5.7: Multiplying Fractions
- Page ID
- 48771
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider the image in Figure \(\PageIndex{1}\), where the vertical lines divide the rectangular region into three equal pieces. If we shade one of the three equal pieces, the shaded area represents \(\frac{1}{3}\) of the whole rectangular region.
We’d like to visualize taking \(\frac{1}{2}\) of \(\frac{1}{3}\). To do that, we draw an additional horizontal line which divides the shaded region in half horizontally. This is shown in Figure \(\PageIndex{2}\). The shaded region that represented \(\frac{1}{3}\) is now divided into two smaller rectangular regions, one of which is shaded with a different color. This region represents \(\frac{1}{2}\) of \(\frac{1}{3}\).
Next, extend the horizontal line the full width of the rectangular region, as shown in Figure \(\PageIndex{3}\).
Note that drawing the horizontal line, coupled with the three original vertical lines, has succeeded in dividing the full rectangular region into six smaller but equal pieces, only one of which (the one representing \(\frac{1}{2}\) of \(\frac{1}{3}\)) is shaded in a new color. Hence, this newly shaded piece represents \(\frac{1}{6}\) of the whole region. The conclusion of our visual argument is the fact that \(\frac{1}{2}\) of \(\frac{1}{3}\) equals \(\frac{1}{6}\). In symbols,
\[ \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} .\nonumber \]
Multiplication Rule
In Figure \(\PageIndex{3}\), we saw that \(\frac{1}{2}\) of \(\frac{1}{3}\) equals \(\frac{1}{6}\). Note what happens when we multiply the numerators and multiply the denominators of the fractions \(\frac{1}{2}\) and \(\frac{1}{3}\).
\[ \begin{aligned} \frac{1}{2} \cdot \frac{1}{3} = \frac{1 \cdot 1}{2 \cdot 3} ~ & \textcolor{red}{ \text{ Multiply numerators; multiply denominators.}} \\ = \frac{1}{6} ~ & \textcolor{red}{ \text{ Simplify numerators and denominators.}} \end{aligned}\nonumber \]
We get \(\dfrac{1}{6}\)!
This example motivates the following definition.
Definition: Multiplication Rule
To find the product of the fractions \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\), multiply their numerators and denominators. In symbols,
\[ \frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}\nonumber \]
Example \(\PageIndex{1}\)
Multiply: \(\dfrac{1}{5} \cdot \dfrac{7}{9}\)
Solution
Multiply numerators and multiply denominators.
\[ \begin{aligned} \frac{1}{5} \cdot \frac{7}{9} = \frac{1 \cdot 7}{5 \cdot 9} ~ & \textcolor{red}{ \text{ Multiply numerators; multiply denominators.}} \\ = \frac{7}{45} ~ & \textcolor{red}{ \text{ Simplify numerators and denominators.}} \end{aligned}\nonumber \]
Try It \(\PageIndex{1}\)
Multiply: \(\dfrac{1}{3} \cdot \dfrac{2}{5}\)
- Answer
-
\( \dfrac{2}{15}\)
Multiply and Reduce
After multiplying two fractions, make sure your answer is reduced to lowest terms (see previous section).
Example \(\PageIndex{2}\)
Multiply \(\dfrac{3}{4} \cdot \dfrac{8}{9}\)
Solution
After multiplying, divide numerator and denominator by the greatest common divisor of the numerator and denominator.
\[ \begin{aligned} \frac{3}{4} \cdot \frac{8}{9} = \frac{3 \cdot 8}{4 \cdot 9} ~ & \textcolor{red}{ \text{ Multiply numerators and denominators.}} \\ = \frac{24}{36} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \\ = \frac{24 \div 12}{36 \div 12} ~ & \textcolor{red}{ \text{ Divide numerator and denominator by GCD.}} \\ = \frac{2}{3} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber \]
Alternatively, after multiplying, you can prime factor both numerator and denominator, then cancel common factors.
\[ \begin{aligned} \frac{3}{4} \cdot \frac{8}{9} = \frac{24}{36} ~ & \textcolor{red}{ \text{ Multiply numerators and denominators.}} \\ = \frac{2 \cdot 2 \cdot 2 \cdot 3}{2 \cdot 2 \cdot 3 \cdot 3} ~ & \textcolor{red}{ \text{ Prime factor numerator and denominator.}} \\ = = \frac{\cancel{2} \cdot \cancel{2} \cdot 2 \cdot \cancel{3}}{\cancel{2} \cdot \cancel{2} \cdot 3 \cdot \cancel{3}} ~ & \textcolor{red}{ \text{ Cancel common factors.}} \\ = \frac{2}{3} ~ \end{aligned}\nonumber \]
Try It \(\PageIndex{2}\)
Multiply: \(\dfrac{3}{7} \cdot \dfrac{14}{9}\)
- Answer
-
\(\dfrac{2}{3}\)
Multiply and Cancel or Cancel and Multiply
When you are working with larger numbers, it becomes a bit harder to multiply, factor, and cancel. Consider the following argument.
