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\usepackage{euscript, float, amsthm, xspace, epsfig, makeidx}
\usepackage{graphics,multicol,appendix,polynom,subfig,pstricks,pst-node,pst-tree}
\usepackage{fraction}
\usepackage{euler}
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\makeindex
\title{Fractions}
\author{Daniel Eric Smith}
\begin{document}
%\baselineskip=36pt
%\tracingparagraphs=1
\frontmatter
\maketitle
\tableofcontents
\listoffigures
\mainmatter
\chapter{Introduction}
\begin{quote}
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.\\
\emph{Count Lev Nikolaevich Tolstoy, (1828-1920)}
\end{quote}
Why do we study fractions?
There are two answers that immediately come to mind.
The first is that they naturally arise when solving equations.
Let's look at an example.
Four times what number gives you two?
Well we can set this up as the following equation
$$4x=2.$$
To solve this we divide both sides by $4$ and get
$$x=\frac{2}{4}.$$
Which reduces to
$$x=\frac{1}{2}.$$
You may say ``That is just a simple little example what does it have to do with anything?''.
That is the point,
if fractions can pop up doing simple little problems you know that sooner or later
they are going to pop up in bigger problems.
The second reason is that one of the mathematical objects you will study
is something called a rational expression.
A rational expression is a polynomial divided by a polynomial.
For example
$$\frac{x^2+5x+6}{x^2-x-6}$$
is a rational expression.
You can add, subtract, multiply and divide rational expressions.
The good news is that if you know how to factor polynomials and
how to add, subtract, multiply and divide fractions you have all the tools
needed to handle rational expressions.
For these two reasons and many others it is important,
even crucial,
to have a solid understanding of fractions.
That is what this book will attempt to do,
give you a solid understanding of fractions and how to use them in calculations.
And on this foundation you will better be able to understand algebra.
So on to the material.
How do we interpret the fraction $a/b$?
If $a**b$.
We do this by first writing $a/b$ as $a\cdot(1/b)$.
This says we have some pies that are divided into $b$ equal pieces and we have $a$ of those pieces.
\begin{ex}
Let us interpret $7/3$ in terms of pies.
First we rewrite $7/3$ as $7\cdot(1/3)$.
Now if we have some pies divided into 3 equal pieces we have 7 of those pieces,
Figure~\ref{frac-7-3}.
\begin{figure}[H]
\centering
\epsfig{file=seven-thirds.eps,height=1in}
\caption{Pie view of the fraction $7/3$.}
\label{frac-7-3}
\end{figure}
\end{ex}
\begin{ex}
Interpret each of the following in terms of pies.
\begin{enumerate}
\begin{multicols}{2}
\item $4/6$
\item $7/10$
\item $4/4$
\item $0/5$
\end{multicols}
\end{enumerate}
\begin{enumerate}
\item The fraction $4/6$ can be drawn as
\begin{figure}[H]
\centering
\epsfig{file=four-sixths.eps,height=1in}
\caption{Pie view of the fraction $4/6$.}
\label{frac-4-6}
\end{figure}
So we have a pie divided into $6$ equal pieces and we have $4$ of them.
\item The fraction $7/10$ can be drawn as
\begin{figure}[H]
\centering
\epsfig{file=seven-tenths.eps,height=1in}
\caption{Pie view of the fraction $7/10$.}
\label{frac-7-10}
\end{figure}
So we have a pie divided into $10$ equal pieces and we have $7$ of those pieces.
\item The fraction $4/4$ can be drawn as
\begin{figure}[H]
\centering
\epsfig{file=four-fourths.eps,height=1in}
\caption{Pie view of the fraction $4/4$.}
\label{frac-4-4}
\end{figure}
So we have a pie divided into $4$ equal pieces and we have all $4$ pieces.
That is we have a whole pie.
\item The fraction $0/5$ can be drawn as
\begin{figure}[H]
\centering
\epsfig{file=zero-fifths.eps,height=1in}
\caption{Pie view of the fraction $0/5$.}
\label{frac-0-5}
\end{figure}
So we have a pie divided into $5$ equal pieces and we have $0$ pieces of the pie.
In this case we do not have any pie at all.
\end{enumerate}
\end{ex}
\section*{Exercises}
\begin{enumerate}
\item Draw the pie representation of each of the following (do not reduce):
\begin{multicols}{6}
\begin{enumerate}
\item $1/2$
\item $1/3$
\item $3/4$
\item $2/4$
\item $1/5$
\item $4/6$
\item $3/7$
\item $3/2$
\item $6/4$
\item $7/5$
\item $11/7$
\item $13/6$
\item $10/3$
\item $15/5$
\item $0/8$
\end{enumerate}
\end{multicols}
\item Write each of the following as a fraction (do not reduce):
%\setlength\columnseprule{0.5pt}
\begin{multicols}{2}
\begin{enumerate}
\item \epsfig{file=frac-ques1.eps,height=0.8in}
\item \epsfig{file=frac-ques2.eps,height=0.8in}
\item \epsfig{file=frac-ques3.eps,height=0.8in}
\item \epsfig{file=frac-ques4.eps,height=0.8in}
\item \epsfig{file=frac-ques5.eps,height=0.8in}
\item \epsfig{file=frac-ques6.eps,height=0.8in}
\item \epsfig{file=frac-ques7.eps,height=0.8in}
\item \epsfig{file=frac-ques8.eps,height=1.4in}
\end{enumerate}
\end{multicols}
\end{enumerate}
\chapter{Integers}
\section{Basics}
\label{integers:basics}
In order to work with fractions it is first necessary to know some facts about integers.
So let us start with what an integer is.
An integer is one of the numbers in the set
$$\{\dots,-2,-1,0,1,2,\dots\}.$$
The are certain integers that are more important to us than the others,
they are called the prime numbers.
Let us define them.
\begin{defn}[Prime number]
\index{prime number}
A prime number is a positive integer that is only divisible by one and itself,
not including the number $1$.
\end{defn}
The first few prime numbers are $\{2,3,5,7,11,13,17,19,\dots\}$.
For more prime numbers see Appendix~\ref{primelist} .
One important property of the positive integers is that they can be factored
into the product of prime numbers in only one way (ignoring the order they are written in).
This is the fundamental theorem of arithmetic.
\begin{thm}[Fundamental Theorem of Arithmetic]
\index{fundamental theorem of arithmetic}
Every positive integer can be factored into a product of primes in only one way
(ignoring the order).
\end{thm}
The process of writing an integer as a product of primes is called finding the
prime factorization of that number.
Since we often need to find the prime factorization of an integer we will spend some time
discussing how to do it as well as looking at two methods of prime factorization.
\subsection{Prime Factorization}
We will start this section off with a definition.
\begin{defn}[Prime Factorization]
\index{prime factorization}
The prime factorization of an integer $m$ is the product of the prime numbers
that equals $m$.
%the factorization of $m$ into its prime constituents.
\end{defn}
%In other words we will rewrite the number as a product of primes.
%Obviously the product of primes must equal the integer we are factoring.
There are two methods for finding the prime factorization of an integer:
\emph{successive division} and the \emph{tree method}.
\begin{description}
\item[Successive division:]
\index{prime factorization!successive division}
In this method we successively divide our number by prime numbers,
usually in order starting with $2$, then $3$, next $5$, etc.
So if we are given the number $m$ we would start by trying to divide $m$ by $2$.
If that works we would take the quotient and try to divide it by $2$.
If it is not divisible by $2$ we would try to divide by $3$.
We continue this process until the final quotient is a prime number.
At each step we would write down the prime number that divided
the number we are working on.
This is much clearer if we looked at some examples.
\begin{ex}
Find the prime factorization of $6$.
First we check to see if $6$ is divisible by $2$.
$$\frac{6}{2}=3$$
Since $6$ is divisible by $2$ we have $6=2 \cdot 3$.
Because $3$ is prime we are done and
we have that the prime factorization of $6$ is $2 \cdot 3$.
\end{ex}
\begin{ex}
Find the prime factorization of $12$.
First check to see if $12$ is divisible by $2$.
$$\frac{12}{2}=6$$
Since $12$ is divisible by $2$ we have $12=2 \cdot 6$.
This time the quotient, $6$, is not prime.
So we need to continue the process.
Check if $6$ is divisible by $2$.
Since it is we have $12=2 \cdot 2 \cdot 3$.
Because $3$ is a prime number we are done and the prime factorization of $12$ is $2^2 \cdot 3$.
This process can be written as\\
\begin{tabular}{rccccc}
$12=$ & $2$ & $\cdot$ & $6$ & &\\
$=$ & $2$ & $\cdot$ & $2$ & $\cdot$ & $3$\\
\end{tabular}\\
Therefore the prime factorization of $12$ is $2^2 \cdot 3$.
\end{ex}
\begin{ex}
Find the prime factorization of $350$.
First check to see if $350$ is divisible by $2$.
$$\frac{350}{2}=175$$
So\\
\begin{tabular}{rccc}
$350=$ & $2$ & $\cdot$ & $175$
\end{tabular}\\
Next we check to see if $175$ is divisible by $2$,
since it is not a prime number.
Clearly it is not divisible by $2$.
So we try $3$ and see that $175$ is not divisible by $3$ either.
Continuing on to the next prime number we try $5$.
$$\frac{175}{5}=35$$
So now we have\\
\begin{tabular}{rccccc}
$350=$ & $2$ & $\cdot$ & $175$ & & \\
$=$ & $2$ & $\cdot$ & $5$ & $\cdot$ & $35$
\end{tabular}\\
We see that $35$ is divisible by $5$ again and get\\
\begin{tabular}{rccccccc}
$350=$ & $2$ & $\cdot$ & $175$ & & & & \\
$=$ & $2$ & $\cdot$ & $5$ & $\cdot$ & $35$ & & \\
$=$ & $2$ & $\cdot$ & $5$ & $\cdot$ & $5$ & $\cdot$ & $7$ \\
\end{tabular}\\
Because $7$ is prime we are done and the prime factorization of $135$ is $2 \cdot 5^2 \cdot 7$.
\end{ex}
\item[Tree method:]
\index{prime factorization!tree method}
This method is the same as the successive division method
except we write each step as a tree instead of as a product.
Before we look at this method we need a little terminology first.
We call the number at the ``top'' of the tree the \textbf{root}\index{tree!root} and
any number at the ``bottom'' of the tree a \textbf{leaf}\index{tree!leaf}.
Naturally we call all of the numbers at the bottom of the tree the
\textbf{leaves}\index{tree!leaves}.
\begin{ex}
Find the prime factorization of $6$.
Since $6$ divided by $2$ is $3$ we can write the tree as:
\begin{center}
\pstree[nodesep=2pt,levelsep=20pt]
{
\TR{$6$}
}
{
\TR{$2$}
\TR{$3$}
}
\end{center}
Because each leaf is a prime number we are done.
Hence the prime factorization is the product of all the leaves:
$6=2 \cdot 3$
\end{ex}
\begin{ex}
Find the prime factorization of $12$.
First we divide $12$ by $2$.
This gives us $6$.
So our tree looks like the following:
\begin{center}
\pstree[nodesep=2pt,levelsep=20pt]
{
\TR{$12$}
}
{
\TR{$2$}
\TR{$6$}
}
\end{center}
Since $6$ is not a prime number we must continue the process.
We know that $6$ factors into $2$ and $3$.
Adding this to our tree we have:
\begin{center}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{$12$}}
{
\TR{$2$}
\pstree{\TR{$6$}}
{
{\TR{$2$}}
{\TR{$3$}}
}
}
\end{center}
Now every leaf is a prime number and we are done.
So the prime factorization of $12$ is $2^2 \cdot 3$, the product of all the leaves.
\end{ex}
\begin{ex}
Find the prime factorization of $350$.
