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\lhead{\color{blue} Lecture Notes}
\chead{\color{black} \Large Fractions -- Part 5: Multiplying and  Dividing Fractions}
\rhead{\small page   \ \thepage}
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\lfoot{\footnotesize \copyright $\;$   Hidegkuti,   2018}
\rfoot{\footnotesize Last revised: September 23, 2018}
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\begin{document}


\begin{center}
{\Large Part 1 - Multiplying Fractions}
\end{center}

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\textbf{Theorem:} \ To multiply two fractions, we multiply one numerator by
the other and one denominator by the other. \ 

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{a}{%
b}\cdot \dfrac{c}{d}=\dfrac{a\cdot c}{b\cdot d}$

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\vspace{0.09in}

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We will refer to multiplying numerator by numerator and denominator by
denominator as mutiplying straight across.

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\textbf{Example 1. }\ Perform each of the multiplications as indicated.%
\vspace{0.05in}

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a) \ $\dfrac{3}{4}\cdot \dfrac{5}{12}$ \ \ \ \ \ \ \ \ \ \ \ b) \ $-2\cdot 
\dfrac{5}{4}$ \ \ \ \ \ \ \ \ \ c) \ $-3\dfrac{1}{3}\cdot 1\dfrac{1}{2}$\ \
\ \ \ \ \ \ \ \ \ \ d) \ $\left( \dfrac{2}{5}\right) ^{2}$\vspace{0.05in}

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\textbf{Solution:} \ a) \ We multiply straight across: numerator by
numerator and denominator by denominator. \ Then we reduce the fraction to
lowest terms before presenting our final answer.\vspace{0.05in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{4}\cdot \dfrac{5}{12}=%
\dfrac{3\cdot 5}{4\cdot 12}=\dfrac{15}{48}$\vspace{0.05in}

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The result is not in lowest terms. \ We reduce it before presentig it as our
answer by dividing both numerator and denominator by $3$.

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{%
15}{48}=\,$\fbox{$\dfrac{5}{16}$ }\vspace{0.05in}

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We often save time and effort by simplifying the fraction before performing
the multiplications. \ Consider the problem we just solved example. \ We
were able to divide both numerator and denominator by $3$, because the
numerator in $\dfrac{3}{4}$ and the denominator in $\dfrac{5}{12}$ are both
divisible by $3$. \ So, we can divide out by that $3$ before we perform the
multiplication. \ We just re-write $12$ as $3\cdot 4$ and cross out a factor
of $3$ from both numerator and denominator. \vspace{0.05in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{4}\cdot \dfrac{5}{12}=\dfrac{3\cdot
5}{4\cdot 12}=\dfrac{\NEG{3}\cdot 5}{4\cdot \NEG{3}\cdot 4}=\dfrac{5}{16}$%
\vspace{0.05in}\vspace{0.05in}

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\textbf{Crossing out the same factor from the products in both numerator and
denominator is the same as dividing both by that factor}. \ This is often
called cancellation.\vspace{0.05in}

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b) \ We can interpret integers as fractions with denominator $1$. \ If there
is a negative sign, we keep it with the numerator for all computations. \
Therefore, we will re-write $-2$ as $\dfrac{-2}{1}$.\vspace{0.05in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ $-2\cdot \dfrac{5}{4}=\dfrac{-2}{1}\cdot \dfrac{5}{4}%
=\dfrac{-2\cdot 5}{1\cdot 4}=\dfrac{-10}{4}=\,$\fbox{$-\dfrac{5}{2}$ }%
\vspace{0.05in}

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c) \ While it is possible to multiply mixed numbers, it takes a lot of work
and it will not be shown here. \ We can simply re-write mixed numbers as
improper fractions. \ If there is a negative sign, we keep it in the
numerator during computations.\vspace{0.05in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $-3\dfrac{1}{3}\cdot 1\dfrac{1}{2%
}=\dfrac{-10}{\NEG{3}}\cdot \dfrac{\NEG{3}}{2}=\dfrac{-10}{2}$\ $=\,$\fbox{$%
-5$}\vspace{0.05in}

\pagebreak

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d) \ If we square a fraction, we must use a pair of parentheses. \ Other
wise, $\dfrac{2}{5}^{2}$ can be interpreted as division between integers $%
\dfrac{2^{2}}{5}$. \ The parentheses guarantees that we all understand that
the entire fraction $\dfrac{2}{5}$ is being suared, and not just its
denominator. \ \vspace{0.05in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\left( \dfrac{2}{5}\right) ^{2}=%
\dfrac{2}{5}\cdot \dfrac{2}{5}$\ $=\,$\fbox{$\dfrac{4}{25}$}\vspace{0.05in}

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As we are defining operations on fractions, let us note that the notation
and order of operations agreement remain the same with fractions as they
were with integers. \ Some issues, such as the difference between $-3^{2}$
and $\left( -3\right) ^{2}$ will also remain.\vspace{0.04in}

