Lesson 16

In this warm-up, students use the structure (MP7) of a set of multiplication equations to see the relationship between two numbers that differ by a factor of a power of 10. Students evaluate the expression \(x\boldcdot 10\) by considering the effect of multiplication by 10. 

Ask students to answer the following questions without writing anything and to be prepared to explain their reasoning. Follow with whole-class discussion.

Ask students to share what they noticed about the first four equations. Record student explanations that connect multiplying a number by 10 with moving the decimal place. 

Discuss how students could use their observations from the first question to multiply 0.81 by a number to get 810. 

In this activity, students continue to think about products of decimals using fractions. They use what they know about \(\frac{1}{10}\) and \(\frac{1}{100}\), as well as the commutative and associative properties, to identify and write multiplication expressions that could help them find the product of two decimals.

While students may be able to start by calculating the value of each decimal product, the goal is for them to look for and use the structure of equivalent expressions (MP7), and later generalize the process to multiply any two decimals.

As students work, listen for the different ways students decide on which expressions are equivalent to \((0.6) \boldcdot (0.5)\). Identify a few students or groups with differing approaches so they can share later.

Arrange students in groups of 2. Give groups 3–4 minutes to work on the first two questions, and then pause for a whole-class discussion. Discuss:

• Why is \((0.6) \boldcdot (0.5)\) equivalent to \(6 \boldcdot \frac{1}{10} \boldcdot 5 \boldcdot \frac{1}{10}\)? (0.6 is 6 tenths, which is the same as \(6 \boldcdot \frac{1}{10}\), and 0.5 is 5 tenths, or \(5 \boldcdot \frac{1}{10}\))
• Why is the expression \(6 \boldcdot \frac{1}{10} \boldcdot 5 \boldcdot \frac{1}{10}\) equivalent to \(6 \boldcdot 5 \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}\)? (We can ‘switch’ the places of 5 and \(\frac{1}{10}\) in the multiplication and not change the product. This follows the commutative property of operations.)
• How did you find the value of \(30 \boldcdot \frac{1}{100}\)? (Multiplying by \(\frac{1}{100}\) means dividing by 100, which moves the decimal point 2 places to the left, so the result is \(0.30\) or \(0.3\).

If students try to use vertical calculation to find the products, ask them to instead do so by thinking of the decimals as fractions and about any patterns they observed.

Select several students to share their responses and reasonings for the last question.

To conclude, ask students to consider how writing \(6 \boldcdot 5 \boldcdot \frac{1}{100}\) might be a favorable way to find \(0.6 \boldcdot 0.5\). Students may respond that using whole numbers and fractions makes multiplication simpler; even if there is division, it is division by a power of 10. In future lessons, students will apply this reasoning to find products of more elaborate decimals, such as \((0.24) \boldcdot (0.011)\).

This activity continues to develop the two methods for computing products of decimals introduced in the previous lesson. The first method uses the idea that multiplying a number by \(\frac{1}{10}\) is the same as dividing the number by 10, multiplying by \(\frac{1}{100}\) is the same as dividing by 100, and so on. The second method is to convert decimals to fractions, compute the product, then convert the product to a decimal. Students make sense of both methods and use one to solve a problem. As they continue to work through examples, students begin to notice a relationship between the location of decimal points in the factors and the product.

As they reason about the placement of the decimal and the relationship between decimals and fractions, students use the structure of the base-ten system (MP7). To reason correctly about the products of decimals, they also need to pay close attention to the digits and their place value.

Arrange students in groups of 2. Give students 5–6 minutes to complete the first set of questions. Ask each student to study one of the two methods and then explain that method to their partner. After making sense of both methods together, each partner applies one method to solve a new problem. After the first question, have students pause for a brief discussion. Invite a student from each camp to share their reasoning. If not already mentioned in students' explanations, ask:

• Why might have Elena multiplied by 2.4 and 1.3 both by 10? (Elena might have multiplied the factors by 10 to get them into whole numbers, which are easier to multiply.) What might be her reason for dividing 312 by 100? (Because she multiplied the original factors by \((10 \boldcdot 10)\) or 100, so the product 312 is 100 times the original product and must therefore be divided by 100). 

• How is Noah's method different than Elena's? (Noah converted each decimal into fractions and multiplied the fractions.)

Then give students another 7–8 minutes of quiet time to work on the second set of questions.

Invite a couple of students to summarize the two different methods used in this activity. Poll the class to see which method they used in the second question. Ask if they preferred one method over the other and, if so, which one and why.

It is important students understand that the two methods presented here are mathematically equivalent. In Noah’s method, the product of the numerators of his fractions (23 and 15) is the same as the product of Elena’s whole numbers. Both of them then move the decimal point three places to the left because they need to divide by 1,000. For Noah, 1,000 is the denominator of his fraction. For Elena, the division by 1,000 reverts the initial multiplication by 1,000 she had performed so she could have whole-number factors.

This activity extends the previous optional activity to products of decimals. It opens with a slight variant of the product \(24 \boldcdot 13\), namely \((2.4) \boldcdot (1.3)\). The side lengths of the area diagrams are multiplied by 0.1, but the reasoning involved is unchanged. While students can find the product using any of the previously developed pathways—i.e., multiplying the factors by powers of 0.1 or \(\frac{1}{10}\), or by using fractions—the focus here is on using partial products and connecting them to the multiplication algorithm. In the next lesson, students will generalize the process and use the algorithm to compute products of other decimals.

Recognition and use of structure (MP7) are once again important here. The use of the same non-zero digits in the problems also gives students a chance to see regularity in the reasoning and to focus on how the decimal point affects the calculation (MP8).

To highlight the role of place value in multiplication, discuss the following questions:

• How does the diagram for the product \((2.4)  \boldcdot (1.3)\) compare to that for \(24  \boldcdot 13\)? (They are the same; the only difference is that the decimal side lengths—the factors—are one-tenth of the whole-number ones.
• How are the two calculations similar? How are they different? (There are no decimals in the partial products of \(24 \boldcdot 13\). There are 0’s at the end of the numbers in the calculation for \(24  \boldcdot 13\) but not in the other calculation. The non-zero numbers are the same in the two calculations.)

Prompt students to look at the close relationship between the partial-product calculations and the technique of using fractions to find products of decimals. Students previously saw that \((2.4) \boldcdot (1.3) = (24 \boldcdot \frac{1}{10}) \boldcdot (13 \boldcdot \frac{1}{10}) = (24  \boldcdot 13) \boldcdot (\frac {1}{10}  \boldcdot \frac {1}{10})= 312 \boldcdot \frac{1}{100} = 3.12\).

The two partial-product calculations, presented side-by-side, validate the previous reasoning. The calculation on the right shows the whole-number product of 24 and 13. The one on the left shows that when each decimal factor is one-tenth of the whole-number factor, the decimal product is one hundredth of the whole-number product, so the decimal point moves two places to the left.