So, why is the result of multiplying two negatives numbers together a positive number? There are various answers to that.

First, let's think a little about why we care.

# The rules of the (arithmetic) game

We start off with the non-negative integers \[ \mathbb{N} = \{0,1,2,3,\ldots\} \] which we all learn to manipulate from a very young age. We know how to add and multiply them, and (at least sometimes) to subtract, at least as long as the number we want to subtract is no bigger than the one we want to subtract it from.Then somebody comes along and says that when you subtract \(2\) from \(1\), the answer is \(-1\): a bit like \(2-1\), but with a minus sign in front of it. This is a new kind of number called a negative number, and they're useful for situations where you have to subtract a bigger number from a smaller one, and in certain unpleasant circumstances to describe the amount of money you have in the bank.

But now we have to ask: when we do arithmetic with them, how do they work?

One solution is just to present some rules, including the rule that the product of two negative numbers is the same as the product of their positive counterparts. In other words:

# That's just the rule, isn't it?

Ooh, not at all satisfactory. The rule seems completely arbitrary. But did we really have any choice? Are there any constraints that determine this? What if we'd chosen the rule that the product of two negative numbers was negative?Let's hold this thought for the moment and think about how \(\mathbb{N}\) behaves.

So, let's take a careful look at the non-negative integers, and what we know about them. In fact, there is a bunch of stuff that we know. I'm too lazy to write in an explicit multiplication symbol, so I'll use \(mn\) to represent the product of \(m\) and \(n\).

In the list below, \(m,n\) and \(p\) are arbitrary elements of \(\mathbb{N}\). Then

- \(m+n = n+m\) (addition is commutative)
- \(m+(n+p) = (m+n)+p\) (addition is associative)
- \(mn = nm\) (multiplication is commutative)
- \(m(np)= (mn)p\) (multiplication is associative)
- \((m+n)p = mp+np\) (multiplication distributes over addition)
- \(0+n = n\) (\(0\) is the additive identity)
- \(1n = n\) (\(1\) is the multiplicative identity)
- If \(m+n = p+n\) then \(m=p\) (cancellation law for addition)
- If \(mn=pn\) and \(n \neq 0\) then \(m=p\) (cancellation law for multiplication)

In particular, there's one very familiar one which is conspicuous by its absence. It is the rule \(0n = 0\). I left it out on purpose, because we can show that it follows from the others, by the following argument:

We know that \(0+0=0\), because \(0\) is the additive identity. But that means that \((0+0)n=0n\), no matter what \(n\) is. By the distributive law, then, \(0n+0n=0n\), so that \(0n+0n=0n+0\), and then by the cancellation law, \(0n=0\).

There is an important lesson hiding in here. It is that our familiar rules aren't independent: once we have agreed on some of them, others are forced upon us. (So it's a good idea to choose a useful collection of rules to start with!)

This should raise an ugly possibility. What if my collection of rules isn't even consistent to begin with? How can I check that?

That's actually quite hard. Let's settle for the moment by accepting that there really is a set of non-negative integers with operations of addition and multiplication which satisfies the above rules. So there are lots of consequences from them that I haven't written above, but at least it all makes sense.

All this doesn't really help yet, but I can make use of it. Whatever the negative integers are, I want arithmetic to keep obeying the same rules as before. I want \(0\) and \(1\) to have their familiar properties, and I want the usual laws of algebra to continue to hold. So for every positive integer \(n\) I introduce a negative integer \(-n\) with the property that \(-n+n=0\). The set of all integers, positive, zero, and negative, is denoted \(\mathbb{Z}\). Now we can investigate what happens if we insist that all the rules above continue to hold.

Then \(\ldots\)

# It follows from the algebra

Once we have all the stuff above, it then follows by a fairly short argument that \((-1)(-1)=1\),**if**we insist that all the above rules continue to hold when we allow \(m,n\) and \(p\) to be negative.

