Properties of Addition and Multiplication

Learning goals

Tell the commutative property (order) apart from the associative property (grouping)
Explain why reordering or regrouping cannot change a sum or a product
Identify $0$ and $1$ as the identities, and say why multiplying by $0$ gives $0$
Reorder and regroup a long sum or product to reach the easiest path
Give a counterexample for each way subtraction and division fail these two rules

What a “property” is claiming

A property of an operation is a statement that holds for every choice of numbers, not just a lucky few. So $a + b = b + a$ is not the single fact ” $3 + 5$ equals $8$ .” It is the promise that the swap works no matter what $a$ and $b$ are. The letters $a$ , $b$ , and $c$ below stand for numbers you pick.

The pictures here use whole numbers, because those are the easiest to draw, so the reasons below are given for whole numbers. Each rule keeps working for the fractions and negative numbers you meet later. An example shows the pattern once; a general reason explains why the pattern keeps working, and every rule below comes with one.

Three words save a lot of repetition. Any calculation written out in symbols, such as $3 + 5$ or $10 - 4 - 2$ , is an expression: it names a value without yet being worked out. The numbers being added in a sum are its terms, so $3$ and $5$ are the terms of $3 + 5$ . The numbers being multiplied in a product are its factors, so $6$ and $9$ are the factors of $6 \times 9$ .

The commutative property: order does not matter

Swapping the two inputs of an addition or a multiplication leaves the result unchanged. Check that on a pair of numbers: $3 + 8 = 11$ and $8 + 3 = 11$ , while $6 \times 9 = 54$ and $9 \times 6 = 54$ . An operation that behaves this way is called commutative, and addition and multiplication both are:

a + b = b + a, \qquad a \times b = b \times a.

The name comes from “commute,” to move from place to place: the numbers may move past each other freely.

Why $a + b = b + a$ #

Lay a stick of length $4$ end to end with a stick of length $9$ , and the row measures $13$ . Now swap the two pieces, so the $9$ -stick comes first and the $4$ -stick second. Nothing was stretched or cut, so it is the same row and it still measures $13$ . That is why $4 + 9$ and $9 + 4$ both come to $13$ .

Nothing in that argument used the lengths $4$ and $9$ . Call the two lengths $a$ and $b$ instead, and every step reads the same. Laying them the other way round gives the same row, measured from the other end. So $a + b = b + a$ for any two lengths.

For multiplication the same idea becomes an array of dots that you can count two ways.

The same dots read two ways. Three rows of five and five rows of three both count to 15, which is why 3 times 5 equals 5 times 3.

Why $a \times b = b \times a$ #

Take the grid drawn above, $3$ rows with $5$ dots in each row. Counting row by row gives $3 \times 5 = 15$ dots. Now read the very same grid by columns: it has $5$ columns with $3$ dots in each, which counts to $5 \times 3 = 15$ . Not one dot moved between the two counts, so both had to land on $15$ .

The grid did not have to be $3$ by $5$ . Call it $a$ rows of $b$ dots, and reading the same grid by columns gives $b$ columns of $a$ dots. Tilting your view adds no dots and removes none, so $a \times b = b \times a$ for any whole numbers $a$ and $b$ .

Those dots settle one pair of numbers, and a property claims something about every pair. So go hunting for a pair that breaks it. In the rectangle below you set the width and the height, and the readout counts the unit squares inside. Look for a width and a height whose swap changes that count.

Swapping width and height leaves the area alone

Width Height

A rectangle 5 units wide and 3 units tall. Area 15 square units, counted as 5 times 3. Turned on its side it is 3 times 5, which is the same 15 squares.

A rectangle built from unit squares. Choose its width and its height; the readout counts the squares and names the same count with the two factors swapped.

No pair breaks it, and the reason is the one the dots gave. A swap stands the rectangle on its end without moving a single square, so the count cannot change.

