larger than the
gain of $19$ and so we obtain a loss.
\end{exa}
\begin{exa}We have,
$$ (-100) + (+210) = +110,   $$since the loss of $100$ is smaller than the
gain of $210$ and so we obtain a gain.
\end{exa}

\bigskip

We now turn to subtraction. We define subtraction in terms of
addition.
\begin{df}
Subtraction is defined as
$$ a - b = a + (-b).  $$
\end{df}
\begin{exa}
We have, $(+8)-(+5)= (+8)+(-5)=3$.
\end{exa}
\begin{exa}
We have, $(-8)-(-5)= (-8)+(+5)=-3$.
\end{exa}
\begin{exa}
We have, $(+8)-(-5)= (+8)+(+5)=13$.
\end{exa}
\begin{exa}
We have, $(-8)-(+5)= (-8)+(-5)=-13$.
\end{exa}

   \section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

\begin{pro}Perform the following operations mentally.
\begin{enumerate}\item $ (-9) - (-17)$ \item $ (-17) - (9)$
\item $ (-9) - (17)$ \item $ (-1) - (2) - (-3)$ \item $ (-100) -
(101) + (-102)$ \item $ |-2| - |-2|$ \item $|-2| - (- |2|)$ \item
$|-100| + (-100) - (- (-100))$

\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}\item $+8$ \item $-25$
\item $-25$ \item $0$ \item $-303$ \item $0$ \item $+4$ \item $-100$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
$1-2+3-4+5-6+7-8+9-10=  $
\begin{answer}
$ -5$
\end{answer}
\end{pro}
\begin{pro}
$-2-4-6-8-10+1+3+5+7+9=  $
\begin{answer}
$ -5$
\end{answer}
\end{pro}
\begin{pro}
$(1-2)-(3-4)+(5-6)-(7-8)+(9-10)=  $
\begin{answer}
$ -1$
\end{answer}
\end{pro}
\begin{pro}
$a-2a+3a-4a+5a=  $
\begin{answer}
$3a $
\end{answer}
\end{pro}
\begin{pro}
Place the nine integers $$\{-4,-3,-2,-1,0,1,2,3,4\}$$ exactly once
in the diagram below  so that every diagonal sum be the same.
\vspace*{2cm}
$$\psset{unit=3pc} \cnodeput(0,0){O}{\hspace{.5cm}} \cnodeput(1;0){A}{\hspace{.5cm}} \cnodeput(1;45){B}{\hspace{.5cm}} \cnodeput(1;90){C}{\hspace{.5cm}}
\cnodeput(1;135){D}{\hspace{.5cm}}
\cnodeput(1;180){E}{\hspace{.5cm}}
\cnodeput(1;225){F}{\hspace{.5cm}}
\cnodeput(1;270){G}{\hspace{.5cm}}
\cnodeput(1;315){H}{\hspace{.5cm}}
\ncline{A}{B}\ncline{B}{C}\ncline{C}{D}\ncline{D}{E}\ncline{E}{F}\ncline{F}{G}\ncline{G}{H}
\ncline{H}{A} \ncline{O}{A} \ncline{O}{B} \ncline{O}{C}
\ncline{O}{D} \ncline{O}{E}  \ncline{O}{F} \ncline{O}{G}
\ncline{O}{H}       $$ \vspace*{1cm}
\begin{answer}\noindent Here is one possible answer.
\vspace*{2cm}
$$\psset{unit=3pc} \cnodeput(0,0){O}{0} \cnodeput(1;0){A}{-4} \cnodeput(1;45){B}{-3} \cnodeput(1;90){C}{-2}
\cnodeput(1;135){D}{-1} \cnodeput(1;180){E}{4}
\cnodeput(1;225){F}{3} \cnodeput(1;270){G}{2} \cnodeput(1;315){H}{1}
\ncline{A}{B}\ncline{B}{C}\ncline{C}{D}\ncline{D}{E}\ncline{E}{F}\ncline{F}{G}\ncline{G}{H}
\ncline{H}{A} \ncline{O}{A} \ncline{O}{B} \ncline{O}{C}
\ncline{O}{D} \ncline{O}{E}  \ncline{O}{F} \ncline{O}{G}
\ncline{O}{H}  $$ \vspace*{1cm}

\end{answer}
\end{pro}


\begin{pro}
Place the nine integers $\{-2,-1,0,1,2,3,4, 5, 6\}$ exactly once in
the diagram below  so that every diagonal sum be the same.
\vspace*{2cm}
$$\psset{unit=3pc,} \cnodeput(0,0){O}{\hspace{.5cm}} \cnodeput(1;0){A}{\hspace{.5cm}} \cnodeput(1;45){B}{\hspace{.5cm}} \cnodeput(1;90){C}{\hspace{.5cm}}
\cnodeput(1;135){D}{\hspace{.5cm}}
\cnodeput(1;180){E}{\hspace{.5cm}}
\cnodeput(1;225){F}{\hspace{.5cm}}
\cnodeput(1;270){G}{\hspace{.5cm}}
\cnodeput(1;315){H}{\hspace{.5cm}}
\ncline{A}{B}\ncline{B}{C}\ncline{C}{D}\ncline{D}{E}\ncline{E}{F}\ncline{F}{G}\ncline{G}{H}
\ncline{H}{A} \ncline{O}{A} \ncline{O}{B} \ncline{O}{C}
\ncline{O}{D} \ncline{O}{E}  \ncline{O}{F} \ncline{O}{G}
\ncline{O}{H}       $$ \vspace*{2cm}
\begin{answer}\noindent Here is one possible answer.
\vspace*{2cm}
$$\psset{unit=3pc} \cnodeput(0,0){O}{2} \cnodeput(1;0){A}{-2} \cnodeput(1;45){B}{-1} \cnodeput(1;90){C}{0}
\cnodeput(1;135){D}{1} \cnodeput(1;180){E}{6} \cnodeput(1;225){F}{5}
\cnodeput(1;270){G}{4} \cnodeput(1;315){H}{3}
\ncline{A}{B}\ncline{B}{C}\ncline{C}{D}\ncline{D}{E}\ncline{E}{F}\ncline{F}{G}\ncline{G}{H}
\ncline{H}{A} \ncline{O}{A} \ncline{O}{B} \ncline{O}{C}
\ncline{O}{D} \ncline{O}{E}  \ncline{O}{F} \ncline{O}{G}
\ncline{O}{H}  $$ \vspace*{1cm}

\end{answer}
\end{pro}
\end{multicols}

\chapter{Integers: Multiplication and Division}
We now explore multiplication of integers. Again, we would like the
multiplication rules to be consistent with those we have studied for
natural numbers. This entails that, of course, that if we multiply
two positive integers, the result will be a positive integer. What
happens in other cases? Suppose $a>0$ and $b<0$. We will
prove that $ab<0$. Observe that
$$ a(b-b)=0 \implies ab-ab=0 \implies ab +a(-b)=0 \implies ab= -a(-b).  $$
Since $-b>0$, $a(-b)$ is the product of two positive integers, and
hence positive. Thus $-a(-b)$ is negative, and so $ab=-a(-b)<0$. We
have proved that the product of a positive integer and a negative
integer is negative. Using the same trick we can prove that
$$(-x)(-y)=xy.
$$If $x<0$, $y<0$, then both $-x>0$, $-y>0$, hence the product of
two negative integers is the same as the product of two positive
integers, and hence positive. We have thus proved the following
rules: $$ (+)(+)=(-)(-)=+, \qquad (+)(-)=(-)(+)=-. $$

\bigskip

Intuitively, you may think of a negative sign as a reversal of
direction on the real line. Thus the product or quotient of two
integers different sign is negative. Two negatives give two
reversals, which is to say, no reversal at all, thus the product or
quotient of two integers with the same sign is positive. The sign
rules for division are obtained from and are identical from those of
division.
\begin{exa}We have,
$$(-2)(5) = -10,\qquad
(-2)(-5) = +10,\qquad (+2)(-5) = -10,\qquad (+2)(+5) = +10.$$
\end{exa}
\begin{exa}We have,
$$(-20)\div (5) = -4,\qquad
(-20)\div (-5) = +4,\qquad (+20)\div (-5) = -4,\qquad (+20)\div (+5)
= +4.$$
\end{exa}
The rules of operator precedence discussed in the section of natural
numbers apply.

\begin{exa}We have,
$$\begin{array}{ccl}\dfrac{(-8)(-12)}{3} + \dfrac{30}{((-2)(3))}  & = &  \dfrac{96}{3}  + \dfrac{30}{(-6)} \\
& = & 32 + (-5) \\
& = & 27.   \end{array}$$
\end{exa}
\begin{exa}We have,
$$\begin{array}{ccl} (5 - 12)^2 - (-3)^3  & = &  (-7)^2 - (-27) \\
& = & 49 + 27 \\
& = & 76.  \end{array}$$
\end{exa}
As a consequence of the rule of signs for multiplication, a product
containing an odd number of minus signs will be negative and a
product containing an even number of minus signs will be positive.
\begin{exa}$$ (-2)^2 = 4, \qquad  (-2)^3 = -8, \qquad  (-2)^{10} = 1024. $$\end{exa}
\begin{rem} Notice the difference between, say, $(-a)^2$ and $-a^2$. $(-a)^2$ is the square of $-a$, and hence it is
always non-negative. On the other hand, $-a^2$ is the opposite of
$a^2$, and therefore it is always non-positive.\end{rem}
\begin{exa} We have, $$ 5 + (-4)^2= 5 + 16 = 21, $$ $$ 5 - 4^2= 5 - 16 = -11, $$
$$ 5 - (-4)^2= 5 - 16 = -11. $$  \end{exa}

   \section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
What is the value of $((1 \cdot  2 \div (3 \cdot  4)-5) \cdot  6-7)\div
(8  \cdot  9)$?
\begin{answer}
$-\dfrac{1}{2}$.
\end{answer}
\end{pro}
\begin{pro}
Calculate:\quad
$\dfrac{(10)^3+(-5)^3+(1)^3-3(10)(-5)(1)}{(10)+(-5)+(1)}$.
\begin{answer}
$171$
\end{answer}
\end{pro}
\begin{pro}Calculate: \quad $(2\cdot ( -3) + 1)^2 - ((-5)\cdot 4+2\cdot 9)^3$.

\begin{answer} $(2\cdot ( -3) + 1)^2 - ((-5)\cdot 4+2\cdot 9)^3= (-6+1)^2-(-20+18)^3=(-5)^2-(-2)^3=25-(-8)=33$.   \end{answer}
\end{pro}
\begin{pro}Calculate: \quad $(2\cdot ( -3) + 1)^2 - ((-5)\cdot 4+2\cdot 9)^3$.

\begin{answer} $(2\cdot ( -3) + 1)^2 - ((-5)\cdot 4+2\cdot 9)^3= (-6+1)^2-(-20+18)^3=(-5)^2-(-2)^3=25-(-8)=33$.   \end{answer}
\end{pro}
\begin{pro}Calculate: $$(2+3+4)(2^2+3^2+4^2-2\cdot 3 -3\cdot 4 - 4\cdot 2).$$
\begin{answer}  $(2+3+4)(2^2+3^2+4^2-2\cdot 3 -3\cdot 4 - 4\cdot 2)=(9)(4+9+16-6-12-8)=(9)(3)=27$  \end{answer}
\end{pro}

\begin{pro}
Complete the ``crossword'' puzzle with $1$'s or $-1$'s.
\vspace*{1cm} \PuzzleSolution
\begin{Puzzle}{5}{5}%
|{$-1$}|{$ \cdot $}|{\empty}|{$=$}|{\empty}|.                      %
|{$ \cdot $}|*|{$ \cdot $}|*|{$ \cdot $}|.                                  %
|{\empty}|{$ \cdot $}|{$-1$}|{$=$}|{\empty}|. %
|{$=$}|*|{$=$}|*|{$=$}|.                                  %
|{\empty}|{$ \cdot $}|{\empty}|{$=$}|{$-1$}|.   %
\end{Puzzle}
\begin{answer}\noindent
Here are two answers. \\
\vspace*{1.5cm}
\begin{minipage}{5cm}
\PuzzleSolution
\begin{Puzzle}{5}{5}%
|{$-1$}|{$ \cdot $}|{$+1$}|{$=$}|{$-1$}|.                      %
|{$ \cdot $}|*|{$ \cdot $}|*|{$ \cdot $}|.                                  %
|{$-1$}|{$ \cdot $}|{$-1$}|{$=$}|{$+1$}|. %
|{$=$}|*|{$=$}|*|{$=$}|.                                  %
|{$+1$}|{$ \cdot $}|{$-1$}|{$=$}|{$-1$}|.   %
\end{Puzzle}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\PuzzleSolution
\begin{Puzzle}{5}{5}%
|{$-1$}|{$ \cdot $}|{$-1$}|{$=$}|{$+1$}|.                      %
|{$ \cdot $}|*|{$ \cdot $}|*|{$ \cdot $}|.                                  %
|{$+1$}|{$ \cdot $}|{$-1$}|{$=$}|{$-1$}|. %
|{$=$}|*|{$=$}|*|{$=$}|.                                  %
|{$-1$}|{$ \cdot $}|{$+1$}|{$=$}|{$-1$}|.   %
\end{Puzzle}
\end{minipage}
\end{answer}

\end{pro}
\end{multicols}

\vspace*{1cm}

\chapter{Rational Numbers}
\QUOTEME{Bridges would not be safer if only people who knew the
proper definition of a real number were allowed to design them.}{N.
David MERMIN}
\begin{df} The set of {\em negative fractions} is the set
$$ \left\{-\dfrac{a}{b}: a\in\naturals, a>0, b\in \naturals, b>0 \right\}.  $$
The set of positive fractions together with the set of negative
fractions and the number $0$ form the set of {\em rational numbers},
which we denote by $\rationals$.
\end{df}
The rules for operations with rational numbers derive from those of
operations with fractions and with integers. Also, the rational
numbers are closed under the four operations of addition,
subtraction, multiplication, and division. A few examples follow.
\begin{exa}
We have $$\begin{array}{lll} \dfrac{2}{5}\cdot \dfrac{15}{12} -
\dfrac{7}{10}\div \dfrac{14}{15} & = & \dfrac{2}{5}\cdot
\dfrac{15}{12} - \dfrac{7}{10}\cdot \dfrac{15}{14}\\
& = & \dfrac{\cancel{2}}{\cancel{5}}\cdot \dfrac{\cancel{5}\cdot
\cancel{3}}{\cancel{2} \cdot 2 \cdot \cancel{3}} -
\dfrac{\cancel{7}}{2\cdot \cancel{5}}\cdot \dfrac{3 \cdot \cancel{5}}{2\cdot \cancel{7}} \\
& = & \dfrac{1}{2}-\dfrac{3}{4}\\
& = & \dfrac{2}{4} - \dfrac{3}{4}\\
& = & -\dfrac{1}{4}.
\end{array}$$
\end{exa}
It can be proved that any rational number has a decimal expansion
which is either periodic (repeats) or terminates, and that
viceversa, any number with either a periodic or a terminating
expansion is a rational number. For example, $\dfrac{1}{4}=0.25$ has
a terminating decimal expansion, and
$\dfrac{1}{11}=0.0909090909\ldots = 0.\overline{09}$ has a repeating
one. By long division you may also obtain
$$ \dfrac{1}{7} = 0.\overline{142857}, \qquad \dfrac{1}{17}=0.\overline{0588235294117647},
$$and as you can see, the periods may be longer than what your
calculator can handle.

