Principle 1.24. Multiplying Fractions.
\begin{equation*}
\frac{a}{b}\cdot \frac{c}{d} = \frac{a\cdot c}{b\cdot d}.
\end{equation*}
Section 1.6 Working with Fractions
To multiply two fractions, just multiply the two numerators and the two denominators.
Example 1.25. Multiplying Fractions.
Multiply \(5/6\) and \(2/5\text{.}\)
\begin{equation*}
\frac{5}{6}\cdot\frac{2}{5} = \frac{10}{30} = \frac{10}{3\cdot 10} = \frac{1}{3}.
\end{equation*}
Of course, we can cancel any common factors as soon as we notice them:
\begin{equation*}
\frac{5}{6}\cdot\frac{2}{5} = \frac{1}{6}\cdot\frac{2}{1} = \frac{2}{6} = \frac{2}{2\cdot 3} = \frac{1}{3}.
\end{equation*}
Warning 1.26. Cancellation in Fractions.
You may cancel common factors in a fraction. For instance, the \(b\)’s cancel in the expression
\begin{equation*}
\frac{ab}{cb} = \frac{a}{c}.
\end{equation*}
However, you must never cancel common terms in a fraction. For instance, the \(b\)’s in the expression \(\displaystyle \frac{a+b}{b}\) do NOT cancel as this would violate PEMDAS Principle 1.9 and Example 1.21.
To divide two fractions, we can multiply by the reciprocal of the denominator.
Principle 1.27. Dividing Fractions.
\begin{equation*}
\frac{a}{b} \div \frac{c}{d} = \frac{a/b}{c/d} = \frac{a}{b} \cdot \frac{d}{c}.
\end{equation*}
Example 1.28. Dividing Fractions.
Divide \(3/2\) by \(3/4\text{.}\)
\begin{equation*}
\frac{3}{2} \div \frac{3}{4} = \frac{3/2}{3/4} = \frac{3}{2} \cdot \frac{4}{3} = \frac{4}{2} = 2.
\end{equation*}
Addition may be performed when the denominators are the same.
Principle 1.29. Adding Fractions.
\begin{equation*}
\frac{a}{b}+\frac{c}{b} = \frac{a+c}{b}.
\end{equation*}
If the denominators are different, we need to find a common denominator. We usually prefer to find the least common denominator (LCD) and multiply each fraction by the appropriate form of one.
Example 1.30. Adding Fractions.
To add \(5/4\) and \(1/6\text{,}\) observe that the least common denominator is \(12\) because this is the smallest multiple of \(4\) and \(6\) they have in common. Then,
\begin{equation*}
\frac{5}{4} + \frac{1}{6} = \frac{5\cdot 3}{4\cdot 3} + \frac{1 \cdot 2}{6\cdot 2}
= \frac{15}{12} + \frac{2}{12} = \frac{15+2}{12} = \frac{17}{12}.
\end{equation*}
This results in an improper fraction which may be written as a
\begin{equation*}
\frac{17}{12} = \frac{12+5}{12} = \frac{12}{12}+\frac{5}{12} = 1 + \frac{5}{12}.
\end{equation*}
Remark 1.31.
The result above is often written as a mixed number \(1 \frac{5}{12}\text{,}\) and while you may encounter this in some subjects, we will prefer the improper fraction \(\frac{17}{12}\) or the sum \(1 + \frac{5}{12}\) in calculus.
Example 1.32. Subtracting Fractions.
What fraction of the circle is unlabeled? 
Solution.
The sum of the portions should add to one, so that the missing fraction is
\begin{align*}
1-\frac{1}{3}-\frac{2}{5} \amp= \frac{15}{15} - \frac{5}{15} - \frac{6}{15} \\
\amp= \frac{15-5-6}{15} \\
\amp= \frac{4}{15}
\end{align*}
Using a common denominator can also be useful when comparing fractions.
Example 1.33. Comparing fractions.
Which is larger \(4/7\) or \(3/5\) ?
Solution.
To avoid any possible confusion, we observe they have a least common denominator of \(35 = 7\times 5\text{.}\) Then,
\begin{equation*}
\frac{4}{7} = \frac{20}{35} \quad \text{and} \quad \frac{3}{5} = \frac{21}{35}.
\end{equation*}
Comparing the numerators, we conclude that \(3/5\) is slightly bigger than \(4/7\)
\begin{equation*}
\frac{4}{7} \lt \frac{3}{5},
\end{equation*}
but only by \(1/21\text{.}\)
Consider the square of a fraction:
\begin{equation*}
\left(\frac{a}{b}\right)^2 = \frac{a}{b}\cdot\frac{a}{b} = \frac{a\cdot a}{b\cdot b} = \frac{a^2}{b^2}.
\end{equation*}
In general, a power of a fraction can be applied to the numerator and denominator individually.
Principle 1.34. Powers of Fractions.
\begin{equation*}
\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}
\end{equation*}
Example 1.35. Power of Fractions.
Evaluate completely
- \(\displaystyle \displaystyle\left(\frac{2}{3}\right)^2\)
- \(\displaystyle \displaystyle\frac{2^2}{3}\)
- \(\displaystyle \displaystyle\frac{-2^2}{3}\)
- \(\displaystyle \displaystyle\frac{(-2)^2}{3}\)
- \(\displaystyle \displaystyle\left(-\frac{2}{3}\right)^2\)
- \(\displaystyle \displaystyle\left(-\frac{2}{3}\right)^3\)
Solution.
Remember that exponents are considered before multiplication or division in PEMDAS 1.9.
- \(\displaystyle \displaystyle\left(\frac{2}{3}\right)^2 = \frac{2^2}{3^2} = \frac{4}{9}\)
- \(\displaystyle \displaystyle\frac{2^2}{3} = \frac{4}{3}\)
- \(\displaystyle \displaystyle\frac{-2^2}{3} = -\frac{4}{3}\)
- \(\displaystyle \displaystyle\frac{(-2)^2}{3} = \frac{4}{3}\)
- \(\displaystyle \displaystyle\left(-\frac{2}{3}\right)^2 = \left(-\frac{2}{3}\right)\cdot\left(-\frac{2}{3}\right) = \frac{4}{9}\)
- \(\displaystyle \displaystyle\left(-\frac{2}{3}\right)^3 = \left(-\frac{2}{3}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{2}{3}\right) = - \frac{8}{27}\)
Checkpoint 1.36.
- Can I cancel the \(5\) in \(\frac{(3+4)\times 5}{15}\text{?}\) Can I cancel the \(3\text{?}\) Why or why not?
- Is \(\frac{2}{3+4}\) the same as \(\frac{2}{3}+\frac{2}{4}\text{?}\)
