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%%%%%%%%%

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%%%%%%%%%%%%%%%%%INTERVALS
%%%%%%%% lo= left open, rc = right closed, etc.
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\def\lorc#1#2{\Big]#1 \ ; #2 \Big]}



%%%%Title Page

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      \math@fontsfalse\selectfont A}%
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} \makeatother


%%%%%%%




%%%%%%%%%%%%%%%%%%%%Two Column Table of Contents
\makeatletter
\newcommand{\twocoltoc}{%
  \begin{multicols}{2}
    \@starttoc{toc}%
  \end{multicols}}
\makeatother
%%%%%%%%%%%%
\makeindex


%%%%%%%%%

%%%%%%%%%%%%%%




\renewcommand{\arraystretch}{2.3}
\setlength{\fboxsep}{.15in} \setlength{\fboxrule}{1pt}




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

\begin{document}
\begin{frontmatter}
\pagenumbering{roman}
\begin{titlepage}
\psset{unit=1in,linewidth=2pt} \psline(-0.75,2)(-0.75,-\paperheight)
\psline(-0.25,2)(-0.25,-\paperheight)
\rput[lb]{90}(-0.3,-0.85\paperheight){\Huge\textcolor{darkgray}{Free
to photocopy and distribute}}
\rput[lb]{90}(-0.4,-0.3\paperheight){\red\Large\textbf{\today\quad
Version}}

\vspace*{3cm}
\centerline{\Huge\textsqrc{Ossifrage and Algebra}}

\vspace*{2cm} \centerline{\Large \textcolor{red}{David A. SANTOS}
\\ \href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu}}
\vfill
 \begin{center}If there should be another flood\\
Hither for refuge fly\\
Were the whole world to be submerged \\
This book would still be dry.\footnote{Anonymous annotation on the
fly leaf of an Algebra book.}\end{center} \vfill

\centerline{\rput{30}(0,0){\pscustom[fillcolor=yellow,fillstyle=solid]{\pspolygon[linearc=.5,linewidth=3pt](1;0)(1;60)(1;120)(1;180)(1;240)(1;300)}
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\vfill
\end{titlepage}
 \clearpage



\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







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

 \begin{center}

       Version 1.2, November 2002


 Copyright \copyright{} 2000,2001,2002  Free Software Foundation, Inc.

 \bigskip

     51 Franklin St, Fifth Floor, Boston, MA  02110-1301  USA

 \bigskip

 Everyone is permitted to copy and distribute verbatim copies
 of this license document, but changing it is not allowed.
\end{center}


\begin{center}
{\bf\large Preamble}
\end{center}

The purpose of this License is to make a manual, textbook, or other
functional and useful document ``free'' in the sense of freedom: to
assure everyone the effective freedom to copy and redistribute it,
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This License is a kind of ``copyleft'', which means that derivative
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complements the GNU General Public License, which is a copyleft
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We have designed this License in order to use it for manuals for
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this License principally for works whose purpose is instruction or
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\begin{center}
{\Large\bf 1. APPLICABILITY AND DEFINITIONS\par} \phantomsection
\addcontentsline{toc}{section}{1. APPLICABILITY AND DEFINITIONS}
\end{center}

This License applies to any manual or other work, in any medium,
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A ``\textbf{Modified Version}'' of the Document means any work
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The ``\textbf{Invariant Sections}'' are certain Secondary Sections
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A section ``\textbf{Entitled XYZ}'' means a named subunit of the
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according to this definition.

The Document may include Warranty Disclaimers next to the notice
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Warranty Disclaimers are considered to be included by reference in
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\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
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Transparent copy along with each Opaque copy, or state in or with
each Opaque copy a computer-network location from which the general
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of added material. If you use the latter option, you must take
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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
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 \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.}

\clearpage






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\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}

These notes started during the Spring of 2002. I would like to
thank Jos\'{e} Mason and John Majewicz for numerous suggestions.
These notes have borrowed immensely from their work and
suggestions.

\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.


\bigskip

I would like to thank Margaret Hitczenko. I have used some of her
ideas from the CEMEC project here. I would also like to thank Iraj
Kalantari, Lasse Skov, and Don Stalk for alerting me of numerous
typos/errors/horrors.


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


\end{frontmatter}



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\pagenumbering{arabic}\setcounter{page}{1}
\part{First Impressions}



\chapter{Why Study Algebra?}
\Opensolutionfile{algebrilla}[algebrilla1] \QUOTEME{ The good
Christian should beware of Mathematicians and all those who make
empty prophesies. The danger already exists that the Mathematicians
have made a covenant with the Devil to darken the spirit and to
confine man in the bonds of Hell.}{\mbox{St.~Augustine}}
\bigskip

\dropping{2}{T}his chapter is optional. It nevertheless tries to
answer two important queries which are often heard:
\begin{itemize}
\item Why am I learning this material?
\item Where will I ever use this material again?
\end{itemize}




Mathematicians can give---what they consider---reasonable
explanations to the questions above. But any reasonable explanation
imposes duties on both the person asking the question and the
mathematician answering it. To understand the given answer, you will
need {\em patience}. If you dislike Mathematics at the outset, no
amount of patience or reasonable explanations will do.



Some of the concepts that we will mention in the history section
might be unfamiliar to you. Do not worry. The purpose of the section
is to give you a point of reference as to  how old many of these
ideas are.

\section{Illiteracy and Innumeracy}
\QUOTEME{No more impressive warning can be given to those who would
confine knowledge and research to what is apparently useful, than
the reflection that conic sections were studied for eighteen hundred
years merely as an abstract science, without regard to any utility
other than to satisfy the craving for knowledge on the part of
mathematicians, and that then at the end of this long period of
abstract study, they were found to be the necessary with which to
attain the knowledge of the most important laws of nature. }{Alfred
North WHITEHEAD}

\bigskip
\dropping{2}{\em{} I}{\em lliteracy} is the lack of ability to read
and write. {\em Innumeracy} is the lack of familiarity with
mathematical concepts and methods. It is a truism that people would
be embarrassed to admit their illiteracy, but no so to admit their
innumeracy.

\bigskip

Mathematicians like to assert the plurality of applications of their
discipline to real world problems, by presenting modelisations from
science, engineering, business, etc. For example, chemical compounds
must obey certain geometric arrangements that in turn specify how
they behave. By means of  mathematical studies called {\em Group
Theory} and {\em P\'{o}lya Theory of Counting}, these geometric
arrangements can be fully catalogued. Modern genetics relies much in
two branches of mathematics---{\em Combinatorics} and {\em
Probability Theory}---in order to explain the multiform combinations
among genes. Some financial firms utilise a mathematical theory
called {\em Brownian Motion} to explain the long term behaviour of
markets. By associating the way virus strands twist using {\em Knot
Theory}, medical research is now better understanding the behaviour
of viruses.\footnote{More such applications can be found in the
popularisation by Keith Devlin [Devl].}


\bigskip
These applications are
perhaps too intricate for a novice to master, and  many years of Mathematics beyond Algebra are required in order to comprehend them.  Hence, why must a
person who  is not planning to become an engineer, a scientist, a
business analyst, etc.,  learn Algebra? A brief answer is the
following:
\begin{itemize}
\item Algebra provides a first example of an {\em abstract system}.
\item Algebra strengthens deductive reasoning.
\item Algebra is a gateway course, used by other disciplines  as a
hurdle for admission.
\item Algebra is part of our cultural legacy, much like Art and
Music, and its mastery is expected by all who want to be considered
educated.
\end{itemize}

\bigskip

What now, if you appreciate all the above reasons, but still claim
that {\em you cannot learn Algebra}? That Algebra {\em is not for
you}? That {\em no grown adult needs} Algebra, since after all,
Algebra does not cure obesity, does not stop poverty, does not stop
war, does not alleviate famine, etc.? Observe that all these reasons
can also be given for learning how to read, or learning a foreign
language. Unlike foreign languages, Algebra is a more universal
language, much like Music, but to fully appreciate its power you
must be willing to learn it. Most of your teachers will assert that
 any non mentally-challenged person can learn Algebra.
Again, most of us will assert that  the major  difficulties in
learning Algebra stem from previous difficulties with Arithmetic.
Hence, if you are reasonably versed in Arithmetic, you should not
confront much trouble with Algebra. All these so-called reasons
against Algebra are balderdash, since lacking clairvoyance, how can
one claim not to need a discipline in the future?

\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Comment on the following assertion: {\em There is no need to learn
Mathematics, since nowadays all calculations can be carried out by
computers.}
\end{pro}
\begin{pro}
Comment on the following assertion: {\em I was never good at Maths.
I will never pass this Algebra class.}
\end{pro}
\begin{pro}
Comment on the following assertions: {\em Only nutritionists should
know about the basics of nutrition since that is their trade. Only
medical doctors should know about the basics of health, since that
is their trade. Only mathematicians should know about Algebra since
that is their trade.}
\end{pro}
\end{multicols}



\section{Brief History}

\QUOTEME{ I have no fault to find with those who teach geometry.
That science is the only one which has not produced sects; it is
founded on analysis and on synthesis and on the calculus; it does
not occupy itself with probable truth; moreover it has the same
method in every country.}{Fredrick the Great}

\bigskip


\dropping{2}{A}lgebra is a very old discipline. Already by 2000 B.C.
the ancient Babylonians were solving quadratic equations by
completing squares \cite[pp. 31-32]{Eves}. \footnote{Equations of
the type $ax^2 + bx + c = 0$, where $x$ is the unknown quantity and
$a, b, c$ are known constants.} The early Egyptians and Babylonians
developed arithmetic and geometry for purely practical
reasons---essentially for the necessities of commerce and land
surveying.

\bigskip

Around  100 BC, in the Chinese mathematics book {\em Jiuzhang
suanshu} (The Nine Chapters on the Mathematical Art),   linear
equations are solved using the method of {\em regula falsa} and
systems of linear equations are solved by the equivalent of modern
matrix methods. Around the same time in India,  {\em Bakhshali
Manuscript} introduces the use of letters and other signs in the
resolution of problems. Cubic and quartic equations are treated, as
well as linear equations with up to five unknowns.




\bigskip
 The Greek civilisation
assimilated the Egyptian and Babylonian practical knowledge and
develops mathematics as an abstract and deductive field. Geometry
and Number Theory are extensively studied by Euclid, whose {\em
Elements} (written approximately 300 BC) is perhaps the most
successful textbook ever written, being one of the very first works
to be printed after the printing press was invented and used as the
basic text on geometry throughout the Western world for about 2,000
years. A first step in the construction of algebra as a formal body
comes with the work of Diophantus of Alexandria, whose {\em
Arithmetic} (written approximately 300 AD) studied what we now call
diophantine equations. One of Diophantus' innovations was to
introduce symbolic notation for arithmetical quantities. For
example, he denoted the square of a quantity by $\Delta \Upsilon$,
the cube of a quantity by $K \Upsilon$, the fourth power of a
quantity by $\Delta \Upsilon\Delta$, and the fifth power by $\Delta
K \Upsilon $. Before him, these quantities were treated rhetorically
(verbally). Diophantus also knew how to manipulate positive and
negative exponents, he represented addition of quantities by
juxtaposition and subtraction with the symbol $:$. Diophantus
methods were not geometrical, like the methods of most Greek
mathematicians, but they used the properties of the numbers
involved. For this and more, Diophantus is considered the father of
algebra.


\bigskip
In 628 AD the  Indian mathematician Brahmagupta, in his treatise
{\em Brahma Sputa Siddhanta},  gives rules for solving linear and
quadratic equations. He also discovers that quadratic equations have
two roots, including both negative as well as irrational roots.


\bigskip
Inspired in the work of Brahmagupta, there appears in what is now
Uzbekistan the mathematician   Muhammad bin Musa al-Khuwarazmi (d.
847 AD), from whose book {\em Kitab al-mukhasar fi hisab al-jabr
wa'l muqabala} (The Book of summary concerning calculating by
transposition and reduction), the word {\em algebra} (\RL{aljabra})
comes from. The word is roughly translated as ``rearranging'' or
``transposing.''\footnote{The word, of course, had a
non-mathematical connotation before the popularity of
al-Khuwarazmi's book. In Moorish Spain an ``algebrista'' was a
bonesetter---a reuniter of broken bones---and so, many barbers of
the time were called {\em algebristas.}} In a way, al-Khuwarazmi's
work was a regression from Diophantus, since his treatment was
rhetorical rather than symbolical. al-Khuwarazmi's however, treated
quantities formally, much like Brhamagupta, rather than
geometrically, like the Greeks. Relying on the work of
al-Khuwarazmi, Abul Kamil  introduces radicals, the use of
irrational quantities, and  systems of equations. Around 1072 AD the
Persian mathematician Omar Khayyam in his {\em Treatise on
Demonstration of Problems of Algebra}, gives a complete
classification of cubic equations with general geometric solutions
found by means of intersecting conic sections.

\bigskip

With the advent of the Renaissance in Europe, the ancient Greek and
Hindu works are known through Arabic translations. Algebra is then
developed at a rapid pace by the Italian mathematicians Cardano,
Tartaglia, Ferrari, and Bombelli. In 1557 Robert Recorde introduces
the sign $=$ for equality. In the same century Widmann introduces
the $+$ for addition and the sign $-$ for subtraction. William
Oughtred in 1631 uses the letter $x$ to denote the unknown of an
equation.


\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Who is called the {\em Father of Algebra} and why?
\begin{answer}
The Alexandrian mathematician Diophantus, for the introduction of
syncopation (symbols to express abstract quantities and procedures
on them) in Algebra.
\end{answer}
\end{pro}
\begin{pro}
Is Algebra an Arab invention?
\begin{answer}
The word {\em algebra} is Arabic, but Algebra was studiend long
before the Arabs had any prominent r\^{o}le in History.
\end{answer}
\end{pro}
\begin{pro}
What is the contraposition alluded to in the title of this book,
{\em Ossifrage and Algebra}?
\begin{answer}
Ossifrage is Latin for {\em breaking bones}. Algebra is Arabic for
{\em bone setting.}
\end{answer}
\end{pro}

\end{multicols}

\section{What is Elementary Algebra All About?}
\QUOTEME{I was just going to say, when I was interrupted, that one
of the many ways of classifying minds is under the heads of
arithmetical and algebraical intellects. All economical and
practical wisdom is an extension of the following arithmetical
formula: $2 + 2 = 4$. Every philosophical proposition has the more
general character of the expression $a + b = c$. We are mere
operatives, empirics, and egotists until we learn to think in
letters instead of figures.}{Oliver Wendell HOLMES} \bigskip

\dropping{2}{E}lementary algebra generalises arithmetic by treating
quantities in the abstract. Thus where in an arithmetic problem you
may assert that $2+3=3+2$ and that $4+1=1+4$, in elementary algebra
you assert that for any two numbers $a, b$ we have
$$a + b = b + a.  $$It allows the formulation of problems and their resolution by treating an unknown quantity
formally. For example, we will learn later on how to calculate
$$123456789^2-(123456787)(123456791)
$$without using a calculator. But the power of algebra goes beyond
these curiosities. Back in 1994, Thomas Nicely, a number theorist,
found an error in the Intel Pentium chip.\footnote{One may find more
information here:
\url{http://www.trnicely.net/pentbug/pentbug.html}.} This means that
computers with this chip were carrying out incorrect calculations.
One example given at the time was the following:
\begin{center}
\begin{tabular}{ll}$4195835.0\div 3145727.0 = 1.333 820 449 136 241 000$ &   (Correct
value)\\   $4195835.0\div 3145727.0 = 1.333 739 068 902 037 589$ &
(Flawed Pentium) \end{tabular}
\end{center}
The lesson here: computers cannot be trusted!
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\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}
\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}

\end{multicols}

\section{Puzzles}
\QUOTEME { Mathematics possesses not only truth, but supreme
beauty---a beauty cold and austere, like that of a sculpture, and
capable of stern perfection, such as only great art can
show.}{Bertrand RUSSELL} \bigskip \dropping{2}{T}he purpose of the
puzzles below is to evince some techniques of mathematical
problem-solving: working backwards, search for patterns, case by
case analysis, etc.



\begin{exa}
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?
\end{exa}\begin{solu} 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{solu}
\begin{exa}
 Dale should have divided a number by $4$, but instead he
subtracted $4$. He got the answer $48$. What should his answer
have been?
\end{exa}
\begin{solu} We work backwards. He obtained $48$ from $48 + 4 = 52$. This means
that he should have performed $52 \div 4 = 13$.
 \end{solu}
\begin{exa} When a number is multiplied by $3$ and
then increased by $16$, the result obtained is $37$. What is the
original number?
\end{exa}
\begin{solu} 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{solu}

\begin{exa}
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 $297$ (!!!). \end{exa} \begin{solu} 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 $3\times 99 =
297$.\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{solu}
\begin{exa}\label{exa:sum1-100-1}
What is the sum $$1 + 2 + 3 + \cdots + 99 + 100$$ of all the
positive integers from $1$ to $100$?
\end{exa}
\begin{solu} Pair up the numbers into the fifty pairs\footnote{This trick is
known as {\em Gau\ss' trick}, after the German mathematician Karl
Friedrich Gau\ss \ (1777-1855). Presumably, when Gau\ss\ was in
first grade, his teacher gave this sum to the pupils in order to
keep them busy. To the amazement of the teacher, Gau\ss\ came up
with the answer almost instantaneously.}
$$(100 + 1) = (99 + 2) = (98 + 3) = \cdots = (50 + 51).
$$Thus we have $50$ pairs that add up to $101$ and so the desired
sum is $101\times 50 = 5050$. Another solution will be given in
example \ref{exa:sum1-100-2}.
\end{solu}



\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
What could St. Augustine mean by mathematicians making {\em
prophecies}? Could he have meant some other profession than
mathematician?
\begin{answer}
St. Augustine is using the word {\em mathematician} in lieu of the
word {\em astrologer}.
\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}
Iblis entered an elevator in a tall building. She went up $4$
floors, down $6$ floors, up $8$ floors and down $10$ floors. She
then found herself on the $23$rd floor. In what floor did she
enter the elevator?
\begin{answer}  $27$th floor. \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} Each square represents a digit.\footnote{A {\em digit}, from the Latin {\em digitum} (finger), is one of the ten numbers
$0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$.} 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?
$$\begin{array}{|c|c|c|} \hline 1 & 2 & 5
\\
\hline 3 & 3 & 2  \\ \hline a & 3 & 1 \\ \hline
\end{array}
$$
\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}
$$\psset{unit=1pc}\psgrid[gridlabels=0,subgriddiv=0](0,0)(3,3)
$$
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:
$$\begin{array}{|c|c|c|} \hline 8 & 1 & 6
\\
\hline 3 & 5 & 7  \\ \hline 4 &  9 & 2 \\ \hline
\end{array}
\qquad \qquad
\begin{array}{|c|c|c|} \hline  4 &  9 & 2
\\
\hline 3 & 5 & 7  \\ \hline  8 & 1 & 6 \\ \hline
\end{array}
$$

\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. This problem is from [Toom].
\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 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}
\end{multicols}




\part{Arithmetic Review}
\chapter{Arithmetic Operations}
\QUOTEME {``Reeling and Writhing, of course, to begin with,'' the
Mock Turtle replied, ``and then the different branches of
Arithmetic: Ambition, Distraction, Uglification, and
Derision.''}{Lewis CARROLL}
\bigskip

\bigskip

\dropping{2}{I}n this chapter we review the operations of addition,
subtraction, multiplication, and division of numbers. We also
introduce exponentiation and root extraction. We expect that most of
the material here will be familiar to the reader. We nevertheless
will present arithmetic operations in such a way so that algebraic
generalisations can be easily derived from them.
\section{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 \dropping{2}{W}e begin our
study of Algebra by interpreting the meaning of its 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.

\bigskip




\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,
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,
$$ 7x, \quad 7\cdot x, \quad 7(x), \quad (7)(x), \quad 7\times x.   $$
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 $5\cdot 6 = 5\times 6 = (5)(6)= 30$ 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 $x/7$. The {\em reciprocal}
of a number $x$ is $\dfrac{1}{x}$.


\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}
The expression $2x-\dfrac{1}{x^2}$ can be translated as ``twice a
number is diminished by the reciprocal of its square.''
\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}
\begin{exa}Find a formula for the $n$-th term of the arithmetic progression
$$ 2, 7, 12, 17, \ldots . $$  \end{exa}
\begin{solu}
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 term is therefore of the form $2+5(n-1)$, where
$n=1,2,3,\ldots$ is a natural number.
\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}
\begin{exa}
Identify the law of formation and conjecture a general formula:
$$ 1 = 1,$$
$$ 1 + 2 = \frac{(2)(3)}{2},$$
$$ 1 + 2 + 3 = \frac{(3)(4)}{2},$$
$$ 1 + 2 + 3 + 4 = \frac{(4)(5)}{2},$$
$$ 1 + 2 + 3 + 4 + 5 = \frac{(5)(6)}{2}.$$
\end{exa}
\begin{solu} Notice that the right hand side consists of the last number on
the left times its successor, and this is then divided by $2$. Thus
we are asserting that
$$ 1 + 2 + 3 + \cdots + (n - 1) + n = \frac{(n)(n + 1)}{2}.$$
\end{solu}


\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. This problem is
taken from [Gard].
\end{answer}
\end{pro}

\begin{pro}
Think of a number. Double it. Add $10$. Half your result. Subtract
your original number. After these five steps, your answer is $5$
regardless of your original number! If $x$ is the original number,
explain by means of algebraic formul\ae\ each step.
\begin{answer}
On the first step you have $x$. On the second step you have $2x$. On
the third step you have $2x+10$. On the four step you have
$\dfrac{2x+10}{2}$. On the fifth step you have $\dfrac{2x+10}{2}-x$.
You are asserting that  $\dfrac{2x+10}{2}-x$ is identically equal to
$5$.
\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}
Find a general formula for the $n$-th term of the arithmetic
progression $$1, 7, 13, 19, 25, \ldots .  $$
\begin{answer} $6n+1$ where $n=0,1,2,3,\ldots$.\end{answer}
\end{pro}

\begin{pro}
Identify the law of formation and conjecture a general formula:
$$ 1^2 = \dfrac{(1)(2)(3)}{6},$$
$$ 1^2 + 2^2 = \frac{(2)(3)(5)}{6},$$
$$ 1^2 + 2^2 + 3^2 = \frac{(3)(4)(7)}{6},$$
$$ 1^2 + 2^2 + 3^2 + 4^2 = \frac{(4)(5)(9)}{6},$$
$$ 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = \frac{(5)(6)(11)}{6}.$$
\begin{answer} First observe that $$3 = 2\times 1 + 1,$$ $$5 = 2\times 2 +
1,$$
$$7 = 2\times 3 + 1,$$ etc. Then it becomes clear that if the last
number on the left is $n$, then the right hand side is $n(n+1)(2n +
1)$ divided by $6$. Hence we are asserting that
$$ 1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 + n^2 = \frac{(n)(n + 1)(2n + 1)}{6}.$$
\end{answer}
\end{pro}
\begin{pro}
Identify the law of formation and conjecture a general formula.
$$ 1 = 1^2,$$
$$ 1 + 3 = 2^2,$$
$$ 1 + 3 + 5 = 3^2,$$
$$ 1 + 3 + 5 + 7 = 4^2,$$
$$ 1 + 3 + 5 + 7 + 9= 5^2.$$
\begin{answer}
The general formula is
$$1 + 3 + \cdots + (2n-1) = n^2.  $$
\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. \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} Let $x$ be an unknown quantity. How would you translate  into
symbols the expression ``the cube of a quantity is reduced by its
square and then what is left is divided by $8$''?  \begin{answer}
The cube is $x^3$, the square is $x^2$. The cube reduced by the
square is $x^3-x^2$. In conclusion, what is left after division by
$8$ is  $$ \dfrac{x^3-x^2}{8}. $$
\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} A camel merchant sold a camel for $a$ dollars  and gained $b$
dollars as profit. What is the real cost of the camel?\footnote{The
answer is not {\em priceless!}}

\begin{answer}The profit is that amount over the real price. If the
real price is $\Gamma$, then $\Gamma +b$ is the real price plus the
profit, that is, what the merchant sold the camel for. In other
words, $\Gamma + b=a$, whence $\Gamma =a-b$.

\end{answer}
\end{pro}
\begin{pro} A merchant started the year with $m$ dollars; the first month
he gained $x$ dollars, the next month he lost $y$ dollars, the third
month he gained $b$ dollars, and the fourth month lost $z$ dollars.
How much had he at the end of that month?
\begin{answer}He starts with $m$, gains $x$ for a total of $m+x$. He
then loses $y$, for a total of $m+x-y$. After he gains $b$ more
dollars he has a running total of $m+x-y+b$, and finally, after
losing $z$ dollars he has a grand total of $m+x-y+b-z$.
\end{answer}
\end{pro}


\end{multicols}

\section{The Natural Numbers}
\QUOTEME { Numbers are the beginning and end of thinking. With
thoughts were numbers born. Beyond numbers thought does not
reach.}{Magnus Gustaf MITTAG-LEFFLER}
\bigskip


\dropping{2}{T}his section gives an overview of the natural numbers.
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
$$ 10= 1+1+1+1+1+1+1+1+1+1, \quad 11= 1+1+1+1+1+1+1+1+1+1+1, \quad  12= 1+1+1+1+1+1+1+1+1+1+1+1, \quad \mathrm{etc.}
$$
\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}
As you know, we use base-10 positional notation. For example, in
$$1234= 1\cdot 1000 + 2\cdot 100 + 3\cdot 10 + 4\cdot 1,  $$ $1$
does not mean ``$1$,'' but $1000$; $2$ does not mean ``$2$,'' but
$200$, etc.

\bigskip

Before positional notation became standard, simple additive systems
(sign-value notation) were used such as the value of the Hebrew
letters, the value of the Greek letters, Roman Numerals, etc.
Arithmetic with these systems was incredibly
cumbersome.\footnote{Try multiplying $123$ by $321$, say, using
Roman numerals!}

\begin{df}The collection of all numbers defined by the recursion
method above is called the set of {\em natural numbers}, and we
represent them by the symbol $\BBN$, that is,
$$ \BBN  = \{0,1,2,3,\ldots \}. $$
\end{df}
\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}

\vspace*{.5cm}

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


\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 $a<b$.
\end{df}






\begin{rem}In what follows,
the symbol $\in$ is used to indicate that a certain element belongs
to a certain set. The negation of $\in$ is $\not\in$. For example,
$1\in \BBN$ because $1$ is a natural number, but
$\dfrac{1}{2}\not\in\BBN$.
\end{rem}

 \vspace*{4cm}
\begin{figure}[h]
\begin{minipage}{5cm}
\centering \psset{unit=1pc}
\psline[linecolor=magenta,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]{$a+b$}\vspace*{1cm}\hangcaption{Addition in
$\BBN$.} \label{fig:add-nat}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=2pc}
\pstGeonode[PosAngle={225,45,0}](0,0){O}(1,0){A}(2.2,0){B}(2.5;60){C}
\pstLineAB[linecolor=magenta,linewidth=2pt]{O}{A}\pstLineAB[linecolor=blue,linewidth=2pt]{A}{B}\pstLineAB[linecolor=red,linewidth=2pt]{O}{C}
\pstLineAB{C}{A}
\pstTranslation[PointName=none,PointSymbol=none]{A}{C}{B}[B']
\pstInterLL[PosAngle=0]{B}{B'}{O}{C}{D}
\pstLineAB{B}{D}\pstLineAB[linecolor=green,linewidth=2pt]{C}{D}
\ncline[offset=-12pt]{|-|}{O}{A}\ncput*{$1$}
\ncline[offset=-12pt]{|-|}{A}{B}\ncput*{$a$}
\ncline[offset=12pt]{|-|}{O}{C}\ncput*[nrot=:U]{$b$}
\ncline[offset=12pt]{|-|}{C}{D}\ncput*[nrot=:U]{$ab$}
\vspace*{1cm}\hangcaption{Multiplication in $\BBN$.}
\label{fig:mult-nat}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=2pc}
\pstGeonode[PointName=none](0,0){O}(1,0){A}(1,3){B}(0,3){C}
\pscustom[fillcolor=magenta,fillstyle=solid]{\pstLineAB{O}{A}\pstLineAB{A}{B}\pstLineAB{B}{C}\pstLineAB{C}{O}}
\ncline[offset=-12pt]{|-|}{O}{A}\ncput*{$a$}
\ncline[offset=-12pt]{|-|}{A}{B}\ncput*{$b$}
\vspace*{1cm}\hangcaption{Multiplication in $\BBN$.}
\label{fig:mult-nat2}
\end{minipage}

\end{figure}


We can think of addition of natural numbers as concatenation of
segment lengths. For example, if we add a segment whose length is
$b$ units to a segment whose length is $a$ units, we obtain a
segment whose length is $a+b$ units. See for example figure
\ref{fig:add-nat}. Multiplication is somewhat harder to interpret.
Form $\triangle OAC$ with $OA=1$ and $OC=b$. Extend the segment
$[OA]$ to $B$, with $AB=a$. Through $B$ draw a line parallel to
$[AC]$, meeting $[OC]$-extended at $D$. By the similarity of
$\triangle OAC$ and $OBD$, $CD=ab$. Another possible interpretation
of multiplication occurs in figure \ref{fig:mult-nat2}, where a
rectangle of area $ab$ (square units) is formed having sides of $a$
units by $b$ units.








\bigskip
Observe that if we add or multiply any two natural numbers, the
result is a natural numbers. We encode this observation in the
following axiom.
\begin{axi}[Closure] $\BBN$ is {\em closed under addition}, that is, if $a\in
\BBN$ and $b\in \BBN$ then also $a+b\in \BBN$. $\BBN$ is {\em closed
under multiplication}, that is, if $x\in \BBN$ and $y\in \BBN$ then
also $xy\in \BBN$.
\end{axi}


If $0$ is added to any natural number, the result is unchanged.
Similarly, if a natural number is multiplied by $1$, the result is
also unchanged. This is  encoded in the following axiom.
\begin{axi}[Additive and Multiplicative Identity] $0\in\BBN$ is the {\em additive identity} of $\BBN$, that it, it has the property that for all
$x\in\BBN$ it follows that $$x=0+x=x+0.  $$ $1\in\BBN$ is the {\em
multiplicative identity} of $\BBN$, that it, it  has the property
that for all $a\in\BBN$ it follows that $$a=1a=a1.  $$
\end{axi}
Again, it is easy to see that when two natural numbers are added or
multiply, the result does not depend on the order. This is encoded
in the following axiom.
\begin{axi}[Commutativity]Let $a\in\BBN$ and $b\in \BBN$. Then
\mbox{$a+b=b+a$} and \mbox{$ab=ba$}.
\end{axi}
Two other important axioms for the natural numbers are now given.
\begin{axi}[Associativity]
Let $a, b, c$ be natural numbers. Then the order of parentheses when
performing addition is irrelevant, that is,
$$ a+(b+c)=(a+b)+c = a+b+c.  $$Similarly, the order of parentheses
when performing multiplication is irrelevant,
$$ a(bc)=(ab)c=abc.  $$
\end{axi}
\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}
We now make some further remarks about addition and multiplication.
The product $mn$ is simply put, stenography for addition. That is,
we have  the equivalent expressions
$$mn = \underbrace{n + n + \cdots + n}_{m\ {\rm times}}  =
 \underbrace{m + m + \cdots + m}_{n\ {\rm times}}.  $$
Thus when we write $(3)(4)$ we mean
$$(3)(4) = 3 + 3 + 3 + 3 = 4 + 4 +4  = 12. $$
Hence if we encounter an expression like $$ (3)(5) + (6)(4)  $$we
must clearly perform the multiplication first and then the addition,
obtaining
$$(3)(5) + (6)(4) = 5 + 5 + 5 + 6 + 6 +6 + 6 = 39,  $$
or more succinctly $$(3)(5) + (6)(4) = 15 + 24 = 39.$$ In turn, a
stenographic form for multiplication by the same number is
exponentiation, which we will now define.
\begin{df}[Exponentiation]\index{exponentiation}
If $n$ is a natural 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}
\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 $2^{10}\approx 1000$, we call  $2^{10}$ a {\em
    kilobit}.\footnote{From the Greek {\em kilo}, meaning {\em thousand}.}
\end{exa}
\begin{exa}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.
\end{exa}
\begin{rem}
Notice that $a^b$ {\bf is not} $ab$. Thus $2^3 = (2)(2)(2) = 8$, and
{\bf not} $(2)(3) = 6$.
\end{rem}

In any expression containing addition and exponentiation, we perform
the exponentiation first, since it is really a shortcut for writing
multiplication.
\begin{exa}We have
$$(2)(4) + 3^3 = 8 + 27 = 35,  $$
$$3^2 + 2^3 = 9 + 8 = 17,  $$
$$(3^2)(4)(5) = (9)(4)(5) = 180.  $$
\end{exa}
The order of operations can be coerced by means of grouping symbols,
like parentheses $(\ )$, brackets $[\ ]$, or braces $\{\ \}$.
\begin{exa}
We have
$$ (3 + 2)(5 + 3) = (5)(8) = 40,  $$
$$ (3 + 2)^2 = (5)^2 = 25,  $$
$$ (5 + (3 + 2(4))^2)^3 = (5 + (3 + 8)^2)^3 = (5 + (11)^2)^3 = (5 + 121)^3 = 126^3 =  2000376.$$
\end{exa}
\begin{rem}
Observe that $ (3 + 2)^2  = 25$ but that $3^2 + 2^2 = 9 + 4 = 13$.
Thus exponentiation does not distribute over addition.