\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Find the value of $\dfrac{5}{6}-\left(-\dfrac{5}{6}\right)^2$.
\begin{answer} $\dfrac{5}{36}$ \end{answer}
\end{pro}
\begin{pro}
Find the value of $$((1\cdot 2 \div (3\cdot 4)-5)\cdot 6 -7)\div (8\cdot 9).$$
\begin{answer} $-\dfrac{1}{2}$ \end{answer}
\end{pro}
\begin{pro}
Find $$(0.1)(-0.3)(0.5)(-0.7)(0.9)-(-0.2)(-0.4)(-0.6)(-0.8).$$
\begin{answer}
$-0.02895$
\end{answer}
\end{pro}
\begin{pro}
Find $\dfrac{\dfrac{3}{4}-0.7}{0.2-\dfrac{2}{5}}$ and convert the
result into a fraction.
\begin{answer}
$-0.25=-\dfrac{1}{4}$
\end{answer}
\end{pro}

\begin{pro} Suppose that you know that $\dfrac{1}{3}=0.333333\ldots = 0.\overline{3}$. What should
$0.1111\ldots = 0.\overline{1}$ be?
\begin{answer} $0.1111\ldots = 0.\overline{1}=\dfrac{1}{3}\left(0.333333\ldots \right) = \dfrac{1}{3}\cdot \dfrac{1}{3}=\dfrac{1}{9}$. \end{answer}
 \end{pro}
\begin{pro}
Find the value of $121 (0.\overline{09})$.
\begin{answer}
From the text, we know that $0.\overline{09}=\dfrac{1}{11}$, and
hence $121 (0.\overline{09}) = 121 \cdot \dfrac{1}{11}=11$.
\end{answer}
\end{pro}
\begin{pro}
Compute:
$$\dfrac{1}{2+\dfrac{3}{4-\dfrac{5}{6+\dfrac{7}{8-\dfrac{9}{10}}}}}.$$
\begin{answer}
$\dfrac{1411}{4214}$
\end{answer}

\end{pro}


\end{multicols}




\chapter{Irrational Numbers}

What about numbers whose decimal expansion is infinite and does not
repeat? This leads us to the following definition.
\begin{df}
A number whose decimal expansion is infinite and does not repeat is
called an {\em irrational number}.
\end{df}
From the discussion above, an irrational number is one that cannot
be expressed as a fraction of two integers.
\begin{exa}
Consider the number $$0.1010010001000010000010000001\ldots,  $$
where the number of $0$'s between consecutive $1$'s
 grows in sequence: $1,2,3,4,5, \ldots$. Since the number of $0$'s is progressively growing, this infinite decimal
 does not have a repeating period and hence must be an irrational number.\end{exa}


Using my computer, when I enter $\sqrt{2}$ I obtain as an answer
$$1.4142135623730950488016887242097.$$
Is this answer exact? Does this decimal repeat?
 It can be proved that the number
$\sqrt{2}$ is irrational, hence the above answer is only an
approximation and the decimal does not repeat. The first proof of
the irrationality of $\sqrt{2}$ is attributed to Hippasus of
Metapontum, one of the disciples of Pythagoras (c 580 BC--c 500
BC).\footnote{The Pythagoreans were akin to religious cults of
today. They forbade their members to eat beans, dedicated their
lives to Mathematics and Music, and believed that the essence of
everything in the world was {\em number}.} The Greek world view at
that time was that all numbers where rational, and hence this
discovery was anathema to the Pythagoreans who decided to drown
Hipassus for his discovery.

\begin{rem}
It can be proved that if $n$ is a natural number that is not a
perfect square, then $\sqrt{n}$ is irrational. Hence $\sqrt{2},
\sqrt{3}, \sqrt{5},\sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{10}$, etc., are all
irrational.
\end{rem}


\bigskip

In 1760, Johann Heinrich Lambert (1728 - 1777) proved that $\pi$ is
irrational.

\begin{rem}
In particular, then, it would be incorrect to write $\pi=3.14$, or
$\pi = \dfrac{22}{7}$, or $\pi = \dfrac{355}{113}$, etc., since
$\pi$ is not rational. All of these are simply approximations, and
hence we must write  $\pi\approx 3.14$, $\pi \approx \dfrac{22}{7}$,
or $\pi \approx \dfrac{355}{113}$, etc.
\end{rem}

\begin{df}
The set of {\em real numbers}, denoted by $\reals$, is the
collection of rational numbers together with the irrational numbers.
\end{df}
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}Use a calculator to round $\sqrt{2}+\sqrt{3}+\sqrt{5}$ to two decimal places.
\begin{answer} $\sqrt{2}+\sqrt{3}+\sqrt{5}\approx 5.38$\end{answer}
\end{pro}
\begin{pro}Use a calculator to round $\sqrt{2}\cdot \sqrt{3}\cdot \sqrt{5}$ to two decimal places.
\begin{answer} $\sqrt{2}\cdot \sqrt{3}\cdot \sqrt{5}\approx 5.48$\end{answer}
 \end{pro}
\begin{pro}Let $A$ be a digit.
The four-digit number $12A6$ is a perfect square. What is the value of $A$?
\begin{answer}
$9$
\end{answer}
\end{pro}
\begin{pro} Let $a$, $b$ be positive real numbers. Is it always true that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$?
\begin{answer} No, this is not always true. For example, $\sqrt{1}+\sqrt{3}=1+\sqrt{3}\approx 2.73$, but
$\sqrt{1+3}=\sqrt{4}=2$. But, is it {\em ever} true? For $a=0$,
$\sqrt{a}+\sqrt{b} = \sqrt{0}+\sqrt{b}=\sqrt{b}$ and
$\sqrt{a+b}=\sqrt{0+b}=\sqrt{b}$, so it is true in this occasion.
\end{answer}
  \end{pro}


\end{multicols}



\Closesolutionfile{aritmetiquilla}


\clearpage

\appendix
\renewcommand{\chaptername}{Appendix}

\chapter{Multiple-Choice Exam Questions}
\begin{enumerate}
\item[] {\bf Ancient Numerals}
\item Convert $123$ to Egyptian numerals. \\ \vfill \psovalbox{A}
\mbox{\egyptify{0}{0}{0}{0}{3}{2}{1}} \hfill  \psovalbox{B}
\mbox{\egyptify{0}{0}{0}{0}{1}{2}{3}} \hfill  \psovalbox{C}
\mbox{\egyptify{0}{0}{0}{0}{1}{3}{2}} \hfill  \psovalbox{D}
\mbox{\egyptify{0}{0}{0}{0}{2}{1}{3}} \hfill  \psovalbox{E} none of
these   \item  Convert $123$ to Roman
numerals. \\ \vfill\psovalbox{A} \Romannum{123} \hfill \psovalbox{B}
\Romannum{132} \hfill  \psovalbox{C}  \Romannum{312} \hfill
\psovalbox{D} \Romannum{213} \hfill  \psovalbox{E} none of these
   \item Convert  the Egyptian numeral
\mbox{\egyptify{0}{0}{0}{0}{3}{2}{1}} to decimal. \\
\vfill\psovalbox{A} $123$ \hfill  \psovalbox{B}  $312$ \hfill
\psovalbox{C} $321$ \hfill \psovalbox{D} $231$ \hfill \psovalbox{E}
none of these    \item  Convert  the Roman
numeral \Romannum{231} to decimal. \\ \vfill\psovalbox{A} $123$
\hfill \psovalbox{B}  $312$ \hfill \psovalbox{C} $321$ \hfill
\psovalbox{D} $231$ \hfill \psovalbox{E} none of these    \item  If two $\heartsuit$'s are equivalent to
one $\clubsuit$ and three $\clubsuit$'s are equivalent to two
$\spadesuit$'s, to how many $\spadesuit$'s is $\heartsuit\heartsuit
\heartsuit \heartsuit \clubsuit\clubsuit\clubsuit\clubsuit$? \\
\vfill\psovalbox{A} $4$ \hfill  \psovalbox{B}  $6$ \hfill
\psovalbox{C} $12$ \hfill  \psovalbox{D} $18$ \hfill  \psovalbox{E}
none of these

  \item[] {\bf Positional Notation}
  \item  Convert $123_4$ to decimal. \\
\vfill \psovalbox{A} $75$ \hfill  \psovalbox{B}  $57$ \hfill
\psovalbox{C} $72$ \hfill  \psovalbox{D} $27$ \hfill  \psovalbox{E}
none of these   \item  Convert $123$ to
base-$4$. \\ \vfill\psovalbox{A} $123_4$ \hfill  \psovalbox{B}
$1331_4$ \hfill \psovalbox{C} $1323_4$ \hfill  \psovalbox{D}
$3231_4$ \hfill \psovalbox{E} none of these
  \item  Convert $A2B_{16}$ to decimal. \\
\vfill \psovalbox{A} $274$ \hfill  \psovalbox{B}  $10211$ \hfill
\psovalbox{C} $2588$ \hfill  \psovalbox{D} $2587$ \hfill
\psovalbox{E} none of these   \item  $32_4+33_4=$ \\
\vfill\psovalbox{A} $110_4$ \hfill \psovalbox{B} $120_4$ \hfill
\psovalbox{C} $131_4$ \hfill \psovalbox{D} $65_4$ \hfill
\psovalbox{E} none of these
 \item  $52_6-33_6=$ \\
\vfill\psovalbox{A} $15_6$ \hfill \psovalbox{B} $19_6$ \hfill
\psovalbox{C} $25_6$ \hfill \psovalbox{D} $14_6$ \hfill
\psovalbox{E} none of these

\item  Which integer follows the hexadecimal integer $A2FF_{16}$? In other words, what is
$A2FF_{16} + 1_{16}$? \\
\vfill \psovalbox{A} $A2A00_{16}$ \hfill  \psovalbox{B}  $A300_{16}$
\hfill \psovalbox{C} $B300_{16}$ \hfill  \psovalbox{D} $A2FF1_{16}$
\hfill
\psovalbox{E} none of these  \item  $11_2+10_2=$ \\
\vfill\psovalbox{A} $100_2$ \hfill \psovalbox{B} $110_2$ \hfill
\psovalbox{C} $111_2$ \hfill \psovalbox{D} $101_2$ \hfill
\psovalbox{E} none of these
 \item  Which of the following is the {\em wrong} description for the `$6$' in
$654321$?\\
 \vfill
\psovalbox{A} $600000$ units \vfill \psovalbox{B} $6000$ hundreds
\vfill \psovalbox{C} $600$ thousands \vfill \psovalbox{D} $60$ ten
thousands \vfill \psovalbox{E} 6 millions \vfill

 \item[] {\bf Symbolical Expression}

\item Let $x$ be the unknown quantity. Translate into symbols: ``the excess of a number over three.'' \\
\vfill \psovalbox{A} $\dfrac{x}{3}$ \hfill  \psovalbox{B}  $x-3$
\hfill \psovalbox{C} $3-x$ \hfill  \psovalbox{D} $\dfrac{3}{x}$
\hfill \psovalbox{E} none of these
\item Let $s$ be the unknown quantity. Translate into symbols: ``thrice a number is reduced by its square.'' \\
\vfill \psovalbox{A} $s^2+3s$ \hfill  \psovalbox{B}  $s^2-3s$ \hfill
\psovalbox{C} $3s-s^2$ \hfill  \psovalbox{D} $2s-s^2$ \hfill
\psovalbox{E} none of these

 \item  If you had $a$ $\$20$'s and $b$ $\$50$'s, how much money, in dollars, do you have? \\
\vfill \psovalbox{A} $a+b$ \hfill  \psovalbox{B}  $70ab$ \hfill
\psovalbox{C} $20a+50b$ \hfill  \psovalbox{D} $70+a+b$ \hfill
\psovalbox{E} none of these
\item  Rod is currently twice as old as Tit\'{\i}. If Tit\'{\i} is currently $t$ years old, how old will  Rod be $10$ years from now? \\
\vfill \psovalbox{A} $t+10$ \hfill  \psovalbox{B}  $2t+10$ \hfill
\psovalbox{C} $\dfrac{t}{2}+10$ \hfill  \psovalbox{D} $2t+20$ \hfill
\psovalbox{E} none of these
\item  Rod is currently twice as old as Tit\'{\i}. If Rod is currently $r$ years old, how old was  Tit\'{\i}  $10$ years ago? \\
\vfill \psovalbox{A} $r-10$ \hfill  \psovalbox{B}  $2r-10$ \hfill
\psovalbox{C} $\dfrac{r}{2}-10$ \hfill  \psovalbox{D} $2r-20$ \hfill
\psovalbox{E} none of these

\item Which of the following expressions best represents:
``the excess of a number over $3$''?
  \\ \psovalbox{A}
$\dfrac{x}{3}$ \hfill\psovalbox{B}$x-3$ \hfill\psovalbox{C}$3x$
\hfill\psovalbox{D}$\dfrac{3}{x}$ \hfill\psovalbox{E}none of these

\item Which of the following expressions best represents:
``the sum of a number plus thrice another is being diminished by
$3$''?
  \\ \psovalbox{A}
$\dfrac{x+3y}{3}$ \hfill\psovalbox{B}$x+3y-3$ \hfill\psovalbox{C}$x
+ 3(y-3)$ \hfill\psovalbox{D}$x+2y-3$ \hfill\psovalbox{E}none of
these

\item The algebraic expression $x^2 -\dfrac{1}{x}$ can be translated as:
  \\ \psovalbox{A} ``The square of a number is increased by
  the reciprocal of the number.'' \\
 \psovalbox{B} ``The square of a number is reduced by
  the reciprocal of the number.''\\ \psovalbox{C} ``Twice a number is increased by
  the reciprocal of the number.'' \\
\psovalbox{D} ``Twice a number is decreased by
  the reciprocal of the number.''\\ \psovalbox{E}none of these

\item You have $t$ \$10 and $w$ \$20 bills. How much money, in dollars, do you have? \\ \vfill
\psovalbox{A} $t+w$ \hfill  \psovalbox{B} $10t+20w$\hfill
\psovalbox{C} $200tw$ \hfill \psovalbox{D} $30+t+w$\hfill
\psovalbox{E} none of these
\item You have $t$ \$10 and $w$ \$ 20 bills. How many bank notes (bills) do you have? \\ \vfill
\psovalbox{A} $t+w$ \hfill  \psovalbox{B} $10t+20w$\hfill
\psovalbox{C} $200tw$ \hfill \psovalbox{D} $30+t+w$\hfill
\psovalbox{E} none of these

