# 7.2 Commutative and associative properties

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By the end of this section, you will be able to:
• Use the commutative and associative properties
• Evaluate expressions using the commutative and associative properties
• Simplify expressions using the commutative and associative properties

Before you get started, take this readiness quiz.

1. Simplify: $7y+2+y+13.$
If you missed this problem, review Evaluate, Simplify and Translate Expressions .
2. Multiply: $\frac{2}{3}·18.$
If you missed this problem, review Multiply and Divide Fractions .
3. Find the opposite of $15.$
If you missed this problem, review Subtract Integers .

In the next few sections, we will take a look at the properties of real numbers. Many of these properties will describe things you already know, but it will help to give names to the properties and define them formally. This way we’ll be able to refer to them and use them as we solve equations in the next chapter.

## Use the commutative and associative properties

Think about adding two numbers, such as $5$ and $3.$

$\begin{array}{cccc}\hfill 5+3\hfill & & & \hfill 3+5\hfill \\ \hfill 8\hfill & & & \hfill 8\hfill \end{array}$

The results are the same. $5+3=3+5$

Notice, the order in which we add does not matter. The same is true when multiplying $5$ and $3.$

$\begin{array}{cccc}\hfill 5·3\hfill & & & \hfill 3·5\hfill \\ \hfill 15\hfill & & & \hfill 15\hfill \end{array}$

Again, the results are the same! $5·3=3·5.$ The order in which we multiply does not matter.

These examples illustrate the commutative properties of addition and multiplication.

## Commutative properties

Commutative Property of Addition : if $a$ and $b$ are real numbers, then

$a+b=b+a$

Commutative Property of Multiplication : if $a$ and $b$ are real numbers, then

$a·b=b·a$

The commutative properties have to do with order. If you change the order of the numbers when adding or multiplying, the result is the same.

Use the commutative properties to rewrite the following expressions:

$\phantom{\rule{0.2em}{0ex}}-1+3=_____$

$\phantom{\rule{0.2em}{0ex}}4·9=_____$

## Solution

 ⓐ $-1+3=_____$ Use the commutative property of addition to change the order. $-1+3=3+\left(-1\right)$
 ⓑ $4·9=_____$ Use the commutative property of multiplication to change the order. $4·9=9·4$

Use the commutative properties to rewrite the following:

1. $\phantom{\rule{0.2em}{0ex}}-4+7=_____\phantom{\rule{0.2em}{0ex}}$
2. $\phantom{\rule{0.2em}{0ex}}6·12=_____$

1. −4 + 7 = 7 + (−4)
2. 6 · 12 = 12 · 6

Use the commutative properties to rewrite the following:

1. $\phantom{\rule{0.2em}{0ex}}14+\left(-2\right)=_____\phantom{\rule{0.2em}{0ex}}$
2. $\phantom{\rule{0.2em}{0ex}}3\left(-5\right)=_____$

1. 14 + (−2) = −2 + 14
2. 3(−5) = (−5)3

What about subtraction? Does order matter when we subtract numbers? Does $7-3$ give the same result as $3-7?$

$\begin{array}{ccc}\hfill 7-3\hfill & & \hfill 3-7\hfill \\ \hfill 4\hfill & & \hfill -4\hfill \\ & \hfill 4\ne -4\hfill & \end{array}$
$\text{The results are not the same.}\phantom{\rule{0.2em}{0ex}}7-3\ne 3-7$

Since changing the order of the subtraction did not give the same result, we can say that subtraction is not commutative.

Let’s see what happens when we divide two numbers. Is division commutative?

$\begin{array}{ccc}\hfill 12÷4\hfill & & \hfill 4÷12\hfill \\ \hfill \frac{12}{4}\hfill & & \hfill \frac{4}{12}\hfill \\ \hfill 3\hfill & & \hfill \frac{1}{3}\hfill \\ & \hfill 3\ne \frac{1}{3}\hfill & \end{array}$
$\text{The results are not the same. So}\phantom{\rule{0.2em}{0ex}}12÷4\ne 4÷12$

Since changing the order of the division did not give the same result, division is not commutative.

Addition and multiplication are commutative. Subtraction and division are not commutative.

Suppose you were asked to simplify this expression.

$7+8+2$

Some people would think $7+8\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}15$ and then $15+2\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}17.$ Others might start with $8+2\phantom{\rule{0.2em}{0ex}}\text{makes}\phantom{\rule{0.2em}{0ex}}10$ and then $7+10\phantom{\rule{0.2em}{0ex}}\text{makes}\phantom{\rule{0.2em}{0ex}}17.$

Both ways give the same result, as shown in [link] . (Remember that parentheses are grouping symbols that indicate which operations should be done first.)

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