Grade 3 · 3.OA.B.5 · 5.MD.C.5A

Properties of multiplication

Four rows of nine seats and nine rows of four seats look like two different rooms. Why do they hold the same number of people, and what other changes leave a product the same?

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Why this exists

Isko helps set up the school theater. The seating plan on the wall shows 4 rows with 9 seats in each row. Nkiru is looking at the same plan from the side of the room, and to her it shows 9 rows with 4 seats in each. Isko works out 4 \times 9; Nkiru works out 9 \times 4. Do they have to count again to be sure they agree?

They do not. Nobody moved a seat, so the number of seats cannot have changed. That simple fact, that a product does not care which factor comes first, is the first of three rules in this lesson. Each rule says that some change to a multiplication leaves its value alone:

swapping the order of two factors,
changing which pair of three factors is multiplied first,
multiplying by 1, which changes nothing, and by 0, which always gives 0.

These rules matter because they cut the work of learning the facts. Once 9 \times 4 is known, 4 \times 9 is known too, so most facts come in pairs and only about half need learning. And when three numbers are multiplied, the rules let you pick the easiest pair to start with, such as a 5 and a 2 that make 10.

Definition

The numbers being multiplied are the factors and the answer is the product, as in Arrays.

The commutative property of multiplication. Two factors can be multiplied in either order and the product is the same. For example,

4 \times 9 = 9 \times 4

The associative property of multiplication. To multiply three factors, multiply any two neighbors first, then multiply by the third. Which pair you choose does not change the product. Parentheses show the pair that is done first:

(2 \times 3) \times 5 = 2 \times (3 \times 5)

The identity property of multiplication. Any number times 1 is that same number, and 1 times any number is that same number: 1 \times 8 = 8 and 8 \times 1 = 8.

The zero property of multiplication. Any number times 0 is 0, and 0 times any number is 0: 0 \times 6 = 0 and 6 \times 0 = 0.

The word commutative comes from an old word meaning “to exchange”: the factors trade places. Associative comes from “associate”, to keep company: the middle factor keeps company first with the one on its left or first with the one on its right.

Intuition

The lesson runs over three days. Part 1 turns an array on its side. Part 2 regroups three factors. Part 3 multiplies by 1 and by 0.

Part 1: turning an array on its side

Here is the theater plan, as Isko sees it and as Nkiru sees it.

Figure 1The same 36 seats, read as 4 \times 9 or, turned on its side, as 9 \times 4.

On the left the seats are 4 rows of 9. Turn the whole plan a quarter turn and the rows become columns: the same seats now stand in 9 rows of 4. Nothing was added and nothing was taken away, so the two counts must agree. Counting gives 9, 18, 27, 36 for the first and 4, 8, 12, \dots, 36 for the second:

4 \times 9 = 36 = 9 \times 4

The same is true of equal groups that are not set out in rows, because they can always be set out in rows. Leota has 6 lunch boxes with 3 apples in each. Put each box’s apples in a row, one row under the next. That is an array of 6 rows of 3. Read down instead of across and it is 3 columns of 6. So 6 groups of 3 hold as many apples as 3 groups of 6: 6 \times 3 = 3 \times 6 = 18.

The two stories are still different. Six boxes of 3 and three boxes of 6 are different ways to pack, and a theater of 4 long rows is a different room from one of 9 short rows. What is the same is only the count. That is all the commutative property claims, and it is exactly what makes it useful: when one order is easier to work out, use it. Nine fives, 9 \times 5, are quick to count by fives; five nines, 5 \times 9, are the same number, so they are quick too.

Part 2: grouping three factors

Leota sets up 2 lunch tables. On each table she puts 3 boxes, and in each box there are 5 apples. How many apples are there?

Figure 2Count boxes first, (2 \times 3) \times 5, or count each table first, 2 \times (3 \times 5). Either way there are 30 apples.

There are two natural ways to count.

Count the boxes first. Two tables of 3 boxes is 2 \times 3 = 6 boxes, and 6 boxes of 5 apples is 6 \times 5 = 30. Written with parentheses for the first step, that is (2 \times 3) \times 5.
Count one table first. One table has 3 boxes of 5, which is 3 \times 5 = 15 apples, and there are 2 such tables, so 2 \times 15 = 30. That is 2 \times (3 \times 5).