 \item  If my age by $31$ December 2020 will be $s$, what was my age by $31$ December $2005$? \\
\vfill \psovalbox{A} $s-15$ \hfill  \psovalbox{B}  $s+15$ \hfill
\psovalbox{C} $15-s$ \hfill  \psovalbox{D} $s-14$ \hfill
\psovalbox{E} none of these
\item  If my age by $31$ December 2005 was $s$, what will my age be by $31$ December $2020$? \\
\vfill \psovalbox{A} $s-15$ \hfill  \psovalbox{B}  $s+15$ \hfill
\psovalbox{C} $15-s$ \hfill  \psovalbox{D} $s-14$ \hfill
\psovalbox{E} none of these
\item If $a$ is an even natural number and $b$ is an odd natural number, how many of the following are even numbers?
$$ I: a+b \qquad II: ab \qquad III: a+b-1 \qquad IV: ab+2b   $$
\vfill \psovalbox{A} none \hfill  \psovalbox{B}  exactly one \hfill
\psovalbox{C} exactly two \hfill  \psovalbox{D} exactly three \hfill
\psovalbox{E} all four

\item  I am thinking of a rule that converts the number $3$ into the number
$20$. Which of the following could {\em not} be my rule? \vfill
\psovalbox{A} add seventeen \vfill \psovalbox{B} multiply by five,
then add five \vfill \psovalbox{C} add one, then multiply by five
\vfill \psovalbox{D} square, then add eleven \vfill \psovalbox{E}
cube, then add three \vfill
\item[] {\bf Addition and Subtraction of Natural Numbers}

\item Collect like terms: $(2a+3b+5c)+(2b+a+2c)$ \\
\vfill \psovalbox{A} $15abc$ \hfill  \psovalbox{B} $4a+4b+7c$ \hfill
\psovalbox{C} $2a+6b+10c$ \hfill  \psovalbox{D} $3a+5b+7c$ \hfill
\psovalbox{E} none of these
\item Collect like terms: $(x^2+3x+1)+(2x^2+2x+3)$ \\
\vfill \psovalbox{A} $3x^2+5x+4$ \hfill  \psovalbox{B} $12x^3$
\hfill \psovalbox{C} $2x^2+6x+3$ \hfill  \psovalbox{D} $2x^2+5x+4$
\hfill \psovalbox{E} none of these
\item In the addition
$$\opadd[voperator=bottom,operandstyle.1.1=\hole,
operandstyle.1.3=\hole, operandstyle.2.4=\hole, resultstyle.2=\hole,
carryadd=false] {1482}{2720}$$what is the sum of the missing digits?\\
\vfill \psovalbox{A} $6$ \hfill  \psovalbox{B} $7$ \hfill
\psovalbox{C} $8$ \hfill  \psovalbox{D} $9 $ \hfill \psovalbox{E}
none of these
\item How many different sums can we obtain when we add two
different numbers from the set $\{1,2,3,4,5\}$? \\
\vfill \psovalbox{A} $5$ \hfill  \psovalbox{B} $6$ \hfill
\psovalbox{C} $7$ \hfill  \psovalbox{D} $8 $ \hfill \psovalbox{E}
none of these
\item You start the day with $A$ dollars. Your uncle Bob gives you enough
money to  double your amount. Your aunt Rita gives you $12$ dollars.
You have to pay $B$ dollars in fines, and spent $10$ dollars fueling
your camel with gas.  How much money do you have now?\\
\psovalbox{A} $2A-B-2$ \hfill  \psovalbox{B} $2A-B+2$ \hfill
\psovalbox{C} $3A-B-2$ \hfill  \psovalbox{D} $3A-B+2$ \hfill
\psovalbox{E} none of these
\item In the {\bf difference}
$$\opsub[voperator=bottom,operandstyle.1.1=\hole,
operandstyle.1.3=\hole, operandstyle.2.4=\hole, resultstyle.2=\hole,
carryadd=false] {5123}{1321}$$what is the sum of the missing digits?\\
\vfill \psovalbox{A} $4$ \hfill  \psovalbox{B} $5$ \hfill
\psovalbox{C} $6$ \hfill  \psovalbox{D} $7$ \hfill \psovalbox{E}
none of these
\item In the magic square, the three numbers in each row, in each column, and in each diagonal add up to the same number. When the magic square below is completed, which of the following numbers
is {\em not} used?
\begin{center}
\PuzzleSolution
\begin{Puzzle}{3}{3}%
|{13}|{\empty }|{\empty }|. |{\empty }|{10}|{\empty }|. |{9}|{\empty
}|{7}|.
\end{Puzzle}
 \end{center}
\vfill \psovalbox{A} $6$ \hfill  \psovalbox{B} $8$ \hfill
\psovalbox{C} $12$ \hfill  \psovalbox{D} $14 $ \hfill \psovalbox{E}
$15$
\item For a given charity, Adam contributes $a$ dollars, Betty contributes $b$ dollars and Carl contributes $c$ dollars. Dwight, Erin, and Frances
are {\bf thieves}, and  so from the
contributions, Dwight steals $d$ dollars, Erin steals $e$ dollars and Frances steals $f$ dollars. What is the available amount of dollars that the charity has?\\
\psovalbox{A} $a+b+c-d-e-f$ \hfill  \psovalbox{B} $a+b+c+d+e+f$
\hfill \psovalbox{C} $abc-def$ \hfill  \psovalbox{D}
$\dfrac{abc}{def}$ \hfill \psovalbox{E} none of these \vfill

\item[] {\bf Arithmetic progressions}  \vfill
 \item  Consider the arithmetic progression  $2, 9, 16, 23, \ldots$, where $2$ is on the first position, $9$ in the second position, etc.
 Which number occupies the $100$th position?
\\ \vfill\psovalbox{A} $695$ \hfill \psovalbox{B}
$702$ \hfill  \psovalbox{C}  $100$ \hfill \psovalbox{D} $700$ \hfill
\psovalbox{E} none of these  \vfill\item Which of the following
numbers does not belong to the
arithmetic progression $2, 9, 16, 23, \ldots$? \\
$$ I: 68  \qquad  II: 30  \qquad III: 72\qquad IV: 702   $$
\vfill\psovalbox{A} $68$ \hfill  \psovalbox{B}  $30$ \hfill
\psovalbox{C} $72$ \hfill \psovalbox{D} $702$ \hfill \psovalbox{E}
all these numbers belong to the progression  \vfill

\item Consider  the arithmetic progression $3, 14, 25, 36, \ldots$, where $3$ is on the first position, $14$ is on the second, etc. Which number
occupies the $101$st position?
\\ \vfill
\psovalbox{A}  $101$\hfill \psovalbox{B} $1092$  \hfill
\psovalbox{C}$1103$ \hfill \psovalbox{D} $1114$\hfill \psovalbox{E}
none of these \vfill
  \item[] {\bf Multiplication}
 \vfill
\item  $2^2\cdot 3 + 2\cdot 3^2=$ \\
\vfill \psovalbox{A} $30$ \hfill  \psovalbox{B}  $24$ \hfill
\psovalbox{C} $126$ \hfill  \psovalbox{D} $216$ \hfill \psovalbox{E}
none of these  \vfill

 \item  Multiply: $(3x)(5x^2)$. \\
\vfill \psovalbox{A} $15x^3$ \hfill  \psovalbox{B}  $8x^3$ \hfill
\psovalbox{C} $15x^2$ \hfill  \psovalbox{D} $15x^4$ \hfill
\psovalbox{E} none of these  \vfill\item  Multiply: $(2x+1)(x+3)$. \\
\vfill\psovalbox{A} $2x^2+7x+3$ \hfill \psovalbox{B} $2x^2+3$ \hfill
\psovalbox{C} $3x+4$ \hfill \psovalbox{D} $2x^2+4$ \hfill
\psovalbox{E} none of these  \vfill
\item  When writing all the natural numbers from $100$ to $500$ inclusive, how many digits have I used? \\
\vfill \psovalbox{A} $1203$ \hfill  \psovalbox{B}  $1200$ \hfill
\psovalbox{C} $1500$ \hfill  \psovalbox{D} $400$ \hfill
\psovalbox{E} none of these  \vfill
\item According to experts the first $4$ moves in a chess game can be played in $197299$
totally different ways. If it takes $30$ seconds to make one move,
what time, in seconds, would it take one player to try every
possible set of $4$ moves?\\
\vfill \psovalbox{A} $5918970$ \hfill  \psovalbox{B}  $23675880$
\hfill \psovalbox{C} $120$ \hfill  \psovalbox{D} $789196$ \hfill
\psovalbox{E} none of these  \vfill

\item Evaluate: $100 - (5 + 1)^2 + 3  \cdot  2.$\\ \psovalbox{A}$70$ \hfill \psovalbox{B}   $94$ \hfill
\psovalbox{C}  $80$ \hfill \psovalbox{D}  $82$ \hfill \psovalbox{E}
none of these

 \item[] {\bf Primes and Factorisation}
 \vfill
\item What is the prime factorisation of $252$? \\
\vfill \psovalbox{A} $2^2\cdot 3^3 \cdot 7$ \hfill  \psovalbox{B}
$2^2\cdot 3^2\cdot 7$ \hfill \psovalbox{C} $2^2\cdot 63$ \hfill
\psovalbox{D} $2\cdot 3 \cdot 7$ \hfill \psovalbox{E} none of these
 \vfill
\item What is the prime factorisation of $360$? \\
\vfill \psovalbox{A} $2^3\cdot 3^2 \cdot 5$ \hfill  \psovalbox{B}
$2^2\cdot 3^2\cdot 5$ \hfill \psovalbox{C} $2^3\cdot 45$ \hfill
\psovalbox{D} $2\cdot 3 \cdot 5$ \hfill \psovalbox{E} none of these
 \vfill
 \item  Find $\gcd (252,360)$. \\
\vfill \psovalbox{A} $36$ \hfill  \psovalbox{B}  $72$ \hfill
\psovalbox{C} $216$ \hfill  \psovalbox{D} $2520$ \hfill
\psovalbox{E} none of these \vfill  \clearpage
 \item  Find $\lcm (252,360)$. \\
\vfill \psovalbox{A} $2530$ \hfill  \psovalbox{B}  $720$ \hfill
\psovalbox{C} $2160$ \hfill  \psovalbox{D} $90720$ \hfill
\psovalbox{E} none of these  \vfill




\item Find: LCM$(14, 20, 36)$.\\ \psovalbox{A}2
\hfill \psovalbox{B}   4 \hfill \psovalbox{C}  630 \hfill
\psovalbox{D}  1260 \hfill \psovalbox{E} none of these  \vfill

\item How many primes are there in the set $\{90,91,92,93,94,95,96,97,98,99\}$?\\ \psovalbox{A} $1$
\hfill \psovalbox{B}  $2$ \hfill \psovalbox{C}  $3$ \hfill
\psovalbox{D}  $4$ \hfill \psovalbox{E} none of these  \vfill

\item[] {\bf Division}
 \vfill
\item  $100\div 50 \div 2=$ \\
\vfill \psovalbox{A} $1$ \hfill  \psovalbox{B}  $4$ \hfill
\psovalbox{C} $2$ \hfill  \psovalbox{D} $8$ \hfill \psovalbox{E}
none of these  \vfill
\item  $100\div 50 \cdot 2=$ \\
\vfill \psovalbox{A} $4$ \hfill  \psovalbox{B}  $1$ \hfill
\psovalbox{C} $8$ \hfill  \psovalbox{D} $10$ \hfill \psovalbox{E}
none of these  \vfill
\item Evaluate: $80 \div (5 - 3)^3$.\\ \psovalbox{A}$8$ \hfill \psovalbox{B}   $10$ \hfill \psovalbox{C}
$13$ \hfill \psovalbox{D}  $11$ \hfill \psovalbox{E} none of these

\item If today is Monday, what day will it be  $300$ days from now? \\
\vfill \psovalbox{A} Sunday \hfill  \psovalbox{B} Saturday \hfill
\psovalbox{C} Tuesday \hfill  \psovalbox{D} Wednesday \hfill
\psovalbox{E} none of these  \vfill
\item Amanda and Brenda were selling girl scout cookies. Each
package they sold was priced the same. Amanda sold $30$ packages,
Brenda sold $50$ packages, and they made \$ $320$. How much money
did
Amanda make? \\
\vfill \psovalbox{A} \$ $120$ \hfill  \psovalbox{B} \$  $200$ \hfill
\psovalbox{C} \$  $4$ \hfill  \psovalbox{D} \$  $300$ \hfill
\psovalbox{E} none of these  \vfill
\item Which value of the digit  $d$ makes the number $1d95$ divisible by  $15$?  \\
\vfill \psovalbox{A} $6$ \hfill  \psovalbox{B} $1$ \hfill
\psovalbox{C} $5$ \hfill  \psovalbox{D} $2$ \hfill \psovalbox{E}
none of these  \vfill
\item  Which of the following numbers is divisible by $6$? \\
$$ I: 54  \qquad  II: 56  \qquad III: 62\qquad IV: 702   $$

\vfill \psovalbox{A} $54$ \hfill  \psovalbox{B}  $56$ \hfill
\psovalbox{C} $62$ \hfill  \psovalbox{D} $702$ \hfill \psovalbox{E}
none of these  \vfill
\item[] {\bf Long Division}
 \vfill
\item   $102030405060708090\div 5=$  \\
\vfill \psovalbox{A} $20406081012141618$ \vfill  \psovalbox{B}
$24681012141618$ \vfill \psovalbox{C} $204060810012141618$ \vfill
\psovalbox{D} $20040060081012141618$ \vfill \psovalbox{E} none of
these  \vfill


\item Divide:  $(20x^6)\div (5x^2)$  \\
\vfill \psovalbox{A} $4x^4$ \hfill  \psovalbox{B} $4x^3$ \hfill
\psovalbox{C} $15x^2$ \hfill  \psovalbox{D} $15x^4$ \hfill
\psovalbox{E} none of these  \vfill
\item Divide:  $\dfrac{6x+2x^2}{2x}$  \\
\vfill \psovalbox{A} $3+x$ \hfill  \psovalbox{B} $3x+x^2$ \hfill
\psovalbox{C} $3x$ \hfill  \psovalbox{D} $3+2x^2$ \hfill
\psovalbox{E} none of these \vfill

\item $\dfrac{3^{4}}{3^{3}}=$\\ \vfill
\psovalbox{A} $3$ \hfill \psovalbox{B} $3^{12}$ \hfill \psovalbox{C}
$\dfrac{1}{3}$  \hfill \psovalbox{D} $3^7$ \hfill \psovalbox{E} none
of these

