Math 208, Section 4.6 Solutions: The Commutative and Associative
Properties of Addition and Mental Math
2. Figure 4.45 indicates a “make-a-10” strategy for adding 6 + 8.
a. Write equations that correspond to the make-a-10 strategy for adding 6 + 8 depicted in Figure
4.45. Your equations should make careful and appropriate use of parentheses. Which property
of arithmetic do you equations and the picture of Figure 4.45 illustrate?
6 + 8 = 6 + (4 + 4) = (6 + 4) + 4 = 10 + 4 = 14
The equations illustrate the associative property of arithmetic because we first view 6 + 8 as 6 + (4 +
4), then we “regroup” and see 6 + 8 as (6 + 4) + 4.
b. Draw a picture for 7 + 5 that illustrates a make-a-10 strategy. Write equations that correspond
to the strategy in your picture, making careful and appropriate use of parentheses.
7 + 5 = 7 + (3 + 2) = (7 + 3) + 2 = 10 + 2 = 12
We have seven black dots and 5 gray dots. We can split the gray dots into a group of 3 and a group of
2. If we add the group of 3 gray dots to the 7 black dots, we get 10 dots plus 2 extra dots. So, we have
12 dots in all.
4. When we use the standard subtraction algorithm to subtract 52 – 19 we must regroup. Which
property or properties of arithmetic do we use when we regroup? Answer this question by
writing 52 in expanded form and by using parentheses carefully to show regrouping.
We almost always use both the associative and commutative properties of addition when we do a
multidigit computation.
52 – 19 = (50 + 2 ) – ( 10 + 9)
= (50 – 10 ) + (2 – 9) = 40 + (2 – 9)
commutative law
then we notice that 9 is too big to take away from 2 (without using negative numbers) so now is
when we regroup:
52 – 19 = (50 + 2 ) – ( 10 + 9) = 40 + (2 – 9)
= (30 + 10) + (2 – 9)
right here is the associative law
= 30 + (10 + 2 – 9)
= 30 + (12 – 9)
= 30 + 3
= 33.
So in the regrouping process itself, we don’t actually use the commutative law.
Page 1 of 3
6. To solve 201 – 88, a student writes the following equations:
88 + 2 = 90 +10 = 100 + 100 = 200 + 1 = 201
2 + 10 = 12 + 100 = 112 + 1 = 113
Although the student has a good idea for solving the problem, his equations are not correct. In
words, describe the student’s solution strategy and explain why it makes sense; then write
correct equations that correspond to this solution strategy.
This student is writing equations as if the equal sign means “go on to the next step” instead of “these
two expressions have the same value.” This student is using the “counting on” method of subtraction.
He wants to know what to add to 88 in order to get 201. So he is really thinking of the equation
88 + ? = 201.
He adds 2 to get 90; then adds 10 to get 100 (aiming for round numbers); then adds 100 to get 200;
then adds 1 to get 201. So he has added 2 and 10 and 100 and 1, which adds up to 113.
The counting on method doesn’t lend itself well to a single series of equations (like A=B=C=D=...) but
you can keep track of it with different equations:
88 +
2 = 90
90 + 10 = 100
100 + 100 = 200
200 +
1 = 201
The numbers in the right column before the equal signs are what we had to add so the total is 2 + 10
+ 100 + 1.
10. Describe a way to solve 304 – 81 mentally, by using reasoning other than the common
subtraction algorithm. Then write a coherent sequence of equations that correspond to your
reasoning.
304 – 81 is very close to 304 – 80, which equals 224; but we needed to subtract one more, so the
answer is 223.
As a set of equations:
304 – 81 = 304 – (80 + 1) = (304 – 80) – 1 = 224 – 1 = 223.
Page 2 of 3
13. Is there an associative property of subtraction? In other words, is A – (B – C) = (A – B) – C
true for all real numbers A, B, C? Explain your answer. If the equation is always true, explain
why; if the equation is not always true, find another expression that is equal to the expression
A – (B – C) and explain why the two expressions are equal.
No, there is not an associative property of subtraction. The following example, where A = 5, B = 3, and
C = 2, shows that an associative property of subtraction is not always true:
5 – (3 – 2) = 5 – 1 = 4, while (5 – 3) – 2 = 2 – 2 = 0.
The equation A – (B – C) = A – (B + (-C)) = A + (-B) + (-(-C)) = A – B + C is always true. To see
why, we can use the distributive property of multiplication and the fact that “minus B” is the same as
“negative B”. First, we distribute the negative outside the parentheses. Then we remember that “minus
B” is the same as “negative B”, and a double negative is positive. So, A – (B – C) = A – B + C.
Page 3 of 3