Last time we looked at explanations for the product of negative numbers in terms of various concrete models or examples. But it really requires a mathematical proof, as we’ll explain and demonstrate here, first with a couple different proofs, then with the bigger picture, giving the context of such proofs.

## It has to be that way

First, from 1998:

Why Does a Negative Times a Negative Equal a Positive? How does a negative number times another negative number equal a positive number?

Doctor Bruce answered:

Hello Jessica, I detect in your question a measure of annoyance at having to learn the rule for multiplying negative numbers. I'll bet it seems like someone just made the rule up out of thin air, with no particular reason why the answer should be positive. I want to reassure you thatthis rule is not just "made up."There is a chain of reasoning -- a mathematical "argument" -- that showswhy the rule *has* to bethat negative times negative equals positive.

If someone did just decree this “rule”, then it would be annoying, wouldn’t it? But math is not about arbitrary rules; it’s about reasoning from basic assumptions or known facts, to less obvious facts.

What follows is a **proof**, presented in a style intended for students who are not familiar with proofs. We have to start with known facts about multiplication (assumptions or axioms):

Mathematical argument takes a little getting used to. This might look rather strange at first. Here's how the reasoning goes: (1)Zero times anythingequals zero. (2) Every number hasexactly one additive inverse. This means if N is a positive number, then -N is its additive inverse, so that N + (-N) = 0. Likewise, the additive inverse of -N is N. (3) We want negative numbers to obey thedistributive law. This says that a*(b+c) = a*b + a*c.

The first and third facts are true of multiplication as we know it; in adding the concept of negative numbers (that is, the additive inverse) to the arithmetic we are already familiar with, we don’t want to change these facts.

### Positive times negative

We can show that these facts imply what multiplication of negative numbers has to look like, in two steps. First:

(4) Now,we are forced to accepta new law, thatnegative times positive equals negative. This is because we can use the distributive law on an expression like 2*(3 + (-3)). This equals 2*(0), which is zero. But by the distributive law, it also equals 2*3 + 2*(-3). So2*(-3) does the job of the additive inverse of 2*3, and therefore 2*(-3)isthe additive inverse of 2*3. But the additive inverse of 6 is just -6. So 2 times -3 equals -6.

There is only one additive inverse of a number; anything that *does* what \(-6\) does must *be* \(-6\).

We can write this in the general case, supposing that *m* and *n* are any two positive numbers: $$0=m\times 0=m\times(n+(-n))=m\times n+m\times(-n)\\ \Rightarrow\;m\times(-n)=-(m\times n)$$ So a positive times a negative is a negative (and, by the commutative property, a negative times a positive is negative).

### Negative times negative

But we can repeat the same process, using the fact just demonstrated:

(5) Next,we are forced to acceptanother new law, thatnegative times negative equals positive. It's a lot like the example in (4). We use the distributive law on, say, -3*(5 + (-5)). This is again equal to zero. But by the distributive law, it also equals -3*5 + (-3)*(-5). We know the first thing, (-3*5) equals -15 because of the law in (4). So (-3)*(-5) is doing the job of the additive inverse of -15. We know -15 has exactly one additive inverse, namely 15. Therefore, (-3)*(-5) = 15.

Again, $$0=(-m)\times 0=(-m)\times(n+(-n))\\=(-m)\times n+(-m)\times(-n)\\=-(m\times n)+(-m)\times(-n)\\ \Rightarrow\;(-m)\times(-n)=-(-(m\times n))=m\times n$$

I hope this doesn't frighten you! The main thing is, keep right on questioning the things that don't make sense. In mathematics, you are always entitled to an explanation of WHY things are the way your teacher (or I) say they are.

## Negative *of* negative is positive

Next, a slightly different question from 2001:

Prove That -(-a) = a Why does -(-a) = a? How do you prove this using the properties of real numbers?

Doctor Bruce had passed over this lightly; we’ll go a little deeper and further here.

Doctor Rick answered:

Hi, Eduardo. We can start with how-ais defined. It is "theadditive inverseof a" - that is, it is the number that, when added to a, gives 0: a + -a = 0 Therefore -(-a) meansthe number that, when added to -a, gives 0. But applying the commutative property of addition, the equation above becomes -a + a = 0 Therefore the number that, when added to -a, gives 0is a; or,-(-a) = a

That’s all it takes; it follows immediately from the definition. But notice that this doesn’t just tell us that the negative of a negative number is positive; our variable *a* could be negative itself! What we have proved is that changing sign twice returns us to where we started, wherever that was.

