[wsxcde]

This is not a proof of why the product of negative numbers is positive. The reason why the product of negative numbers is positive is that we define multiplication to be that way.

Also, this post conflates the unary negation operator with negative numbers. The two are not the same. In so far as this post constitutes a proof (which IMO it does not), it is a proof about the behavior of the negation operator.

A good question to ask is why we made this specific choice of definition. Why should multiplication be defined such that -2*-3 = 6? This is a question that the post does shed some light on. If we'd chosen some other definition of multiplication, a lot of the "intuitive" properties of multiplication that hold over the natural numbers (such as the distributivity of multiplication over addition and subtraction) would no longer be true over the integers.

[thaumasiotes]

> If we'd chosen some other definition of multiplication, a lot of the "intuitive" properties of multiplication... would no longer be true

Well, sure, if you change the definition of something, then it may end up having different properties. What's your point?

[wsxcde]

My point is that you cannot prove something that is true by definition. The OP trying to prove that the product of two negative numbers is positive is like asking to prove that 0 + 1 = 1 in Peano arithmetic.

The OP thinks that his "proof" is showing why multiplying negative values yields a positive result. But the proof is a load of nonsense because it assumes facts like distributivity of multiplication over addition and subtraction. It is literally impossible to prove that $\forall a, b, c \in Z. (a - b) * c = (a * c - b * c)$ -- distributivity of multiplication over subtraction -- without having already defined the meaning of a * b for all integers! This leads to a circular reasoning loop that the OP's "proof" can't get out of.

The thing to realize is that multiplication is not some magic operation handed down to us by god. It is just a binary total function defined over the integers. What the OP is trying to confusedly get at is the following:

1. There is an intuitive definition of multiplication as repeated addition over natural numbers.

2. It is not clear what the corresponding definition of multiplication over negative numbers is.

3. If we want to define multiplication as a total function over the integers, we need to define what the result should be when multiplying negative integers.

4. Specifically, with (3), we are taught in school that the result of multiplying two negative numbers should be positive, but it is not clear why this seemingly arbitrary choice was made.

Unfortunately, the OP is going about this all backwards. One cannot prove what the OP wants to prove. What one can instead do is argue that the specific (but seemingly arbitrary) definition that one has chosen for multiplication is a "good" choice because it has the same properties (distributivity etc.) as multiplication over natural numbers. At its core, this is a stylistic appeal about the "naturalness" of the definition.

[yori]

Not every arithmetic property needs to be proved from Peano axioms. One can but it is tedious and unnecessary. A much better starting point is the set of field axioms where the distributivity property is already available as an axiom.

The assumptions made in the article are perfectly fine as per field axioms. Granted it would have been nicer if distributivity over addition was used instead of distributivity over subtraction. But it is not a big leap to derive distributivity over substraction from field axioms by distributing multiplication over a positive number and the additive inverse of another positive number.

Wherever you see an assumption made about negative number, just mentally replace it with additive inverse of a positive number and you would be fine.

[wsxcde]

You are wrong in several ways.

1. You literally cannot prove this fact from the Peano axioms because Peano arithmetic operates on natural numbers, not integers.

2. As I said in my original post at the top, negative numbers are different from the unary subtraction operator (the additive inverse in the field). The number -2 is an entity that exists by itself regardless of whether you've defined an additive inverse. It turns out that the additive inverse of every positive integer is the corresponding negative integer, but this follows from the definition of +, not the other way around.

3. Even if you give OP the benefit of the doubt regarding his dodgy proof, it is saying something about the additive inverse and its relation to multiplication. It is not saying why the result of multiplying two negative values must be positive.

4. The multiplicative operator over the field must already be defined for you to be able to prove distributivity over addition. You can't assume distributivity over an operator that is only partially defined.

--

Think about how you'd define a field. First, you need a set (let's call it Z), then you need two total operators over the set (+ and ), and two elements of the set (0 and 1) and each of these must satisfy specific properties (aka the field axioms). In particular, + and must be defined for all members of Z, not just Z+ and further + and * must be distributive. These are all facts you need to prove about Z, *, +, and 1 and only then do you have a field. You cannot work backward by assuming the field axioms (which are unfortunately named because they are not axioms at all but properties) to derive the definition o the field operators.

[carloswilson]

> You literally cannot prove this fact from the Peano axioms because Peano arithmetic operates on natural numbers, not integers.