\end{rem}
\begin{exa}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?\end{exa} \begin{solu} 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{solu}
\begin{df}
To {\em evaluate} an expression with letters means to substitute the
values of its letters by  the equivalent values given.
\end{df}

\begin{exa}
Evaluate $a^3+b^3+c^3+3abc$ when $a=1$, $b=2$, $c=3$.
\end{exa}
\begin{solu}
Substituting,
$$1^3+2^3+3^3+3(1)(2)(3)=1+8+27+18=54.  $$
\end{solu}

\bigskip
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}


Having now an idea of what it means to add and multiply natural
numbers, we define subtraction and division of natural numbers by
means of those operations. This is often the case in Mathematics: we
define a new procedure in terms of old procedures.
\begin{df}[Definition of Subtraction]
Let $m, n, x$ be natural numbers. Then the statement $m-n=x$ means
that $m=x+n$.
\end{df}
\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}
\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}
Thus 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}
Neither subtraction nor division are closed in $\BBN$. For example,
$3-5$ is not a natural number, and neither is $3\div 5$. Again, the
operations of subtraction and division misbehave in the natural
numbers, they are not commutative. For example, $5-3$ is not the
same as $3-5$ and $20\div 4$ is not the same as $4\div 20$.
\end{rem}

\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Find the numerical value of\quad  $1^12^23^3$.
\begin{answer}
$108$.
\end{answer}
\end{pro}
\begin{pro}
Find the numerical value of  $$(\sqrt{36}-\sqrt{25})^2.$$
\begin{answer}
$1$.
\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}
Evaluate $(a+b)(a-b)$ when $a=5$ and $b=2$.
\begin{answer}
$21$
\end{answer}
\end{pro}
\begin{pro}
Evaluate $(a^2+b^2)(a^2-b^2)$ when $a=2$ and $b=1$.
\begin{answer}
$15$
\end{answer}
\end{pro}
\begin{pro}
If today is Thursday, what day will it be $100$ days from now?
\begin{answer}Since $100 = 7\times 14 + 2$, $98$ days from today will be a
Thursday, and $100$ days from today will be a Saturday.
\end{answer}
\end{pro}

\begin{pro}
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 $3\times 5 \times 6 = 90.$
\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. This problem is taken from [Toom].
\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}
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}
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}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 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. This problem is taken from [Gard].
\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
$19\times 4 = 76 < 77$; She could not answer more than $20$, because
her score would have been at least $21\times 4 - 3=81.77$). 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}
A certain calculator  gives as the result of the product
$$ 987654 \times 745321 $$ 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:
$$321\times 654=209934,   $$
$$321\times 987 = 316827,  $$
$$745\times 654 = 487230,  $$
$$745 \times 987 = 735315.  $$
Thus
$$\begin{array}{llll}
  & & \psframebox{987} &\psframebox{654}\\
\times & & \psframebox{745} &\psframebox{321}\\
\hline  &  & 209 & 934 \\
& 316 & 827 & \\
& 487 & 230 & \\
735 & 315 & & \\
\hline
736& 119 & 266 & 934 \\
\end{array}   $$
\end{answer}
\end{pro}

\begin{pro}
 How many digits
does $4^{16}5^{25}$ have?
\begin{answer} There are $28$ digits, since
$$4^{16}5^{25} = 2^{32}5^{25} = 2^{7}2^{25}5^{25} = 128\times
10^{25},
$$which is the $3$ digits of $128$ followed by $25$ zeroes.
\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}
Create a new arithmetic operation $\oplus$ by letting $a\oplus b =
1+ ab$.
\begin{enumerate}
\item Compute $1 \oplus (2\oplus 3)$.
\item Compute $(1\oplus 2) \oplus 3$.
\item Is your operation associative. Explain.
\item Is the operation commutative? Explain.
\end{enumerate}
\begin{answer}
Using the definition of $\oplus$,
$$1 \oplus (2\oplus 3) = 1 \oplus (1+2(3)) = 1 \oplus 7 = 1+1(7)=8,   $$
but
$$ (1\oplus 2) \oplus 3 = (1+1(2)) \oplus 3 = 3\oplus 3 = 1+3(3)=10. $$
Since these two answers differ, the operation is not associative.



Now, for the operation to be commutative, we must have, in every
instance, $a\oplus b = b\oplus a$. Using the fact  that
multiplication of numbers is commutative, we deduce
$$ a\oplus b= 1+ab=1+ba= b\oplus a, $$so our operation is indeed
commutative.
\end{answer}

\end{pro}

\end{multicols}

\section{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.
\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} of 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){$\frac{\n}{2}$}}
 \vspace*{1cm}\hangcaption{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){$\frac{\n}{3}$}}
 \vspace*{1cm}\hangcaption{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){$\frac{\n}{4}$}}
 \vspace*{1cm}\hangcaption{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}
By doing more of the 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}
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}

\section*{Homework}
 \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
 \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 $8+8+4=20$. This problem is from [Gard].
\end{answer}
\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}
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}
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}



 \end{multicols}

\section{Operations with 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}
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.

\bigskip

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}
\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}
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}





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

\begin{pro}Complete the ``fraction puzzle'' below. \vspace*{1cm}
 \PuzzleSolution
\begin{Puzzle}{5}{5}%
|{$\frac{1}{3}$}|{$+$}|{\empty}|{$=$}|{2}|. |{$+$}|*|{$-$}|*|*|.
|{$\frac{3}{4}$}|{$\times$}|{\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}|.
|{$+$}|*|{$-$}|*|*|.
|{$\frac{3}{4}$}|{$\times$}|{$1$}|{$=$}|{$\frac{3}{4}$}|.
|{$=$}|*|{$=$}|*|*|.
|{$\frac{13}{12}$}|{$\div$}|{$\frac{2}{3}$}|{$=$}|{$\frac{13}{8}$}|.
\end{Puzzle}

\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}
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?
\begin{answer}
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{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}
 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}
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}



\end{multicols}

\section{The Integers}
\QUOTEME { God created the integers. Everything else is the work of
Man.}{\mbox{Leopold KROENECKER}}
\bigskip
\dropping{2}{T}he introduction of fractions in the preceding section
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
$$n + (-n)=(-n)+n=0.  $$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}.}
 $\BBZ$.
\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)
\vspace*{.5cm}\hangcaption{The Integers $\BBZ$.}
\label{fig:integers}
\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 $a\in \BBZ$ and $b\in\BBZ$. 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 $a\in\BBZ$,
$-(-a)=a$. This leads to the following definition.
\begin{df}
Let $a\in\BBZ$. 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 again  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]{$a$}\vspace*{.5cm}\hangcaption{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.)



\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}](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)\vspace*{.5cm}\hangcaption{$a+b$
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=: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)\vspace*{.5cm}\hangcaption{$3+2$.}
\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}

\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=: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)\vspace*{.5cm}\hangcaption{$a+b$
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)\vspace*{.5cm}\hangcaption{$-3+(-2)$.}
\label{fig:-3+(-2)}
\end{figure}




\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}

\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)\vspace*{.5cm}\hangcaption{$a+b$
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)\vspace*{.5cm}\hangcaption{$-3+2$.}
\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)\vspace*{.5cm}\hangcaption{$a+b$
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)\vspace*{.5cm}\hangcaption{$3+(-2)$.}
\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}


\bigskip
We now explore multiplication. 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 would like to
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}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 $-26$
\item $-26$ \item $0$ \item $-303$ \item $0$ \item $+4$ \item $-100$
\end{enumerate}
\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}

\columnbreak

\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}

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

\end{pro}

\begin{pro}
Evaluate the expression $$
\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca} $$when $a=2$, $b=-3$,
and $c=5$.
\begin{answer}
We have,
$$ \begin{array}{lll} \dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}  & = &
\dfrac{(2)^3+(-3)^3+(5)^3-3(2)(-3)(5)}{(2)^2+(-3)^2+(5)^2-(2)(-3)-(-3)(5)-(5)(2)}\\[2ex]
& = & \dfrac{8-27+125+90}{4+9+25+6+15-10}\\[2ex]
& = & \dfrac{196}{49}\\[2ex]
& = & 4.  \end{array}$$
\end{answer}
\end{pro}

\end{multicols}


\section{Rational, Irrational, and Real 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
$$ \left\{-\dfrac{a}{b}: a\in\BBN, a>0, b\in \BBN, b>0 \right\}.  $$
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 $\BBQ$.
\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.

\bigskip

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 $\sqrt{2},
\sqrt{3}, \sqrt{5},\sqrt{6}, \sqrt{7}, \sqrt{8}$, 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 $\BBR$, 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}
 Evaluate the expression $$\dfrac{x}{y+z}+\dfrac{y}{z+x} +\dfrac{z}{x+y}-\left(\dfrac{x}{y}+ \dfrac{y}{z}+\dfrac{z}{x} \right)^2$$when $x=-1$, $y=2$,
and $z=-3$.
\begin{answer}
We have,
$$ \begin{array}{lll}\dfrac{x}{y+z}+\dfrac{y}{z+x} +\dfrac{z}{x+y}-\left(\dfrac{x}{y}+ \dfrac{y}{z}+\dfrac{z}{x}
\right)^2\\[2ex]
& = &
\dfrac{-1}{2+(-3)}+\dfrac{2}{-3+(-1)}+\dfrac{-3}{-1+2}+\left(\dfrac{-1}{2}+\dfrac{2}{-3}+\dfrac{-3}{-1}\right)^2\\[2ex]
& = & 1-\dfrac{1}{2}-3 + \left(-\dfrac{1}{2}-\dfrac{2}{3}+3\right)^2              \\[2ex]
& = & -2-\dfrac{1}{2} + \left(-\dfrac{1}{2}-\dfrac{2}{3}+3\right)^2              \\[2ex]
& = & -\dfrac{4}{2}-\dfrac{1}{2} + \left(-\dfrac{3}{6}-\dfrac{4}{6}+\dfrac{18}{6}\right)^2              \\[2ex]
& = & -\dfrac{5}{2} -\left(\dfrac{11}{6}\right)^2\\[2ex]
& = &-\dfrac{5}{2}-\dfrac{121}{36} \\[2ex]
& = &-\dfrac{90}{36}-\dfrac{121}{36} \\[2ex]
 & = & -\dfrac{211}{36}  \end{array}$$
\end{answer}
\end{pro}
\begin{pro}
Evaluate the following expressions when $x=-\dfrac{2}{3}$ and
$y=\dfrac{3}{5}$.
\begin{enumerate}
\item $2x+3y$
\item $xy-x-y$
\item $x^2+y^2$
\end{enumerate}
\begin{answer}
\noindent
\begin{enumerate}
\item $2x+3y= 2\left(-\dfrac{2}{3}\right)+3\left(\dfrac{3}{5}\right) = -\dfrac{4}{3}+\dfrac{9}{5}=-\dfrac{20}{15}+\dfrac{27}{15}=\dfrac{7}{15}.$
\item $xy-x-y= \left(-\dfrac{2}{3}\cdot \dfrac{3}{5}\right)-\left(-\dfrac{2}{3}\right)-\dfrac{3}{5}=-\dfrac{2}{5}+\dfrac{2}{3}-\dfrac{3}{5}=-\dfrac{6}{15}
+\dfrac{10}{15}-\dfrac{9}{15}=-\dfrac{5}{15}=-\dfrac{1}{3}.$
\item $x^2+y^2=\left(-\dfrac{2}{3}\right)^2+\left(\dfrac{3}{5}\right)^2=\left(-\dfrac{2}{3}\right)\left(-\dfrac{2}{3}\right)+\left(\dfrac{3}{5}\right)\left(\dfrac{3}{5}\right)=
\dfrac{4}{9}+\dfrac{9}{25} =
\dfrac{100}{225}+\dfrac{81}{225}=\dfrac{181}{225}.$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
In this problem, you are allowed to use any of the operations  $+,
-, \div, \cdot, !$, and exponentiation. You must use exactly four
$4$'s. Among your fours you may also use $.4$. The $n!$ (factorial)
symbol means that you multiply all the integers up to $n$. For
example, $1!=1$, $2! = 1\cdot 2 =2$, $3! = 1\cdot 2 \cdot 3 = 6$,
$4! = 1\cdot 2 \cdot 3 \cdot 4 = 24$. {\bf With these rules, write
every integer, from $1$ to $20$ inclusive.} For example,


$$ 11 = \dfrac{4}{.4} + \dfrac{4}{4}, \qquad 15 = \dfrac{44}{4} + 4, \qquad 20 =\dfrac{4}{.4} + \dfrac{4}{.4}, \qquad 13 = 4!-\dfrac{44}{4}. $$

\begin{answer}

Here are some possible answers. In some cases, there might be more
than one answer, as for example, $1 =
\dfrac{44}{44}=\dfrac{4+4}{4+4}=\dfrac{4^4}{4^4}$, etc.
$$\renewcommand{\arraystretch}{2.3}\begin{array}{|l|l|}
\hline 1 = \dfrac{44}{44} & 2 = \dfrac{4\cdot 4}{4+4} \\
\hline 3 = \dfrac{4+4+4}{4} & 4 =  4 + 4\cdot (4-4) \\
\hline 5 = \dfrac{4\cdot 4 + 4}{4} & 6 = 4 + \dfrac{4+4}{4} \\
\hline 7 = \dfrac{44}{4}-4 & 8 = 4\cdot 4 -4-4 \\
\hline 9 =  \dfrac{4}{.4}-\dfrac{4}{4} & 10 = \dfrac{44-4}{4} \\
\hline 11 = \dfrac{4}{.4} + \dfrac{4}{4} & 12 = \dfrac{44+ 4}{4} \\
\hline 13 = 4!-\dfrac{44}{4} & 14 = 4\cdot (4-.4)-.4 \\
\hline 15 = \dfrac{44}{4}+4 & 16 = 4\cdot 4 + 4-4 \\
\hline 17 = 4\cdot 4 + \dfrac{4}{4} & 18 = \dfrac{4}{.4}+4+4 \\
\hline 19 = 4!-4-\dfrac{4}{4} & 20 = \dfrac{4}{.4}+\dfrac{4}{.4} \\
\hline
\end{array}
$$


\end{answer}

\end{pro}
\begin{pro} Without using any of the signs $+, -, \div, \cdot$, but
exponentiation being allowed, what is the largest number that you
can form using three $4$'s? Again, you must explain your reasoning.

\begin{answer}To do this problem correctly, you must compare all
combinations of numbers formed with three fours.  Observe that
$$ (4^4)^4 =256^4=4294967296, $$
$$ 44^4 = 3748096, \quad 4^{44} = 309485009821345068724781056, $$
and that $$\begin{array}{lll}4^{(4^4)} & = & 4^{256}\\
& = &
134078079299425970995740249982058461274793658205923933777235614437217640300735 \to \\
\qquad & & \to
46976801874298166903427690031858186486050853753882811946569946433649006084096
\end{array}$$where we have broken the number so that it fits the
paper.

\bigskip

Note: This last calculation I did with a computer programme. How
does one actually check that the computer is right? At any rate,
since each time we multiply a given quantity by $4$ the quantity
increases, we must have $4^{44}<4^{256}= 4^{4^4}$.
\end{answer}
\end{pro}

\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} 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}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 $a$, $b$ 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}


\part{Algebraic Operations}
\chapter{Addition and Subtraction}

\section{Terms and Algebraic Expressions}
\QUOTEME{ The study of non-Euclidean Geometry brings nothing to
students but fatigue, vanity, arrogance, and imbecility.
``Non-Euclidean'' space is the false invention of demons, who gladly
furnish the dark understanding of the "non-Euclideans" with false
knowledge. The ``Non-Euclideans,'' like the ancient sophists, seem
unaware that their understandings have become obscured by the
promptings of the evil spirits.} {\mbox{Matthew~RYAN}}\bigskip

\dropping{2}{I}f in one room we had a group of $7$ Americans, $8$
Britons, and $3$ Canadians, and if in another room we had $4$
Americans, $4$ Britons and $1$ Canadian, we could write---in
shorthand--addressing the total of people by nationality, that we
have
$$(7A + 8B+3C) + (4A+4B+1C) = 11A + 12B+4C.$$The procedure of {\em collecting like
terms} that will be shortly explained, draws essentially from the
concept utilised in this example.

\bigskip

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:
$$(7\cdot 10^2 + 3\cdot 10 + 1)  +(6\cdot 10^2 + 9\cdot 10 + 5)= 13\cdot 10^2+12\cdot 10 + 6=1300+120+6=1426. $$


\begin{df} An {\em algebraic expression} is a collection of
symbols (letters and/or numbers).  An algebraic expression may have
one or more {\em terms}, which are separated from each other by the
signs $+$ (plus) or $-$ (minus).
\end{df}
\begin{exa}
The expression $18a + 3b - 5$ consists of three terms.
\end{exa}
\begin{exa}
The expression $18ab^2c$ consists of one term. Notice in this case
that since no sign precedes $18ab^2c$, the $+$ sign is tacitly
understood. In other words, $18ab^2c = +18ab^2c$.
\end{exa}
\begin{exa}
The expression $a + 3a^2 - 4a^3 -8ab + 7$ consists of five terms.
\end{exa}
\begin{df}
When one of the factors of a term is a number we call this number
the {\em numerical coefficient} (or {\em coefficient} for short) of
the term. In the expression $a^b$, $a$ is the {\em base} and $b$ is
the exponent.\index{exponent}
\end{df}
\begin{exa}
In the expression $13a^2b^3$, $13$ is the numerical coefficient of
the term, $2$ is the exponent of $a$ and $3$ is the exponent of $b$.
\end{exa}
\begin{exa}
In the expression $a^2b^d$, $1$ is the numerical coefficient of the
term, $2$ is the exponent of $a$ and $d$ is the exponent of $b$.
\end{exa}
\begin{exa}
In the expression $-a^c$, $-1$ is the numerical coefficient of the
term, and $c$ is the exponent of $a$. Notice then that  the
expression $-1a^c$ is equivalent to  the expression $-a^c$.
\end{exa}
\begin{df}If two terms
\begin{enumerate}
\item agree in their letters, and
\item for each letter appearing in them the exponent is the same,
\end{enumerate}then we say that the terms are  {\em like terms}.
\end{df}
\begin{exa}
The terms $-a, 7a, 5a$, are all like terms.
\end{exa}
\begin{exa}
The terms $4a^2b^3, 7b^3a^2, 5a^2b^3$, are all like terms. Notice
that by commutativity of multiplication, $7b^3a^2 = 7a^2b^3$.
\end{exa}
\begin{exa}
The terms $4a$ and $4a^2$, are unlike terms, since the exponent of
$a$ in $4a$ is $1$ and the exponent of $a$ in  $4a^2$ is $2$.
\end{exa}
\begin{exa}
The terms $4ab^2$ and $4ab^3$, are unlike terms, since the exponent
of $b$ in $4ab^2$ is $2$ and the exponent of $b$ in $4ab^3$ is $3$.
\end{exa}
To add or collect like terms, we simply add the coefficients of
the like terms.
\begin{exa}We have,$$(a + b) + (a + b) + (a + b) =  a + a + a + b + b + b = 3a +
3b$$
\end{exa}
\begin{rem}
By the commutative law, we may also write this as $3b+3a$.
\end{rem}
\begin{exa}We have,
$$\begin{array}{lcl} (-8a) + 9a + (-15a) + 4a & = & a + (-15a) + 4a \\
& = & (-14a) + 4a \\
& = & -10a \end{array}$$
\end{exa}
\begin{exa} We have, $$\begin{array}{lll}(5\clubsuit - 7\spadesuit )  + (-5\clubsuit - 7\spadesuit) &
= & (5\clubsuit) + (-5\clubsuit) + (-7\spadesuit) +(-7\spadesuit)\\
& = & -14\spadesuit.\end{array}$$
\end{exa}
\begin{exa}We have,
$$\frac{5a}{7} + \frac{(-6a)}{7} + \frac{(-18a)}{7} = \frac{5a + (-6a) + (-18a)}{7} = \frac{-19a}{7}.  $$
\end{exa}
\begin{exa}We have,
$$\frac{5a}{7} + \frac{6a}{7} + \frac{(-18a)}{7} = \frac{5a + 6a + (-18a)}{7} = \frac{-7a}{7} = -a.  $$
\end{exa}
\begin{exa}We have,
$$\begin{array}{lcl} (-13a + 5b) + (18a + (-18b)) & = & -13a + 18a + 5b + (-18b) \\
& = & 5a + (-13b)\\
& = & 5a -13b\\
\end{array}$$
\end{exa}
\begin{rem}
By the commutative law, we may also write this as $-13b+6a$.
\end{rem}
\begin{exa}We have,
$$\begin{array}{lcl} (5 - 4x + 3x^2) + (-1 -8x -3x^2)& = &
5 + (-1) -4x + (-8x) + 3x^2 + (-3x^2)  \\
& = & 4 + (-12x) + 0 \\
& = & 4 + (-12x)\\
& = & 4 -12x. \end{array}$$
\end{exa}
\begin{rem}
By the commutative law, we may also write this as $-12x+4$.
\end{rem}
\begin{exa}We have,
$$\begin{array}{lcl} (x^2 + xy + y^2 + xy^2) + (-x^2 + 2yx - 3y^2 - x^2y)
& = &  3xy - 2y^2 + xy^2 - x^2y. \\
 \end{array}$$
 This last expression is one of the many possible ways to write the
 final result, thanks to the commutative law.
\end{exa}

  \section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}A boy buys $a$ marbles, wins $b$, and loses $c$. How many
marbles has he then? Write one or two sentences explaining your
solution. \begin{answer} After buying the marbles, he has $+a$
marbles. After winning $b$ marbles, he now has $a+b$ marbles. After
losing $c$ marbles, he ends up with $a+b-c$. Conclusion: he has
$a+b-c$ marbles in the end.
\end{answer}
\end{pro}
\begin{pro}Collect like terms {\em mentally}.
\begin{enumerate}\item $ (-9a) + (+17a) - 8a$ \item $ (-17a) + (-9a)+ 20a $
\item $ -a-2a-3a-4a$ \item $10a-8a+6a-4a+2a$ \item $-5a
-3b-9a-10b+10a+17b$
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}\item $0$ \item $-6a $
\item $-10a$ \item $6a$ \item $-4a+4b$
\end{enumerate}
\end{answer}
\end{pro}
 \begin{pro} Collect like terms:
$$(a + b^2 + c + d^2) + (a^2 - b - c^2 - d) + (a^2 - b^2 + c - d) +
(a - b - c^2 - d^2).$$
\begin{answer} $2a-2b+2c-2d+2a^2-2c^2$.
   \end{answer}
   \end{pro}
\begin{pro} Collect like terms.
\begin{enumerate} \item $(a + 2b + 3c) + (3a + 2b + c)$
\item $(a + 2b + 3c) + (3a - 2b - c)$ \item $(a + 2b + 3c) + (3a +
2b + c) + (a + b + c)$ \item $(a + b + c) + (a - b + c) + (-a - b -
c)$ \item $(x^2 - 2x + 1) + (x^2 + 2x + 1)$ \item $(x^3 - x^2 + x -
1) + (2x^2 - x + 2)$ \item $(2x^3 - 2x^2 - 2) + (2x^3 - 2x - 1) +
(2x^3 - x^2 - x + 1)$  \item $\dfrac{(-5a)}{8} + \dfrac{19a}{8} +
\dfrac{(-23a)}{8}$
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate} \item $4a+4b+4c$
\item $4a+2c$ \item $5a+5b+5c$ \item $a-b+c$
\item $2x^2+2$ \item $x^3+x^2+1$ \item $6x^3-3x^2-3x-2$  \item $-\dfrac{9a}{8}$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
Find the value of the expression $$ (2x^2+x-2)+(-x^2-x+2) $$ when
$x=2$.
\begin{answer} Since $ (2x^2+x-2)+(-x^2-x+2) = x^2, $ the value is $2^2=4$. \end{answer}
\end{pro}
\begin{pro} Is it always  true that $a+a=2a$?
\begin{answer} Yes, this is always true.  \end{answer}
 \end{pro}
 \begin{pro} Is it always true that $a+a=2a^2$?
\begin{answer}No, this is not always true. For example, if $a=10$, then $a+a=10+10=20$, but $2a^2=2(10)^2=2\cdot 100 = 200$. Observe, however, that
$0+0=2\cdot 0^2$ and that $1+1=2\cdot 1^2$, so the assertion is true
in these two cases. That is it {\em only} true in these two cases is
more difficult to prove, and we will have to wait till we study
quadratic equations.
\end{answer}
 \end{pro}
 \begin{pro} Is it always true that $a+a^2=2a^3$?
\begin{answer}No, this is not always true. For example, if $a=10$, then $a+a^2=10+100=110$, but $2a^3=2(10)^3=2\cdot 1000 = 2000$.
Observe, however, that $0+0^2=2\cdot 0^3$, that $1+1^2=2\cdot 1^3$,
and that $\left(-\frac{1}{2}\right)+
\left(-\frac{1}{2}\right)^2=2\cdot \left(-\frac{1}{2}\right)^3$ so
the assertion is true in these three cases. That is it {\em only}
true in these three cases is more difficult to prove, and we will
have to wait till we study quadratic equations.
\end{answer}
 \end{pro}


\begin{pro}
If $$(8x - 5)^5 = Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F,$$  find $A + B
+ C + D + E + F$.
\begin{answer}
Letting $x = 1$ one gathers that $$243 = 3^5 = (8(1) - 5)^5 = A + B
+ C + D + E + F.$$
\end{answer}
\end{pro}

\begin{pro}
Give three different equivalent forms for the expression $3yx-x^2y$.
\begin{answer}
Here are a few. Many more are possible.
\begin{enumerate}
\item $3xy-x^2y   $
\item $-x^2y+3yx   $
\item $3yx -yx^2  $
\item $3xy-yx^2   $
\item $ -yx^2-3xy  $
\end{enumerate}
\end{answer}

\end{pro}
\end{multicols}

\section{More Suppression of Parentheses}
\QUOTEME{ He is unworthy of the name of {\em Man} who is ignorant of
the fact that the diagonal of a square is incommensurable with its
side.}{Plato}

\bigskip

A minus sign preceding an expression in parentheses negates every
term of the expression. Thus
$$ -(a+b)=-a-b.$$
\begin{exa}We have,
$$\begin{array}{lcl} (a - 2b) - (-a + 5b) & = & a - 2b + a - 5b \\
& = & a + a -2b -5b \\
& = & 2a - 7b  \end{array}  $$
\end{exa}
\begin{exa}We have,
$$\begin{array}{lcl} (3- 2x - x^2) - (5 - 6x - 7x^2) & = &
3-2x-x^2-5+6x+7x^2\\
& = & 3-5-2x+6x-x^2+7x^2 \\
& = & -2+4x+6x^2  \end{array}  $$
\end{exa}
\begin{exa}We have,
$$\begin{array}{lcl}  (a - b) - (2a -3b) - (-a +3b) & = & a - b
-2a+3b+a-3b\\
& = & a-2a+a-b+3b-3b \\
& = & -b \end{array}  $$
\end{exa}
\begin{exa}We have,
$$\begin{array}{lcl} (x^2 - x - 1) - (-x -3x^2) + (2x^2 - 8)
& = & x^2-x-1+x+3x^2+2x^2-8\\
& = & x^2+3x^2 +2x^2-x+x-1-8 \\
& = &  6x^2 - 9.  \end{array}  $$
\end{exa}
Recall that the distributive law states that for real numbers $a, b,
c$, $$ a(b+c) = ab+ac. $$We now apply this to algebraic expressions.
\begin{exa}
Multiply term by term: \quad $2(a+2b-3c).$
\end{exa}
\begin{solu}
We have, $$ 2(a+2b-3c) = 2a + 2(2b)+2(-3c)=2a+4b-6c. $$
\end{solu}
\begin{exa}
Multiply term by term: \quad $-2(a+2b-3c).$
\end{exa}
\begin{solu}
We have, $$ -2(a+2b-3c) = -2a + (-2)(2b)+(-2)(-3c)=-2a-4b+6c. $$
\end{solu}
\begin{exa}
Divide term by term: \quad $\dfrac{2a -4b+7c}{2}.$
\end{exa}
\begin{solu}
We have, $$\dfrac{2a -4b+7c}{2} =
\dfrac{2a}{2}+\dfrac{-4b}{2}+\dfrac{7c}{2}=a-2b+\dfrac{7c}{2} . $$
\end{solu}
\begin{rem}We prefer to use improper fractions and write $\dfrac{7c}{2}$ rather than $3\tfrac{1}{2}c$. Also, the expressions $\dfrac{7}{2}c$ and $\dfrac{7c}{2}$
are identical, that is, $\dfrac{7}{2}c=\dfrac{7c}{2}$.
\end{rem}
We may combine the distributive law and combining like terms.
\begin{exa}
Combine like terms: \qquad $2(a-2b)+3(-a+2b)$.
\end{exa}
\begin{solu}
We have, $$ \begin{array}{lll}2(a-2b)+3(-a+2b) & = & 2a -4b +
(-3a)+6b \\
& = & -a + 2b.
\end{array}$$
\end{solu}
\begin{exa}
Combine like terms: \qquad $2(a-2b)-3(-a+2b)$.
\end{exa}
\begin{solu}
We have, $$ \begin{array}{lll}2(a-2b)-3(-a+2b) & = & 2a -4b +
3a-6b \\
& = & 5a-10b.
\end{array}$$
\end{solu}
\begin{exa}
Combine like terms: \qquad $-3(1-2x+4x^2)+\dfrac{9x^2-3x+6}{-3}$.
\end{exa}
\begin{solu}
We have, $$ \begin{array}{lll}-3(1-2x+4x^2)+\dfrac{9x^2-3x+6}{-3} &
= & -3+6x-12x^2-3x^2+x-2 \\
& = & -15x^2+7x-5.
\end{array}$$
\end{solu}
\begin{exa}
Combine like terms: \qquad
$\dfrac{3}{2}\left(\dfrac{x}{3}-2\right)-2\left(\dfrac{x}{12}+2\right)$.
\end{exa}
\begin{solu}
We have, $$
\begin{array}{lll}\dfrac{3}{2}\left(\dfrac{x}{3}-2\right)-2\left(\dfrac{x}{12}+2\right)
& = & \dfrac{3x}{6} -\dfrac{6}{2} -\dfrac{2x}{12}-4 \\
& = & \dfrac{6x}{12}-\dfrac{2x}{12}-3-4\\
& = & \dfrac{4x}{12}-7\\
& = & \dfrac{x}{3}-7.
\end{array}$$
\end{solu}
  \section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
From a rod $a+b$ inches long, $b-c$ inches are cut off; how much
remains? Write one or two sentences explaining your solution.
\begin{answer}
Since $b-c$ inches are cut of from $a+b$, we must subtract $b-c$
from $a+b$, giving
$$ (a+b)-(b-c) = a+b-b+c = a+c, $$whence there remain $a+c$ inches.
\end{answer}
\end{pro}

\begin{pro}
 A group of $x$ people is going on an excursion and decides to
charter a plane. The total price of renting the plane is $D$
dollars, and everyone will pay an equal share. What is this share?
Now, the day of departure it turns out that $p$ people are no-shews.
How much more will the people who shew up have to pay in order to
cover the original cost?
\begin{answer} Since there are $x$ people
to share the cost of $D$, the share of each person is
$\dfrac{D}{x}$. If $p$ persons disappear, there remain $x-p$
persons, and the share of each would increase to $\dfrac{D}{x-p}$.
This means they will have to pay
$$ \dfrac{D}{x-p}-\dfrac{D}{x}  $$more.
\end{answer}
\end{pro}

\begin{pro} Suppress parentheses and collect like terms:
$$(a + b^2 + c + d^2) - (a^2 - b - c^2 - d) + (a^2 - b^2 + c - d) -
(a - b - c^2 - d^2).$$
\begin{answer}
$2b+2c+2c^2+2d^2 $
\end{answer}
\end{pro}
\begin{pro}Expand and simplify: $$2(a + b - 2c) - 3(a - b - c).$$
\begin{answer} $-a+5b-c$ \end{answer}
\end{pro}
\begin{pro} Suppress parentheses and collect like terms.
\begin{enumerate}\item $(a + 2b + 3c) - (3a + 2b + c)$
\item $(a + 2b + 3c) + (3a + 2b + c) - (a + b + c)$ \item $(a + b
+ c) - (a - b + c) - (-a - b - c)$ \item $(x^3 - x^2 + x - 1) -
(2x^2 - x + 2)$ \item $(2x^3 - 2x^2 - 2) - (2x^3 - 2x - 1) - (2x^3 -
x^2 - x + 1)$ \item $(8.1a - 9.2b) - (-0.2a - 3.1b) - (-a +
3b)$\item $-\left(-\dfrac{14b}{a} +\dfrac{15b}{a} -
\dfrac{1}{a}\right) -\left( -\dfrac{4b}{a}+ \dfrac{5}{a} -
\dfrac{8b}{a}\right)$
\item $-\left(\dfrac{4}{a} -\dfrac{5}{a} - 2\right) -\left(
\dfrac{4}{a}+ \dfrac{5}{a} - 8\right)$ \item $(5\clubsuit -
7\spadesuit )  - (-5\clubsuit - 7\spadesuit)$
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}\item $-2a+2c$
\item $3a+3b+3c$ \item $a+3b+c$ \item $x^3-3x^2+2x-3$
\item $-2x^3-x^2+3x-2$
\item $9.3a-3.1b$\item
$ \dfrac{11b-4}{a}$ \item $ 10-\dfrac{8}{a}$
\item $10\clubsuit$
\end{enumerate}