### Multiplying by negative 1

From that basic fact, we can take a step toward multiplication:

A closely related, but different, question is how we can prove that -1 * -1 = 1 The theorem linking these two is this:-1 * a = -a

We’re working our way up, bit by bit, to the general fact. To some people it can seem obvious that multiplying by -1 changes the sign, but it needs to be proved; and proving this little fact will make everything else easy. We’ll use ideas very much like Doctor Bruce’s:

Let's prove this. Start with the fact thatzero times any number is zero: 0 * a = 0 Write 0 as (1 + -1), which follows from the definition of -1. (1 + -1)*a = 0 Apply thedistributive property: 1*a + -1*a = 0 Use the fact that 1 times any number is the same number: a + -1*a = 0 Now,the number that, when added to a, gives 0is -a. Therefore -a = -1*a Using this theorem, we can easily prove that -1 times -1 is 1: -1 * -1 = -(-1) = 1

From here we could go all the way to the general fact about negatives time negatives, by just bringing in the commutative and associative properties: $$(-a)\times(-b)=((-1)\times a)\times((-1)\times b)\\=(-1)\times a\times(-1)\times b\\=((-1)\times(-1))\times(a\times b)\\=1\times(a\times b)\\=a\times b$$

## Discarding old models

These proofs can be a little dizzying. The next question, from 2003, asks us to back up and think about what’s happening.

Why Does a Negative Times a Negative Make a Positive? Why is a negative times a negative a positive? When you answered this question to other people, the way you explained it was over my head. When you have a negative integer, in order to multiply by another negative numberwhy would you have to go to the positive side of the number line?

Doctor Ian answered:

Hi Brea, It's really hard to understand this in terms of a number line. One of the ways math works is that we make up simple ways of thinking about some set of numbers; but thosebreak downwhen we try to think of other sets of numbers. So then we have to make upnew ways of visualizing our expanded sets.

We aren’t yet talking about the question itself, but about the need to change our thinking when we apply old ideas to a new kind of number. We’re used to positive numbers; expanding to negative numbers requires a new perspective.

First, another example of such an expansion:

The easiest example isfractions. At first, you learn to think of something like 3/4 as meaning "break an item into 4 pieces, and keep 3 of them". That works okay as long as you're dealing with integers. But what would sqrt(2) ------- pi mean in those terms? How would you break something into 'pi' pieces? And how would you keep sqrt(2) of them? Eventually,you have to let go of the 'dividing into pieces' model, and just start thinking of fractions asdivisions that you haven't done yet.

This was discussed in Fractions: What Are They, and Why?, and in How to Convert a Fraction to a Decimal – and Why.

Similarly, thenumber lineis pretty good for thinking about addition and subtraction, but it'snot so good for thinking about multiplication. To be honest, I've never been able to think of a way that is bothintuitiveandrigorously correct, to explain why the product of two negatives is positive. There are nice ways to visualize the result, but they don't explain why it _has_ to be that way.

The examples we looked at last time are attempts to make this idea intuitive, but they can all leave a student unconvinced. What we need is … a proof.

### A compact proof

Doctor Ian here gives a proof stripped of the words of explanation we’ve used until now, so you have to think more carefully about why each step is true, and even why he would start the way he does; this is typical of mathematicians’ proofs. From the proofs we’ve seen before, you should be able to see most of the reasons; what you’ll discover is essentially that he’s combined the two steps of Doctor Bruce’s version into one.

There is a pretty simple proof, which goes like this: Let a and b be any two real numbers. Consider the number x defined by x = ab + (-a)(b) + (-a)(-b). We can write x = ab + (-a)[ (b) + (-b) ] (factor out -a) = ab + (-a)(0) = ab + 0 = ab. Also, x = [ a + (-a) ]b + (-a)(-b) (factor out b) = 0 * b + (-a)(-b) = 0 + (-a)(-b) = (-a)(-b). So we have x = ab and x = (-a)(-b) Two things that are equal to the same thing are equal to each other, so ab = (-a)(-b)

This is an interesting style of proof, where we evaluate one expression in two different ways and get different results, which must therefore be equal. But such “elegance” can be very hard to follow; and you could only invent it by first exploring the ideas we’ve seen above.

### Changing signs

He followed the proof with an attempted model, which I will omit; I wouldn’t find it very convincing.

But he closes with something useful:

So what about the remaining possibility? Well, so far, we've seen thatchanging the sign of _either_ factor changes the sign of the product. That is, in the diagram below, (+) | II | I | (-)-------- --------(+) | III | IV | (-) as we move from quadrant I to quadrant II, we change signs; and as we move from quadrant I to quadrant IV, we change signs. So we should also change signs as we move from II to III, or from IV to III, right? But if we do that, we end up with a positive area in quadrant III... which is just another way of saying that theproduct of two negatives should be a positive.

Changing the sign of one factor changes the sign of the product; so changing the sign of *both* should change it back to positive. That’s not proof, but plausibility.

So maybe that's another way to think about it. If we take _anything_, and multiply it by -1, we should end up with a different sign, shouldn't we? So let's start with something like a * b = c where a, b, and c are all positive. If we multiply that by -1, we have to end up with a different sign: -1 * (a * b) = -c and if we multiply by -1 again, we have to end up with a different sign again: -1 * (-1 * (a * b)) = c Does that make sense so far? If we _don't_ do this, thensometimes multiplying by -1 would change the sign, and sometimes it wouldn't. That would be pretty strange.

This assumes the facts that Doctor Rick proved in our second proof, both easy to believe.