Since $2$ divides into $350$ we can write it as $350=2 \cdot 175$.
This gives us the tree:
\begin{center}
\pstree[nodesep=2pt,levelsep=20pt]
{
\TR{$350$}
}
{
\TR{$2$}
\TR{$175$}
}
\end{center}
Now we find that $175=5 \cdot 35$ and $35= 5 \cdot 7$.
So we have the trees:
\begin{center}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{$350$}}
{
\TR{$2$}
\pstree{\TR{$175$}}
{
{\TR{$5$}}
{\TR{$35$}}
}
}
\hspace{.5in}
and then
\hspace{.5in}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{$350$}}
{
\TR{$2$}
\pstree{\TR{$175$}}
{
{\TR{$5$}}
\pstree{\TR{$35$}}
{
{\TR{$5$}}
{\TR{$7$}}
}
}
}
\end{center}
In the last tree all the leaves are prime numbers so we are finished factoring.
The prime factorization of $350$ is $2 \cdot 5^2 \cdot 7$.
Note that you could have factored $350$ in one of the following ways:
\begin{center}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{$350$}}
{
{\TR{7}}
\pstree{{\TR{50}}}
{
{\TR{2}}
\pstree{{\TR{25}}}
{
{\TR{5}}
{\TR{5}}
}
}
}
\hspace{.75in} or \hspace{.75in}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{$350$}}
{
\pstree{\TR{35}}
{
{\TR{5}}
{\TR{7}}
}
\pstree{\TR{10}}
{
{\TR{2}}
{\TR{5}}
}
}
\end{center}
It does not matter in which order you break down the number
as long as each leaf is a prime number the prime factorization
will be the same: $2 \cdot 5^2 \cdot 7$.
\end{ex}
\end{description}
So now that we know how to find the prime factorization of a number
let us see some more examples.
\begin{ex}\hfill
\begin{enumerate}
\item The integer $6$ can be factored as $6=2\cdot3$.
\item We can factor $12$ as $12=2^2\cdot3$.
\item We factor $25$ as $25=5^2$.
\item Factoring $30$ gives us $30=2\cdot3\cdot5$.
\item Since $41$ is a prime number its factorization is just $41$.
Note that if we tried to factor $41$ using the tree method all we would have
is the root of the tree,
$$41$$
\item Finally, factoring $99$ gives $99=3^2\cdot11$.
\end{enumerate}
\end{ex}
To factor a negative integer, for example $-4$, we only need to add one step.
First we factor out the negative one, $(-1)\cdot 4$.
Then we factor the positive integer, $(-1)\cdot 2^2$.
\subsection{Least Common Multiple}
When working with fractions we often have to find a common multiple of a set of numbers.
So let us define common multiple and see some examples of how to find it.
\begin{defn}[Common Multiple]
\index{common multiple}
A common multiple of the two positive integers
$a$ and $b$ is a positive integer, call it $c$,
that is divisible by both $a$ and $b$,
i.e. both $c \div a$ and $c \div b$ are integers.
\end{defn}
\begin{ex}
Examples of common multiples.
\begin{enumerate}
\item Some common multiples of $2$ and $3$ are $6,12,18,\dots$.
\item Some common multiples of $4$ and $6$ are $12,24,36,\dots$.
\item We can find common multiples of more than two numbers at a time.
Some common multiples of the set of numbers $\{2,3,5\}$ are $30,60,90,\dots$.
\end{enumerate}
\end{ex}
Notice that for $\{2,3\}$ there is no common multiple smaller than 6 and
for $\{4,6\}$ the smallest common multiple is 12.
This leads us to another definition.
\begin{defn}[Least Common Multiple]
\index{least common multiple}
The least common multiple of a set of positive integers is the
smallest positive integer that is divisible by all the numbers in that set,
i.e. it is what it says it is: the smallest common multiple.
\end{defn}
\begin{ex}
Some examples of least common multiples.
\begin{enumerate}
\item The least common multiple, \lcm, of $2$ and $3$ is $6$,
i.e. $\lcm\{2,3\}=6$
\item The \lcm\ of $4$ and $6$ is $\lcm\{4,6\}=12$.
\item $\lcm\{2,3,5\}=30$.
\end{enumerate}
\end{ex}
So, how do we find the LCM of a set of numbers?
\begin{algor}[Least Common Multiple]\hfill
\index{least common multiple!algorithm}
\begin{description}
\hfill\parbox{4.5in}{
\item[Step 1] Find the prime factorization of each of our numbers.
\item[Step 2] List each prime number that occurs in any of the factorizations.
\item[Step 3] For each prime number in our list,
find the largest power it is raised to in any of the factorizations.
Then write each prime to that power in a list.
\item[Step 4] Multiply the results in the previous step.}
\end{description}
\end{algor}
\begin{ex} Let us find the least common multiple of $4$ and $6$, $\lcm\{4,6\}$.
\begin{description}\hfill\parbox{4.5in}{
\item[Step 1] Find the prime factorizations: $4=2^2$ and $6=2\cdot3$.
\item[Step 2] List the primes: $2,3$.
\item[Step 3] Raise the primes to their highest power: $2^2,3$.
\item[Step 4] Multiply the numbers found in the last step: $2^2\cdot3=12=\lcm\{4,6\}$.}
\end{description}
So the least common multiple of $4$ and $6$ is $12$.
\end{ex}
\begin{ex} Let us find the $\lcm\{90,300\}$.
\begin{description}\hfill\parbox{4.5in}{
\item[Step 1] $90=2^2\cdot3\cdot5^2$ and $300=2\cdot3^2\cdot5$.
\item[Step 2] $2,3,5$.
\item[Step 3] $2^2,3^2,5^2$
\item[Step 4] $2^2\cdot3^2\cdot5^2=900=\lcm\{90,300\}$}
\end{description}
So the least common multiple of $90$ and $300$ is $900$.
\end{ex}
\section{Long Division}
\label{integers:longdivision}
When we add, subtract, or multiply two integers we get an integer for an answer.
If we divide two integers we may not get an integer for an answer.
For example dividing $6$ by $2$ gives us $6 \div 2=3$,
but $5 \div 2$ is not an integer.
The reason why is that dividing the integer $a$ by the integer $b$,
$a \div b$,
is asking how many groups of $b$ items can $a$ be made into.
So for $6 \div 2$, we can divide $6$ into $3$ groups of $2$.
If we try this with $5 \div 2$,
there will be $2$ groups of $2$ and $1$ item left over.
All is not lost though.
We can still divide $5$ by $2$ if we are willing to accept that sometimes we will have
left over items (\emph{remainders}).
The method we will use to divide two integers is called \emph{long division}.
Let us look at two examples, Figure~\ref{div-6-2-5-2}.
\begin{figure}[H]
\centering
\mbox{\longdiv{6}{2}} \hfil
\mbox{\longdiv{5}{2}}
\caption{Long division: $6 \div 2$ and $5 \div 2$.}
\label{div-6-2-5-2}
\end{figure}
Let us look at the division $6 \div 2$.
The largest number that we can multiply $2$ times and
remain less than or equal to $6$ is $3$.
So this is put above the $6$.
Since $3$ times $2$ is $6$ it is put under the $6$.
Subtracting $6$ from $6$ give $0$.
So $6$ divided by $2$ is $3$ with a remainder of $0$.
Doing the same thing for $5 \div 2$ gives us the following.
The largest number that can be multiplied times $2$ and
remain less than or equal to $5$ is $2$.
We place the $2$ above the $5$.
Now $2 \cdot 2 = 4$, so place the $4$ below the $5$.
The difference of $4$ from $5$ is $1$,
so place a $1$ under the $4$.
Since $2$ is greater than $1$ we are done.
Therefore $5$ divided by $2$ is $2$ with a remainder of $1$.
Let us look at another example $257 \div 3$,
see Figure~\ref{div-257-3}.
\begin{figure}[H]
\centering
\mbox{\longdiv{257}{3}}
\caption{Long division: $257 \div 3$.}
\label{div-257-3}
\end{figure}
Our first step is to find the largest integer that when multiplied times $3$ is less than $2$.
There is no such number,
so we find the largest integer that when multiplied times $3$ is less than $25$.
That number is $8$.
Now we multiply $8$ and $3$ to get $24$.
Subtract $24$ from $25$ to get $1$.
Next take the digits after the $25$ in the dividend,
$7$ in this example,
and write them down behind the $1$.
This gives us $17$.
Now look for the largest integer that when multiplied times $3$ is less than $17$,
since $3$ is larger than $1$.
This is $5$.
Now $3 \cdot 5 =15$ and $17-15=2$.
Since there are no other numbers to drop down and $2$ is less than $3$ we are done.
So $257 \div 3$ is equal to $85$ with a remainder of $2$.
%Let us look at a algorithm for long division.
%First we need to look at some notation.
%Suppose we let $a$ be the number $987$.
%We will let $a_i$ be the $i$th digit from the left.
%So in our example $a=987$: $a_1=9$, $a_2=8$, $a_3=7$.
%\begin{algor}[Integer Long Division]
% \index{integer long division}
% Let $a$ and $b$ be integers where $a$ can be expressed by $a=a_1a_2\ldots a_n$.
% To divide $a$ by $b$, where $bb.$$
We know how to determine if two fractions are equal,
but how can we determine which of the two fractions $a/b$ and $c/d$ is larger?
If they have a common denominator it is easy.
\begin{defn}[Inequality of Fractions, common denominator]
\index{fraction!inequality, common denominator}
The fraction $a/b$ is said to be less than $c/b$ if $a 5$,
we have $113/8 > 5/8$.
\end{enumerate}
\end{ex}
If the denominators are different,
then we have to find a common denominator.
So the obvious question is how do we find the common denominator of two or more fractions?
It turns of to be easy to do this if we know how to find the least common multiple of a set of numbers.
\begin{defn}[Least Common Denominator]
\index{least common denominator}
The least common denominator (\lcd) of a set of fractions is the least common multiple of their denominators.
\end{defn}
Let us look at an example.
\begin{ex}
Find the least common denominator (LCD) of\/ $\left\{2/3,1/5\right\}$.\\
\\
The set of denominators for our fractions is $\{3,5\}$.
We find that $\lcm\{3,5\}=3 \cdot 5 =15$.
So the $\lcdd{2/3}{1/5}=15$.
\end{ex}
How do we give two fractions a common denominator?
Lets look at the fractions $a/b$ and $c/d$.
Let us assume that the $\lcm\{b,d\}=bd$.
Now to give $a/b$ and $c/d$ common denominators we take the denominator of the first fraction,
$b$, and divide the least common denominator of the two fractions.
So, $(bd)/b=d$.
Next we multiply the numerator and denominator of the first fraction by $d$,
i.e. we get
$$\frac{d}{d}\cdot\frac{a}{b}=\frac{da}{db}.$$
Doing the same thing for the second fraction we get:\\
$$\frac{bd}{d}=b$$
$$\frac{b}{b}\cdot\frac{c}{d}=\frac{bc}{bd}.$$
Now we have two fractions with a common denominator
$$\frac{ad}{bd}, \quad \frac{cb}{bd}.$$
This is easier to understand using a concrete example.
\begin{ex}
Give the fractions $2/3$ and $1/5$ a common denominator.
The least common denominator of the fractions is $\lcd\left\{2/3,1/5\right\}=15$.
First we divide the denominator of $2/3$ into $15$,
so $15\div3=5$.
Then we multiply the numerator and denominator of $2/3$ by $5$,
$$\frac{5}{5}\cdot\frac{2}{3}=\frac{10}{15}.$$
Doing the same thing for the second fraction we have:
$$\frac{15}{5}=3$$
and
$$\frac{3}{3}\cdot\frac{1}{5}=\frac{3}{15}.$$
Therefore our new fractions are
$$\frac{10}{15} \quad \mbox{and} \quad \frac{3}{15}.$$
\end{ex}
\begin{ex}
Let us compare the fractions $2/3$ and $1/5$.
