Multiplying and Dividing Integers: Free Response

The product $4 \times (-9)$ counts four groups, each of them $-9$. Writing that out as a sum,

$4 \times (-9) = (-9) + (-9) + (-9) + (-9).$

Every term is negative, so every step goes the same way along the number line, to the left. Adding numbers that share a sign adds the sizes and keeps the shared sign, so the total is $9 + 9 + 9 + 9 = 36$ units to the left of zero:

$4 \times (-9) = -36.$

Notice what did not happen. The two numbers were not combined into $-5$ or $-13$: the $4$ counts groups, and only the $-9$ is the thing being added.

Think about what the long sum would do rather than writing it. There would be fifteen steps, each of them six units to the left. The steps never turn around, so the sizes simply pile up:

$15 \times 6 = 90.$

That is the total distance travelled, and the direction of travel is left, so the total is ninety units to the left of zero:

$15 \times (-6) = -90.$

The shortcut agrees with the sum because it is a description of that sum: multiplying the sizes counts the total distance the equal steps cover, and the single negative factor is what points every step the same way. Multiply the sizes, then settle the sign.

Only one of the two orders can be read as groups. Fifteen groups of $-6$ is a sum you could in principle write out; six negative groups of anything is not something you can count.

The rescue is a property you already have. Multiplication is commutative, which says that for any two numbers the order of the factors does not change the product. That was stated for all numbers, not only for positive ones, so it applies here:

$(-6) \times 15 = 15 \times (-6) = -90.$

The value of the order that has no meaning as groups is therefore fixed by the order that does.

The conclusion is general. Take any product of a negative number and a positive number. Exchange the factors if necessary so that the positive one counts the groups. What you then have is a repeated sum of negative terms. That sum is negative, and its size is the product of the two sizes. So the answer never depends on which side the negative factor is written:

$(-6) \times 15 = -90 \qquad \text{and} \qquad 15 \times (-6) = -90.$

Inside the parentheses, $8$ and $-8$ are opposites, so they form a zero pair:

$8 + (-8) = 0.$

Any number times zero is zero, so the whole expression is worth nothing at all:

$(-5) \times \bigl(8 + (-8)\bigr) = (-5) \times 0 = 0.$

Now take the other route. The distributive property says a factor standing outside a sum may be multiplied across each term of that sum. Multiplying the $-5$ across the two terms, and leaving both products alone,

$(-5) \times \bigl(8 + (-8)\bigr) = (-5) \times 8 + (-5) \times (-8).$

Both routes work out the same expression, so they have to give the same number. The first route already told us that number: it is $0$.

The two expressions of part A are two readings of one and the same thing, so they are equal, and the first reading gave zero:

$(-5) \times 8 + (-5) \times (-8) = 0.$

The first product has exactly one negative factor, so it is negative, with size $5 \times 8 = 40$:

$(-5) \times 8 = -40.$

Put that in, and only one product is left whose value you do not know yet:

$-40 + (-5) \times (-8) = 0.$

Read that line as a question. It says that $(-5) \times (-8)$ is the number you must add to $-40$ to reach zero. Only one number does that to $-40$, and that number is its opposite, $40$. So

$(-5) \times (-8) = 40.$

Nothing about the answer was assumed in advance, and no chart was consulted. Three facts you already trust force the value: a number and its opposite add to zero, any number times zero is zero, and multiplication distributes over addition.

Put the student's value into the distributed form and work it out:

$(-5) \times 8 + (-5) \times (-8) = -40 + (-40) = -80.$

Now compare that with the other route from part A. Adding inside the parentheses first gives $(-5) \times 0$, which is $0$. So the student's value requires

$-80 = 0,$

which is false.

The rule it disagrees with is the distributive property, since that is the rule saying the two routes must match. The other two rules used along the way both held up: a number and its opposite add to zero, and any number times zero is zero. Only $40$ leaves all three standing, which is why the value is not something anyone gets to choose.

Eight minutes each contribute the same change of $-11$ feet, so the total change is that amount taken eight times over:

$8 \times (-11) = -88 \text{ feet}.$

Exactly one factor is negative, so the total change is negative, and its size is $8 \times 11 = 88$.

The craft started at the surface, where the reading is $0$ feet, so the reading afterwards is $0 + (-88) = -88$ feet. In the language of the situation, it is 88 feet below the surface. The minus sign is not decoration: it is what records the direction of the change.

The whole change is split into six equal pieces, one per minute, so the question is which number taken six times rebuilds $-96$. That is a division:

$(-96) \div 6 = -16 \text{ feet per minute}.$

The two numbers being divided have different signs, so the quotient is negative, and the sizes divide as usual: $96 \div 6 = 16$.

Checking by multiplying back is the definition of division at work:

$6 \times (-16) = -96.$

The original total change comes back, so $-16$ feet per minute is right. The sign is telling you the craft is going down, which is what the story said.