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\textbf{Example 2. }\ Perform the indicated operations. \ \ \ \ $-\dfrac{2}{3%
}-\dfrac{4}{5}\left( -2\dfrac{6}{7}\right) $\vspace{0.05in}

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\textbf{Solution:} \ We see a multiplication and a subtraction. \ We start
with the multiplication, but first re-write the mixed\vspace{0.05in} number
as an improper fraction. \ We keep the negative sign in the numerator. \ $2%
\dfrac{6}{7}=\dfrac{20}{7}$ ad so $-2\dfrac{6}{7}=\dfrac{-20}{7}$.\vspace{%
0.05in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $-\dfrac{2}{3}-\dfrac{4}{5}\left( -2%
\dfrac{6}{7}\right) =-\dfrac{2}{3}-\dfrac{4}{5}\left( \dfrac{-20}{7}\right)
=-\dfrac{2}{3}-\dfrac{4\left( -20\right) }{5\cdot 7}=-\dfrac{2}{3}-\dfrac{%
4\left( -4\right) \cdot \NEG{5}}{\NEG{5}\cdot 7}=-\dfrac{2}{3}-\dfrac{-16}{7}
$\vspace{0.05in}

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Next we bring the fractions to the least common denominator and perform the
subtraction of signed numbers in the numerator. \ \ The least common
denominator is $21$.\vspace{0.05in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ $-\dfrac{2}{3}-\dfrac{-16}{7}=-\dfrac{14}{21}-\dfrac{-48}{21}=\dfrac{%
-14-\left( -48\right) }{21}=\dfrac{-14+48}{21}=\,$\fbox{$\dfrac{34}{21}$}%
\vspace{0.05in}\vspace{0.2in}

\begin{center}
{\Large Part 2 - Multiplying by One}
\end{center}

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It is a frequently occuring task in algebra to re-write an expression
without changing its value. \ There are just about two operations that
guarantee this in case of any quantity: we can always \textit{add zero} to
it or \textit{multiply} it \textit{by one}. \ (Actually, we can also
subtract zero and divide by one.) \ In abstract algebra, an element that
does not make any change under the operation to any quantity, is called an
identity element. \ Zero is called the \textbf{additive identity}, as it
does nothing in addition. The number $1$ is called the \textbf{%
multiplicative identity} because it does nothing in multiplication.

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For every real number\textbf{\ }$x$, \ \ $x+0=x$. \ This is the \textbf{%
additive identity} property of zero.

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For every real number\textbf{\ }$x$, \ \ $x\cdot 1=x$. \ This is the \textbf{%
multiplicative identity} property of one.

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\vspace{0.09in}

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Many algebraic techniques are based zero or multiplication (or division) by $%
1$. With fractions, multiplication by $1$ is extremely useful. \ We can even
explain the fundamental property of fractions in terms of multiplication by $%
1.$ \ Recall the fundamental property of fractions: we can multiply both
numerator and denominator by the same non-zero number and the value of the
fraction would remain the same. \ In the addition $\dfrac{2}{3}+\dfrac{1}{2}$%
, we would need to re-write $\dfrac{2}{3}$ with a denominator of $6$.%
\begin{equation*}
\dfrac{2}{3}=\dfrac{2}{3}\cdot 1=\dfrac{2}{3}\cdot \dfrac{2}{2}=\dfrac{%
2\cdot 2}{3\cdot 2}=\dfrac{4}{6}
\end{equation*}

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\textbf{Theorem:} \ The fractions $-\dfrac{2}{3}$, $\dfrac{-2}{3}$, and $%
\dfrac{2}{-3}$ all have the same value. \ \vspace{0.04in}

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\textit{Proof.} \ We already knew that $\dfrac{-2}{3}$ \ and $\dfrac{2}{-3}$
are\ \vspace{0.04in} equivalent fractions. \ We can get from one to the
other one by multiplying both numerator and denominator by $-1$.\ \ As a
general habit, we should rarely tolerate a negative denominator.\vspace{%
0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ $\dfrac{2}{-3}=\dfrac{2\cdot \left( -1\right) }{-3\cdot \left(
-1\right) }=\dfrac{-2}{3}$\vspace{0.04in}

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But only now can we connect $\dfrac{-2}{3}$ to $-\dfrac{2}{3}$.\ \vspace{%
0.04in}

$-\dfrac{2}{3}$ is the opposite of $\dfrac{2}{3}$. \ That means
multiplication by $-1$, and we will express $-1$ as $\dfrac{-1}{1}$.\vspace{%
0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $-\dfrac{2}{3}=-1\cdot \dfrac{2%
}{3}=\dfrac{-1}{1}\cdot \dfrac{2}{3}=\dfrac{-1\cdot 2}{1\cdot 3}=\dfrac{-2}{3%
}$ \ \ \ \ \ \ \ $\blacksquare $ \ (End of Proof)\vspace{0.04in}\vspace{%
0.04in}

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The negative sign is not the only thing that can freely move in a fraction.
\ Other factors can only move between being a factor in the numerator to
being a multiplyer of a fraction. \ 

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\textbf{Theorem:} \ The algebraic expressions $\dfrac{3}{5}x$ and $\dfrac{3x%
}{5}$ are equivalent. \ \vspace{0.04in}

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\vspace{0.09in}

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\textit{Proof.} \ The key here is that we can re-write $x$ as $\dfrac{x}{1}$%
. \ Division by one never changes the value of a number.