The associative property: grouping does not matter

Commutativity is about the order of two numbers. The next property is about something different. When you combine three numbers, you can only do one $+$ (or one $\times$ ) at a time. So you must choose which pair to combine first. The parentheses say “do this pair first.” Below, both choices are tried on $2$ , $7$ , and $3$ . The numbers hold their places; only the pair you combine first changes.

Two groupings of the same three numbers. Combining 2 and 7 first, or 7 and 3 first, sends you through 9 or through 10, and both routes land on 12.

Two different routes, the same total. An operation is associative when that choice of first pair never changes the result:

(a + b) + c = a + (b + c), \qquad (a \times b) \times c = a \times (b \times c).

Why regrouping is safe

Why $(a + b) + c = a + (b + c)$ #

Lay three sticks in a row: one of length $2$ , then one of length $7$ , then one of length $3$ . Fusing the first pair gives $(2 + 7) + 3 = 9 + 3 = 12$ . Fusing the last pair instead gives $2 + (7 + 3) = 2 + 10 = 12$ . The sticks never moved, so the row was $12$ long before either fusing began.

Nothing there depended on $2$ , $7$ , and $3$ . Lay sticks of length $a$ , $b$ , and $c$ in that order. The row has a total length already, before you decide how to add. Fusing $a$ to $b$ first, or $b$ to $c$ first, glues the same three pieces at the same two joints. Gluing joints in a different order cannot lengthen or shorten the row, so

(a + b) + c = a + (b + c).

Multiplication regroups for the same reason, with equal groups in place of a row of sticks. The figure below holds three bags, two boxes to a bag, and five marbles to a box.

Three bags of two boxes, five marbles to a box. Count the boxes first and you get 6 boxes of 5; count a bag first and you get 3 bags of 10. Either way there are 30 marbles.

Why $(a \times b) \times c = a \times (b \times c)$ #

Both routes in the figure count one fixed pile of marbles, so $(3 \times 2) \times 5 = 3 \times (2 \times 5)$ , with $30$ at the end of each.

Nothing there depended on $3$ , $2$ , and $5$ . Take $a$ bags, $b$ boxes to a bag, and $c$ marbles to a box. Grouping $a$ with $b$ counts the boxes first, and grouping $b$ with $c$ counts one bag first. The marbles are in the bags before you pick a route, so neither route can reach a different total.

Because the grouping never matters, we are allowed to drop the parentheses entirely and write $a + b + c$ or $a \times b \times c$ . Every way of putting them back gives the same value, so none of them is needed.

Check your understanding

Which of these equations shows the associative property?

7 \times 5 = 5 \times 7

(7 \times 5) \times 2 = 7 \times (5 \times 2)

7 \times 1 = 7

7 \times 0 = 0

Reshuffling a calculation

Used together, the two rules give you permission. You may reorder a sum or a product and then regroup it into whatever form is easiest, and the value will not change.

Worked example 1 Compute $4 \times 17 \times 25$ the easy way

Multiplying in the printed order gives $4 \times 17 = 68$ , then the clumsy $68 \times 25$ . Instead, move the $25$ next to the $4$ and group that pair first:

\begin{aligned} 4 \times 17 \times 25 &= 4 \times 25 \times 17 &&\text{(commutative: swap the order)}\\ &= (4 \times 25) \times 17 &&\text{(associative: group the easy pair)}\\ &= 100 \times 17\\ &= 1700. \end{aligned}

The two properties are what guarantee this rearranged product equals the original one, so nothing is lost by choosing the convenient route.

Worked example 2 Add $28 + 47 + 2$ by making a round number

The numbers $28$ and $2$ pair into a clean $30$ , so bring them together first:

\begin{aligned} 28 + 47 + 2 &= 28 + 2 + 47 &&\text{(commutative property of addition)}\\ &= (28 + 2) + 47 &&\text{(associative property of addition)}\\ &= 30 + 47\\ &= 77. \end{aligned}

Reordering to create the round $30$ first turns an awkward sum into one you can finish in your head.