Since
$$\frac{2}{3}=\frac{10}{15} \quad \mbox{and} \quad \frac{1}{5}=\frac{3}{15}$$
we have that $10/15 > 3/15$,
hence $2/3 > 1/5$.
\end{ex}
Now we can compare fractions by getting common denominators and comparing numerators.
\begin{ex}
Which is smaller $5/7$ or $2/3$?
First we find their least common denominator $\lcdd{2/3}{5/7}=21$.
Give each fraction the common denominator $21$,
$$\frac{3}{3}\cdot\frac{5}{7}=\frac{15}{21}$$
and
$$\frac{7}{7}\cdot\frac{2}{3}=\frac{14}{21}.$$
Since $14<15$ we have $14/21 < 15/21$ and therefore $2/3 < 5/7$.
\end{ex}
We also know that the fraction $1/2$ is larger than $1/3$.
This is because the denominator of $1/2$ is smaller than the denominator of $1/3$.
A smaller denominator means the pie has been cut fewer times and therefore each piece is larger.
So another way to compare fractions is to get a common numerator and then compare denominators.
The fraction with the smaller denominator is the larger of the fractions.
\begin{defn}[Inequality of Fractions, common numerator]
\index{fraction!inequality, common numerator}
The fraction $a/b$ is said to be less than $a/c$ if $c****2/7$ and therefore
$$\frac{1}{3}>\frac{2}{7}.$$
\end{ex}
\begin{ex}
Let us look at some more examples of comparing fractions by getting a common numerator.
\begin{enumerate}
\item Compare the fractions $1/2$ and $1/3$.
Since the numerators are already common all we need to do is compare the denominators.
Since $2 < 3$ then $1/3 < 1/2$.
\item Compare the fractions $3/7$ and $2/5$.
The least common numerator is $6$.
Giving each fraction a common numerator we get,
$$\frac{6}{14} \quad \mbox{and} \quad \frac{6}{15}.$$
Since $14 < 15$ we have
$$\frac{6}{15} < \frac{6}{14}$$
so $2/5 < 3/7$.
\end{enumerate}
\end{ex}
\begin{ex}
Some more examples of comparing fractions.
\begin{enumerate}
\item Compare $5/6$ and $7/12$.
For this example it will be easiest to get a common denominator,
which is $12$.
So we have
$$\frac{2}{2}\cdot \frac{5}{6}=\frac{10}{12}.$$
Since $7 < 10$ we have
$$\frac{10}{12}>\frac{7}{12} \quad \mbox{and therefore} \quad \frac{5}{6}>\frac{7}{12}.$$
\item Compare $3/5$ and $5/7$.
We can use either common denominators or common numerators with equal ease to
compare these fractions.
So let us do both methods.
The least common denominator is $35$.
So giving each fraction a common denominator we have
$$
\frac{7}{7} \cdot \frac{3}{5}=\frac{21}{35}
\quad \mbox{and} \quad
\frac{5}{5} \cdot \frac{5}{7}=\frac{25}{35}.
$$
Since $21/35<25/35$,
because $21<25$,
we have that $3/5<5/7$.
Using common numerators,
which is $15$,
we have
$$\frac{5}{5}\cdot\frac{3}{5}=\frac{15}{25}
\quad \mbox{and} \quad
\frac{3}{3}\cdot\frac{5}{7}=\frac{15}{21}.$$
Recall to do the comparison we find the fraction with the largest denominator is the
smaller fraction.
So
$$\frac{15}{25}<\frac{15}{21}.$$
Therefore $3/5<5/7$.
\item Compare $1/30$ and $3/70$.
In this case it will be easiest to find a common numerator to compare the fractions.
The common numerator is $3$.
Since $3/70$ already has the common numerator we just need to find one for $1/30$.
$$\frac{3}{3} \cdot \frac{1}{30}=\frac{3}{90}.$$
Since $90 > 70$ we have $3/90<3/70$.
Therefore $1/30<3/70$.
\item Compare $16/24$ and $2/5$.
Before we try to compare these fractions we notice that $16/24$ can be reduced.
This will make the computations easier,
so
$$\frac{16}{24}=\frac{8\cdot 2}{8\cdot 3}=\frac{2}{3}.$$
So now we are comparing $2/3$ and $2/5$.
We note that they already have a common numerator.
Using this we have $5>3$ which means $2/5<2/3$.
Therefore $2/5<16/24$.
\end{enumerate}
\end{ex}
\section*{Exercises}
\begin{enumerate}
\item Determine which of the following sets of fractions are equal.
\begin{multicols}{3}
\begin{enumerate}
\item $\left\{\dfrac{2}{3}, \dfrac{4}{6} \right\}$
\item $\left\{\dfrac{10}{14}, \dfrac{25}{35} \right\}$
\item $\left\{\dfrac{3}{8}, \dfrac{11}{32} \right\}$
\item $\left\{\dfrac{21}{36}, \dfrac{28}{48} \right\}$
\item $\left\{\dfrac{10}{14}, \dfrac{40}{49} \right\}$
\item $\left\{\dfrac{41}{101},\dfrac{123}{303} \right\}$
\item $\left\{\dfrac{-6}{8}, \dfrac{15}{-20} \right\}$
\item $\left\{\dfrac{33}{-15},\dfrac{-66}{30} \right\}$
\item $\left\{\dfrac{-20}{35},\dfrac{-32}{-56} \right\}$
\end{enumerate}
\end{multicols}
\item Determine which of the following fractions is greater than the other.
\begin{multicols}{3}
\begin{enumerate}
\item $\left\{\dfrac{2}{3}, \dfrac{3}{4} \right\}$
\item $\left\{\dfrac{5}{7}, \dfrac{8}{15} \right\}$
\item $\left\{\dfrac{1}{8}, \dfrac{3}{23} \right\}$
\item $\left\{\dfrac{11}{17}, \dfrac{23}{50} \right\}$
\item $\left\{\dfrac{5}{3}, \dfrac{20}{13} \right\}$
\item $\left\{\dfrac{16}{7}, \dfrac{8}{3} \right\}$
\item $\left\{\dfrac{15}{4}, \dfrac{41}{12} \right\}$
\item $\left\{\dfrac{7}{19}, \dfrac{17}{5} \right\}$
\end{enumerate}
\end{multicols}
\end{enumerate}
\newpage
\section{Addition and Subtraction}
\label{fractions:addition}
The next operation we will look at is the addition of fractions.
Addition of fractions requires a little more work to perform than multiplication,
but it is not hard.
The first thing we will do is to understand addition by looking at pies.
Let us start by looking at the sum of $2/5$ and $1/5$, i.e.
$$\frac{2}{5}+\frac{1}{5}.$$
Viewing these as pies we have the following, Figure~\ref{frac-2-5+1-5}.
\begin{figure}[H]
\centering
\epsfig{file=twofifths-plus-onefifth.eps,height=1in}
\caption{Pies $\frac{2}{5}$ plus $\frac{1}{5}$}
\label{frac-2-5+1-5}
\end{figure}
So it is easy to see that $2/5+1/5=3/5$.
What do we do if the denominators are different?
Suppose we have the fractions $1/2$ and $1/3$,
Figure \ref{frac-1-2-1-3}.
\begin{figure}[H]
\centering
\epsfig{file=half-third.eps,height=1in}
\caption{Pies $\frac{1}{2}$ and $\frac{1}{3}$}
\label{frac-1-2-1-3}
\end{figure}
If we add these pieces of pie together how much pie do we have?
See Figure \ref{frac-1-2+1-3a}.
\begin{figure}[H]
\centering
\epsfig{file=halfthird-pacman.eps,height=1in}
\caption{Pie $\frac{1}{2}+\frac{1}{3}$}
\label{frac-1-2+1-3a}
\end{figure}
We cannot talk about the sum in terms of halves since one piece is a third and
we cannot talk about it in terms of thirds since one piece is a half.
What we need to do is to be able to compare the two pieces of pie.
We recall that this can be done by finding a common denominator of the fractions.
We find the least common denominator of the fractions $1/2$ and $1/3$ is $6$.
So let us divide the pies into sixths, Figure \ref{frac-1-2-1-3-6}.
\begin{figure}[H]
\centering
\epsfig{file=half-third-sixths.eps,height=1in}
\caption{Pies $\frac{1}{2}$ and $\frac{1}{3}$ divided into sixths.}
\label{frac-1-2-1-3-6}
\end{figure}
After adding them we have:
\begin{figure}[H]
\centering
\epsfig{file=half-third-sum.eps,height=1in}
\caption{The sum of $\frac{1}{2}$ and $\frac{1}{3}$.}
\label{frac-1-2+1-3b.eps}
\end{figure}
It is now easy to see that if we have half of a pie and one third of a pie we have
five sixths of a pie.
So
$$\frac{1}{2}+\frac{1}{3}=\frac{5}{6}.$$
To add fractions we do algebraically what we just did graphically.
\begin{defn}[Addition of Fractions]
\index{fraction!addition}
Let $a/b$ and $c/b$ be fractions.
Then the sum of $a/b$ and $c/b$ is
$$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}.$$
\end{defn}
Notice that in the definition of addition the denominators are the same.
This is because, as we saw with pies, we can only make sense of the sum of pieces of pie
if the pies are divided into the same number of pieces.
\begin{ex}
\begin{enumerate}
\item Find the sum of $2/5$ and $1/5$.
Since the denominators are equal we can just add the numerators and leave the
denominators alone.
So
$$\frac{2}{5}+\frac{1}{5}=\frac{2+1}{5}=\frac{3}{5}.$$
\item Find the sum of $3/11$ and $5/11$.
Again the denominators are equal so we just add the numerators
$$\frac{3}{11}+\frac{5}{11}=\frac{3+5}{11}=\frac{8}{11}.$$
\item Find the sum of $3/7$ and $11/7$.
The denominators are equal so we add the numerators
$$\frac{3}{7}+\frac{11}{7}=\frac{3+11}{7}=\frac{14}{7}.$$
Now we notice that we can simplify this fraction using Theorem \ref{frac-simp} so,
$$\frac{14}{7}=\frac{2\cdot7}{1\cdot7}=2.$$
\end{enumerate}
\end{ex}
What do we do if the denominators are different?
In the last section we learned how to give two fractions a common denominator,
so we first apply that procedure and then add the new fractions.
\begin{ex}
\begin{enumerate}
\item Find the sum of $1/2$ and $1/3$.
Our first step is to find the least common denominator:
$$\lcdd{\frac{1}{2}}{\frac{1}{3}}=6.$$
Next we divide the least common denominator by $2$ and
multiply the numerator and denominator of $1/2$ by this
(we divided by $2$ since it is the denominator of this fraction).
$$\frac{6}{2}=3, \quad \mbox{then}\quad \frac{3}{3}\cdot\frac{1}{2}=\frac{3}{6}.$$
Now do the same thing with $1/3$:
$$\frac{6}{3}=2, \quad \mbox{then}\quad \frac{2}{2}\cdot\frac{1}{3}=\frac{2}{6}.$$
Finally we can add the fractions:
$$\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$$
\item Find the sum of $5/6$ and $3/4$.
What we will do is shown graphically in Figure \ref{frac-5-6+3-4}.
\begin{figure}[H]
\centering
\epsfig{file=fivesixths-threefourths-sum.eps,height=2in}
\caption{The sum of $\frac{5}{6}$ and $\frac{3}{4}$.}
\label{frac-5-6+3-4}
\end{figure}
The least common denominator of $5/6$ and $3/4$ is $\lcdd{5/6}{3/4}=12$.
