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Why the product of two negative numbers is a positive number? We all learn about it, and I'm not sure why it is like that. Is it a convention or what?

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  • $\begingroup$ So: you can define "Bre multiplication" in another way if you like. No one is stopping you. But then (see the answers) your number system will fail to have one or more of the usual properties that we like to use. $\endgroup$ – GEdgar Oct 26 '13 at 12:54
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Suppose that you have a number $a.$ There exists a number $b$ such that $a+b=0$--called the additive inverse of $a$--and we denote this number $b$ by $-a$. A particular example is the additive inverse of $1,$ that is, $-1$. Note that the number $1$ has the special property that $1\cdot a=a$ for all $a$. Then there are a few quick facts that we can use.

Fact 1: $0\cdot a=0$ for all $a$. Indeed, $$0=0\cdot a+-(0\cdot a)=(0+0)\cdot a+-(0\cdot a)=0\cdot a+0\cdot a+-(0\cdot a)=0\cdot a+0=0\cdot a.$$

Fact 2: $-1\cdot a=-a$ for all $a$. Indeed, $$-a=-a+0\cdot a=-a+(1+-1)\cdot a=-a+1\cdot a+-1\cdot a=-a+a+-1\cdot a=-1\cdot a.$$

Finally, since $1+-1=0,$ then $1=-(-1)=(-1)\cdot(-1)$, as desired. Because of commutativity of multiplication, it follows from this, together with fact 2, that the product of two negative numbers is positive, since the product of two positive numbers is positive.

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It follows from the basic axioms ("conventions" if you like) for addition and multiplication, especially for all $x,y,z$ $$\tag0 (x+y)+z=x+(y+z)$$ $$\tag1 (x+y)z=xz+yz$$ $$ \tag2 0+x=x+0=x$$ $$ \tag3 x+(-x)=(-x)+x=0$$ $$\tag 4 1\cdot x=x\cdot 1=x$$ (associativity of addition, distributivity of multiplication over addition, neutrality of $0$ under addition, definition of additive inverse, neutrality of $1$ under multiplication). With these we are forced to accept $$\begin{align} 0 &\stackrel {(3)}= 0\cdot x+(-(0\cdot x))\\ &\stackrel {(2)}=(0+0)\cdot x +(-(0\cdot x))\\ &\stackrel{(1)}=(0\cdot x+0\cdot x)+(-(0\cdot x))\\ &\stackrel{(0)}=0\cdot x+(0\cdot x+(-(0\cdot x)))\\ &\stackrel{(3)}=0\cdot x+0\\ &\stackrel{(2)}=0\cdot x\\ \end{align}$$ i.e. multiplying anything by zero gives zero: $$\tag5 0\cdot x=0.$$ Next, using this result we find $$ \begin{align} (-1)\cdot x &\stackrel{(2)}=(-1)\cdot x+0\\ &\stackrel{(3)}=(-1)\cdot x+(x+(-x))\\ &\stackrel{(0)}=((-1)\cdot x+x)+(-x)\\ &\stackrel{(4)}=((-1)\cdot x+1\cdot x)+(-x)\\ &\stackrel{(1)}=((-1)+1)\cdot x+(-x)\\ &\stackrel{(3)}=0\cdot x+(-x)\\ &\stackrel{(5)}=0+(-x)\\ &\stackrel{(2)}=-x,\\ \end{align}$$ i.e. multiplying with $-1$ gives the additive inverse. Note that the inverse of the inverse is the original again: $$ -(-x) \stackrel{(2)}=-(-x)+0 \stackrel{(3)}=-(-x)+((-x)+x) \stackrel{(0)}=(-(-x)+(-x))+x \stackrel{(3)}=0+x \stackrel{(2)}=x$$ Therefore $$ (-1)\cdot(-1)= -(-1) = 1.$$

Of course one may ask why $(1),(2),(3),(4)$ hold in the first, but these rules can indeed be shown for the natural numbers by induction except (3), which does not make sense for natural numbers), and then for integers, rational numbers, real numbers, complex numbers by the very way we define these extensions of the number concept from the previous ones.

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Because it can't be any other way. It's a logical consequence of the definitions of addition and multiplication. Don't look for any deeper metaphysical meaning.

It follows from the fact that $(-a)b=-ab$ and $a(-b)=-ab$ for all $a$ and $b$:

$$(-1)(-1) = - (1)(-1) = +(1)(1) = 1$$

so really it's just a consequence of the fact that the opposite of a negative is a positive.