Three quantities are in play on any of these dives: the change per minute, the number of minutes, and the total change. The division names two of them and leaves out the number of minutes, so that is the question it asks:

$(-84) \div (-12) = 7 \text{ minutes}.$

Read it aloud as a question about the dive: how many changes of $-12$ feet does it take to make a change of $-84$ feet? Seven of them. The check is the multiplication it undoes:

$7 \times (-12) = -84.$

Now the sign. The quotient counts minutes. There is no such thing as a negative number of minutes, so the answer has to be positive. The general rule agrees, because the two numbers being divided share a sign. Here you can also see why it has to. Both negatives describe the same downward direction. Asking how many of one make up the other throws that shared direction away and leaves a plain count.

The quotient $(-72) \div 9$ is by definition the number which, multiplied by $9$, gives $-72$. So the multiplication to solve is

$(\text{missing factor}) \times 9 = -72.$

The divisor $9$ is positive, and the product $-72$ is negative, so the missing factor has to be negative: a positive times a positive could never land below zero. Its size comes from dividing the sizes, $72 \div 9 = 8$. Therefore

$(-72) \div 9 = -8.$

Multiplying back is not a ritual, it is the definition being checked:

$(-8) \times 9 = -72.$

The original dividend comes back, so the quotient is right.

A multiplication has three numbers in it: two factors and a product. Hide one factor and you have a division question, so one multiplication yields two of them.

Hide the $-6$ and ask which number multiplied by $-7$ gives $42$. Hide the $-7$ and ask which number multiplied by $-6$ gives $42$. The given fact answers both:

$42 \div (-7) = -6, \qquad 42 \div (-6) = -7.$

Nothing here was memorized, and notice what has just been established as a by-product: a positive number divided by a negative number is negative. That was not a new rule to learn. It is the multiplication fact you were handed, read from the other end.

Start with the named case:

$42 \div (-7).$

The missing factor multiplied by $-7$ has to give $42$, a positive product. A positive factor times $-7$ would give a negative product, and a factor of zero would give zero, so the missing factor can be neither. A negative factor times $-7$ does give a positive product, which is what is wanted:

$(-6) \times (-7) = 42, \qquad \text{so} \qquad 42 \div (-7) = -6.$

That verdict was reached without consulting any rule about division. Now say why that always happens. Every division question asks which number, multiplied by the divisor, rebuilds the dividend. So a quotient is always a factor in some multiplication, and its sign was settled the moment the multiplication chart was settled. A division chart can only be the multiplication chart read from the other end: for nonzero numbers, matching signs give a positive result and differing signs a negative one, for products and quotients alike.

The question is which number multiplied by $-13$ gives $0$. Zero does:

$0 \div (-13) = 0, \qquad \text{because } 0 \times (-13) = 0.$

Nothing else does. Two nonzero numbers multiplied together always give a nonzero product, so only a factor of $0$ can bring the product down to $0$:

$1 \times (-13) = -13, \qquad (-2) \times (-13) = 26, \qquad 0 \times (-13) = 0.$

So this division behaves perfectly well. There is a missing factor, and there is exactly one of it. Dividing zero by a nonzero number is ordinary; it is the other arrangement that causes trouble.

By the definition, $(-13) \div 0$ is the number which, multiplied by $0$, gives $-13$. Write down what that demands:

$(\text{the quotient}) \times 0 = -13.$

Now test it. Any number times zero is zero, whatever the number is and whichever sign it carries:

$5 \times 0 = 0, \qquad (-5) \times 0 = 0, \qquad 0 \times 0 = 0.$

Every candidate produces $0$, and $0$ is not $-13$. There is no number left to try. So $(-13) \div 0$ names no number, and we say it is undefined.

Notice why it is undefined. Nobody banned it. The question it asks simply has no answer.

Take the dividend $-13$ and put the proposed value back into the multiplication it claims to solve. If $(-13) \div 0$ were $0$, then multiplying that quotient by the divisor would have to rebuild the dividend:

$0 \times 0 = 0, \qquad \text{but the dividend is } -13.$

The proposal fails its own test. Agreeing to it would not make it true. It would only mean writing down an answer that the check by multiplying back rejects, when every other division passes that check.

Now $0 \div 0$, where the proposal looks harmless. Ask which number multiplied by $0$ gives $0$:

$7 \times 0 = 0, \qquad (-4) \times 0 = 0, \qquad 0 \times 0 = 0.$

Every number passes. So the trouble here is the opposite of the trouble in part B. There, no number worked. Here, every number works, and nothing picks out one of them. A quotient has to be a single number, so again no value can be given.

Both arrangements fail the same requirement from opposite sides: division by zero never picks out exactly one number, which is why it is left undefined in every case.