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# Fractions, Percents, and Ratios Part A: Models for the Multiplication and Division of Fractions (45 minutes)

In This Part: Area Model for Multiplication
In the past, you may have learned particular algorithms for the multiplication and division of fractions. We are now going to use some of the visual models we’ve employed earlier in this course to better understand what is actually happening when we perform these operations. See Note 2 below.

First we’ll use an area model — one that superimposes squares that are partitioned into the appropriate number of regions, and shaded as needed — to clarify what happens when you multiply fractions. For example, here’s how we would use the area model to demonstrate the problem 3/8 • 2/3:

 Shade one square, partitioned vertically, to represent 3/8 (shown below in pink): Shade another square, partitioned horizontally, to represent 2/3 (shown below in blue): Superimpose the two squares. The product is the area that is double-shaded (shown below in purple):      What is the value of this purple area? There are 3 • 2, or 6, purple parts out of 8 • 3, or 24, parts in all, so the value of the purple area is 6/24.

This model visually demonstrates the familiar algorithm: To multiply two fractions, multiply the numerators and then multiply the denominators. This algorithm “counts” both the purple parts (the product of the two numerators) and the total number of parts (the product of the two denominators).

We can also use this model to “reduce” the fraction. First we swap the positions of some of the purple parts. Two of the purple parts can be moved to the top, and thus, two of the eighths are now shaded. These two eighths are the same area as one quarter: In This Part: Try It Yourself

Make your own square transparencies to superimpose, and model the solution for each step in Problem A1 below.

Problem A1
An aerial photo of farmland shows the dimensions of three fields in fractions of a mile. Use the area model you’ve just learned to model the area in square miles of each of these fields:
a. 3/4 • 1/3
b. 3/5 • 2/3
c. 1/4 • 8/9

Problem A2
Describe how the area model shows that the product of two positive fractions, each less than 1, must be smaller than either of the fractions. Video Segment
In this segment, Jeanne and Liz use the area model to multiply fractions. They relate this model to the multiplication algorithm and check for ways to visually reduce fractions. Watch this video segment after you’ve completed Problems A1 and A2.

You can find this segment on the session video approximately 4 minutes and 32 seconds after the Annenberg Media logo.

In This Part: Area Model for Division
We can apply the area model for the multiplication of fractions to visualize the division of two fractions when each is less than 1. To model division with fractions, we more or less reverse the process used for multiplication. We start with an area we’re looking for, and we find one of the missing factors that makes up that area. See Note 3 below.

For example, here’s how we would use the model to demonstrate the problem 1/4 2/3:

 Shade one square, partitioned vertically, to represent 1/4 (as in the multiplication model, it’s shaded purple): Superimpose a square partitioned into thirds, positioned horizontally, onto the fourths square, and draw a bracket to the right of the thirds square to show the size of 2/3:    What you see now is the purple (1/4) area and the size of one of the factors that made that area.

We know from the multiplication model that the product of 2/3 and another factor (the quotient) defines an area equivalent in size to 1/4. To find the quotient, we need to move the top part of the purple area so that it’s the same height as the 2/3 factor.

 Subdivide the fourths square to make an eighths square: Move the top two purple pieces into the 2/3 height area (the area within the 2/3 bracket): Now shade the rectangles immediately to the right and immediately above the purple area:      This shows that there are 3 • 2, or 6, purple parts out of 8 • 3, or 24, parts in all. The purple area equals 1/4, and it came from the product of 2/3 multiplied by what? We can see that the other factor is 3/8.

Problem A3
A town plans to build a community garden that will cover 2/3 of a square mile. They would like to situate it on a pasture of an old horse farm. One dimension of the garden area will be determined by a fence that is 3/4 of a mile long. Use the area model for division to determine the other dimension of the new garden area.

Problem A4
Describe how the area model shows that the quotient of two positive fractions, each less than 1, must be larger than the first fraction.

In This Part: The Common Denominator Model for Division
The area model for the division of fractions does not help to illustrate why the algorithm we’re most familiar with (invert the divisor and then multiply) works. Unfortunately, no model can show that. See Note 4 below.

But here is a different division algorithm, one that we can explain with a model: Find the common denominator, find the equivalent fractions, and divide the numerators.

In order to understand the model for this algorithm, let’s first go back to review some of the concepts of division. It is usually easier to compute if you think about division in a quotative way. Thus, you can say that 6 3 asks, “How many 3s are there in 6?”

Next, we need to understand the role of units in division.

Problem A5
Which, if any, of these questions yields a different answer?
• How many 3s are there in 6?
• How many groups of 3 tens are there in 6 tens?
• How many groups of 3 fives are there in 6 fives?
• How many groups of 3 tenths are there in 6 tenths?How many groups of 3 @s are there in 6 @s?
• How many groups of 3 anythings are there in 6 anythings (as long as both anythings refer to the same unit)?

The point here is that the units of the problem do not matter — if the units are the same entity, they disappear when you divide.

This brings us back to the new algorithm for division with fractions: To divide two fractions, find a common denominator and then divide the numerators.

Let’s try a visual version of the problem we did before: 1/4 2/3. First, find a common denominator:

1/4 2/3 = 3/12 8/12

Next, divide the numerators:

3 8 = 3/8

Here is the model for this problem, called the common denominator model: In this problem, in effect, the original question was “How many 2/3s are there in 1/4?”By finding a common denominator, we changed the question to “How many 8/12s are there in 3/12?” — which is the same as asking “How many 8s are there in 3?” The answer to both questions is the same: “There is 3/8 of an 8 in 3.”