Giving the fractions common denominators we have
$$\frac{12}{6}=2 \quad \mbox{then} \quad \frac{2}{2}\cdot\frac{5}{6}=\frac{10}{12}$$
$$\frac{12}{4}=3 \quad \mbox{then} \quad \frac{3}{3}\cdot\frac{3}{4}=\frac{9}{12}$$
Now we can add these fractions
$$\frac{10}{12}+\frac{9}{12}=\frac{19}{12}.$$
Since this fraction cannot be reduced we are done.
\end{enumerate}
\end{ex}
Now that we know how to add fractions, subtraction is easy.
\begin{defn}[Subtraction of Fractions]
\index{fraction!subtraction}
Let $a/b$ and $c/b$ be fractions.
Then the difference of $c/b$ from $a/b$ is
$$\frac{a}{b}-\frac{c}{b}=\frac{a-c}{b}.$$
\end{defn}
So to subtract fractions we just subtract the numerators if the denominators are the same.
If the denominators are different we follow the same steps we did when adding fractions
by first finding a common denominator,
then subtracting the fractions.
\begin{ex}
\begin{enumerate}
\item Find the difference of $1/5$ from $3/5$.
Since they have a common denominator we can just subtract the numerators.
$$\frac{3}{5}-\frac{1}{5}=\frac{3-1}{5}=\frac{2}{5}$$
Since we can not reduce this fraction we are done.
\item Find the difference of $1/3$ from $1/2$.
First we need to find a common denominator,
$\lcdd{1/2}{1/3}=6$.
Next we rewrite each fraction in terms of the least common denominator:
$$\frac{6}{2}=3 \Rightarrow \frac{3}{3}\cdot\frac{1}{2}=\frac{3}{6}$$
$$\frac{6}{3}=2 \Rightarrow \frac{2}{2}\cdot\frac{1}{3}=\frac{2}{6}.$$
Finally we subtract the fractions:
$$\frac{3}{6}-\frac{2}{6}=\frac{1}{6}.$$
\end{enumerate}
\end{ex}
Sometimes it is easier to reduce the fractions before we work with them.
Let look at some examples where we reduce the fractions first,
then we add or subtract them.
\begin{ex}
\begin{enumerate}
\item Let's add $3/12$ and $1/5$.
If we don't reduce first,
then the least common denominator will be $\lcd=5\cdot12=60$.
But if we reduce first then we are adding
$$\frac{3}{12}+\frac{1}{5}=\frac{1}{4}+\frac{1}{5}$$
and the least common denominator is $\lcd=4\cdot5=20$.
This gives us
$$\frac{1}{4}+\frac{1}{5}=\frac{5}{20}+\frac{4}{20}=\frac{9}{20}.$$
\item Subtract $5/40$ from $7/21$.
First we reduce the fractions.
$$\frac{5}{40}=\frac{1}{8} \quad \mbox{and} \quad \frac{7}{21}=\frac{1}{3}$$
The least common denominator of these fractions is $\lcd=8\cdot3=24$.
Subtracting the fractions gives us
$$\frac{7}{21}-\frac{5}{40}=\frac{1}{3}-\frac{1}{8}=\frac{8}{24}-\frac{3}{24}=\frac{5}{24}.$$
\end{enumerate}
\end{ex}
You should try the above examples without first simplifying the fractions.
This will convince you that first simplifying the fractions can be a good idea.
The second example has an $\lcd=840$.
One thing to watch out for is shown in the following example.
\begin{ex}
Let us add the fractions $1/7$ and $7/21$.
If we simplify $7/21$ this gives us $1/3$.
But the least common denominator of $1/7$ and $1/3$ is $21$.
So in this case it would be better not to simplify.
Just find the least common denominator and add.
$$\frac{1}{7}+\frac{7}{21}=\frac{3}{21}+\frac{7}{21}=\frac{10}{21}$$
If you simplify first it will result in more work,
but you will still get the correct answer.
\end{ex}
Recall that $a+b=b+a$ and that $(a+b)+c=a+(b+c)$.
So that $2+3=3+2=6$ and $(2+3)+5=2+(3+5)=10$.
The addition of fractions also follow those two rules.
This means we can do the following:
$$\frac{1}{2}+\frac{1}{3}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$$
and
$$
\left(\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{5}
=\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{5}\right)
=\frac{31}{30}
.$$
This allows us to write the sum of $1/2$, $1/3$, and $1/5$ as
$$\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30}.$$
The following \emph{will not work}: $a-b=b-a$.
But this is easy to see this since $2-3=-1$ and $3-2=1$.
Now guess what, flipping the numbers when subtracting fractions wont work either.
The best you can do is the following:
$$a-b=a+(-b)$$
and this can be rewritten as
$$a+(-b)=-b+a.$$
So we can do the following
$$3-2=3+(-2)=-2+3=1.$$
The following point about notation is \emph{extremely important}.
You can never write the following:
$$3+-2$$
\textbf{THIS NEVER MAKES SENSE!}
You must put the negative number in parenthesis if it follows either
an addition or subtraction sign.
So it is illegal to write $3+-2$ and $3--2$.
You must write them as $3+(-2)$ and $3-(-2)$ respectively.
Writing either $3+-2$ or $3--2$ would be like writing the following sentence:
\begin{center}
Thedogchasedthecatintothemousehole.
\end{center}
\section*{Exercises}
\begin{enumerate}
\item Perform the following operations:
\begin{multicols}{3}
\begin{enumerate}
\item $\dfrac{1}{2}+ \dfrac{3}{2}$
\item $\dfrac{2}{5}+ \dfrac{4}{5}$
\item $\dfrac{2}{11}+\dfrac{7}{11}$
\item $\dfrac{1}{3}+ \dfrac{3}{2}$
\item $2+\dfrac{1}{2}$
\item $3+\dfrac{1}{5}$
\item $\dfrac{4}{7}+ \dfrac{3}{5}$
\item $\dfrac{4}{3}+ \dfrac{5}{11}$
\item $\dfrac{1}{3}+7$
\item $\dfrac{3}{12}+\dfrac{1}{11}$
\item $\dfrac{1}{7}+11$
\item $\dfrac{5}{12}+\dfrac{2}{14}$
\item $\dfrac{7}{18}+\dfrac{11}{30}$
\item $\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}$
\item $\dfrac{3}{2}+\dfrac{1}{4}+\dfrac{1}{8}$
\item $\dfrac{2}{5}+\dfrac{7}{2}+\dfrac{5}{3}$
\item $\dfrac{3}{11}+\dfrac{2}{3}+\dfrac{1}{2}$
\item $\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}$
\item $\dfrac{1}{2}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{5}$
\end{enumerate}
\end{multicols}
\item Perform the following operations:
\begin{multicols}{3}
\begin{enumerate}
\item $\dfrac{1}{2}- \dfrac{3}{2}$
\item $\dfrac{2}{5}- \dfrac{4}{5}$
\item $\dfrac{2}{11}-\dfrac{7}{11}$
\item $\dfrac{1}{3}- \dfrac{3}{2}$
\item $\dfrac{4}{7}- \dfrac{3}{5}$
\item $\dfrac{4}{3}- \dfrac{5}{11}$
\item $\dfrac{3}{12}-\dfrac{1}{11}$
\item $\dfrac{5}{12}-\dfrac{2}{14}$
\item $\dfrac{7}{18}-\dfrac{11}{30}$
\item $\dfrac{-1}{2}+\dfrac{1}{3}$
\item $-\dfrac{1}{7}+\dfrac{1}{5}$
\item $\dfrac{2}{-3}+\dfrac{4}{6}$
\item $-\dfrac{11}{13}+\dfrac{1}{3}$
\end{enumerate}
\end{multicols}
\item Simplify each of the following:
\begin{multicols}{3}
\begin{enumerate}
\item $\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{5}$
\item $\dfrac{2}{3}+\dfrac{1}{5}-\dfrac{8}{15}$
\item $-\dfrac{2}{5}+\dfrac{1}{7}-\dfrac{1}{2}$
\item $-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{5}$
\item $\dfrac{6}{7}+\dfrac{19}{28}-\dfrac{13}{2}$
\end{enumerate}
\end{multicols}
\end{enumerate}
\newpage
\section{Mixed Numbers}
\label{fractions:mixed}
You may recall that if you have a fraction and the numerator is larger than the denominator,
called an \emph{improper fraction},
then you can write that fraction as a mixed number.
An example of this is
$$\frac{3}{2}=1\frac{1}{2}.$$
We will not write mixed numbers this way.
There are two main reasons for this.
First,
the symbol $1\frac{1}{2}$ could be mistaken for the product of $1$ and $1/2$.
The second reason is that it is not obvious how you would multiply two numbers
written in mixed number form.
How do you multiply $(1\frac{1}{2})(2\frac{1}{3})$?
There is another reason why we do not do this.
In mathematics numbers like $3/2$ and $11/5$ are perfectly valid.
So there is usually no reason to rewrite them as mixed numbers.
In fact we will not even call fractions whose numerator is larger than its
denominator an improper fraction.
We will just call it a fraction.
But it is useful to have some kind of notation for mixed numbers.
If we can't write $3/2$ as $1\frac{1}{2}$,
what do we do?
This notation really means $1+\frac{1}{2}$,
since
$$1+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}=\frac{3}{2}.$$
This means the way to write $11/5$ as a mixed number is $2+1/5$.
The question now is how do we put a fraction in mixed number form?
Let us convert $3/2$ to a mixed number.
Recall that right after Example~\ref{frac-ex}
we discussed that there is a relationship between fractions and division.
We will use that relationship to rewrite $3/2$ as a mixed number.
If we divide $3$ by $2$,
$3 \div 2$,
then we get a quotient of $1$ and a remainder of $1$.
The mixed number is
$$\mbox{quotient}+\frac{\mbox{remainder}}{2}=1+\frac{1}{2}.$$
Doing the same thing for $11/5$ we get:
$11 \div 5$ has a quotient of $2$ and a remainder of $1$.
So
$$\mbox{quotient}+\frac{\mbox{remainder}}{5}=2+\frac{1}{5}.$$
This leads us to the following definition.
\begin{defn}[Mixed number]
\index{mixed number}
Suppose we have the fraction $a/b$, $a > b$.
If the quotient of $a \div b$ is $q$ and the remainder is $r$,
then we can write $a/b$ as the mixed number
$$q+\frac{r}{b}.$$
\end{defn}
\begin{ex}
\begin{enumerate}
\item Rewrite $17/7$ as a mixed number.
Since the quotient of $17 \div 7$ is $q=2$ and the remainder is $r=3$,
$$\frac{17}{7}=2+\frac{3}{7}.$$
\item Rewrite $54/5$ as a mixed number.
The quotient of $54 \div 5$ is $q=10$ and the remainder is $r=4$.
So
$$\frac{54}{5}=10+\frac{4}{5}.$$
\item Rewrite $9/11$ as a mixed number.
Since the numerator is already smaller than the denominator we are done.
\item Rewrite $-5/2$ as a mixed number.
We will do this in two steps.
First we convert $5/2$ to mixed number form.
So $5/3=2+1/2$.
Then we multiply the mixed number by $-1$,
so
$$(-1)(2+1/2)=-2-1/2$$
\end{enumerate}
\end{ex}
Since we are writing mixed numbers as $q+r/b$ it is easy to add, subtract, and multiply them.
Let us add $1+1/2$ and $2+1/3$.