Both counts are of the same apples, so they agree. This is the associative property: in a product of three factors, it does not matter which neighboring pair is multiplied first.

Its everyday use is making a ten. To find 7 \times 5 \times 2, multiplying 7 \times 5 first gives 35, and then 35 \times 2 is a double. Grouping the other way, 7 \times (5 \times 2), the 5 and the 2 make 10 at once, and 7 \times 10 = 70 is a tens fact from Facts for two, five and ten. When the 5 and the 2 are not neighbors, the commutative property can swap factors until they are, and then the associative property groups them.

Part 3: multiplying by one and by zero

Figure 3One row of 8 is 8. Six tables with nothing on them hold 0.

The front row of the theater is a single row of 8 seats. One group of 8 is just 8, so 1 \times 8 = 8. Read by columns, the same row is 8 columns of 1 seat, so 8 \times 1 = 8 as well. Multiplying by 1 keeps a number exactly as it was, which is why 1 is called the identity for multiplication: the number keeps its identity.

Zero is the opposite extreme. 0 \times 6 means 0 groups of 6: Leota has not carried in any tables, so there are no boxes at all. And 6 \times 0 means 6 groups with nothing in each: six tables are set out, but no boxes are on any of them. Either way the count is 0. Turning the array on its side agrees, since 0 \times 6 and 6 \times 0 must be equal by the commutative property.

Compare these with adding. Adding 0 changes nothing, 0 + 6 = 6, but multiplying by 0 wipes the number out, 0 \times 6 = 0. Adding 1 moves up one, 1 + 8 = 9, but multiplying by 1 keeps the number, 1 \times 8 = 8. For multiplication it is 1 that changes nothing.

Results

Commutative property. For any two whole numbers a and b,

a \times b = b \times a

Associative property. For any three whole numbers a, b and c,

(a \times b) \times c = a \times (b \times c)

Identity property. For any whole number a, \;1 \times a = a and a \times 1 = a.

Zero property. For any whole number a, \;0 \times a = 0 and a \times 0 = 0. The other way around is also true: if a product of whole numbers is 0, then one of its factors is 0.

Any order, any grouping. Using the first two properties again and again, the factors of a product can be put in any order and grouped in any way without changing the product. So 2 \times 4 \times 5 may be found as 4 \times (2 \times 5).

Why it is true

Commutative. Set out a rows of b objects. Every column takes one object from each row, so every column has a objects, and there is one column for each object in a row, so there are b columns. Counting by rows gives a \times b; counting by columns gives b \times a. It is one set of objects counted two ways, so the two products are equal.

Associative. Take a tables, b boxes on each table and c objects in each box. Counting boxes first gives a \times b boxes of c, which is (a \times b) \times c. Counting one table first gives b \times c objects on each table and a tables, which is a \times (b \times c). Again it is one set of objects counted two ways.

Identity. 1 \times a is one group of a, which is a. By the commutative property, a \times 1 is the same number.

Zero. 0 \times a is no groups at all, so there is nothing to count. a \times 0 is a groups that are each empty, so again there is nothing. For the last claim: if neither factor is 0, there is at least one group and at least one object in it, so the product is at least 1. So a product of 0 needs a factor of 0. \blacksquare

Worked examples

Examples 1 and 2 go with Part 1, Examples 3 to 5 with Part 2, and Examples 6 and 7 with Part 3.

Example 1

The theater plan both ways

The school theater has 4 rows of 9 seats. Write the number of seats as a product read by rows and as a product read by columns, and find it.

Figure 4The same 36 seats, read as 4 \times 9 or, turned on its side, as 9 \times 4.

Solution

Read by rows: 4 rows with 9 seats in each is 4 groups of 9. Count by nines, one number per row: 9, 18, 27, 36. So 4 \times 9 = 36.

Read by columns: each column takes one seat from every row, so each column has 4 seats, and there are 9 columns. That is 9 \times 4. Count by fours: 4, 8, 12, 16, 20, 24, 28, 32, 36.

Both products count the same seats, as the commutative property says they must:

4 \times 9 = 9 \times 4 = 36

The theater has 36 seats.

