[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.

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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.