We get
\begin{eqnarray*}
\left(1+\frac{1}{2}\right)+\left(2+\frac{1}{3}\right) & = & 1+2+\frac{1}{2}+\frac{1}{3}\\
& = & 3+\frac{3}{6}+\frac{2}{6}\\
& = & 3+\frac{5}{6}
\end{eqnarray*}
If we subtract $2+1/3$ from $1+1/2$ we get
\begin{eqnarray*}
\left(1+\frac{1}{2}\right)-\left(2+\frac{1}{3}\right) & = & 1+\frac{1}{2}-2-\frac{1}{3}\\
& = & 1-2+\frac{1}{2}-\frac{1}{3}\\
& = & -1+\frac{3}{6}-\frac{2}{6}\\
& = & -1+\frac{1}{6}
\end{eqnarray*}
Finally multiplication gives us
\begin{eqnarray*}
\left(1+\frac{1}{2}\right)\left(2+\frac{1}{3}\right) & = & 2+\frac{1}{3}+\frac{2}{2}+\frac{1}{6}\\
& = & 2+1+\frac{1}{3}+\frac{1}{6}\\
& = & 3+\frac{2}{6}+\frac{1}{6}\\
& = & 3+\frac{3}{6}=\frac{1}{2}
\end{eqnarray*}
This makes sense,
because when we have a binomial times a binomial we \textsc{foil},
see Appendix~\ref{foil}.
\section*{Exercises}
\begin{enumerate}
\item Write the following fractions as mixed numbers:
\begin{multicols}{2}
\begin{enumerate}
\item $\dfrac{3}{2}$
\item $\dfrac{5}{3}$
\item $\dfrac{12}{5}$
\item $\dfrac{5}{4}$
\item $\dfrac{24}{7}$
\item $\dfrac{13}{12}$
\item $\dfrac{125}{3}$
\item $\dfrac{33}{3}$
\item $\dfrac{247}{13}$
\item $\dfrac{5}{8}$
\item $\dfrac{1331}{24}$
\item $\dfrac{1223}{123}$
\end{enumerate}
\end{multicols}
\item Write the following mixed numbers as fractions:
\begin{multicols}{2}
\begin{enumerate}
\item $1 + \dfrac{1}{3}$
\item $3 + \dfrac{3}{5}$
\item $2 + \dfrac{13}{57}$
\item $6 + \dfrac{7}{8}$
\item $4 + \dfrac{6}{12}$
\item $13 + \dfrac{7}{9}$
\item $11 + \dfrac{3}{13}$
\item $7 + \dfrac{7}{7}$
\item $5 + \dfrac{14}{23}$
\item $3 + \dfrac{3}{10}$
\item $121 + \dfrac{2}{15}$
\item $1 + \dfrac{122}{357}$
\end{enumerate}
\end{multicols}
\end{enumerate}
\newpage
\section{Division}
\label{fractions:division}
We will now learn how to divide fractions.
Many people find this conceptually to be the hardest operation to understand.
It is not as bad as it seems.
First let look at what it means to divide integers.
Let's divide $a$ by $b$, $a \div b$.
Suppose its equal to $c$,
so $a \div b=c$.
This can be transformed into a multiplication as
$$c \cdot b=a.$$
Thinking of $a \div b=c$ as a multiplication allows us to interpret it as
the number of $b$'s we need to get the number $a$ is $c$.
Again a concrete example will be easier to follow.
Let us divide $6$ by $2$, $6 \div 2=c$.
Rewriting this in terms of multiplication we have
$$c \cdot 2=6.$$
so the number $c$ must be $3$.
So we need, for example, $3$ pairs of socks to get six socks.
One way to interpret $a \div b = c$ is
\emph{we can divide $a$ into $c$ groups each having $b$ members.}
Since I said this is one way to interpret division, there must be another way to interpret it.
We can interpret it as a fraction provided $a$ and $b$ are integers
$$a \div b = \frac{a}{b}.$$
Next we divide a integer by a fraction.
\begin{ex}
Let us divide $3$ by $1/2$.
So we want to compute
$$\frac{3}{\frac{1}{2}}=3 \div \frac{1}{2}=c.$$
Let us rewrite this as
$$c \cdot \frac{1}{2}=3.$$
It easy to see that $c=6$.
We can interpret this (in terms of multiplication) as saying we need six
halves of a pie to get three pies.
Or in terms of division as saying three pies contain $6$ halves, Figure \ref{frac-pie-half2-b}.
\begin{figure}[H]
\centering
\epsfig{file=three-divided-half.eps,height=1in}
\caption{Three pies, each divided in half.}
\label{frac-pie-half2-b}
\end{figure}
\end{ex}
Next we will divide a fraction by a integer
$$\frac{\frac{b}{c}}{a}.$$
Let us look at an example.
\begin{ex}
Let us divide $1/3$ by $2$.
We first rewrite the division as a multiplication:
$$\frac{\frac{1}{3}}{2}=c\quad \Leftrightarrow \quad c \cdot 2=\frac{1}{3}.$$
Then we find that $c=1/6$.
So the expression $1/3 \div 2$ asks what is half of $1/3$.
\end{ex}
So if $a$, $b$, and $c$ are integers then we interpret $b/c \div a$ as
\begin{description}
\item[First] We have a pie that is divided into $c$ equal pieces and
we have $b$ of those pieces.
\item[Second] We take those $c$ pieces and divide them into $a$ pieces.
So the whole pie is now divided into $c \cdot a$ pieces.
Of those $c \cdot a$ pieces we have $b$ of them.
\end{description}
\begin{ex}
Let us perform the division $2/3 \div 5$.
The $2/3$ says divide a pie into $3$ equal pieces and take $2$ of those pieces.
\begin{figure}[H]
\centering
\epsfig{file=two-thirds.eps,height=1in}
\caption{Two thirds of a pie.}
\label{frac-two-thirds}
\end{figure}
Now take each third of the pie and divide each of them into $5$ pieces.
\begin{figure}[H]
\centering
\epsfig{file=thirds-divided-five.eps,height=1in}
\caption{Each third divided into five pieces.}
\label{frac-thirds-into-fifths}
\end{figure}
This will divide the pie into $15$ pieces.
We have $2$ pieces of this pie.
\begin{figure}[H]
\centering
\epsfig{file=two-fifteenths.eps,height=1in}
\caption{Two fifteenths of a pie}
\label{frac-two-fifteenths}
\end{figure}
So we have $2/15$ of the pie.
\end{ex}
Now we will divide a fraction by another fraction.
Let us start with some concrete examples.
The first example is
$$\frac{\frac{1}{2}}{\frac{1}{2}}.$$
We know that this is equal to $1$ since any nonzero number divided by itself is $1$.
This is interpreted as there is one half of a pie in one halve of a pie.
Next let us look at
$$\frac{\frac{1}{2}}{\frac{1}{4}}.$$
This asks how many quarters of a pie are there in one halve of a pie,
Figure~\ref{one-fourth-of-one-half}.
\begin{figure}[H]
\centering
\epsfig{file=half-to-halfhalf.eps,height=1in}
\caption{$\frac{1}{2}\div\frac{1}{4}$}
\label{one-fourth-of-one-half}
\end{figure}
We can easily see that there are 2 quarters of a pie in one halve of a pie.
How can we see these answers arithmetically?
Let us do the following:
\begin{eqnarray*}
\frac{\frac{1}{2}}{\frac{1}{4}} & = & c\\
\frac{1}{2} & = & c\frac{1}{4} \qquad
\mbox{by converting to multiplication}\\
4 \cdot \frac{1}{2} & = & c\frac{1}{4}\cdot 4\\
\frac{4}{2} & = & c\\
2 & = & c
\end{eqnarray*}
Now this was a lot of work to get $c=2$.
If we carefully look at what we did we will realize that we can,
as a shortcut,
flip the denominator fraction and multiply it with the numerator fraction.
So if we wanted to calculate
$$\frac{\frac{1}{2}}{\frac{1}{8}}$$
all we have to do is
$$\frac{\frac{1}{2}}{\frac{1}{8}}=\frac{1}{2}\cdot\frac{8}{1}=\frac{8}{2}=4.$$
This says that the number of eighths in a halve of a pie is $4$.
This gives us the following definition.
\begin{defn}[Division of Fractions]
\index{fraction!division}
Let $a/b$ and $c/d$ be fractions.
Then the quotient of $a/b$ divided by $c/d$ is
$$\frac{\dfrac{a}{b}}{\dfrac{c}{d}}
=\frac{a}{b}\div\frac{c}{d}
=\frac{a}{b}\cdot\frac{d}{c}
=\frac{ad}{bc}$$
\end{defn}
It is cumbersome to say
``multiply the fraction in the numerator by the fraction in the denominator
after it has been flipped''.
So let us make a new definition that means the flipped version of a fraction.
\begin{defn}[Reciprocal]
\index{reciprocal}
The reciprocal of the fraction $a/b$ is $b/a$.
\end{defn}
Now we can say
``multiply the fraction in the numerator by the reciprocal of the fraction
in the denominator''.
``And now for something completely different'',
some examples.
\begin{ex}
\begin{enumerate}
\item Let us perform the division $1/2 \div 1/3$.
By our definition of division of fractions we have
$$\frac{\frac{1}{2}}{\frac{1}{3}}=\frac{1}{2}\cdot\frac{3}{1}=\frac{3}{2}.$$
So there are $1+1/2$ thirds of a pie in one halve of a pie, Figure~\ref{half-divby-third}.
\begin{figure}[H]
\centering
\epsfig{file=half-dividedby-third.eps,height=1.5in}
\caption{$\frac{1}{2}\div\frac{1}{3}$.}
\label{half-divby-third}
\end{figure}
\item Divide $1/3$ by $1/2$.
Using the definition we have
$$\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{1}{3}\cdot\frac{2}{1}=\frac{2}{3}.$$
We interpret $(1/3)/(1/2)=2/3$ as $1/3$ of a pie makes up two-thirds of $1/2$ of a pie,
Figure~\ref{third-divby-half}.
\begin{figure}[H]
\centering
\epsfig{file=third-divby-half.eps,height=1.5in}
\caption{$\frac{1}{3}\div\frac{1}{2}$.}
\label{third-divby-half}
\end{figure}
\item Divide $2/5$ by $1/3$.
Using the definition we have
$$\frac{\frac{2}{5}}{\frac{1}{3}}=\frac{2}{5}\cdot\frac{3}{1}=\frac{6}{5}.$$
At this point it becomes hard to interpret the result of dividing the fractions
$2/5$ by $1/3$ as pieces of a pie.
So don't worry if the following is not clear.
We can interpret dividing $2/5$ by $1/3$ as
$2/5$ of a pie is six-fifths of $1/3$ of a pie,
Figure~\ref{twofifths-divby-third}.
\begin{figure}[H]
\centering
\epsfig{file=twofifths-divby-third.eps,height=1.5in}
\caption{$\frac{2}{5}\div\frac{1}{3}$.}
\label{twofifths-divby-third}
\end{figure}
\end{enumerate}
\end{ex}
Let us get away from fractions for a moment and look at why division by zero is illegal.
We know that division can be re-expressed in terms of multiplication:
$$\frac{a}{b}=c \mbox{ is equivalent to } a=c \cdot b.$$
So let us assume for the moment that it makes sense to divide by zero,
i.e. that the following is valid:
$$\frac{a}{0}=c.$$
There are two cases to look at: when $a=0$ and $a \neq 0$.
\begin{description}
\item[Case: $a \neq 0$:] If $a \neq 0$ then we have the following
$$\frac{a}{0}=c \mbox{ is equivalent to } a=c \cdot 0, \quad a \neq 0.$$
So we need to figure out what value $c$ can be multiplied by zero and
give us a nonzero answer.
But we know that anything times zero is zero,
so there is no value $c$ that can be multiplied by zero and get a nonzero answer.
Therefore division of a nonzero number by zero does not make sense.
And hence this is undefined.
\item[Case: $a=0$:] Now suppose that $a = 0$ then we have the following
$$\frac{0}{0}=c \mbox{ is equivalent to } 0=c \cdot 0.$$
Can we find a value for $c$ to make the statement $0 = c \cdot 0$ true?