Example 2

Choosing the easier order

Nkiru needs 5 \times 9. Counting by nines is slow for her, but she knows her fives well. Find the product. Then fill in the boxes: 8 \times 6 = 6 \times \square and \square \times 7 = 7 \times 3.

Solution

By the commutative property, 5 \times 9 = 9 \times 5, so Nkiru may count nine fives instead of five nines: 5, 10, 15, 20, 25, 30, 35, 40, 45. So

5 \times 9 = 45

For the boxes, each side has the same two factors in the other order, so the box holds the factor that is missing. In 8 \times 6 = 6 \times \square, the box is 8. In \square \times 7 = 7 \times 3, the box is 3. Neither needs the product worked out at all, which is the point: the property says the two sides agree whatever the product is.

Example 3

Boxes on lunch tables

Leota sets up 2 lunch tables, puts 3 boxes on each table and 5 apples in each box. Find the number of apples as (2 \times 3) \times 5 and as 2 \times (3 \times 5).

Figure 5Count boxes first, (2 \times 3) \times 5, or count each table first, 2 \times (3 \times 5). Either way there are 30 apples.

Solution

(2 \times 3) \times 5 counts the boxes first. Two tables of 3 boxes is 2 \times 3 = 6 boxes. Six boxes of 5 apples is 6 \times 5 = 30 apples.

2 \times (3 \times 5) counts one table first. One table has 3 boxes of 5, so 3 \times 5 = 15 apples. Two such tables have 2 \times 15, which is the double 15 + 15 = 30.

\begin{aligned} (2 \times 3) \times 5 &= 6 \times 5 = 30 \\ 2 \times (3 \times 5) &= 2 \times 15 = 30 \end{aligned}

There are 30 apples. The two groupings agree, which is the associative property.

Example 4

Making a ten first

Find 3 \times 5 \times 2. Which grouping is easier?

Solution

Grouping from the left, (3 \times 5) \times 2 = 15 \times 2, and 15 + 15 = 30.

Grouping from the right, 3 \times (5 \times 2). The 5 and the 2 make 10, and 3 \times 10 = 30 is a tens fact:

3 \times (5 \times 2) = 3 \times 10 = 30

The second grouping is easier because it makes a ten. The associative property says the two groupings must give the same product, and they do: 30.

Example 5

When the five and the two are apart

Isko needs 2 \times 4 \times 5. Find it by making a ten, and name each property you use.

Solution

The 2 and the 5 make a ten, but the 4 sits between them. First swap the 2 and the 4, which the commutative property allows:

2 \times 4 \times 5 = 4 \times 2 \times 5

Now the 2 and the 5 are neighbors, and the associative property lets us multiply them first:

4 \times 2 \times 5 = 4 \times (2 \times 5) = 4 \times 10 = 40

Check by grouping from the left with no swap: (2 \times 4) \times 5 = 8 \times 5 = 40. Both give 40.

Example 6

One row and no tables

(a) The front row of the theater is 1 row of 8 seats. Find 1 \times 8 and 8 \times 1. (b) Find 0 \times 6 and 6 \times 0 with a lunch table story for each.

Figure 6One row of 8 is 8. Six tables with nothing on them hold 0.

Solution

(a) 1 \times 8 is one group of 8, the single row, so it is 8. Read by columns, the row is 8 columns of 1 seat each, so 8 \times 1 = 8 too. This is the identity property:

1 \times 8 = 8 \times 1 = 8

(b) 0 \times 6 is 0 groups of 6: no tables carried in, so no boxes, even though each table would have held 6. 6 \times 0 is 6 groups of 0: six tables set out with no boxes on any of them. Both give 0, which is the zero property:

0 \times 6 = 6 \times 0 = 0

Example 7

Using all the properties

Find 5 \times 1 \times 9 \times 2 and 7 \times 3 \times 0 with as little work as you can, and name the property behind each step.

Solution

For 5 \times 1 \times 9 \times 2: by the identity property, 5 \times 1 = 5, so the product is 5 \times 9 \times 2. Swap the 5 and the 9 (commutative), then group the 5 with the 2 (associative):

5 \times 9 \times 2 = 9 \times 5 \times 2 = 9 \times (5 \times 2) = 9 \times 10 = 90

For 7 \times 3 \times 0: group the last two first (associative). By the zero property, 3 \times 0 = 0, and then 7 \times 0 = 0:

7 \times 3 \times 0 = 7 \times (3 \times 0) = 7 \times 0 = 0

There was no need to find 7 \times 3 at all. Any product with a factor of 0 is 0.