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The distrubutive law can demonstrate this. First observe that for any $x$ we have $$(-1)x + x = (-1 + 1)x = 0x = 0, $$ so $-x = (-1)x$. Next notice that $$ -1 + (-1)(-1) = -1(1 + (-1)) = -1(0) = 0.$$ By the cancellation laws, $-(-1) = 1$.

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All the answers so far are good, but I add this one because it only uses the fact that $0$ times anything is $0$, and $1$ times anything is that thing again:

$0=0 \times 0 = (1+(-1))\times (1+(-1)) = (1\times 1) +(1\times (-1))+((-1)\times 1)+((-1)\times(-1))$

$0= 1+ -1 + -1 +(-1)\times (-1)=-1+(-1)\times (-1)$

Therefore, $1 = (-1)\times (-1)$

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I was a bit disappointed to see other answers suggesting that $(-1)(-1) = 1$ is just a consequence of some axioms and has no deeper meaning. On the contrary, the meaning of this formula shows up in many places (as often as negative numbers and multiplication do, anyway.)

One interpretation of the product $ab$ of two numbers $a$ and $b$ is that if you start with $1$, multiply it by $a$, and then multiply it by $b$, you get $ab$. More generally, multiplying any amount by $a$ and then by $b$ is the same as multiplying it by $ab$. If you have an lucky investment, say, or a population of rabbits, that doubles in the first year and then triples in the second year, it has increased by a factor of $2 \times 3$ (${}=6$) over the span of both years.

To get a similar example involving negative numbers, we first need to assign some meaning to negative numbers. One way would be to use them in measuring position. Suppose we have a measuring stick that is two yards long, has the center point marked with zero, has markings emanating $1,2,3,\ldots$ emanating to the right at intervals of one inch, and similarly has markings $-1,-2,-3,\ldots$ emanating to the left.

Such a measuring stick would be useful in order to measure distances to the right of, and also to the left of, a fixed object without moving the stick. If you don't like the measuring stick you can just think of an abstract number line.

Now if we were to flip the measuring stick around its center point, the negative numbers would be on the right hand side and the positive numbers on the left hand side. More specifically, $1$ and $-1$ would switch places, $2$ and $-2$ would switch places, $3$ and $-3$ would switch places, and so on. Each number would go to the space previously occupied by its negative. Equivalently, each number would go to the space previously occupied by itself times $-1$.

The flipping represents multiplication by $-1$, just as if the measuring stick were a stretchy tape, stretching it to three times its original length (while leaving the center point fixed) would send $1$ to the space previously occupied by $3$, send $2$ to the space previously occupied by $6$, and so on, corresponding to multiplication by $3$.

To see what $(-1)(-1)$ should mean, let's see what happens if we multiply by $-1$ and then multiply by $-1$ again. We flip the stick around its center point, and then flip it again. Every number ends up back where it started! So multiplication by $(-1)(-1)$ is the same as multiplication by $1$, or in other words doing nothing. This is why we say that $(-1)(-1)=1$.

If your measuring stick is stretchy then you can see what $(-2)(-3)$ is, etc.

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    $\begingroup$ Many people simply don't care about "real life"; whatever that is. $\endgroup$ – Pedro Tamaroff Oct 26 '13 at 18:19
  • $\begingroup$ @PedroTamaroff Why are you pretending that you don't know what real life is? $\endgroup$ – Trevor Wilson Oct 26 '13 at 18:22
  • $\begingroup$ Well, you're using the word "real" on "life". Is the a "fake" life? A "complex" life? A "quaternionic" life"? What is the point of such use of words? =) $\endgroup$ – Pedro Tamaroff Oct 26 '13 at 18:23
  • $\begingroup$ @PedroTamaroff To distinguish real life from fantasy life. But if you don't like it I can change "real life" to "life." $\endgroup$ – Trevor Wilson Oct 26 '13 at 18:27
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    $\begingroup$ @Trevor I admit that the argument you've given does provide some intuition, and that particular intuition may be what the OP was asking for or could benefit from. (Based on their apparent level of sophistication, that does seem to be the case.) On the other hand, this example, like every metaphor, has its limits. They include: what is "left" and "right"? what is the meaning of arbitrarily large numbers in a finite universe? what is the effect of other physics on the meaning of arithmetic? They must be addressed by explaining that you only use some essential features, i.e. axioms. $\endgroup$ – Ryan Reich Oct 26 '13 at 20:56

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