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{5}x=%
\dfrac{3}{5}\cdot x=\dfrac{3}{5}\cdot \dfrac{x}{1}=\dfrac{3x}{5}$. \ $\ \ \
\ \ \ \ \blacksquare $ \ (End of Proof)\vspace{0.04in}\vspace{0.04in}\vspace{%
0.04in}\vspace{0.04in}\vspace{0.04in}\vspace{0.04in}\vspace{0.04in}\vspace{%
0.04in}

Units behave the same way as $x$ does. \ Kilogramm is a measure of mass, 1
kilogram is about $2.2$ pounds. \ If we measure mass in kilograms, (denoted
by $\unit{kg}$), we can have the unit either in the numerator, or after the
number. \ For example, \ $\dfrac{2}{3}\unit{kg}$ is the same as $\dfrac{2%
\unit{kg}}{3}$. \ Although these fractions express slightly different
things, they are equaivalent, and we can freely move between the two forms.
\ Inside computations we often prefer $\dfrac{2\unit{kg}}{3}$ and we would
present the final answer as $\dfrac{2}{3}\unit{kg}$.\bigskip

\pagebreak

\begin{center}
{\Large Part 3 - The Reciprocal}
\end{center}

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\textbf{Example 3.} \ Perform the multiplication \ $\dfrac{3}{8}\cdot \left(
2\dfrac{2}{3}\right) $.

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Solution: \ Before multiplying stratight across, we re-write the mixed
number as an improper fraction.\vspace{0.05in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{8}\cdot \left( 2\dfrac{2}{3}\right) =%
\dfrac{3}{8}\cdot \dfrac{8}{3}=\dfrac{24}{24}=\,$\fbox{$1$}\vspace{0.05in}

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\textbf{Definition:} \ If the product of two numbers is $1$, we call\vspace{%
0.04in}\vspace{0.04in} such a pair of numbers \textbf{reciprocal}s of each
other. \ For example, the reciprocal of $\dfrac{2}{3}$ is $\dfrac{3}{2}$,%
\vspace{0.04in}\vspace{0.04in} and the reciprocal of $\dfrac{3}{2}$ is $%
\dfrac{2}{3}$. \ The reciprocal is also called the \textbf{multiplicative
inverse}.\vspace{0.04in}

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\vspace{0.09in}

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There is an obvious symmetry between $\dfrac{2}{3}$ and $\dfrac{3}{2}$. 
\vspace{0.04in}\vspace{0.04in} It is easy to see why those two fractions
would multiply to $1$. \ This is because the product of them is\vspace{0.04in%
}\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{a}{b}\cdot 
\dfrac{b}{a}=\dfrac{ab}{ab}=1$. \ \vspace{0.04in}\vspace{0.04in}

So, we simply flip a fraction upside down to get to its reciprocal.\vspace{%
0.06in} \ However, we should always remember that the definition of the
reciprocal of a number $x$ is another number so that their product is $1$.%
\vspace{0.04in}\vspace{0.04in}

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\textbf{Example 4.} \ Find the reciprocal for each of the following.\vspace{%
0.04in}

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a) \ $\dfrac{2}{5}$ \ \ \ \ \ \ b) \ $-\dfrac{3}{7}$\ \ \ \ \ \ \ \ c) \ $2$
\ \ \ \ \ \ \ \ d) \ $x$ \ \ \ \ \ \ \ e) \ $1\dfrac{3}{5}$\vspace{0.04in}%
\vspace{0.04in}

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\textbf{Solution:} \ a) \ To find the reciprocal of \ $\dfrac{2}{5}$, we
need to flip it upside down.\vspace{0.04in} \ So the reciprocal of $\dfrac{2%
}{5}$ is \fbox{$\dfrac{5}{2}$}. \ Indeed, the product of these two fractions
is $\dfrac{2}{5}\cdot \dfrac{5}{2}=\dfrac{10}{10}=1$. \vspace{0.04in}\vspace{%
0.04in}

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b) \ How does the flipping work with the negative sign? \ The short story is
that the reciprocal of $\ -\dfrac{3}{7}$ \ is \ \fbox{$-\dfrac{7}{3}$}. \ 
\vspace{0.04in}

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The long story is that we can re-write $-\dfrac{3}{7}$ as $\dfrac{-3}{7}$.%
\vspace{0.04in} \ Then we flip for the reciprocal: we get from $\dfrac{-3}{7}
$ to $\dfrac{7}{-3}.$ \ Negative signs are not acceptable in the
denominator, so we\vspace{0.04in} immediately lift it up to the numerator. \
(Just mutiply numerator and denominator by $-1$.) \ So the reciprocal of $%
\dfrac{-3}{7}$ is $\dfrac{7}{-3}$, which is $\dfrac{-7}{3}$. \ That's the
same as \ $-\dfrac{7}{3}$.

