Adding and Subtracting Integers: Free Response

Both walks stand on $-11$, which is eleven units to the left of zero. Only the direction of the second step is different.

Adding $-9$ steps nine units further left, so the two leftward moves pile up and the distances add:

$(-11) + (-9) = -(11 + 9) = -20.$

Adding $9$ steps nine units to the right instead. Those nine right-units cancel nine of the eleven left-units on the way back toward zero, and two left-units survive:

$(-11) + 9 = -(11 - 9) = -2.$

The second walk never reaches zero, which is why the landing point is still to the left of it.

A step is fixed by two things, how long it is and which way it points, and the two ends of a walk supply both.

From $14$ to $6$ the walk covers eight units and moves toward zero, that is to the left. A step to the left is what adding a negative does, so the number added is $-8$:

$14 + (-8) = 6.$

From $-3$ to $5$ the walk again covers eight units, since three of them reach zero and five more carry on past it, but this time the movement is to the right, so the number added is $8$:

$-3 + 8 = 5.$

The two steps are the same length and point opposite ways, which is exactly the relationship between $-8$ and $8$.

Pair one unit of the leftward move against one unit of the rightward move. Each such pair is a step out and a step straight back, so it returns the walk to where it began and contributes nothing. That is the zero-pair fact $a + (-a) = 0$ applied one unit at a time.

The pairing continues until one of the two moves is used up, and the one that runs out first is the shorter one. So every surviving unit belongs to the longer move, that is to whichever number is farther from zero, and every surviving unit points that number's way. The walk therefore finishes on that number's side of zero, whatever the other number is.

How far from zero it finishes is the same count read as a distance: the longer move, less the part of it that was cancelled. A single pair of examples shows both halves at once:

$(-18) + 7 = -(18 - 7) = -11, \qquad 18 + (-7) = 18 - 7 = 11.$

If the two distances happen to be equal, nothing survives the pairing at all. The walk ends on zero, which is neither to the left nor to the right, and the sum has no sign to inherit.

Read each one as a walk that has to stop exactly on zero.

Start at $-13$, thirteen units to the left. Only a rightward step of thirteen units arrives, and a rightward step of thirteen units is what adding $13$ does:

$-13 + 13 = 0.$

Start at $7$, seven units to the right. Now the step must point left and must be seven units long, so the number added is $-7$:

$7 + (-7) = 0.$

In each pair the two numbers sit the same distance from zero on opposite sides, which is what it means for one to be the opposite of the other. Two numbers that add to zero in this way are called a zero pair.

The point of splitting is to make a zero pair appear. The partner of $-9$ is $9$, so take a $9$ out of the $15$:

$15 = 9 + 6.$

Addition may be regrouped freely, so the pair can be put together and added first:

$-9 + 15 = -9 + (9 + 6) = (-9 + 9) + 6 = 0 + 6 = 6.$

The walk says the same thing. The first nine of the fifteen rightward units are spent climbing back to zero, and the remaining six carry on past it. That is the whole of the different-signs case: cancel what cancels, then read off what is left.

Look through the four terms for two that are opposites. Here $-12$ and $12$ are a zero pair.

Additions may be taken in any order and grouped in any way, so those two terms can be brought together and added first:

$-12 + 7 + 12 + (-3) = (-12 + 12) + (7 + (-3)).$

The pair adds to nothing, so only the other two terms are left. Their signs disagree, the distances are $7$ and $3$, and the number farther from zero is positive:

$0 + (7 + (-3)) = 0 + 4 = 4.$

What the zero pair saved is two stops on the walk. Going straight through from the left reaches the same landing point by a longer route: $-12 + 7 = -5$, then $-5 + 12 = 7$, then $7 + (-3) = 4$.

Each logged change is a step along the line: a fall steps left, a rise steps right.

Start at $-7$ degrees Celsius. A fall of six degrees adds $-6$, and since the position and the step are both on the left of zero, the distances add:

$-7 + (-6) = -13.$

From $-13$ degrees Celsius a rise of four degrees adds $4$, a step of four units to the right. The signs disagree now, so four of the thirteen left-units cancel and nine survive:

$-13 + 4 = -9.$

So the station reads $-13$ degrees Celsius after the fall and $-9$ degrees Celsius after the rise. The later reading is the warmer of the two even though both are below zero, because it lies nearer zero on the line.

A change is what has to be added to the earlier reading to arrive at the later one, so it is the later reading with the earlier one taken away:

$8 - (-5).$

Subtracting a number is adding its opposite, and the opposite of $-5$ is $5$:

$8 - (-5) = 8 + 5 = 13.$

The change is a rise of thirteen degrees Celsius, and the number line confirms it directly: the walk from $-5$ to $8$ climbs five units to reach zero and eight more beyond it, and those two distances add because the walk crosses zero.

Taking the readings in the other order, $-5 - 8$, answers a different question, where the earlier reading stands relative to the later one, and it reports a fall of thirteen degrees instead.

A change is a step, and one signed number carries both things a step has. Its sign is the direction: a positive change means the reading moved to the right along the line, toward warmer, and a negative change means it moved to the left, toward colder. Its size is how far the reading travelled, counted in degrees.

