Properties of Addition and Multiplication: Additional Practice (Set 2)

Compare the two sides: the grouping is identical, and only the order inside the parentheses flipped.

$b + c \;\longrightarrow\; c + b$

Reversing the order of those two terms is the commutative property applied to the inner sum.

Compare the first two expressions. The grouping into $a$ and the block $(b \times c)$ is identical; only their order is swapped.

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

Moving the factor $a$ from the front to the back is the commutative property of multiplication.

Reading $0 \times a$ as groups, it asks for zero groups of $a$. You never pick up a group at all, so there is nothing there.

$0 \times a = 0$

The zero property works from either side, so $a \times 0 = 0$ as well. The other three expressions all hand back $a$ unchanged.

The pair $8 \times 125 = 1000$ is round, and it is already the leading pair, so reading left to right hands it to you with nothing moved or regrouped.

$8 \times 125 \times 7 = (8 \times 125) \times 7 = 1000 \times 7 = 7000$

Associativity is why a product may be written without parentheses at all. Here you did not have to use that freedom, because the convenient pair came first already.

Adding $0$ leaves a value unchanged, and multiplying by $1$ leaves a value unchanged.

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

So $0$ is the additive identity and $1$ is the multiplicative identity.

Both routes reach the same total.

$(6 \times 25) \times 4 = 150 \times 4 = 600, \qquad 6 \times (25 \times 4) = 6 \times 100 = 600$

That is one case agreeing, which is what the associative property predicts. Her check confirms this case; what makes it certain for every choice of factors is the general reason, not this one pair of calculations. No factor moved order, so commutativity is not involved.

Division is not associative, so the grouping is fixed by the left-to-right convention, not chosen freely.

$16 \div 4 \div 2 = (16 \div 4) \div 2 = 4 \div 2 = 2$

The student grouped the right pair first, computing $4 \div 2 = 2$ and then $16 \div 2 = 8$, which changes the value.

Rewrite $32$ as $4 \times 8$, then use the associative property to pair $25$ with the $4$.

$25 \times (4 \times 8) = (25 \times 4) \times 8 = 100 \times 8 = 800$

We need the factor that turns $50 \times \square$ into $100$.

$50 \times 2 = 100, \qquad (50 \times 2) \times 7 = 100 \times 7 = 700$

So the blank is $2$.

Move the $25$ next to the $4$ (commutative), then group the two pairs (associative).

$4 \times 7 \times 25 \times 3 = (4 \times 25) \times (7 \times 3) = 100 \times 21 = 2100$

Addition and multiplication commute, and addition is associative, so the first three hold for every choice of numbers.

$9 - 4 = 5, \qquad 4 - 9 \ne 5$

Since $a$ and $b$ are different, one of them is the larger, as $9$ and $4$ are here. Taking the smaller from the larger leaves something above zero, while the reversed subtraction drops below zero. The two sides could only agree if $a$ and $b$ were equal, and they are not, so $a - b = b - a$ must be false.

Associativity claims that regrouping never changes a sum, so the only thing that can contradict it is a case where the two groupings disagree.

$(a + b) + c \ne a + (b + c)$

A disagreement between $a + b$ and $b + a$ would be about order rather than grouping, and a single sum that agrees confirms nothing on its own. What rules the counterexample out is the argument in the lesson: the three sticks cover the same row whichever joint you glue first.