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ The\ \ reciprocal of $-%
\dfrac{3}{7}$ is the flip of $\dfrac{-3}{7}.$ \ That\ is $\dfrac{7}{-3}=%
\dfrac{-7}{3}=\,$\fbox{$-\dfrac{7}{3}$}\vspace{0.04in}\vspace{0.04in}

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This algebraic gymnastics with the negative sign works, but we can also
think of reciprocals in terms of the definition. \ A number and its
reciprocal multiply to $1$. \ That's a positive product. \ Therefore, the
reciprocal of a negative number is negative so that the product can be
positive. \ For the reciprocal of $-\dfrac{3}{7}$, we do two things:\ flip
the fraction $\dfrac{3}{7}$ and decide that the reciprocal of $-\dfrac{3}{7}$
must be negative. \ The reciprocal is then $-\dfrac{7}{3}$.

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c) \ How do we take the reciprocal of something that is not necessarily a
fraction?\vspace{0.04in} \ \ We can re-write $2$ as $\dfrac{2}{1}$ and then
we can flip. \ The reciprocal of $2$ is the same as the reciprocal of $%
\dfrac{2}{1}$, which is \fbox{$\dfrac{1}{2}$}.\vspace{0.04in}\vspace{0.04in}

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d) \ Let $x$ be any non-zero number. \ We re-write it as $\dfrac{x}{1}$ and
flip. \ So the reciprocal of $x$ is \fbox{$\dfrac{1}{x}$}\vspace{0.04in}. \
This expression is meaningful as long as $x$ is not zero. \ \ Therefore,
every non-zero number has a unique \vspace{0.04in}reciprocal and is denoted
by $\dfrac{1}{x}$. \ To check if $x$ and $\dfrac{1}{x}$ are really
reciprocals, we use the definition. \ The \vspace{0.04in}product of a number
and its reciprocalis $1$. \ Indeed,\vspace{0.04in}\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ $x\cdot \dfrac{1}{x}=\dfrac{x}{1}\cdot \dfrac{1}{%
x}=\dfrac{x}{x}=1$ \ \ \ \ \ This is\ true as long as $x$ is not zero.%
\vspace{0.04in}\vspace{0.04in}\vspace{0.04in}

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e) \ We can only find the reciprocal of a mixed number by converting it to an%
\vspace{0.04in} improper fraction and then flip it. \ \ $1\dfrac{3}{5}=%
\dfrac{8}{5},$ so

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
the reciprocal of $1\dfrac{3}{5}$ is the reciprocal of $\dfrac{8}{5}$, which
is \fbox{$\dfrac{5}{8}$}. \ \vspace{0.04in}\vspace{0.04in}\vspace{0.04in}

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Notice that separately flipping the integer- and fraction parts in a mixed
number does not pruduce the right results: $1\dfrac{3}{5}$ \ and $1\dfrac{5}{%
3}$ are not reciprocals. \ Their product is \vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $1\dfrac{%
3}{5}\cdot 1\dfrac{5}{3}=\dfrac{8}{5}\cdot \dfrac{8}{3}=\dfrac{64}{15}$ and
not $1$.\vspace{0.04in}

To get the reciprocal, we \textit{have to} convert mixed numbers to improper
fractions.\ \vspace{0.25in}

\pagebreak

\begin{center}
{\Large Part 4 - Dividing Fractions}
\end{center}

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As it turns out, we never divide fractions. \ Every non-zero number has \ a
reciprocal. \ We interpret division as multiplication by the reciprocal.

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\textbf{Theorem:} \ \textbf{To divide is to multiply by the reciprocal.}%
\vspace{0.04in}\vspace{0.04in}

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\vspace{0.09in}

\textit{Proof}. \ Let $x$ be any non-zero number. Let us multiply $y$ by $%
\dfrac{1}{x}$.

