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Consider the following excerpts from Ask Dr. Math :

Excerpt 1

So the real question is, $$(-1)(-1) = ?$$ and the answer is that the following convention has been adopted: $$(-1)(-1) = (+1)$$ This convention has been adopted for the simple reason that any other convention would cause something to break.

Excerpt 2

Let $a$ and $b$ be any two real numbers. Consider the number $x$ defined by $$x = ab + (-a)(b) + (-a)(-b)$$ We can write $$x = ab + (-a)[ (b) + (-b) ]$$ $$ = ab + (-a)(0)$$ $$ = ab + 0$$ $$ = ab$$ Also, $$x = [ a + (-a) ]b + (-a)(-b)$$ $$ = 0 * b + (-a)(-b)$$ $$ = 0 + (-a)(-b)$$ $$ = (-a)(-b)$$ So we have $$x = ab$$ and $$x = (-a)(-b)$$ Hence, by the transitivity of equality, we have $$ab = (-a)(-b)$$

The two excerpts seem to contradict each other- the first suggests that the property is just a convention, while the second suggests it's an intrinsic mathematical property belonging to negative numbers.

Convention is defined as something that has been adopted for convenience ($0! = 1$) while a "intrinsic mathematical property" is something that cannot be defined otherwise ($e^{\pi i} = -1$).

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I agree that "convention" is not a very good choice of a word. The fact that $(-a)(-b)=ab$ is proved below, so it is not a convention, it is a theorem. –  Giuseppe Negro Feb 1 '13 at 1:39
Previously: Why negative times negative = positive? –  Rahul Feb 1 '13 at 1:56
It is neither a convention nor a fundamental property but a derived property, a property deduced from fundamental properties. –  Gerry Myerson Feb 1 '13 at 2:07

5 Answers 5

up vote 3 down vote accepted

"$1$" is a symbol used to define the number $e$ such that $ex=x$ for all $x$. "$-1$" is a symbol we use to represent the value $x$ for which $x+1 = 0$. We have to assume that such an $x$ exists, and there are systems where it doesn't, for example the natural numbers. We know that there can only be one such number because if there were some $x'$ such that $x'+1=0$ we would have $$x+1=x'+1\\$x+1+x=x'+1+x\\x'+0=x+0\\x'=x$$ meaning that if there are two such numbers then they are the same.

But what would $(-1)(-1)$ be? Well, we have $(-1) = (-1)(1)$ by the definition of $1$, so we think about what we want $(-1)(-1)+(-1)(1)$ to be. If we assume the distributive law, we have: $$(-1)(-1)+(-1)(1) = \\ (-1)(-1+1)=\\ (-1)(0)=\\ 0$$

So $(-1)(-1)+-1 = 0$, but we have already shown that the only number for which this is true (or at least have analogously shown) is $1$. If this weren't the case, then at least one of the laws we used would be false and that wouldn't make much sense.

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Law of Signs proof: $\rm\,\ (-x)(-y) = (-x)(-y) + x(-y + y) = (-x+x)(-y) + xy = xy$

Equivalently, evaluate $\rm\:\overline{(-x)(-y) +} \overline{ \underline {x(-y)}} \underline{ +xy_{\phantom{.}}}\ $ in two ways, noting each over/under term $ = 0$.

Said more conceptually, $\rm\:(-x)(-y)\ $ and $\rm\:xy\:$ are both inverses of $\rm\ x(-y)\ $ so they are equal by uniqueness of inverses.

Note that the proof uses only ring laws (most notably the distributive law), so the law of signs holds true in every ring. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring would simply be a set with two completely unrelated additive and multiplicative structures. So, in a certain sense, the distributive law is the keystone of the ring structure.

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Most of the answers so far give the algebraic viewpoint. But there's a dual geometric viewpoint that's more intuitive.

Think of a number $k \in \mathbb R$ as an operator acting on the real number line $\mathbb R$ as $k: \mathbb R \to \mathbb R$ as $k: x \mapsto kx$.

Then the action of -1 on $\mathbb R$ is $(-1) \mathbb R= -\mathbb R$, ie, the real line is reflected about the origin: $0 \mapsto 0$, $1 \mapsto -1$, $2 \mapsto -2$, $- \pi \mapsto \pi$ etc.

From this it's easy to see that $-1$, as an operator is an involution: the product $(-1)(-1)$ acting on $\mathbb R$ reflects the real line twice, sending it back to the original $\mathbb R$.

So if it's a convention it's consistent with this symmetry, which is as fundamental a concept as anything in math or physics.

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So negative numbers come up as we need additive inverses, ie we need a number $x$ such that $x+1=0$. We will call this $x$ $-1$. Similar for all other negative numbers.

It can then be shown that taking the additive inverse twice yields the name number. That is if $-n+x=0$ adding $n$ to both sides that $n-n+x=n$, so $0+x=n$, so $x=n$.

Also note that we can show that taking the additive inverse is the same as multiplying by $-1$ if we want addition and multiplication to behave well. Therefore, $n+(-1)n=n(1+(-1))=n\cdot 0$ by distributivity. This shows that $-n=(-1)$.

As taking the additive inverse twice yields the same number, this means that $-(-1)=(-1)(-1)=1$, and in general $(-n)(-m)=(-1)n(-1)m=(-1)(-1)nm=1nm=nm$.

I hope that makes sense and helps at least why this is true. It is true because for addition and multiplication to work as we want them to (with associativity, commutativity, and distributivity), this must be true.

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The reason is: it is highly desired that what we call "real numbers" be a Field.

Mathematically, we define the real numbers to be that way (see the Construction by Dedekind Cuts, for instance)

Intuitevily, we "know" that what we call addition is associative, commutative, has a neutral element and a inverse element, and multiplication is associative, commutative, has a neutral element and has an inverse element (for every non-zero element). Based on those, we show that:


Proof: $(-a).b+(a.b)=(-a+a).b=0.b=0$

Therefore, it is proved $(-a).b=-(a.b)$

But then: $(-a).(-b)=-(a.(-b))=-(-(a.b))=a.b$

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