\[ \begin{aligned} \frac{18}{30} \cdot \frac{35}{6} = \frac{630}{180} ~ & \textcolor{red}{ \text{ Multiply numerators; multiply denominators.}} \\ = \frac{2 \cdot 3 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 2 \cdot 3 \cdot 3 \cdot 5} ~ & \textcolor{red}{ \text{ Prime factor numerators and denominators.}} \\ = \frac{ \cancel{2} \cdot \cancel{3} \cdot \cancel{3} \cdot \cancel{5} \cdot 7}{2 \cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{3} \cdot \cancel{5}} ~ & \textcolor{red}{ \text{ Cancel common factors.}} \\ = \frac{7}{2} ~ & \textcolor{red}{ \text{ Remaining factors.}} \end{aligned}\nonumber \]
There are a number of difficulties with this approach. First, you have to multiply large numbers, and secondly, you have to prime factor the even larger results.
One possible workaround is to not bother multiplying numerators and denominators, leaving them in factored form.
\[ \begin{aligned} \frac{18}{30} \cdot \frac{35}{6} = \frac{18 \cdot 35}{30 \cdot 6} ~ & \textcolor{red}{ \text{ Multiply numerators; multiply denominators.}} \end{aligned}\nonumber \]
Finding the prime factorization of these smaller factors is easier.
\[ \begin{aligned} = \frac{(2 \cdot 3 \cdot 3) \cdot (5 \cdot 7)}{(2 \cdot 3 \cdot 5) \cdot (2 \cdot 3)} ~ & \textcolor{red}{ \text{ Prime factor.}} \end{aligned}\nonumber \]
Now we can cancel common factors. Parentheses are no longer needed in the numerator and denominator because both contain a product of prime factors, so order and grouping do not matter.
\[ \begin{aligned} = \frac{ \cancel{2} \cdot \cancel{3} \cdot \cancel{3} \cdot \cancel{5} \cdot 7}{ \cancel{2} \cdot \cancel{3} \cdot \cancel{5} \cdot 2 \cdot \cancel{3}} ~ & \textcolor{red}{ \text{ Cancel common factors.}} \\ = \frac{7}{2} ~ & \textcolor{red}{ \text{ Remaining factors.}} \end{aligned}\nonumber \]
Another approach is to factor numerators and denominators in place, cancel common factors, then multiply.
\[ \begin{aligned} \frac{18}{30} \cdot \frac{35}{6} = \frac{2 \cdot 3 \cdot 3}{2 \cdot 3 \cdot 5} \cdot \frac{5 \cdot 7}{2 \cdot 3} ~ & \textcolor{red}{ \text{ Factor numerators and denominators.}} \\ = \frac{ \cancel{2} \cdot \cancel{3} \cdot \cancel{3}}{ \cancel{2} \cdot \cancel{3} \cdot \cancel{5}} \cdot \frac{ \cancel{5} \cdot 7}{2 \cdot \cancel{3}} ~ & \textcolor{red}{ \text{ Cancel common factors.}} \\ = \frac{7}{2} ~ & \textcolor{red}{ \text{ Remaining factors.}} \end{aligned}\nonumber \]
Note that this yields exactly the same result, \(\dfrac{7}{2}\).
Cancellation Rule
When multiplying fractions, cancel common factors according to the following rule: “Cancel a factor in a numerator for an identical factor in a denominator.”
Example \(\PageIndex{3}\)
Multiply: \(\dfrac{14}{15} \cdot \dfrac{30}{140}\)
Solution
Multiply numerators and multiply denominators. Prime factor, cancel common factors, then multiply.
\[ \begin{aligned} \frac{14}{15} \cdot \frac{30}{140} = \frac{14 \cdot 30}{15 \cdot 140} ~ & \textcolor{red}{ \text{ Multiply numerators; multiply denominators.}} \\ = \frac{(2 \cdot 7) \cdot (2 \cdot 3 \cdot 5)}{(3 \cdot 5) \cdot (2 \cdot 2 \cdot 5 \cdot 7)} ~ & \textcolor{red}{ \text{ Prime factor numerators and denominators.}} \\ = \frac{ \cancel{2} \cdot \cancel{7} \cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{5}}{ \cancel{3} \cdot 5 \cdot \cancel{2} \cdot \cancel{2} \cdot \cancel{5} \cdot \cancel{7}} ~ & \textcolor{red}{ \text{ Cancel common factors.}} \\ = \frac{1}{5} ~ & \textcolor{red}{ \text{ Multiply.}} \end{aligned}\nonumber \]
Note: Everything in the numerator cancels because you’ve divided the numerator by itself. Hence, the answer has a 1 in its numerator.
Try It \(\PageIndex{3}\)
Multiply: \( \dfrac{6}{35} \cdot \dfrac{70}{36} \)
- Answer
-
\(\dfrac{1}{3}\)
When Everything Cancels
When all the factors in the numerator cancel, this means that you are dividing the numerator by itself. Hence, you are left with a 1 in the numerator. The same rule applies to the denominator. If everything in the denominator cancels, you’re left with a 1 in the denominator.