\end{answer}
\end{pro}
\begin{pro}
Collect like terms and write as a single fraction:\qquad\qquad  $$
\dfrac{a}{3} - \dfrac{a^2}{5}+\dfrac{a}{2} + \dfrac{5a^2}{6}. $$

\begin{answer}
We have
$$  \dfrac{a}{3} - \dfrac{a^2}{5}+\dfrac{a}{2} + \dfrac{5a^2}{6} =  \dfrac{a}{3}+\dfrac{a}{2} - \dfrac{a^2}{5} + \dfrac{5a^2}{6}
=  \dfrac{2a}{6}+
\dfrac{3a}{6}-\dfrac{6a^2}{30}+\dfrac{25a^2}{30}=\dfrac{5a}{6}+\dfrac{19a^2}{30}=
\dfrac{25a}{30}+\dfrac{19a^2}{30}=\dfrac{25a+19a^2}{30}$$

\end{answer}

\end{pro}

\begin{pro}
Combine like terms.
\begin{enumerate}
\item $2(-x+2y)+3(x-2y)-(x+y)+(y-2x)$
\item $\dfrac{14x-7}{7}-\dfrac{1}{3}\left(-3x+12\right)$
\item $\dfrac{4x-12x^2+6}{-2}-2(1-2x+x^2)$
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}
\item $-2x-2y $
\item $3x-5$
\item $4x^2+2x-5$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
Match each expression on the left with the equivalent expression on
the right.
\begin{center}
\begin{tabular}{|l|l|}
\hline
\rule{1cm}{2pt} 1. \quad $x-y$  & A. \quad $y-x$ \\
\rule{1cm}{2pt} 2. \quad  $x+y$  & B. \quad  $-y+x$ \\
\rule{1cm}{2pt} 3.\quad   $-(x-y)$  & C. \quad  $x-(-y)$ \\
\hline
\end{tabular}
\end{center}
\begin{answer}
$x-y$ by commutativity is $-y + x$, and so 1 matches with B. By the
subtraction rule, $x-(-y)=x+(+y)=x+y$, and so 2 matches with C.
Finally, using the distributive law and the commutative law,
$-(x-y)=-x+y=y-x$, and so 3 matches with A.
\end{answer}
\end{pro}
\begin{pro}
You start the day with $A$ dollars. Your uncle Bob gives you enough
money to  double your amount. Your aunt Rita gives you $10$ dollars.
You have to pay $B$ dollars in fines, and spent $12$ dollars fueling
your camel with gas.  How much money do you have now?

\begin{answer}
You start with $A$, your uncle Bob gives you $A$ more so that you
now have $A+A=2A$. Your Aunt Rita gives you $10$ more, so that now
you have $2A+10$. You pay $B$ dollars in fines, hence you are left
with $2A+10-B$. You spent $12$ in fuel, so now you have $$
2A+10-B-12 = 2A-B-2,$$in total.
\end{answer}
\end{pro}
\begin{pro}
Simplify each of the following expressions as much as possible
\begin{enumerate}
\item $(2-4)x -5(-6)x$
\item $5-2(2x)$
\item $(5-2)(1-2^2)^3x$
\end{enumerate}
\begin{answer}
\noindent
\begin{enumerate}
\item $(2-4)x -5(-6)x=-2x+30x=28x$
\item $5-2(2x)=5-4x$
\item $(5-2)(1-2^2)^3x=(3)(1-4)x=(3)(-3)^3x=(3)(-27)x=-81x$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
Remove all redundant parentheses. If removing parentheses changes
the expression, state so.
\begin{enumerate}
\item $(6t)+x$
\item $6(t+x)$
\item $(ab)^2$
\item $(ab-c)^2$
\end{enumerate}
\begin{answer}
\noindent
\begin{enumerate}
\item $(6t)+x=6t+x$, so the parentheses are redundant.
\item In $6(t+x)$, the parentheses are needed.
\item In $(ab)^2$, the parentheses are needed.
\item In $(ab-c)^2$, the parentheses are needed.
\end{enumerate}

\end{answer}

\end{pro}

\begin{pro}
Fill in the box to make the expression true.
\begin{enumerate}
\item $\dfrac{a-2b}{c} = \dfrac{a}{c}+\fbox{}$
\item $\dfrac{t+3}{t-3} =(t+3)\fbox{}$
\end{enumerate}
\begin{answer}
\noindent
\begin{enumerate}
\item In $\dfrac{a-2b}{c} = \dfrac{a}{c}+\fbox{}$,  the $\fbox{}$ is $-\dfrac{2b}{c}$
\item In $\dfrac{t+3}{t-3} =(t+3)\fbox{}$, the $\fbox{}$ is $\dfrac{1}{t-3}$.
\end{enumerate}
\end{answer}
\end{pro}

\begin{pro}
Fill in the box to make the expression true.
\begin{enumerate}
\item $1-t = \fbox{} t+1$
\item $3x+y-t = \fbox{}+3x$
\end{enumerate}

\begin{answer}
\noindent
\begin{enumerate}
\item In $1-t = \fbox{} t+1$, the $\fbox{}$ is $ -1$ or simply, $-$.
\item In $3x+y-t = \fbox{}+3x$ the $\fbox{}$ is $y-t .$
\end{enumerate}
\end{answer}

\end{pro}

\begin{pro}
What is the opposite of $4-y+2$?
\begin{answer}
The opposite sought is $$-(4-y+2)=-4+y-2.  $$
\end{answer}
\end{pro}

\begin{pro}
What is the additive inverse of $\dfrac{-a}{-(b-c)}$?
\begin{answer}
The additive inverse sought is
$$-\left(\dfrac{-a}{-(b-c)}\right)=-\left(\dfrac{a}{(b-c)}\right)=-\dfrac{a}{b-c}.
$$
\end{answer}
\end{pro}

\end{multicols}
\chapter{Multiplication}
\section{Laws of Exponents}
\QUOTEME{Anyone who cannot cope with mathematics is not fully human.
At best he is a tolerable subhuman who has learned to wear shoes,
bathe, and not make messes in the house. }{Robert A. HEINLEIN}

\bigskip
\index{exponents!laws} Recall that if $a$ is a real number and if
$n$ is a natural number then
$$a^n = \underbrace{a\cdot a \cdots a}_{n\ a\mathrm{'s}}, $$
with the interpretation that  $a^0=1$ if $a\neq 0$.

\bigskip

Suppose we wanted to compute $a^2a^5$. We proceed as follows, using
the fact that $a^2 =aa$ and that $a^5=aaaaa$:
$$(a^2)(a^5)=(aa)(aaaaa)=aaaaaaa=a^7,
$$since the penultimate expression consists of seven $a$'s. In
general, we have the following.
\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}$(ab^2c^3)(a^3b^3c) = a^{1 + 3}b^{2 + 3}c^{3 + 1} =
a^4b^5c^4$.
\end{exa}
\begin{exa}$(a^x)(b^{2y})(a^{2x}b^{2y}) = a^{x + 2x}b^{2y + 2y} = a^{3x}b^{4y}.$
\end{exa}\begin{exa}$(a^{x -
2y + z})(a^{2x - y - z})(a^{x + y + z}) = a^{x - 2y + z + 2x - y - z
+ x + y + z} = a^{4x - 2y + z}.$
\begin{exa}
$$5^5 + 5^5 + 5^5 + 5^5 + 5^5 = 5(5^5) = 5^{1 + 5} = 5^6 =  15625. $$
\end{exa}
We may use associativity and commutativity in the case when the
numerical coefficients are different from $1$.
\begin{exa} $(5x^2)(6x^3)=30x^5$.\end{exa}
\begin{exa} $(-2ab^2c^3)(3abc^4)=-6a^2b^3c^7$.\end{exa}
We now tackle division. As a particular case, observe that
$$\dfrac{a^5}{a^2} = \dfrac{aaaaa}{aa} = \dfrac{aaa\cancel{a}\cancel{a}}{\cancel{a}\cancel{a}} = a^3,  $$
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}
$$\dfrac{-24a^{12}b^9c^5d^2}{2a^6b^3c^4d^2} = -12a^{12 - 6}b^{9 - 3}c^{5 - 4}d^{2- 2} = -12a^6b^6c^1d^0 = -12a^6b^6c.  $$
\end{exa}
\begin{exa}$\dfrac{(5^{a})(5^b)}{5^{a + b - 2}} = 5^{a + b - (a + b - 2)} = 5^2 = 25.$ \end{exa}
\begin{exa}$\dfrac{15x^5y^7z^4}{5x^2y^2z^2} = 3x^3y^5z^2.$\end{exa}
\begin{exa}$\dfrac{16b^2yx^2}{-2xy} = -8b^2x$\end{exa}
\begin{exa}$\dfrac{7a^2bc}{-7a^2bc} = -1.$\end{exa}
\begin{exa}$\dfrac{(a + b)^4(2 + a + b)^6}{(a + b)^3(2 + a + b)^4} = (a + b)(2 + a + b)^2$ \end{exa}
\begin{exa}$\dfrac{a^5b^4(a + b)^3}{a^3b^4(a + b)} = a^2(a + b)^2.$ \end{exa}
Suppose we wanted to compute $(a^2)^5$. Since a quantity raised to
the fifth power is simply the quantity multiplied by itself five
times, we have:
$$(a^2)^5=(a^2)(a^2)(a^2)(a^2)(a^2) = a^{2+2+2+2+2}=a^{10},
$$upon using the first law of exponents. In
general we have the following law dealing with ``exponents of
exponents.''
\begin{thm}[Third Law of Exponents]Let $a$ be a real number and $m, n$ natural numbers. Then
$$(a^m)^n = a^{mn}.  $$
\end{thm}
\begin{pf}
We have $$\begin{array}{lll}(a^m)^n & = & \underbrace{a^m\cdot a^m
\cdots a^m}_{n\ a^m\mathrm{'s}} \\[5mm]
& = &  a^{\underbrace{m+m+\cdots +m}_{n\ a\mathrm{'s}}}\\[3mm]
 & = & a^{mn}.
\end{array}$$
\end{pf}
\begin{thm}[Fourth Law of Exponents]Let $a$ and $b$ be a real numbers and let $m$ be a natural number. Then
$$(ab)^m = a^{m}b^m.  $$
\end{thm}
\begin{pf}
We have $$\begin{array}{lll}(ab)^m & = & \underbrace{ab\cdot ab
\cdots ab}_{m\ ab\mathrm{'s}} \\
& = & \underbrace{a\cdot a \cdots a}_{m\ a\mathrm{'s}}
\underbrace{b\cdot b \cdots b}_{m\ b\mathrm{'s}} \\
 & = & a^{m}b^m.
\end{array}$$
\end{pf}
\begin{exa} We have, $$\begin{array}{lll}(2ab^2c^3)^2(-3a^3bc^2)^3 & = &
(2^2(a^1)^2(b^2)^2(c^3)^2)((-3)^3(b^1)^3(c^2)^3)\\ &  = &
(4a^2b^4c^6)(-27a^9b^3c^6) \\ & = &  -108a^{11}b^7c^{12}.
\end{array}$$
\end{exa}
\begin{exa}$(a^{x^3})^{2x} = a^{(x^3)(2x)} = a^{2x^4}$ \end{exa}
\end{exa}
\begin{exa}
$25^34^3 = (25\cdot 4)^3 = 100^3 = 1000000. $
 \end{exa}
 \section*{Homework}

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

\begin{pro}Simplify: $\dfrac{24xyz^3}{-3z^2}$
\begin{answer} $-8xyz$  \end{answer}
\end{pro}
 \begin{pro}Simplify: $\dfrac{-168a^2b^2cx^2}{-7abx^2}$
\begin{answer} $24abc$ \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}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}

\begin{pro}Demonstrate that $$(a^xb^y)\left(\dfrac{b^{2x}}{a^{-y}}\right) = a^{x + y}b^{y +
2x}.$$
\begin{answer} $(a^xb^y)\left(\dfrac{b^{2x}}{a^{-y}}\right)
=a^{x-(-y)}b^{y+2x} = a^{x + y}b^{y + 2x}  $ \end{answer}
\end{pro}
\begin{pro}Divide: $\dfrac{1}{x} \div \dfrac{1}{x^2}$.

\begin{answer} $x$   \end{answer}
 \end{pro}

\begin{pro}Multiply: $\left(\dfrac{1}{x}\right)\left( \dfrac{1}{x^2}\right)$.
\begin{answer} $\dfrac{1}{x^3}$  \end{answer}
\end{pro}
\begin{pro}Divide: $\dfrac{1}{s - 1} \div \dfrac{1}{(s - 1)^2}$.
\begin{answer}  $s-1 $\end{answer}

\end{pro}\begin{pro}Multiply: $\left(\dfrac{1}{s - 1} \right)\left(  \dfrac{1}{(s -
1)^2}\right)$.
\begin{answer} $ \dfrac{1}{(s-1)^3} $ \end{answer}

\end{pro}\begin{pro}Divide: $\dfrac{a^2}{x} \div
\dfrac{x}{a^2}$.

\begin{answer} $ \dfrac{a^4}{x^2} $ \end{answer}

\end{pro}\begin{pro}Multiply: $\left(\dfrac{a^2}{x} \right)\left(
\dfrac{x}{a^2}\right)$.

\begin{answer} $ 1 $ \end{answer}

\end{pro}
\begin{pro}
Simplify and write as a single number:
$$\dfrac{6^5+6^5+6^5+6^5+6^5+6^5+6^5+6^5+6^5+6^5}{3^3+3^3+3^3+3^3+3^3}.$$
\begin{answer}
We have,
$$\begin{array}{lll}\dfrac{6^5+6^5+6^5+6^5+6^5+6^5+6^5+6^5+6^5+6^5}{3^3+3^3+3^3+3^3+3^3} & = &
\dfrac{10\cdot 6^5}{5\cdot 3^3} \\
& = & \dfrac{2\cdot 6^5}{3^3}\\
& = & \dfrac{2\cdot 2^5\cdot 3^5}{3^3}\\
& = & 2\cdot 2^5\cdot 3^2\\
& = & 2\cdot 32 \cdot 9 \\
& = & 576. \end{array}$$
\end{answer}

\end{pro}
\begin{pro}
If $\dfrac{a^8}{b^9}\div \dfrac{(a^2b)^3}{b^{20}} =a^mb^n$, then
\mbox{$m= \rule{2cm}{2pt}$} and \mbox{$n=\rule{2cm}{2pt}$}.
\begin{answer}
We have,
$$ \begin{array}{lll} \dfrac{a^8}{b^9}\div \dfrac{(a^2b)^3}{b^{20}}& = &
\dfrac{a^8}{b^9}\cdot \dfrac{b^{20}}{(a^2b)^3}\\ \vspace{2mm} & = &
\dfrac{a^8}{b^9}\cdot \dfrac{b^{20}}{a^6b^3}\\ \vspace{2mm} & = &
\dfrac{a^8b^{20}}{a^6b^{12}}\\ \vspace{2mm} & = & a^{8-6}b^{20-12}\\
\vspace{2mm} & = & a^2b^{8}.
 \end{array}$$
 So $m=2$ and $n=11$.
\end{answer}
\end{pro}

\end{multicols}




\section{Negative Exponents}
\QUOTEME{One of the big misapprehensions about mathematics that we
perpetrate in our classrooms is that the teacher always seems to
know the answer to any problem that is discussed. This gives
students the idea that there is a book somewhere with all the right
answers to all of the interesting questions, and that teachers know
those answers. And if one could get hold of the book, one would have
everything settled. That's so unlike the true nature of
mathematics.}{Leon HEINKIN}


\bigskip
Let $a\neq 0$ be a real number and let $n$ be a natural
number.\index{exponents!negative} Then
$$ 1=a^0=a^{n-n}=a^na^{-n} \qquad \implies\qquad  a^{-n} = \dfrac{1}{a^n}. $$This
gives an interpretation to negative exponents. Observe also that
taking reciprocals,
$$a^{-n}=\dfrac{1}{a^n} \qquad \implies\qquad  \dfrac{1}{a^{-n}} = \dfrac{1}{\dfrac{1}{a^n}} = a^n.  $$
\begin{exa} $2^{-3} = \dfrac{1}{2^3}=\dfrac{1}{8}$. \end{exa}
\begin{exa}
$$ \dfrac{2^4}{2^9} = 2^{4 - 9} = 2^{-5} = \dfrac{1}{2^5} = \dfrac{1}{32}. $$
\end{exa}

\begin{exa} $\dfrac{2^{-3}}{3^{-2}} = \dfrac{3^2}{2^3}=\dfrac{9}{8}$. \end{exa}
\begin{exa} $\left(\dfrac{2^4}{5^{-3}}\right)^{-2} = \dfrac{2^{-8}}{5^6}=\dfrac{1}{256\cdot 15625} = \dfrac{1}{4000000}$. \end{exa}
\begin{exa}
Simplify, using positive exponents only: $$ \dfrac{x^{-6}}{y^9}\cdot
\dfrac{y^{-8}}{x^{-12}}
$$
\end{exa}\begin{solu} $$
\dfrac{x^{-6}}{y^9}\cdot \dfrac{y^{-8}}{x^{-12}}
 = \dfrac{x^{-6}y^{-8}}{y^9x^{-12}} = x^{-6-(-12)}y^{-8-9} = x^{6}y^{-17} = \dfrac{x^6}{y^{17}}.$$
\end{solu}
\begin{exa}
Simplify, using positive exponents only: $$
\left(\dfrac{x^{-6}}{y^9}\right)^{-3}\cdot
\left(\dfrac{y^{-1}}{x^{-2}}\right)^2.
$$
\end{exa}
\begin{solu}We have, $$ \left(\dfrac{x^{-6}}{y^9}\right)^{-3}\cdot
\left(\dfrac{y^{-1}}{x^{-2}}\right)^2
 = \dfrac{x^{18}}{y^{-27}}\cdot\dfrac{y^{-2}}{x^{-4}} = x^{18-(-4)}y^{-2-(-27)} = x^{22}y^{25}.$$
 \end{solu}
\section*{Homework}

\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}Simplify, using positive exponents
only.
\begin{enumerate}\item $\dfrac{x^{-5}}{x^4}$
\item $\dfrac{x^4}{x^{-5}}$ \item $x^{-5}x^4$ \item
$(x^{-1}y^2z^3)^3$ \item $\dfrac{a^3b^{-2}}{a^{-6}b^4}$ \item
$\dfrac{(ab)(ab^2)}{(a^{-1}b^{-2})^2(a^2b^2)}$
\end{enumerate}
\begin{answer} \noindent
\begin{enumerate}\item $\dfrac{1}{x^9}$
\item $x^9$ \item $\dfrac{1}{x}$ \item
$\dfrac{y^6z^9}{x^3}$ \item $ \dfrac{a^9}{b^6}$
\item $a^2b^5$
\end{enumerate}

 \end{answer}
\end{pro}
\begin{pro}Demonstrate that $$(a^{x + 3}b^{y - 4})(a^{2x -
3}b^{-2y + 5}) = a^{3x}b^{1 - y}.$$
\begin{answer}
We have,
$$(a^{x + 3}b^{y - 4})(a^{2x -
3}b^{-2y + 5}) =a^{x+3+2x-3}b^{y-4-2y+5} = a^{3x}b^{1 - y}.$$
\end{answer}
\end{pro}

\begin{pro}Demonstrate that $\dfrac{a^xb^y}{a^{2x}b^{-2y}} = a^{-x}b^{3y}$. \end{pro}
\begin{answer}
We have,
$$\dfrac{a^xb^y}{a^{2x}b^{-2y}}=a^{x-2x}b^{y-(-2y)} = a^{-x}b^{3y}
.$$
\end{answer}
\begin{pro}
Simplify and express with positive exponents only: \quad
$\dfrac{x^6}{y^9}  \div \dfrac{x^{-2}}{y^{-3}} $
\begin{answer}
$\dfrac{x^8}{y^{12}}$
\end{answer}

\end{pro}
\begin{pro}
What is the exact numerical value of
$\dfrac{2^{20}6^{10}}{4^{11}3^{10}}$?
\begin{answer}
$2^8=256$
\end{answer}

\end{pro}
\begin{pro}
What is the exact numerical value of
$\left(\left(\dfrac{2}{5}\right)^{-1}-\left(\dfrac{1}{2}\right)^{-1}\right)^{-2}$?
\begin{answer}
$4$
\end{answer}

\end{pro}
\begin{pro} What is the exact numerical value of $\dfrac{(2^4)^8}{(4^8)^2}$.
\begin{answer} $\dfrac{(2^4)^8}{(4^8)^2}=\dfrac{2^{32}}{((2^2)^8)^2}=\dfrac{2^{32}}{2^{32}} =1.$ \end{answer}
\end{pro}
\end{multicols}

\section{Distributive Law}
\QUOTEME{You treat world history as a mathematician does
mathematics, in which nothing but laws and formul\ae\ exist, no
reality, no good and evil, no time, no yesterday, no tomorrow,
nothing but an eternal, shallow, mathematical present.}{Hermann
HESSE}

\bigskip
Recall that the distributive law says that if $a, b,$ and $c$ are
real numbers, then
$$a(b + c) = ab + ac,
\qquad (a+ b)c = ac + bc.  $$ We now apply the distributive law to
multiplication of terms.
\begin{exa}We have,
$$\begin{array}{lll}-3xy^2(2x^3 - 5xy^3) &  = & (-3xy^2)(2x^3) -
(-3xy^2)(5xy^3)\\
& = &  -6x^4y^2 + 15x^2y^5. \end{array}$$
\end{exa}
\begin{exa}We have,
$$\begin{array}{lll}(-3a^2b)(-3ab^2 + 2b) - (a^2 - 3a^2b)(ab^2) &  = &
9a^3b^3 - 6a^2b^2 - (a^3b^2  - 3a^3b^3). \\
& = & 9a^3b^3 - 6a^2b^2 + (-a^3b^2)  + 3a^3b^3 \\
& = & 12a^3b^3 -6a^2b^2 - a^3b^2.
\end{array}$$
\end{exa}
From the distributive law we deduce that
$$(a + b)(c + d) = (a + b)c + (a + b)d = ac + bc + ad + bd.    $$
\begin{exa}We have,$$\begin{array}{lll}(2x + 3)(5x + 7) &  = & (2x + 3)(5x) + (2x + 3)(7) \\
& = & 10x^2 + 15x + 14x + 21 \\
& = & 10x^2 + 29x + 21. \end{array} $$
\end{exa}
\begin{rem}
Because of commutativity, we may also proceed as follows,
$$\begin{array}{lll}(2x + 3)(5x + 7) &  = & (2x)(5x+7)+3(5x+7) \\
& = & 10x^2+14x+15x+21 \\
& = & 10x^2 + 29x + 21. \end{array} $$
\end{rem}
\begin{exa}We have,$$\begin{array}{lll}(2x - 3)(x - 4) &  = & (2x - 3)(x) + (2x - 3)(-4) \\
& = & 2x^2 - 3x - 8x + 12 \\
& = & 2x^2 - 11x + 12. \end{array} $$
\end{exa}
\begin{rem}
Because of commutativity, we may also proceed as follows,
$$\begin{array}{lll}(2x - 3)(x - 4) &  = & (2x)(x-4)-3(x-4)\\
& = &   2x^2-8x -3x+12    \\ & = & 2x^2 - 11x + 12. \end{array} $$
\end{rem}
We also recall the meaning of exponentiation.
\begin{exa}
We have,
$$\begin{array}{lll} (x-4)^2  & = &  (x-4)(x-4)\\ & = &  x(x-4)-4(x-4)\\ & = & x^2-4x-4x+16\\ & = & x^2-8x+16. \end{array}$$
\end{exa}

\begin{exa} To expand $(x+1)^3$, we first find
$$ (x+1)^2=(x+1)(x+1)=x(x+1)+1(x+1)=x^2+x+x+1=x^2+2x+1. $$
We use this result and find,
$$
\begin{array}{lll} (x+1)^3 & = & (x+1)(x+1)^2 \\
& = & (x+1)(x^2+2x+1) \\
& = & x(x^2+2x+1)+1(x^2+2x+1)\\
& = & x^3+2x^2+x+x^2+2x+1\\
& = & x^3+3x^2+3x+1.
  \end{array}$$
\end{exa}

 \section*{Homework}
 \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Expand and simplify.
\begin{enumerate}
\item $x(x^2+x+1)$
\item $2x(x^2-2x+3)$
\item $(2x+1)(2x+1)$
\item $(2x+1)(2x-1)$
\item $(2x-1)(2x-1)$
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}
\item $x^3+x^2+x$
\item $2x^3-4x^2+6x$
\item $4x^2+4x+1$
\item $4x^2-1$
\item $4x^2-4x+1$
\end{enumerate}
\end{answer}

\end{pro}

\begin{pro}Expand and
simplify: $$(x^2 + 1) + 2x(x^2 + 2) - 3x^3(x^2 + 3).$$
\begin{answer}
$ x^2+1-7x^3+4x-3x^5  $
\end{answer}
\end{pro}
\begin{pro}Expand and simplify: $$4x^3(x^2
- 2x + 2) - 2x^2(x^2 - x^2 - 1).$$
\begin{answer} $4x^5-8x^4+8x^3+2x^2$  \end{answer}

\end{pro}

\begin{pro}Multiply and collect like terms.
\begin{enumerate}
\item $-2x(x + 1) + 3x^2(x - 3x^2)$ \item $-2x(x + 1)^2$ \item
$a^2c(ab - ac) + ac^2(a^2 - 2b)$ \item $(x + 2y + 3z)(x - 2y)$

\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}
\item $-9x^4+3x^3-2x^2-2x$ \item $-2x^3-4x^2-2x$ \item
$a^3cb-2ac^2b$ \item $x^2-4y^2+3zx-6zy$

\end{enumerate}
\end{answer}


\end{pro}


\begin{pro}
Expand the product $(2x + 2y + 1)(x - y + 1).$
\begin{answer}We have,$$\begin{array}{lll}(2x + 2y + 1)(x - y + 1) &  = &
(2x  + 2y + 1)(x) + (2x + 2y + 1)(-y) + (2x + 2y + 1)(1) \\
& = & 2x^2 + 2xy + x - 2xy - 2y^2 - y + 2x + 2y + 1\\
& = & 2x^2 + 3x - 2y^2 + y + 1. \end{array} $$
\end{answer}
\end{pro}

\begin{pro}
Expand and collect like terms:  $$(x+2)(x+3)-(x-2)(x-3).$$
\begin{answer}
$10x$.
\end{answer}

\end{pro}
\begin{pro}
Expand and collect like terms: \quad
$$(x-1)(x^2+x+1)-(x+1)(x^2-x+1).$$
\begin{answer}
$-2$.
\end{answer}

\end{pro}

\begin{pro}Expand and simplify: $$(x + y - z)(x - y + 2z).$$
\begin{answer}
$ x^2+zx-y^2+3zy-2z^2  $
\end{answer}
\end{pro}

\begin{pro}Expand the product $$(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).$$
\begin{answer} We have,
$$\begin{array}{lll}(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)& =
& a^3 + ab^2 + ac^2 - a^2b -
abc - ca^2 \\ & & \quad + ba^2 + b^3 + bc^2 - ab^2 - b^2c - bca \\
& & \quad + a^2c + b^2c + c^3
-abc - bc^2 - c^2a        \\
& = & a^3 + b^3 + c^3 - 3abc\end{array}$$
\end{answer}
\end{pro}

\begin{pro}
Expand the product $(a + b + c)(x + y + z)$.
\begin{answer}From the distributive law we deduce that $$(a + b + c)(x + y +
z) = ax + ay + az + bx + by + bz + cx + cy + cz.$$
\end{answer}
\end{pro}
\begin{pro}
Prove that the product of two even integers is even.
\begin{answer} Let $2a$ be one of the integers and let $2b$ be the
other integer. Then $(2a)(2b)=2(2ab)$, is twice the integer $2ab$
and hence it is even.
\end{answer}
\end{pro}
\begin{pro}
Prove that the product of two odd integers is odd.
\begin{answer} Let $2a + 1$ be one of the integers and let $2b + 1$ be the
other integer. Then
$$\begin{array}{lcl}
(2a + 1)(2b + 1) & = & 4ab + 2a + 2b + 1 \\
& = & 2(2ab) + 2a + 2b + 1 \\
& = & 2(2ab + a + b) + 1 \\
& = & 2m + 1,
\end{array}$$where $m = 2ab + a + b$. Since $m$ is an
integer, the equality $(2a + 1)(2b + 1) = 2m + 1$ shews that the
product $(2a + 1)(2b + 1)$ leaves remainder $1$ upon remainder by
$2$, that is, the product  is odd. \end{answer}
\end{pro}
\begin{pro}
Prove that the product of two integers that leave remainder $3$ upon
division by $4$ leaves remainder $1$ upon division by $4$.
\begin{answer} Let $4a + 3$ be one of the integers and let $4b +
3$ be the other integer. Then
$$\begin{array}{lcl}
(4a + 3)(4b + 3) & = & 16ab + 12a + 12b + 9 \\
& = & 16ab + 12a + 12b + 8 + 1 \\
& = & 4(4ab) + 4(3a) + 4(3b) + 4(2) + 1 \\
& = & 4(4ab + 3a + 3b + 2) + 1\\
& = & 4m + 1,
\end{array}$$where $m = 4ab + 3a + 3b + 1$. Since $m$ is an
integer, the equality $(4a + 3)(4b + 3) = 4m + 1$ shews that the
product $(4a + 3)(4b + 3)$ leaves remainder $1$ upon remainder by
$4$.
\end{answer}
\end{pro}
\begin{pro}Prove that the product of any two integers
leaving remainder $2$ upon division by $3$ leaves remainder $1$
upon division by $3$.
\begin{answer}
Let $3a+2$ and $3b+2$ be two numbers leaving remainder $2$ upon
division by $3$. Then
$$(3a+2)(3b+2) = 9ab+6a+6b+4=3(3ab+2a+2b+1)+1,  $$which is a number
leaving remainder $1$ upon division by $3$.
\end{answer}
\end{pro}
\begin{pro}
Prove that the product of two integers leaving remainder $2$ upon
division by $5$ leaves remainder $4$ upon division by $5$.
\begin{answer}
Let $5a+2$ and $5b+2$ be two numbers leaving remainder $2$ upon
division by $5$. Then
$$(5a+2)(5b+2) = 25ab+10a+10b+4=5(5ab+2a+2b)+4,  $$which is a number
leaving remainder $4$ upon division by $5$.
\end{answer}
\end{pro}
\begin{pro}
Prove that the product of two integers leaving remainder $3$ upon
division by $5$ leaves remainder $4$ upon division by $5$.
\begin{answer}
Let $5a+3$ and $5b+3$ be two numbers leaving remainder $3$ upon
division by $5$. Then
$$(5a+3)(5b+3) = 25ab+15a+15b+9=5(5ab+2a+2b+1)+4,  $$which is a number
leaving remainder $4$ upon division by $5$.
\end{answer}
\end{pro}
\begin{pro}
If $x^2 + x - 1 = 0$, find $x^4 + 2x^3 + x^2.$
\begin{answer} Observe that $x^2 + x = 1.$ Hence $x^4 + x^3 = x^2$ and $x^3 + x^2 = x.$
Thus $$x^4 + 2x^3 + x^2 = x^4 + x^3 + x^3 + x^2 = x^2 + x = 1. $$
\end{answer}
\end{pro}
\begin{pro}
Expand and collect like terms: $$2x (x+1)-x^2-x(x-1).$$ Then use
your result to evaluate $$2\cdot (11112)\cdot
(11113)-11112^2-(11111)\cdot (11112)$$ {\em without a calculator.}
\begin{answer}
We have$$\begin{array}{lll}2x (x+1)-x^2-x(x-1) & = & 2x^2+2x-x^2-x^2+x \\
& = & 3x.\end{array}$$ Putting $x=11112$ one obtains from the
preceding part that  $$2\cdot (11112)\cdot
(11113)-11112^2-(11111)\cdot (11112)=3\cdot 11112 = 33336.$$
\end{answer}
\end{pro}


