Product of '-' and '-' is '+'! In School, students were taught that "if $m$ and $n$ are two non-negative integer, then their product $m \times n$ is equal to the sum of $n$, $m$-times or sum of $m$, $n$-times". If one of them is negative, then the principle still work. After this they were taught that "$ - \times - = +$", so the product of two negative integers is well defined. Does this follow from the above principle? Some people told me that it follows from the above principle, but i don't know how it follow. If someone ask you that how "$ - \times - = +$", what will be your answer?
Thanks in advance
 A: $5\times 2$: I give you two five-dollar bills.  You are \$10 richer.
$(-5)\times 2$: I give you two five-dollar IOU's.  You are \$10 poorer.
$5\times (-2)$: I take away two five-dollar bills.  You are \$10 poorer.
$(-5)\times (-2)$: I take away two five-dollar IOU's.  You are \$10 richer.
A: Suppose you do not know this to be true, but you know $a+-a = 0$. We ask ourselves then what $b \times -a$ yields. Well to do this think about what $b \times (a+-a)$ yields. Well obviously this is zero. So using the distributive property we get $ba + -ba = 0$ so we know that if we multiply a positive number by a negative number we get the opposite. To figure out what $-b \times -a$ yields, we can do the same form of thought experiment. For suppose we want to find $-b(a+-a)$ obviously this is once again zero. By our previous thought experiment, we know that if we distribute this, we get $-ba + (-b \times -a) = 0$ and since we have established that $-ba +ba = 0$ we can now infer that $-b \times -a $ must indeed equal $ba$
A: For any real number $a$ define $a'$ to be the additive inverse of $a$, that is
$$a+a'=0.$$
Observe that obviously $(-1)'=1$ as $-1+1=0$.
From $a'=a-2a=-1\cdot a$ we see: we get the additive inverse of $a$ by multiplying $a$ by $-1$.  So what's $-1\cdot (-1)$?  Well, $-1\cdot$ anything is the additive inverse of anything, so 
$$-1\cdot(-1)=\text{additive inverse of}-1=1.$$
Michael
