Commutativity or Distributivity - Which One to Use to DEFINE Multiplication of Negative Numbers?

Question

It's easy to calculate $3 \times (-4)$, using the meaning of multiplication: $3 \times (-4)=(-4)+(-4)+(-4)=-12$. But it's not the case about $(-4)\times 3$!

To DEFINE $(-4)\times 3$ we can choose one of the following ways:

1. We can generalize the multiplication so that it remains commutative as before. So $(-4) \times 3 = 3 \times (-4)$.
2. We can generalize the multiplication so that it remains distributive as before. So $(-4) \times 3 + 4 \times 3= 0 \times 3 =0$, hence $(-4) \times 3=-(4\times 3)=-12$.

Which one is preferable? Why?

Answer 1

$-4\times3$ works just as well as $3\times -4$; you have -4 instances of 3, or $-(3)-(3)-(3)-(3)=-12$. There is no need to make a choice between commutativity and distributivity as you have both when it comes to multiplication in $\mathbb{Z}$.

What you do need to prove for negative multiplication, though, is that $-1\times-1=1$, which can be done using the distributive property, commutativity, additive inverses, and the zero property of multiplication.

Answer 2

From your comments to your question, I think I might be able to offer an answer.

In Abstract Algebra, one works with something called rings. In a ring you have multiplication and addition defined. The multiplication is always assumed distributive over a sum, so: $a\times(b+c) =a\times b + a\times c$, but you don't assume commutativity. (I use $\times$ here only because your do). So I would say that if you were trying to define somethin and you had to choose between commutative and distributive, then I would go with distributive.

You see this for example in the ring of matrices.

Now, for the case of integers. Here you probably also would want $-a = -1\times a$. So you would want $(-a)\times b = ((-1)\times a )\times b$. Then if you insist on associativity, then this would give you $(-1)\times (a\times b)$. My point here is that if you want associativity, and you want to product of positive integers to commute, then you are pretty much forced to have commutativity when multiplying by negative integers.

If nothing else, this example explains why, in your question, you need to be very clear about what you are assuming. What is your definition of multiplication for example?

Answer 3

The example of integers is not very useful for this question, since there is no conflict at all between requiring commutativity of the multiplication and distributivity over addition: the usual extension of multiplication from natural numbers to all integers gives you both properties.

But in other situations requiring one might violate the other. In this case it is as far as I know invariably distributivity that is preferable. It is not even clear that commutativity will give you a strong enough requirement to define your operation at all. An example having distributivity without commutativity is the quaternions, another is multiplication of square matrices. It shoudl be said that in both cases even before applying distributivity to extend, some basic instances of multiplication are fixed that already violate commutativity. For the quaternions the basic definitions include $ij=k=-ji$ for instance.