Sure you can. With definitions! Define integers from natural numbers. Define rationals from integers. And so on. And so on.

> The number -2 is an entity that exists by itself regardless of whether you've defined an additive inverse.

I see a serious misunderstanding of this topic. Please read upon the field axioms and ring axioms if you haven't so already. Then please check https://math.stackexchange.com/a/878844 which is arguably more rigorous than this post. But the essence is the same. This is more rigorous because the subtraction operator is not used anywhere. Only addition, multiplication and additive inverses have been used. Like another commenter said, if you just replace subtraction with addition with an additive inverse in the OP's post, things fall in place.

[yori]

> You cannot work backward by assuming the field axioms (which are unfortunately named because they are not axioms at all but properties) to derive the definition o the field operators.

Of course, when we say they are field axioms we mean those properties hold true for the elements of the field. If you see those properties, they talk about distributivity over the elements of the field and additive inverses of the elements also belong to the field, so the distributivity automatically applies to additive inverses too.

After that with a little algebra, "product of additive inverses of two elements is equal to the product of the two elements" comes out as a result (not a definition).

Of course, by "product" we mean whatever * represents. It is not necessarily the multiplication operator we see in numbers.

[carloswilson]

While this product of additive inverses property does follow from the field axioms, there is a short discussion in the blog comments at https://susam.in/blog/product-of-negatives/comments/ which shows that this property holds true for all rings too.

So it is not just numbers for which this property holds true but for all elements of fields and rings too. Quite simply, (-a)(-b) = (a)(b) in all rings where (-a) and (-b) are the additive inverses of a and b respectively.

[yori]

> If we'd chosen some other definition of multiplication, a lot of the "intuitive" properties of multiplication that hold over the natural numbers (such as the distributivity of multiplication over addition and subtraction) would no longer be true over the integers.

This is backward reasoning. The chosen definition of multiplication is not to keep things "intuitive". If you start with the field axioms, the chosen definition of multiplication is pretty much dictated by the axioms. If you choose another definition of multiplication, you would end with contradictions like 1 = 0 and such nonsense! And mathematicians abhor contradictions!

"Product of additive inverses of two elements is equal to the product of the two elements" is dictated by the field axioms in all fields.

[carloswilson]

> Also, this post conflates the unary negation operator with negative numbers.

-a is a standard way to represent additive inverse of an element in field.

The point about "unary negation operator" seems irrelevant.

In the real number field, additive inverse of a positive real number is indeed the negative of that number. The negative of that number is also obtained by the application of unary negation operator on the positive number.

The additive inverse of 3.14 is -3.14. Unary negation operator applied to 3.14 gives us -3.14. I don't see how conflating unary negation operator with negative numbers here is any issue here.

[wsxcde]

> In the real number field, additive inverse of a positive real number is indeed the negative of that number. The negative of that number is also obtained by the application of unary negation operator on the positive number.

This is a fact that follows from the definition of +. But + needs to be defined before you can start making assumptions about what the additive inverse is. The set over which the field is defined (Z or R) already contains -3, -2 etc. and -3 * -2 or -3 + -2 needs to be defined when you're constructing the field. It then turns out that -3 is the additive inverse of 3. You can't use this when arguing about the definition of why applying * on negative 2 and negative 3 gives you the result positive 6. Because you need to define * over all members of the field before you construct a field in the first place.

[carloswilson]

> The set over which the field is defined (Z or R)

What? The set of all integers, Z, is not a field! R is. But Z isn't. Z is a ring, a commutative ring. I doubt you understand what a field is!

> already contains -3, -2 etc.

Yes, and those elements are literally the additive inverses of their positive counterparts. If you disagree with this, then the numbers -3, -2, etc. literally have no meaning.

> It then turns out that -3 is the additive inverse of 3.

Are you making this all up with your original research or do you have any proper literature written by a professional mathematician to back it up?

[Koshkin]

To be fair, you can have a field that only has integers. For example, Z mod 5 is a field.

[carloswilson]

I am aware. It is typically represented as Z_5. They are called prime fields.

I highly doubt wsxcde meant prime fields in their comment though. wsxcde seemed to be talking about the set of all integers and the set of all real numbers in their comment. Only the latter is a field (and a ring) whereas the former is only a ring.

And (-a)(-b) = ab holds in rings (and thus fields).