It turns out that any value of $c$ will satisfy this statement.
That is why we call $0/0$ indeterminate.
So the question is can we arbitrarily pick some value for $c$ and
use this in our calculations?
The answer is no.
Sometime there are ways to determine what value, if any,
the expression $0/0$ represents,
but this requires calculus.
Obviously we will not discuss this in any more depth.
We will also call division of zero by zero undefined.
\end{description}
This is the reason we say division by zero is illegal.
\section*{Exercises}
\begin{multicols}{4}[Do the following divisions:]
\begin{enumerate}
\item $\fracf{\dfrac{1}{2}}{\dfrac{1}{3}}$
\item $\fracf{\dfrac{3}{5}}{\dfrac{2}{15}}$
\item $\fracf{\dfrac{7}{3}}{\dfrac{21}{4}}$
\item $\fracf{\dfrac{11}{9}}{\dfrac{13}{27}}$
\item $\fracf{\dfrac{2}{3}}{\dfrac{42}{69}}$
\item $\fracf{\dfrac{16}{17}}{\dfrac{64}{34}}$
\item $\fracf{\dfrac{9}{8}}{\dfrac{123}{15}}$
\item $\fracf{\dfrac{264}{157}}{\dfrac{67}{321}}$
\item $\fracf{3}{\dfrac{5}{7}}$
\item $\fracf{8}{\dfrac{3}{10}}$
\item $\fracf{\dfrac{3}{4}}{5}$
\item $\fracf{\dfrac{8}{7}}{11}$
\end{enumerate}
\end{multicols}
\newpage
\section{Complex Fractions}
\label{fractions:complexfractions}
Another object that occurs in mathematics is the complex fraction.
Some examples of complex fractions are given below.
$$\frac{\frac{1}{2}}{1},
\quad \frac{2}{\frac{1}{3}},
\quad \frac{\frac{1}{2}}{\frac{1}{3}},
\quad \frac{\frac{1}{2}+\frac{1}{3}}{\frac{2}{3}+\frac{2}{5}}$$
From these examples we can see what the definition of a complex fraction should be.
\begin{defn}[Complex Fractions]
\index{complex fractions}
A complex fractions is a fractional form whose numerator or denominator is a fraction.
\end{defn}
The question that comes to mind is how do we work with complex fractions?
We have seen how to work with some of them.
If the are in the form of a fraction divided by an integer,
an integer divided by a fraction,
or a fraction divided by another fraction we just have the division of two fractions
as we saw in section~\ref{fractions:division}.
But our last example above requires more work.
Suppose we have the fraction
$$\frac{\frac{1}{2}+\frac{1}{3}}{\frac{2}{3}+\frac{2}{5}}.$$
Then there are two ways of simplifying this into a fraction.
The first way is to reduce the numerator and then the denominator into a single fraction,
by adding the fractions in the numerator and then the fractions in the denominator.
Then we have the division of two fractions which we already know how to work with.
The second way is to multiply the numerator and denominator of the complex fraction
by the least common denominator of all the fractions in it.
Let us look at some examples to see how this works.
We will start with the first method first.
We simplify the numerator and denominator of
$$\frac{\frac{1}{2}+\frac{1}{3}}{\frac{2}{3}+\frac{2}{5}}$$
which gives us
$$\frac{\frac{1}{2}+\frac{1}{3}}{\frac{2}{3}+\frac{2}{5}}
=\frac{\frac{3+2}{6}}{\frac{10+6}{15}}
=\frac{\frac{5}{6}}{\frac{16}{15}}.$$
Now we divide the fraction $5/6$ by $16/15$,
so we get
$$\frac{\frac{5}{6}}{\frac{16}{15}}=\frac{5}{6}\cdot\frac{15}{16}=
\frac{75}{96}=\frac{25}{32}.$$
In the second method we first find the least common denominator of all the fractions
that make up
$$\frac{\frac{1}{2}+\frac{1}{3}}{\frac{2}{3}+\frac{2}{5}}.$$
So we find the least common denominator of $1/2$, $1/3$, $2/3$, and $2/5$.
So the least common denominator is $30$.
We multiply the numerator and denominator of our complex fraction by $30$:
$$\frac{\frac{1}{2}+\frac{1}{3}}{\frac{2}{3}+\frac{2}{5}}\cdot\frac{30}{30}
=\frac{\frac{30}{2}+\frac{30}{3}}{\frac{60}{3}+\frac{60}{5}}
=\frac{15+10}{20+12}
=\frac{25}{32}.
$$
\begin{ex}
\begin{enumerate}
\item Simplify
$$\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{3}+\frac{1}{5}}.$$
We will simplify this complex fraction using both methods.
Recall that in the first method we find a \lcd\ for the numerator and denominator
fractions separately and then perform the indicated operations.
This will give us a division of fractions which we then perform.
Whew that is a mouthful,
it is easier to just do the calculations.
The \lcd\ of the numerator fractions is $6$ and of the denominator fractions is $15$.
So
$$
\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{3}+\frac{1}{5}}
=\frac{\frac{3+2}{6}}{\frac{5+3}{15}}
=\frac{\frac{5}{6}}{\frac{8}{15}}
=\frac{5}{6}\cdot\frac{15}{8}
=\frac{5}{2}\cdot\frac{5}{8}
=\frac{25}{16}
.$$
In the second method we find the \lcd\ for all the fractions and
multiply the numerator and denominator fractions that \lcd.
The \lcd\ for all the fractions is $30$,
so
$$
\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{3}+\frac{1}{5}} \cdot \frac{30}{30}
=\frac{\frac{30}{2}+\frac{30}{3}}{\frac{30}{3}+\frac{30}{5}}
=\frac{15+10}{10+6}
=\frac{25}{16}
.$$
\item Simplify
$$\frac{\frac{2}{3}-\frac{1}{2}}{\frac{5}{6}+\frac{1}{2}}.$$
First method
$$\frac{\frac{2}{3}-\frac{1}{2}}{\frac{5}{6}+\frac{1}{2}}
=\frac{\frac{4-3}{6}}{\frac{5+3}{6}}
=\frac{\frac{1}{6}}{{8}{6}}
=\frac{1}{6}\cdot\frac{6}{8}
=\frac{1}{8}
.$$
Second method
$$\frac{\frac{2}{3}-\frac{1}{2}}{\frac{5}{6}+\frac{1}{2}}\cdot\frac{6}{6}
=\frac{4-3}{5+3}
=\frac{1}{8}
.$$
\item Simplify
$$\frac{\frac{1}{2}+5}{\frac{1}{7}+\frac{1}{3}}.$$
First method
$$
\frac{\frac{1}{2}+5}{\frac{1}{7}+\frac{1}{3}}
=\frac{\frac{1+10}{2}}{\frac{3+7}{21}}
=\frac{\frac{11}{2}}{\frac{10}{21}}
=\frac{11}{2}\cdot\frac{21}{10}
=\frac{231}{20}
.$$
Second method
$$
\frac{\frac{1}{2}+5}{\frac{1}{7}+\frac{1}{3}}\cdot\frac{42}{42}
=\frac{21+210}{6+14}
=\frac{231}{20}
.$$
\end{enumerate}
\end{ex}
\section*{Exercises}
\begin{multicols}{2}[Simplify the following:]
\begin{enumerate}
\item $\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{\dfrac{1}{4}+\dfrac{1}{5}}$
\item $\dfrac{\dfrac{2}{3}+\dfrac{3}{2}}{\dfrac{3}{7}+\dfrac{3}{2}}$
\item $\dfrac{\dfrac{4}{3}+\dfrac{7}{6}}{\dfrac{11}{4}+\dfrac{1}{12}}$
\item $\dfrac{\dfrac{1}{11}+\dfrac{2}{13}}{\dfrac{5}{13}-\dfrac{7}{11}}$
\item $\dfrac{\dfrac{1}{2}-\dfrac{1}{3}}{\dfrac{3}{2}-\dfrac{5}{3}}$
\item $\dfrac{\dfrac{23}{5}+\dfrac{6}{4}}{\dfrac{11}{2}-\dfrac{9}{10}}$
\item $\dfrac{\dfrac{2}{3}+5}{\dfrac{6}{7}+\dfrac{4}{3}}$
\item $\dfrac{7-\dfrac{4}{3}}{\dfrac{5}{6}-2}$
\item $\dfrac{\dfrac{2}{3}+\dfrac{7}{3}}{5}$
\end{enumerate}
\end{multicols}
\newpage
\section{Putting It All Together}
\begin{quote}
Science is built up with facts, as a house is with stones.
But a collection of facts is no more a science than a heap of stones is a house. \\
\emph{Jules Henri Poincar\`{e} (1854-1912) }
\end{quote}
Well,
that's all there is to learn about fractions.
So what is left to do?
We need to be able to simplify expressions that involve any or all of the arithmetic operations.
First we need to recall our order of operations:
parentheses, exponents, multiplication and division, addition and subtraction.
Recall that multiplication and division and addition and subtraction is performed left to right.
As an example,
If we wanted to simplify $5-2+6$ we would first do the subtraction, $5-2=3$ and
then add the $6$ to the $3$, $3+6=9$.
So if we wanted to use parentheses to show the order of operations we would write $(5-2)+6$.
Using parentheses can help us read an expression.
Compare the expression,
$$2+8+7+3+4+8+1+5$$
to the expression with parentheses added,
$$(2+8)+(7+3)+(4+8)+(1+5).$$
But be careful about adding to many parentheses.
While the following is mathematically correct and give us the same answer as the two expressions above,
it is harder to read:
$$((((2+8)+(7+3))+(4+8))+(1+5)).$$
You also must be careful about removing parentheses.
You can change the meaning of an expression if you blindly remove parentheses.
All of the following parentheses are necessary:
$$(5+6)\cdot(2+8-4)\div 2+(6-5+3)\cdot 5.$$
In fact you may want to add one set of parentheses around the $(2+8-4)\div 2$.
This give us
$$(5+6)\cdot((2+8-4)\div 2)+(6-5+3)\cdot 5.$$
But the following has some unnecessary parentheses:
$$((((5+3)-7)+8)+1+2)-(6-5).$$
So let us remove all of the parentheses that are not necessary:
$$5+3-7+8+1+2-(6-5).$$
Now I think that this is a little bit hard to read,
so I will add a few parentheses to make it a easier to read
$$(5+3)-7+(8+1+2)-(6-5).$$
You may have chosen to put more or less parentheses in the expression or
chosen to group some of the numbers differently than I did.
The point here is that you should add as many parentheses as you need
to make it readable and no more,
as long as you do not change the meaning of the expression.
Let me break away from talking specifically about fractions and talk a little about mathematics in general.
When someone who is good at arithmetic sees an expression like
$$5+2+3+7+9+3+6+1+4+8$$
what they normally do is to group the numbers together into chunks,
e.g.
$$(5+2+3)+(7+3)+(9+1)+(6+4)+8.$$
Now there are two reasons for grouping the expression in this way.
The first is that each group sums to $10$ which makes the expression easier to compute.
The second reason,
and this is the main reason I am showing this example,
is that the calculation is easier to perform when it is broken down into smaller pieces that can
be quickly worked.
This idea of breaking down a more complex expression (or idea)
into several simpler expressions (or ideas),
is the key to understanding mathematics.
In fact,
it is the key to understanding any field that consists of several complex connected ideas.
Take a look at the table of contents for this book or the table of contents for any book that
involves the collection of complex ideas into a single subject,
such as textbooks in science, music theory, grammar, etc.
Each book starts with some basic facts and terminology.
Then it will proceed to use these facts and terminology to build up more and more complex ideas.
And finally all of these ideas are combined into a coherent whole.
If you fail to grasp a single ideal you will not be able to combine them into that coherent whole.