\end{multicols}
\section{Square of a Sum}
\QUOTEME{Galileo was no idiot. Only an idiot could believe that
science requires martyrdom - that may be necessary in religion, but
in time a scientific result will establish itself. }{David HILBERT}
\bigskip In this section we study some special products that often
occur in algebra.
\begin{thm}[Square of a Sum]
For all real numbers $a,b$ the following identity holds
$$ (a\pm b)^2 = a^2 \pm 2ab + b^2.  $$
\label{thm:square-of-a-sum}
\end{thm}
\begin{pf}
Using the distributive law,
$$\begin{array}{lll}(a + b)^2  & = &  (a + b)(a + b) \\
& = & a(a+b)+b(a+b)\\ & = &  a^2 + ab + ba + b^2 \\ &  = &  a^2 +
2ab + b^2,
\end{array}
$$and so $(a + b)^2 = a^2 + 2ab + b^2$. Putting $-b$ instead of
$b$ in this last identity,
$$(a - b)^2 = (a + (-b))^2 = a^2 + 2a(-b) + (-b)^2 = a^2 - 2ab + b^2,
$$proving both identities.
\end{pf}

\vspace*{2cm}

\begin{figure}[h]
\centering \psset{unit=1.5pc}
\pstGeonode[PointName=none,PointSymbol=none](-2,-2){A}(2,-2){B}(2,2){C}(-2,2){D}(1,-2){e}(1,1){f}(-2,1){g}(2,1){h}(1,2){i}
\pstLineAB[linewidth=2pt]{A}{B}\pstLineAB[linewidth=2pt]{B}{C}\pstLineAB[linewidth=2pt]{C}{D}\pstLineAB[linewidth=2pt]{D}{A}
\pscustom[fillcolor=red,fillstyle=solid]{\pstLineAB{A}{e}\pstLineAB{e}{f}\pstLineAB{f}{g}\pstLineAB{g}{A}}
\pscustom[fillcolor=blue,fillstyle=solid]{\pstLineAB{f}{h}\pstLineAB{h}{C}\pstLineAB{C}{i}\pstLineAB{i}{f}}
\ncline[offset=-8pt]{|-|}{A}{B}\ncput*[nrot=:U]{$a+b$}
\ncline[offset=8pt]{|-|}{A}{g}\ncput*[nrot=:U]{$a$}
\ncline[offset=-8pt]{|-|}{h}{C}\ncput*[nrot=:U]{$b$}
\vspace*{2cm}\hangcaption{$(a+b)^2=a^2+2ab+b^2$
.}\label{fig:square-of-a-sum}

\end{figure}


The identity $(a+b)^2=a^2+2ab+b^2$ has a pictorial justification in
terms of areas of squares, as seen in figure
\ref{fig:square-of-a-sum}. The area of the large $(a+b)\times (a+b)$
square is $(a+b)^2$. In terms, this area can be decomposed into two
square regions, one of area $a^2$ and the other of area $b^2$, and
two rectangular regions each of area $ab$.

\begin{exa}We have,
$$ (2x + 3y)^2  = (2x)^2 + 2(2x)(3y) + (3y)^2 =   4x^2 + 12xy + 9y^2.$$
\end{exa}
\begin{exa}We have,
$$ (2 - x^2)^2  = (2)^2 - 2(2)(x^2) + (x^2)^2 =   4 -4x^2 + x^4.$$
\end{exa}
The principle expounded here can be used to facilitate mental
arithmetic.
\begin{exa}
One may compute mentally $52^2$ as follows:
$$52^2 = (50+2)^2=50^2+2(50)(2)+2^2=2500+200+4=2704.  $$
\end{exa}


The formula $(a+b)^2=a^2+2ab+b^2=a^2+b^2+2ab$ gives the square of a
sum in terms of the sum of squares and the product of the numbers.
Thus knowing any two of them will determine the other.

\begin{exa}
If the sum of two numbers is $3$ and their product $4$, find the sum
of their squares.
\end{exa}
\begin{solu}
Let the two numbers be $x, y$. Then $x+y=3$ and $xy=4$. Hence
$$ 9=(x+y)^2=x^2+2xy+y^2 = x^2+8+y^2 \implies x^2+y^2=1. $$
\end{solu}

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

\begin{pro}Expand and simplify: $(3 - y)^2$.
\begin{answer}
$ 9-6y+y^2 $
\end{answer}
\end{pro}
\begin{pro}Expand and simplify: $(5 - 2x^2)^2$.
\begin{answer}
$25-20x^2+4x^4  $
\end{answer}
\end{pro}\begin{pro}Expand and simplify: $(2ab^2 - 3c^3d^4)^2$.
\begin{answer}
$ 4a^2b^4-12ab^2c^3d^4+9c^6d^8 $
\end{answer}
\end{pro}
\begin{pro}Expand and simplify: $(x + 2y)^2 + 4xy$.
\begin{answer}
$ x^2+8xy+4y^2 $
\end{answer}
\end{pro}\begin{pro}Expand and simplify: $(x + 2y)^2 -
4xy$.
\begin{answer}
$ x^2+4y^2 $
\end{answer}
\end{pro}
\begin{pro}Expand $(xy - 4y)^2 .$
\begin{answer}We have,
$$ (xy - 4y)^2  = (xy)^2 - 2(xy)(4y) + (4y)^2 =   x^2y^2 - 8xy^2 + 16y^2.$$
\end{answer}
\end{pro}
\begin{pro}Expand $(ax + by)^2 $.
\begin{answer}We have,
$$ (ax + by)^2  = (ax)^2 + 2(ax)(by) + (by)^2 =   a^2x^2 + 2abxy + b^2y^2.$$
\end{answer}
\end{pro}

\begin{pro}Expand and simplify: $(a + 2)^2 + (a  - 2)^2$.
\begin{answer}
$ 2a^2+8 $
\end{answer}
 \end{pro}
\begin{pro}Expand and simplify: $(a + 2)^2 - (a  -
2)^2$.
\begin{answer}
$ 8a $
\end{answer}
\end{pro}
\begin{pro}Expand
and simplify: $(a + 2b + 3c)^2$.
\begin{answer}
$a^2+4ab+6ac+4b^2+12bc+9c^2  $
\end{answer}
\end{pro}\begin{pro}Expand and
simplify: $(a + 2b - c)^2$.
\begin{answer}
$ a^2+4ab-2ac+4b^2-4bc+c^2 $.
\end{answer}
\end{pro}

\begin{pro} If $x+\dfrac{1}{x}=6$, find $x^2+\dfrac{1}{x^2}$.
\begin{answer}
Observe that
$$ 36=\left(x+\dfrac{1}{x}\right)^2=x^2+2+\dfrac{1}{x^2}\implies x^2+\dfrac{1}{x^2} =36-2=34. $$
\end{answer}
\end{pro}

\begin{pro}Find the value of $n$ in
$$(10^{2002} + 25)^2 - (10^{2002} - 25)^2 = 10^n.$$
\begin{answer} We have, $$\begin{array}{lll}(10^{2002} + 25)^2 - (10^{2002} -
25)^2  & = & 10^{4004} + 2\cdot 25\cdot 10^{2002} +
25^2 \\ & & \quad - (10^{4004} - 2\cdot 25\cdot 10^{2002} + 25^2) \\
&  = & 4\cdot 25\cdot 10^{2002} \\
&  = &  100 \cdot 10^{2002} = 10^{2004},
\end{array}
$$whence $n = 2004$.
\end{answer}
\end{pro}
\begin{pro} Given that $2x + 3y = 3$ and $xy = 4$, find $4x^2 + 9y^2$.
\begin{answer} We have $$\begin{array}{lll} 4x^2 + 9y^2 & = & 4x^2 +
12xy + 9y^2 - 12xy \\
& = & (2x + 3y)^2 - 12xy \\
& = & 3^2 - 12(4) \\
& = & 9 - 48 \\
& = & -39. \end{array}$$
\end{answer}
 \end{pro}

\begin{pro}
If the difference of two numbers is $3$ and their product $4$, find
the sum of their squares.
\begin{answer}
Let the two numbers be $x, y$. Then $x-y=3$ and $xy=4$. Hence
$$ 9=(x-y)^2=x^2-2xy+y^2 = x^2-8+y^2 \implies x^2+y^2=17. $$
\end{answer}
\end{pro}

\begin{pro}
Given that $x + y = 4$ and $xy = -3$, find $x^2 + y^2$.
\begin{answer}
We have $$16=(x+y)^2=x^2+2xy+y^2 = x^2-6+y^2 \implies x^2+y^2 = 22.
$$
\end{answer}
\label{exa:sum-of-squares}\end{pro}
\begin{pro}
Given that $x + y = 4$ and $xy = -3$, find $x^4 + y^4$.
\begin{answer}
From problem \ref{exa:sum-of-squares}, $x^2+y^2 = 22.$ Observe that
$2x^2y^2=2(xy)^2=2(-3)^2=18$. Hence
$$484= (x^2+y^2)^2 = x^4+2x^2y^2+y^4 = x^4+18+y^4 \implies x^4+y^4 = 466.  $$
\end{answer}
\end{pro}
\begin{pro}Expand $(x + y + z)^2$.\begin{answer}We have,$$\begin{array}{lll}(x + y + z)^2 & = & ((x + y) + z)^2
\\ & = & (x + y)^2 + 2(x + y)z + z^2 \\
& = & x^2 + 2xy + y^2 + 2xz + 2yz + z^2 \\
& = &  x^2 + y^2  + z^2 + 2xy + 2yz + 2zx.\end{array}$$ \end{answer}
\end{pro}
\begin{pro}Demonstrate that $$\begin{array}{lll}(x + y + z + w)^2 & = &  x^2 + y^2 + z^2 + w^2\\
& & \qquad  +
2xy + 2xz + 2xw + 2yz + 2yw + 2zw.\end{array}$$
\begin{answer}
We have $$\begin{array}{lll} (x + y + z + w)^2 & = &
(x+y)^2+2(x+y)(z+w)+(z+w)^2\\
& = & x^2+2xy+y^2+2xz+2xw+2yz+2yw+z^2+2zw+w^2\\
& = &  = x^2 + y^2 + z^2 + w^2 + 2xy + 2xz + 2xw + 2yz + 2yw + 2zw.
\end{array}$$
\end{answer}
\end{pro}

\end{multicols}
\section{Difference of Squares}\QUOTEME{Science, being human enquiry, can hear no answer except an answer couched somehow in human tones. Primitive man stood in the mountains and shouted against a cliff; the echo brought back his own voice, and he believed in a disembodied spirit. The scientist of today stands counting out loud in the
face of the unknown. Numbers come back to him - and he believes in the Great Mathematician.
}{Richard HUGHES} \bigskip In this section we continue our study of
special products.
\begin{thm}[Difference of Squares]
For all real numbers $a,b$ the following identity holds
$$ (a +  b)(a - b) = a^2 - b^2.  $$
\end{thm}
\begin{pf}
Using the distributive law,
$$ \begin{array}{lll}(a + b)(a - b) & = & a(a-b)+b(a-b)\\ &  = &  a^2 - ab + ba - b^2 \\ & = &  a^2 - b^2, \end{array}$$proving the
identity.
\end{pf}
\begin{exa}
$$(4x - 5y)(4x + 5y) = 16x^2 - 25y^2.  $$
\end{exa}
\begin{exa}
$$(xy - 5y)(xy + 5y) = x^2y^2 - 25y^2.  $$
\end{exa}
\begin{exa}Shew that $$ (a - b - c)(a + b + c) = a^2 - b^2 - c^2 - 2bc. $$

\end{exa}\begin{solu} We have $$\begin{array}{lcl} (a - b - c)(a + b + c)& = & (a - (b + c))(a + (b + c))\\ & = &
a^2 - (b + c)^2 \\
& = & a^2 - (b^2 + 2bc + c^2) \\
& = & a^2 - b^2 - c^2 - 2bc,  \end{array}$$as we wanted to shew.
\end{solu}
\begin{exa}
$$ \begin{array}{lll} (x + 1)^2 - (x - 1)^2 & = & ((x + 1) + (x - 1))((x + 1) - (x - 1)) \\
& = & (2x)(2) \\
& = & 4x  \end{array}$$
\end{exa}
\begin{exa}
We have $$\begin{array}{lll}(a - 1)(a + 1)(a^2 + 1)(a^4 + 1)(a^8 +
1)  & = & (a^2 - 1)(a^2 + 1)(a^4 + 1)(a^8 + 1) \\
& = & (a^4 - 1)(a^4 + 1)(a^8 + 1)  \\
& = & (a^8 - 1)(a^8 + 1)  \\
& = & a^{16} - 1. \end{array}$$
\end{exa}
\begin{exa}We have
$$ \begin{array}{lll}(x^2 - x + 1)(x^2 + x + 1) &  =  &
((x^2 + 1) - x)((x^2 + 1) + x) \\ & = &  (x^2 + 1)^2 - x^2\\ &  =
&  x^4 + 2x^2 + 1 - x^2 \\ & = &  x^4 + x^2 + 1. \end{array}$$
\end{exa}
\begin{exa}Prove that $$x^2 - (x + 2)(x - 2) = 4.$$ Explain how to find
the exact value of $$(987654321)(987654321) -
(987654323)(987654319),$${\em mentally}.
\end{exa}\begin{solu} We have
$$\begin{array}{ccl}x^2 - (x + 2)(x - 2) & = & x^2 - (x^2 - 4) \\
& = & x^2 - x^2 + 4 \\
& = & 4.   \end{array}$$Now put $x =987654321$. Then $x+2 =
987654323$ and $x-2 = 987654319$, from where
$$(987654321)(987654321) - (987654323)(987654319) = 4$$ follows.
\end{solu}

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

\begin{pro}
Multiply and collect like terms: $$(a + 4)^2 - (a + 2)^2.  $$
\begin{answer}
We have $$\begin{array}{lll}(a + 4)^2 - (a + 2)^2 & = &
((a+4)-(a+2))((a+4)+(a+2))\\
& = & (2)(2a+6)\\
& = & 4a+12.
\end{array}$$
We may also use the square of a sum identity:
$$\begin{array}{lll}(a + 4)^2 - (a + 2)^2 & = &
(a^2+8a+16)-(a^2+4a+4)\\
 & = & 4a+12,
\end{array}$$like before.
\end{answer}
\end{pro}
\begin{pro}
Prove that  $$(x - y)(x + y)(x^2 + y^2) = x^4 - y^4.  $$
\begin{answer}
We have, $$\begin{array}{lll}(x - y)(x + y)(x^2 + y^2) & = &
(x^2-y^2)(x^2+y^2)\\
& = & x^4-y^4.
\end{array}$$
\end{answer}
\end{pro}
\begin{pro}
Multiply and collect like terms: $$(a + 1)^4 - (a - 1)^4.  $$
\begin{answer}
We have $$\begin{array}{lll} (a + 1)^4 - (a - 1)^4 & = &
((a+1)^2-(a-1)^2)((a+1)^2+(a-1)^2)\\
& = & ((a+1)-(a-1))((a+1)+(a-1))((a+1)^2+(a-1)^2)\\
& = & (2)(2a)((a^2+2a+1)+(a^2-2a+1))\\
& = & 4a(2a^2+2)\\
& = & 8a^3+8a.
\end{array}$$
\end{answer}
\end{pro}
\begin{pro}Shew, without expressly computing any power of $2$, that $$(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)(2^{16} + 1) = 2^{32} - 1.$$
\begin{answer}
Multiplying by $1=2-1$ does not alter the product, and so,
$$\begin{array}{lll} (2-1)(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 +
1)(2^{16} + 1) & = & (2^2 - 1)(2^2 + 1)(2^4 + 1)(2^8 +
1)(2^{16} + 1) \\
& = & (2^4- 1)(2^4 + 1)(2^8 +
1)(2^{16} + 1) \\
& = & (2^8 - 1)(2^8 +
1)(2^{16} + 1) \\
& = & (2^{16} -
1)(2^{16} + 1) \\
& = & 2^{32} - 1
\end{array}$$
\end{answer}
\end{pro}
\begin{pro} Find $$1^2 - 2^2 + 3^2 - 4^2 + \cdots + 99^2 - 100^2.$$
\begin{answer}
We have,$$\begin{array}{lll} (1^2 - 2^2) + (3^2 - 4^2) + \cdots +
(99^2 - 100^2) & = & (1-2)(1+2)+(3-4)(3+4)+\cdots + (99-100)(99+100)
\\
& = & -1(1+2)-1(3+4)+\cdots + -1(99+100)\\
& = & -(1+2+3+\cdots + 100)\\
 & = &-5050,
\end{array}$$using the result of Example \ref{exa:sum1-100-1}.
\end{answer}
\end{pro}
\begin{pro}
{\em Without using a calculator}, what is the exact numerical value
of $$(123456789)^2-(123456787)(123456791)\ ?$$
\begin{answer}
Put $x=123456789$. Then
$$(123456789)^2-(123456787)(123456791)=x^2-(x-2)(x+2)=x^2-(x^2-4)=4.$$
\end{answer}

\end{pro}
\begin{pro}
{\em Without using a calculator}, what is the exact numerical value
of $(666\ 666\ 666)^2 - (333\ 333\ 333)^2$?
\begin{answer}
Using $x^2 - y^2 = (x - y)(x + y)$,
$$
\begin{array}{lll}
(666\ 666\ 666)^2 - (333\ 333\ 333)^2 & = & (666\ 666\ 666 - 333\
333\ 333)(666\ 666\ 666 +  333\ 333\ 333) \\
& = & (333\ 333\ 333)(999\ 999\ 999) \\
& = & (333\ 333\ 333)(10^9 - 1)\\
& = & 333\ 333\ 333\ 000\ 000\ 000 - 333\ 333\ 333\\
& = & 333\ 333\ 332\ 666\ 666\ 667
\end{array}
$$\end{answer}

\end{pro}
\end{multicols}
\section{Cube of a Sum}
\QUOTEME{If we take in our hand any volume; of divinity or school
metaphysics, for instance; let us ask, `Does it contain any abstract
reasoning concerning quantity or number?' No. `Does it contain any
experimental reasoning concerning matter of fact and existence?' No.
Commit it then to the flames: for it can contain nothing but
sophistry and illusion.}{David HUME}
\bigskip
We introduce now a special product formula for the cube of a sum.
\begin{thm}\label{thm:cube_of_a_sum}
Let $a$ and $b$ be real numbers. Then $$ (a + b)^3 = a^3 + 3a^2b +
3ab^2 + b^3.$$
\end{thm}
\begin{pf}
Using the distributive law and the square of a sum identity,
$$\begin{array}{lll}(a + b)^3 & = & (a + b)^2(a + b) \\
& = & (a^2 + 2ab + b^2)(a + b) \\
& = & a^2(a+b)+2ab(a+b)+b^2(a+b)\\
& = & a^3 + a^2b + 2a^2b + 2ab^2 + b^2a + b^3 \\
& = & a^3 + 3a^2b + 3ab^2 + b^3,  \end{array}$$as claimed.
\end{pf}
\begin{cor}
Let $a$ and $b$ be real numbers. Then $$ (a - b)^3 = a^3 - 3a^2b +
3ab^2 - b^3.$$
\end{cor}
\begin{pf}Replace $b$ by $-b$ in Theorem \ref{thm:cube_of_a_sum},
obtaining
$$\begin{array}{lll}(a - b)^3 & = & (a + (-b))^3 \\
& = &  a^3 + 3a^2(-b) + 3a(-b)^2 + (-b)^3 \\
& = & a^3 - 3a^2b + 3ab^2 - b^3,
\end{array}$$as claimed.
\end{pf}
%Theorem  \ref{thm:cube_of_a_sum} has a pictorial interpretation in terms of volumes of cubes, as seen in figures
%\ref{fig:cube-of-a-sum-a} and \ref\ref{fig:cube-of-a-sum-b}.

%\vspace*{2cm}
%\begin{figure}[h]
%\begin{minipage}{7cm}
%\psset{CubeColorFaceOne=1 1 1,%
%       CubeColorFaceTwo=1 0 0,%
%       CubeColorFaceThree=0 1 0,%
%       CubeColorFaceFour=0 0 1,%
%       CubeColorFaceFive=1 1 0,%
%       CubeColorFaceSix=0 1 1,%
 %      unit=2pc}
%\centering
%\CubeThreeD[A=3,B=3,C=3](6,6,6)
%\CubeThreeD[A=2,B=2,C=2](-6,-6,-3)
%\CubeThreeD[A=2,B=2,C=3](-6,-6,6)
%\CubeThreeD[A=2,B=3,C=2](0,10,0)
%\CubeThreeD[A=3,B=2,C=2](0,13,0)
%\CubeThreeD[A=3,B=3,C=2](0,14,0)
%\CubeThreeD[A=3,B=2,C=3](0,18,0)
%\CubeThreeD[A=2,B=3,C=3](0,21,0)
%\vspace*{1cm}\hangcaption{ $a^3 + 3a^2b +
%3ab^2 + b^3.$}\label{fig:cube-of-a-sum-a}
%\end{minipage}
%\hfill
%\end{figure}

\begin{exa}We have,
$$(x + 2y)^3 = x^3 + 3x^2(2y) + 3x(2y)^2 + (2y)^3  = x^3 + 6x^2y + 12xy^2 + 8y^3.  $$
\end{exa}
\begin{exa}We have,
$$(3x - 2y)^3 = (3x)^3 - 3(3x)^2(2y) + 3(3x)(2y)^2 - (2y)^3  = 27x^3 - 54x^2y + 36xy^2 - 8y^3.  $$
\end{exa}
Observe that $$ 3ab(a + b) = 3a^2b + 3ab^2, $$hence we may write
the cube of a sum identity as
\begin{equation} (a + b)^3 = a^3 + b^3 + 3ab(a + b). \end{equation}
\begin{exa}
Given that $a + b = 2$ and $ab = 3$, find $a^3 + b^3$.
\end{exa}
\begin{solu} We have $$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 2^3 - 3(3)(2) =
8 - 18 = -10.
$$
\end{solu}
 \section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

  \begin{pro}
Expand and simplify: $(5x + 1)^3$.
\begin{answer}
$125x^3+75x^2+15x+1  $
\end{answer}
\end{pro}

\begin{pro}
Multiply and collect like terms: $$(x + 1)^3 + (x + 1)^2.$$
\begin{answer}
$x^3+4x^2+5x+2  $
\end{answer}
\end{pro}
\begin{pro}
Multiply and collect like terms: $$(x - 1)^3 - (x + 2)^2.$$
\begin{answer}
$ x^3-4x^2-x-5 $
\end{answer}
\end{pro}
\begin{pro}
Multiply and collect like terms: $$(x - 2)^3 - (x - 2)^2.$$
\begin{answer}
$ x^3-7x^2+16x-12 $
\end{answer}
\end{pro}
\begin{pro}\label{pro:sum-of-cubes}
Given that $a - b = 6$ and $ab = 3$, find $a^3 - b^3$.
\begin{answer}We have,
$$ 216=(a-b)^3 = a^3-3a^2b+3ab^2-b^3 = a^3-b^3-3ab(a-b)=a^3-b^3-3(3)(6)\implies a^3-b^3=270. $$
\end{answer}
\end{pro}
\begin{pro}
Given that $a + 2b = 6$ and $ab = 3$, find $a^3 + 8b^3$.
\begin{answer}We have,
$$ 216=(a+2b)^3 = a^3+6a^2b+12ab^2+8b^3 = a^3+8b^3+6ab(a+2b)=a^3+8b^3+6(3)(6)\implies a^3+8b^3=108. $$
\end{answer}
\end{pro}
\end{multicols}
\section{Sum and Difference of Cubes}
\QUOTEME{Take from all things their number and all shall
perish.}{Saint Isidore of Seville}
\bigskip
In this section we study our last special product.
\begin{thm}[Sum and Difference of Cubes]
For all real numbers $a,b$ the following identity holds
$$ (a \pm  b)(a^2 \mp ab +  b^2) = a^3 \pm b^3.  $$
\end{thm}
\begin{pf}
Using the distributive law,
$$ \begin{array}{lll} (a + b)(a^2 - ab + b^2) & = & a^3 - a^2b +
ab^2 + ba^2 - ab^2 + b^3 \\
& = & a^3 + b^3,
\end{array}$$from where the sum of cubes identity is deduced.
We obtain $$(a - b)(a^2 + ab + b^2) = a^3 - b^3  $$upon replacing
$b$ by $-b$ in the above sum of cubes identity.
\end{pf}
\begin{exa}
$$ (a - 2b)(a^2 + 2ab + 4b^2) = a^3 - 8b^3. $$
\end{exa}
\begin{exa}
$$ (5a + 2b)(25a^2 - 10ab + 4b^2) = 125a^3 - 8b^3. $$
\end{exa}

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

\begin{pro}
Expand and simplify: $$(x - 2)(x^2 + 2x + 4).  $$
\begin{answer} $x^3-8$ \end{answer}
\end{pro}
\begin{pro}
Expand and simplify: $$(x + 8)(x^2 - 8x + 64).  $$
\begin{answer} $ x^3+512$ \end{answer}
\end{pro}
\begin{pro}
Given that $a - b = 6$ and $ab = 3$, find $a^3 - b^3$.

\begin{answer}
Notice that this is problem \ref{pro:sum-of-cubes}. We will solve it
here in a different way. We have
$$ a^3-b^3=(a-b)(a^2+ab+b^2)=6(a^2+3+b^2), $$so the problem bois
down to finding $a^2+b^2$. Now,
$$36=(a-b)^2=a^2-2ab+b^2=a^2-6+b^2 \implies a^2+b^2=42,  $$
whence
$$ a^3-b^3=6(a^2+3+b^2) = 6(42+3)=270, $$as was obtained in the
solution of problem \ref{pro:sum-of-cubes}.
\end{answer}
\end{pro}
\end{multicols}
\chapter{Division}
\QUOTEME{From the intrinsic evidence of his creation, the Great
Architect of the Universe now begins to appear as a pure
mathematician. }{Sir James JEANS}
\section{Term by Term Division}
\QUOTEME{The science of mathematics presents the most brilliant
example of how pure reason may successfully enlarge its domain
without the aid of experience. }{\mbox{Emmanuel KANT}}
\bigskip
 From the distributive law we deduce $$\frac{b +
c}{a} = \frac{1}{a}\left(b + c\right) = \frac{b}{a} + \frac{c}{a}.
$$
\begin{exa}$\dfrac{x^2 -
2xy}{x} = \dfrac{x^2}{x} - \dfrac{2xy}{x} = x -
2y.$\end{exa}
\begin{rem}
Be careful not to confuse an expression of the form $\dfrac{a+b}{c}$
with one of the form $\dfrac{ab}{c}$. We also remark that an
expression of the form $\dfrac{ab}{c}$ can be evaluated in various
ways, for example, one may evaluate first the product $ab$ and then
divide this by $c$, or, one may divide $a$ by $c$ and then multiply
this by $b$, etc.
\end{rem}

\begin{exa}We have, $$\dfrac{(x^2)(-2xy)}{x} = \dfrac{-2x^3y}{x} =-2x^2y.$$ This may also be evaluated in the following manner,
$$\dfrac{(x^2)(-2xy)}{x} = \dfrac{x^2}{x}(-2xy) =x(-2xy)=-2x^2y,$$as
before.
\end{exa}
\begin{exa}$\dfrac{-24x^6 - 32x^4}{-8x^3} =
\dfrac{-24x^6}{-8x^3} - \dfrac{32x^4}{-8x^3} = 3x^3 - (-4x) = 3x^3 +
4x.$\end{exa}

\begin{exa}We have, $$\dfrac{(-24x^6)(-32x^4)}{-8x^3} =
\dfrac{768x^{10}}{-8x^3}=-96x^7.$$ We may also proceed as follows,
$$\dfrac{(-24x^6)(-32x^4)}{-8x^3} =
\dfrac{-24x^6}{-8x^3}(-32x^4)=(3x^3)(-32x^4) = -96x^7.$$
\end{exa}



When a negative exponent arises in a term, we prefer to express the
term with positive exponents.
\begin{exa}
$$\begin{array}{lll} \dfrac{12x^2y^3 + 6x^4y^2}{3x^3y^3} & = &
\dfrac{12x^2y^3}{3x^3y^3} + \dfrac{6x^4y^2}{3x^3y^3}\\
& = & 4x^{-1} + 2xy^{-1} \\
& = & \dfrac{4}{x} + \dfrac{2x}{y}.\end{array}$$
\end{exa}
 \section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro} Perform the division: $\dfrac{q^2 -pq -
pqr}{-q}$.
\begin{answer}
$$\begin{array}{lll}\dfrac{q^2 -pq -
pqr}{-q} & = & \dfrac{q^2}{-q} - \dfrac{pq}{-q} -
\dfrac{-pqr}{-q}\\
&  = &  -q + p + pr \\
& = &  -q + p +pr. \\
\end{array}$$\end{answer}
\end{pro}
\begin{pro}Divide term by term: $\dfrac{34x^3y^2 +
51x^2y^3}{17xy}.$
\begin{answer}
$2x^2y+3xy^2$
\end{answer}
\end{pro}
\begin{pro}Divide: $\dfrac{(34x^3y^2)
(51x^2y^3)}{17xy}.$
\begin{answer}
$102x^4y^4$
\end{answer}
\end{pro}

\begin{pro}Divide term by term: $$\dfrac{6aaaaxxx + 4aaaxxx}{2aaxx}.$$
\begin{answer}
$3a^2x+2ax$
\end{answer}
\end{pro}
\begin{pro}Divide: $\dfrac{(6aaaaxxx)(4aaaxxx)}{2aaxx}$.
\begin{answer}
$12a^5x^4$
\end{answer}
\end{pro}
\begin{pro}Divide term by term: $\dfrac{x^2 - xy -
xz}{-x}$.
\begin{answer}
$-x+y+z$
\end{answer}

