What if you had two numbers like \begin{align*}\frac{5}{6}\end{align*} and \begin{align*}\frac{2}{5}\end{align*}? How could you multiply them so that your answer was in simplest form? After completing this Concept, you'll be able to solve multiplication problems like this one.

### Watch This

CK-12 Foundation: 0205S Multiplying Rationals

### Try This

For more practice multiplying fractions, try playing the fraction game at http://www.aaamath.com/fra66mx2.htm, or the one at http://www.mathplayground.com/fractions_mult.html.

### Guidance

Whenever we multiply a number by negative one, the sign of the number changes. In more mathematical terms, multiplying by negative one maps a number onto its opposite. The number line below shows two examples: \begin{align*}3 \cdot -1 = 3\end{align*} and \begin{align*}-1 \cdot -1 = 1\end{align*}.

When we multiply a number by negative one, the absolute value of the new number is the same as the absolute value of the old number, since both numbers are the same distance from zero.

The product of a number “\begin{align*}x\end{align*}” and negative one is \begin{align*}-x\end{align*}. This does not mean that \begin{align*}-x\end{align*} is necessarily less than zero! If \begin{align*}x\end{align*} itself is negative, then \begin{align*}-x\end{align*} will be positive because a negative times a negative (negative one) is a positive.

When you multiply an expression by negative one, remember to multiply the **entire expression** by negative one.

#### Example A

*Multiply the following by negative one.*

a) 79.5

b) \begin{align*}\pi\end{align*}

c) \begin{align*}(x + 1)\end{align*}

d) \begin{align*}| x |\end{align*}

**Solution**

a) -79.5

b) \begin{align*}-\pi\end{align*}

c) \begin{align*}-(x + 1) \ \text{or} \ -x - 1\end{align*}

d) \begin{align*}-| x |\end{align*}

Note that in the last case the negative sign **outside** the absolute value symbol applies **after** the absolute value. Multiplying the **argument** of an absolute value equation (the term inside the absolute value symbol) does not change the absolute value. \begin{align*}| x |\end{align*} is always positive. \begin{align*}| -x |\end{align*} is always positive. \begin{align*}-| x |\end{align*} is always negative.

Whenever you are working with expressions, you can check your answers by substituting in numbers for the variables. For example, you could check part \begin{align*}d\end{align*} of Example 1 by letting \begin{align*}x = -3\end{align*}. Then you’d see that \begin{align*}| -3 | \neq -| 3 |\end{align*}, because \begin{align*}| -3 | = 3\end{align*} and \begin{align*}-| 3 | = -3\end{align*}.

Careful, though—plugging in numbers can tell you if your answer is wrong, but it won’t always tell you for sure if your answer is right!

**Multiply Rational Numbers**

#### Example B

*Simplify* \begin{align*}\frac{1}{3} \cdot \frac{2}{5}\end{align*}.

One way to solve this is to think of money. For example, we know that *one third of sixty dollars* is written as \begin{align*}\frac{1}{3} \cdot \$ 60\end{align*}. We can read the above problem as *one-third of two-fifths*. Here is a visual picture of the fractions ** one-third** and

**.**

*two-fifths*

If we divide our rectangle into thirds one way and fifths the other way, here’s what we get:

Here is the intersection of the two shaded regions. The whole has been divided into five pieces width-wise and three pieces height-wise. We get two pieces out of a total of fifteen pieces.

**Solution**

\begin{align*}\frac{1}{3} \cdot \frac{2}{5} = \frac{2}{15}\end{align*}

Notice that \begin{align*}1 \cdot 2 = 2\end{align*} and \begin{align*}3 \cdot 5 = 15\end{align*}. This turns out to be true in general: when you multiply rational numbers, the numerators multiply together and the denominators multiply together. Or, to put it more formally:

When multiplying fractions: \begin{align*}\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}\end{align*}

This rule doesn’t just hold for the product of two fractions, but for any number of fractions.

#### Example C

*Evaluate and simplify* \begin{align*}\frac{12}{25} \cdot \frac{35}{42}\end{align*}.

**Solution**

We can see that 12 and 42 are both multiples of six, 25 and 35 are both multiples of five, and 35 and 42 are both multiples of 7. That means we can write the whole product as \begin{align*}\frac{6 \cdot 2}{5 \cdot 5} \cdot \frac{5 \cdot 7}{6 \cdot 7} = \frac{6 \cdot 2 \cdot 5 \cdot 7}{5 \cdot 5 \cdot 6 \cdot 7}\end{align*}. Then we can cancel out the 5, the 6, and the 7, leaving \begin{align*}\frac{2}{5}\end{align*}.