That is why a change is not a reading and must not be read as one. A word such as rose would give the direction and lose the size; a bare size would give the distance and lose the direction. The signed number keeps both.

A change farther from zero than both of its readings is then no puzzle. When the two readings lie on opposite sides of zero, the walk between them is made of two legs, one up to zero and one beyond it, and the legs add. From $-5$ degrees Celsius to $8$ degrees Celsius those legs are five units and eight units:

$5 + 8 = 13.$

No reading of thirteen degrees ever appeared on the thermometer. The thirteen is a distance travelled, not a place on the scale, and only a change that crosses zero can finish farther from zero than both of the readings it connects.

Trade each subtraction for an addition of the opposite, then use what you already know about adding.

The opposite of $6$ is $-6$, so the first becomes an addition of two negatives and the distances add:

$-14 - 6 = -14 + (-6) = -20.$

The opposite of $-17$ is $17$, so the second becomes an addition of two positives:

$9 - (-17) = 9 + 17 = 26.$

The opposite of $-6$ is $6$, so the third becomes an addition whose signs disagree. The distances are $22$ and $6$, and the number farther from zero is the negative one:

$-22 - (-6) = -22 + 6 = -16.$

The second and the third both take away a negative, and both therefore step to the right. Whether the landing point ends up positive depends on where the walk started, not on the two minus signs.

Make the rewrite first. The opposite of $-7$ is $7$:

$12 - (-7) = 12 + 7 = 19.$

Now the check. Subtracting $-7$ undoes a step of adding $-7$, so putting that step back onto the answer should return the walk to where it began. Add $-7$ to $19$: the signs disagree, the distances are $19$ and $7$, and the number farther from zero is positive.

$19 + (-7) = 12.$

The walk comes home to $12$, so $19$ is right.

The check is worth running because a wrong answer fails it. A student who drops the inner minus sign and computes $12 - 7 = 5$ would find that $5 + (-7)$ gives $-2$, not $12$.

Both ingredients are sound on their own. Subtraction really can be rewritten, and an addition really can be taken in either order. What fails is the last step, and the reason is what reordering is allowed to move.

Make the rewrite:

$a - b = a + (-b).$

The terms of that addition are $a$ and $-b$. Reordering moves those two terms and nothing else, so it produces $(-b) + a$, which has the same value and is still $a - b$. To arrive at $b - a$, which is $b + (-a)$, each term would have to be exchanged for its opposite: $a$ for $-a$ and $-b$ for $b$. That is a different move, and no rule about adding allows it.

One pair of numbers settles the claim:

$4 - 15 = 4 + (-15) = -11, \qquad 15 - 4 = 15 + (-4) = 11.$

The two results are $-11$ and $11$, so the student's conclusion is wrong: swapping the two numbers has not left the answer alone, it has moved it to the other side of zero.

Each verdict comes from the two directions and the two distances alone.

$(-23) + 18$: the signs disagree, so the units pair off, and the number farther from zero is $-23$. The survivors lie to the left, so the sum is negative.

$17 + (-17)$: the two numbers are opposites, a zero pair, so every unit finds a partner and none survives. The walk stops on zero, which is neither positive nor negative.

$(-12) + (-31)$: both steps go left, so they pile up and the walk can only finish farther left than it began. The sum is negative.

$25 + (-19)$: the signs disagree and the number farther from zero is $25$, so the survivors lie to the right and the sum is positive.

Evaluating afterwards confirms all four calls, though none of them needed the values:

$-5, \qquad 0, \qquad -43, \qquad 6.$

Turn every subtraction into an addition of the opposite, so the chain becomes one string of steps in the same operation:

$-11 + 20 - 7 - (-3) = -11 + 20 + (-7) + 3.$

Now the sign can be settled before any adding. The rightward steps are $20$ and $3$, which come to twenty-three units of travel to the right. The leftward steps are $-11$ and $-7$, which come to eighteen units to the left. More of the walk goes right than left, so it finishes to the right of zero and the answer is positive.

The pairing off then leaves the difference of those two totals:

$23 + (-18) = 5.$

A left-to-right check agrees: $-11 + 20 = 9$, then $9 + (-7) = 2$, then $2 + 3 = 5$.

Work out the three sums first.

$(-30) + 12 = -18, \qquad 30 + (-12) = 18, \qquad (-12) + (-31) = -43.$

In the first two the signs disagree, and $18$ is nearer zero than $30$ is, so neither lands farther out than both of its numbers. In the third both numbers are negative, and $-43$ is farther from zero than $-12$ and farther than $-31$.

Now the reason behind the first two. When the signs disagree, one step goes right and the other left, so their units pair off, and each pair leaves the walk where it was. The pairing stops when the shorter move runs out. What survives is part of the longer move and never more than all of it, so the landing point is never farther from zero than the farther of the two numbers. Two numbers at equal distances are the extreme case: every unit finds a partner, and the walk stops on zero.

Matching signs behave the other way, which is what the third sum shows. Two steps that go the same way pile up, so the walk can only finish farther out than either number.