Arithmetic with negative numbers

[Arta]

This is probably a fairly simple question, but I'm pretty clueless about maths in general, so here goes.

Why is the result of a negative multiplied by a negative positive? I know that's the rule, but -2 * -2 = +4 just seems unintuitive to me. If 2 + 2 = 4, and 2 * 2 = 4, shouldn't -2 + -2 = -4 (correct) be the same as -2 * -2 = -4 (incorrect)?

My question isn't about how negative numbers should be calculated, I understand that fine. My question is more why do they behave the way they do? Was this system arbitrarily invented so that negative numbers make sense, or is there some theory/number system behind it?

[Adron]

Well... Multiplying by -1 changes sign.

It wouldn't make sense to have -1 * 1, 1 * -1 and -1 * -1 all be -1?

Think about division: What should 1 / -1 be? Undefined?

[Arta]

I agree, it would make no sense at all! But why? Did someone just decide that -1 * -1 should be 1, so that it doesn't contradict 1 * -1? If so isn't that rather arbitrary?

[Yoni]

Good question!
I will prove it to you.

Read this first, this outlines what I'm going to show to make it simpler to follow.

1. I'm going to define the concept of a "field".
2. The set of real numbers is a field. I won't prove this here, but take it for granted.
3. I'm going to prove the following:
-(-a) = a
a * 0 = 0
(-a) = (-1) * a
(-1) * (-1) = 1
And finally: (-a) * (-b) = a * b

If you don't want to read through lots of boring(?) math skip to the last part with the conclusion now.
Here is the definition of a field: (This slightly varies depending on the source you take it from.)

Assumptions:
A. "F" is a set containing at least 2 items. [1]
B. (+) is a binary operator in F. That is, for every a, b in F, "a (+) b" is an unambiguous member of F. [2]
C. (*) is another binary operator in F.

F is called a field if the following 9 axioms apply:

"Sum" axioms:
1. For every a, b in F: a (+) b = b (+) a
2. For every a, b, c in F: a (+) (b (+) c) = (a (+) b) (+) c
3. There exists a unique member in F, called (0), so that for every a in F: a (+) (0) = a.
4. For every a in F, there exists a unique member in F, called (-a), so that: a (+) (-a) = 0.

"Product" axioms:
5. For every a, b in F: a (*) b = b (*) a
6. For every a, b, c in F: a (*) (b (*) c) = (a (*) b) (*) c
7. There exists a unique member in F, called (1), so that for every a in F: a (*) (1) = a.
8. For every a, except a = (0), in F, there exists a unique member in F, called (a^-1), so that: a (*) (a^-1) = 1.

The axiom that binds them together:
9. For every a, b, c in F: a (*) (b (+) c) = (a (*) b) (+) (a (*) c)

[1] I'm not sure the requirement of at least 2 items is necessary, but if it isn't, it can be later proven that all fields have at least 2 items, so it doesn't matter.
[2] I named the operators (+) and (*) instead of + and * to emphasize that they don't have to be addition and multiplication as you know them. All they have to do is satisfy their axioms.
It can be proven (but I won't do this here) that R (the set of real numbers), together with the operations + and * as you learned them in elementary school, is a field.

Here are some theorems that apply to all fields:
1. If F with (+), (*) is a field, then for every a in F: -(-a) = a.
Proof:
Let a be a member of F.
From axiom #4 we have: a (+) (-a) = (0)
From axiom #1 we have: (-a) (+) a = a (+) (-a)
Therefore: (-a) (+) a = (0)
I will set b = (-a) to clear possible confusion.
b (+) a = (0)
According to axiom #4, the member (-b) is *unique*. Since b (+) a = (0), and (-b) is unique, I conclude that: a = (-b).
Since b = (-a), that means a = (-(-a)). Q.E.D.
2. If F with (+), (*) is a field, then for every a in F: a (*) (0) = (0).
Proof:
Let a be a member of F.
1. From axiom #7 we have: a (*) (1) = a
2. From axiom #3 we have: (1) (+) (0) = (1)
3. Replacing (1) with (1) (+) (0) in the equation from line 1 gives: a (*) ((1) (+) (0)) = a
4. From axiom #9 we have: a (*) ((1) (+) (0)) = (a (*) (1)) (+) (a (*) (0))
5. And since a (*) (1) = a, replacing a (*) (1) with a in line 4 gives: a (*) ((1) (+) (0)) = a (+) (a (*) (0))
6. From line 3, a (*) ((1) (+) (0)) = a, so replacing a (*) ((1) (+) (0)) with a in line 5 gives:

