Properties of Addition and Multiplication: Free Response

Read the tray one way first. There are 7 rows, and each row holds 12 muffins, so the count is 7 groups of 12.

$7 \times 12 = 84.$

Now turn the tray a quarter turn. Each of the original 7 rows is standing as a column of 12 muffins, so the tray now reads as 12 rows with 7 muffins in each row, which is 12 groups of 7.

$12 \times 7 = 84.$

The tray holds 84 muffins, and both readings report that same 84.

The turn moved the tray, not the muffins on it. Not one muffin was added, taken away, or split in two, so after the turn you are looking at exactly the collection you were looking at before.

A collection of objects has one definite size. Counting by rows and counting by columns are two routes through that collection, and each route visits every muffin exactly once, so each route has to arrive at that one size.

$7 \times 12 = 84 = 12 \times 7.$

The equality is guaranteed rather than a lucky feature of these two numbers, and the giveaway is that the argument never used their values. Any rectangular arrangement can be read along either direction, so reversing the order of two factors can never change a product.

Lay the 9-foot board against the wall starting at the corner, and the 4-foot board immediately after it. Mark the wall at the far end of the pair. The two boards now cover one definite stretch of wall, and the length of that stretch is what $9 + 4$ measures.

Now swap the two boards, keeping them end to end and keeping the same starting corner. Neither board was cut, stretched or overlapped, and no gap was opened between them, so the pair still runs from the corner out to the same mark. The stretch of wall is the stretch you had before, so its length is unchanged, and that length is what $4 + 9$ measures.

$9 + 4 = 13 = 4 + 9.$

The feature carrying the argument is that swapping rearranges the pieces without altering any piece or the space they fill between them. Because the argument never used the numbers 9 and 4, it works for any two lengths, which is exactly what a property has to do.

Keep the three numbers in the order they are written, and choose which pair to combine first.

Combining the first pair:

$(47 + 3) + 68 = 50 + 68 = 118.$

Combining the last pair:

$47 + (3 + 68) = 47 + 71 = 118.$

The two routes pass through different intermediate values, 50 and 71, and still arrive at the same total of 118.

Combining the first pair:

$(5 \times 20) \times 43 = 100 \times 43 = 4300.$

Combining the last pair:

$5 \times (20 \times 43) = 5 \times 860 = 4300.$

Both groupings give 4300. The first is much lighter work by hand, and the reason is specific: pairing the 5 with the 20 produces a round 100, and multiplying by 100 is a single step. The other grouping forces the awkward $20 \times 43$ first and then a multiplication by 5 on top of it.

Nothing about the three numbers changed between the two routes, and neither did their order. Only the choice of which multiplication to perform first changed, so choosing the convenient route costs nothing in accuracy.

A reader given $47 + 3 + 68$ can put the parentheses in either place:

$(47 + 3) + 68 = 118, \qquad 47 + (3 + 68) = 118.$

Both land on 118. That is what the plain form needs: every way of filling the parentheses in gives the same total, so the three symbols name one definite number.

Now suppose addition were not like that, and the two choices could disagree. The same three symbols would name two different numbers at once. A reader would have to be told which pair to combine first before the sum meant anything, in the way that $10 - 4 - 2$ has to be read from left to right. The value would then rest on knowing that rule rather than on the three numbers, and regrouping would no longer be safe.

The same holds for a product of three numbers, which is why $5 \times 20 \times 43$ can be written plainly too. The parentheses are not left out because a reader stops caring about them. They are left out because there is nothing left for them to say.

"The garden added no new plots" says the number of plots added over the month was 0, so the count after the additions is the old count with 0 added to it.

$58 + 0 = 58.$

The garden still has 58 plots.

"Exactly one water tap in every plot" says the taps come as 58 groups of one, which is a product.

$58 \times 1 = 58.$

There are 58 taps. The two calculations do different arithmetic to the same 58 and both hand it straight back. That is what it means to call 0 the additive identity and 1 the multiplicative identity, seen in a real count.

An identity has to work for every number, so the useful tests are on numbers unlike 58, including the awkward case where the number itself is 0.