\ \ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y\cdot \dfrac{1}{x}=\dfrac{y}{1}%
\cdot \dfrac{1}{x}=\dfrac{y\cdot 1}{1\cdot x}=\dfrac{y}{x}$ \ \ is the same
as division by $x$. \ $\blacksquare \vspace{0.08in}$

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\textbf{Example 5. }\ Perform the divisions of the fractions as indicated.%
\vspace{0.04in}

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a) \ $\dfrac{5}{6}\div \dfrac{2}{3}$ \ \ \ \ \ \ \ \ b) \ $\dfrac{2}{5}\div
\left( -4\right) $ \ \ \ \ \ \ \ \ c) \ $4\dfrac{1}{5}\div 1\dfrac{1}{6}$ \
\ \ \ \ \ \ d) \ $-2\div \left( -\dfrac{3}{8}\right) $\vspace{0.04in}

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\textbf{Solution:} \ a) \ To \ divide is to multiply by the reciprocal.
Instead of dividing by $\dfrac{2}{3}$, we will multiply by its reciprocal, $%
\dfrac{3}{2}$.\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{5}{6}\div \dfrac{2}{3}=%
\dfrac{5}{6}\cdot \dfrac{3}{2}=\dfrac{5\cdot 3}{6\cdot 2}=\dfrac{5\cdot \NEG%
{3}}{2\cdot \NEG{3}\cdot 2}=\,$\fbox{$\dfrac{5}{4}$}\vspace{0.04in}

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b) \ To \ divide is to multiply by the reciprocal. Instead of dividing by $%
-4 $\vspace{0.04in}, we multiply by its reciprocal.\vspace{0.04in} $-4=%
\dfrac{-4}{1}$, therefore, its reciprocal is $\dfrac{1}{-4}$.\vspace{0.04in}
\ However, we immediately lift the negative sign and as always, we keep it
in the numerator. \ Thus, the reciprocal of $-4$ is $-\dfrac{1}{4}$ and we
will write it as $\dfrac{-1}{4}$.\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{2}{5}\div \left( -4\right) =%
\dfrac{2}{5}\cdot \dfrac{-1}{4}=\dfrac{-2}{20}=\,$\fbox{$-\dfrac{1}{10}$}%
\vspace{0.04in}

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c) \ We re-write the subtraction with one big fraction bar and a subtraction
of integers the same numerator. \ Then, if we can, we simplify the answer.%
\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{7}-%
\dfrac{9}{7}=\dfrac{3-9}{7}=\,$\fbox{$\dfrac{-6}{7}$ or $-\dfrac{6}{7}$}%
\vspace{0.04in}

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d) \ We can not take the reciprocal of a mixed number. \ Therefore, we
immediately convert each mixed number to an improper fraction, and instead
of division, we multiply by the reciprocal.\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ $4\dfrac{1}{5}\div 1\dfrac{1}{6}=\dfrac{21}{5}\div 
\dfrac{7}{6}=\dfrac{21}{5}\cdot \dfrac{6}{7}=\dfrac{\left( \NEG{7}\cdot
3\right) \cdot 6}{5\cdot \NEG{7}}=\,$\fbox{$\dfrac{18}{5}$ }\vspace{0.04in}

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\textbf{Example 6. }\ a) \ Perform the division $10\div \dfrac{1}{4}$ \ \ \
\ \ \ \ \ b) \ How many quarters are there in a $10-$dollar roll of quarters?%
\vspace{0.05in}

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\textbf{Solution:} \ a) \ To divide is to multiply by the reciprocal. \ The
reciprocal of $\dfrac{1}{4}$ is $\dfrac{4}{1}$ or $4$.

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ $10\div \dfrac{1}{4}=10\cdot 4=\,$\fbox{$40$ }\vspace{0.04in}

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b) \ Each dollar can be exchanged for four quarters, so ten dollars would be
the same as ten times four, or forty quarters.

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From this last example we see that division still has its same old meaning:
how many times can we fit $\dfrac{1}{4}$ into $10$? \ \vspace{0.04in}\vspace{%
0.04in}

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\textbf{Example 7. }\ Simplify the given expression: $\dfrac{~\dfrac{2}{3}+%
\dfrac{3}{4}~}{\dfrac{5}{6}-\dfrac{1}{3}}$\vspace{0.04in}\vspace{0.04in}

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\textbf{Solution: \ }Recall that if a fraction bar (or division bar) is
stretching under entire expressions, then it also serves as (invisible)
parentheses. \ \ Thus \vspace{0.04in}\vspace{0.04in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $%
\dfrac{~\dfrac{2}{3}+\dfrac{3}{4}~}{\dfrac{5}{6}-\dfrac{1}{3}}$ \ \ is the
same as $\left( \dfrac{2}{3}+\dfrac{3}{4}\right) \div \left( \dfrac{5}{6}-%
\dfrac{1}{3}\right) $\vspace{0.04in}\vspace{0.04in}

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Therefore, we will first perform the addition in the numerator and the
subtraction in the denominator. \ We will finally divide. \ For additions
and subtractions, we need to use a common denominator.

