Suppose a box of cereal is \begin{align*}\frac{4}{5}\end{align*} full, and you want to divide the remaining cereal into portions so that each portion is \begin{align*}\frac{1}{5}\end{align*} of the full box. In this case, you would have to divide a fraction by a fraction to come up with the number of portions you could make. After completing this Concept, you'll be able to use reciprocals to perform division problems such as these.

### Guidance

**Division of Rational Numbers**

Previously, you have added, subtracted, and multiplied rational numbers. It now makes sense to learn how to divide rational numbers. We will begin with a definition of **inverse operations.**

Inverse operations "undo" each other.

For example, addition and subtraction are inverse operations because addition cancels subtraction and vice versa. The additive identity results in a sum of zero. In the same sense, multiplication and division are inverse operations. This leads into the next property: The Inverse Property of Multiplication.

**The Inverse Property of Multiplication:** For every nonzero number \begin{align*}a\end{align*}, there is a multiplicative inverse \begin{align*}\frac{1}{a}\end{align*} such that \begin{align*}a \left ( \frac{1}{a} \right ) = 1\end{align*}.

This means that the multiplicative inverse of \begin{align*}a\end{align*} is \begin{align*}\frac{1}{a}\end{align*}. The values of \begin{align*}a\end{align*} and \begin{align*}\frac{1}{a}\end{align*} are called also called **reciprocals.** In general, two nonzero numbers whose product is 1 are multiplicative inverses or reciprocals.

**Reciprocal:** The reciprocal of a nonzero rational number \begin{align*}\frac{a}{b}\end{align*} is \begin{align*}\frac{b}{a}\end{align*}.

Note: The number zero does not have a reciprocal.

**Using Reciprocals to Divide Rational Numbers**

When dividing rational numbers, use the following rule:

**“When dividing rational numbers, multiply by the ‘right’ reciprocal.”**

In this case, the “right” reciprocal means to take the reciprocal of the fraction on the right-hand side of the division operator.

#### Example A

*Simplify* \begin{align*}\frac{2}{9} \div \frac{3}{7}\end{align*}.

**Solution:**

Begin by multiplying by the “right” reciprocal.

\begin{align*}\frac{2}{9} \times \frac{7}{3} = \frac{14}{27}\end{align*}

#### Example B

*Simplify* \begin{align*}\frac{7}{3} \div \frac{2}{3}\end{align*}.

**Solution:**

Begin by multiplying by the “right” reciprocal.

\begin{align*}\frac{7}{3} \div \frac{2}{3} = \frac{7}{3} \times \frac{3}{2} = \frac{7 \cdot 3} {2 \cdot 3} = \frac{7}{2}\end{align*}

Instead of the division symbol \begin{align*}\div\end{align*}, you may see a large fraction bar. This is seen in the next example.

#### Example C

*Simplify* \begin{align*}\frac{\frac{2}{3}}{\frac{7}{8}}\end{align*}.

**Solution:**

The fraction bar separating \begin{align*}\frac{2}{3}\end{align*} and \begin{align*}\frac{7}{8}\end{align*} indicates division.

\begin{align*}\frac{2}{3} \div \frac{7}{8}\end{align*}

Simplify as in Example B:

\begin{align*}\frac{2}{3} \times \frac{8}{7} = \frac{16}{21}\end{align*}

### Video Review

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### Guided Practice

1. Find the multiplicative inverse of \begin{align*}\frac{5}{7}\end{align*}.

2. Simplify \begin{align*} 5\div \frac{3}{2}\end{align*}.

**Solutions:**

1. The multiplicative inverse of \begin{align*}\frac{5}{7}\end{align*} is \begin{align*}\frac{7}{5}.\end{align*} We can see that by multiplying them together:

\begin{align*}\frac{5}{7}\times \frac{7}{5}= \frac{5\times 7}{7\times 5}=\frac{35}{35}=1.\end{align*}

2. When we are asked to divide by a fraction, we know we can rewrite the problem as multiplying by the reciprocal:

\begin{align*} 5\div \frac{3}{2}=5 \times \frac{2}{3}=\frac{5\times 2}{3}=\frac{10}{3}\end{align*}

### Explore More

- Define
*inverse.* - What is a multiplicative inverse? How is this different from an additive inverse?

In 3 – 11, find the multiplicative inverse of each expression.

- 100
- \begin{align*}\frac{2}{8}\end{align*}
- \begin{align*}-\frac{19}{21}\end{align*}
- 7
- \begin{align*}- \frac{z^3}{2xy^2}\end{align*}
- 0
- \begin{align*}\frac{1}{3}\end{align*}
- \begin{align*}\frac{-19}{18}\end{align*}
- \begin{align*}\frac{3xy}{8z}\end{align*}

In 12 – 20, divide the rational numbers. Be sure that your answer is in the simplest form.

- \begin{align*}\frac{5}{2} \div \frac{1}{4}\end{align*}
- \begin{align*}\frac{1}{2} \div \frac{7}{9}\end{align*}
- \begin{align*}\frac{5}{11} \div \frac{6}{7}\end{align*}
- \begin{align*}\frac{1}{2} \div \frac{1}{2}\end{align*}
- \begin{align*}- \frac{x}{2} \div \frac{5}{7}\end{align*}
- \begin{align*}\frac{1}{2} \div \frac{x}{4y}\end{align*}
- \begin{align*}\left ( - \frac{1}{3} \right ) \div \left ( - \frac{3}{5} \right )\end{align*}
- \begin{align*}\frac{7}{2} \div \frac{7}{4}\end{align*}
- \begin{align*}11 \div \left ( - \frac{x}{4} \right ) \end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 2.10.