 \end{pro}\begin{pro}Divide term by term: $\dfrac{5a^2b -
7ab^3}{-ab}$.
\begin{answer}
$-5a+7b^2 $
\end{answer}
\end{pro}

\begin{pro}Divide: $\dfrac{(5a^2b)(-7ab^3)}{-ab}$.
\begin{answer}
$35a^2b^3 $
\end{answer}
\end{pro}
\begin{pro}Divide: $\dfrac{(x^2)(-xy)(-xz)}{-x}$.
\begin{answer}
$-x^3yz$
\end{answer}
\end{pro}
\end{multicols}
\section{Long Division}
\QUOTEME{Medicine makes people ill, mathematics make them sad and
theology makes them sinful.}{Martin LUTHER}
\bigskip
 We must now confront the problem when the divisor
consists of more than one term. The algorithm for algebraic long
division resembles that for long division of natural numbers.
\begin{exa}Find \quad $(-x+x^2-6)\div (x+2)$.
\end{exa}
\begin{solu}
First, we rewrite $-x+x^2-6$ so that the exponents of each of the
terms are listed in decreasing order: $x^2-x-6$. Next display
dividend and divisor as follows.
 $$\polylongdiv[stage=1]{-x+x^2-6}{x+2} $$
Now, think, by what must $x$ (the term with the largest degree in
the divisor) must be multiplied to give $x^2$ (the term with the
largest degree in the dividend? The answer is of course
$\dfrac{x^2}{x}=x$, and so we write $x$ in the quotient,
 $$\polylongdiv[stage=2]{-x+x^2-6}{x+2} $$
Multiply this $x$ of the quotient by the divisor, obtaining
$x(x+2)=x^2+2x$ and change all signs:
$$\polylongdiv[stage=3]{-x+x^2-6}{x+2} $$
Add, and obtain
$$\polylongdiv[stage=4]{-x+x^2-6}{x+2} $$
The new dividend is now $-3x-6$. Again, think, by what must $x$ (the
term with the largest degree in the divisor) must be multiplied to
give $-3x$ (the term with the largest degree in the dividend? The
answer is of course $\dfrac{-3x}{x}=-3$, and so we write $-3$ in the
quotient,
 $$\polylongdiv[stage=5]{-x+x^2-6}{x+2} $$
 Multiply this $-3$ of the quotient by the divisor, obtaining
$-3(x+2)=-3x-6$ and change all signs:
$$\polylongdiv[stage=6]{-x+x^2-6}{x+2} $$
Add, and obtain
$$\polylongdiv[stage=7]{-x+x^2-6}{x+2} $$
Since the remainder is $0$, the division ends and we conclude that
$$(-x+x^2-6)\div (x+2)= \dfrac{x^2-x-6}{x+2}=x+3.$$
\end{solu}

\begin{exa}Find \quad $(3x^4-2x^3-9x^2+5x-6)\div (x-2)$.
\end{exa}
\begin{solu}
The terms of $3x^4-2x^3-9x^2+5x-6$ are already written with
exponents in decreasing order. Next display dividend and divisor as
follows.
 $$\polylongdiv[stage=1]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
Now, think, by what must $x$ (the term with the largest degree in
the divisor) must be multiplied to give $3x^4$ (the term with the
largest degree in the dividend? The answer is of course
$\dfrac{3x^4}{x}=3x^3$, and so we write $x$ in the quotient,
 $$\polylongdiv[stage=2]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
Multiply this $3x^3$ of the quotient by the divisor, obtaining
$3x^3(x-2)=3x^4-6x^3$ and change all signs:
$$\polylongdiv[stage=3]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
Add, and obtain
$$\polylongdiv[stage=4]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
The new dividend is now $4x^3-9x^2$. Again, think, by what must $x$
(the term with the largest degree in the divisor) must be multiplied
to give $4x^3$ (the term with the largest degree in the dividend?
The answer is of course $\dfrac{4x^3}{x}=4x^2$, and so we write
$+4x^2$ in the quotient,
 $$\polylongdiv[stage=5]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
 Multiply this $+4x^2$ of the quotient by the divisor, obtaining
$+4x^2(x-2)=4x^3-8x^2$ and change all signs:
$$\polylongdiv[stage=6]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
Add, and obtain
$$\polylongdiv[stage=7]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
The new dividend is now $-x^2+5x$. Again, think, by what must $x$
(the term with the largest degree in the divisor) must be multiplied
to give $-x^2$ (the term with the largest degree in the dividend?
The answer is of course $\dfrac{-x^2}{x}=-x$, and so we write $-x$
in the quotient,
 $$\polylongdiv[stage=8]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
 Multiply this $-x$ of the quotient by the divisor, obtaining
$-x(x-2)=-x^2+2x$ and change all signs:
$$\polylongdiv[stage=9]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
Add, and obtain
$$\polylongdiv[stage=10]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
The new dividend is now $3x-6$. Again, think, by what must $x$ (the
term with the largest degree in the divisor) must be multiplied to
give $3x$ (the term with the largest degree in the dividend? The
answer is of course $\dfrac{3x}{x}=3$, and so we write $3$ in the
quotient,
 $$\polylongdiv[stage=11]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
 Multiply this $3$ of the quotient by the divisor, obtaining
$3(x-2)=3x-6$ and change all signs:
$$\polylongdiv[stage=12]{3x^4-2x^3-9x^2+5x-6}{x-2} $$
Add, and obtain
$$\polylongdiv[stage=13]{3x^4-2x^3-9x^2+5x-6}{x-2} $$


Since the remainder is $0$, the division ends and we conclude that
$$(3x^4-2x^3-9x^2+5x-6)\div (x-2)= \dfrac{3x^4-2x^3-9x^2+5x-6}{x-2}=3x^3+4x^2-x+3.$$
\end{solu}
We now give an example where there is a remainder.
\begin{exa}Find \quad $(2x^2+1)\div (x+1)$.
\end{exa}
\begin{solu}
Display dividend and divisor as follows.
 $$\polylongdiv[stage=1]{2x^2+1}{x+1} $$
Now, think, by what must $x$ (the term with the largest degree in
the divisor) must be multiplied to give $2x^2$ (the term with the
largest degree in the dividend? The answer is of course
$\dfrac{2x^2}{x}=2x$, and so we write $2x$ in the quotient,
 $$\polylongdiv[stage=2]{2x^2+0x+1}{x+1} $$
Multiply this $2x$ of the quotient by the divisor, obtaining
$2x(x+1)=2x^2+2x$ and change all signs:
$$\polylongdiv[stage=3]{2x^2+0x+1}{x+1} $$
Add, and obtain
$$\polylongdiv[stage=4]{2x^2+1}{x+1} $$
The new dividend is now $-2x+1$. Again, think, by what must $x$ (the
term with the largest degree in the divisor) must be multiplied to
give $-2x$ (the term with the largest degree in the dividend? The
answer is of course $\dfrac{-2x}{x}=-2$, and so we write $-2$ in the
quotient,
 $$\polylongdiv[stage=5]{2x^2+1}{x+1} $$
 Multiply this $-2$ of the quotient by the divisor, obtaining
$-2(x+1)=-2x-2$ and change all signs:
$$\polylongdiv[stage=6]{2x^2+1}{x+1} $$
Add, and obtain
$$\polylongdiv[stage=7]{2x^2+1}{x+1} $$
Now, the remainder is $3$, which has lower degree than the divisor
$x+1$, hence the division ends. We write
$$(2x^2+1)\div (x+1)= \dfrac{2x^2+1}{x+1}=2x-2+\dfrac{3}{x+1}.$$
\end{solu}

We provide below some assorted examples, without explanations.


\begin{exa}We have,
\polylongdiv{x^3+27}{x+3}
\end{exa}
\begin{exa}We have,
\polylongdiv{x^4-16}{x-2}
\end{exa}
\begin{exa}We have,
$$\polylongdiv{x^4+3*x^3-2*x^2+x-1}{x^2-x-2}$$ and hence we write
$$ \dfrac{x^4+3x^3-2x^2+x-1}{x^2-x-2} = x^2+4x+4+\dfrac{13x+7}{x^2-x-2}. $$
\end{exa}


\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Perform the division:\quad  $(x^2-2x+1)\div (x-1)$.
\begin{answer} $x-1$ \end{answer}
\end{pro}
\begin{pro}
Perform the division:\quad  $(x^2+5x+6)\div (x+2)$.
\begin{answer} $x+3$ \end{answer}
\end{pro}
\begin{pro}
Perform the division:\quad  $(x^3+1)\div (x+1)$.
\begin{answer} $x^2-x+1$ \end{answer}
\end{pro}
\begin{pro}
Expand and collect like terms: \quad $$\dfrac{x^3-8}{x-2} - x^2.$$
\begin{answer}
$2x+4$.
\end{answer}

\end{pro}
\begin{pro}
Expand and collect like terms: $$\dfrac{6x^3-x^2-4x-1}{2x+1}
-\dfrac{3x^3+3x-x^2-1}{3x-1}.$$
\begin{answer}
$2x^2-2x-2$.
\end{answer}

\end{pro}
\end{multicols}

\section{Factoring I}
\QUOTEME{Even fairly good students, when they have obtained the
solution of the problem and written down neatly the argument, shut
their books and look for something else. Doing so, they miss an
important and instructive phase of the work. \ldots A good teacher
should understand and impress on his students the view that no
problem whatever is completely exhausted. One of the first and
foremost duties of the teacher is not to give his students the
impression that mathematical problems have little connection with
each other, and no connection at all with anything else. We have a
natural opportunity to investigate the connections of a problem when
looking back at its solution.}{George P\'{O}LYA}

\bigskip
\begin{df}To {\em factor} an algebraic expression is to express it as a product.  \end{df}
The idea behind factoring is essentially that of using the
distributive law backwards,
$$ ab+ac = a(b+c),  $$where the dextral quantity is a decomposition
into factors of the sinistral quantity.


\bigskip
We will begin by giving examples where the greatest common divisor
of the terms is different from $1$ and hence it may be removed. Thus
an expressions with two terms will be factored as $$ a+b =
\blacksquare (\square _1 + \square _2), $$ one with three terms will
be factored as
$$ a+b+c =
\blacksquare (\square _1 + \square _2+\square _3), $$etc.

\begin{exa} Factor $20a^2b^3+24a^3b^2$. \end{exa}
\begin{solu}
The greatest common divisor of $20$ and $24$ is $4$, hence it can be
removed. Also for the two expressions $a^2$ and $a^3$, their
greatest common divisor is the one with the least exponent, that is
$a^2$. Similarly, the greatest common divisor of $b^2$ and $b^3$ is
$b^2$. Thus $4a^2b^2$ is a common factor of both terms and the
desired factorisation has the form,
$$20a^2b^3+24a^3b^2=4a^2b^2 (\square _1 + \square _2).
$$We need to determine $\square _1$ and $\square _2$. We have
$$ 20a^2b^3 = 4a^2b^2 \square _1\implies \square _1 = 5b, $$
and
$$ 24a^3b^2 = 4a^2b^2 \square _2\implies \square _1 = 6a. $$
The desired factorisation is thus
$$20a^2b^3+24a^3b^2=4a^2b^2 (5b+6a).
$$
\end{solu}
\begin{rem}
Multiplying, $4a^2b^2 (5b+6a)=20a^2b^3+24a^3b^2. $
\end{rem}
\begin{exa}
Factor $-30x^2yza^2b+36xy^2za+16x^3y^2z^2b$.
\end{exa}
\begin{solu}
The greatest common divisor of $-30, 36, 16$ is $2$. The letters
$x,y,z$ appear in all three terms, but not so for $a$ and $b$. The
least power of $x$ appearing in all three terms is $1$, and same for
$y$ and $z$. Thus the greatest common divisor of the three terms is
$2xyz$ and the desired factorisation is thus $$
2xyz(-15xa^2b+18ya+8x^2yzb) .$$
\end{solu}
\begin{exa}
Factor $ab+a^2b+ab^2$.
\end{exa}
\begin{solu}
The desired factorisation is clearly
$$ab+a^2b+ab^2=ab(1+a+b). $$
\end{solu}
\begin{exa}
Prove that the sum of two even integers is even.
\end{exa}
\begin{solu}
Let $2a$ and $2b$ be two even integers. Then $$2a+2b=2(a+b),  $$that
is, twice the integer $a+b$, and hence an even integer.
\end{solu}
\begin{exa}
Factor $14a^3b^2(x+y)^2(x-y)^3+20ab^3(x+y)^3(x-y)^4$.
\end{exa}
\begin{solu}
The desired factorisation is clearly
$$14a^3b^2(x+y)^2(x-y)^3+20ab^3(x+y)^3(x-y)^4=2ab^2(x+y)^2(x-y)^3(7a^2+10b(x+y)(x-y)).  $$
\end{solu}
\begin{exa}Resolve into factors: $ax^2+a+bx^2+b$. \end{exa}
\begin{solu}
In this case, no factor greater than $1$ is common to all terms. We
have, however,
$$ax^2+a+bx^2+b=a(x^2+1)+b(x^2+1).   $$Notice now that $x^2+1$ is a
common factor of both summands, and finally,
$$ax^2+a+bx^2+b=a(x^2+1)+b(x^2+1)=(x^2+1)(a+b).   $$
\end{solu}
\begin{exa}Resolve into factors: $x^3+x^2+x+1$. \end{exa}
\begin{solu}
Like in the preceding problem, we have no factor greater than $1$
being common to all terms. Observe that
$$x^3+x^2+x+1=x^2(x+1)+1(x+1).   $$Notice now that $x+1$ is a
common factor of both summands, and finally,
$$x^3+x^2+x+1=x^2(x+1)+1(x+1)=(x+1)(x^2+1).   $$
\end{solu}
\begin{exa}Resolve into factors: $x^2-ax+bx-ab $. \end{exa}
\begin{solu}
We have,
$$x^2-ax+bx-ab =x(x-a)+b(x-a)=(x-a)(x+b).  $$
\end{solu}
Sometimes it is necessary to rearrange the summands.
\begin{exa}Resolve into factors: $12a^2-4ab-3ax^2+bx^2$. \end{exa}
\begin{solu}
We have, $$12a^2-4ab-3ax^2+bx^2=
(12a^2-3ax^2)-(4ab-bx^2)=3a(4a-x^2)-b(4a-x^2)=(4a-x^2)(3a-b).
$$
\end{solu}
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

\begin{pro}
Factor $-2a^2b^3$ from $-4a^6b^5+6a^3b^8 -12a^5b^3-2a^2b^3$.
\begin{answer}
We have
$$ -4a^6b^5+6a^3b^8 -12a^5b^3-2a^2b^3= -2a^2b^3(2a^4b^2-3ab^5+6a^3+1). $$
\end{answer}

\end{pro}
\begin{pro}
Factor $2a^3b^3$ from $4a^3b^4-10a^4b^3$.
\begin{answer}
$4a^3b^4-10a^4b^3=2a^3b^3(2b-5a)$.
\end{answer}
\end{pro}
\begin{pro}
Factor $\dfrac{3}{4}x$ from $\dfrac{9}{16}x^2-\dfrac{3}{4}x$.
\begin{answer}
$\dfrac{9}{16}x^2-\dfrac{3}{4}x=\dfrac{3}{4}x\left(\dfrac{3}{4}x-1\right)$.
\end{answer}
\end{pro}
\begin{pro}
Factor $-1$ from $-x+2y-3z$.
\begin{answer}
 $-x+2y-3z=-1(x-2y+3z)$.
\end{answer}
\end{pro}
\begin{pro}Prove that the sum of two odd integers is even.
\begin{answer}Let $2a+1$ and $2b+1$ be odd integers. Then $$ 2a+1+2b+1 = 2a+2b+2 = 2(a+b+1), $$that is, twice the
integer $a+b+1$, and hence even. \end{answer}
\end{pro}
\begin{pro}
Resolve into factors: \quad $x^3-x^2 $.
\begin{answer}
$x^2(x-1)$
\end{answer}
\end{pro}
\begin{pro} Factor $$125a^4b^5c^5 - 45a^5b^3c^4 + 5a^3b^2c^4 - 300a^4b^2c^8 - 10a^3b^2c^5.$$
\begin{answer} $5a^3b^2c^4(25cb^3a-9ba^2+1-60c^4a-2c)  $ \end{answer}
\end{pro}
\begin{pro} Resolve into factors: $5x^5-10a^7x^3-15a^3x^3$.
\begin{answer} $5x^3(x^2-2a^7-3a^3)  $ \end{answer}
\end{pro}
\begin{pro} Factor $3x^3y + 4x^2y^3 - 6x^6 - 10x^4$.
\begin{answer} $ x^2(3xy+4y^3-6x^4-10x^2) $ \end{answer}
 \end{pro}
\begin{pro}  Factor $$3m^6p^4q^2 - 9m^5p^2qx + 3m^7p^3qx + 3m^4p^2q - 6m^5p^4qx^2y.$$
\begin{answer} $3m^4p^2q(m^2p^2q-2p^2x^2ym+m^3px-3mx+1)  $ \end{answer}
\end{pro}
\begin{pro} Resolve into factors: $38a^2x^5+57a^4x^2$.
\begin{answer} $ 19a^2x^2(2x^3+3a^2) $ \end{answer}
\end{pro}
\begin{pro}Decompose into factors: $a^2+ab+ac+bc$.
\begin{answer} $(a + c)(a + b)$ \end{answer}
 \end{pro}
\begin{pro}Decompose into factors: $a^2-ab+ac-bc$.
\begin{answer} $ (a+c)(a-b) $ \end{answer}
 \end{pro}
\begin{pro}Decompose into factors: $y^3-y^2+y-1$.
\begin{answer} $ (y-1)(1+y^2) $ \end{answer}
\end{pro}
\begin{pro}Decompose into factors: $2x^3+3x^2+2x+3$.
\begin{answer} $(2x+3)(x^2+1)  $ \end{answer}
 \end{pro}
\begin{pro}Decompose into factors: $$a^2x+abx+ac+aby+b^2y+bc.$$
\begin{answer} $ (a+b)(xa+c+by) $ \end{answer}
 \end{pro}
\end{multicols}
\section{Factoring II}
\QUOTEME{The simplest schoolboy is now familiar with facts for which
Archimedes would have sacrificed his life. }{Ernst RENAN}
\bigskip
In this section we study factorisations of expressions of the form
$ax^2+bx+c$, the so-called {\em quadratic trinomials }.

\bigskip

Let us start with expressions of the form $x^2+bx+c$. Suppose that
$$ x^2+bx+c=(x+p)(x+q). $$Then, upon multiplying,
$$ x^2+bx+c=(x+p)(x+q)=x^2+(p+q)x+pq.  $$
Thus $c=pq$ and $b=p+q$, that is, the constant term is the product
of two numbers, and the coefficient of $x$ is the sum of these two
numbers.
\begin{exa} To factor $x^2+5x+6$ we look for two numbers whose product is $6$ and whose sum is $5$. Clearly $2$ and $3$ are these two numbers and so,
$$x^2+5x+6=(x+2)(x+3).  $$You should multiply the last product to verify the equality. \end{exa}
\begin{exa} To factor $x^2+7x+6$ we look for two numbers whose product is $6$ and whose sum is $7$. Clearly $1$ and $6$ are these two numbers and so,
$$x^2+7x+6=(x+1)(x+6).  $$You should multiply the last product to verify the equality. \end{exa}
\begin{exa} To factor $x^2-5x+6$ we look for two numbers whose product is $6$ and whose sum is $-5$. Clearly $-2$ and $-3$ are these two numbers and so,
$$x^2-5x+6=(x-2)(x-3).  $$ \end{exa}
\begin{exa} To factor $x^2-x-6$ we look for two numbers whose product is $-6$ and whose sum is $-1$. Clearly $-3$ and $+2$ are these two numbers and so,
$$x^2-x-6=(x-3)(x+2).  $$ \end{exa}
\begin{exa} To factor $x^2+x-6$ we look for two numbers whose product is $-6$ and whose sum is $+1$. Clearly $+3$ and $-2$ are these two numbers and so,
$$x^2+x-6=(x+3)(x-2).  $$ \end{exa}
\begin{exa}
Here are some more examples which you should inspect:
$$x^2+10x+24=(x+4)(x+6),   $$ $$x^2+11x+24=(x+3)(x+8),   $$
$$x^2+14x+24=(x+2)(x+12),   $$$$x^2+25x+24=(x+1)(x+24).   $$
\end{exa}
\begin{exa}
Here are some more examples which you should inspect:
$$x^2-10x+24=(x-4)(x-6),   $$ $$x^2-11x+24=(x-3)(x-8),   $$
$$x^2-14x+24=(x-2)(x-12),   $$$$x^2-25x+24=(x-1)(x-24).   $$
\end{exa}
\begin{exa}
Here are some more examples which you should inspect:
$$x^2+2x-24=(x-4)(x+6),   $$ $$x^2-5x-24=(x+3)(x-8),   $$
$$x^2+10x-24=(x-2)(x+12),   $$$$x^2-10x-24=(x+2)(x-12).   $$
\end{exa}
In the case when the coefficient of the $x^2$ is different from $1$,
more cases must be considered. The problem is more difficult, but
the method is essentially the same.
\begin{exa}
By inspection,
$$5x^2+22x+8= (5x+2)(x+4), $$
$$5x^2-22x+8= (5x-2)(x-4),  $$
$$5x^2-18x-8= (5x+2)(x-4),$$
$$5x^2+18x-8= (5x-2)(x+4),$$
$$5x^2+39x-8= (5x-1)(x+8).$$
\end{exa}
\begin{exa}
By inspection,
$$20x^2+43x+21=(5x+7)(4x+3),$$
$$20x^2+47x+21=(5x+3)(4x+7),$$
$$20x^2+76x+21=(10x+3)(2x+7),$$
$$20x^2+143x+21=(20x+3)(x+7).$$
\end{exa}
The inquiring reader may wonder whether all quadratic trinomials
$ax^2+bx+c$ factor in the manner above, that is, into linear factors
all whose coefficients are integers. The answer is {\em no}. For
example if $x^2-2$ does not factor into linear factors whose
coefficients are integers, since there are no two integers whose
product is $-2$ and that add up to $0$. If we augment our choice for
the coefficients for linear factors, then we may write
$$ x^2-2=(x-\sqrt{2})(x+\sqrt{2}), $$but we will not consider these
cases in this course.
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}Decompose into factors: $a^2-11a+30$.
\begin{answer}  $(a-5)(a-6)  $ \end{answer}
 \end{pro}
\begin{pro}Decompose into factors: $ a^2-38a+361$.
\begin{answer}  $ (a-19)^2 $ \end{answer}
 \end{pro}
\begin{pro}Decompose into factors: $$ a^4b^4+37a^2b^2+300.$$
\begin{answer}  $ (a^2b^2+25)(a^2b^2+12) $ \end{answer}
 \end{pro}
\begin{pro}Decompose into factors: $x^2-23x+132$.
\begin{answer}  $ (x-11)(x-12) $ \end{answer}
\end{pro}
\begin{pro}Decompose into factors: $x^4-29x^2+204$.
\begin{answer}  $ (x^2-12)(x^2-17) $ \end{answer}
 \end{pro}
\begin{pro}Decompose into factors: $ x^2+35x+216$.
\begin{answer}  $ (x+27)(x+8) $ \end{answer}
 \end{pro}
\begin{pro}
Resolve into factors: \quad $5x^2+17x+6 $.
\begin{answer}
$(5x+2)(x+3) $
\end{answer}
\end{pro}

\begin{pro}
Resolve into factors: \quad $14x^2+29x-15$.
\begin{answer}
$(7x-3)(2x+5)$
\end{answer}

\end{pro}

\end{multicols}

\section{Special Factorisations}
\QUOTEME{At the age of eleven, I began Euclid, with my brother as my
tutor. This was one of the great events of my life, as dazzling as
first love. I had not imagined there was anything so delicious in
the world. From that moment until I was thirty-eight, mathematics
was my chief interest and my chief source of
happiness.}{\mbox{Bertrand RUSSELL}}
\bigskip
Recall the special products we studied in the Multiplication
chapter. We list them here for easy reference.

\bigskip
\noindent {\bf Difference of Squares}
\begin{equation} x^2-y^2=(x-y)(x+y).  \label{eq:diff-of-squares}\end{equation}
\bigskip
\noindent {\bf Difference of Cubes}
\begin{equation} x^3-y^3=(x-y)(x^2+xy+y^2).  \label{eq:diff-of-cubes}\end{equation}
\bigskip
\noindent {\bf Sum of Cubes}
\begin{equation} x^3+y^3=(x+y)(x^2-xy+y^2).  \label{eq:sum-of-cubes}\end{equation}
\bigskip
\noindent {\bf Square of a Sum}
\begin{equation} x^2+2xy+y^2=(x+y)^2.  \label{eq:square-of-a-sum}\end{equation}
\bigskip
\noindent {\bf Square of a Difference}
\begin{equation} x^2-2xy+y^2=(x-y)^2.  \label{eq:square-of-a-diff}\end{equation}

\bigskip

We will use these special factorisations and the methods of the
preceding sections in order to treat more complicated factorisation
problems.
\begin{exa}
To factor  $x^4-9y^2$ observe that it is a difference of squares,
and so  $x^4-9y^2=(x^2-3y)(x^2+3y)$.
\end{exa}
\begin{exa}
To factor  $x^3-1$ observe that it is a difference of cubes, and so
$x^3-1=(x-1)(x^2+x+1)$.
\end{exa}
Sometimes we need to use more than one method.
\begin{exa}To factor $x^3-4x$ factor a common factor and then the
difference of squares: $$ x^3-4x = x(x^2-4)=x(x-2)(x+2). $$
\end{exa}
\begin{exa}To factor $x^4-81$,  observe that there are two difference of squares:
$$ x^4-81 = (x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9). $$
\end{exa}
\begin{exa}To factor $x^6-1$,  observe that there is a   difference of
squares and then a sum and a difference of cubes:
$$ x^6-1 = (x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1). $$
\end{exa}
The following method, called {\em Sophie Germain's
trick}\footnote{Sophie Germain  (1776--1831) was an important French
mathematician of the French Revolution. She  pretended to be a man
in order to study Mathematics. At the time, women were not allowed
to matriculate at the \'{E}cole Polytechnique, but she posed as a M.
Leblanc in order to obtain lessons from Lagrange.} is useful to
convert some expressions into differences of squares.\index{Sophie
Germain}\index{Sophie Germain's trick}
\begin{exa}
We have $$ \begin{array}{lll}x^4+x^2+1 & = & x^4+2x^2+1 -x^2 \\
& = & (x^2+1)^2-x^2\\
& = & (x^2+1-x)(x^2+1+x).
\end{array}$$
\end{exa}
\begin{exa}
We have $$ \begin{array}{lll}x^4+4 & = & x^4+4x^2 + 4-4x^2\\
& = & (x^2+2)^2-4x^2\\
& = & (x^2+2-2x)(x^2+2+2x).
\end{array}$$
\end{exa}
Sophie Germain's trick is often used in factoring quadratic
trinomials, where it is often referred to as the technique of {\em
completing the square}. We will give some examples of factorisations
that we may also obtain with the trial an error method of the
preceding section.\index{completing the square}
\begin{exa}
We have $$ \begin{array}{lll} x^2 -8x-9 & =  & x^2-8x +16-9-16 \\
& = & (x-4)^2-25\\
& = & (x-4)^2-5^2\\
& = & (x-4-5)(x-4+5)\\
& =& (x-9)(x+1).
\end{array}$$
\end{exa}
\begin{exa}
We have $$ \begin{array}{lll} x^2+4x-117 & = & x^2+4x+4-117-4\\
& = & (x+2)^2-11^2\\
& = & (x+2-11)(x+2+11)\\
& = & (x-9)(x+13).
\end{array}$$
\end{exa}


The techniques learned may be used to solve some purely arithmetic
problems.
\begin{exa} Given that
$$ 1,\ 000, \ 002,\ 000, \ 001$$ has a prime factor greater than $9000$, find
it.\end{exa} \begin{solu} We have
$$ \begin{array}{lll}1,\ 000, \ 002,\ 000, \ 001  & =  & 10^{12} + 2\cdot 10^6 + 1\\
&  =  & (10^6 + 1)^2 \\
& = & ((10^2)^3 + 1)^2 \\
& = & (10^2 + 1)^2((10^2)^2 - 10^2 + 1)^2 \\
& = & 101^29901^2,\\
\end{array}$$whence the prime sought is $9901$.
\end{solu}

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

\begin{pro}
Given that $x + 2y = 3$ and $x - 2y = -1$, find $x^2 - 4y^2$.
\begin{answer}
We have $$x^2 - 4y^2=(x+2y)(x-2y)=(3)(-1)=-3.  $$
\end{answer}
\end{pro}
\begin{pro}
Resolve into factors: \quad $x^4-16 $.
\begin{answer}
$(x-2)(x+2)(x^2+4)$
\end{answer}

\end{pro}

\begin{pro}
Resolve into factors: \quad $(a+b)^2-c^2 $
\begin{answer}
$(a+b-c)(a+b+c)$
\end{answer}

\end{pro}
\begin{pro}
Resolve into factors: \quad $x^3-x $.
\begin{answer}
$x(x-1)(x+1)$
\end{answer}

\end{pro}
\begin{pro}
Find all positive primes of the form $n^3 - 8$, where $n$ is a
positive integer. \begin{answer}Since $n^3 - 8 = (n - 2)(n^2 + 2n +
2)$ and since $n - 2 < n^2 + 2n + 4$, we must have $n - 2 = 1
\implies n = 3$ and $n^3 - 8 = 19$.