Check your understanding

You want to work out $37 + 45 + 3$ in your head. Which pair should you bring together first?

37

and

45

, the first pair as written B

37

and

3

, because they make a round

40

45

and

3

, the last pair as written D It makes no difference, every grouping is equally quick

The identity elements: the numbers that change nothing

Adding nothing to a quantity leaves it as it was, so $17 + 0 = 17$ . Taking exactly one copy of a quantity is the quantity itself, so $17 \times 1 = 17$ . A number you can combine with any number and change nothing is called an identity element for that operation. The rule saying so is the identity property. Addition and multiplication each have one identity:

a + 0 = a, \qquad a \times 1 = a.

Zero is the additive identity and one is the multiplicative identity.

The zero property of multiplication

Multiplying any number by $0$ collapses it to $0$ . Keep the reading from the dot array: the first factor counts the groups, and the second counts what sits in each group. So $7 \times 0$ is seven groups with nothing in them, and seven helpings of nothing is nothing. Turn it round: $0 \times 7$ is zero groups of seven. You never pick up a group at all, so again you have nothing. The same reading works for every number:

a \times 0 = 0.

This is a separate fact from the identity property, and it is easy to mix them up. The identity $1$ leaves $a$ untouched; the number $0$ wipes $a$ out.

Check your understanding

Which equation shows the multiplicative identity?

9 + 0 = 9

9 \times 0 = 0

9 \times 1 = 9

9 \times 9 = 81

Both rules earn their keep when they let you skip work:

Worked example 3 Simplify $(53 \times 0) + (1 \times 38)$

Use the zero property on the first product and the multiplicative identity on the second:

\begin{aligned} (53 \times 0) + (1 \times 38) &= 0 + 38 &&\text{(zero property; multiplicative identity)}\\ &= 38 &&\text{(additive identity)}. \end{aligned}

The entire first product collapses to $0$ , multiplying $38$ by $1$ leaves it alone, and adding $0$ to $38$ keeps it at $38$ . Three properties, one line.

Why subtraction and division do not get these properties

The order rule and the grouping rule were both stated for $+$ and $\times$ only, and that restriction is not an accident. Subtraction and division are neither commutative nor associative. To prove a “for every number” claim false you only need a single example where it breaks.

Order matters for both. Start from $5 - 3 = 2$ . Reversing it asks for $3 - 5$ , which means taking $5$ away from just $3$ . You run out before you finish, dropping below zero, so the result cannot also be $2$ . Division reverses just as badly. From $8 \div 2 = 4$ , swapping to $2 \div 8$ asks how many $8$ s fit inside $2$ . Not even one does, so the answer is less than a whole, nowhere near $4$ :

5 - 3 \ne 3 - 5, \qquad 8 \div 2 \ne 2 \div 8.

Grouping matters too, which is why an expression like $10 - 4 - 2$ would be genuinely ambiguous without a rule. The two groupings disagree:

Subtraction is not associative#

Take $a = 10$ , $b = 4$ , and $c = 2$ .

Grouping the first pair:

(a - b) - c = (10 - 4) - 2 = 6 - 2 = 4.

Grouping the second pair:

a - (b - c) = 10 - (4 - 2) = 10 - 2 = 8.

Since $4 \ne 8$ , there is a choice of numbers for which $(a - b) - c$ and $a - (b - c)$ disagree. One counterexample is enough, so subtraction is not associative. (The same numbers also break commutativity: $10 - 4 = 6$ , but $4 - 10$ reverses which number is taken from which and cannot give $6$ .)

Division breaks in exactly the same way. Grouping the first pair, $(16 \div 4) \div 2$ is $4 \div 2 = 2$ . Grouping the second pair, $16 \div (4 \div 2)$ is $16 \div 2 = 8$ . The same three numbers gave $2$ one way and $8$ the other, so division is not associative either.