\item $(3^3)^4=$\\ \vfill
\psovalbox{A} $3$ \hfill \psovalbox{B} $3^{12}$ \hfill \psovalbox{C}
$\dfrac{1}{3}$  \hfill \psovalbox{D} $3^7$ \hfill \psovalbox{E} none
of these
\item $3^3\cdot 3^4=$\\ \vfill
\psovalbox{A} $3$ \hfill \psovalbox{B} $3^{12}$ \hfill \psovalbox{C}
$\dfrac{1}{3}$  \hfill \psovalbox{D} $3^7$ \hfill \psovalbox{E} none
of these
\item $\dfrac{1111^5+1111^5+1111^5+1111^5}{1111^4+1111^4}=$\\
\\ \vfill \psovalbox{A} $1111$ \hfill \psovalbox{B}$2222$
\hfill \psovalbox{C} $4444$ \hfill \psovalbox{D} $\dfrac{1}{2222}$
\hfill \psovalbox{E} none of these

\item[] {\bf Roots}  \vfill
 \item  $\sqrt{100}+\sqrt{64}$
\\ \vfill\psovalbox{A} $18$ \hfill \psovalbox{B}
$80$ \hfill  \psovalbox{C}  $1600$ \hfill \psovalbox{D} $82$ \hfill
\psovalbox{E} none of these  \vfill
\item  $\sqrt{100}\cdot \sqrt{64}$
\\ \vfill\psovalbox{A} $18$ \hfill \psovalbox{B}
$80$ \hfill  \psovalbox{C}  $1600$ \hfill \psovalbox{D} $82$ \hfill
\psovalbox{E} none of these  \vfill

\item  $\sqrt{9}+\sqrt{16}$
\\ \vfill\psovalbox{A} $\sqrt{25}$ \hfill \psovalbox{B}
$7$ \hfill  \psovalbox{C}  $12$ \hfill \psovalbox{D}
$12\tfrac{1}{2}$ \hfill \psovalbox{E} none of these  \vfill
\item  $\sqrt{9}\cdot \sqrt{16}$
\\ \vfill\psovalbox{A} $36$ \hfill \psovalbox{B}
$7$ \hfill  \psovalbox{C}  $12$ \hfill \psovalbox{D}
$12\tfrac{1}{2}$ \hfill \psovalbox{E} none of these  \vfill

  \item[] {\bf Fractions}
 \vfill
\item $\dfrac{3^{3}}{3^{4}}=$\\ \vfill
\psovalbox{A} $3$ \hfill \psovalbox{B} $3^{12}$ \hfill \psovalbox{C}
$\dfrac{1}{3}$  \hfill \psovalbox{D} $3^7$ \hfill \psovalbox{E} none
of these
\item  How many of the following statements are {\bf true}?
$$I: \dfrac{3}{4}=\dfrac{5}{6}\qquad II: \dfrac{3}{4}<\dfrac{5}{6}\qquad III: \dfrac{6}{8}=\dfrac{15}{20}\qquad
IV: 4\tfrac{1}{3}=\dfrac{4}{3}  $$ \vfill \psovalbox{A} none \hfill
\psovalbox{B} exactly one \hfill \psovalbox{C} exactly two \hfill
\psovalbox{D} exactly three \hfill \psovalbox{E} all four  \vfill
 \item  What fraction of an hour  is  $18$ minutes? \\
\vfill \psovalbox{A} $\dfrac{3}{4}$ \hfill  \psovalbox{B}
$\dfrac{1}{2}$ \hfill \psovalbox{C} $\dfrac{3}{10}$ \hfill
\psovalbox{D} $\dfrac{5}{6}$ \hfill \psovalbox{E} none of these
 \vfill
\item  Express  $\dfrac{6}{132}$ in least terms. \\
\vfill\psovalbox{A} $\dfrac{1}{11}$ \hfill \psovalbox{B}
$\dfrac{1}{22}$ \hfill \psovalbox{C} $\dfrac{1}{44}$ \hfill
\psovalbox{D} $\dfrac{1}{66}$ \hfill \psovalbox{E} none of these
\vfill
\item  Convert the improper fraction to a mixed number: $\dfrac{43}{8}$ \\
\vfill \psovalbox{A} $8\tfrac{5}{8}$ \hfill  \psovalbox{B}
$8\tfrac{3}{5}$ \hfill \psovalbox{C} $5\tfrac{3}{8}$ \hfill
\psovalbox{D} $8\tfrac{3}{8}$ \hfill \psovalbox{E} none of these
 \vfill
\item Convert the mixed number to an improper fraction: $2\tfrac{3}{4}$\\
\vfill \psovalbox{A} $\dfrac{6}{4}$ \hfill  \psovalbox{B}
$\dfrac{10}{4}$ \hfill \psovalbox{C} $\dfrac{11}{4}$ \hfill
\psovalbox{D} $\dfrac{4}{11}$ \hfill \psovalbox{E} none of these
 \vfill

\item How many  of the following assertions  are {\bf false}?
$$I: \frac{1}{2} = \frac{2}{1} \qquad II: \frac{3}{5} < \frac{4}{7} \qquad III: \frac{1}{2} <
\frac{1}{3} \qquad IV: \frac{3}{1} < \frac{2}{2}  $$

 \psovalbox{A} exactly one \hfill \psovalbox{B}
exactly two \hfill \psovalbox{C}  exactly three \hfill \psovalbox{D}
all four   \hfill \psovalbox{E} none of them  \vfill
\item  How many of the following statements are {\bf true}?
$$I: \dfrac{3}{4}=\dfrac{15}{20}\qquad II: \dfrac{3}{4}>\dfrac{2}{5}\qquad III: \dfrac{2}{4}=\dfrac{3}{6}\qquad
IV: 5\tfrac{2}{3}=\dfrac{17}{3}  $$ \vfill \psovalbox{A} none \hfill
\psovalbox{B} exactly one \hfill \psovalbox{C} exactly two \hfill
\psovalbox{D} exactly three \hfill \psovalbox{E} all four  \vfill
 \item  What fraction of a day is  $18$ hours? \\
\vfill \psovalbox{A} $\dfrac{3}{4}$ \hfill  \psovalbox{B}
$\dfrac{1}{2}$ \hfill \psovalbox{C} $\dfrac{2}{3}$ \hfill
\psovalbox{D} $\dfrac{5}{6}$ \hfill \psovalbox{E} none of these
 \vfill
\item  Express  $\dfrac{104}{140}$ in least terms. \\
\vfill\psovalbox{A} $\dfrac{26}{35}$ \hfill \psovalbox{B}
$\dfrac{52}{70}$ \hfill \psovalbox{C} $\dfrac{13}{7}$ \hfill
\psovalbox{D} $\dfrac{10}{14}$ \hfill \psovalbox{E} none of these
\vfill
\item  Convert the improper fraction to a mixed number: $\dfrac{100}{7}$ \\
\vfill \psovalbox{A} $14\tfrac{2}{7}$ \hfill  \psovalbox{B}
$13\tfrac{6}{7}$ \hfill \psovalbox{C} $\dfrac{6}{5}$ \hfill
\psovalbox{D} $\dfrac{200}{14}$ \hfill \psovalbox{E} none of these
 \vfill
\item Convert the mixed number to an improper fraction: $3\tfrac{2}{5}$\\
\vfill \psovalbox{A} $\dfrac{17}{5}$ \hfill  \psovalbox{B}
$\dfrac{15}{5}$ \hfill \psovalbox{C} $\dfrac{13}{5}$ \hfill
\psovalbox{D} $\dfrac{5}{17}$ \hfill \psovalbox{E} none of these
 \vfill

\item How many  of the following assertions  are {\bf false}?
$$I: \frac{3}{13} = \frac{51}{169} \qquad II: \frac{2}{3} < \frac{4}{5} \qquad III: \frac{1}{15} <
\frac{1}{16} \qquad IV: \frac{7}{8} > \frac{1}{2}  $$

 \psovalbox{A} exactly one \hfill \psovalbox{B}
exactly two \hfill \psovalbox{C}  exactly three \hfill \psovalbox{D}
all four   \hfill \psovalbox{E} none of them  \vfill
\item How many of the following expressions are undefined?
$$ I: \dfrac{5}{5-5}\qquad II: \dfrac{5-5}{5} \qquad III: \dfrac{1}{1^3-2^0} \qquad IV: \dfrac{0}{0}$$
\psovalbox{A} exactly one \hfill \psovalbox{B}   exactly two \hfill
\psovalbox{C} exactly three \hfill \psovalbox{D}  all four \hfill
\psovalbox{E} none of them
\item How many of the following expressions are zero?
$$ I: \dfrac{5}{5-5}\qquad II: \dfrac{5-5}{5} \qquad III: \dfrac{1}{1^3-2^0} \qquad IV: \dfrac{0}{0}$$
\psovalbox{A} exactly one \hfill \psovalbox{B}   exactly two \hfill
\psovalbox{C} exactly three \hfill \psovalbox{D}  all four \hfill
\psovalbox{E} none of them


\item Expressed in least terms $\dis{\frac{24}{52}}$ is \\ \psovalbox{A}$\frac{8}{13}$
\hfill \psovalbox{B}   $\frac{6}{13}$ \hfill \psovalbox{C}
$\frac{2}{13}$ \hfill \psovalbox{D} $\frac{12}{26}$ \hfill
\psovalbox{E} none of these
\item Find the missing numerator: $\dis{\frac{7}{12} = \frac{?}{60}}$
\\ \psovalbox{A}7
\hfill \psovalbox{B}   5 \hfill \psovalbox{C}  35 \hfill
\psovalbox{D}  12 \hfill \psovalbox{E}  none of these

\item Expressed in least terms $\dis{\frac{221}{247}}$ is \\ \psovalbox{A}$\frac{21}{47}$
\hfill \psovalbox{B}   $\frac{73}{82}$ \hfill \psovalbox{C}
$\frac{17}{19}$ \hfill \psovalbox{D}  $\frac{2}{3}$ \hfill
\psovalbox{E} none of these
\item Find the missing numerator: $\dis{\frac{7}{12} = \frac{?}{60}}$ \\ \psovalbox{A}7
\hfill \psovalbox{B}   5 \hfill \psovalbox{C}  35 \hfill
\psovalbox{D}  12 \hfill \psovalbox{E} none of these
\item How many  of the following fractions are {\bf not} equivalent to $\frac{3}{7}$?
$$I: \frac{9}{21}\qquad II: \frac{15}{35}\qquad III: \frac{2}{5} \qquad IV: \frac{51}{119}  $$
 \psovalbox{A} exactly one \hfill \psovalbox{B}
exactly two \hfill \psovalbox{C}  exactly three \hfill \psovalbox{D}
all four   \hfill \psovalbox{E} none of them
\item Three of the following fractions represent the same number. Which is the odd one out?\\ \psovalbox{A}$\frac{9}{12}$ \hfill \psovalbox{B}
$\frac{2}{3}$  \hfill \psovalbox{C}  $\frac{150}{225}$
 \hfill \psovalbox{D}  $\frac{14}{21}$ \hfill \psovalbox{E} none of these
\item Change into an improper fraction: \[2\tfrac{5}{7}\]
\\ \psovalbox{A}$\frac{17}{7}$ \hfill \psovalbox{B}   $\frac{19}{7}$
\hfill \psovalbox{C}  $\frac{25}{7}$ \hfill \psovalbox{D}
$\frac{14}{7}$ \hfill \psovalbox{E} none of these
\item Express as a mixed numeral:  $\dis{\frac{24}{7}}$
\\ \psovalbox{A}$3\frac{3}{7}$ \hfill \psovalbox{B}   $2\frac{3}{7}$
 \hfill \psovalbox{C}  $\frac{7}{24}$
\hfill \psovalbox{D} 3 \hfill \psovalbox{E} none of these
\item Write a mixed fraction for $33 \div 4$.\\ \psovalbox{A}$8\frac{1}{4}$
\hfill \psovalbox{B}   $8.14$ \hfill \psovalbox{C}  $\frac{33}{4}$
\hfill \psovalbox{D}  $\frac{4}{33}$ \hfill \psovalbox{E} none of
these
 \item[] {\bf Addition and Subtraction of Fractions}

\item Add the fractions:  $\dfrac{2}{5} + \dfrac{3}{4}$ \\
\vfill \psovalbox{A} $\dfrac{5}{9}$ \hfill  \psovalbox{B}
$\dfrac{1}{4}$ \hfill \psovalbox{C} $\dfrac{5}{20}$ \hfill
\psovalbox{D} $\dfrac{23}{20}$ \hfill \psovalbox{E} none of these
\item At the beginning of the week, a bookstore sold $\dfrac{3}{7}$ of its inventory of the latest
{\em Hairy Puker} book. In the middle of the week, it sold
$\dfrac{3}{8}$ further. No more such books were sold during the rest
of the week. What fraction of the inventory of {\em Hairy Puker}
books remain at the end of the week? \\ \vfill \psovalbox{A}
$\dfrac{45}{56}$ \hfill \psovalbox{B} $\dfrac{11}{56}$ \hfill
\psovalbox{C} $\dfrac{3}{56}$ \hfill \psovalbox{D} $\dfrac{6}{15}$
\hfill \psovalbox{E} none of these

\item Subtract the fractions: $\dfrac{11}{20}-\dfrac{2}{15}$ \\
\vfill \psovalbox{A} $\dfrac{9}{5}$ \hfill  \psovalbox{B}
$\dfrac{5}{12}$ \hfill \psovalbox{C} $\dfrac{9}{12}$ \hfill
\psovalbox{D} $\dfrac{3}{4}$ \hfill \psovalbox{E} none of these
 \item  Calculate: $3\tfrac{3}{4}-2\tfrac{1}{2}$. \\
\vfill \psovalbox{A} $\dfrac{14}{15}$ \hfill  \psovalbox{B}
$1\tfrac{3}{5}$ \hfill \psovalbox{C} $\dfrac{3}{2}$ \hfill
\psovalbox{D} $\dfrac{5}{4}$ \hfill \psovalbox{E} none of these

 \item  Calculate: $\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{9}$. \\
\vfill \psovalbox{A} $\dfrac{1}{12}$ \hfill  \psovalbox{B}
$\dfrac{7}{18}$ \hfill \psovalbox{C} $\dfrac{1}{18}$ \hfill
\psovalbox{D} $\dfrac{5}{18}$ \hfill \psovalbox{E} none of these