First, we all agree that whatever we mean by \(-1\), it has the property that \(-1+1=0\). Then we know that \[ (-1+1)(-1) = 0(-1) = 0 \] and so that \[ \begin{split} (-1+1)(-1)&=(-1)(-1)+1(-1)\\ &=(-1)(-1)+(-1)\\ &=0 \end{split} \] But now adding \(1\) to each side gives \[ (-1)(-1)+(-1)+1=0+1 \] so that \[ (-1)(-1)+ 0 = (-1)(-1)= 1 \] So

**if**we insist that all the algebraic rules for non-negative numbers continue to hold, we have no option. A minus times a minus is a plus.

Well, now we can see that if we decide that a minus times a minus is minus, then the ordinary rules of algebra must break somewhere. So this isn't really an arbitrary rule, it's forced upon us by the requirement that the algebraic properties of \(\mathbb{Z}\) are the familiar algebraic properties of \(\mathbb{N}\).

We can see various other things, too: one very important one is that \((-1)n = -n\). The argument is fairly similar: \[ \begin{split} (-1)n & =(-1)n + 0\\ &= (-1)n + n + (-n)\\ &= (-1)n+1n + (-n)\\ &=(-1+1)n + (-n)\\ &=0n + (-n) \\ &= -n \end{split} \]

So, this tells us that if we extend the non-negative integers to include negative ones, and insist that algebra works as before, there are certain logical consequences, including the familiar

*a minus times a minus is a plus*. If. But how do we know that these rules describe anything at all? What are these negative numbers that we have introduced, and how do we know that it is even possible to introduce negative numbers while preserving the algebraic rules?

You might be tempted to think I'm about to kick up a lot of dust and then complain about how hard it is to see, but when the dust settles the view is wonderful.

So here's the awkward question \(\ldots\)

# What is a negative number anyway?

Negative numbers are there to be the solutions to equations of the form \[ m+x = n \] when \(m>n\). In particular, \(-n\) is the solution to \(n+x=0\),**if**this makes sense.

So, let's make sense of it.

We're going to do this a bit indirectly, thinking about pairs of integers \((m,n)\). Secretly we want to think of this pair as \(m-n\), but we don't yet have a meaning for this when \(m \lt n\).

So we're going to lump together certain pairs.

We say that the pairs \((m,n)\) and \((p,q)\) are

**equivalent**if \(m+q=n+p\). A collection of pairs which are all equivalent to each other is called an

**equivalence class**.

This is actually quite a familiar idea: it's analogous to thinking of \(3/4\) and \(6/8\) as different ways of representing the same fraction. There are infinitely many pairs of integers \((m,n)\) which we can use to do it, namely all the ones satisfying \(3n=4m\). We don't (often) talk explicitly about a fraction being an equivalence class of pairs of integers, but that's what's really going on.

We're going to do the same trick, but with pairs where it's the difference than matters, not the ratio.

We need to do three things now:

- Define addition on these equivalence classes of pairs.
- Define multiplication on these equivalence classes of pairs.
- Show how we can think of these equivalence classes as the familiar integers.

## Addition

This is the easy one.We define addition on pairs by \[ (m,n)+(p,q)=(m+p,n+q) \]

But we're not done yet, because the collection that the answer lies in should only depend on the collections of the original pairs, not on the specific pairs. We can check this.

So suppose that \((m,n)\) and \((M,N)\) are equivalent to each other, as are \((p,q)\) and \((P,Q)\). Then we need to check that \((m+p,n+q)\) is equivalent to \((M+P,N+Q)\).

First, we know that \[ m+N=n+M \] and \[ p+Q=q+P \] so adding these together we have \[ m+N+p+Q=n+M+q+P \] i.e. \[ (m+p)+(N+Q)=(n+q)+(M+P) \] which is exactly what we need.

So this operation of addition works in the sense that the equivalence class of the sum depends only on the equivalence classes of the pairs we add.

Multiplication is a bit tougher.