a (+) (a (*) (0)) = a

According to axiom #3, the member (0) such that a (+) (0) = a for every a is *unique*.
I found another member of F that does this: a (*) (0).
Because (0) is unique, I conclude: a (*) (0) = (0). Q.E.D.
3. If F with (+), (*) is a field, then for every a in F: (-a) = (-1) (*) a.
Proof:
Let a be a member of F.
According to theorem #2: a (*) (0) = (0)
According to axiom #4: (0) = (1) (+) (-1)
Therefore: a (*) ((1) (+) (-1))
From axiom #9: (a (*) (1)) (+) (a (*) (-1)) = (0)
From axiom #7: a (*) (1) = a, so:

a (+) (a (*) (-1)) = (0)

According to axiom #4, the member (-a) such that a (+) (-a) = (0) for every a is *unique*.
I found another member of F that does this: a (*) (-1).
Because (-a) is unique, I conclude: (-a) = a (*) (-1).
From axiom #5 I get: (-a) = (-1) (*) a. Q.E.D.
4. If F with (+), (*) is a field, then (-1) (*) (-1) = (1).
Proof:
According to theorem #3: (-1) (*) (-1) = -(-1).
According to theorem #1: -(-1) = (1).
Therefore: (-1) (*) (-1) = (1). Q.E.D.
5. If F with (+), (*) is a field, then for every a, b in F: (-a) (*) (-b) = a (*) b.
Proof:
Let a, b be members of F.
According to theorem #3:
(-a) = (-1) (*) a
(-b) = (-1) (*) b
Then:
(-a) (*) (-b) = ((-1) (*) a) (*) ((-1) (*) b)
By axioms #5 and #6 I continue to:
(-a) (*) (-b) = ((-1) (*) (-1)) (*) (a (*) b)
According to theorem #4 this becomes:
(-a) (*) (-b) = (1) (*) (a (*) b)
And from axioms #5 and #7 I finally get:
(-a) (*) (-b) = a (*) b
Q.E.D.
The last theorem is what you wanted. Hope this clears things up instead of making them more confusing (which appears to me as more likely).

As you see, there is a very concrete theory behind number systems and fields.
The rules/axioms are defined (arbitrarily, you may say, but intuitively!), and the results such as -2 * -2 = +4 are not arbitrarily defined, but proven from the above rules.
The axioms are not dependent on each other, so anything you prove based on these axioms does not contradict them. There are no contradictions here.

Note that a slightly unintuitive rule, "except for (0)", was added in axiom #8. This was indeed added so it wouldn't conflict with other axioms: I proved, without using axiom #8, that a * 0 = 0 (theorem #2). If there was a number 0^-1, then 0^-1 * 0 = 1, and from theorem #2, 0^-1 * 0 = 0, so 1 = 0, and this is a contradiction (I didn't prove that 1 is always different from 0, but it can be proven).

Also note that all this formality is pretty recent. Old age mathematicians were more like engineers than scientists. They were very sloppy, and didn't care about this at all. You may be surprised, but it was only the 17th century A.D. when widespread use of negative numbers started! Also, it was only the 19th and 20th centuries when people started actually caring about the axiomization of the number systems. To this day, it has no practical uses; it is pure mathematics at its best.

Edit: Something cool I wanted to mention and forgot. Fields are a very beautiful mathematical structure and have extremely neat results. Take this theorem for example, I won't prove it but hopefully you'll see its inner beauty:

If F is a finite field (that is, there is a finite amount of elements in it), then the size of the field (the amount of elements in it) is a prime number.

Enjoy.

[jigsaw]

Aha that was so much harder then just saying... Arta, you dumbshit. Dont worry about it