\end{answer}
\end{pro}

\end{multicols}

\section{Rational Expressions}
\QUOTEME{For all their wealth of content, for all the sum of history
and social institution invested in them, music, mathematics, and
chess are resplendently useless (applied mathematics is a higher
plumbing, a kind of music for the police band). They are
metaphysically trivial, irresponsible. They refuse to relate
outward, to take reality for arbiter. This is the source of their
witchery.}{G. STEINER}

To reduce an algebraic fraction, we must first factor it.
\begin{exa}
Reduce the fraction $\dfrac{x^2-1}{x^2-2x+1}$.
\end{exa}
\begin{solu}
We have $\dfrac{x^2-1}{x^2-2x+1}=\dfrac{(x-1)(x+1)}{(x-1)(x-1)}=\dfrac{x+1}{x-1}$.
\end{solu}

\begin{exa}
Express  the fraction $\dfrac{x^2+5x+6}{x^2-5x-14}$ in lowest terms.
\end{exa}
\begin{solu}
We have $\dfrac{x^2+5x+6}{x^2-5x-14}=\dfrac{(x+2)(x+3)}{(x+2)(x-7)}=\dfrac{x+3}{x-7}$.
\end{solu}


\bigskip

To add  or subtract algebraic fractions, we first write them in common denominators. We will prefer to expand all products and collect like terms.



\begin{exa}
Add: $$\dfrac{3}{x-3} + \dfrac{2}{x+2}.  $$
\end{exa}\begin{solu} We have,
$$\begin{array}{lll} \dfrac{3}{x-3} + \dfrac{2}{x+2} & = & \dfrac{3(x+2)}{(x-3)(x+2)} + \dfrac{2(x-3)}{(x+2)(x-3)}\\
& = & \dfrac{3x+6}{(x-3)(x+2)} + \dfrac{2x-6)}{(x+2)(x-3)}\\
& = & \dfrac{5x}{(x-3)(x+2)}\\
& = & \dfrac{5x}{x^2-x-6}.\\
 \end{array}$$
\end{solu}



\begin{exa}
Add: $$\dfrac{x}{x^2+2x+1} + \dfrac{x}{x^2-1}.  $$\end{exa} \begin{solu} We have,
$$\begin{array}{lll} \dfrac{x}{x^2+2x+1} + \dfrac{x}{x^2-1} & = & \dfrac{x}{(x+1)^2} + \dfrac{x}{(x-1)(x+1)} \\
 & = &  \dfrac{x(x-1)}{(x+1)^2(x-1)} + \dfrac{x(x+1)}{(x-1)(x+1)^2} \\
 & = & \dfrac{x^2-x+x^2+x}{(x+1)^2(x-1)}\\
  & = & \dfrac{2x^2}{x^3+x^2-x-1}
 \end{array}$$
\end{solu}
\begin{exa}
Gather all the fractions: $$ \dfrac{a}{b}+\dfrac{c}{d} - \dfrac{e}{f}. $$
\end{exa}
\begin{solu} We have,
$$\begin{array}{lll}  \dfrac{a}{b}+\dfrac{c}{d} - \dfrac{e}{f} & = &  \dfrac{adf}{bdf}+\dfrac{cbf}{dbf} - \dfrac{ebd}{fbd}\\
& = & \dfrac{adf+cbf-ebd}{fbd}.
 \end{array}$$
\end{solu}
Multiplication and division of algebraic fractions is carried out in a manner similar to the operations of arithmetical fractions.
\begin{exa}
Multiply: $$ \dfrac{x-1}{x+1}\cdot \dfrac{x+2}{x-2}. $$
\end{exa}
\begin{solu} We have,
$$\begin{array}{lll}  \dfrac{x-1}{x+1}\cdot \dfrac{x+2}{x-2} & = & \dfrac{(x-1)(x+2)}{(x+1)(x-2)} \\
& = & \dfrac{x^2+x-2}{x^2-x-2}. \end{array}$$
\end{solu}
\begin{exa}
Divide: $$ \dfrac{x-1}{x+1}\div \dfrac{x+2}{x-2}. $$
\end{exa}
\begin{solu} We have,
$$\begin{array}{lll}  \dfrac{x-1}{x+1}\div \dfrac{x+2}{x-2} & = &  \dfrac{x-1}{x+1}\cdot \dfrac{x-2}{x+2}\\
& = & \dfrac{(x-1)(x-2)}{(x+1)(x+2)}\\
& = & \dfrac{x^2-3x+2}{x^2+3x+2}. \end{array}$$
\end{solu}



\begin{exa}
The sum of two numbers is $7$ and their product $21.$ What is the
sum of their reciprocals?
\end{exa}\begin{solu} Let $x, y$ be the numbers. One has $x + y = 7, xy = 21$ whence
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{y + x}{xy} = \dfrac{7}{21} =
\dfrac{1}{3}.$$
\end{solu}

\begin{exa}
Prove that $$ \dfrac{1}{1+\dfrac{1}{x+\dfrac{1}{x}}}
=\dfrac{x^2+1}{x^2+x+1}.
$$
\end{exa}
\begin{solu}
Proceeding from the innermost fraction
$$ \begin{array}{lll}  \dfrac{1}{1+\dfrac{1}{x+\dfrac{1}{x}}} & = & \dfrac{1}{1+\dfrac{1}{\dfrac{x^2}{x}+\dfrac{1}{x}}} \\
& = & \dfrac{1}{1+\dfrac{1}{\dfrac{x^2+1}{x}}} \\
& = & \dfrac{1}{1+\dfrac{x}{x^2+1}} \\
& = & \dfrac{1}{\dfrac{x^2+1}{x^2+1}+\dfrac{x}{x^2+1}} \\
& = & \dfrac{1}{\dfrac{x^2+x+1}{x^2+1}} \\
& = & \dfrac{x^2+1}{x^2+x+1}.

 \end{array}$$

\end{solu}

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

\begin{pro}
Add: $\dfrac{1}{x-1}+\dfrac{1}{x+1}$
\begin{answer}
$\dfrac{2x}{x^2-1}  $
\end{answer}

\end{pro}

\begin{pro}
Subtract: $\dfrac{1}{x-1}-\dfrac{1}{x+1}$
\begin{answer}
$\dfrac{2}{x^2-1}  $
\end{answer}
\end{pro}
\begin{pro}
Add: $\dfrac{x}{x-2}+\dfrac{2}{x+2}$
\begin{answer}
$\dfrac{x^2+4x-4}{x^2-4}  $
\end{answer}

\end{pro}

\begin{pro}
Subtract: $\dfrac{x}{x-2}-\dfrac{2}{x+2}$
\begin{answer}
$\dfrac{x^2+4}{x^2-4}  $
\end{answer}
\end{pro}
\begin{pro}
Gather the fractions: $\dfrac{1}{x}+\dfrac{1}{x-1}-\dfrac{2}{x+1}$
\begin{answer}
$\dfrac{3x-1}{x^3-x}  $
\end{answer}

\end{pro}


\begin{pro}Gather the fractions:  $\dfrac{3x}{2} + \dfrac{3x}{2a} - \dfrac{4x}{8a^2}.$
\begin{answer}
$ \dfrac{3a^2x+3ax-x}{2a^2} $
\end{answer}

\end{pro}
\begin{pro}Gather the fractions: $\dfrac{1}{s - 1} - \dfrac{s}{(s - 1)(s + 1)}.$
\begin{answer}
$ \dfrac{1}{s^2-1} $
\end{answer}

\end{pro}
\begin{pro}Gather the fractions: $$\dfrac{x^2 - xy}{x^2y} - \dfrac{y - z}{yz} - \dfrac{2z^2 -
xz}{z^2x}.$$
\begin{answer}
$ \dfrac{2x-3y}{xy} $
\end{answer}

\end{pro}
\begin{pro} Gather the fractions: $\dfrac{2s}{s + 2} - \dfrac{3}{s - 2}.$
\begin{answer}
$ \dfrac {2s^2-7s-6}{s^2-4} $
\end{answer}

\end{pro}

\begin{pro}Gather the fractions: $\dfrac{2x}{a} + \dfrac{3x}{a^2} - \dfrac{4x}{a^2}.$
\begin{answer}
$ \dfrac {x \left( 2\,a-1 \right) }{{a}^{2}} $
\end{answer}

\end{pro}
\begin{pro}
The sum of two numbers is $7$ and their product $21.$ What is the
sum of the squares of their  reciprocals?
\begin{answer}
We have, $$x+y=7 \implies x^2+2xy+y^2 = 49\implies x^2+y^2 =
49-2xy=49-2(21)=7.$$Observe also that $x^2y^2=(xy)^2=21^2$. Hence
$$\dfrac{1}{x^2}+\dfrac{1}{y^2} = \dfrac{y^2+y^2}{x^2y^2}=\dfrac{7}{21^2}=\dfrac{1}{63}.  $$
\end{answer}
\end{pro}
\end{multicols}

\part{Equations}

\chapter{Linear Equations in One Variable}
\QUOTEME{ Mathematics is an obscure field, an abstruse science,
complicated and exact; yet so many have attained perfection in it
that we might conclude almost anyone who seriously applied himself
would achieve a measure of success.}{Cicero}
\section{Simple Equations}
\QUOTEME{The mathematician is entirely free, within the limits of
his imagination, to construct what worlds he pleases. What he is to
imagine is a matter for his own caprice; he is not thereby
discovering the fundamental principles of the universe nor becoming
acquainted with the ideas of God. If he can find, in experience,
sets of entities which obey the same logical scheme as his
mathematical entities, then he has applied his mathematics to the
external world; he has created a branch of science.}{\mbox{John
William Navin SULLIVAN}}
\bigskip
\begin{df}
An {\em identity} is an assertion that is true for all values of
the variables involved.
\end{df}
\begin{exa}From the square of the sum formula we get $(x + 1)^2 = x^2 + 2x + 1$. Since this is valid for all real
values of $x$, this is an identity.  \end{exa}
\begin{exa}\label{eq:linear_eq_1}
The assertion $4 + x = 6 - x$ is not an identity. For example, if
$x = 0$ we have $4 + 0 = 6 - 0,$ which is patently false. For $x =
1$ we have, however, $4 + 1 = 6 - 1$, which is true. Thus the
assertion is sometimes true and sometimes false.
\end{exa}
\begin{exa}
The assertion $x = x + 1$ is never true. How could a real number
be equal to itself plus $1$?
\end{exa}
\begin{df}An {\em equation of condition} or {\em equation} for short, is an assertion that is true for only particular values
of the variables employed.
\end{df}
Example \ref{eq:linear_eq_1} is an example of  an equation.
\begin{df}
The variable whose value is required to find is called the {\em
unknown variable}. The process of finding a value that satisfies
the equation is called {\em solving the equation.}
\end{df}
In this section our main preoccupation will be linear equations in
one unknown.
\begin{df}
A {\em linear equation} in the unknown $x$ is an equation of the
form $ax+b = c$, where $a\neq 0$, $b$, $c$ are real numbers.
\end{df}


In order to solve simple equations, we will appeal to the
following axioms.
\begin{axi}If to equals we add equals, the sums are equal.     \end{axi}
\begin{axi} If equals are multiplied by equals, the products are equal.    \end{axi}
In what follows we will use the symbol $\implies$, which is read
``implies.''

\begin{exa} Solve for $x$: $x - 3 = -9$ \end{exa}
\begin{solu} $$\begin{array}{lll} x - 3 = -9 & \implies & x = -9 + 3
\\
 & \implies & x = -6.
\end{array}$$
\flushleft{{\em Verification:}} $x -3 = -6 - 3 \checks -9$.
\end{solu}
\begin{exa} Solve for $x$: $x + 3 = -9$ \end{exa}
\begin{solu} $$\begin{array}{lll} x + 3 = -9 & \implies & x = -9 - 3
\\
 & \implies & x = -12.
\end{array}$$
\flushleft{{\em Verification:}} $x + 3 = -12 + 3 \checks -9$.
\end{solu}
\begin{exa}Solve for $x$: $x + 3a = -9a$ \end{exa}\begin{solu} $$\begin{array}{lll}x + 3a = -9a & \implies & x = -3a - 9a \\
& \implies & x = -12a. \end{array}$$ \flushleft{{\em
Verification:}} $x + 3a = -12a + 3a \checks -9a$. \begin{exa}Solve
for $x$: $x - a + b= 2a +3b$
\end{exa}\begin{solu} $$\begin{array}{lll}x -a+ b = 2a + 3b & \implies & x = 2a + 3b + a - b \\
& \implies & x = 3a + 2b. \end{array}$$ \flushleft{{\em
Verification:}} $x -a + b = 3a + 2b -a + b \checks 2a + 3b$.
\end{solu}
\begin{exa}Solve for $x$: $2x = -88$ \end{exa}
\begin{solu} $$\begin{array}{lll} 2x = -88 & \implies & x =
\dfrac{-88}{2} \\ & \implies & x = -44.
\end{array}$$
\end{solu}
\flushleft{{\em Verification:}} $2x = 2(-44) \checks -88$.
\end{solu}
\begin{exa}Solve for $x$: $2ax = 4a$ \end{exa}
\begin{solu} $$\begin{array}{lll} 2ax = 4a & \implies & x =
\dfrac{4a}{2a} \\ & \implies & x = 2.
\end{array}$$
\flushleft{{\em Verification:}} $2ax = 2a(2) \checks 4a$.
\end{solu}
\begin{exa}Solve for $x$: $2ax = 4a^2$ \end{exa}
\begin{solu} $$\begin{array}{lll} 2ax = 4a^2 & \implies & x =
\dfrac{4a^2}{2a} \\ & \implies & x = 2a.
\end{array}$$
\flushleft{{\em Verification:}} $2ax = 2a(2a) \checks 4a^2$.
\end{solu}

\begin{exa}Solve for $x$: $\dfrac{x}{3} = -9$ \end{exa}
\begin{solu} $$\begin{array}{lll} \dfrac{x}{3} = -9 & \implies & x =
-9(3) \\ & \implies & x = -27.
\end{array}$$
\flushleft{{\em Verification:}} $\dfrac{x}{3} = \dfrac{-27}{3}
\checks -9$.
\end{solu}
\begin{exa}Solve for $x$: $\dfrac{x}{a} = a^4$  \end{exa}
\begin{solu} $$\begin{array}{lll} \dfrac{x}{a} = a^4 & \implies & x =
a^4(a) \\ & \implies & x = a^5.
\end{array}$$
\flushleft{{\em Verification:}} $\dfrac{x}{a} = \dfrac{a^5}{a}
\checks a^4$.
\end{solu}
\begin{exa}Solve for $x$: $\dfrac{ax}{b} = \dfrac{b}{a}$\end{exa}\begin{solu} $$\begin{array}{lll} \dfrac{ax}{b} = \dfrac{b}{a} & \implies & x =
\dfrac{b}{a}\cdot \dfrac{b}{a} \\ & \implies & x =
\dfrac{b^2}{a^2}.
\end{array}$$
\flushleft{{\em Verification:}} $\dfrac{ax}{b} = \dfrac{a}{b}\cdot
\dfrac{b^2}{a^2} \checks \dfrac{b}{a}$.
\end{solu}
\begin{exa}Solve for $x$: $\dfrac{ax}{b} =
a^2b^3$ \end{exa} \begin{solu} $$\begin{array}{lll} \dfrac{ax}{b}  =
a^2b^3 & \implies & x = (a^2b^3)\dfrac{b}{a} \\ & \implies & x =
ab^4.
\end{array}$$
\flushleft{{\em Verification:}} $\dfrac{ax}{b}  =
(ab^4)\dfrac{a}{b} \checks a^2b^3$.
\end{solu}
\begin{exa}Solve for $x$: $0 =
12x$\end{exa}\begin{solu} $$\begin{array}{lll} 0 = 12x & \implies &
\dfrac{0}{12} = x \\ & \implies & x = 0.
\end{array}$$
\flushleft{{\em Verification:}} $12x = 12(0) \checks 0$.
\end{solu}
\begin{exa}Solve for $x$: $3x - 5 = 28$
\end{exa}\begin{solu} $$\begin{array}{lll} 3x - 5 = 28 & \implies & 3x =
 28 + 5\\ & \implies & x = \dfrac{33}{3} \\
 & \implies & x = 11.
\end{array}$$
\flushleft{{\em Verification:}} $3x - 5 = 3(11) - 5 = 33 - 5
\checks 28$.
\end{solu}
\begin{exa}Solve for $x$: $ax - 2a = 2ab - a$
\end{exa}\begin{solu}
$$\begin{array}{lll} ax - 2a = 2ab - a & \implies & ax = 2ab - a + 2a\\ &
\implies & x = \dfrac{2ab + a}{a} \\
& \implies & x = 2b + 1. \\
\end{array}$$
\flushleft{{\em Verification:}} $ax - 2a = a(2b + 1) - 2a = 2ab +
a - 2a\checks 2ab - a$.
\end{solu}

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

\begin{pro}Solve for the variable $z$: $z + 4 = -4.$
\begin{answer} $ z=-8 $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $z - 4 = -4.$
\begin{answer} $z=0  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $4z = -44.$
\begin{answer} $ z=-11 $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $4z = -3.$
\begin{answer} $ z=-\dfrac{3}{4} $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $\dfrac{z}{4}= -4.$
\begin{answer} $z=-16  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $\dfrac{2z}{3} = -22.$
\begin{answer} $ z=-33 $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $\dfrac{2z}{3} + 2 = -22.$
\begin{answer} $z=-36  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $\dfrac{z}{3} = a + 1.$
\begin{answer} $z=3a+3  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $5z - 3 = 22.$
\begin{answer} $ z=5 $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $z$: $\dfrac{bz}{a + 1} - b^3 = ab.$
\begin{answer} $ z=(b^2+a)(a+1)=b^2a+b^2+a^2+a $ \end{answer}

\end{pro}

\begin{pro}Solve for the variable $N$: $3N = 3A.$
\begin{answer} $N=A  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $N$: $\dfrac{N}{A} = A^2 + 1.$
\begin{answer} $ N=A^3+A $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $N$: $$\dfrac{XN}{Y} = \frac{Y}{X}.$$
\begin{answer} $N=\dfrac{Y^2}{X^2}  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $N$: $N - 5A = 7A + A^2.$
\begin{answer} $N=12A+A^2  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $N$: $N + 8X = -9X.$
\begin{answer} $N=-17X  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $N$: $\dfrac{N}{A + 2} = A + 1.$
\begin{answer} $N=A^2+3A+2  $ \end{answer}

\end{pro}
\begin{pro}Solve for the variable $N$: $$\dfrac{A^2N}{B^3} = A^2B^4.$$
\begin{answer} $ N=  B^7  $ \end{answer}

 \end{pro}

\begin{pro}Solve for the variable $N$: $\dfrac{7N}{A} =
\dfrac{14}{A^2}.$
\begin{answer} $ N=\dfrac{2}{A} $ \end{answer}

\end{pro}\begin{pro}Solve for the variable $N$: $N
+ 2 = A + 14.$
\begin{answer} $ N=A+12 $ \end{answer}

 \end{pro}
\end{multicols}
\section{Miscellaneous Linear Equations}
\QUOTEME{If they would, for Example, praise the Beauty of a Woman,
or any other Animal, they describe it by Rhombs, Circles,
Parallelograms, Ellipses, and other geometrical terms \ldots
.}{Jonathan SWIFT}


\begin{exa}Solve for $x$: $\dfrac{x}{2} -
\dfrac{x}{3} = \dfrac{x}{4} + \dfrac{1}{2}$
\end{exa}\begin{solu} The denominators are $3, 4$ and $2$, and their least common multiple is
$\lcm{3,4,2} = 12$. We multiply both sides of the equation by this
least common multiple, and so
$$\begin{array}{lll} \dfrac{x}{2} -
\dfrac{x}{3} = \dfrac{x}{4} + \dfrac{1}{2} & \implies &
12\left(\dfrac{x}{2} -
\dfrac{x}{3}\right) = 12\left(\dfrac{x}{4} + \dfrac{1}{2}\right) \\ & \implies & 6x - 4x = 3x + 6  \\
 & \implies & 2x = 3x + 6\\
  & \implies & 2x - 3x = 6\\
   & \implies & -x = 6\\
    & \implies & x=-6\\
\end{array}$$
\end{solu}
\begin{exa}
Solve for $a$: $\dfrac{b - a}{6} = \dfrac{b - a}{15} - b.$
\end{exa}\begin{solu} The denominators are $6$ and $15$, and their least common multiple is
$\lcm{6,15} = 30$. We multiply both sides of the equation by this
least common multiple, and so
$$\begin{array}{lll}
\dfrac{b - a}{6} = \dfrac{b - a}{15} - b & \implies &
30\left(\dfrac{b - a}{6}\right) = 30\left(\dfrac{b - a}{15} -
b\right) \\
& \implies & 5(b - a) = 2(b - a) - 30b \\
& \implies & 5b - 5a = 2b - 2a - 30b \\
& \implies &  -5a + 2a = 2b - 5b  - 30b \\
& \implies & -3a = -33b \\
& \implies & a = 11b.
\end{array}$$
\end{solu}
\begin{exa}Solve for $x$: $\dfrac{2x-1}{3}=\dfrac{1-3x}{2}$.\end{exa}
\begin{solu}
Cross-multiplying,
$$\begin{array}{lll}\dfrac{2x-1}{3}=\dfrac{1-3x}{2} & \implies & 2(2x-1)= 3(1-3x) \\
& \implies &  4x-2= 3-9x \\
& \implies & 4x+9x = 3+2 \\
& \implies & 13x=5 \\
& \implies & x=\dfrac{5}{13}. \\
  \end{array}$$

\end{solu}


\begin{exa} Write the infinitely repeating decimal $0.3\overline{45}
= 0.345454545\ldots$ as the quotient of two natural numbers.
\end{exa}\begin{solu} The trick is to obtain multiples of
$x=0.345454545\ldots$ so that they have the same infinite tail, and
then subtract these tails, cancelling them out.\footnote{That this
cancellation is meaningful depends on the concept of {\em
convergence}, which will be studied in later courses.} So observe
that
$$10x = 3.45454545\ldots; \qquad 1000x= 345.454545\ldots \qquad \implies \qquad1000x - 10x = 342 \qquad\implies\qquad x = \dfrac{342}{990} = \dfrac{19}{55}.  $$
\end{solu}
\begin{exa}
Find the fraction that represents the repeating decimal $$
4.3\overline{21} = 4.32121212121\ldots .$$
\end{exa}
\begin{solu} Put $x = 4.32121212121 \ldots$. Then $10x = 43.21212121\ldots$
and $1000x = 4321.2121212121 \ldots.$ Hence
$$1000x - 10x =  4321.2121212121 \ldots - 43.212121212121 \ldots  4278.$$
Finally $$x = \dfrac{4278}{990} =   \dfrac{2139}{495} =
\dfrac{713}{165}.$$
\end{solu}
\begin{exa}\label{exa:sum1-100-2}
Find the sum of the first hundred positive integers, that is, find
$$1+2+3+\cdots + 99 + 100.  $$
\end{exa}
\begin{solu}Notice that this is example \ref{exa:sum1-100-1}.
Put $$x=1+2+3+\cdots + 99 + 100.  $$The crucial observation is that
adding the sum forwards is the same as adding the sum backwards,
hence
$$x=100+99+\cdots +3 + 2+1.  $$
Adding,
$$ \begin{array}{lllllllllll} x & = &  1 & + &  2 & + & \cdots & + & 99 & + & 100 \\
  x & = & 100 & + &  99 & + & \cdots & + & 2 & + & 1 \\
\hline
 2x & = & 101 & + &  101 & + & \cdots & + & 101 & + & 101\\
   & = & 100\cdot 101, &  &   &  &  &  &  &  &
 \end{array} $$
 whence $$x=\dfrac{100\cdot 101}{2} = 50\cdot 101=5050. $$
\end{solu}
\begin{exa}
Consider the arithmetic progression $$ 2, 7, 12, 17, \ldots . $$ Is
$302$ in this progression? Is $303$ in this progression?
\end{exa}
\begin{solu}
Observe that we start with $2$ and keep adding $5$. Thus
$$2=2+5\cdot 0, \quad 7=2+5\cdot 1, \quad 12=2+5\cdot 2, \quad 17=2+5\cdot 3, \quad \ldots
$$so the general term has the form $5n+2$, which are the numbers
leaving remainder $2$ when divided by $5$. Since $$ 5n+2=302
\implies 5n=300 \implies n=60,
$$ is an integer, $302$ is in this progression. Since
$$ 5n+2=303
\implies 5n=301 \implies n=\dfrac{301}{5},
$$ is an not integer, $303$ is not in this progression.
\end{solu}

Some equations which are in appearance not linear, may be reduced to
a linear equation. Sometimes it is necessary to do some
simplification of the expressions involved before solving an
equation. In the examples that follow, we leave verification to the
reader.
\begin{exa}Solve for $x$: $(3x + 1)^2 + 6 + 18(x +
1)^2 = 9x(3x - 2) + 65$
\end{exa}
\begin{solu} $$\begin{array}{lll} (3x + 1)^2 + 6 + 18(x + 1)^2 =
9x(3x - 2) + 65 & \implies & 9x^2 + 6x + 1 + 6 + 18x^2 + 36x + 18\\ & & \qquad = 27x^2 - 18x + 65\\
& \implies & 27x^2 + 42x + 25 = 27x^2 - 18x + 65 \\
 & \implies & 27x^2 + 42x - 27x^2 + 18x = 65 - 25\\
 & \implies & 60x = 40 \\
 & \implies & x = \dfrac{2}{3}.
\end{array}$$
\end{solu}
\begin{exa}Solve for $x$: $(2x + 1)(2x + 6) - 7(x - 2) = 4(x + 1)(x - 1)
- 9x$ \end{exa}\begin{solu} $$\begin{array}{lll} (2x + 1)(2x + 6) -
7(x - 2) = 4(x + 1)(x - 1) - 9x & \implies & 4x^2 + 14x + 6 -7x + 14 \\
& & \qquad = 4x^2 - 4 - 9x\\
& \implies & 4x^2 + 7x + 20 = 4x^2 - 9x - 4 \\
 & \implies & 4x^2 + 7x - 4x^2 + 9x = -4 - 20 \\
 & \implies & 16x = -24 \\
 & \implies & x = -\dfrac{3}{2}.
\end{array}$$
\end{solu}

\begin{exa}
Solve for $x$: $\dfrac{2}{x+1} = \dfrac{3}{x}$.
\end{exa}
\begin{solu}
Cross-multiplying,
$$\dfrac{2}{x+1} = \dfrac{3}{x} \implies 2x = 3(x+1) \implies 2x=3x+3 \implies 2x-3x=3 \implies -x=3 \implies x=-3.  $$
\end{solu}
  \section*{Homework}
  \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

  \begin{pro}
Solve for $x$: \quad $2(3x-4)-4(2-3x)=1$.
\begin{answer}
$x=\dfrac{17}{18}$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $x - \dfrac{x}{2}-\dfrac{x}{3}=1$.
\begin{answer}
$x=6$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{x-2}{2}=\dfrac{3-x}{3}$.
\begin{answer}
$x=\dfrac{12}{5}$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{x}{a}-1=2$.
\begin{answer}
$x=3a$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{ax}{b}=a$.
\begin{answer}
$x=b$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $ax+b=c$.
\begin{answer}
$x=\dfrac{c-b}{a}$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{x+a}{2}= 2x+1$.
\begin{answer}
$x=\dfrac{a-2}{3}$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{x+1}{2}-\dfrac{x+2}{3}=
\dfrac{x-1}{4}$.
\begin{answer}
$x=1$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{a}{x}=b$.
\begin{answer}
$x=\dfrac{a}{b}$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{ab}{cx}=d$.
\begin{answer}
$x=\dfrac{ab}{cd}$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{3}{x-2}=1$.
\begin{answer}
$x=5$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: \quad $\dfrac{3}{x-2}=\dfrac{2}{x+3}$.
\begin{answer}
$x=-13$
\end{answer}
\end{pro}
\begin{pro}Solve for $x$: $2(3 - x) = 3x - 4$
\begin{answer}
$x=2$
\end{answer}
 \end{pro}
\begin{pro}Solve for $x$: $(2
- x)(x + 3) = -x(x - 4)$
\begin{answer}
$x=\dfrac{6}{5}$
\end{answer}
\end{pro}
\begin{pro}Solve for $x$: $(x -
a)b = (b - x)a$.
\begin{answer} We have,
$$ (x -
a)b = (b - x)a \implies bx-ab=ab-ax \implies bx=-ax \implies
(a+b)x=0 \implies x=0.
$$
\end{answer}
\end{pro}
\begin{pro}
Solve for $x$: $\dfrac{x - 3}{3} - \dfrac{x - 2}{2} = 6.$
\end{pro}
\begin{pro}
Write the infinitely repeating decimal $0.\overline{123} =
0.123123123\ldots$ as the quotient of two positive integers.
\begin{answer}
If $x = 0.123123123\ldots$ then $1000x = 0.123123123\ldots $ giving
$1000x-x =  123  $, since the tails cancel out. This results in $x =
\dfrac{123}{999} = \dfrac{41}{333}$.
\end{answer}
\end{pro}
\end{multicols}



\section{Word Problems}
\QUOTEME{ ``When I use a word,'' Humpty Dumpty said in rather a
scornful tone, ``it means just what I choose it to mean - neither
more nor less.''}{Lewis CARROLL}

\begin{exa}Find two numbers whose sum is $28$, and whose difference is
$4$. \end{exa} \begin{solu} Let $x$ be one of the numbers, then the
other number is $28 - x$. Then we have $$\begin{array}{lll}x - (28
- x) = 4 & \implies & 2x - 28 = 4\\
& \implies & x = 16. \end{array}$$ The numbers are $x = 16$ and
$28 - x = 28 - 16 = 12$.
\end{solu}
\begin{exa}Divide $\$47$ between Peter, Paul, and
Mary, so that Peter may have $\$10$ more than Paul, and Paul $\$8$
more than Mary. \end{exa} \begin{solu} Let $p$ be Paul's amount in
dollars. Then Peter has $p + 10$ dollars and Mary has $p - 8$
dollars. Then we have $$\begin{array}{lll}p + (p + 10) + (p - 8) =
47 & \implies & 3p + 2 = 47 \\
\implies & 3p = 45 \\
\implies & p = 15. \\
\end{array}$$
Thus Paul has $\$15$, Peter has $\$25$ and Mary has $\$7$.
\end{solu}

\begin{exa}The sum of three consecutive odd integers
is $609$. Find the numbers.
\end{exa}\begin{solu} Let the numbers be $x -2, x, x + 2$. Then
$$\begin{array}{lll}(x - 2) + x + (x + 2) =
609 & \implies & 3x = 609 \\
& \implies & x = 203 \\
\end{array}$$
The numbers are $x - 2 = 201,$ $x = 203$, and $x + 2 = 205$.
\end{solu}

\begin{exa}A glass of beer
costs $40$ cents more than a loaf of bread but 50 cents less than
a glass of wine. If the cost of the three items, in cents, is
$730$, what is the price of each item, in cents?
\end{exa}\begin{solu} Let $b$ be the price of a glass of beer in cents. Then
bread costs $b - 40$ cents and wine costs $b + 50$ cents. This
gives $$b + (b - 40) + (b + 50) = 730 \implies 3b + 10 = 730
\implies b = 240.
$$
Thus beer costs $240$ cents, bread costs $200$ cents and wine
costs $290$ cents.
\end{solu}


\begin{exa}Currently, the age of a father is four times
the age of his son, but in $24$ years from now it will only be
double. Find their ages. \end{exa} \begin{solu} Let $s$ be the
current age of the son. Then the current age of the father is $4s$.
In $24$ years the son will be $s + 24$ and the father will be $4s +
24$ and we will have $$ 4s + 24 = 2(s + 24). $$ This give
$$4s + 24 = 2s + 48 \implies 4s - 2s = 48 - 24 \implies s = 12.  $$
Thus the son is currently $12$ years-old and the father is
currently $48$ years-old.
\end{solu}
  \section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small

\begin{pro}Six times a number increased
by $11$ is equal to $65$. Find it.
\begin{answer}
If $x$ is the number, then $$6x+11 = 65 \implies 6x=54 \implies x=9,
$$so the number is $9$.
\end{answer}



\end{pro}\begin{pro}Find a number
which when multiplied by $11$ and then diminished by $18$ is equal
to $15$.
\begin{answer}
If $x$ is the number, then $$11x-18 = 15 \implies 11x=33 \implies
x=3,
$$so the number is $3$.
\end{answer}

 \end{pro}\begin{pro}If $3$ is added to a number, and the
sum multiplied by $12$, the result is $84$. Find the number.
\begin{answer}
If $x$ is the number, then $$12(x+3) = 84 \implies 12x+36=84
\implies 12x=48 \implies x=4,
$$so the number is $4$.
\end{answer}


\end{pro}
\begin{pro} The sum of eleven consecutive integers is $2002$. Find them.
\begin{answer} Let $$ x-5,x-4,x-3,x-2,x-1,x,x+1,x+2,x+3,x+4,x+5, $$ be the eleven consecutive integers. Then
$$ x-5+x-4+x-3+x-2+x-1+x+x+1+x+2+x+3+x+4+x+5=2002 \implies 11x=2002 \implies x
=182.$$The numbers are
$$177, 178,179,180,181,182,183,184,185,186,187.  $$
 \end{answer}
 \end{pro}

\begin{pro}One number exceeds another by $3$, and their
sum is $27$; find them.
\begin{answer}
If the two numbers be $x$ and $x+3$, then
$$ x+x+3=27 \implies x=12, $$so the numbers are $12$ and $15$.
\end{answer}
\end{pro}

\begin{pro}Find two numbers whose
sum is $19$, such that one shall exceed twice the other by $1$.
\begin{answer}
If $x$ is one number, the other is $19-x$. If $x$ exceeds twice
$19-x$ by $1$, then
$$ x-2(19-x)=1 \implies 3x-38=1 \implies 3x=39 \implies x=13. $$
Thus one number is $13$ and the other is $6$.
\end{answer}
\end{pro}


\begin{pro}Split \$380 among Peter, Paul and Mary, so
that Paul has \$30 more than Peter, and Mary has \$20 more than
Paul.
\begin{answer}
If Paul has $x$ dollars, then Mary has $x+20$ and Peter has $x-30$.
Therefore
$$x + x+20 + x-30 = 380 \implies 3x-10 = 380 \implies 3x = 390
\implies x=130.
$$Hence Paul has \$130, Mary \$150 and Peter \$100.
\end{answer}


\end{pro}
\begin{pro}
How much pure water should be added to $100$ gram of $60$\% acid
solution to make a $20$\% acid solution?
\begin{answer}
The amount of acid does not change. Hence
$$ 0.60 = 0.20(100+x)\implies x=200, $$thus $200$ grams should be
added.
\end{answer}
\end{pro}
\begin{pro}Jane's age is twice Bob's age increased by $3$.
Bill's age is Bob's age decreased by $4$. If the sum of their ages
is $27$, how old is Bill?
\begin{answer}
If Bob's age is $x$, then Bill's age is $x-4$ and Jane's $2x+3$.
Therefore
$$2x+3 +x-4+x = 27 \implies 4x-1 = 27 \implies 4x=28 \implies x=7.  $$
Thus Bob is $7$, Bill is $3$, and Jane is $17$.
\end{answer}

\end{pro}
\begin{pro}
The average of six numbers is $4$. A seventh number is added and the
new average increases to $5$. What was the seventh number?
\begin{answer} The sum of the original 6 numbers is $S = 6\cdot 4 =
24.$ If the 7th number is $x$, then $$\frac{24 + x}{7} = 5,$$whence
$x = 11.$
\end{answer}

\end{pro}


\begin{pro}
Bill currently has five times as much money as Bob. If he gives \$20
to Bob, then Bill will only have four times as much. Find their
current amounts.
\begin{answer}
If Bob currently has $b$ dollars, then Bill has $5b$ dollars. After
giving \$20 to Bob, Bill now has $5b-20$ and Bob now has $b+20$. We
are given that $$5b-20=4(b+20) \implies 5b-20 = 4b+80 \implies b =
100.
$$So currently, Bill has \$500 and Bob has \$100.
\end{answer}

\end{pro}

\begin{pro}Find a number so that
six sevenths of it exceed four fifths of it by $2$.
\begin{answer}
If $x$ is the number then $$\dfrac{6x}{7}-\dfrac{4x}{5}=2.
$$Multiplying both sides of this equation by $35$ we obtain
$$35 \left(\dfrac{6x}{7}-\dfrac{4x}{5}\right)=2(35)\implies
30x-28x=70 \implies 2x=70 \implies x=35. $$
\end{answer}

\end{pro}
\begin{pro}The difference between two numbers is $8$. If we add $2$ to the
largest we obtain $3$ times the smaller one. Find the numbers.
\begin{answer}
If one number is $x$, the larger is $x+8$. Thus
$$x+8+2 = 3x \implies x+10 = 3x \implies 10 =2x \implies 5 =x,  $$so
the smallest number is $5$ and the larger is $13$.
\end{answer}

\end{pro}

\begin{pro}Find two numbers whose difference is $10$, and
whose sum equals twice their difference.
\begin{answer}
If one of the numbers is $x$, the other is $x+10$. Thus
$$x+x+10 = 2(10) \implies 2x+10=20 \implies 2x =10 \implies x=5. $$
The smaller number is $5$ and the larger is $15$.
\end{answer}

\end{pro}

\begin{pro}
I bought a certain amount of avocados at four for \$2; I kept a
fifth of them, and then sold the rest at three for \$2. If I made a
profit of \$2, how many avocados did I originally buy?
\begin{answer}
If $x$ is the amount originally bought, then I spent
$2\left(\dfrac{x}{4}\right) = \dfrac{x}{2}$ dollars. Since I kept
$\dfrac{x}{5}$, I must have sold $\dfrac{4x}{5}$ of them, making
$2\left(\dfrac{\dfrac{4x}{5}}{3}\right) = \dfrac{8x}{15}$ dollars on
this sale. My net gain is thus
$$\dfrac{8x}{15}- \dfrac{x}{2} =2. $$Multiplying both sides by $30$,
we have
$$30\left(\dfrac{8x}{15}- \dfrac{x}{2}\right) =2(30)\implies 16x-15x=60. $$
I originally bought sixty avocados.

