# 1.9 Properties of real numbers

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By the end of this section, you will be able to:
• Use the commutative and associative properties
• Use the identity and inverse properties of addition and multiplication
• Use the properties of zero
• Simplify expressions using the distributive property

A more thorough introduction to the topics covered in this section can be found in the Prealgebra chapter, The Properties of Real Numbers .

## Use the commutative and associative properties

Think about adding two numbers, say 5 and 3. The order we add them doesn’t affect the result, does it?

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

The results are the same.

As we can see, the order in which we add does not matter!

What about multiplying $5\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}3?$

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

Again, the results are the same!

The order in which we multiply does not matter!

These examples illustrate the commutative property . When adding or multiplying, changing the order gives the same result.

## Commutative property

$\begin{array}{ccccccccc}\mathbf{\text{of Addition}}\hfill & & & \text{If}\phantom{\rule{0.2em}{0ex}}a,b\phantom{\rule{0.2em}{0ex}}\text{are real numbers, then}\hfill & & & \hfill a+b& =\hfill & b+a\hfill \\ \mathbf{\text{of Multiplication}}\hfill & & & \text{If}\phantom{\rule{0.2em}{0ex}}a,b\phantom{\rule{0.2em}{0ex}}\text{are real numbers, then}\hfill & & & \hfill a·b& =\hfill & b·a\hfill \end{array}$

When adding or multiplying, changing the order gives the same result.

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

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

$\begin{array}{c}\hfill \begin{array}{cc}\hfill 7-3\hfill & \hfill \phantom{\rule{1em}{0ex}}3-7\hfill \\ \hfill 4\hfill & \hfill \phantom{\rule{1em}{0ex}}-4\hfill \end{array}\hfill \\ \\ \\ \hfill \phantom{\rule{0.5em}{0ex}}4\ne \text{−}4\hfill \\ \hfill 7-3\ne 3-7\hfill \end{array}$

The results are not the same.

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

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

$\begin{array}{}\\ \hfill \begin{array}{cc}\hfill 12÷4\hfill & \hfill \phantom{\rule{1em}{0ex}}4÷12\hfill \\ \hfill \frac{12}{4}\hfill & \hfill \phantom{\rule{1em}{0ex}}\frac{4}{12}\hfill \\ \hfill 3\hfill & \hfill \phantom{\rule{1em}{0ex}}\frac{1}{3}\hfill \end{array}\hfill \\ \hfill 3\ne \frac{1}{3}\hfill \\ \hfill 12÷4\ne 4÷12\hfill \end{array}$

The results are not the same.

Since changing the order of the division did not give the same result, division is not commutative . The commutative properties only apply to addition and multiplication!

• Addition and multiplication are commutative.
• Subtraction and Division are not commutative.

If you were asked to simplify this expression, how would you do it and what would your answer be?

$7+8+2$

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 $7+10\phantom{\rule{0.2em}{0ex}}\text{makes}\phantom{\rule{0.2em}{0ex}}17.$

Either way gives the same result. Remember, we use parentheses as grouping symbols to indicate which operation should be done first.

$\begin{array}{c}\begin{array}{ccc}& & \hfill \phantom{\rule{4em}{0ex}}\left(7+8\right)+2\hfill \\ \text{Add}\phantom{\rule{0.2em}{0ex}}7+8.\hfill & & \hfill \phantom{\rule{4em}{0ex}}15+2\hfill \\ \text{Add.}\hfill & & \hfill \phantom{\rule{4em}{0ex}}17\hfill \\ \\ \\ & & \hfill \phantom{\rule{4em}{0ex}}7+\left(8+2\right)\hfill \\ \text{Add}\phantom{\rule{0.2em}{0ex}}8+2.\hfill & & \hfill \phantom{\rule{4em}{0ex}}7+10\hfill \\ \text{Add.}\hfill & & \hfill \phantom{\rule{4em}{0ex}}17\hfill \end{array}\hfill \\ \\ \\ \left(7+8\right)+2=7+\left(8+2\right)\hfill \end{array}$

When adding three numbers, changing the grouping of the numbers gives the same result.

This is true for multiplication, too.