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{~\dfrac{2}{3}+\dfrac{3}{4}~%
}{\dfrac{5}{6}-\dfrac{1}{3}}=\dfrac{~\dfrac{8}{12}+\dfrac{9}{12}~}{\dfrac{5}{%
6}-\dfrac{2}{6}}=\dfrac{~\dfrac{8+9}{12}~}{\dfrac{5-2}{6}}=\dfrac{~\dfrac{17%
}{12}~}{\dfrac{3}{6}}=\dfrac{17}{12}\cdot \dfrac{6}{3}=\dfrac{17\cdot 6}{%
\left( 6\cdot 2\right) \cdot 3}=\dfrac{17}{2\cdot 3}=\,$\fbox{$\dfrac{17}{6}$
}\vspace{0.04in}\vspace{0.04in}\vspace{0.04in}

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\textbf{Example 8. }\ Perform each of the given divisions. \ \ \ \ \ \ \ \ \
\ a) \ $\dfrac{~~\dfrac{2}{3}~~}{5}$\vspace{0.04in}\ \ \ \ \ \ b) \ $\dfrac{%
~~2~~}{\dfrac{3}{5}}\vspace{0.04in}$

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\textbf{Solution: \ }a) \textbf{\ }If needed, we can always re-write an
integer with a denominator of $1$.$\vspace{0.04in}\vspace{0.04in}$\ \ \ \ \
\ \ \ \ $\dfrac{~~\dfrac{2}{3}~~}{5}=\dfrac{~~\dfrac{2}{3}~~}{\dfrac{5}{1}}=%
\dfrac{2}{3}\cdot \dfrac{1}{5}=\,\fbox{$\dfrac{17}{6}$ }\vspace{0.04in}$

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b) \ \ This time we will need to re-write $2$ as a fraction.$\vspace{0.04in}%
\vspace{0.04in}$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{~~2~~}{\dfrac{3}{5}}%
\vspace{0.04in}=\dfrac{~~\dfrac{2}{1}~~}{\dfrac{3}{5}}=\dfrac{2}{1}\cdot 
\dfrac{5}{3}=\,\fbox{$\dfrac{10}{3}$ }\vspace{0.04in}$.

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As the previous example shows, $\ \dfrac{~~\dfrac{a}{b}~~}{c}$ \ and \ $%
\dfrac{a}{~~\dfrac{b}{c}~~}$ are different \ expressions, with different
values. \ Because of this, we have to be careful with expressions with more
than one fraction bar (or division bar). \ Order of operations must be
clearly indicated by the different sizes of the fraction bars. \ \ A
fraction or division problem with more than one fraction bar is called a 
\textbf{complex fraction.} Complex fractions can always be simplified to a
form with just one fraction bar.

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\textbf{Example 9. }\ Simplify each of the given expressions. \ \ \ \ \ \ \
\ \ \ a) \ $\dfrac{~~\dfrac{a}{b}~~}{c}$\vspace{0.04in}\ \ \ \ \ \ b) \ $%
\dfrac{~~a~~}{\dfrac{b}{c}}\vspace{0.04in}\vspace{0.04in}$

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\textbf{Solution:\vspace{1in} }%
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\ a) \ In this case, we need to re-write $c$ as $\dfrac{c}{1}$.$\vspace{%
0.04in}$

Then we multiply by the reciprocal.$\vspace{0.04in}\vspace{0.04in}\vspace{%
0.04in}$

\ \ \ \ \ \ $\dfrac{~~\dfrac{a}{b}~~}{c}=\dfrac{~~\dfrac{a}{b}~~}{\dfrac{c}{1%
}}=\dfrac{a}{b}\cdot \dfrac{1}{c}=\,\fbox{$\dfrac{a}{bc}$}\vspace{0.04in}$%
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b) \ In this case, we need to re-write $a$ as $\dfrac{a}{1}$.$\vspace{0.04in}
$

Then we multiply by the reciprocal.$\vspace{0.04in}\vspace{0.04in}\vspace{%
0.04in}$

$\ \ \ \ \dfrac{~~a~~}{\dfrac{b}{c}}=\dfrac{~~\dfrac{a}{1}~~}{\dfrac{b}{c}}=%
\dfrac{a}{1}\cdot \dfrac{c}{b}=\,\fbox{$\dfrac{ac}{b}$}$%
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\pagebreak

\begin{center}
{\Large Part 5 - An Application: \ Conversion Factors}
\end{center}

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In physics, in some applications of mathematics, and even in every day life,
we dealing with numbers with units. \ For example, length can not be
determined or communicated by a number only. \ $1$ mile is much much longer
than $1$ inch. \ Also, $5$ minutes is much shorter than $5$ years. \ 

The cancellation of factors in numerator and denominator of fractions can be
used to convert a quantity from one unit to another one.\ \ Recall that we
never change the value of any number if we multiply it by $1$. \ For this
reason, $1$ is sometimes called the \textbf{multiplicative identity}. \ 

Consider now the statement $1$ hour$\,=\,60$ minutes. \ In physics, hour is
denoted by $\unit{h}$ and minutes by $\unit{min}$. If we divide any non-zero
quantity by itself, the result is $1$. \ Thus, using the equality $1\unit{h}%
=60\unit{min}$, we can write two fractions, both of value $1$.\vspace{0.04in}%
\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{1\unit{h}}{60\unit{%
min}}$ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ $\dfrac{60\unit{min}}{1\unit{h}}$ \ 
\vspace{0.04in}\vspace{0.04in}

These two fractions are called \textbf{conversion factors} or unit
multiplyers, and we use them to converting time measurements from minutes to
hours or backward, from hours to minutes.