But if you did grasp each simpler idea then with a little more work you can combine those simpler
ideas into the complex topic of,
in this case,
fractions.
But just knowing each simpler idea is not enough.
This is the point to the quote at the beginning of this section.
To claim that you understand fractions you must be able to combine all of the ideas together
into a coherent whole.
So let us get back to the task at hand,
combining all of the topics together.
To do this let us look at some examples.
\begin{ex}
Simplify $\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)$.
The order of operations tells us to do the addition first and then to multiply.
\begin{eqnarray*}
\frac{1}{2}\left(\frac{1}{3}+\frac{1}{5}\right)&=&\frac{1}{2}\left(\frac{5}{15}+\frac{3}{15}\right)\\
& = & \frac{1}{2}\cdot\frac{8}{15}\\
& = & \frac{4}{15}
\end{eqnarray*}
\end{ex}
\begin{ex}
Simplify $\dfrac{1}{2}\cdot \dfrac{1}{3} - \dfrac{2}{3}\cdot\dfrac{1}{5}$.
The order of operations tells us that we should do each multiplication first and
then perform the subtraction last.
\begin{eqnarray*}
\frac{1}{2}\cdot \frac{1}{3} - \frac{2}{3}\cdot\frac{1}{5}
& = & \frac{1}{6}-\frac{2}{15}\\
& = & \frac{5}{30}-\frac{4}{30}\\
& = & \frac{1}{30}
\end{eqnarray*}
\end{ex}
\begin{ex}
Simplify $\left(\dfrac{2}{3}+\dfrac{3}{5}\right) \div \dfrac{2}{3}$.
Recall that there two ways to simplify this expression.
The first method is to simplify the numerator and denominator fractions and
then perform the division.
\begin{eqnarray*}
\dfrac{\dfrac{2}{3}+\dfrac{3}{5}}{\dfrac{2}{3}}
& = & \dfrac{\dfrac{10+19}{15}}{\dfrac{2}{3}}\\
& = & \dfrac{\dfrac{19}{15}}{\dfrac{2}{3}}\\
& = & \dfrac{19}{15}\cdot \dfrac{3}{2}\\
& = & \dfrac{19}{10}
\end{eqnarray*}
In the second method we find the least common denominator of all the fractions and
multiply the numerator and denominator of the fractional form by it.
\begin{eqnarray*}
\dfrac{\dfrac{2}{3}+\dfrac{3}{5}}{\dfrac{2}{3}}
& = & \dfrac{\dfrac{2}{3}+\dfrac{3}{5}}{\dfrac{2}{3}}\cdot \dfrac{15}{15}\\
& = & \dfrac{10+9}{10}\\
& = & \dfrac{19}{10}
\end{eqnarray*}
\end{ex}
\begin{ex}
Is $\dfrac{2}{3}+\dfrac{3}{4} < \dfrac{9}{5}-\dfrac{2}{10}$ a true statement?
To see if it is true or false we need to simplify each side and then compare.
\begin{eqnarray*}
\frac{2}{3}+\frac{3}{4} & < & \frac{9}{5}-\frac{2}{10} \\
\frac{8+9}{12} & < & \frac{18-2}{10} \\
\frac{17}{12} & < & \frac{16}{10} \\
\frac{17}{12} & < & \frac{8}{5} \\
\end{eqnarray*}
Giving them a common denominator we get
$$\frac{85}{60} < \frac{96}{60}.$$
So it is a true statement.
\end{ex}
\begin{ex}
Is
$\dfrac{9}{5}\cdot\left(\dfrac{11}{3}-\dfrac{3}{7}\right)<\dfrac{4}{5}+\dfrac{12}{7}-\dfrac{3}{2}$
a true statement?
Simplifying each side:
\begin{eqnarray*}
\dfrac{9}{5}\cdot\left(\dfrac{11}{3}-\dfrac{3}{7}\right)& < & \dfrac{4}{5}+\dfrac{12}{7}-\dfrac{3}{2}\\
\frac{9}{5}\cdot\frac{77-9}{21} & < & \frac{28+60}{35}-\frac{3}{2}\\
\frac{9}{5}\cdot\frac{69}{21} & < & \frac{88}{35}-\frac{3}{2}\\
\frac{9}{5}\cdot\frac{23}{7} & < & \frac{176-105}{70}\\
\frac{207}{35} & < & \frac{71}{70}\\
\end{eqnarray*}
Getting a common denominator of $70$ we have
$$\frac{414}{70} < \frac{71}{70}.$$
So it is false.
\end{ex}
\newpage
\section*{Exercises}
\begin{enumerate}
\item Remove the excess parentheses:
\begin{enumerate}
\item
$\left(\left(\left(\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\dfrac{1}{5}\right)+\dfrac{1}{7}\right)
+\dfrac{1}{11}\right)+\dfrac{1}{13}$
\item
$\dfrac{1}{2}+\left(\dfrac{1}{3}-\left(\dfrac{1}{5}+\left(\dfrac{1}{7}-
\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\right)\right)\right)$
\item
$\left(\left(\left(\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{7}{5}\right)+\dfrac{13}{7}-
\left(\dfrac{3}{5}+\dfrac{7}{11}\right)\right)$
\item
$\left(\dfrac{1}{2}+\dfrac{11}{13}\right)\cdot\dfrac{1}{6}+\dfrac{4}{7}+
\left(\left(\dfrac{1}{3}\right)-\dfrac{1}{7}\right)+
\dfrac{5}{11}\left(\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\dfrac{1}{5}\right)$
\item
$\left(\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\div\left(\dfrac{1}{13}+\dfrac{3}{8}\right)\right)
+\left(\dfrac{6}{8}-\dfrac{13}{5}+\left(\dfrac{11}{30}+\left(\dfrac{1}{5}+
\left(\dfrac{11}{13}-\dfrac{11}{12}\right)\right)\right)\right)$
\item
$25+\left(\dfrac{1}{3}-\dfrac{1}{7}\right)+\left(\left(\left(\dfrac{1}{5}-\dfrac{1}{3}\right)
\div\dfrac{3}{8}\right)\div\left(\dfrac{16}{7}+\dfrac{15}{5}\right)+7-\dfrac{11}{50}\right)+2$
\end{enumerate}
\item Simplify the following:
\begin{multicols}{2}
\begin{enumerate}
\item $\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{2}{7}$
\item $\dfrac{1}{3}\left(\dfrac{3}{2}+\dfrac{5}{7}\right)$
\item $\dfrac{5}{2}\left(\dfrac{1}{3}-\dfrac{5}{3}\right)+\left(\dfrac{2}{3}+\dfrac{3}{2}\right)
\div\dfrac{13}{2}$
\item $\dfrac{1}{2}\left[\dfrac{1}{3}\left(\dfrac{1}{5}+\dfrac{2}{5}\right)-
\dfrac{4}{3}\left(\dfrac{3}{5}+\dfrac{1}{4}\right)\right]$
\item $\dfrac{2}{3}\left[\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}\right)-
\left(\dfrac{3}{2}-\dfrac{2}{5}\right)\div\dfrac{3}{7}\right]$
\item $\dfrac{2}{5}\cdot\dfrac{3}{4}+\left(\dfrac{3}{10}-\dfrac{5}{2}\right)\dfrac{1}{4}+
\dfrac{1}{6}\div\dfrac{1}{5}$
\end{enumerate}
\end{multicols}
\item Which of the following are true:
\begin{multicols}{2}
\begin{enumerate}
\item $\dfrac{1}{2}+\dfrac{1}{3} < \dfrac{1}{5}+1$
\item $\dfrac{2}{3}+\dfrac{7}{11} < \dfrac{5}{12}+\dfrac{3}{4}$
\item $\dfrac{2}{3}\left(\dfrac{13}{5}-\dfrac{5}{6}\right) <
\dfrac{2}{5}\cdot\dfrac{3}{4}+\dfrac{1}{3}\cdot\dfrac{5}{7}$
\item $\left(\dfrac{1}{3}+\dfrac{1}{2}\right)\div\dfrac{6}{5} < \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{24}{6}$
\item $\left(\dfrac{2}{3}+\dfrac{3}{5}\right)\dfrac{30}{7} <
\dfrac{2}{3}\div\left(\dfrac{1}{5}+\dfrac{1}{3}\right)$
\item $\dfrac{49}{5}-\dfrac{11}{2} < \dfrac{1}{2}\div\dfrac{1}{3}\cdot\dfrac{1}{5}\div\dfrac{1}{7}$
\end{enumerate}
\end{multicols}
\end{enumerate}
\newpage
\appendix
\appendixpage
\addappheadtotoc
\chapter{Glossary}
\begin{description}
\item[Common multiple-]A common multiple of the two positive integers
$a$ and $b$ is a positive integer, call it $c$,
that is divisible by both $a$ and $b$,
i.e. both $c \div a$ and $c \div b$ are integers.
\item[Denominator-] Given a fraction $a/b$ we call the number $b$ the denominator.
\item[Fraction-] A fraction is a number of the form $a/b$ where $a$ and $b$ are integers and $b \neq 0$.
\item[Fractional form-] A fractional form is any expression in the form of
$$\frac{a}{b}$$
where $a$ and $b$ can be any valid mathematical expression.
\item[Leaf-] The numbers at the bottom of a tree.
\item[Least common denominator (LCD)-] The least common denominator of a set of fractions is
the least common multiple of their denominators.
\item[Least common multiple (LCM)-] The least common multiple of a set of positive integers is the
smallest positive integer that is divisible by all the numbers in that set,
i.e. it is what it says it is: the smallest common multiple.
\item[Mixed number-] Suppose we have the fraction $a/b$, $a > b$.
If the quotient of $a \div b$ is $q$ and the remainder is $r$,
then we can write $a/b$ as the mixed number
$$q+\frac{r}{b}.$$
\item[Numerator-] Given a fraction $a/b$ we call the number $a$ the numerator
\item[Prime number-] A prime number is a positive integer that is only divisible by one and itself,
not including the number $1$.
\item[Reciprocal-] The reciprocal of the fraction $a/b$ is $b/a$.
\item[Root-] The number at the top of a tree.
\item[Tree-] A tree is a diagram that shows the relationship between the root and the leaves
such as a family tree or the prime factorization.
\begin{figure}[H]
\begin{center}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{root}}
{
{\TR{leaf}}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{node}}
{
{\TR{leaf}}
{\TR{leaf}}
}
}
\end{center}
\caption{General Tree}
\end{figure}
\begin{figure}[H]
\begin{center}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{Noah}}
{
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{Japheth}}
{
{\TR{Gomer}}
{\TR{...}}
{\TR{Tiras}}
}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{Ham}}
{
{\TR{Cush}}
{\TR{...}}
{\TR{Canaan}}
}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{Shem}}
{
{\TR{Arpachshad}}
}
}
\end{center}
\caption{Noah's Family Tree}
\end{figure}
\begin{figure}[H]
\begin{center}
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{20}}
{
\pstree[nodesep=2pt, levelsep=20pt]
{\TR{4}}
{
{\TR{2}}
{\TR{2}}
}
{\TR{5}}
}
\end{center}
\caption{Prime Factorization of $20$}
\end{figure}
\end{description}
\chapter{Prime Numbers}
\label{primelist}
Here is a list of all the prime numbers less than $300$.\\
$$
\begin{array}{||r|r|r|r|r|r|r|r|r|r|r|r|r|r|r||}\hline
2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37 & 41 & 43 & 47 \\
53 & 59 & 61 & 67 & 71 & 73 & 79 & 83 & 89 & 97 & 101 & 103 & 107 & 109 & 113 \\
127 & 131 & 137 & 139 & 149 & 151 & 157 & 163 & 167 & 173 & 179 & 181 & 191 & 193 & 197 \\
199 & 211 & 223 & 227 & 229 & 233 & 239 & 241 & 251 & 257 & 263 & 269 & 271 & 277 & 281 \\
283 & 293 & & & & & & & & & & & & & \\
\hline
\end{array}
$$
\chapter{Division Tricks}
Let $a_i$ denote the digits of a number,
e.g. $542$ is represented by $a_3a_2a_1$ where $a_3=5$, $a_2=4$, and $a_1=2$.