$\begin{array}{c}\begin{array}{ccc}& & \hfill \phantom{\rule{2em}{0ex}}\left(5·\frac{1}{3}\right)·3\hfill \\ \text{Multiply.}\phantom{\rule{1em}{0ex}}5·\frac{1}{3}\hfill & & \hfill \phantom{\rule{2em}{0ex}}\frac{5}{3}·3\hfill \\ \text{Multiply.}\hfill & & \hfill \phantom{\rule{2em}{0ex}}5\hfill \\ \\ \\ & & \hfill \phantom{\rule{2em}{0ex}}5·\left(\frac{1}{3}·3\right)\hfill \\ \text{Multiply.}\phantom{\rule{1em}{0ex}}\frac{1}{3}·3\hfill & & \hfill \phantom{\rule{2em}{0ex}}5·1\hfill \\ \text{Multiply.}\hfill & & \hfill \phantom{\rule{2em}{0ex}}5\hfill \end{array}\hfill \\ \\ \\ \left(5·\frac{1}{3}\right)·3=5·\left(\frac{1}{3}·3\right)\hfill \end{array}$

When multiplying three numbers, changing the grouping of the numbers gives the same result.

You probably know this, but the terminology may be new to you. These examples illustrate the associative property .

## Associative property

$\begin{array}{cccc}\mathbf{\text{of Addition}}\hfill & & & \text{If}\phantom{\rule{0.2em}{0ex}}a,b,c\phantom{\rule{0.2em}{0ex}}\text{are real numbers, then}\phantom{\rule{0.2em}{0ex}}\left(a+b\right)+c=a+\left(b+c\right)\hfill \\ \mathbf{\text{of Multiplication}}\hfill & & & \text{If}\phantom{\rule{0.2em}{0ex}}a,b,c\phantom{\rule{0.2em}{0ex}}\text{are real numbers, then}\phantom{\rule{0.2em}{0ex}}\left(a·b\right)·c=a·\left(b·c\right)\hfill \end{array}$

When adding or multiplying, changing the grouping gives the same result.

Let’s think again about multiplying $5·\frac{1}{3}·3.$ We got the same result both ways, but which way was easier? Multiplying $\frac{1}{3}$ and $3$ first, as shown above on the right side, eliminates the fraction in the first step. Using the associative property can make the math easier!

Priam has pennies and dimes in a cup holder in his car. The total value of the coins is $4.21 . The number of dimes is three less than four times the number of pennies. How many pennies and how many dimes are in the cup? Cecilia Reply Arnold invested$64,000 some at 5.5% interest and the rest at 9% interest how much did he invest at each rate if he received $4500 in interest in one year Heidi Reply List five positive thoughts you can say to yourself that will help youapproachwordproblemswith a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often. Elbert Reply Avery and Caden have saved$27,000 towards a down payment on a house. They want to keep some of the money in a bank account that pays 2.4% annual interest and the rest in a stock fund that pays 7.2% annual interest. How much should they put into each account so that they earn 6% interest per year?
324.00
Irene
1.2% of 27.000
Irene
i did 2.4%-7.2% i got 1.2%
Irene
I have 6% of 27000 = 1620 so we need to solve 2.4x +7.2y =1620
Catherine
I think Catherine is on the right track. Solve for x and y.
Scott
next bit : x=(1620-7.2y)/2.4 y=(1620-2.4x)/7.2 I think we can then put the expression on the right hand side of the "x=" into the second equation. 2.4x in the second equation can be rewritten as 2.4(rhs of first equation) I write this out tidy and get back to you...
Catherine
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Joshua
will every polynomial have finite number of multiples?
a=# of 10's. b=# of 20's; a+b=54; 10a + 20b=$910; a=54 -b; 10(54-b) + 20b=$910; 540-10b+20b=$910; 540+10b=$910; 10b=910-540; 10b=370; b=37; so there are 37 20's and since a+b=54, a+37=54; a=54-37=17; a=17, so 17 10's. So lets check. $740+$170=$910. David Reply . A cashier has 54 bills, all of which are$10 or $20 bills. The total value of the money is$910. How many of each type of bill does the cashier have?
whats the coefficient of 17x
the solution says it 14 but how i thought it would be 17 im i right or wrong is the exercise wrong
Dwayne
17
Melissa
wow the exercise told me 17x solution is 14x lmao
Dwayne
thank you
Dwayne
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90 minutes