Number (mult/div)

UK Primary (3-6)

Properties of multiplication (10x10)

Lesson

Before we get too far into this one, have a play with the applet below.

- Create an array by using the sliders
- Then rotate the array around to a different orientation
- Does the size of the array change?
- Why do we write the multiplication differently?
- Are the answers to the multiplication the same?

With multiplication, it doesn't matter which way two numbers are multiplied together, we get the same answer each time. This is called the commutative property of multiplication or turnarounds.

You might like to refresh your memory on using arrays for multiplication (and division), as this is what we are going to use to prove that we can multiply numbers in any order.

Now that we know we can multiply numbers in any order, let's look at how we can use them to solve some multiplication questions. We will look at multiplying by 0 and multiplying by 1.

Did you know?

Not only can you multiply two numbers in any order, you can multiply any quantity of numbers, in any order.

Why don't you see if you can prove that this works? Perhaps you can solve $3\times4\times5$3×4×5 and see if you get the same answer as you do when you solve $5\times4\times3$5×4×3. Can you write this problem in a different order again, and solve it?

The *commutative property* of multiplication says that the order of numbers doesn't matter in multiplication. For example, $3\times7=7\times3$3×7=7×3.

Use the commutative property to complete the following statements:

If $5\times3=15$5×3=15, then $3\times5=\editable{}$3×5=.

If $5\times4=20$5×4=20, then $4\times5=\editable{}$4×5=.

If $5\times\editable{}=35$5×=35, then $\editable{}\times5=35$×5=35.

Answer the following questions:

When $16$16 is multiplied by $1$1, the answer is:

$1$1

A$16$16

B$0$0

C$1$1

A$16$16

B$0$0

CWhen $1195$1195 is multiplied by $1$1, the answer is:

$1$1

A$1195$1195

B$0$0

C$1$1

A$1195$1195

B$0$0

CWhen any number (that isn't zero) is multiplied by $1$1, the answer is:

$0$0

A$1$1

BItself

C$0$0

A$1$1

BItself

C

Let's see if we can work out $4\times5\times3$4×5×3 by multiplying any pair of the numbers first.

First find $4\times5=\editable{}$4×5=.

Then multiply the result by $3$3, to get $\editable{}\times3=\editable{}$×3=.

Now find $5\times3=\editable{}$5×3=.

Then multiply the result by $4$4, to get $4\times\editable{}=\editable{}$4×=.

This time, find $4\times3=\editable{}$4×3=.

Then multiply the result by $5$5, to get $\editable{}\times5=\editable{}$×5=.

Does the order of multiplication matter?

No

AYes

BNo

AYes

B