$7 + 0 = 7, \qquad 7 \times 1 = 7.$

$200 + 0 = 200, \qquad 200 \times 1 = 200.$

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

Every test hands the number back unchanged. What that does is make the claim believable and rule out the possibility that 58 was special. What it does not do is establish the claim, because whatever list of numbers you try, there are always numbers you did not try. A single failure, on the other hand, would have finished the claim on the spot, and none of the three fails.

Check the arithmetic first. Adding 3 to 0 does give 3, so the equation $0 + 3 = 3$ is true and there is nothing to correct in it.

The conclusion is what fails. An additive identity is a number that leaves every number unchanged when it is added, so for 3 to be one, the equation would have to say that some number came back unchanged after 3 was added to it. This one says the opposite: the 3 is the number that survived, and the 0 is the number that was added to it and did nothing.

And 3 fails the test outright on the garden's own count:

$58 + 3 = 61,$

which is not 58. Adding 3 changed the count, so 3 is not an additive identity. The number doing nothing in the student's own equation is the 0.

Both counts are of the form "this many groups, each of that size", so both are products. What differs is which number plays which role.

Seats in service counts the 52 seats on a bus once for every bus that runs, and no bus runs, so there are no groups of 52 to add up at all:

$0 \times 52 = 0.$

There are 0 seats in service. Passengers at the depot counts the passengers aboard each of the 8 parked buses, and every one of those buses holds none:

$8 \times 0 = 0.$

There are 0 passengers at the depot.

Reading multiplication as repeated addition, a product of two whole numbers adds up equal groups. The first number says how many groups there are, and the second says how much sits in each group.

In $0 \times 52$ there are no groups at all, so there is nothing to add up:

$0 \times 52 = 0.$

In $8 \times 0$ there are 8 groups, but every one of them holds nothing. Adding 0 eight times keeps handing back the same nothing:

$8 \times 0 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0.$

The two stories really are different. The first is a depot with no bus in service to count seats on; the second is 8 real buses, each with nobody aboard. They reach the value 0 for different reasons, and that is why it is worth stating the property for both positions: a product with a factor of 0 is 0 whichever side the 0 sits on.

Try both multipliers on one of the company's counts that is not itself zero, the 52 seats on a bus.

$52 \times 1 = 52, \qquad 52 \times 0 = 0.$

Multiplying by 1 handed the 52 straight back, so the claim is right about 1: fifty-two groups of one is fifty-two, and that is what makes 1 the multiplicative identity. Multiplying by 0 returned 0, not 52, so the claim is wrong about 0. Far from changing nothing, 0 replaces whatever it meets with 0, and part A showed it doing that from either side of the product.

The two roles are opposites, not variations on one idea. The identity 1 preserves a number; a factor of 0 collapses it. The claim runs them together because 0 and 1 are both small and familiar, but a single pair of calculations separates them.

The parentheses say which pair to combine first, so do the bracketed subtraction and then finish.

$(20 - 7) - 5 = 13 - 5 = 8.$

$20 - (7 - 5) = 20 - 2 = 18.$

The same three numbers, in the same order, gave 8 under one grouping and 18 under the other.

Choose the three numbers so that nothing but whole numbers appears. Taking 36, 6 and 3 in that order:

$(36 \div 6) \div 3 = 6 \div 3 = 2.$

$36 \div (6 \div 3) = 36 \div 2 = 18.$

One grouping gives 2 and the other gives 18, from the same three numbers written in the same order, so the choice of first pair is not free for division. Plenty of other triples work just as well.

"Regrouping never changes the value" is a claim about every choice of three numbers. Your pair from part B is one choice where the two groupings disagree:

$(36 \div 6) \div 3 = 2, \qquad 36 \div (6 \div 3) = 18.$

The claim said every, and here is one that fails, so it is contradicted exactly as it stands.

A pair that agreed would do nothing of the kind. Whatever triples you check, endlessly many are left untried, and the next one is exactly where a failure would be hiding. Agreeing examples make a claim believable; they cannot make it certain.

The addition claim is settled by the row of sticks in the lesson instead. Gluing the first joint before the second, or the second before the first, leaves the same row covering the same length, and no step of that argument depends on how long the sticks are.