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.

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