Café Math : A Question of (First) Principles

Yesterday my friend Fiza told me about an annoying question that her students asked her, namely,
*Can we prove that minus multiplied by minus gives plus ?*
Mmm... that's the kind of question with no clear answer. As this is merely a postulate...
But the conversation took an interesting turn as we talked about how to construct the negative numbers from the non-negative ones.

In what I know of how the system of numbers is constructed, the negative numbers are introduced by the following trick,

Step 0. In the beginning you start with a set $A$ of positive numbers that you can add and multiply but you can't subtract because there is no negative number in your system.

Step 1. You generalize your system $A$ by taking as new numbers, the set couples of numbers from $A$: $$ B = \{\, (a_1, a_2) \,|\, a_1, a_2 \in A \,\} $$

Step 2. Ok nice, but how do you add, multiply etc... well, \begin{align*} (a_1, a_2) + (b_1, b_2) &= (a_1 + b_1, a_2 + b_2) \\ (a_1, a_2) \times (b_1, b_2) &= (a_1 a_2 + b_1 b_2, a_1 b_2 + a_2 b_1) \end{align*}

Step 3. There is too much numbers in $B$... so we introduce the number system $C$ elements of $C$ are equivalence class of elements of $B$ under the following equivalence relation

Two elements in $B$ say $(a_1, a_2)$ and $(b_1, b_2)$ are said to be equivalent if, $$ a_1 + b_2 = a_2 + b_1, $$ So for example in our new system $(1, 2) = (3, 4)$ because $1 + 4 = 2 + 3$, but $(1, 2)$ is different from $(4, 3)$ because $1 + 3 \neq 2 + 4$.

This very much resembles the calculus of fraction.
If it's too abstract... remember,
$$
1/2 = 2/4 = ... = 100/200 = ... = 30/60 = \text{etc...}
$$
This is the same thing here,
A couple $(a_1, a_2)$ is said to be a representative of a number in our new system of number. But another couple $(b_1, b_2)$ is also a representative of the *same* number, if it is equivalent to $(a_1, a_2)$. Just like $6/3$ and $10/5$ are both representatives of the same number $2$.

Step 4.

Fiza. Where are the original numbers from $A$ in this system ?

Samuel. A given number $a$ in $A$ is represented by the couple, $(a, 0)$. Consequence: The number $(a, b)$ represents the difference $a - b$...

Fiza. Ho ! I get it ! this is where those funny laws of addition and multiplication comes from and why you said that $(a_1, a_2)$ is equivalent to $(b_1, b_2)$ if, $a_1 + b_2 = a_2 + b_1$ ha ha

Samuel. Yes very clever of you.

Fiza. And so about the original question ? I fell we've digressed a bit far ;-)

Samuel. Sure but that was interesting, wasn't it ?

Fiza. Sure it was ;-)

Samuel. Ok now we can prove your original statement, the product of two negative numbers is positive, $$ (0, a) \times (0, b) = (00 + ab, 0b + 0a) = (ab, 0) $$ mmm no need to say more.

Fiza. Ok nice and simple as I like it ;-)

Some exercises for those who want to learn more. As a student, I always read math book with a paper and a pen. When I found something unclear or mysterious, I always turned to my paper and pen and tried to figure out the thing myself. I recommend that practice warmly for those who sometime feel uncomfortable with some parts of a math text. By doing that, not that the Magic evaporates... instead I think it's the way one becomes a math Wizard oneself.

Reading the above text I've found some of the point that deserve to be worked with pen and paper.

- Can you recover the law of addition and multiplication, \begin{align*} (a_1, a_2) + (b_1, b_2) &= (a_1 + b_1, a_2 + b_2) \\ (a_1, a_2) \times (b_1, b_2) &= (a_1 a_2 + b_1 b_2, a_1 b_2 + a_2 b_1) \end{align*} from the comment "The number $(a, b)$ represents the difference $a - b$..." like Fiza did ?
- Prove that the notion of addition and multiplication is invariant by equivalence. Giving a precise meaning to that statement is part of the exercise.
- Give the definition of the subtraction in the new system. Hint: solve the equation $(a_1, a_2) + (x, y) = (b_1, b_2)$ for $(x,y)$.
- We supposed that positive numbers are represented in the new system by $(a, 0)$ and negative one by $(0, b)$. I said that a number $(a, b)$ is considered positive whenever $a > b$. Now can you give an invariant definition of the order relation, given that you can always compare two numbers from the original system $A$.

Have FUN ;-)