The number $a_na_{n-1}\cdots a_2a_1$ is divisible\\
$
\begin{array}{||c|l|l|l||}
\hline
\hbox{by} & \hbox{if} & \hbox{is divisible by} & or \\ \hline\hline
2 & a_1 & 2 & \\ \hline
3 & a_n+a_{n-1}+\cdots+a_2+a_1 & 3 & \\ \hline
4 & a_2a_1 & 4 & =00 \\\hline
5 & a_1 & 5 & \\\hline
6 & a_na_{n-1}\cdots a_2a_1 & 2 \hbox{ and } 3 & \\\hline
7 & a_na_{n-1}\cdots a_2 - 2a_1 & 7 & \\\hline
8 & a_3a_2a_1 & 8 & =000 \\\hline
9 & a_n+a_{n-1}+\cdots+a_2+a_1 & 9 & \\\hline
10 & a_1=0 & & \\\hline
11 & (a_1+a_3+a_5+\cdots)-(a_2+a_4+a_6+\cdots) & 11 & \\\hline
12 & a_na_{n-1}\cdots a_2a_1 & 3 \hbox{ and } 4 & \\\hline
13 & a_na_{n-1}\cdots a_2-9a_1 & 13 & \\\hline
\end{array}
$
\begin{ex}\hfill
\begin{description}
\item[2:] Since the last digit of $23578$,
which is $8$,
is divisible by $2$,
then $23578$ is divisible by $2$.
The last digit of $2365$ is $5$.
Since $5$ is not divisible by $2$,
$2365$ is not divisible by $2$.
\item[3:] Let us see if $231$ is divisible by $3$.
Summing up the digits of $231$ we get $2+3+1=6$.
Since $6$ is divisible by $3$ then $231$ is divisible by $3$.
Let us check if $23574$ is divisible by $3$.
Summing up the digits of $23574$ gives us $2+3+5+7+4=21$.
Since $21$ is divisible by $3$,
then $23574$ is divisible by $3$.
But suppose that you could not remember that $21$ is divisible by $3$.
Then you could sum up the digits of $21$ to get $2+1=3$.
Clearly $3$ is divisible by $3$.
So $21$ is divisible by $3$ and therefore $23574$ is divisible by $3$.
Is $27934576592477693217$ divisible by $3$?
Let us sum up the digits again $2+7+9+3+4+5+7+6+5+9+2+4+7+7+6+9+3+2+1+7=105$.
If we are not sure if $105$ is divisible by three we can sum up its digits.
So $1+0+5=6$.
Now we know $6$ is divisible by $3$,
so $105$ is divisible by $3$ and therefore $27934576592477693217$ is divisible by $3$.
The number $16$ is not divisible by $3$ since $1+6=7$ is not divisible by $3$.
Let us try $23761$.
It's digits sum to $2+3+7+6+1=19$ and $19$ is not divisible by $3$
since $1+9=10$ is not divisible by $3$.
\item[4:] The number $116$ is divisible by $4$ since the last two digits of
$116$ is $16$ and $16$ is divisible by $4$.
How about the number $1352$?
The last two digits of $1352$ is $52$.
Now $4$ will divide $52$ ($52 \div 4 = 13$).
So $4$ divides $1352$.
Will $4$ divide $125487100$?
Yes since the last two digits are $00$.
\item[5] It is easy to see that $5$ will divide $25$, $40$, $14520$, $785425$,
and $789875412545$,
since each one ends in a $0$ or a $5$.
\item[6:] To check if $132$ is divisible by $6$ requires two steps.
First is $132$ divisible by $2$?
Yes since the last digit is divisible by $2$.
Next we sum up its digits: $1+3+2=6$.
Since $6$ is divisible by $3$,
$132$ is divisible by $3$.
Therefore $132$ is divisible by $6$.
Let us try $1764$.
Clearly $1764$ is divisible by $2$
and it is divisible by $3$ since $1+7+6+4=18=9$.
So $1764$ is divisible by $6$.
The number $1371$ is not divisible by $6$ since $1371$ is not divisible by $2$.
Note that it is divisible by $3$, $1+3+7+1=12$,
but that does not matter since it is not divisible by $2$.
The number $1678$ is not divisible by $6$ since it is not divisible by $3$.
This can be seen since $1+6+7+8=22$.
It does not matter that it is divisible by $2$.
It has to be divisible by both.
We can see that $3211$ is not divisible by $6$ since it is not divisible by either $2$
or $3$.
\item[7:] Let us start off with some easy examples.
First $14$ is divisible by $7$ since $1-2(4)=1-8=-7$ and $-7$ is divisible by $7$.
Next $21$ is divisible by $7$ since $2-2(1)=2-2=0$ and $0$ is divisible by $7$.
Is $924$ divisible by $7$?
We apply the rule to get $92-2(4)=92-8=84$.
If we are not sure whether $7$ divides $84$ we can apply the rule again.
Doing this gives us $8-2(4)=8-8=0$ and $0$ is divisible by $7$.
So $924$ is divisible by $7$.
Our next example is to determine if $2632$ is divisible by $7$.
Apply the rule we get $263-2(2)=263-4=259$.
Applying the rule again $25-2(9)=25-18=7$.
So $7$ divides $2632$.
Is $7532$ divisible by $7$?
Let us apply the rule: $753-2(2)=753-4=749$.
Applying the rule again we get $74-2(9)=74-18=56$.
If we do not recognize that $56=7 \cdot 8$ we can apply the rule another time to get
$5-2(6)=5-12=-7$.
So $7$ divides $7532$.
Does $7$ divide $72$?
Well $7-2(2)=7-4=3$,
so $7$ does not divide $72$.
\item[8:] This is a little tricker to apply since you have to look at the
last three digits.
For example, does $8$ divide $3120$?
Since $8$ divides $120$, $8$ divides $3120$.
But you have to know that $8$ divides $120$ to apply this trick.
So normally I just divide the number by $2$ and
if I can do this three times the number is divisible by $8$.
Of course I would already know that since I have the quotient.
There are situations where this rule is actually useful for me.
First if the number ends in three $0$'s I know it is divisible by $8$.
For example, the number $1548000$ is divisible by $8$.
The other time I find this trick useful is if the third digit is a $0$.
Then I can just check to see if the last two digits are divisible by $8$.
As an example let us look at $547024$.
To check this number for divisibility by $8$ we look at $024$ or
as we usually write it $24$.
Since $24$ is divisible by $8$, so is $547024$.
\item[9:] Is $234$ divisible by $9$?
Yes since $2+3+4=9$ and $9$ is divisible by $9$.
Also $31275$ is divisible by $9$ since $3+1+2+7+5=18$ and $18$
is divisible by $9$ ($1+8=9$).
Let us try $573$.
Since $5+7+3=15$ and $15$ is not divisible by $9$,
$573$ is not divisible by $9$.
\item[10:] The following are all divisible by $10$ since they end in a $0$:
$20$, $25640$, $87951880$,and $603020105040708090201020$.
\item[11:] Does $11$ divide $143$?
To check we do the following calculation:
$(1+3)-4=4-4=0$.
Since $11$ divides $0$ it divides $143$.
To check to see if $11$ divides $2607$ we make the calculation:
$(2+0)-(6+7)=2-13=-11$.
Since $11$ divides $-11$ it divides $2607$.
Now try $38236$.
So $(3+2+6)-(8+3)=11-11=0$.
So $11$ divides $38236$.
Finally does $11$ divide $3042149$?
Well since $11$ divides $(3+4+1+9)-(0+2+4)=17-6=11$ it divides $3042149$.
\item[12:] Is $132$ divisible by $12$?
If it is it has to be divisible by both $3$ and $4$.
Since $1+3+2=6$ it is divisible by $3$ and since $32$ is divisible by $4$,
$132$ is divisible by $12$.
What about $264$?
Again we test for divisibility by $3$ and $4$.
Since $2+6+4=12$ is divisible by $3$ and $64$ is divisible by $4$,
$264$ is divisible by $12$.
Can we divide $1800$ by $12$?
First we add the digits of $1800$ to get $1+8+0+0=9$.
So $1800$ is divisible by $3$.
The last two digits of $1800$ are $00$,
so $1800$ is divisible by $4$.
Therefore $1800$ is divisible by $12$.
\item[13:] Is $65$ divisible by $13$?
By the rule we compute
$$6-9(5)=6-45=-39.$$
Since $39$ is divisible by $13$,
$65$ is divisible by $13$.
Lets try $377$.
$$37-9(7)=37-63=-26$$
So $377$ is divisible by $13$,
since $26$ is divisible by $13$.
Is $2041$ divisible by $13$?
$$204-9(1)=195$$
$$19-9(5)=19-45=-26$$
So $2041$ is divisible by $13$.
\end{description}
\end{ex}
\chapter{\textsc{foil}}
\label{foil}
Anytime you have the sum of two numbers multiplied by the sum of two numbers
you can use a trick called \textsc{foil} to do the multiplication.
\textsc{foil} is an acronym for: First, Outer, Inner, Last;
it is the order that you do the multiplication in.
Let us look at an example of \textsc{foil}ing
$$(a+b)(c+d)$$
\begin{description}
\item[First] The first item in each parenthesis is $a$ and $c$, so $ac$.
\item[Outer] The outer two items are $a$ and $d$, so $ad$.
\item[Inner] The inner two items are $b$ and $c$, so $bc$.
\item[Last] The last two items are $b$ and $d$, so $bd$.
\end{description}
Adding these together gives us $ac+ad+bc+bd$.
Therefore the \textsc{foil}ing $(a+b)(c+d)$ we get
$$(a+b)(c+d)=ac+ad+bc+bd$$
The reason that the \textsc{foil} trick works is that $(c+d)$ is a number.
So we can distribute the number $(c+d)$ over the sum $a+b$.
Therefore
$$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$
\chapter{Pies}
\begin{figure}[H]
\centering
\subfloat[$2$]{
\epsfig{file=app-one-half.eps,height=1.2in} }
\qquad
\subfloat[$3$]{
\epsfig{file=app-one-third.eps,height=1.2in} }
\qquad
\subfloat[$4$]{
\epsfig{file=app-one-fourth.eps,height=1.2in}}
\caption{Pie divided into $2$--$4$ pieces.}
%\label{frac-two-fifteenths}
\end{figure}
\begin{figure}[H]
\centering
\subfloat[$5$]{
\epsfig{file=app-one-fifth.eps,height=1.2in} }
\qquad
\subfloat[$6$]{
\epsfig{file=app-one-sixth.eps,height=1.2in} }
\qquad
\subfloat[$7$]{
\epsfig{file=app-one-seventh.eps,height=1.2in}}
\caption{Pie divided into $5$--$7$ pieces.}
%\label{frac-two-fifteenths}
\end{figure}
\begin{figure}[H]
\centering
\subfloat[$8$]{
\epsfig{file=app-one-eighth.eps,height=1.2in} }
\qquad
\subfloat[$9$]{
\epsfig{file=app-one-ninth.eps,height=1.2in}}
\qquad
\subfloat[$10$]{
\epsfig{file=app-one-tenth.eps,height=1.2in}}
\caption{Pie divided into $8$--$10$ pieces.}
%\label{frac-two-fifteenths}
\end{figure}
\backmatter
\printindex
\end{document}
$\displaystyle{}$
\begin{figure}[H]
\centering
\epsfig{file=,height=1in}
\caption{}
\label{}
\end{figure}
**