**Identify and Apply Properties of Multiplication**

The four mathematical properties which involve multiplication are the **Commutative, Associative, Multiplicative Identity** and **Distributive Properties**.

**Commutative property:** When two numbers are multiplied together, the product is the same regardless of the order in which they are written.

**Example:** \begin{align*}4 \cdot 2 = 2 \cdot 4\end{align*}

*We can see a geometrical interpretation of* **The Commutative Property of Multiplication** *to the right. The Area of the shape \begin{align*}(length \times width)\end{align*} is the same no matter which way we draw it.*

**Associative Property:** When three or more numbers are multiplied, the product is the same regardless of their grouping.

**Example:** \begin{align*}2 \cdot (3 \cdot 4) = (2 \cdot 3) \cdot 4\end{align*}

**Multiplicative Identity Property:** The product of one and any number is that number.

**Example:** \begin{align*}5 \cdot 1 = 5\end{align*}

**Distributive Property:** The product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For expressions \begin{align*}a, b, \end{align*} and \begin{align*}c\end{align*}, \begin{align*}a(b+c)=ab+ac\end{align*}.

**Example:** \begin{align*}4(6 + 3) = 4 \cdot 6 + 4 \cdot 3\end{align*}

#### Example D

*A gardener is planting vegetables for the coming growing season. He wishes to plant potatoes and has a choice of a single \begin{align*}8 \times 7\end{align*} meter plot, or two smaller plots of \begin{align*}3 \times 7 \end{align*} and \begin{align*}5 \times 7 \end{align*} meters. Which option gives him the largest area for his potatoes?*

**Solution**

In the first option, the gardener has a total area of \begin{align*}(8 \times 7)\end{align*} or 56 square meters.

In the second option, the gardener has \begin{align*}(3 \times 7)\end{align*} or 21 square meters, plus \begin{align*}(5 \times 7)\end{align*} or 35 square meters. \begin{align*}21 + 35 = 56\end{align*}, so the area is the same as in the first option.

Watch this video for help with the Examples above.

CK-12 Foundation: Multiplying Rational Numbers

### Guided Practice

*Multiply the following rational numbers:*

a) \begin{align*}\frac{2}{5} \cdot \frac{5}{9}\end{align*}

b) \begin{align*}\frac{1}{3} \cdot \frac{2}{7} \cdot \frac{2}{5}\end{align*}

c) \begin{align*}\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5}\end{align*}

**Solution**

a) With this problem, we can cancel the fives: \begin{align*}\frac{2}{5} \cdot \frac{5}{9} = \frac{2 \cdot 5}{5 \cdot 9} = \frac{2}{9}\end{align*}.

b) With this problem, we multiply **all the numerators** and **all the denominators**:

\begin{align*}\frac{1}{3} \cdot \frac{2}{7} \cdot \frac{2}{5} = \frac{1 \cdot 2 \cdot 2}{3 \cdot 7 \cdot 5} = \frac{4}{105}\end{align*}

c) With this problem, we multiply all the numerators and all the denominators, and then we can cancel most of them. The 2’s, 3’s, and 4’s all cancel out, leaving \begin{align*}\frac{1}{5}\end{align*}.

With multiplication of fractions, we can simplify before or after we multiply. The next example uses factors to help simplify before we multiply.

### Explore More

In 1-4, multiply the following expressions by negative one.

- 25
- -105
- \begin{align*}x^2\end{align*}
- \begin{align*}( 3 + x)\end{align*}

In 5-10, multiply the following rational numbers. Write your answer in the **simplest form**.

- \begin{align*}\frac{5}{12} \times \frac{9}{10}\end{align*}
- \begin{align*}\frac{2}{3} \times \frac{1}{4}\end{align*}
- \begin{align*}\frac{3}{4} \times \frac{1}{3}\end{align*}
- \begin{align*}\frac{15}{11} \times \frac{9}{7}\end{align*}
- \begin{align*}\frac{1}{13} \times \frac{1}{11}\end{align*}
- \begin{align*}\frac{12}{15} \times \frac{35}{13} \times \frac{10}{2} \times \frac{26}{36}\end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 2.5.