\item Add the fractions:  $\dfrac{3}{7} + \dfrac{1}{4}$ \\
\vfill \psovalbox{A} $\dfrac{19}{28}$ \hfill  \psovalbox{B}
$\dfrac{4}{11}$ \hfill \psovalbox{C} $\dfrac{3}{28}$ \hfill
\psovalbox{D} $\dfrac{4}{28}$ \hfill \psovalbox{E} none of these
\vfill
\item Subtract the fractions: $\dfrac{5}{6}-\dfrac{2}{15}$ \\
\vfill \psovalbox{A} $\dfrac{7}{10}$ \hfill  \psovalbox{B}
$\dfrac{3}{9}$ \hfill \psovalbox{C} $\dfrac{1}{9}$ \hfill
\psovalbox{D} $\dfrac{29}{30}$ \hfill \psovalbox{E} none of these
 \vfill
 \item  Calculate: $2\tfrac{1}{3}-1\tfrac{2}{5}$. \\
\vfill \psovalbox{A} $\dfrac{14}{15}$ \hfill  \psovalbox{B}
$1\tfrac{1}{2}$ \hfill \psovalbox{C} $1\tfrac{1}{14}$ \hfill
\psovalbox{D} $1\tfrac{1}{15}$ \hfill \psovalbox{E} none of these
 \vfill
 \item  Calculate: $\dfrac{2}{5}-\dfrac{1}{3}+\dfrac{1}{10}$. \\
\vfill \psovalbox{A} $\dfrac{1}{6}$ \hfill  \psovalbox{B}
$\dfrac{1}{12}$ \hfill \psovalbox{C} $\dfrac{1}{75}$ \hfill
\psovalbox{D} $\dfrac{1}{3}$ \hfill \psovalbox{E} none of these
 \vfill
\item Write as a mixed numeral in lowest terms: $$
4\tfrac{35}{50} +  3\tfrac{15}{60} $$
\\ \psovalbox{A}$\frac{159}{20}$
\hfill \psovalbox{B}   $7\frac{190}{200} $ \hfill \psovalbox{C}
$7\frac{1900}{3000} $ \hfill \psovalbox{D}  $7\frac{19}{20}$ \hfill
\psovalbox{E} none of these \vfill
\item Write as a mixed numeral in lowest terms: $$5\tfrac{35}{50}   -
2\tfrac{15}{60}$$
\\ \psovalbox{A}$3\frac{9}{20}$
\hfill \psovalbox{B}   $\frac{69}{20}$ \hfill \psovalbox{C}
$3\frac{20}{10} $ \hfill \psovalbox{D}  $3\frac{1350}{3000} $ \hfill
\psovalbox{E} none of these \vfill

\item[] {\bf Multiplication and Division of Fractions}
\vfill
\item Reduce to simplest form: $\dis{3\frac{1}{3}\div 24}$
 \vspace{2mm}
\\ \psovalbox{A}$\frac{1}{80}$
\hfill \psovalbox{B}   $\frac{5}{36}$ \hfill \psovalbox{C}
$7\frac{1}{5}$ \hfill \psovalbox{D}  $\frac{6}{43}$ \hfill
\psovalbox{E}  none of these \vfill
\item What is the product of a non-zero fraction and its reciprocal?
\\ \psovalbox{A}$0$
\hfill \psovalbox{B}   $1$ \hfill \psovalbox{C} $\frac{1}{2} $
\hfill \psovalbox{D}  $2 $ \hfill \psovalbox{E} impossible to
determine with the given information \vfill

\item Find the reciprocal of $2\frac{3}{4}$.
\\ \psovalbox{A}$2\frac{4}{3}$
\hfill \psovalbox{B}   $\frac{4}{11}$ \hfill \psovalbox{C}
$\frac{11}{4} $ \hfill \psovalbox{D}  $3\frac{4}{11} $ \hfill
\psovalbox{E} none of these  \vfill

\item Simplify as a mixed numeral in lowest terms:
$$\left(2\frac{2}{3}\right) \cdot  \left(3\frac{3}{4}\right).$$
\\ \psovalbox{A}$10$ \hfill \psovalbox{B}
$6\frac{6}{12} $ \hfill \psovalbox{C}  $\frac{5}{4}$ \hfill
\psovalbox{D}  $8\frac{3}{4}$ \hfill \psovalbox{E} none of these
 \vfill

\item Calculate: $\dfrac{4}{5}\cdot \dfrac{5}{6}+\dfrac{4}{5}\div \dfrac{5}{6}$   \\
\vfill \psovalbox{A} $\dfrac{4}{3}$ \hfill  \psovalbox{B}
$\dfrac{16}{75}$ \hfill \psovalbox{C} $\dfrac{49}{15}$\hfill
\psovalbox{D} $\dfrac{122}{75}$ \hfill \psovalbox{E} none of these
\vfill
\item Calculate: $\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{\dfrac{1}{2}-\dfrac{1}{3}}$ \\
\vfill \psovalbox{A} $5$ \hfill  \psovalbox{B} $\dfrac{1}{5}$\hfill
\psovalbox{C} $1$ \hfill \psovalbox{D} $\dfrac{2}{5}$\hfill
\psovalbox{E} none of these  \vfill
\item Calculate: $\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}$ \\
\vfill \psovalbox{A} $\dfrac{3}{2}$ \hfill  \psovalbox{B}
$\dfrac{2}{5}$ \hfill \psovalbox{C} $\dfrac{5}{3}$ \hfill
\psovalbox{D} $\dfrac{3}{5}$ \hfill \psovalbox{E} none of these
 \vfill



 \vfill
\item Find the reciprocal of $5\frac{2}{3}$.
\\ \psovalbox{A}$5\frac{3}{2}$
\hfill \psovalbox{B}   $6\frac{1}{2}$ \hfill \psovalbox{C}
$\frac{17}{3} $ \hfill \psovalbox{D}  $\frac{3}{17} $ \hfill
\psovalbox{E} none of these  \vfill

\item Simplify as a mixed numeral in lowest terms: $$\left(8\frac{1}{8}\right) \cdot  \left(1\frac{1}{13}\right).$$\\ \psovalbox{A}$\frac{1}{13}$ \hfill \psovalbox{B}
$8\frac{1}{104} $ \hfill \psovalbox{C}  $\frac{35}{4}$ \hfill
\psovalbox{D}  $8\frac{3}{4}$ \hfill \psovalbox{E} none of these
 \vfill

\item Calculate: $\dfrac{2}{3}\cdot \dfrac{3}{5}+\dfrac{2}{3}\div \dfrac{5}{4}$   \\
\vfill \psovalbox{A} $\dfrac{14}{15}$ \hfill  \psovalbox{B}
$\dfrac{16}{75}$ \hfill \psovalbox{C} $\dfrac{37}{30}$\hfill
\psovalbox{D} $\dfrac{377}{180}$ \hfill \psovalbox{E} none of these
 \vfill
\item Calculate: $\dfrac{\dfrac{3}{4}-\dfrac{2}{3}}{\dfrac{3}{4}+\dfrac{2}{3}}$ \\
\vfill \psovalbox{A} $\dfrac{1}{17}$ \hfill  \psovalbox{B}
$\dfrac{19}{36} $\hfill \psovalbox{C} $1$ \hfill \psovalbox{D}
$\dfrac{17}{144}$\hfill \psovalbox{E} none of these  \vfill
\item Calculate: $\dfrac{1}{2-\dfrac{1}{3-\dfrac{1}{4}}}$ \\
\vfill \psovalbox{A} $\dfrac{11}{18}$ \hfill  \psovalbox{B}
$\dfrac{3}{2}$ \hfill \psovalbox{C} $\dfrac{7}{18}$ \hfill
\psovalbox{D} $\dfrac{13}{18}$ \hfill \psovalbox{E} none of these
 \vfill

\item[] {\bf Ratio and Proportion}
\item Solve the proportion for $n$ $$\frac{5}{6} = \frac{60}{n} $$
\\ \psovalbox{A} $72$  \hfill \psovalbox{B}   $50$\ \hfill
\psovalbox{C}  $\dfrac{1}{2}$ \hfill \psovalbox{D}  $12$ \hfill
\psovalbox{E} none of these  \vfill
\item  A recipe for five people takes four eggs. Assuming proportional ingredients available, how many people will twenty eggs serve?
\vfill \psovalbox{A} $21$ \hfill  \psovalbox{B} $10$ \hfill
\psovalbox{C} $25$ \hfill \psovalbox{D} $15$ \hfill \psovalbox{E}
none of these  \vfill
\item John takes $8$ hours to fill  a hole,   Atish takes $4$ hours to fill the same hole, and
David takes $2$ hours to fill the same hole. If the three are
working together, how many hours will it take them to fill the hole? \\
\vfill \psovalbox{A} $1\tfrac{1}{8}$ hours \hfill  \psovalbox{B}
$1\tfrac{1}{2}$ hours \hfill \psovalbox{C} $\dfrac{7}{8}$ hours
\hfill \psovalbox{D} $1\tfrac{1}{7}$ hours \hfill \psovalbox{E} none
of these \vfill
\item 120 ft = ? in
\\ \psovalbox{A}10
\hfill \psovalbox{B}   1440 \hfill \psovalbox{C}  144 \hfill
\psovalbox{D}  100 \hfill \psovalbox{E}  none of these


\item 15840 ft = ? mi
\\ \psovalbox{A}3
\hfill \psovalbox{B}   $\frac{1}{3}$ \hfill \psovalbox{C}  9 \hfill
\psovalbox{D}  27878400 \hfill \psovalbox{E}  none of these
\item 120 ft = ? yd
\\ \psovalbox{A}10
\hfill \psovalbox{B}   40 \hfill \psovalbox{C}  1440 \hfill
\psovalbox{D}  360 \hfill \psovalbox{E}  none of these


\item Solve the proportion for $n$ $$\frac{39}{42} = \frac{13}{n} $$
\\ \psovalbox{A} $12\tfrac{7}{100}$  \hfill \psovalbox{B}   $3$\ \hfill
\psovalbox{C}  $14$ \hfill \psovalbox{D}  $126$ \hfill \psovalbox{E}
none of these  \vfill
\item  A recipe for seven people takes six eggs. Assuming proportional ingredients available, how many people will eighteen eggs serve?
\vfill \psovalbox{A} $21$ \hfill  \psovalbox{B} $10$ \hfill
\psovalbox{C} $15$ \hfill \psovalbox{D} $16$ \hfill \psovalbox{E}
none of these  \vfill

\item John takes $3$ hours drive her crazy and Atish takes $2$ hours to drive her crazy. If John and Atish are working together, how long will it take them
to drive her crazy, assuming proportional time?  \\
\vfill \psovalbox{A} $1\tfrac{1}{5}$ hours \hfill  \psovalbox{B}
$1\tfrac{1}{2}$ hours \hfill \psovalbox{C} $1\tfrac{1}{3}$ hours
\hfill \psovalbox{D} $1\tfrac{1}{4}$ hours \hfill \psovalbox{E} none
of these \vfill


\item Find $n$ if $\dis{\frac{8}{7} = \frac{n}{20}}$
\\ \psovalbox{A}19\ \ \hfill \psovalbox{B}   1120\ \ \hfill
\psovalbox{C} $22\frac{6}{7}$\ \ \hfill \psovalbox{D}  17.5 \hfill
\psovalbox{E} none of these

\vfill
\item Express the ratio $40:112$ in lowest terms.
\\ \psovalbox{A}$\frac{20}{56}$  \hfill \psovalbox{B}
$\frac{10}{28}$\ \ \hfill \psovalbox{C}  $\frac{5}{14}$\ \ \hfill
\psovalbox{D} $\frac{80}{224}$ \hfill \psovalbox{E}  none of these
\vfill

\item[] {\bf Sets and Inclusion-Exclusion}
\item  If $X = \{1,2,3,4,5\}$, $Y = \{3,4,6,8,9\}$, what is $X\setminus Y$?\\
\vfill \psovalbox{A}   $\{1,2,5\}$\hfill \psovalbox{B}
$\{6,8,9\}$\hfill \psovalbox{C} $\{1,2,5,6,8,9\}$\hfill
\psovalbox{D} $\varnothing$ \hfill \psovalbox{E} none of these
\item  If $X = \{1,2,3,4,5\}$, $Y = \{3,4,6,8,9\}$, what is $Y\setminus X$?\\
\vfill \psovalbox{A}   $\{1,2,5\}$\hfill \psovalbox{B}
$\{6,8,9\}$\hfill \psovalbox{C} $\{1,2,5,6,8,9\}$\hfill
\psovalbox{D} $\varnothing$ \hfill \psovalbox{E} none of these
\item Given the  sets
$$A = \{a,c,d\},  \ B = \{c,d, e,g\},  \ C = \{a,b,e,f\} ,
$$ find  $A\cap B \cap C$.\\

\vfill \psovalbox{A} $\varnothing$ \hfill\psovalbox{B}
$\{c,d\}$\hfill\psovalbox{C}  $\{a\}$\hfill \psovalbox{D}
$\{b,c,e,f,g,h\}$ \hfill \psovalbox{E}  none of these
 \vfill
\item[] {\bf
Situation:} Of the $15$ students in a class, $5$ speak Chinese, $7$
speak Spanish, $7$ speak French, $3$ speak Spanish and Chinese, $4$
speak Spanish and French, and $3$ speak Chinese and French. One
student speaks all three languages.
 Questions \ref{pro:inclu_exclu_begin2} to
\ref{pro:inclu_exclu_end2} refer to this situation, and you may wish
to fill in the regions of the diagram below in order to aid your
reasoning. \vspace{2cm}

\begin{center}
\psset{unit=1pc} \pscircle(-1,0){2} \pscircle(1,0){2}
\pscircle(0,1.41421356){2} \rput(-4.5, 0){\tiny French} \rput(4.5,
0){\tiny Spanish} \rput(0, 4.2){\tiny Chinese}
\end{center} \vspace{1cm}

 \vfill

\item \label{pro:inclu_exclu_begin2} How many students speak French and Spanish, but not Chinese? \\

\psovalbox{A}   $1$\hfill \psovalbox{B}   $2$ \hfill
\psovalbox{C} $3$\hfill \psovalbox{D}$4$ \hfill \psovalbox{E} none
of these


 \item  How many students speak only French?\\
 \psovalbox{A}   $1$\hfill \psovalbox{B}   $2$ \hfill
\psovalbox{C} $3$\hfill \psovalbox{D}$4$ \hfill \psovalbox{E} none
of these
\item  How many students speak exactly two of these three languages?\\

 \psovalbox{A}   $4$\hfill \psovalbox{B}   $5$ \hfill
\psovalbox{C} $6$\hfill \psovalbox{D}$7$ \hfill \psovalbox{E} none
of these
\item  How many students speak exactly one of these three languages?\\

\psovalbox{A}   $0$\hfill \psovalbox{B}   $1$ \hfill
\psovalbox{C} $2$\hfill \psovalbox{D}$3$ \hfill \psovalbox{E} none
of these

\item  \label{pro:inclu_exclu_end2}  How many students speak none of these three languages?\\

 \psovalbox{A}   $10$\hfill \psovalbox{B}   $5$ \hfill
\psovalbox{C} $2$\hfill \psovalbox{D}$3$ \hfill \psovalbox{E} none
of these






\item[] {\bf Decimals}
\item  Convert to a decimal: $2\tfrac{3}{4}$.
\\ \vfill\psovalbox{A} $2.34$ \hfill \psovalbox{B}
$2.25$ \hfill  \psovalbox{C}  $2.75$ \hfill \psovalbox{D} $2.034$
\hfill \psovalbox{E} none of these
\item  Convert to a decimal: $\dfrac{9}{11}$.
\\ \vfill\psovalbox{A} $0.8\overline{1}$ \hfill \psovalbox{B}
$0.\overline{911}$ \hfill  \psovalbox{C}  $0.81818181818$ \hfill
\psovalbox{D} $0.\overline{81}$ \hfill \psovalbox{E} none of these