\end{answer}
\end{pro}

\begin{pro}
Find a number whose fourth, sixth, and eighth add up to $13$.
\begin{answer}
If $x$ is the number, then
$$\dfrac{x}{4}+\dfrac{x}{6}+\dfrac{x}{8}=13.  $$Multiplying both
sides by $24$, we have
$$24\left(\dfrac{x}{4}+\dfrac{x}{6}+\dfrac{x}{8}\right)=13(24) \implies  6x+4x+3x =312 \implies 13x=312 \implies x=24,  $$
whence the number is $24$.
\end{answer}
\end{pro}
\begin{pro}
A fifth of the larger of two consecutive integers exceeds a seventh
of the smaller by $3$. Find the integers.
\begin{answer}
Let $x$ and $x+1$ be the integers. Then
$$\dfrac{x+1}{5} - \dfrac{x}{7}=3 \implies 35\left(\dfrac{x+1}{5} - \dfrac{x}{7}\right) = 3(35)
\implies 3(x+1)- 5x=105 \implies 3-2x =105 \implies 2x=102 \implies
x=51.
$$The integers are $51$ and $52$.
\end{answer}
\end{pro}
\begin{pro}
I bought a certain number of oranges at three for a dollar and five
sixths of that number at four for a dollar. If I sold all my oranges
at sixteen for six dollars, I would make a profit of three and a
half dollars. How many oranges did I buy?
\begin{answer}
Let $x$ be total amount of oranges bought at  three for a dollar. On
these I spent $\dfrac{1}{3}\cdot x\cdot 1 = \dfrac{x}{3}$ dollars. I
bought $\dfrac{5x}{6}$ oranges at four for a dollar, thus spending
$\dfrac{1}{4}\cdot\left(\dfrac{5x}{6}\right)\cdot 1 =
\dfrac{5x}{24}$ dollars. Notice that I have bought a total of
$x+\dfrac{5x}{6}=\dfrac{11x}{6}$ oranges. If I sell all of them at
sixteen for six dollars, I make $6\cdot\dfrac{11x}{6}\cdot
\dfrac{1}{16}=\dfrac{11x}{16}$ dollars. Thus
$$ \dfrac{11x}{16} -\left(\dfrac{x}{3}+\dfrac{5x}{24}\right) =\dfrac{7}{2}
\implies x=24.$$ Hence I bought $\left(\dfrac{11}{6}\right)24=44$
oranges.
\end{answer}
\end{pro}

\begin{pro}A piece of equipment is bought by a factory. During its first year,
the equipment depreciates a fifth of its original price. During its
second year, it depreciates a sixth of its new value. If its current
value is \$56, 000, what was its original price?
\begin{answer}
Let $x$ be its original price. After a year, its new price will
$x-\dfrac{x}{5}=\dfrac{4x}{5}$. After another year, its new price
will be $\dfrac{4x}{5}- \dfrac{1}{6}\cdot \dfrac{4x}{5} =
\dfrac{2x}{3}$. Hence we must have
$$ \dfrac{2x}{3} = 56000 \implies x = 56000\cdot \dfrac{3}{2} = 84000. $$
Thus the original price was \$84,000.
\end{answer}
\end{pro}
\begin{pro} The difference of the squares of two consecutive integers is
$665$. Find the integers.
\begin{answer}
Let the integers be $n$ and $n+1$. Then $$ (n+1)^2-n^2=665 \implies
2n+1 = 665 \implies n=332.
$$Thus the integers are $332$ and $333$.

\end{answer}
\end{pro}
\begin{pro}
In how many different ways can one change $50$ cents using nickels,
dimes or quarters? \begin{answer}There are ten different ways. We
want the number of solutions of
$$5x + 10y + 25z = 50,
$$that is, of $$x + 2y + 5z = 10,     $$ with integer $0 \leq x \leq 10$,
$0 \leq y \leq 5$, $0 \leq z \leq 2$. The table below exhausts all
ten  possibilities.
$$ \begin{array}{|c|c|c|}
\hline z & y & x \\
\hline 2 & 0 & 0 \\
\hline 1 & 2 & 1 \\
\hline 1 & 1 & 3 \\
\hline 1 & 0 & 5 \\
\hline 0 & 5 & 0 \\
\hline 0 & 4 & 2 \\
\hline 0 & 3 & 4 \\
\hline 0 & 2 & 6 \\
\hline 0 & 1 & 8 \\
\hline 0 & 0 & 10 \\
\hline
\end{array}         $$





\end{answer}
\end{pro}

\begin{pro}
Five burglars stole a purse with gold coins. The five burglars took
each different amounts according to their meanness, with the meanest
among the five taking the largest amount of coins and the meekest of
the five taking the least amount of coins. Unfair sharing caused a
fight that was brought to an end by an arbitrator. He ordered that
the meanest burglar  should  double the shares of the other four
burglars below him.  Once this was accomplished, the second meanest
burglar should double the shares of the other three burglars below
him. Once this was accomplished, the third meanest burglar should
double the shares of the two burglars below him. Once this was
accomplished, the fourth meanest burglar should double the shares of
the meekest burglar.  After this procedure was terminated, each
burglar received the same amount of money. How many coins were in a
purse if the meanest of the burglars took $240$ coins initially?
\begin{answer}
Let there be $a$ coins in the purse. There are five stages. The
fifth stage is when all the burglars have the same amount of money.
Let $$a_k, b_k, c_k, d_k, e_k$$ be the amount of money that each
burglar has, decreasing lexicographically, with the $a$'s denoting
the amount of the meanest burglar and $e_k$ denoting the amount of
the meekest burglar. Observe that for all $k$ we have $$
a_k+b_k+c_k+d_k+e_k=a. $$ On stage five we are given that
$$ a_5=b_5=c_5=d_5=e_5 =\dfrac{a}{5}.
$$ On stage four
$$a_4=b_4=c_4=\dfrac{a}{5},\quad  d_4= \dfrac{a}{5}+\dfrac{a}{10}=\dfrac{3a}{10} \quad e_4=
\dfrac{a}{10}.$$ On stage three we have
$$a_3=b_3=\dfrac{a}{5},\quad \quad c_3= \dfrac{a}{5}+\dfrac{3a}{20}+ \dfrac{a}{20} = \dfrac{2a}{5} \quad  d_3= \dfrac{3a}{20} \quad e_3=
\dfrac{a}{20}.$$
 On stage two  we have
$$a_2=\dfrac{a}{5}, \quad b_2=\dfrac{a}{5}+\dfrac{a}{5}+\dfrac{3a}{40}+\dfrac{a}{40}=\dfrac{a}{2},\quad  c_2= \dfrac{a}{5} \quad  d_2= \dfrac{3a}{40}
\quad e_2= \dfrac{a}{40}.$$ On stage one we have
$$a_1=\dfrac{a}{5}+\dfrac{a}{4}+\dfrac{a}{10}+\dfrac{3a}{80}+\dfrac{a}{80}=\dfrac{3a}{5}, \quad b_1=\dfrac{a}{4},\quad  c_1= \dfrac{a}{10} \quad  d_1= \dfrac{3a}{80}
\quad e_1= \dfrac{a}{80}.$$ Since $a_1=240$, we deduce
$\dfrac{3a}{5}=240 \implies a=400$.

\bigskip

Check: On the first stage the distribution is (from meanest to
meekest):
$$240, \quad 100, \quad 40, \quad 15,\quad 5.  $$
On the second stage we have
$$ 80, \qquad  200, \quad 80\quad 30, \quad 10. $$
On the third stage we have
$$ 80, \quad 80, \quad 160, \quad 60, \quad 20. $$
On the fourth stage we have
$$ 80, \quad 80, \quad 80, \quad 120, \quad 40. $$
On the fifth stage we have
$$ 80, \quad 80, \quad 80, \quad 80, \quad 80. $$

  \end{answer}
\end{pro}



\end{multicols}


\chapter{Quadratic Equations in One Variable}

\section{Quadratic Equations}
\begin{df}
A {\em quadratic equation} in the unknown $x$ is an equation of the
form $$ax^2+bx+c =  0, $$where $a\neq 0$, $b$, $c$ are real numbers.
\end{df}
We will circumscribe ourselves to the study of quadratic trinomials
$ax^2+bx+c$ that are amenable to the factorisations studied earlier.
The crucial observation here is that if the product of two real
numbers is zero, then at least one of them must be zero. In symbols,
$$ ab=0 \implies a = 0 \quad \mathrm{or} \quad b =0.$$
\begin{exa}
Solve the quadratic equation $x^2-x=0$ for $x$.
\end{exa}
\begin{solu}
We first factor the expression $x^2-x=x(x-1)$. Now, $x(x-1)=0$ is
the product of two real numbers giving $0$, hence either $x=0$ or
$x-1=0$. Thus either $x=0$ or $x=1$. One can easily verify now that
$0^2-0 = 0$ and $1^2-1=0$.
\end{solu}
\begin{exa}
Solve the quadratic equation $x^2+2x-3=0$ for $x$.
\end{exa}
\begin{solu}Factoring,
$$\begin{array}{lll} x^2+2x-3=0 & \implies & (x+3)(x-1) =0 \\
 &\implies & x+3 =0 \quad \mathrm{or} \quad x-1=0\\
  &\implies & x=-3 \quad \mathrm{or} \quad x=1.\\
 \end{array}$$
\end{solu}
\begin{exa}
Solve the quadratic equation $8x^2-2x=15$ for $x$.
\end{exa}
\begin{solu}First, transform the equation into $8x^2-2x-15=0$. Now, factoring,
$$\begin{array}{lll} 8x^2-2x-15 & \implies & (2x-3)(4x+5) =0 \\
 &\implies & 2x-3 =0 \quad \mathrm{or} \quad 4x+5=0\\
 & \implies & x=\dfrac{3}{2} \quad \mathrm{or} \quad x=-\dfrac{5}{4}.\\
 \end{array}$$
\end{solu}

Some equations can be reduced to quadratic equations after proper massaging.
\begin{exa}
Solve for $x$: $\dfrac{x-4}{3} = \dfrac{4}{x}$.
\end{exa}
\begin{solu}
Cross-multiplying,
$$\begin{array}{lll}  \dfrac{x-4}{3} = \dfrac{4}{x}   & \implies  & x(x-4)=12\\
& \implies  & x^2-4x-12=0 \\
& \implies  & (x-6)(x+2)=0\\
& \implies  & x=6 \qquad \mathrm{or} \qquad x=-2.\end{array}  $$
\end{solu}

The argument that a product of real numbers is zero if and only if at least one of its factors can be used with equations of
higher degree.

\begin{exa}
Solve for $x$: $x^3-4x=0$.
\end{exa}
\begin{solu}
Factoring a common factor and then a difference of squares,
$$\begin{array}{lll}  x^3-4x=0   & \implies  & x(x^2-4)=0\\
& \implies  & x(x-2)(x+2)=0 \\
& \implies  & x=0 \qquad \mathrm{or} \qquad x=+2 \qquad \mathrm{or} \qquad x=-2.\end{array}  $$
\end{solu}

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

\begin{pro}
Solve for $x$: $x^2-4 =0$.

\begin{answer} $x=2$ or $x=-2$ \end{answer}

\end{pro}

\begin{pro}
Solve for $x$: $ x^2-x-6=0$.

\begin{answer} $x=3$ or $x=-2$ \end{answer}

\end{pro}

\begin{pro}
Solve for $x$: $ x^2+x-6=0$.

\begin{answer} $x=-3$ or $x=2$ \end{answer}

\end{pro}

\begin{pro}
Solve for $x$: $x^2-4x =5$.

\begin{answer} $x=5$ or $x=-1$ \end{answer}

\end{pro}

\begin{pro}

Solve for $x$: $x^2 =1$

\begin{answer} $x=1$ or $x=-1$ \end{answer}

\end{pro}

\begin{pro}
If eggs had cost $x$ cents less per dozen, it would have cost 3  \
cents less for $x+3$ eggs  than if they had cost $x$ cents more per
dozen. What is $x$?
\begin{answer}
If eggs had cost $x$ cents less per dozen, I would have saved
$\frac{x}{12}$ cents per egg. If eggs had cost $x$ cents more per
dozen, I would have lost  $\frac{x}{12}$ cents per egg. The
difference between these two prices is $2(\frac{x}{12})$ per egg.
Thus if I buy $x+3$ eggs and the total difference is 3 cents, I can
write
$$(x + 3)2\left(\frac{x}{12}\right) = 3,$$which is to say $x^2 - 3x - 18 =  (x - 3)(x + 6) = 0$.
Since $x$ has to be a positive number, $x=3$.
\end{answer}
\end{pro}
\begin{pro}A group of people rents a bus for an excursion, at the price of
\$2,300, with each person paying an equal share. The day of the
excursion, it is found that six participants do not shew up, which
increases the share of those present by \$7.50. What was the
original number of people renting the bus?
\begin{answer}
Let $x$ be the original number of people renting the bus. Each
person must originally pay $\dfrac{2300}{x}$ dollars. After six
people do not shew up there are $x-6$ people, and each must pay
$\dfrac{2300}{x}+7.50$ dollars. We need
$$ (x-6)\left(\dfrac{2300}{x}+7.50\right)=2300.$$
Rearranging,
$$ (x-6)(2300+7.5x) = 2300x \implies 7.5x^2-45x-13800=0 \implies 75x^2-450x-138000=0 \implies 75(x+40)(x-46)=0 \implies x\in\{-40,46\}.  $$
Since $x$ must be positive, there were $46$ people on the bus
originally.
\end{answer}
\end{pro}

\begin{pro}\label{pro:continued-fraction}
Let
$$ x = 6 + \dfrac{1}{6 + \dfrac{1}{6 + \dfrac{1}{6 + \dfrac{1}{6 + \dfrac{1}{\vdots}}}}},$$
where there are an infinite number of fractions. Prove that $x^2 -
6x - 1 = 0.$
\begin{answer} One has $x = 6 + \dfrac{1}{x}$ or $x^2 - 6x - 1 = 0.$
Later on you will learn how to solve this equation and see that
 $x = 3 + \sqrt{10}$. There is some deep mathematics going
on here. That the infinite expression makes sense depends on the
concept of {\em convergence}, which will be dealt with in Calculus
courses.


\end{answer}

\end{pro}
\begin{pro}
Assume that there is a positive real number $x$ such that
$$x^{x^{x^{.^{.^{.}}}}} = 2,$$where there is an infinite number of
$x$'s. What is the value of $x$? \begin{answer} We have $x^2 = 2,$
and since $x$ is positive, $x = \sqrt{2}$. Like problem
\ref{pro:continued-fraction}, this problem is deep and depends on
the concept of convergence.


\end{answer}
\end{pro}

\end{multicols}

\part{Inequalities}
\chapter{Linear Inequalities}
\section{Intervals}
\dropping{2}{L}et us start this chapter by recalling some properties
of the order symbols \begin{center}\begin{enumerate} \item[] $<$ (is
less than), \item[] $>$ (is greater than), \item[] $\leq$ (is less
than or equal to) and, \item[]  $\geq$ (is greater than or equal
to).
\end{enumerate}\end{center}
We will simply state the useful properties for the symbol $<$, as
the same properties hold for the other three symbols {\em mutatis
mutandis}.
\begin{axi}
Let $a, b, c$ be real numbers. Then
\begin{enumerate}
\item If $a<b$ and $b<c$, then $a<c$. That is, inequalities are   {\em
transitive}.
\item If $a<b$ and $c<d$, then $a+c<b+d$, that is, addition preserves the
inequality sense.
\item If $a<b$ and $c>0$, then $ac<bc$, that is, multiplication by a positive
factor preserves the inequality sense.
\item If $a<b$ and $c<0$, then $ac>bc$, that is, multiplication by a negative
factor reverses the inequality sense.
\end{enumerate}
\end{axi}

\begin{df}
A  {\em non-degenerate interval}\footnote{A {\em degenerate}
interval is one consisting of no points or  exactly one point. We
will not consider such cases in these notes.} {\em interval} for
short, is a set $I$ of real numbers with the following property: for
any two different elements $x,y$ in $I$, any number $a$ in between
$x$ and $y$ also belongs to $I$.
\end{df}

\begin{rem}
From the definition we see that non-degenerate intervals have
infinitely many elements. For example consider the interval in
figure \ref{fig:interval-last}. Every number  between $-2$ and $1$
is in this interval, so in particular, $-1.5$ belongs to it, so does
$-1.51$, $-1.501$, $-1.5001$, etc.\end{rem}



Some examples of intervals are shewn in figures
\ref{fig:interval-first} through \ref{fig:interval-last}. If we
define the {\em length} of the interval to be the difference between
its final point and its initial point, then intervals can be finite,
like those in figures \ref{fig:intv-finite-first}
 through \ref{fig:interval-last}
 or infinite, like those in figures \ref{fig:interval-first}
 through \ref{fig:intv-last-infi}. In either
case, intervals contain an infinite number of elements.
\vspace*{2cm}
\begin{figure}[h]
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{<-o}(-3,0)(-1,0)
\vspace*{1cm}\footnotesize\hangcaption{$x < -1$}
\label{fig:interval-first}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{o->}(-1,0)(3,0)\vspace*{1cm}\footnotesize\hangcaption{$x
> -1$}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering\psset{unit=1.5pc} \psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{<-o}(-3,0)(1,0)\vspace*{1cm}\footnotesize\hangcaption{$x
< 1$}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{o->}(1,0)(3,0)\vspace*{1cm}\footnotesize\hangcaption{$x
> 1$}
\end{minipage}

\end{figure}

\vspace*{2cm}

\begin{figure}[h]
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{<-*}(-3,0)(-1,0)
\vspace*{1cm}\footnotesize\hangcaption{$x \leq -1$}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{*->}(-1,0)(3,0)\vspace*{1cm}\footnotesize\hangcaption{$x
\geq -1$}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering\psset{unit=1.5pc} \psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{<-*}(-3,0)(1,0)\vspace*{1cm}\footnotesize\hangcaption{$x
\leq 1$}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{*->}(1,0)(3,0)\vspace*{1cm}\footnotesize\hangcaption{$x
\geq 1$}\label{fig:intv-last-infi}
\end{minipage}

\end{figure}
\vspace*{2cm}

\begin{figure}[h]
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{*-*}(-1,0)(2,0)
\vspace*{1cm}\footnotesize\hangcaption{$-1 \leq x \leq 2$}
\label{fig:intv-finite-first}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{o-o}(-1,0)(2,0)\vspace*{1cm}\footnotesize\hangcaption{$-1<x
<2$}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering\psset{unit=1.5pc} \psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{o-*}(-2,0)(1,0)\vspace*{1cm}\footnotesize\hangcaption{$-2<x
\leq 1$}
\end{minipage}
\hfill
\begin{minipage}{4cm}
\centering \psset{unit=1.5pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{*-o}(-1,0)(1,0)\vspace*{1cm}\footnotesize\hangcaption{$-1\leq
x < 1$} \label{fig:interval-last}
\end{minipage}
\end{figure}

\section{One-Variable Linear Inequalities}
\dropping{2}{W}e are now interested in solving linear inequalities
in one variable. The procedure will closely resemble to that for
solving linear equations in one variables, with the caveat that if
we ever divide or multiply by a negative quantity, then the sense of
the inequality changes.

\begin{exa}\label{exa:ineq-1}
Solve the inequality $$ 2x-3 < -13 $$ and graph its solution set.
\end{exa}
\begin{solu}
Observe that $$2x-3 < -13 \implies 2x < -13 + 3 \implies 2x < -10.
$$The next stage would be to divide by $2$, which is a positive
quantity. The sense of the inequality is retained, and we gather,
$$2x<-10 \implies x < \dfrac{-10}{2} \implies x<-5. $$The solution
set appears in figure \ref{fig:ineq-exa1}.
\end{solu}

\begin{exa}\label{exa:ineq-2}
Solve the inequality $$ -2x-3 \leq -13 $$ and graph its solution
set.
\end{exa}
\begin{solu}
Observe that $$-2x-3\leq   -13 \implies -2x \leq  -13 + 3 \implies
-2x \leq -10.
$$The next stage would be to divide by $-2$, which is a negative
quantity. The sense of the inequality is reversed, and we gather,
$$-2x\leq -10 \implies x \geq \dfrac{-10}{-2} \implies x\geq -5. $$The solution
set appears in figure \ref{fig:ineq-exa2}.
\end{solu}



\vspace*{2cm}

\begin{figure}[h]
\begin{minipage}{7cm}
\centering \psset{unit=1pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-8,0)(8,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{<-o}(-8,0)(-5,0)
\vspace*{1cm}\footnotesize\hangcaption{Example \ref{exa:ineq-1}.
$x<-5$.} \label{fig:ineq-exa1}
\end{minipage}
\hfill
\begin{minipage}{7cm}
\centering \psset{unit=1pc}\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-8,0)(8,
0)\psline[linewidth=2pt,linecolor=red,arrowscale=2]{*->}(-5,0)(8,0)
\vspace*{1cm}\footnotesize\hangcaption{Example \ref{exa:ineq-2}.
$x\geq-5$.} \label{fig:ineq-exa2}
\end{minipage}

\end{figure}



\section*{Homework}

\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Find the set of all real numbers $x$ satisfying the inequality
$$\dfrac{x}{3} - \dfrac{x-2}{4}<2+x,$$ and graph its solution set.
\begin{answer}
$x>-\dfrac{18}{11}$.
\end{answer}
\end{pro}
\begin{pro}
Find the set of all real numbers $x$ satisfying the inequality
$$4(x+1)\leq x-5,$$ and graph its solution set.
\begin{answer}
$x\leq -3$.
\end{answer}
\end{pro}
\begin{pro}
Find the set of all real numbers $x$ satisfying the inequality
$$\dfrac{x-2}{3}-\dfrac{1-x}{2}>0,$$ and graph its solution set.
\begin{answer}
$x> \dfrac{7}{5}$.
\end{answer}
\end{pro}

\end{multicols}
\Closesolutionfile{algebrilla}

\appendix
\renewcommand{\chaptername}{Appendix}

\chapter{Old Exam Questions}
\begin{enumerate}
\item[] {\bf Multiple Choice Problems}

\item[] {\bf Chapter II Problems}

\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 $\dfrac{5^3-(-3)^3}{5^2-(-3)^2} =$\\  %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 $\dfrac{3}{4} - \dfrac{11}{12}=$\\
\\  \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 $\dfrac{3}{14} \div \dfrac{21}{8}=$\\
\\  \psovalbox{A} $\dfrac{9}{16}$ \hfill \psovalbox{B}$\dfrac{4}{49}$
\hfill \psovalbox{C} $\dfrac{24}{22}$ \hfill \psovalbox{D}
$\dfrac{49}{4}$ \hfill \psovalbox{E} none of these
\item $\dfrac{(1 + 2\cdot 3)^2 - 9}{(2\cdot 5 - 2)(2^2+1)}=$\\
\\  \psovalbox{A} $1$ \hfill \psovalbox{B}$-1$
\hfill \psovalbox{C} $\dfrac{29}{30}$ \hfill \psovalbox{D}
$\dfrac{2}{3}$ \hfill \psovalbox{E} none of these
\item $1024\div 512\times 2 = $
  \\ \psovalbox{A}
$1$ \hfill\psovalbox{B}$2$ \hfill\psovalbox{C}$4$
\hfill\psovalbox{D}$8$ \hfill\psovalbox{E}none of these


\item $\dfrac{5^3 - 3^3}{5^2 + 15 + 3^2} = $
  \\ \psovalbox{A}
$1$ \hfill\psovalbox{B}$\dfrac{1}{2}$ \hfill\psovalbox{C}$2$
\hfill\psovalbox{D}$\dfrac{8}{49}$ \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 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 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

\item $\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{6}{7}=$\\
\\ \vfill \psovalbox{A} $\dfrac{13}{42}$ \hfill \psovalbox{B}$-3$
\hfill \psovalbox{C} $-\dfrac{1}{14}$ \hfill \psovalbox{D}
$-\dfrac{53}{60}$ \hfill \psovalbox{E} none of these
\item $-5^2+4-9-(-7)+(-9+5\cdot 3)^2=$\\
\\ \vfill \psovalbox{A} $-13$ \hfill \psovalbox{B}$-1$
\hfill \psovalbox{C} $63$ \hfill \psovalbox{D} $13$ \hfill
\psovalbox{E} none of these
\item Perform the calculation: \quad  $1-2+3-4+5-6+7-8+9-10$.\\
\\ \vfill \psovalbox{A} $5$ \hfill \psovalbox{B}$0$
\hfill \psovalbox{C} $-5$ \hfill \psovalbox{D} $1$ \hfill
\psovalbox{E} none of these
\item  Perform the calculation: \quad  $\dfrac{\frac{3}{4}}{\frac{5}{6}} $.\\
\\ \vfill \psovalbox{A} $\dfrac{9}{10}$ \hfill \psovalbox{B}$\dfrac{10}{9}$
\hfill \psovalbox{C} $\dfrac{19}{12}$ \hfill \psovalbox{D}
$\dfrac{5}{8}$ \hfill \psovalbox{E} none of these
\item  Perform the calculation: \quad  $\dfrac{3}{4}+\dfrac{5}{6} $.\\
\\ \vfill \psovalbox{A} $\dfrac{9}{10}$ \hfill \psovalbox{B}$\dfrac{10}{9}$
\hfill \psovalbox{C} $\dfrac{19}{12}$ \hfill \psovalbox{D}
$\dfrac{5}{8}$ \hfill \psovalbox{E} none of these

\item[] {\bf Chapter III Problems}
\item Collect like terms:\qquad  $(x+y+2z)-(x-2y+z)$. \\  %A
\psovalbox{A} $3y+z$  \hfill \psovalbox{B} $2x-y+3z$ \hfill
\psovalbox{C} $3y+3z$  \hfill \psovalbox{D} $3y-z$ \hfill
\psovalbox{E} none of these

\item Which of the following are like terms?\\  %B
\psovalbox{A}  $x^6$ and $6x$\hfill \psovalbox{B} $2x^2$ and
$-x^2$\hfill \psovalbox{C}$j^2kl$ and $jk^2l$ \hfill\psovalbox{D}
$c$ and $c^2$ \hfill\psovalbox{E} none of these
\item $2(a-2b)-3(b-2a) =$ \\  %C
\psovalbox{A} $3a-b$ \hfill  \psovalbox{B} $-5a-10b$\hfill
\psovalbox{C} $8a-7b$ \hfill \psovalbox{D} $8a+7b$\hfill
\psovalbox{E} none of these
\item $\left(\dfrac{20x-10}{5}\right)-\left(\dfrac{6-9x}{3}\right)=$\\
\\  \psovalbox{A} $7x$ \hfill \psovalbox{B}$7x-4$
\hfill \psovalbox{C} $x$ \hfill \psovalbox{D} $x-4$ \hfill
\psovalbox{E} none of these
\item Collect like terms: $2(a-2b)-(2a-b)$\\
  \psovalbox{A} $-3b$ \hfill \psovalbox{B} $-b$ \hfill
\psovalbox{C} $4a-5b$ \hfill \psovalbox{D} $4a-3b$ \hfill
\psovalbox{E} none of these


\item Collect like terms: $-2(x^2-1)+4(2-x-x^2)$
  \\ \psovalbox{A}
$-6x^2-4x+10$ \hfill\psovalbox{B}$6x^2+4x+10$
\hfill\psovalbox{C}$-6x^2+4x+10$ \hfill\psovalbox{D}$-2x^2-4x+10$
\hfill\psovalbox{E}none of these
\item Collect like terms: $(-2a+3b-c)+(5a-8b+c)=$ \\
 \psovalbox{A}  $7a-11b-2c$\hfill  \psovalbox{B}
$3a-5b-2c$\hfill \psovalbox{C} $-7a+11b-2c$\hfill \psovalbox{D}
$3a-5b$\hfill \psovalbox{E} none of these


\item $-a+2b+3a-4b=$
  \\ \vfill\psovalbox{A}
$-2a+2b$ \hfill\psovalbox{B} $2a-2b$ \hfill\psovalbox{C} $-4a-6b$
\hfill\psovalbox{D} $2a+2b$ \hfill\psovalbox{E}none of these

\item $\dfrac{a}{3}-\dfrac{a}{5}=$
  \\ \vfill\psovalbox{A}
 $\dfrac{a}{15}$ \hfill\psovalbox{B} $-\dfrac{a}{2}$
\hfill\psovalbox{C} $\dfrac{2a}{15}$ \hfill\psovalbox{D}
$\dfrac{a}{8}$ \hfill\psovalbox{E}none of these

