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For example, $\frac{2}{3}$ and $\frac{4}{6}$ represent the same part of a whole quantity and are therefore equivalent. Several more collections of equivalent fractions are listed below.
$\frac{15}{25},\frac{12}{20},\frac{3}{5}$
$\frac{1}{3},\frac{2}{6},\frac{3}{9},\frac{4}{12}$
$\frac{7}{6},\frac{14}{12},\frac{21}{18},\frac{28}{24},\frac{35}{30}$
We can reduce a fraction to lowest terms by
Reduce each fraction to lowest terms.
$$\begin{array}{lllll}\frac{6}{18}\hfill & =\hfill & \frac{2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}3}{2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}3}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{\overline{)2}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}\overline{)3}}{\overline{)2}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}\overline{)3}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}3}\hfill & \hfill & \text{2\hspace{0.17em}and\hspace{0.17em}3\hspace{0.17em}are\hspace{0.17em}common\hspace{0.17em}factors}.\hfill \\ \hfill & =\hfill & \frac{1}{3}\hfill & \hfill & \hfill \end{array}$$
$$\begin{array}{lllll}\frac{16}{20}\hfill & =\hfill & \frac{2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}2}{2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{\overline{)2}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}\overline{)2}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}2}{\overline{)2}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}\overline{)2}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5}\hfill & \hfill & \text{2\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}only\hspace{0.17em}common\hspace{0.17em}factor}\text{.}\hfill \\ \hfill & =\hfill & \frac{4}{5}\hfill & \hfill & \hfill \end{array}$$
$$\begin{array}{lllll}\frac{56}{70}\hfill & =\hfill & \frac{2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}7}{2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}7}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{\overline{)2}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}\overline{)7}}{\overline{)2}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}\overline{)7}}\hfill & \hfill & \text{2\hspace{0.17em}and\hspace{0.17em}7\hspace{0.17em}are\hspace{0.17em}common\hspace{0.17em}factors}.\hfill \\ \hfill & =\hfill & \frac{4}{5}\hfill & \hfill & \hfill \end{array}$$
$$\begin{array}{l}\begin{array}{lllll}\frac{8}{15}\hfill & =\hfill & \frac{2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}2}{3\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5}\hfill & \hfill & \text{There\hspace{0.17em}are\hspace{0.17em}no\hspace{0.17em}common\hspace{0.17em}factors}.\hfill \end{array}\\ \text{Thus},\text{\hspace{0.17em}}\frac{8}{15}\text{\hspace{0.17em}is\hspace{0.17em}reduced\hspace{0.17em}to\hspace{0.17em}lowest\hspace{0.17em}terms}.\end{array}$$
Equally important as reducing fractions is
raising fractions to higher terms. Raising a fraction to higher terms is the process of constructing an equivalent fraction that has higher values in the numerator and denominator. The higher, equivalent fraction is constructed by multiplying the original fraction by 1.
Notice that
$\frac{3}{5}$ and
$\frac{9}{15}$ are equivalent, that is
$\frac{3}{5}=\frac{9}{15}.$ Also,
This observation helps us suggest the following method for raising a fraction to higher terms.
For example,
$\frac{3}{4}$ can be raised to
$\frac{24}{32}$ by multiplying both the numerator and denominator by 8, that is, multiplying by 1 in the form
$\frac{8}{8}.$
$$\frac{3}{4}=\frac{3\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}8}{4\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}8}=\frac{24}{32}$$
How did we know to choose 8 as the proper factor? Since we wish to convert 4 to 32 by multiplying it by some number, we know that 4 must be a factor of 32. This means that 4 divides into 32. In fact,
$32\xf74=8.$ We divided the original denominator into the new, specified denominator to obtain the proper factor for the multiplication.
Determine the missing numerator or denominator.
$\begin{array}{l}\begin{array}{lllll}\frac{3}{7}\hfill & =\hfill & \frac{?}{35}.\hfill & \hfill & \text{Divide\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator},\text{\hspace{0.17em}7},\text{\hspace{0.17em}into\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}denominator},\hfill \end{array}\text{\hspace{0.17em}}35.\text{\hspace{0.17em}}35\xf77=5.\\ \text{Multiply\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}numerator\hspace{0.17em}by\hspace{0.17em}5}.\\ \begin{array}{lllll}\frac{3}{7}\hfill & =\hfill & \frac{3\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5}{7\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5}\hfill & =\hfill & \frac{15}{35}\hfill \end{array}\end{array}$
$\begin{array}{l}\begin{array}{lllll}\frac{5}{6}\hfill & =\hfill & \frac{45}{?}.\hfill & \hfill & \text{Divide\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}numerator},\text{\hspace{0.17em}5},\text{\hspace{0.17em}into\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}numerator},\hfill \end{array}\text{\hspace{0.17em}}45.\text{\hspace{0.17em}}45\xf75=9.\\ \text{Multiply\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}by\hspace{0.17em}9}.\\ \begin{array}{lllll}\frac{5}{6}\hfill & =\hfill & \frac{5\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}9}{6\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}9}\hfill & =\hfill & \frac{45}{54}\hfill \end{array}\end{array}$
For the following problems, reduce, if possible, each fraction lowest terms.
$\frac{5}{10}$
$\frac{4}{14}$
$\frac{20}{8}$
$\frac{14}{4}$
$\frac{32}{28}$
$\frac{26}{60}$
$\frac{18}{27}$
$\frac{32}{40}$
$\frac{17}{51}$
$\frac{16}{42}$
$\frac{44}{11}$
$\frac{30}{105}$
For the following problems, determine the missing numerator or denominator.
$\frac{1}{3}=\frac{?}{12}$
$\frac{3}{3}=\frac{?}{9}$
$\frac{5}{6}=\frac{?}{18}$
$\frac{1}{2}=\frac{4}{?}$
$\frac{3}{2}=\frac{18}{?}$
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