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# 2.6: Division of Rational Numbers

Difficulty Level: At Grade Created by: CK-12

So far in this chapter, 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*}.

The values of \begin{align*}a\end{align*} and \begin{align*}\frac{1}{a}\end{align*} are called reciprocals. In general, two nonzero numbers whose product is 1 are 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 1: 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 2: 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 3: 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 2:

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

## Using Reciprocals to Solve Real-World Problems

The need to divide rational numbers is necessary for solving problems in physics, chemistry, and manufacturing. The following example illustrates the need to divide fractions in physics.

Example 4: Newton’s Second Law relates acceleration to the force of an object and its mass: \begin{align*}a = \frac{F}{m}\end{align*}. Suppose \begin{align*}F = 7\frac{1}{3}\end{align*} and \begin{align*}m= \frac{1}{5}\end{align*}. Find \begin{align*}a\end{align*}, the acceleration.

Solution: Before beginning the division, the mixed number of force must be rewritten as an improper fraction.

Replace the fraction bar with a division symbol and simplify: \begin{align*}a = \frac{22}{3} \div \frac{1}{5}.\end{align*}

\begin{align*}\frac{22}{3} \times \frac{5}{1} = \frac{110}{3} = 36 \frac{2}{3}\end{align*}. Therefore, the acceleration is \begin{align*}36 \frac{2}{3} \ m/s^2.\end{align*}

Example 5: Anne runs a mile and a half in one-quarter hour. What is her speed in miles per hour?

Solution: Use the formula \begin{align*}speed = \frac{distance}{time}\end{align*}.

\begin{align*}s = 1.5 \div \frac{1}{4}\end{align*}

Rewrite the expression and simplify: \begin{align*}s = \frac{3}{2} \cdot \frac{4}{1} = \frac{4 \cdot 3} {2 \cdot 1} = \frac{12}{2} = 6 \ mi/hr.\end{align*}

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Division of Rational Numbers (8:20)

1. Define inverse.
2. What is the multiplicative inverse? How is this different from the additive inverse?

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

1. 100
2. \begin{align*}\frac{2}{8}\end{align*}
3. \begin{align*}-\frac{19}{21}\end{align*}
4. 7
5. \begin{align*}- \frac{z^3}{2xy^2}\end{align*}
6. 0
7. \begin{align*}\frac{1}{3}\end{align*}
8. \begin{align*}\frac{-19}{18}\end{align*}
9. \begin{align*}\frac{3xy}{8z}\end{align*}

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

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

In 21 – 23, evaluate the expression.

1. \begin{align*}\frac{x}{y}\end{align*} for \begin{align*}x = \frac{3}{8}\end{align*} and \begin{align*}y= \frac{4}{3}\end{align*}
2. \begin{align*}4z \div u\end{align*} for \begin{align*}u = 0.5\end{align*} and \begin{align*}z = 10\end{align*}
3. \begin{align*}\frac{-6}{m}\end{align*} for \begin{align*}m= \frac{2}{5}\end{align*}
4. The label on a can of paint states that it will cover 50 square feet per pint. If I buy a \begin{align*}\frac{1}{8}\end{align*}-pint sample, it will cover a square two feet long by three feet high. Is the coverage I get more, less, or the same as that stated on the label?
5. The world’s largest trench digger, “Bagger 288,” moves at \begin{align*}\frac{3}{8}\end{align*} mph. How long will it take to dig a trench \begin{align*}\frac{2}{3}\end{align*}-mile long?
6. A \begin{align*}\frac{2}{7}\end{align*} Newton force applied to a body of unknown mass produces an acceleration of \begin{align*}\frac{3}{10} \ m/s^2\end{align*}. Calculate the mass of the body. Note: \begin{align*}\text{Newton} = kg \ m/s^2\end{align*}
7. Explain why the reciprocal of a nonzero rational number is not the same as the opposite of that number.
8. Explain why zero does not have a reciprocal.

Mixed Review

Simplify.

1. \begin{align*}199 - (-11)\end{align*}
2. \begin{align*}-2.3 - (-3.1)\end{align*}
3. \begin{align*}|16-84|\end{align*}
4. \begin{align*}|\frac{-11}{4}|\end{align*}
5. \begin{align*}(4 \div 2 \times 6 + 10-5)^2\end{align*}
6. Evaluate \begin{align*}f(x)= \frac{1}{9} (x-3); f(21)\end{align*}.
7. Define range.

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