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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

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  • $\begingroup$ That two negatives multiply to a positive follows from the distributive property. $\endgroup$ – anon Sep 16 '13 at 4:42
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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 $ba$

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$5\times 2$: 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.

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  • $\begingroup$ Should be UOI's technically! But good analogy. $\endgroup$ – texasflood Feb 16 '15 at 20:05
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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

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