\item  Convert to an improper fraction: $2.34$.
\\ \vfill\psovalbox{A} $\dfrac{117}{100}$ \hfill \psovalbox{B}
$\dfrac{234}{50}$ \hfill  \psovalbox{C}  $\dfrac{3}{2}$ \hfill
\psovalbox{D} $\dfrac{117}{50}$ \hfill \psovalbox{E} none of these

\item  Determine the largest number among the following:
$$ A: \dfrac{1}{6}\qquad B: 0.16\overline{7}\qquad C: 0.\overline{17} \qquad D: 0.\overline{16} \qquad E: 0.1\overline{7} $$
\\ \vfill\psovalbox{A} \hfill \psovalbox{B}
\hfill  \psovalbox{C}  \hfill \psovalbox{D}  \hfill \psovalbox{E}

\item  Convert to a decimal: $3\tfrac{3}{5}$.
\\ \vfill\psovalbox{A} $3.6$ \hfill \psovalbox{B}
$3.35$ \hfill  \psovalbox{C}  $3.3$ \hfill \psovalbox{D} $3.06$
\hfill \psovalbox{E} none of these
\item  Convert to a decimal: $2\tfrac{1}{4}\%$.
\\ \vfill\psovalbox{A} $0.0225$ \hfill \psovalbox{B}
$2.25$ \hfill  \psovalbox{C}  $2.14$ \hfill \psovalbox{D} $0.00225$
\hfill \psovalbox{E} none of these
\item  Convert to a decimal: $\dfrac{2}{15}$.
\\ \vfill\psovalbox{A} $0.1\overline{3}$ \hfill \psovalbox{B}
$0.\overline{13}$ \hfill  \psovalbox{C}  $0.1333$ \hfill
\psovalbox{D} $0.13333333$ \hfill \psovalbox{E} none of these
\item  Convert to a fraction: $1.12$.
\\ \vfill\psovalbox{A} $\dfrac{28}{25}$ \hfill \psovalbox{B}
$\dfrac{26}{25}$ \hfill  \psovalbox{C}  $\dfrac{3}{2}$ \hfill
\psovalbox{D} $\dfrac{6}{5}$ \hfill \psovalbox{E} none of these

\item  Determine the largest number among the following:
$$ A: \dfrac{19}{50}\qquad B: 0.3\overline{7}\qquad C: 0.\overline{37} \qquad D: \dfrac{37}{100} \qquad E: 0.377 $$
\\ \vfill\psovalbox{A} \hfill \psovalbox{B}
\hfill  \psovalbox{C}  \hfill \psovalbox{D}  \hfill \psovalbox{E}
\item Which of the following numbers is {\bf One hundred thousand five and three ten thousandths}?
\\ \psovalbox{A}$105003.0010$
\hfill \psovalbox{B}   $105000.0003$ \hfill \psovalbox{C}
$100005.0003$ \hfill \psovalbox{D}  $100005.00003$ \hfill
\psovalbox{E} none of these


\item Write as a decimal: $\dfrac{3}{16}$.
\\ \psovalbox{A}0.316 \hfill \psovalbox{B}   3.16
\hfill \psovalbox{C}  0.1875 \hfill \psovalbox{D}
$0.\overline{1875}$ \hfill \psovalbox{E} none of these
\item Write as a decimal: $\dfrac{3}{11}$.
\\ \psovalbox{A}0.311 \hfill \psovalbox{B}   3.11
\hfill \psovalbox{C}  0.272727273 \hfill \psovalbox{D}
$0.\overline{27}$ \hfill \psovalbox{E} none of these
\item Write $1.03$ as a percent.\\ \psovalbox{A}$1.03\%$\ \ \hfill \psovalbox{B}   $10.3\%$\ \
\hfill \psovalbox{C} $0.103\%$\ \ \hfill \psovalbox{D}  $103\%$
\hfill \psovalbox{E} none of these

\item Write $\frac{5}{7}$ as a percent.
\\ \psovalbox{A}$\frac{5}{700}\%$\ \ \hfill \psovalbox{B}
$\frac{5}{70}\%$\ \ \hfill \psovalbox{C}  $\frac{5}{7}\%$\ \ \hfill
\psovalbox{D} $71\frac{3}{7}\%$ \hfill \psovalbox{E}  none of these
\item $12$ minutes is what percent of an hour?
\\ \psovalbox{A}$12\%$  \hfill \psovalbox{B}   $20\%$  \hfill
\psovalbox{C} $1.2\% $  \hfill \psovalbox{D}  $5\%$

\item What percent of $552$ is $24$?
\\ \psovalbox{A}$23\%$\ \ \hfill \psovalbox{B}   $\frac{1}{23}\%$\ \
\hfill \psovalbox{C}  $4\frac{8}{23}\%$\ \ \hfill \psovalbox{D}
$2300\%$  \hfill \psovalbox{E}  none of these
\item Four of these are equal. Which one is the odd one out?
$$I: \frac{1}{3} + \frac{1}{6} \qquad  II: 50\% \qquad  III:0.5 \qquad  IV: \frac{3}{6} \qquad  V: \frac{1}{3} + \frac{2}{3}   $$
\\ \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C}   \hfill \psovalbox{D}  \hfill \psovalbox{E}
\item Write as a decimal: $\dis{\frac{3}{16}}$.
\\ \psovalbox{A}0.316 \hfill \psovalbox{B}   3.16
\hfill \psovalbox{C}  0.1875
\hfill \psovalbox{D}  $0.\overline{1875}$ \hfill \psovalbox{E}  none
of these



\item Write $\frac{1}{6}$ as a percent.\\
 \psovalbox{A}$.167\%$
\hfill \psovalbox{B}   $60\%$
\hfill \psovalbox{C}  $16\frac{2}{3}\%$
\hfill \psovalbox{D}  $600\%$ \hfill \psovalbox{E}  none of these


\item[] {\bf Addition, Subtraction, and Multiplication of Decimals}

 \item  $1.234+ 5.67=$
\\ \vfill\psovalbox{A} $6.904$ \hfill \psovalbox{B}
$6.94$ \hfill  \psovalbox{C}  $6.094$ \hfill \psovalbox{D} $6.0094$
\hfill \psovalbox{E} none of these
 \item  $45\% + 9.9\%=$
\\ \vfill\psovalbox{A} $0.549$ \hfill \psovalbox{B}
$54.9$ \hfill  \psovalbox{C}  $5.49$ \hfill \psovalbox{D} $0.0549$
\hfill \psovalbox{E} none of these


\item A janitor found that   $8$ out of $48$ light bulbs had to be replaced. What percent of
the lightbulbs needed replacing? \\ \psovalbox{A}$33\frac{1}{3}\%$\
\ \hfill \psovalbox{B} $16\frac{1}{3}\%$\ \ \hfill \psovalbox{C}
$16\frac{2}{3}\%$\ \ \hfill \psovalbox{D} $\frac{1}{6}\%$ \hfill
\psovalbox{E}  none of these

\item What percent of $408$ is $34$?
\\ \psovalbox{A}$\frac{1}{12}\%$
\hfill \psovalbox{B}   $\frac{1}{120}\%$ \hfill \psovalbox{C}
$\frac{1}{1200}\%$ \hfill \psovalbox{D}  $8\frac{1}{3}\%$ \hfill
\psovalbox{E}  none of these
 \item  $9.0909+ 2.0202=$
\\ \vfill\psovalbox{A} $10.1111$ \hfill \psovalbox{B}
$11.0111$ \hfill  \psovalbox{C}  $11.111$ \hfill \psovalbox{D}
$11.1111$ \hfill \psovalbox{E} none of these
 \item Convert to a decimal: $11\% + 11\%$.
\\ \vfill\psovalbox{A} $0.22$ \hfill \psovalbox{B}
$22.0$ \hfill  \psovalbox{C}  $2.2$ \hfill \psovalbox{D} $0.0022$
\hfill \psovalbox{E} none of these
\item Convert to a decimal: $(11\%) \cdot (11\%)$.
\\ \vfill\psovalbox{A} $0.121$ \hfill \psovalbox{B}
$121.0$ \hfill  \psovalbox{C}  $1.21$ \hfill \psovalbox{D} $0.0121$
\hfill \psovalbox{E} none of these


\item Convert to a decimal: $\dfrac{1}{8}+0.25=$ \\
\psovalbox{A} $0.43$ \ \hfill \psovalbox{B} $0.37$\ \ \hfill
\psovalbox{C} $0.375$\ \ \hfill \psovalbox{D} $0.5$ \hfill
\psovalbox{E} none of these

 \item Convert to a decimal: $(0.1)^2\cdot 0.2+0.1\cdot (0.2)^2=$ \\
\psovalbox{A} $0.06$ \ \hfill \psovalbox{B} $0.006$\ \ \hfill
\psovalbox{C} $0.0028$\ \ \hfill \psovalbox{D} $0.6$ \hfill
\psovalbox{E} none of these

\item What percent of $44$ is $11$?
\\ \psovalbox{A}$11\%$
\hfill \psovalbox{B}   $484\%$ \hfill \psovalbox{C} $400\%$ \hfill
\psovalbox{D}  $25\%$ \hfill \psovalbox{E} none of these

\item Find $30\%$ of $810$.
\\ \psovalbox{A} $24.3$
\hfill \psovalbox{B}   $243$
\hfill \psovalbox{C}  $2430$
\hfill \psovalbox{D}  $24300$
\hfill \psovalbox{E}  none of these
\item  After a discount of $25\%$, a robotic dog sells for $\$1 500$. What was the
original selling price?
\\ \psovalbox{A}$\$1875$\ \ \hfill \psovalbox{B}   $\$1 525$\ \
\hfill \psovalbox{C} $\$2 000$\ \ \hfill \psovalbox{D}  $\$1 750$
\hfill \psovalbox{E}  none of these

\item $840$ is $28\%$ of what number?\\
 \psovalbox{A} $3000$
\hfill \psovalbox{B}   $300$
\hfill \psovalbox{C}  $30$
\hfill \psovalbox{D}  $235\frac{1}{5}$ \hfill \psovalbox{E}  none of
these
\item Find $37\%$ of $500$.
\\ \psovalbox{A}18500\ \ \hfill \psovalbox{B}   185\ \ \hfill
\psovalbox{C} $13\frac{19}{37}$\ \ \hfill \psovalbox{D}
$13\frac{19}{3700}$ \hfill \psovalbox{E}  none of these
\item Find $5\frac{1}{3}\%$ of $936$.
\\ \psovalbox{A}4992\ \ \hfill \psovalbox{B}   499200\ \ \hfill
\psovalbox{C} $175\frac{1}{2}$\ \ \hfill \psovalbox{D}
$49\frac{23}{25}$ \hfill \psovalbox{E}  none of these

\item Abdullah buys a sandwich for $\$2.30$. The sales tax is $8\%$. How must tax
must he pay?
\\ \psovalbox{A}$\$1.84$
\hfill \psovalbox{B}   $\$18.40$
\hfill \psovalbox{C}  $\$18.40$
\hfill \psovalbox{D}  $\$2.48$ \hfill \psovalbox{E}  none of these

\item Adeena's car  cost $\$32 000$ brand new. The following year it
depreciated to $\$22 080$. Find the percent of decrease.
\\ \psovalbox{A}$99.20\%$\ \ \hfill \psovalbox{B}   $31\%$\ \ \hfill
\psovalbox{C} $169\%$\ \ \hfill \psovalbox{D}  $69\%$  \hfill
\psovalbox{E}  none of these
\item Trisha gets a  $15\%$ commission on sales. If she was able to sell
$\$2 040$ on Monday, what was her commission? \\ \psovalbox{A}$\$2
025$\ \ \hfill \psovalbox{B}   $\$2346$\ \ \hfill \psovalbox{C}
$\$306$\ \ \hfill \psovalbox{D}  $\$1734$\ \ \hfill \psovalbox{E}
none of these

\item  Ann Arbor MI, has a $4\%$ sales tax. Max bought a book at {\em Dawn Treader}  costing
$\$19.95$. What was his total bill? \\ \psovalbox{A}$\$20.75$\ \
\hfill \psovalbox{B}   $\$19.99$\ \ \hfill \psovalbox{C} $\$0.80$\ \
\hfill \psovalbox{D}  $\$19.15$ \hfill \psovalbox{E} none of these


\item Tuition at {\em Gosh Knows Where College} went from $\$320$ per credit to
$\$368$ per credit. Find the percent of increase. \\ \psovalbox{A}
$48\%$\ \ \hfill \psovalbox{B}   $0.48\%$\ \ \hfill \psovalbox{C}
$13.04\%$\ \ \hfill \psovalbox{D}  $15\%$ \hfill \psovalbox{E}  none
of these
\item $0.1^2+0.3^4+0.5\cdot 0.6=$\\ \psovalbox{A}
$3.0181$\ \ \hfill \psovalbox{B}   $3.018$\ \ \hfill \psovalbox{C}
$3.181$\ \ \hfill \psovalbox{D}  $.00030181\%$ \hfill \psovalbox{E}
none of these
\item $0.1\cdot 0.2+0.3\cdot 0.4+0.5\cdot 0.6+0.7\cdot 0.8+0.9=$\\ \psovalbox{A}
$1.9$\ \ \hfill \psovalbox{B}   $1.09$\ \ \hfill \psovalbox{C}
$0.19$\ \ \hfill \psovalbox{D}  $19\%$ \hfill \psovalbox{E} none of
these

\item[] {\bf Division of Decimals}

 \item  $0.102 \div 0.03=$
\\ \vfill\psovalbox{A} $3.4$ \hfill \psovalbox{B}
$34$ \hfill  \psovalbox{C}  $0.034$ \hfill \psovalbox{D} $0.34$
\hfill \psovalbox{E} none of these
\item  $1.1 \div 0.3=$
\\ \vfill\psovalbox{A} $3.\overline{06}$ \hfill \psovalbox{B}
$3.\overline{6}$ \hfill  \psovalbox{C}  $3.0\overline{6}$ \hfill
\psovalbox{D} $3.0\overline{36}$ \hfill \psovalbox{E} none of these


\item $81$ is $27\%$ of what number?
\\ \psovalbox{A} $3000$\ \ \hfill \psovalbox{B}   $300$\ \ \hfill
\psovalbox{C}  $30$\ \ \hfill \psovalbox{D}  $10\frac{1}{3}$ \hfill
\psovalbox{E}  none of these
 \item  $0.045 \div 0.3=$
\\ \psovalbox{A} $0.15$ \hfill \psovalbox{B}
$1.5$ \hfill  \psovalbox{C}  $0.015$ \hfill \psovalbox{D} $0.015$
\hfill \psovalbox{E} none of these
\item  $3.32 \div 0.33=$
\\ \psovalbox{A} $10.\overline{03}$ \hfill \psovalbox{B}
$10.03$ \hfill  \psovalbox{C}  $10.0303$ \hfill \psovalbox{D}
$10.0\overline{3}$ \hfill \psovalbox{E} none of these