## Multiplication

This is not so obvious, so let's cheat a little. We are thinking of \((m,n)\) and \(m-n\), and \((p,q)\) as \(p-q\). Then the product ought to be \[ \begin{split} (m-n)(p-q) &= mp-mq-np+nq\\ & = (mp+nq)-(mq+np) \end{split} \] So we use the definition \[ (m,n)(p,q)= (mp+nq,mq+np) \] We can quickly check that \[ \begin{split} (p,q)(m,n) &= (pm+qn,qm+pn)\\ &=(mp+nq,mq+np)\\ &= (m,n)(p,q) \end{split} \] so that this is commutative, but now we have to check that the result is also independent of the choice of pairs.So we consider \((m,n)\), \((p,q)\) and \((P,Q)\) where \((p,q)\) and \((P,Q)\) are equivalent. Then \[ (m,n)(p,q) = (mp+nq,mq+np) \] and \[ (m,n)(P,Q) = (mP+nQ,mQ+nP) \] and we see that \[ \begin{split} mp+nq+mQ+nP &= m(p+Q)+n(q+P)\\ &=m(P+q)+n(Q+p)\\ &=mP+nQ+mq+np \end{split} \] so that the two answers are equivalent.

Nearly there.

In just the same way, if \((m,n)\) is equivalent to \((M,N)\), then \((m,n)(p,q)\) is equivalent to \((M,N)(p,q)\). Putting this together, \((m,n)(p,q)\) is equivalent to \((M,N)(p,q)\), which is equivalent to \((M,N)(P,Q)\), so the equivalence class of the product only depends on the equivalence classes of the pairs we are multiplying.

Now we know that we have a well-defined arithmetic on our (equivalence classes of) pairs. But we still need to see how we can think of this as supplementing the non-negative integers with the negative integers.

We make use of the fact that we can choose any representative we like for each equivalence class. So we use the representative of the form \((n,0)\) for all pairs of the form \((p+n,q)\) where \(n \geq 0\), and the representative of the form \((0,n)\) for the pairs of the form \((p,q+n)\) where \(n \gt 0\).

Using this, we can now see that \((m,0)+(n,0)=(m+n,0)\) and \((m,0)(n,0)=(mn,0)\), so the non-negative integers live inside this strange new object.

We also see that \((n,0)+(0,n)=(n,n)\), which is equivalent to \((0,0)\). If we think of \((n,0)\) representing \(n\), then \((0,n)\) represents \(-n\).

It is then

*tedious but straightforward*to check that all the rules of algebra for the non-negative integers still work. (This is mathematics code for 'I can't be bothered to write it all out, but I don't see any reason why you shouldn't suffer.')

So, now we know that is is, indeed, possible to add negative integers into the mix while preserving the well-known algebraic structure of the non-negative integers, and that the argument up above for why \((-1)(-1)=1\) does actually apply.

Of course, now that we have defined multiplication of our pairs of positive numbers, we can also do this explicitly and see the same answer from an entirely different perspective. Given than \(-1\) is represented by \((0,1)\), we have \[ (0,1)(0,1)=(0+1,0+0)=(1,0) \] and \((1,0)\) represents \(1\) in our model.

# What was the point of all that again?

After all this, you might wonder what the point of all that peculiar stuff about equivalence classes was. It just made a complicated picture of the positive and negative numbers, which everybody was happy with anyway. The snag is that you can't always just add a new kind of number into the mix and preserve the algebra that you like, but now we know that it can be done in this particular case of interest.Maybe the most famous example of this is Hamilton's search for a three-dimensional version of complex numbers. You can start off with the real numbers, then include a new quantity \(i\) which satisfies \(i^2=-1\), and just do algebra with combinations of this \(i\) and the real numbers, and it all works: complex numbers make sense, and we can think of their algebra as a way of adding and multiplying two-dimensional vectors.

Hamilton spent some time trying to add in another quantity like \(i\) so get a three-dimensional version of algebra. In fact, that isn't possible. He eventually realised that it could almost be done in four dimensions, but you had to give up on the multiplication being commutative.

So there are problems with adding the new kind of numbers in and just assuming that the old rules of algebra will continue to hold. But now, after quite a lot of work, we can say with confidence that it is possible to introduce negative integers in a way which is compatible with the algebra of the non-negative ones. This is a Good Thing.

## Food for thought

- I wrote down a multiplication rule which was motivated by me knowing how the answer should behave. Could I have chosen another rule with difference consequences?
- I wrote down a bunch of rules above, which are followed by the non-negative integers. But are there other sets of quantities which satisfy those same rules? What happens if I try this trick on these other sets?

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