[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.