\item[] {\bf Integer Addition and Subtraction}

\item If the following numbers were written from smallest to largest, which one would be in the
middle?
$$A: -7 \qquad B: -11 \qquad C: -18 \qquad D: 18 \qquad E: 0  $$
\\ \psovalbox{A} \hfill \psovalbox{B}    \hfill
\psovalbox{C}   \hfill \psovalbox{D}  \hfill \psovalbox{E}
\item $1 - 2 + 4
- 8 + 16 - 32 = $ \\ \psovalbox{A}$20$  \hfill \psovalbox{B} $-20$
\hfill \psovalbox{C}  $21$  \hfill \psovalbox{D}  $-21$
  \hfill \psovalbox{E}  none of these


\item $1 - |-3| =$
\\ \psovalbox{A}$4$  \hfill \psovalbox{B}   $-4$  \hfill
\psovalbox{C}  $-2$  \hfill \psovalbox{D}  $2$  \hfill \psovalbox{E}
none of these
\item If the following numbers were written from smallest to largest, which one would be in the
middle?
$$A: 7 \qquad B: -3 \qquad C: -4 \qquad D: 0 \qquad E: -1  $$
\\ \psovalbox{A} \hfill \psovalbox{B}    \hfill
\psovalbox{C}   \hfill \psovalbox{D}  \hfill \psovalbox{E}
\item $3-9+27-81 = $ \\ \psovalbox{A} $60$  \hfill \psovalbox{B} $-42$
\hfill \psovalbox{C}  $-60$  \hfill \psovalbox{D}  $-114$
  \hfill \psovalbox{E}  none of these


\item $-1 + |-2| =$
\\ \psovalbox{A} $-3$  \hfill \psovalbox{B}   $1$  \hfill
\psovalbox{C}  $-1$  \hfill \psovalbox{D}  $3$  \hfill \psovalbox{E}
none of these
\item $ 3 - (5 - (5 - 3)) = $
\\ \psovalbox{A}$-10$  \hfill \psovalbox{B}   $-4$  \hfill
\psovalbox{C}  $4$  \hfill \psovalbox{D}  $0$  \hfill \psovalbox{E}
none of these
\item[] {\bf Integer Multiplication and Division}
\item $\dfrac{5^3-(-3)^3}{5^2-(-3)^2} =$\\  %D
\psovalbox{A} $\dfrac{49}{2}$ \hfill \psovalbox{B} $\dfrac{3}{2}$
\hfill \psovalbox{C} $6$  \hfill \psovalbox{D} $\dfrac{19}{2}$
\hfill \psovalbox{E} none of these
\item $(-100)\div (10)\div (-10) = $\\  %A
\psovalbox{A}$1$ \hfill \psovalbox{B} $10$ \hfill \psovalbox{C}
$100$ \hfill \psovalbox{D} $-1$\hfill  \psovalbox{E} none of these
\item $\dfrac{5^3-(-3)^3}{5^2-(-3)^2} =$\\  %D
\psovalbox{A} $\dfrac{49}{2}$ \hfill \psovalbox{B} $\dfrac{3}{2}$
\hfill \psovalbox{C} $6$  \hfill \psovalbox{D} $\dfrac{19}{2}$
\hfill \psovalbox{E} none of these
\item $ (2)(-8) - (-2)(-2)(3) =  $
\\ \psovalbox{A} $4$  \hfill \psovalbox{B}   $-4$  \hfill
\psovalbox{C}  $28$  \hfill \psovalbox{D}  $-28$  \hfill
\psovalbox{E} none of these
\item $ 5-4^2 =$
\\ \psovalbox{A} $-11$  \hfill \psovalbox{B}   $11$  \hfill
\psovalbox{C}  $21$  \hfill \psovalbox{D}  $-21$  \hfill
\psovalbox{E} none of these
\item $1 - (-2)^3=$ \\  \psovalbox{A} $-7 $ \hfill \psovalbox{B} $9  $ \hfill
\psovalbox{C} $-9 $ \hfill \psovalbox{D} $7 $ \hfill \psovalbox{E}
none of these
\item $2-\left(-\dfrac{1}{2}\right)^3 =$\\
\\  \psovalbox{A} $\dfrac{15}{8}$ \hfill \psovalbox{B}$\dfrac{11}{6}$
\hfill \psovalbox{C} $\dfrac{17}{8}$ \hfill \psovalbox{D}
$\dfrac{13}{6}$ \hfill \psovalbox{E} none of these
\item $\dfrac{2^3-(-1)^3}{2 - (-1)}$
 \\ \psovalbox{A} $3$\ \
 \hfill \psovalbox{B}   $1$\ \
 \hfill \psovalbox{C}  $-3$\ \
 \hfill \psovalbox{D}  $-1$\ \
 \hfill \psovalbox{E}  none of these


\item $ (2)(-9) - (-1)(-8)(-3) =  $
\\ \psovalbox{A}$-6$  \hfill \psovalbox{B}   $42$  \hfill
\psovalbox{C}  $-42$  \hfill \psovalbox{D}  $6$  \hfill
\psovalbox{E}  none of these
\item $ (-3)^2 + (-2)^3 =$
\\ \psovalbox{A}$-9$  \hfill \psovalbox{B}   $-8$  \hfill
\psovalbox{C}  $1$  \hfill \psovalbox{D}  $-17$  \hfill
\psovalbox{E}  none of these
\item[] {\bf Rational, Irrational, and Real Numbers}

\item $\dfrac{3}{4} - \dfrac{11}{12}=$\\
\\  \psovalbox{A} $\dfrac{1}{6}$ \hfill \psovalbox{B}$-\dfrac{1}{6}$
\hfill \psovalbox{C} $-1$ \hfill \psovalbox{D} $\dfrac{11}{16}$
\hfill \psovalbox{E} none of these
\item $-\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{2}{3}\right)^3=$\\
\\  \psovalbox{A} $-\dfrac{20}{27}$ \hfill \psovalbox{B}$-\dfrac{4}{27}$
\hfill \psovalbox{C} $\dfrac{20}{27}$ \hfill \psovalbox{D}
$\dfrac{4}{27}$ \hfill \psovalbox{E} none of these
\item $\dfrac{\dfrac{2}{3}-\dfrac{3}{4}}{-\dfrac{5}{6}+\dfrac{7}{8}}=$\\
\\  \psovalbox{A} $2$ \hfill \psovalbox{B}$-2$
\hfill \psovalbox{C} $\dfrac{1}{2}$ \hfill \psovalbox{D}
$-\dfrac{1}{2}$ \hfill \psovalbox{E} none of these
\item $2+(-0.5)(0.1)^2=$
\\ \psovalbox{A}$1.995$  \hfill \psovalbox{B}   $0.015$  \hfill
\psovalbox{C}  $2.005$  \hfill \psovalbox{D}  $-1.995$  \hfill
\psovalbox{E} none of these

\item $\dfrac{(-8)(-9)}{-6} + 7 =$
\\ \psovalbox{A}$-19$  \hfill \psovalbox{B}   $5$  \hfill
\psovalbox{C}  $-5$  \hfill \psovalbox{D}  19  \hfill \psovalbox{E}
none of these
\item $\dfrac{14 - 24}{12 - (-5 + 7)}$
 \\ \psovalbox{A}$1$\ \
 \hfill \psovalbox{B}   $10$\ \
 \hfill \psovalbox{C}  $-1$\ \
 \hfill \psovalbox{D}  $-10$\ \
 \hfill \psovalbox{E}  none of these

\end{enumerate}





\chapter{Essay Exam Questions}
\begin{enumerate}
\item[] {\bf Natural Numbers}
\item Convert $123_4$ to decimal.
\item Convert the decimal number $123$ to base-$4$.
\item Compute: $10\cdot (9\cdot 7-8\cdot 6)^2-2^3\cdot 3^2$
\item Consider the arithmetic progression  $8, 19, 30,  \ldots$, where $8$ is on the first position, $19$ in the second position, etc.
 Which number occupies the $1001$st position?
 \item  When writing all the natural numbers from $666$ to $999$ inclusive, how many digits have I used?
 \item Write the prime factorisation of $272$ and $170$.  Find $\gcd (272,170)$. Find $\lcm (272,170)$.
\item If today is Wednesday, what day will it be $1000$ days from
today? \item Determine a digit $d$ so that the $3$-digit integer
$10d$ be divisible by $9$.
\item Compute: $\dfrac{2^6+2^6+2^6+2^6+2^6+2^6}{2^3+2^3+2^3}$.
\item Compute: $\dfrac{1111^5+1111^5+1111^5+1111^5}{1111^4+1111^4}$.
\item[] {\bf Fractions}
\item What is one third of a half of a fifth of a quarter of $99000$?
\item Compute and express your answer as an improper fraction:
$\dfrac{2}{5}+\dfrac{3}{4}+\dfrac{3}{15}$.
\item Compute and express your answer as a mixed number:
$\left(1\tfrac{2}{3}\right)\left(3\tfrac{4}{5}\right)$.

\item Compute: $\dfrac{1}{2+\dfrac{3}{4}}$
\item Compute: $\dfrac{1}{2}\cdot \dfrac{3}{4}+\dfrac{5}{6}\div  \dfrac{8}{7}$

\item What fraction of a non-leap year does a $100$ days constitute?
\item David spent $\dfrac{2}{5}$ of his money on a storybook. The
storybook cost $\$20$. How much money did he have at first?
\item Tito was pigging-out on cookies, and in the course of three
days, he ate $420$ cookies. On the first day, he ate $\dfrac{2}{5}$
of the cookies. On the second day, he ate $\dfrac{5}{6}$ of the
remaining cookies. How many cookies did he eat on the third day?
\item Find $n$ if $\dfrac{2n}{3}=\dfrac{14}{21}$.
\item A recipe for sangr\'{\i}a uses $3$ measures of lemonade for
$2$ measures of red wine. How many cups of lemonade and of red wine
are required if $40$ cups of sangr\'{\i}a are desired? \clearpage
\item[] {\bf Decimals}
\item Compute and write your answer as a decimal: $\dfrac{1}{2}+3.45 + 67\%.$
\item Compute and write your answer as a decimal:
$\left(\dfrac{1}{10}\right)\left(\dfrac{2}{100}\right)\left(\dfrac{3}{1000}\right)\left(\dfrac{4}{10000}\right)$.
\item Compute: $(0.1)(0.2)+(0.3)^2$.
\item Convert to a fraction: $\dfrac{1}{0.1 + \dfrac{1}{0.1}}$.
\item Back in the good old days of 1999, a euro was worth \$1.05.
Now, the value of the dollar has fallen, and a euro is  worth $20\%$
more of  what it was in 1999. How much is worth a euro now, in
dollars?
\item How much does it cost to fill a $12$ gallon tank, if gas costs
$\$3.16$ a gallon?
\item Divide: $123.123\div 0.06$. Is the result a periodic or a
finite decimal?
\item Divide: $123.123 \div 0.9$.  Is the result a periodic or a
finite decimal?
\item T-shirts are now at a 20\% discount at the local T-shirt
store. What is the discounted price of a T-shirt of original price
$\$19.90$? \item The area of an aerobic hall in a tri-complex is
70\% of the total area (aerobic hall, tennis court, and pool). The
area of a tennis court is 30\% of the aerobic hall area. Find the
area of a pool, if the aerobic hall area is $476$ sq. feet.
\item[] {\bf Integers}
\item Compute: $(1-2+3)(4-5+6)- (-8+9-10)$
\item Compute: $|(-1)^2+(-2)^3-(-4)(-5)|$
\item Calculate: $(6-4^2)(4+(-4)^2)$
\item When a number is halved  and to this result we add $5$, we
obtain $4$. What was the original number?
\item A number is trebled, and then the result is increased by $10$.
One obtains $4$ as the final answer. What was the original number?
\item[] {\bf Rational Numbers}

\item Calculate: $2-\left(-\dfrac{3}{2}\right)^2$.
\item Calculate: $1+(-0.1)^3$.
\item Calculate: $(1-1.1)^3+1$.
\item Calculate:
$\dfrac{-\dfrac{2}{3}\div\dfrac{3}{4}}{-\dfrac{2}{3}+\dfrac{3}{4}}$.
\item Calculate:
$\dfrac{(-2008)^1+(-2)^2+(-3)^3}{(-2008)^0+(-2)^1+(-3)^2}$.
\end{enumerate}




\chapter{Answers and Hints}\addcontentsline{toc}{section}{Answers
and Hints}\markright{Answers and Hints}
 {\small \input{aritmetiquilla1}}




\clearpage







{\tiny\chapter*{\rlap{GNU Free Documentation License}}
\phantomsection  % so hyperref creates bookmarks
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%\label{label_fdl}

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The ``\textbf{Invariant Sections}'' are certain Secondary Sections
whose titles are designated, as being those of Invariant Sections,
in the notice that says that the Document is released under this
License.  If a section does not fit the above definition of
Secondary then it is not allowed to be designated as Invariant.  The
Document may contain zero Invariant Sections.  If the Document does
not identify any Invariant Sections then there are none.

The ``\textbf{Cover Texts}'' are certain short passages of text that
are listed, as Front-Cover Texts or Back-Cover Texts, in the notice
that says that the Document is released under this License.  A
Front-Cover Text may be at most 5 words, and a Back-Cover Text may
be at most 25 words.

A ``\textbf{Transparent}'' copy of the Document means a
machine-readable copy, represented in a format whose specification
is available to the general public, that is suitable for revising
the document straightforwardly with generic text editors or (for
images composed of pixels) generic paint programs or (for drawings)
some widely available drawing editor, and that is suitable for input
to text formatters or for automatic translation to a variety of
formats suitable for input to text formatters.  A copy made in an
otherwise Transparent file format whose markup, or absence of
markup, has been arranged to thwart or discourage subsequent
modification by readers is not Transparent. An image format is not
Transparent if used for any substantial amount of text.  A copy that
is not ``Transparent'' is called ``\textbf{Opaque}''.

Examples of suitable formats for Transparent copies include plain
ASCII without markup, Texinfo input format, LaTeX input format, SGML
or XML using a publicly available DTD, and standard-conforming
simple HTML, PostScript or PDF designed for human modification.
Examples of transparent image formats include PNG, XCF and JPG.
Opaque formats include proprietary formats that can be read and
edited only by proprietary word processors, SGML or XML for which
the DTD and/or processing tools are not generally available, and the
machine-generated HTML, PostScript or PDF produced by some word
processors for output purposes only.

The ``\textbf{Title Page}'' means, for a printed book, the title
page itself, plus such following pages as are needed to hold,
legibly, the material this License requires to appear in the title
page.  For works in formats which do not have any title page as
such, ``Title Page'' means the text near the most prominent
appearance of the work's title, preceding the beginning of the body
of the text.

A section ``\textbf{Entitled XYZ}'' means a named subunit of the
Document whose title either is precisely XYZ or contains XYZ in
parentheses following text that translates XYZ in another language.
(Here XYZ stands for a specific section name mentioned below, such
as ``\textbf{Acknowledgements}'', ``\textbf{Dedications}'',
``\textbf{Endorsements}'', or ``\textbf{History}''.) To
``\textbf{Preserve the Title}'' of such a section when you modify
the Document means that it remains a section ``Entitled XYZ''
according to this definition.