Boundaries

Division is not commutative. 12 \div 3 = 4: twelve apples shared among 3 tables gives 4 apples to each table. But 3 \div 12 means 3 apples shared among 12 tables, and no table gets a whole apple. So 12 \div 3 and 3 \div 12 are not equal, and the order in a division matters. Subtraction is not commutative either: 9 - 4 = 5, but 4 - 9 has no whole number answer.
Division is not associative. (16 \div 4) \div 2 = 4 \div 2 = 2, but 16 \div (4 \div 2) = 16 \div 2 = 8. Regrouping changes a division.
The count is the same, the situation is not. 4 rows of 9 seats and 9 rows of 4 seats are different rooms. The properties say the products are equal; they do not say the two stories are the same story.
The properties are about multiplying only. In 2 \times 3 + 4 the 2 may not be regrouped with the 3 + 4: (2 \times 3) + 4 = 10 but 2 \times (3 + 4) = 14. Splitting a product over a sum is a different rule, the The distributive property.
No dividing by zero. 0 \times 6 = 0 is fine, but asking for 6 \div 0, six apples shared among no tables, has no answer. That comes up in Division facts.

Where this goes next

The commutative property means each fact is learned once and used twice, and the table of facts looks the same flipped across its diagonal, which Patterns in the multiplication table explores. The associative property, together with the The distributive property (cutting an array into two smaller arrays), is how the harder facts in Facts for three, four and six and Facts for seven, eight and nine are built from easy ones. Regrouping and reordering to make a ten or a hundred become a whole way of working in Doubling, halving and quick products.

A product of three factors also has a picture of its own: a box built from cubes, in layers of rows. Counting it layer by layer or column by column is the associative property again, and it is how the volume of a box is found. Addition has matching rules, in Properties of addition, and later both are stated for every number in Commutative, associative and distributive properties.

Exercises

Exercises 1 to 7 go with Part 1, 8 to 13 with Part 2, and 14 to 20 with Part 3. The reasoning group draws on all three.

Part 1: turning an array on its side

A small theater has 5 rows of 6 seats. Write a product for the seats read by rows and another read by columns, and find the number of seats.
Fill in the boxes: 7 \times 4 = 4 \times \square and \square \times 9 = 9 \times 2.
The theater balcony has 8 rows of 2 seats. A side section has 2 rows of 8 seats. Which holds more people?
Leota knows 4 \times 10 = 40. Which other fact does she know without working anything out?
Find 9 \times 2 two ways: as nine twos counted by twos, and as two nines added. Which is quicker?
A seating plan of 4 rows of 7 seats is turned on its side. How many rows does it show now, how many seats in each row, and how many seats in all?
True or false: (a) 6 \times 3 = 3 \times 6 (b) 6 \times 3 = 6 + 3.

Part 2: grouping three factors

Nkiru sets up 3 lunch tables with 2 boxes on each table and 4 sandwiches in each box. Find the number of sandwiches as (3 \times 2) \times 4 and as 3 \times (2 \times 4).
Find 6 \times 5 \times 2 by making a ten first.
Find 2 \times 8 \times 5 by making a ten. Name each property you use.
Fill in the box: (2 \times 2) \times 7 = 2 \times (2 \times \square). Then find the product.
The theater has 2 sections, and each section has 3 rows of 6 seats. Find the number of seats in two ways.
Find 5 \times 7 \times 2. Name each property you use.

Part 3: multiplying by one and by zero

Find 1 \times 9, 6 \times 1, 0 \times 4 and 7 \times 0.
Leota sets out 5 lunch tables, but no boxes have arrived. How many boxes are on the tables? Write the product.
A narrow balcony has 7 rows with 1 seat in each row. How many seats is that?
Find 1 + 5 and 1 \times 5, then 0 + 5 and 0 \times 5.
Fill in the boxes: \square \times 8 = 8 and 9 \times \square = 0.
Find 3 \times 1 \times 9 and 8 \times 7 \times 0.
Find 1 \times 3 \times 2 \times 1 \times 4 with as little work as you can.