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\textbf{Example 10. }\ Convert $\dfrac{7}{12}$ hours to minutes.\vspace{%
0.04in}

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\textbf{Solution:} \ We will re-write $\dfrac{7}{12}$ hours as $\dfrac{7%
\unit{h}}{12}$\vspace{0.04in}\vspace{0.04in} and multiply it by one of the
conversion factors shown above. \ Since we would like to get rid of \textit{%
hours}, we will select the conversion factor that has \textit{hour} in its
denominator.\vspace{0.04in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ $\dfrac{7}{12}\unit{h}=\dfrac{7\unit{h}}{12}\cdot 1=\dfrac{7\unit{h}}{12}%
\cdot \dfrac{60\unit{min}}{1\unit{h}}$\ \vspace{0.04in}\vspace{0.04in}

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Now the \textit{hour} unit is a factor in both numerator and denominator, so
we can cancel it out, and we will be left with just minutes in the numerator.%
\vspace{0.04in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{7\unit{h}}{12}\cdot \dfrac{60\unit{min}}{1%
\unit{h}}=\dfrac{7\cdot 60\unit{min}}{12}=\dfrac{7\cdot \left( 5\cdot
12\right) \unit{min}}{12}=\dfrac{35\unit{min}}{1}=\,$\fbox{$35\unit{min}$}%
\vspace{0.04in}\vspace{0.04in}

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\textbf{Example 11. }\ Convert $9000$ minutes to hours.\vspace{0.04in}

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\textbf{Solution:} \ We first re-write $9000$ minutes as a fraction, and
keep the unit with the denominator. \ If needed, we can always write a
denominator $1$. \ Then we multiply it by $1$.\vspace{0.04in}\vspace{0.04in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ $9000\ \unit{min}=\dfrac{9000\ \unit{min}}{1}=\dfrac{9000\ \unit{min}}{1}%
\cdot 1$\vspace{0.04in}\vspace{0.04in}

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The fraction expressing $1$ is going to be the conversion factor that has
minutes in its denominator. \ The minutes unit is cancelled out, and we are
left with hours instead.\vspace{0.04in}\vspace{0.04in}

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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ $9000\ \unit{min}=\dfrac{9000\ \unit{min}}{1}\cdot \dfrac{1\unit{h}}{60%
\unit{min}}=\dfrac{9000\unit{h}}{60}=\dfrac{150\unit{h}}{1}=\,$\fbox{$150$ $%
\unit{h}$}\vspace{0.04in}\vspace{0.04in}

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Conversion factors can also be used in groups.\vspace{0.04in}

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\textbf{Example 12. }\ Convert $10$ years to seconds.\vspace{0.04in}

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\textbf{Solution:} \ We will go from years to days, then from days to hours,
then to minutes and finally to seconds. \ The entire computation can be done
in a single line. \ \ Let us assume that a year is $365$ days long.\vspace{%
0.04in}\vspace{0.04in}

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$\ \ \ \ \ \ \ \ \ \ \ \ \ 10\unit{y}=\dfrac{10\unit{y}}{1}=\dfrac{10\unit{y}%
}{1}\cdot \dfrac{365\unit{d}}{1\unit{y}}\cdot \dfrac{24\unit{h}}{1\unit{d}}%
\cdot \dfrac{60\unit{min}}{1\unit{h}}\cdot \dfrac{60\unit{s}}{1\unit{min}}=%
\dfrac{10\cdot 365\cdot 24\cdot 60\cdot 60\unit{s}}{1}=\,\fbox{$315\,360\,000%
\unit{s}$}$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vspace{0.2in}

\pagebreak

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\ \ {\large Practice Problems}

\begin{enumerate}
\item Perform the indicated operations. \ Present your answer as an integer
or a reduced fraction. \ You do not need to convert improper numbers to
mixed numbers.%
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\begin{enumerate}
\item[a)] $-\dfrac{8}{15}\cdot \dfrac{3}{4}\vspace{0.09in}$

\item[b)] $\left( -\dfrac{1}{2}\right) ^{2}\vspace{0.09in}$

\item[c)] $\dfrac{4}{15}\cdot \left( 2\dfrac{1}{7}\right) \vspace{0.09in}$

\item[d)] $-\left( -\dfrac{2}{3}\right) ^{2}$ $\vspace{0.09in}$

\item[e)] $\dfrac{1}{6}-\dfrac{8}{25}\left( -1\dfrac{2}{3}\right) \vspace{%
0.09in}$

\item[f)] $\dfrac{2}{7}-\dfrac{2}{3}\div \left( 1\dfrac{2}{5}\right) \vspace{%
0.09in}$

\item[g)] $-\dfrac{5}{6}-\left( 3-\dfrac{2}{5}\right) \left( -\dfrac{1}{2}%
\right) \vspace{0.09in}$

\item[h)] $\left( 3\dfrac{1}{3}\right) \div \left( 2\dfrac{1}{2}\right) 
\vspace{0.09in}$

\item[i)] $\dfrac{\dfrac{5}{4}-\left( \dfrac{1}{6}\right) ^{2}}{\left( -1%
\dfrac{1}{3}\right) -\dfrac{1}{2}}\vspace{0.09in}$

\item[j)] $\dfrac{1-\dfrac{2}{3}}{2-\dfrac{1}{3}}\vspace{0.09in}$
\end{enumerate}

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\item Find the perimeter and area of the object shown on the picture.