The Document may include Warranty Disclaimers next to the notice
which states that this License applies to the Document.  These
Warranty Disclaimers are considered to be included by reference in
this License, but only as regards disclaiming warranties: any other
implication that these Warranty Disclaimers may have is void and has
no effect on the meaning of this License.


\begin{center}
{\Large\bf 2. VERBATIM COPYING\par} \phantomsection
\addcontentsline{toc}{section}{2. VERBATIM COPYING}
\end{center}

You may copy and distribute the Document in any medium, either
commercially or noncommercially, provided that this License, the
copyright notices, and the license notice saying this License
applies to the Document are reproduced in all copies, and that you
add no other conditions whatsoever to those of this License.  You
may not use technical measures to obstruct or control the reading or
further copying of the copies you make or distribute.  However, you
may accept compensation in exchange for copies.  If you distribute a
large enough number of copies you must also follow the conditions in
section~3.

You may also lend copies, under the same conditions stated above,
and you may publicly display copies.


\begin{center}
{\Large\bf 3. COPYING IN QUANTITY\par} \phantomsection
\addcontentsline{toc}{section}{3. COPYING IN QUANTITY}
\end{center}


If you publish printed copies (or copies in media that commonly have
printed covers) of the Document, numbering more than 100, and the
Document's license notice requires Cover Texts, you must enclose the
copies in covers that carry, clearly and legibly, all these Cover
Texts: Front-Cover Texts on the front cover, and Back-Cover Texts on
the back cover.  Both covers must also clearly and legibly identify
you as the publisher of these copies.  The front cover must present
the full title with all words of the title equally prominent and
visible.  You may add other material on the covers in addition.
Copying with changes limited to the covers, as long as they preserve
the title of the Document and satisfy these conditions, can be
treated as verbatim copying in other respects.

If the required texts for either cover are too voluminous to fit
legibly, you should put the first ones listed (as many as fit
reasonably) on the actual cover, and continue the rest onto adjacent
pages.

If you publish or distribute Opaque copies of the Document numbering
more than 100, you must either include a machine-readable
Transparent copy along with each Opaque copy, or state in or with
each Opaque copy a computer-network location from which the general
network-using public has access to download using public-standard
network protocols a complete Transparent copy of the Document, free
of added material. If you use the latter option, you must take
reasonably prudent steps, when you begin distribution of Opaque
copies in quantity, to ensure that this Transparent copy will remain
thus accessible at the stated location until at least one year after
the last time you distribute an Opaque copy (directly or through
your agents or retailers) of that edition to the public.

It is requested, but not required, that you contact the authors of
the Document well before redistributing any large number of copies,
to give them a chance to provide you with an updated version of the
Document.


\begin{center}
{\Large\bf 4. MODIFICATIONS\par} \phantomsection
\addcontentsline{toc}{section}{4. MODIFICATIONS}
\end{center}

You may copy and distribute a Modified Version of the Document under
the conditions of sections 2 and 3 above, provided that you release
the Modified Version under precisely this License, with the Modified
Version filling the role of the Document, thus licensing
distribution and modification of the Modified Version to whoever
possesses a copy of it.  In addition, you must do these things in
the Modified Version:

\begin{itemize}
\item[A.]
   Use in the Title Page (and on the covers, if any) a title distinct
   from that of the Document, and from those of previous versions
   (which should, if there were any, be listed in the History section
   of the Document).  You may use the same title as a previous version
   if the original publisher of that version gives permission.

\item[B.]
   List on the Title Page, as authors, one or more persons or entities
   responsible for authorship of the modifications in the Modified
   Version, together with at least five of the principal authors of the
   Document (all of its principal authors, if it has fewer than five),
   unless they release you from this requirement.

\item[C.]
   State on the Title page the name of the publisher of the
   Modified Version, as the publisher.

\item[D.]
   Preserve all the copyright notices of the Document.

\item[E.]
   Add an appropriate copyright notice for your modifications
   adjacent to the other copyright notices.

\item[F.]
   Include, immediately after the copyright notices, a license notice
   giving the public permission to use the Modified Version under the
   terms of this License, in the form shown in the Addendum below.

\item[G.]
   Preserve in that license notice the full lists of Invariant Sections
   and required Cover Texts given in the Document's license notice.

\item[H.]
   Include an unaltered copy of this License.

\item[I.]
   Preserve the section Entitled ``History'', Preserve its Title, and add
   to it an item stating at least the title, year, new authors, and
   publisher of the Modified Version as given on the Title Page.  If
   there is no section Entitled ``History'' in the Document, create one
   stating the title, year, authors, and publisher of the Document as
   given on its Title Page, then add an item describing the Modified
   Version as stated in the previous sentence.

\item[J.]
   Preserve the network location, if any, given in the Document for
   public access to a Transparent copy of the Document, and likewise
   the network locations given in the Document for previous versions
   it was based on.  These may be placed in the ``History'' section.
   You may omit a network location for a work that was published at
   least four years before the Document itself, or if the original
   publisher of the version it refers to gives permission.

\item[K.]
   For any section Entitled ``Acknowledgements'' or ``Dedications'',
   Preserve the Title of the section, and preserve in the section all
   the substance and tone of each of the contributor acknowledgements
   and/or dedications given therein.

\item[L.]
   Preserve all the Invariant Sections of the Document,
   unaltered in their text and in their titles.  Section numbers
   or the equivalent are not considered part of the section titles.

\item[M.]
   Delete any section Entitled ``Endorsements''.  Such a section
   may not be included in the Modified Version.

\item[N.]
   Do not retitle any existing section to be Entitled ``Endorsements''
   or to conflict in title with any Invariant Section.

\item[O.]
   Preserve any Warranty Disclaimers.
\end{itemize}

If the Modified Version includes new front-matter sections or
appendices that qualify as Secondary Sections and contain no
material copied from the Document, you may at your option designate
some or all of these sections as invariant.  To do this, add their
titles to the list of Invariant Sections in the Modified Version's
license notice. These titles must be distinct from any other section
titles.

You may add a section Entitled ``Endorsements'', provided it
contains nothing but endorsements of your Modified Version by
various parties--for example, statements of peer review or that the
text has been approved by an organization as the authoritative
definition of a standard.

You may add a passage of up to five words as a Front-Cover Text, and
a passage of up to 25 words as a Back-Cover Text, to the end of the
list of Cover Texts in the Modified Version.  Only one passage of
Front-Cover Text and one of Back-Cover Text may be added by (or
through arrangements made by) any one entity.  If the Document
already includes a cover text for the same cover, previously added
by you or by arrangement made by the same entity you are acting on
behalf of, you may not add another; but you may replace the old one,
on explicit permission from the previous publisher that added the
old one.

The author(s) and publisher(s) of the Document do not by this
License give permission to use their names for publicity for or to
assert or imply endorsement of any Modified Version.


\begin{center}
{\Large\bf 5. COMBINING DOCUMENTS\par} \phantomsection
\addcontentsline{toc}{section}{5. COMBINING DOCUMENTS}
\end{center}


You may combine the Document with other documents released under
this License, under the terms defined in section~4 above for
modified versions, provided that you include in the combination all
of the Invariant Sections of all of the original documents,
unmodified, and list them all as Invariant Sections of your combined
work in its license notice, and that you preserve all their Warranty
Disclaimers.

The combined work need only contain one copy of this License, and
multiple identical Invariant Sections may be replaced with a single
copy.  If there are multiple Invariant Sections with the same name
but different contents, make the title of each such section unique
by adding at the end of it, in parentheses, the name of the original
author or publisher of that section if known, or else a unique
number. Make the same adjustment to the section titles in the list
of Invariant Sections in the license notice of the combined work.

In the combination, you must combine any sections Entitled
``History'' in the various original documents, forming one section
Entitled ``History''; likewise combine any sections Entitled
``Acknowledgements'', and any sections Entitled ``Dedications''. You
must delete all sections Entitled ``Endorsements''.

\begin{center}
{\Large\bf 6. COLLECTIONS OF DOCUMENTS\par} \phantomsection
\addcontentsline{toc}{section}{6. COLLECTIONS OF DOCUMENTS}
\end{center}

You may make a collection consisting of the Document and other
documents released under this License, and replace the individual
copies of this License in the various documents with a single copy
that is included in the collection, provided that you follow the
rules of this License for verbatim copying of each of the documents
in all other respects.

You may extract a single document from such a collection, and
distribute it individually under this License, provided you insert a
copy of this License into the extracted document, and follow this
License in all other respects regarding verbatim copying of that
document.


\begin{center}
{\Large\bf 7. AGGREGATION WITH INDEPENDENT WORKS\par}
\phantomsection \addcontentsline{toc}{section}{7. AGGREGATION WITH
INDEPENDENT WORKS}
\end{center}


A compilation of the Document or its derivatives with other separate
and independent documents or works, in or on a volume of a storage
or distribution medium, is called an ``aggregate'' if the copyright
resulting from the compilation is not used to limit the legal rights
of the compilation's users beyond what the individual works permit.
When the Document is included in an aggregate, this License does not
apply to the other works in the aggregate which are not themselves
derivative works of the Document.

If the Cover Text requirement of section~3 is applicable to these
copies of the Document, then if the Document is less than one half
of the entire aggregate, the Document's Cover Texts may be placed on
covers that bracket the Document within the aggregate, or the
electronic equivalent of covers if the Document is in electronic
form. Otherwise they must appear on printed covers that bracket the
whole aggregate.


\begin{center}
{\Large\bf 8. TRANSLATION\par} \phantomsection
\addcontentsline{toc}{section}{8. TRANSLATION}
\end{center}


Translation is considered a kind of modification, so you may
distribute translations of the Document under the terms of
section~4. Replacing Invariant Sections with translations requires
special permission from their copyright holders, but you may include
translations of some or all Invariant Sections in addition to the
original versions of these Invariant Sections.  You may include a
translation of this License, and all the license notices in the
Document, and any Warranty Disclaimers, provided that you also
include the original English version of this License and the
original versions of those notices and disclaimers.  In case of a
disagreement between the translation and the original version of
this License or a notice or disclaimer, the original version will
prevail.

If a section in the Document is Entitled ``Acknowledgements'',
``Dedications'', or ``History'', the requirement (section~4) to
Preserve its Title (section~1) will typically require changing the
actual title.


\begin{center}
{\Large\bf 9. TERMINATION\par} \phantomsection
\addcontentsline{toc}{section}{9. TERMINATION}
\end{center}


You may not copy, modify, sublicense, or distribute the Document
except as expressly provided for under this License.  Any other
attempt to copy, modify, sublicense or distribute the Document is
void, and will automatically terminate your rights under this
License.  However, parties who have received copies, or rights, from
you under this License will not have their licenses terminated so
long as such parties remain in full compliance.


\begin{center}
{\Large\bf 10. FUTURE REVISIONS OF THIS LICENSE\par} \phantomsection
\addcontentsline{toc}{section}{10. FUTURE REVISIONS OF THIS LICENSE}
\end{center}


The Free Software Foundation may publish new, revised versions of
the GNU Free Documentation License from time to time.  Such new
versions will be similar in spirit to the present version, but may
differ in detail to address new problems or concerns.  See
http://www.gnu.org/copyleft/.

Each version of the License is given a distinguishing version
number. If the Document specifies that a particular numbered version
of this License ``or any later version'' applies to it, you have the
option of following the terms and conditions either of that
specified version or of any later version that has been published
(not as a draft) by the Free Software Foundation.  If the Document
does not specify a version number of this License, you may choose
any version ever published (not as a draft) by the Free Software
Foundation. }




 \QUOTEME{Que a quien robe este libro, o lo tome prestado y no lo
devuelva, se le convierta en una serpiente en las manos y lo venza.
Que sea golpeado por la par\'{a}lisis y todos sus miembros
arruinados. Que languidezca de dolor gritando por piedad, y que no
haya coto a su agon\'{\i}a hasta la \'{u}ltima disoluci\'{o}n. Que
las polillas roan sus entra\~{n}as y, cuando llegue al final de su
castigo, que arda en las llamas del Infierno para siempre.}
{Maldici\'{o}n an\'{o}nima contra los ladrones de libros en el
monasterio de San Pedro, Barcelona.}

\section*{Updates} The most current version of this book (both the
.pdf and the .tex source) can be found here:

\centerline{\url{faculty.ccp.edu/faculty/dsantos/lecture_notes.html}}

\clearpage
\begin{thebibliography}{mmm.m}
\bibitem[Aha]{Aha} Ron AHARONI,
\href{http://www.math.technion.ac.il/~ra/education.html}{http://www.math.technion.ac.il/~ra/education.html},
Education Papers

\bibitem[Chen]{Chen} William CHEN, \href{http://www.maths.mq.edu.au/~wchen/lnemfolder/lnem.html}{http://www.maths.mq.edu.au/~wchen/lnemfolder/lnem.html}, Elementary
Mathematics Notes.
\bibitem[Devl]{Devl} Keith DEVLIN, ``Life by the Numbers'', Wiley, 1998.

\bibitem[Eves]{Eves} Howard EVES, ``An Introduction to the History of Mathematics'', 3rd ed., 1969.
\bibitem[HaKn]{HaKn} H.S. HALL, and S.R. KNIGHT, ``Algebra for Beginners'', Macmillan, 1st ed., 1893.
\bibitem[Gard]{Gard} Tony GARDINER, ``Mathematical Challenge'', Cambridge, 1996.
\bibitem[GeSh]{GeSh} I.M. GELFAND and A. SHEN, ``Algebra'', Birkhauser, 1993.
\bibitem[Penn]{Penn} Philip PENNANCE, \href{http://math.uprrp.edu/profs/pennance/courses/3041.php}{http://math.uprrp.edu/profs/pennance/courses/3041.php}, Matem\'{a}tica I para maestros.
\bibitem[Toom]{Toom} Andrei TOOM, \href{http://www.de.ufpe.br/~toom/articles/index.htm}{http://www.de.ufpe.br/~toom/articles/index.htm}, Teaching Articles.
\bibitem[Woo]{Woo} Mark R. WOODARD, \href{http://math.furman.edu/~mwoodard/data.html}{http://math.furman.edu/~mwoodard/data.html}, Quotes from the Mathematical Quotations Server.
\bibitem[Wu1]{Wu1} Hung-Hsi WU, \href{http://math.berkeley.edu/~wu/Algebrasummary.pdf}{http://math.berkeley.edu/~wu/Algebrasummary.pdf}, Introduction to School Algebra.
\bibitem[Wu2]{Wu2} Hung-Hsi WU, \href{http://math.berkeley.edu/~wu/EMI1c.pdf}{http://math.berkeley.edu/~wu/EMI1c.pdf}, Whole Numbers.
\bibitem[Wu3]{Wu3} Hung-Hsi WU, \href{http://math.berkeley.edu/~wu/EMI2a.pdf}{http://math.berkeley.edu/~wu/EMI2a.pdf}, Fractions.

\end{thebibliography}








\end{document}
Gordon, 267.980.7098
