# 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!

how many typos can we find...?
5
Joseph
In the LCM Prime Factors exercises, the LCM of 28 and 40 is 280. Not 420!
4x+7y=29,x+3y=11 substitute method of linear equation
substitute method of linear equation
Srinu
Solve one equation for one variable. Using the 2nd equation, x=11-3y. Substitute that for x in first equation. this will find y. then use the value for y to find the value for x.
bruce
I want to learn
Elizebeth
help
Elizebeth
I want to learn. Please teach me?
Wayne
1) Use any equation, and solve for any of the variables. Since the coefficient of x (the number in front of the x) in the second equation is 1 (it actually isn't shown, but 1 * x = x), use that equation. Subtract 3y from both sides (this isolates the x on the left side of the equal sign).
bruce
2) This results in x=11-3y. x is note in terms of y. Use that as the value of x and substitute for all x in the first equation. The first equation becomes 4(11-3y)+7y =29. Note that the only variable left in the first equation is the y. If you have multiple variable, then something is wrong.
bruce
3) Distribute (multiply) the 4 across 11-3y to get 44-12y. Add this to the 7y. So, the equation is now 44-5y=29.
bruce
4) Solve 44-5y=29 for y. Isolate the y by subtracting 44 from birth sides, resulting in -5y=-15. Now, divide birth sides by -5 (since you have -5y). This results in y=3. You now have the value of one variable.
bruce
5) The last step is to take the value of y from Step 4) and substitute into the 2nd equation. Therefore: x+3y=11 becomes x+3(3)=11. Then multiplying, x+9=11. Finally, solve for x by subtracting 9 from both sides. Therefore, x=2.
bruce
6) The ordered pair of (2, 3) is the proposed solution. To check, substitute those values into either equation. If the result is true, then the solution is correct. 4(2)+7(3)=8+21=29. TRUE! Finished.
bruce
At 1:30 Marlon left his house to go to the beach, a distance of 5.625 miles. He rose his skateboard until 2:15, and then walked the rest of the way. He arrived at the beach at 3:00. Marlon's speed on his skateboard is 1.5 times his walking speed. Find his speed when skateboarding and when walking.
divide 3x⁴-4x³-3x-1 by x-3
how to multiply the monomial
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike? Got questions? Get instant answers now!
how do u solve that question
Seera
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
Seera
Speed=distance ÷ time
Tremayne
x-3y =1; 3x-2y+4=0 graph
Brandon has a cup of quarters and dimes with a total of 5.55\$. The number of quarters is five less than three times the number of dimes
app is wrong how can 350 be divisible by 3.
June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?
Susanna if the first cooler holds five times the gallons then the other cooler. The big cooler holda 40 gallons and the 2nd will hold 8 gallons is that correct?
Georgie
@Susanna that person is correct if you divide 40 by 8 you can see it's 5 it's simple
Ashley
@Geogie my bad that was meant for u
Ashley
Hi everyone, I'm glad to be connected with you all. from France.
I'm getting "math processing error" on math problems. Anyone know why?
Can you all help me I don't get any of this
4^×=9