\item $\dfrac{a}{3}+\dfrac{a^2}{3}+\dfrac{a}{4} + \dfrac{3a^2}{4} = $
  \\ \vfill\psovalbox{A}
 $\dfrac{4a}{7}+\dfrac{4a^2}{7}$ \hfill\psovalbox{B}
$\dfrac{7a}{12}+\dfrac{13a^2}{12}$ \hfill\psovalbox{C}
$\dfrac{a}{12}+\dfrac{a^2}{12}$ \hfill\psovalbox{D}
$\dfrac{a^6}{48}$ \hfill\psovalbox{E}none of these



\item $(-a+2b-3c)-(-a+2b+3c) = $
  \\ \vfill\psovalbox{A}
$-6c$ \hfill\psovalbox{B} $4b-6c$ \hfill\psovalbox{C} $2a+4b+6c$
\hfill\psovalbox{D} $2a-6c$ \hfill\psovalbox{E}none of these

\item Collect like terms: $\dfrac{3x-6}{3}+x-1$.
\\ \vfill\psovalbox{A} $2x-2$ \hfill\psovalbox{B}$2x-3$
\hfill\psovalbox{C}$x-3$ \hfill\psovalbox{D}$2A-B-2$
\hfill\psovalbox{E}none of these
\item  How many of the following expressions are equivalent to $5+x^2y$?
$$I: x^2y+5, \qquad II: 5+yx^2, \qquad III: 5+xy^2, \qquad IV: yx^2+5.    $$
\psovalbox{A} exactly one \hfill\psovalbox{B} exactly two
  \hfill\psovalbox{C} exactly three
\hfill\psovalbox{D} all four \hfill\psovalbox{E}none
\item Collect like terms: $\left(\dfrac{-35x+20}{5}\right)+\left(\dfrac{12-24x}{6}\right)$. \\
\vfill \psovalbox{A}  $-11x+6$\hfill  \psovalbox{B} $3x+2$\hfill
\psovalbox{C} $3x+6$\hfill \psovalbox{D} $-3x+2$\hfill \psovalbox{E}
none of these

\item[] {\bf Chapter IV Problems}
\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 $\dfrac{3^{3}}{3^{4}}=$\\ \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 $(3^3)^4=$\\ \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 $\dfrac{3^3}{3^{-4}}=$\\ \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 $\dfrac{1111^5+1111^5+1111^5+1111^5}{1111^4+1111^4}=$\\
\\ \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 Simplify and write with positive exponents only: $(ab^2)^2(a^3b^{-4})$.\\
\\ \vfill \psovalbox{A} $a^5$ \hfill \psovalbox{B} $a^5b^8$
\hfill \psovalbox{C} $\dfrac{b^8}{a}$ \hfill \psovalbox{D} $a^6$
\hfill \psovalbox{E} none of these
\item Simplify and write with positive exponents only: $(ab^2)^2\div (a^3b^{-4})$.\\
\\ \vfill \psovalbox{A} $a^5$ \hfill \psovalbox{B} $a^5b^8$
\hfill \psovalbox{C} $\dfrac{b^8}{a}$ \hfill \psovalbox{D} $a^6$
\hfill \psovalbox{E} none of these
\item Perform the calculation: $\dfrac{2^{-3}}{3^{-2}} - \dfrac{3^2}{2^2}$.\\
\\ \vfill \psovalbox{A} $\dfrac{9}{8}$ \hfill \psovalbox{B}$\dfrac{9}{4}$
\hfill \psovalbox{C} $-\dfrac{9}{8}$ \hfill \psovalbox{D}
$-\dfrac{9}{4}$ \hfill \psovalbox{E} none of these
\item Perform the calculation: $\dfrac{1}{2^{-5}}-5^{-2}$.\\
\\ \vfill \psovalbox{A} $0$ \hfill \psovalbox{B} $20$
\hfill \psovalbox{C} $\dfrac{799}{25}$ \hfill \psovalbox{D}
$\dfrac{2}{25}$ \hfill \psovalbox{E} none of these \vfill




\item Multiply and write with positive exponents only:\quad
$(x^4y^7)(x^{-2}y^{-8})$. \\
 \psovalbox{A} $x^6y^{15}$ \hfill \psovalbox{B} $x^2y$ \hfill
\psovalbox{C} $\dfrac{x^2}{y}$  \hfill \psovalbox{D}
$\dfrac{y}{x^2}$ \hfill \psovalbox{E} none of these
\item Multiply and collect like terms:\quad
$(x-2)(x^2+2x+4)$. \\
 \psovalbox{A} $x^3-8$ \hfill \psovalbox{B} $x^3-4x^2-4x-8$
\hfill \psovalbox{C} $x^3+8$  \hfill \psovalbox{D} $x^3-4x^2-8$
\hfill \psovalbox{E} none of these
\item Divide and write with positive exponents only:\quad
$\dfrac{x^4y^7}{x^{-2}y^{-8}}$. \\
 \psovalbox{A} $x^6y^{15}$ \hfill \psovalbox{B} $x^2y$ \hfill
\psovalbox{C} $\dfrac{x^2}{y}$  \hfill \psovalbox{D}
$\dfrac{y}{x^2}$ \hfill \psovalbox{E} none of these
\item  $\dfrac{2^{-3}}{3^{-2}2^3}=$\\ \vfill \psovalbox{A} $\dfrac{9}{64}$ \hfill \psovalbox{B}$9$
\hfill \psovalbox{C} $\dfrac{64}{9}$ \hfill \psovalbox{D}
$\dfrac{1}{9}$ \hfill \psovalbox{E} none of these

 \item  $(x^{-1}y^{-2}z^3)^{-3}=$\\ \vfill \psovalbox{A} $x^4y^5z^0$ \hfill \psovalbox{B}$\dfrac{z^9}{x^3y^6}$
\hfill \psovalbox{C} $\dfrac{x^3y^6}{z^9}$ \hfill \psovalbox{D}
$\dfrac{x^3z^9}{y^6}$ \hfill \psovalbox{E} none of these

\item  $(x^4y^3z^2)(x^2y^4z^2)=$\\ \vfill \psovalbox{A} $\dfrac{x^2}{y}$ \hfill \psovalbox{B}$x^2y$
\hfill \psovalbox{C} $\dfrac{y}{x^2}$ \hfill \psovalbox{D}
$x^6y^7z^4$ \hfill \psovalbox{E} none of these
\item Multiply and
collect like terms: $(x+2)(x-3)=$\\ \psovalbox{A} $x^2-6$\hfill
\psovalbox{B} $x^2-x-6$\hfill \psovalbox{C} $x^2 +x -6$\hfill
\psovalbox{D} $x^2-5x-6$\hfill \psovalbox{E} none of these

\item Multiply and
collect like terms: $(2x-1)(2x+1) - 2(2x -1)=$\\ \psovalbox{A} $4x^2
-4x$\hfill \psovalbox{B} $4x^2 - 4x+1$\hfill \psovalbox{C} $4x^2 -4x
-1$\hfill \psovalbox{D} $1$\hfill \psovalbox{E} none of these

\item Multiply and
collect like terms: $(a+2)(a-2)-(a-1)(a+1)=$\\ \psovalbox{A}
$2a^2-3$\hfill \psovalbox{B} $-3$\hfill \psovalbox{C} $-5$\hfill
\psovalbox{D} $2a^2-5$\hfill \psovalbox{E} none of these
\item Multiply and
collect like terms: $(a+2)^2+(a-2)^2=$\\ \psovalbox{A}
$2a^2+8$\hfill \psovalbox{B} $8a$\hfill \psovalbox{C} $2a^2+4$\hfill
\psovalbox{D} $2a^2-4a+8$\hfill \psovalbox{E} none of these
\item Multiply and
collect like terms: $(a+2)^2-(a-2)^2=$\\ \psovalbox{A}
$2a^2+8$\hfill \psovalbox{B} $8a$\hfill \psovalbox{C} $2a^2+4$\hfill
\psovalbox{D} $2a^2-4a+8$\hfill \psovalbox{E} none of these
\item Expand and collect like terms: $(x+2)(x^2-4x+1)$ \\ \vfill
\psovalbox{A} $x^3-6x^2+9x-2$ \hfill\psovalbox{B} $x^3-2x^2-7x+2$
\hfill\psovalbox{C} $x^3+2x^2-9x+2$ \hfill\psovalbox{D} $x^4-1$
\hfill\psovalbox{E}none of these
\item[] {\bf Chapter V Problems}
\item Perform the division: \quad $(x^3-8x^2) \div (-x)$.
\\
 \psovalbox{A} $x^2-8x$ \hfill \psovalbox{B} $x^2+8x$ \hfill
\psovalbox{C} $-x^2+8x$  \hfill \psovalbox{D} $-x^2-8x$ \hfill
\psovalbox{E} none of these


\item Factor: \quad $2x^2-3x$. \\
 \psovalbox{A} $x(2x-3)$ \hfill \psovalbox{B} $2x(x-3)$ \hfill
\psovalbox{C} $x^2(2x-3)$  \hfill \psovalbox{D} $x(2x+3)$ \hfill
\psovalbox{E} none of these
\item Factor: \quad $x^2+3x+2$. \\
 \psovalbox{A} $(x+1)(x-2)$ \hfill \psovalbox{B} $(x+1)(x+2)$
\hfill \psovalbox{C} $(x-1)(x+2)$  \hfill \psovalbox{D} $(x-1)(x-2)$
\hfill \psovalbox{E} none of these
\item Factor: \quad $x^2-3x+2$. \\
 \psovalbox{A} $(x+1)(x-2)$ \hfill \psovalbox{B} $(x+1)(x+2)$
\hfill \psovalbox{C} $(x-1)(x+2)$  \hfill \psovalbox{D} $(x-1)(x-2)$
\hfill \psovalbox{E} none of these

\item Factor: \quad $x^2-x-2$. \\
 \psovalbox{A} $(x+1)(x-2)$ \hfill \psovalbox{B} $(x+1)(x+2)$
\hfill \psovalbox{C} $(x-1)(x+2)$  \hfill \psovalbox{D} $(x-1)(x-2)$
\hfill \psovalbox{E} none of these

\item Perform the division: \quad $(2x^3-3x^2-4x-1) \div (2x+1)$.
\\
 \psovalbox{A} $x^2+2x+1$ \hfill \psovalbox{B} $x^2-2x-1$
\hfill \psovalbox{C} $x^2-2x+1$  \hfill \psovalbox{D} $x^2+2x-1$
\hfill \psovalbox{E} none of these
\item $\dfrac{6x^2+2x}{2x}=$\\ \vfill
\psovalbox{A} $3x+1$ \hfill \psovalbox{B} $3x$ \hfill \psovalbox{C}
$3x+2$  \hfill \psovalbox{D} $6x^2$ \hfill \psovalbox{E} none of
these
\item $\dfrac{(6x^2)(2x)}{2x}=$\\ \vfill
\psovalbox{A} $3x+1$ \hfill \psovalbox{B} $3x$ \hfill \psovalbox{C}
$3x+2$  \hfill \psovalbox{D} $6x^2$ \hfill \psovalbox{E} none of
these
\item $(9a^3b^3-6a^2b^4+3a^2b^3)\div(3a^2b^3)$\\ \vfill
\psovalbox{A} $3a-2b$ \hfill \psovalbox{B} $3a-2b+1$ \hfill
\psovalbox{C} $-6ab$  \hfill \psovalbox{D} $3a^5b^6-2a^4b^7+a^4b^6$
\hfill \psovalbox{E} none of these
\item $(x^2+5x-14)\div (x-2)=$\\ \vfill
\psovalbox{A} $x-7$ \hfill \psovalbox{B} $x+7$ \hfill \psovalbox{C}
$x+2$ \hfill \psovalbox{D} $x+12$ \hfill \psovalbox{E} none of these

\item $(6x^3-5x^2+7x-2)\div (3x-1)=$\\ \vfill
\psovalbox{A} $2x^2+x+2$ \hfill \psovalbox{B} $2x^2-x-2$ \hfill
\psovalbox{C} $2x^2-x+2$ \hfill \psovalbox{D} $2x^2+x-2$ \hfill
\psovalbox{E} none of these

\item $(x^3+27)\div (x+3)=$\\
\\ \vfill \psovalbox{A} $x^2-9x+9$ \hfill \psovalbox{B}$x^2+9x+9$
\hfill \psovalbox{C} $x^2+3x+9$ \hfill \psovalbox{D} $x^2-3x+9$
\hfill \psovalbox{E} none of these
\item $\dfrac{x^9+x^6}{x^3}=$\\
\\ \vfill \psovalbox{A} $x^3+x^2$ \hfill \psovalbox{B} $3x^3+2x^2$
\hfill \psovalbox{C} $x^6+x^3$ \hfill \psovalbox{D} $x^{12}$ \hfill
\psovalbox{E} none of these
\item $\dfrac{(x^9)(x^6)}{x^3}=$\\
\\ \vfill \psovalbox{A} $x^3+x^2$ \hfill \psovalbox{B} $3x^3+2x^2$
\hfill \psovalbox{C} $x^6+x^3$ \hfill \psovalbox{D} $x^{12}$ \hfill
\psovalbox{E} none of these
\item Factor: $xy^2-x^2y$.\\
\\ \vfill \psovalbox{A} $xy(x-y)$ \hfill \psovalbox{B}$xy(y-x)$
\hfill \psovalbox{C} $x^2y^2(x-y)$ \hfill \psovalbox{D} $xy(x+y)$
\hfill \psovalbox{E} none of these
\item Factor: $x^2-x-12$.\\
\\ \vfill \psovalbox{A} $(x-3)(x+4)$ \hfill \psovalbox{B}$(x-6)(x+2)$
\hfill \psovalbox{C} $(x+3)(x-4)$ \hfill \psovalbox{D} $(x-12)(x+1)$
\hfill \psovalbox{E} none of these
\item Factor: $x^2-12x+27$.
  \\ \vfill\psovalbox{A}
$(x-3)(x-9)$ \hfill\psovalbox{B} $(x+3)(x+9)$ \hfill\psovalbox{C}
$(x-3)(x+9)$ \hfill\psovalbox{D} $(x-1)(x-27)$
\hfill\psovalbox{E}none of these
\item Factor: $x^3-12x^2+27x$.
  \\ \vfill\psovalbox{A}
$x(x-3)(x-9)$ \hfill\psovalbox{B} $x(x+3)(x+9)$ \hfill\psovalbox{C}
$x(x-3)(x+9)$ \hfill\psovalbox{D} $x(x-1)(x-27)$
\hfill\psovalbox{E}none of these

\item Factor: $a^2+16a+64$.
  \\ \vfill\psovalbox{A}
$(a+8)^2$ \hfill\psovalbox{B} $(a-8)(a+8)$ \hfill\psovalbox{C}
$(a-8)^2$ \hfill\psovalbox{D} $(a+16)(a+4)$ \hfill\psovalbox{E}none
of these
\item Factor: $a^3+16a^2+64a$.
  \\ \vfill\psovalbox{A}
$a(a+8)^2$ \hfill\psovalbox{B} $a(a-8)(a+8)$ \hfill\psovalbox{C}
$a(a-8)^2$ \hfill\psovalbox{D} $a(a+16)(a+4)$
\hfill\psovalbox{E}none of these
\item Factor: $a^2-64$.
  \\ \vfill\psovalbox{A}
$(a+8)^2$ \hfill\psovalbox{B} $(a-8)(a+8)$ \hfill\psovalbox{C}
$(a-8)^2$ \hfill\psovalbox{D} $(a+16)(a+4)$ \hfill\psovalbox{E}none
of these


\item Divide:  \qquad     $(6x^3+x^2-11x-6)\div (3x+2)$ \\
\vfill\psovalbox{A} $2x^2-x+3$ \hfill\psovalbox{B} $2x^2-x-3$
\hfill\psovalbox{C} $2x^2+x-3$ \hfill \psovalbox{D}
$2x^2+x+3$\hfill\psovalbox{E}none of these

\item Reduce the fraction:  \qquad     $\dfrac{x^3-x}{x^2-1}$ \\
\vfill\psovalbox{A} $x$ \hfill\psovalbox{B} $\dfrac{1}{x}$
\hfill\psovalbox{C} $x+1$ \hfill \psovalbox{D} $x-1$,
\hfill\psovalbox{E}none of these
\item Add:  \qquad     $\dfrac{2}{x-2}+\dfrac{3}{x+3}$ \\
\vfill\psovalbox{A} $\dfrac{5x}{x^2+x+6}$ \hfill\psovalbox{B}
$\dfrac{5x}{x^2+x-6}$ \hfill\psovalbox{C} $\dfrac{2x+1}{x^2+x-6}$
\hfill \psovalbox{D} $\dfrac{12-x}{x^2+x-6}$\hfill\psovalbox{E}none
of these
\item Reduce the fraction: $\dfrac{a^2-a}{a^2+a}$.
  \\ \vfill\psovalbox{A}
$\dfrac{a^2-1}{a^2+1}$ \hfill\psovalbox{B} $-1$ \hfill\psovalbox{C}
$\dfrac{a-1}{a+1}$ \hfill\psovalbox{D} $\dfrac{a}{a-1}$
\hfill\psovalbox{E}none of these
\item Reduce the fraction: $\dfrac{a^2}{a^2+a}$.
  \\ \vfill\psovalbox{A}
$\dfrac{a}{a+1}$ \hfill\psovalbox{B} $1$ \hfill\psovalbox{C}
$\dfrac{1}{a+1}$ \hfill\psovalbox{D} $\dfrac{a}{a-1}$
\hfill\psovalbox{E}none of these

\item Add the fractions: $\dfrac{2}{x-3}+\dfrac{3}{x+2}$.
  \\ \vfill\psovalbox{A}
$\dfrac{5x-5}{x^2-x-6}$ \hfill\psovalbox{B} $\dfrac{13+x}{x^2+x-6}$
\hfill\psovalbox{C} $\dfrac{13-x}{x^2-x-6}$ \hfill\psovalbox{D}
$\dfrac{5x+5}{x^2+x-6}$ \hfill\psovalbox{E}none of these

\item Perform the subtraction: $\dfrac{2}{x-3}-\dfrac{3}{x+2}$.
  \\ \vfill\psovalbox{A}
$\dfrac{5x-5}{x^2-x-6}$ \hfill\psovalbox{B} $\dfrac{13+x}{x^2+x-6}$
\hfill\psovalbox{C} $\dfrac{13-x}{x^2-x-6}$ \hfill\psovalbox{D}
$\dfrac{5x+5}{x^2+x-6}$ \hfill\psovalbox{E}none of these

\item[] {\bf Chapter VI Problems}

\item Solve the following equation for $x$: \quad $ax = 2$.\\
\psovalbox{A} $x=\dfrac{2}{a}$ \hfill \psovalbox{B} $x=2-a$ \hfill
\psovalbox{C} $x=\dfrac{a}{2}$  \hfill \psovalbox{D} $x=a$ \hfill
\psovalbox{E} none of these
\item Solve the following equation for $x$: \quad $\dfrac{ax}{b} = 2a$.\\
\psovalbox{A} $x=b$ \hfill \psovalbox{B} $x=\dfrac{2a}{b}$ \hfill
\psovalbox{C} $x=2b$  \hfill \psovalbox{D} $x=\dfrac{2}{b}$ \hfill
\psovalbox{E} none of these

\item Solve the following equation for $x$: \quad $ax+2b = 5b$.\\
\psovalbox{A} $x=7ab$ \hfill \psovalbox{B} $x=\dfrac{7b}{a}$ \hfill
\psovalbox{C} $x=3ab$  \hfill \psovalbox{D} $x=\dfrac{3b}{a}$ \hfill
\psovalbox{E} none of these

\item Solve the following equation for $x$: \quad $2x-1 = x$.\\
\psovalbox{A} $x=1$ \hfill \psovalbox{B} $x=-1$ \hfill \psovalbox{C}
$x=\dfrac{1}{2}$  \hfill \psovalbox{D} $x=\dfrac{1}{3}$ \hfill
\psovalbox{E} none of these

\item Solve the following equation for $x$: \quad $1-x = 2$.\\
\psovalbox{A} $x=3$ \hfill \psovalbox{B} $x=-1$ \hfill \psovalbox{C}
$x=-3$  \hfill \psovalbox{D} $x=\dfrac{1}{3}$ \hfill \psovalbox{E}
none of these









\item Solve the following equation for $x$: \quad $2x-1 = 1-3x$.\\
\psovalbox{A} $x=\dfrac{5}{2}$ \hfill \psovalbox{B} $x=\dfrac{2}{5}$
\hfill \psovalbox{C} $x=0$  \hfill \psovalbox{D} $x=-\dfrac{2}{5}$
\hfill \psovalbox{E} none of these

\item Solve the following equation for $x$: \quad $2(x-1) = 3(1-3x)$.\\
\psovalbox{A} $x=\dfrac{5}{11}$ \hfill \psovalbox{B}
$x=\dfrac{1}{7}$ \hfill \psovalbox{C} $x=\dfrac{5}{7}$  \hfill
\psovalbox{D} $x=-\dfrac{11}{5}$ \hfill \psovalbox{E} none of these

\item Solve the following equation for $x$: \quad $\dfrac{x-1}{3} = \dfrac{1-3x}{2}$.\\
\psovalbox{A} $x=\dfrac{5}{11}$ \hfill \psovalbox{B}
$x=\dfrac{1}{7}$ \hfill \psovalbox{C} $x=\dfrac{5}{7}$  \hfill
\psovalbox{D} $x=-\dfrac{11}{5}$ \hfill \psovalbox{E} none of these


\item Solve the following equation for $x$: \quad $ax-2b = 5b$.\\
\psovalbox{A} $x=7ab$ \hfill \psovalbox{B} $x=\dfrac{7b}{a}$ \hfill
\psovalbox{C} $x=3ab$  \hfill \psovalbox{D} $x=-\dfrac{3b}{a}$
\hfill \psovalbox{E} none of these

\item Solve the following equation for $x$: \quad $\dfrac{x}{2}-\dfrac{3}{4} =\dfrac{9}{8}$.\\
\psovalbox{A} $x=\dfrac{3}{4}$ \hfill \psovalbox{B} $x=\dfrac{1}{8}$
\hfill \psovalbox{C} $x=\dfrac{15}{8}$  \hfill \psovalbox{D}
$x=\dfrac{15}{4}$ \hfill \psovalbox{E} none of these


\item Adam, Eve, Snake, and Jake share $\$240$ between them. Adam  has twice as much as Eve, Eve has twice as much as Snake, and Snake has twice as much as
Jake. How much money does Eve have? \\  %B
\psovalbox{A} $\$15$ \hfill \psovalbox{B} $\$60$ \hfill
\psovalbox{C} $\$64$\hfill \psovalbox{D} $\$128$\hfill
  \psovalbox{E} none of these



\item Peter, Paul, and Mary share $\$153$ between them. Peter has twice as much as Paul, and Mary has thrice as much as Peter. How much money does Peter have? \\
\psovalbox{A} $\$17$ \hfill \psovalbox{B} $\$34$ \hfill
\psovalbox{C} $\$51$\hfill \psovalbox{D} $\$102$\hfill
  \psovalbox{E} none of these

\item The sum of three consecutive odd integers is $909$. Which one is the
last number? \\  \psovalbox{A} $301$\hfill \psovalbox{B} $302$\hfill
\psovalbox{C} $303$\hfill \psovalbox{D} $305$\hfill \psovalbox{E}
none of these
\item Jane's age is twice Bob's age increased by $3$.
Bill's age is Bob's age decreased by $4$. If the sum of their ages
is $27$, how old is Bill?
\\  \psovalbox{A} $3$\hfill \psovalbox{B}
$7$\hfill \psovalbox{C} $4$\hfill \psovalbox{D} $17$\hfill
\psovalbox{E} none of these

\item A jar of change contains nickels, dimes, and quarters. There are twice as many dimes as nickels, and twice as many quarters as dimes.
If the jar has a total of \$8.75, how many quarters are there?
 \\
\psovalbox{A} $5$ \hfill \psovalbox{B} $7$ \hfill \psovalbox{C}
$14$\hfill \psovalbox{D} $28$\hfill
  \psovalbox{E} none of these




\item Currently, the age of a father is four times
the age of his son, but in $24$ years from now it will only be
double. Find the son's current age. \\  \psovalbox{A} $12$\hfill
\psovalbox{B} $36$\hfill \psovalbox{C} $48$\hfill \psovalbox{D}
$6$\hfill \psovalbox{E} none of these



\item[] {\bf Chapter VII Problems}
\item Solve for $x$: \quad $2x^2-3x=0$. \\
 \psovalbox{A} $x=0$ or $x=\dfrac{3}{2}$ \hfill \psovalbox{B}
$x=0$ or $x=\dfrac{2}{3}$ \hfill \psovalbox{C} $x=0$ or
$x=-\dfrac{3}{2}$ \hfill \psovalbox{D} $x=0$ or $x=-\dfrac{2}{3}$
\hfill \psovalbox{E} none of these

\item Solve for $x$: \quad $x^2+3x+2=0$. \\
 \psovalbox{A} $x=-1$ or $x=2$ \hfill \psovalbox{B} $x=-1$ or
$x=-2$ \hfill \psovalbox{C} $x=1$ or $x=-2$ \hfill \psovalbox{D}
$x=1$ or $x=2$ \hfill \psovalbox{E} none of these

\item Solve for $x$: \quad $x^2-3x+2=0$. \\
 \psovalbox{A} $x=-1$ or $x=2$ \hfill \psovalbox{B} $x=-1$ or
$x=-2$ \hfill \psovalbox{C} $x=1$ or $x=-2$ \hfill \psovalbox{D}
$x=1$ or $x=2$ \hfill \psovalbox{E} none of these

\item Solve for $x$: \quad $x^2-x-2=0$. \\
 \psovalbox{A} $x=-1$ or $x=2$ \hfill \psovalbox{B} $x=-1$ or
$x=-2$ \hfill \psovalbox{C} $x=1$ or $x=-2$ \hfill \psovalbox{D}
$x=1$ or $x=2$ \hfill \psovalbox{E} none of these



\item[] {\bf Chapter VIII Problems}
\item If $2(x-1) \geq x+3$ then \\  %B
\psovalbox{A} $x \geq 1$ \hfill \psovalbox{B} $x\geq 5$ \hfill
\psovalbox{C} $x\leq 1$\hfill \psovalbox{D} $x \leq 5$\hfill
  \psovalbox{E} none of these


\item If $-2x > x+3$ then \\  %B
\psovalbox{A} $x > -1$ \hfill \psovalbox{B} $x > 1$ \hfill
\psovalbox{C} $x < -1$\hfill \psovalbox{D} $x < 1$\hfill
  \psovalbox{E} none of these



\item Which graph gives the correct solution to \hspace{1cm} $2-5x \leq -3$?\\


\psovalbox{A}\hspace{4cm} $\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red]{<-*}(-3,0)(-1,0)$ \\
\psovalbox{B} \hspace{4cm}  $\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red]{*->}(-1,0)(3,0)$ \\
\psovalbox{C} \hspace{4cm}   $\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red]{<-*}(-3,0)(1,0)$ \\
\hfill\psovalbox{D} \hspace{4cm} $\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red]{*->}(1,0)(3,0)$ \\
\psovalbox{E} \hspace{1cm} none of these

\item Which graph gives the correct solution to \hspace{1cm} $2+5x \leq -3$?\\

\psovalbox{A}\hspace{4cm} $\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red]{<-*}(-3,0)(-1,0)$\\
\psovalbox{B} \hspace{4cm}  $\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red]{*->}(-1,0)(3,0)$\\
\psovalbox{C} \hspace{4cm}   $\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red]{<-*}(-3,0)(1,0)$\\
\hfill\psovalbox{D} \hspace{4cm} $\psaxes[yAxis=false,
linewidth=.7pt]{<->}(0, 0)(-3,0)(3,
0)\psline[linewidth=2pt,linecolor=red]{*->}(1,0)(3,0)$\\
\psovalbox{E} \hspace{1cm} none of these

\item[] {\bf True or False Problems.}
\item \fbox{} $(2+3)^{10}=2^{10}+3^{10}$.
\item \fbox{} $(2\cdot 3)^{10}=2^{10}\cdot 3^{10}$.
\item \fbox{} $(2^5)^{2}=2^{7}$.
\item \fbox{} It is always true that $x(2x+3) = x(2x)(3)$.
\item \fbox{}  $(1^1+2^2+3^3+4^4-5^5)^0+4$ equals $5$.


\item[] {\bf In each of the items below, select the
choice, or choices, that is---or are---equivalent to the given
algebraic expression. Denominators are presumed not to vanish. }
\item $-a=$\\
\\ \vfill \psovalbox{A} $a(-1) $ \hfill \psovalbox{B} $-1\cdot a $
\hfill \psovalbox{C} $ \dfrac{a}{-1}$ \hfill \psovalbox{D}
$\dfrac{-a}{1} $ \hfill \psovalbox{E} $1-2a $
\item $xy-z=$\\
\\ \vfill \psovalbox{A} $z-xy $ \hfill \psovalbox{B} $yx-z $
\hfill \psovalbox{C} $xy+(-z) $ \hfill \psovalbox{D} $x(y-z) $
\hfill \psovalbox{E} $-z+xy $
\item $\dfrac{a-b}{2}=$\\
\\ \vfill \psovalbox{A} $a-b\div 2 $ \hfill \psovalbox{B} $(a-b)\div 2 $
\hfill \psovalbox{C} $ \dfrac{a}{2}-\dfrac{b}{2}$ \hfill
\psovalbox{D} $\dfrac{-b+a}{2} $ \hfill \psovalbox{E}
$\dfrac{1}{2}(a-b)$
\item $10x\left(\dfrac{y}{5} +x\right)=$
\\ \vfill \psovalbox{A} $\dfrac{xy}{2}+10x^2$ \hfill \psovalbox{B} $2xy+10x$
\hfill \psovalbox{C} $10\left(\dfrac{y}{5} +x\right)x$ \hfill
\psovalbox{D} $5x\cdot \left(x+\dfrac{y}{5}\right)\cdot 2$ \hfill
\psovalbox{E} $2xy+10x^2$
\item $(a-2b)^2=$\\
\\ \vfill \psovalbox{A} $(a-2b)(a-2b)$ \hfill \psovalbox{B} $a^2-4b^2$
\hfill \psovalbox{C} $a^2+4b^2$ \hfill \psovalbox{D} $a^2-4ab+4b^2$
\hfill \psovalbox{E} $-(2b-a)(a-2b)$
\item $\dfrac{x-y}{y-x}=$\\
\\ \vfill \psovalbox{A} $1$ \hfill \psovalbox{B}  $-1$
\hfill \psovalbox{C} $-2$ \hfill \psovalbox{D} $(x-y)^2$ \hfill
\psovalbox{E} $\dfrac{y-x}{x-y}$
\item  $(2x^2y^2)(3xy^4)=$\\
\\ \vfill \psovalbox{A} $6x^3y^6$ \hfill \psovalbox{B} $6(xy^4)^2$
\hfill \psovalbox{C} $6(xy^2)^3$ \hfill \psovalbox{D}
$\dfrac{12x^9y^{12}}{2x^3y^2}$ \hfill \psovalbox{E}
$\dfrac{12x^6y^7}{2x^3y}$
\item $x\left(x^3-1\right)+2\left(x^3-1\right)=$\\
\\ \vfill \psovalbox{A} $(x^3-1)x-2(-x^3+1)$ \vfill \psovalbox{B} $ 2(x^3-1)+x(-x^3+1)$
\vfill \psovalbox{C} $(x+2)(x^3-1)$ \vfill \psovalbox{D}
$x^4-x+2x^3-2$ \vfill \psovalbox{E} $x^4-x+1(x^3-2)$ \vfill


\end{enumerate}


\chapter{Answers and Hints}\addcontentsline{toc}{section}{Answers and Hints}\markright{Answers and Hints}
 {\small \input{algebrilla1}}

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\end{thebibliography}



\renewcommand{\chaptername}{Index}


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\end{document}