Reasoning

Explain with an array why 8 \times 4 = 4 \times 8.
Is 10 \div 2 the same as 2 \div 10? Explain with boxes shared among lunch tables.
Explain with equal groups why 0 \times 7 and 7 \times 0 are both 0.
Explain with boxes on lunch tables why (3 \times 3) \times 2 = 3 \times (3 \times 2), and find the product.
Isko multiplies two whole numbers and gets 0. Must one of the numbers be 0? Explain.

Answers

30 seats. By rows, 5 \times 6 = 30; by columns, 6 columns of 5, so 6 \times 5 = 30.
7 \times 4 = 4 \times 7, so the first box is 7. 2 \times 9 = 9 \times 2, so the second box is 2.
Neither. 8 \times 2 = 16 and 2 \times 8 = 16: the side section is the balcony plan turned on its side.
10 \times 4 = 40, by the commutative property.
18 both ways. Counting by twos takes nine numbers, 2, 4, \dots, 18; adding two nines is one double, 9 + 9 = 18, which is quicker.
7 rows of 4 seats, and 7 \times 4 = 28 seats, the same as 4 \times 7 = 28 before the turn.
(a) True, both are 18. (b) False: 6 \times 3 = 18 but 6 + 3 = 9.
24 sandwiches. (3 \times 2) \times 4 = 6 \times 4 = 24 (count boxes first), and 3 \times (2 \times 4) = 3 \times 8 = 24 (count one table first).
60. 6 \times (5 \times 2) = 6 \times 10 = 60, by the associative property.
80. Swap the 2 and the 8 (commutative): 8 \times 2 \times 5. Group the 2 with the 5 (associative): 8 \times (2 \times 5) = 8 \times 10 = 80.
The box is 7, by the associative property. The product is 28: 4 \times 7 = 28, and 2 \times 14 = 28.
36 seats. (2 \times 3) \times 6 = 6 \times 6 = 36 (count rows first), or 2 \times (3 \times 6) = 2 \times 18 = 36 (count one section first).
70. Swap the 5 and the 7 (commutative): 7 \times 5 \times 2. Group the 5 with the 2 (associative): 7 \times (5 \times 2) = 7 \times 10 = 70.
1 \times 9 = 9 and 6 \times 1 = 6 (identity property); 0 \times 4 = 0 and 7 \times 0 = 0 (zero property).
0 boxes: 5 \times 0 = 0, five groups with nothing in each.
7 seats: 7 \times 1 = 7.
1 + 5 = 6 but 1 \times 5 = 5. 0 + 5 = 5 but 0 \times 5 = 0. For adding, 0 changes nothing; for multiplying, 1 changes nothing.
1 \times 8 = 8, so the first box is 1. 9 \times 0 = 0, so the second box is 0.
3 \times 1 \times 9 = 3 \times 9 = 27, since 3 \times 1 = 3. 8 \times 7 \times 0 = 8 \times (7 \times 0) = 8 \times 0 = 0.
24. Drop both 1s (identity): 3 \times 2 \times 4. Then (3 \times 2) \times 4 = 6 \times 4 = 24.
Set out 8 rows of 4 dots; that is 8 \times 4. Each column takes one dot from every row, so each column has 8 dots, and there are 4 columns, which is 4 \times 8. Both count the same 32 dots, so the products are equal.
No. 10 \div 2 = 5: ten boxes shared between 2 tables gives 5 boxes to each. But 2 \div 10 means 2 boxes shared among 10 tables, and no table gets a whole box. Division is not commutative.
0 \times 7 is 0 groups of 7: there are no groups, so nothing to count. 7 \times 0 is 7 groups with 0 in each: every group is empty, so again there is nothing. Both are 0.
18. Take 3 tables with 3 boxes on each and 2 apples in each box. Counting boxes first gives (3 \times 3) \times 2 = 9 \times 2 = 18. Counting one table first gives 3 \times 2 = 6 apples per table and 3 \times (3 \times 2) = 3 \times 6 = 18. They count the same apples, so they are equal.
Yes. If neither number is 0, there is at least 1 group with at least 1 thing in it, so the product is at least 1, not 0. So a product of 0 needs a factor of 0.