Because the grouping really does change the value, $10 - 4 - 2$ is not safe to regroup the way a sum is. We rescue it with a convention: read a chain of subtractions (or divisions) left to right, so $10 - 4 - 2$ means $(10 - 4) - 2 = 4$ . The parentheses there are doing real work and cannot be moved for free. Treat the rules of this lesson as belonging to $+$ and $\times$ alone.

What a chain of removals still allows

Losing both properties does not mean that nothing in a chain may ever move. Read $80 - 25 - 15$ left to right: it takes $25$ off $80$ , then takes $15$ off what is left. That is two removals from one starting amount, and the amount gone is $25 + 15 = 40$ whichever removal you make first.

80 - 25 - 15 = 40, \qquad 80 - 15 - 25 = 40.

A chain of divisions behaves the same way, with the two divisors multiplying instead of adding. Dividing by $12$ and then by $3$ cuts the number into $12 \times 3 = 36$ equal parts, and so does dividing by $3$ and then by $12$ :

720 \div 12 \div 3 = 20, \qquad 720 \div 3 \div 12 = 20.

Read that narrowly, because it is easy to over-claim. It is not commutativity, which would let the $80$ and the $25$ trade places. It is not associativity, which would let the last two numbers be grouped: $80 - (25 - 15)$ is $70$ , not $40$ . The first number of a chain never moves. Only the numbers being removed may trade places, and likewise the numbers being divided by. The chain depends on them just through their total, or through their product. Working in whole numbers, the removals must not add up past the number they are taken from, and each division must come out exactly.

Check your understanding

Is the statement $12 \div 4 = 4 \div 12$ true?

A True, division is commutative B False, because

12 \div 4 = 3

but

4 \div 12

is less than

1

C True, because both sides use the same two numbers D It depends on the numbers

Common mistakes

Assuming subtraction or division commute. $5 - 3 \ne 3 - 5$ and $8 \div 2 \ne 2 \div 8$ . A sum or a product can be reordered freely. A subtraction or a division gives no such licence, and the first number of a chain never moves.
Confusing commutative with associative. Commutative moves the order of two numbers ( $a + b = b + a$ ); associative changes the grouping of three ( $(a + b) + c = a + (b + c)$ ).
Thinking $a \times 0 = a$ . Multiplying by $0$ gives $0$ , which is only the same as $a$ when $a$ was already $0$ . The number that leaves every number unchanged under multiplication is the identity $1$ , not $0$ .
Mixing up the two identities. The additive identity is $0$ ( $a + 0 = a$ ); the multiplicative identity is $1$ ( $a \times 1 = a$ ). Adding $1$ or multiplying by $0$ does not leave a number alone.

Key takeaways

Commutative (order): $a + b = b + a$ and $a \times b = b \times a$ .
Associative (grouping): $(a + b) + c = a + (b + c)$ and $(a \times b) \times c = a \times (b \times c)$ . This is why $a + b + c$ needs no parentheses, and the two rules together let you reorder and regroup any sum or product into the easiest path.
Identity and zero: $a + 0 = a$ and $a \times 1 = a$ leave a number alone, while $a \times 0 = 0$ wipes it out. The zero property is a separate rule from the identities.
Subtraction and division are neither commutative nor associative; a single counterexample such as $(10 - 4) - 2 \ne 10 - (4 - 2)$ settles it, and we read them left to right by convention. A chain still lets the numbers being removed trade places, and likewise the divisors. It feels them only through their total or their product.

Properties of Addition and Multiplication

Learning goals

What a “property” is claiming

The commutative property: order does not matter

The associative property: grouping does not matter

Why regrouping is safe

Reshuffling a calculation

The identity elements: the numbers that change nothing

The zero property of multiplication

Why subtraction and division do not get these properties

What a chain of removals still allows

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)