\item Evaluate the expression $x^{2}-6x-1$ \ if\vspace{0.06in}

a) \ $x=-\dfrac{1}{2}$ \ \ \ \ \ b) \ $x=\dfrac{5}{2}$ \ \ \ \ \ c) \ $x=-%
\dfrac{2}{3}$\vspace{0.06in}

\item Evaluate the expression \ \ $\dfrac{-2x^{2}+x+3}{2x-3}$ \ if \vspace{%
0.06in}

a) \ $x=\dfrac{1}{2}$ \ \ \ b) \ $x=\dfrac{3}{2}$ \ \ \ \ c) \ $x=-\dfrac{2}{%
3}$ \ \ \ \ d) \ $x=-\dfrac{5}{6}$%
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\ 

\item Perform each of the following conversions.

\begin{enumerate}
\item[a)] $6000$ inches to feet, given that $12\unit{in}=1\unit{ft}$

\item[b)] $900\,000$ square-inches to square-feet. \ (Hint: $1\unit{ft}^{2}=1%
\unit{ft}\cdot 1\unit{ft}$)

\item[c)] $81\dfrac{\unit{mi}}{\unit{h}}$ (miles per hour) to meter per
second $\left( \dfrac{\unit{m}}{\unit{s}}\right) $. \ Hint: \ $1\unit{mi}%
\approx $ $1600\unit{m}$
\end{enumerate}
\end{enumerate}

\vspace{0.2in}

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\ \ \ \ 
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{\LARGE Enrichment}

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\begin{enumerate}
\item (Enrichment) \ Contributed by Prof. Abdallah Shuaibi. \ 

Two travelers meet a third one, who is very hungry. He offers the two
travelers 8 dollars for a meal. One traveler has three pieces of bread, the
other one has five. So the hungry man gives them the 8 dollars, they all sit
down and eat all 8 pieces of bread together. \ Afterwards, the two get into
an argument about how to divide up the money. The one who contributed 5
pieces of bread wants to split it to 5 and 3. The other wants to divide the
money evenly, 4 and 4. They go to a wise man for advice. They tell him their
story and ask him to divide the money between them. The wise man gives the
man who had 3 pieces of bread 1 dollar and 7 to the man with 5 pieces of
bread. Is this a just or even reasonable decision?%
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\end{enumerate}

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\ \ \ {\large Answers}

\begin{enumerate}
\item[1.] a)\ \ $-\dfrac{2}{5}$ \ \ \ \ \ \ b) \ $\dfrac{1}{4}$ \ \ \ \ \ c)
\ $\dfrac{4}{7}$ \ \ \ \ d) \ $-\dfrac{4}{9}$ \ \ \ \ \ e) \ $\dfrac{7}{10}$
\ \ \ \ \ f) \ $-\dfrac{4}{21}$ \ \ \ \ \ g) \ $\dfrac{7}{15}$ \ \ \ \ h) \ $%
\dfrac{4}{3}$ \ \ \ \ \ i) \ $-\dfrac{2}{3}$ \ \ \ \ j) \ $\dfrac{1}{5}$

\item $P=\dfrac{62}{3}\unit{in}$ \ \ $A=\dfrac{148}{9}\unit{in}^{2}$ \ \ \ \
\ \ \ \ \ \ \ 3. \ a) \ $\dfrac{9}{4}$ \ \ \ \ \ b) \ $-\dfrac{39}{4}$ \ \ \
\ c) \ $\dfrac{31}{9}$

\item[4.] a) \ $-\dfrac{3}{2}$ \ \ \ \ \ b) \ undefined \ \ \ \ d) \ $-%
\dfrac{1}{3}$ \ \ \ \ \ \ d) \ $-\dfrac{1}{6}$ \ \ \ \ \ \ \ \ \ \ \ 5. \ a)
\ $500\unit{ft}$ \ \ \ \ b) \ $6250\unit{ft}^{2}$ \ \ \ \ \ c) \ $36\dfrac{%
\unit{m}}{\unit{s}}$\vspace{0.26in}\vspace{4in}\vspace{0.2in}

\vspace{1.5in}\vspace{0.75in}
\end{enumerate}

{\Large 
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}

\href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{%
For more documents like this, visit our page at\
https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail
questions or comments to mhidegkuti@ccc.edu.}

\end{document}