Properties of Addition and Multiplication: Additional Practice (Set 1)

The four terms $p$, $q$, $r$, $s$ stay in the same left-to-right order; only the parentheses are rearranged.

$(p + q) + (r + s) = p + (q + (r + s))$

Changing which terms are grouped first, with the order fixed, is the associative property of addition.

First simplify the second factor: $6 - 6 = 0$. The zero property of multiplication says any number times $0$ is $0$, so the product is $0$ without ever computing the first factor.

$(37 + 58) \times (6 - 6) = (37 + 58) \times 0 = 0$

The trap answer $95$ is just $37 + 58$; multiplying by the zero factor collapses the whole product to $0$.

Adding $0$ leaves a number unchanged, since $0$ is the additive identity.

$0 + 84 = 84$

In contrast $84 \times 0 = 0$, while $84 + 1 = 85$ and $84 + 84 = 168$ both change the value.

Division is not associative, so a chain of divisions is read left to right, not regrouped freely.

$48 \div 8 \div 2 = (48 \div 8) \div 2 = 6 \div 2 = 3$

The student grouped the right pair first, getting $8 \div 2 = 4$ and then $48 \div 4 = 12$, which changes the answer.

First the parentheses appear around $4 \times 5$ with the order unchanged (a regrouping), then the inner factors swap from $4 \times 5$ to $5 \times 4$ (a reorder).

$9 \times 4 \times 5 \;\xrightarrow{\text{associative}}\; 9 \times (4 \times 5) \;\xrightarrow{\text{commutative}}\; 9 \times (5 \times 4)$

So the steps are associative, then commutative.

Associativity keeps the three terms $36$, $14$, $25$ in the same order and only shifts the parentheses.

$(36 + 14) + 25 = 36 + (14 + 25)$

So the blank is $14$.

Apply the multiplicative identity, the zero property, and the additive identity in turn.

$(99 \times 1) + (37 \times 0) + (0 + 0) = 99 + 0 + 0 = 99$

Move the factors so the round pair $50 \times 2 = 100$ sits together (commutative), then group the two pairs (associative).

$50 \times 17 \times 2 \times 4 = (50 \times 2) \times (17 \times 4) = 100 \times 68 = 6800$

A single counterexample settles each claim. For subtraction, both order and grouping change the value.

$(9 - 5) - 1 = 3, \qquad 9 - (5 - 1) = 5$

The grouping changes the result ($3 \ne 5$), so subtraction is not associative, and reversing $5 - 3$ cannot give the same value as $3 - 5$, so it is not commutative either. Division fails on both counts in the same way, which makes the other options false.

A counterexample must make the two groupings give different results.

$(16 \div 4) \div 2 = 2, \qquad 16 \div (4 \div 2) = 8$

Since $2 \ne 8$, division is not associative. The other options are single quotients and say nothing about grouping.

The same eggs are counted twice, once as rows of $8$ and once as rows of $6$.

$6 \times 8 = 48 = 8 \times 6$

Turning the tray moves no eggs, so both counts have to land on $48$. Two factors traded places and no grouping changed, which makes this the commutative property, not the associative one.

Multiplying by $0$ always gives $0$, and multiplying by $1$ always gives $n$ back, so the two agree only when $n$ is itself $0$.

$0 \times 0 = 0 = 0 \times 1$

For any other number the zero property gives $0$ while the identity gives $n$, and those differ. This is why the zero property and the identity property are two rules, not one.