Problem A6
Use the common denominator model to divide 3/5 by 3/4.

Problem A7
Why does 0.6 0.2 have the same answer as 6 2?

In This Part: Translating the Process to Decimals
You can extend what you’ve learned about operations and fractions to decimals as well. Remember that a terminating decimal can be thought of as a fraction with a power of 10 as the denominator (e.g., 0.4 = 4/10). See Note 5 below.

Why do we need to line up the decimal points when we add or subtract decimals?
We need to line up the decimal points because we can only add or subtract if the units are the same. By aligning the decimal points, we make sure that we are adding or subtracting digits that have the same place values, just as we do when we add or subtract whole numbers.

Why do we count the decimal places when we multiply decimals?
From the section on exponents in Session 3, we know that multiplication with exponents requires adding, or “counting,” the exponents. So, for example, 0.2 • 0.03 in the exponential form is the following:

0.2 • 0.03 = 2/10 • 3/100 = 2 •10-1 • 3 • 10-2

The exponents are -1 and -2, which are one and two places, respectively, to the right of the decimal point. The product will then have an exponent that is the sum of -1 and -2 (i.e., -3), and is three places to the right of the decimal point. The product of 0.2 • 0.03 is 0.006:

0.2 • 0.03 = 6 • 10-3 = 0.006

Why do we move the decimal points when dividing with decimals?
This process is related to finding equivalent fractions. You can think of the division as a fraction. Since the problem 2.5 0.05 is hard to visualize, write it as 2.5/0.05. You need a whole number in the denominator, so multiply by 100 to get a whole number. To compensate for multiplying the denominator by 100, you must also multiply the numerator by 100. That means that you actually multiplied by 100/100, or 1, which doesn’t change the value of the fraction. Here’s what the process looks like:

2.5/0.05 = (2.5 • 100)/(0.05 • 100) = 250/5 = 50

### Notes

Note 2
You probably remember most, if not all, of the rules (or algorithms) for the multiplication and division of fractions. But can you actually remember why those rules work? As you examine the models used to demonstrate these operations, think about how the models are connected to paper-and-pencil computations.

Note 3
In this case, we are thinking of division as a missing-factor problem. We know the product and one of the factors; we need to find the other factor.

Note 4
We can, however, show you why the algorithm works: As you can see, we first wrote the division problem as a fraction with fractions for both its numerator and its denominator. Next, to change the messy denominator to a nice, tidy denominator, we multiplied by 1 in the form of (4/3)/(4/3). We then showed the multiplication problem as multiplying numerators and multiplying denominators. We computed the denominator, which was 1, and then divided by 1, which didn’t change the numerator. We are left with 2/3 • 4/3, exactly as the algorithm tells us: The division problem 2/3 3/4 can be changed to the multiplication problem 2/3 • 4/3, and both will produce the same answer (8/9).

Note 5
We often forget that terminating decimals are fractions too. And actually, the word “decimal” is something of a misnomer; we should not call them decimals unless we are referring specifically to the digits to the right of the decimal point. They should really be called decimal fractions. (Did you know that fractions that were not decimal fractions used to be called vulgar fractions? Perhaps our forebears didn’t like fractions with denominators that were not powers of 10!)

### Solutions

Problem A1
a.
The area is 3/12 (or 1/4) of a square mile. b. The area is 6/15 (or 2/5) of a square mile. c. The area is 8/36 (or 2/9) of a square mile. Problem A2
If the factors are each less than 1, the product cannot be larger than its factors. The part that overlaps the horizontally shaded area and the vertically shaded area is the area of the product, and each of the original areas will be larger than the area shaded by both.

Note that this only works with “proper” fractions, where the numerator is less than the denominator (and therefore the fraction is less than 1).

Problem A3
In essence, you are looking to find the length of a rectangle whose one dimension is 3/4 and whose area is equal to 2/3 — in other words, 3/4 • x = 2/3, or 2/3 3/4 = x.

 You can start by marking 2/3 vertically. Next, mark 3/4 horizontally.  Then subdivide everything into ninths.  Then rearrange the top (purple) pieces from the original 2/3 area to fit the 3/4 height. The result will be the purple rectangle, whose length is 8/9: 2/3 3/4 = 8/9  Problem A4
Since we model division as the reverse of multiplication, the first fraction is represented as an overlap of two areas. It will be smaller than the quotient for the same reason that the multiplication area is smaller than either of the fractions multiplied. In some cases, the first fraction will be smaller than both the quotient and second fraction. (The illustration in the solution to Problem A3 is an example of such a case.)

If the first fraction happens to be larger than the second one in the division problem, the result will be an improper fraction (where the numerator is larger than the denominator).

Problem A5
All the answers are the same, 2, because the units are the same for each division problem.

Problem A6
The original question is “How many 3/4s are there in 3/5?” The common denominator is 20: 3/4 = 15/20, and 3/5 = 12/20. The question is now “How many 15/20s are there in 12/20?,” which is easier to answer because the units are the same: 12/20 15/20 = 12 15 = 12/15 = 4/5. The answer is 4/5.

Problem A7
Both 0.6 and 0.2 can be expressed in units of tenths: “How many 2/10s are there in 6/10?” The answer to this question must be the same as “How many 2s are there in 6?” The answer to both questions will always be 3, regardless of the particular units involved.