So if c = -1 * (-1 * (a * b)) then, since we can group and order multiplications any way we want, c = -1 * (-1 * (a * b)) = -1 * -1 * a * b = -1 * a * -1 * b = (-1 * a) * (-1 * b) = (-a) * (-b) Do any of these explanations work for you? If not, I can try to find another one that does.

## Model with tiles? Or not

I’ll close with one more question, from 2001, asking for a model like those we used last time:

Algebra Tiles and Negatives How can we use a model (algebra tiles) to demonstrate that a negative times a negative = a positive? I tried to show repeated addition but that doesn't work. For example, -3(-5). How do you represent-3 sets of -5at a 7th grade level?

Algebra tiles (see here) are little squares representing positive numbers by one color (say, yellow) and negative by another (red). (They also represent variables, but that is not involved at this level.) A \(+1\) cancels a \(-1\) in a way that resembles antimatter annihilation, but with less energy (depending on the kid who is using them). You could represent \(3\times-5\) as three sets of 5 “\(-1\)”s. But how can you make a negative number of groups?

This question would belong in last week’s post on models … but Doctor Ian didn’t keep the focus on the model:

Hi David, The truth is, models only go so far, and it's not clear that there is a way to model the fact that -1 * -1 = +1 that doesn't create more confusion than it clears up. One thing you might try is modeling 'this times that' in the following way: When 'this' is positive,add copies. When 'this' is negative,take away copies. When 'that' is positive,use items. When 'that' is negative,use holes. So, 3 times 4: 1. Start with 4 items * * * * 2.Add copies* * * * * * * * * * * * -3 times 4: 1. Start with 4 items * * * * 2.Take away copies, leaving holeso o o o o o o o o o o o 3 times -4: 1. Start with 4 holes o o o o 2.Add copieso o o o o o o o o o o o -3 times -4 1. Start with 4 holes o o o o 3.Take away holes... by adding copies!* * * * * * * * * * * *

His “items” and “holes” correspond to the yellow and red tiles in the set, and that terminology is a nice way to think of the tiles. But adding and taking away, as in some of our examples last time, can just be confusing. (In the site I linked about algebra tiles, subtraction is not modeled by *taking away*, but by explicitly *adding the opposite*; and multiplication is done by *changing the sign of the result –* “flipping” tiles, if they had red on only one side; so no attempt is made to motivate those rules.)

But even this isn't very satisfying, because it's not any easier to think of removing holes by adding copies than it is to remember that -1 * -1 = +1. (You could useelectrical chargesas a model -- removing a negative charge has the same effect as adding a positive one -- butmany students find electricity even more confusing than math.) And even if it were satisfying, by trying to reduce this to something that can be modeled with physical items,you would be missing the chance to make a very important point about one of the ways in which mathematics grows.

This is a key insight. Math goes beyond models, taking us into the world of the mind.

### Don’t break anything!

Rather than trying to justify the rules by the model, he turns to an abstract proof:

If we can agree that a negative number is justa positive number multiplied by -1, then we can always write the product of two negative numbers this way: (-a)(-b) = (-1)(a)(-1)(b) = (-1)(-1)ab For example, -2 * -3 = (-1)(2)(-1)(3) = (-1)(-1)(2)(3) = (-1)(-1) * 6

This is Doctor Rick’s approach starting with \((-1)\times a = -a\), though without formally proving that. In effect, he is taking it as an *axiom*, because it is easy to agree to.

So the real question is, (-1)(-1) = ? and the answer is that the following convention has been adopted: (-1)(-1) = +1This convention has been adopted for the simple reason that any other convention would cause something to break.For example, if we adopted the convention that (-1)(-1) = -1, the distributive property of multiplication wouldn't work for negative numbers: (-1)(1 + -1) = (-1)(1) + (-1)(-1) (-1)(0) = -1 + -1 0 = -2 As Sherlock Holmes observed,"When you have excluded the impossible, whatever remains, however improbable, must be the truth."Since everything except +1 can be excluded as impossible, it follows that, however improbable it seems, (-1)(-1) = +1.

He emphasizes more strongly than we have done elsewhere, that the “rule” is just a **convention**, an agreement we make as a way of **defining** a multiplication that, until we do this, has no definition! But the convention is not arbitrary; it is chosen in order not to break the existing behavior of operations, just as we have said before.

If that seems too abstract, try to think of it in terms of a murder mystery. We've considered all the other possible suspects, and we know that none of them could have committed the murder; and we have no reason for believing that '1' didn't commit the murder; and _someone_ did it; so'1' must be guilty. Also, if you just want to teach your students a good way to remember that a negative times a negative must be positive, they can use this trick: -a(b + -b) = (-a)(b) + (-a)(-b) \________/ \_____/ \______/ 0 neg must be pos

This is a quick version of the distributive property proofs we’ve seen.

David replied:

Thank you for your response to my question I thought that I was going to go insane. I also thought that I was the only person having problems explaining why a negative times a negative was a positive. It helps me to rest a little easier knowing that there are a lot of people that have wrestled with this question.

We can attest to that! This is one of our more frequently asked questions!