# The matter of signs in multiplication

I was browsing through the non-mandatory tasks on my college's website and I stumbled upon one which cracks my head quite a bit. It goes as follows:

Prove why multiplying two numbers with different signs gives us a negative number while multiplying two of the same sign - positive.

Thing is... I don't even know where to start. How is one supposed to prove thing like that? I mean - it seems so natural, uncontested and taken as granted from one's early years of life that I can't even figure how one could prove it. Is there some proof of such thing I could read or the question itself is somehow tricky?

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You will not be able to prove something if you don't have a definition. That need not be a formal definition, but you need to have some inkling of what you think the multiplication with negative numbers is. (You might want to start with multiplication by positive integers, fractions and reals before tackling negative numbers.) There are many approaches and you could just read them, but it would be better to first consider what you already know. –  Phira Nov 7 '12 at 18:11
Please specify what type of "numbers" you refer to. The notion of "sign" does not make sense for all types of numbers. –  Bill Dubuque Nov 7 '12 at 18:14
That's all that was in the task description. Keeping in mind we didn't cover a lot of material, though, I'd say they refer to real numbers. –  Straightfw Nov 7 '12 at 18:16
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## 4 Answers

Use the distributive property: $$-a\cdot b = (0-a)\cdot b = 0\cdot b - a\cdot b.$$

(Here it is assumed $a > 0$ and $b > 0$, although it is not a necessary condition -- it just allows us to consider one case without loss of generality).

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Seems great but I have a question. Say we have $-a\cdot(-b)$. Then, we would end with $(0-a)(0-b)=-a\cdot(-b)$ and we're stuck as we can't make it just $ab$ since that's what we are to prove. How should one prove it, then? –  Straightfw Nov 7 '12 at 17:58
In this case simply use $-a = -1 \cdot a$ and then apply the commutative property: $-a\cdot -b = (-1\cdot a) \cdot (-1 \cdot b) = -1\cdot -1 \cdot a\cdot b$. Then, employ the identity $-1\cdot -1 = 1$. –  Arkamis Nov 7 '12 at 18:00
OK, but am I allowed to assume $-1\cdot(-1)$ just like that? I mean - isn't such an identity what I'm to prove rather than what I can safely assume? –  Straightfw Nov 7 '12 at 18:01
Yes, because that is an axiomatic property of the real field. In other words, $-1\cdot -1 = -(-1)$. We have to axiomatically admit the definition of negation. The negation of an element $x$ is such that $x + (-x) = 0$, so $(-1)+(-(-1)) = 0$ means that $-(-1)$ is the negation of $-1$, and since the only negation of $-1$ is $1$, then $-(-1) = 1$. –  Arkamis Nov 7 '12 at 18:07
You could probably extend that property generally to all $a, b$ in the real field, and then prove $-a \cdot -b > 0$ that way, but it's much, much cleaner to leverage commutativity and apply that property only to the multiplicative identity element $1$. –  Arkamis Nov 7 '12 at 18:08
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To complement the other answers, here is a proof that $(-1)(-1) = 1$.

We know $(-1) \cdot 1 = -1$, since this is the definition of $1$ (its use in multiplication does not change the value of the other number).

We know also that $(-1) \cdot 0 = 0$.

Now, \begin{align*} 0 &= (-1) \cdot 0\\ &= (-1) (1 + -1)\\ &= (-1)(1) + (-1)(-1)\\ &= (-1) + (-1)(-1). \end{align*} Adding $1$ to both sides gives $$1 + 0 = 1 + (-1) + (-1)(-1),$$ which simplifies to $$1 = (-1)(-1).$$

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After all that, I am now concerned that the proof of $(-1)\cdot 0 = 0$ begs the question. –  Austin Mohr Nov 7 '12 at 18:23
Haha, not really but other thing does :) I mean: in the passage from $-1\cdot(1+-1)$ to $-1\cdot1+(-1)(-1)$, it's assuming that $(-1)(-1)=1$ which we later prove, isn't it? Thank you anyway, Ed Gorcenski provided a good explanation for $(-1)(-1)=1$ :) –  Straightfw Nov 7 '12 at 18:27
@Straightfw I use only distributivity at that step. Notice I never simplify the two factors of $(-1)(-1)$. They are present until the end of the proof. –  Austin Mohr Nov 7 '12 at 18:28
Oh my, didn't notice it. Great one, then. Thank you a lot :) –  Straightfw Nov 7 '12 at 18:29
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There's the algebraic point of view, but there's also the following. If I'm not mistaken, negative numbers were introduced in Italy in the middle ages to represent debts. If you have $\$30$and you owe$\$20$; your net worth is $\$10$:$\$30+(-\$20)=\$10$. If you have $\$30$and you owe$\$50$; your net worth is $\$30+(-\$50)=-\$20$. So you gain$5$debts of$\$7$ each; this changes your net worth by $5\cdot(-\$7)=-\$35$.

Then suppose $5$ of your debts of $\$7$each are forgiven. To your total number of debts,$-5$is added; your net worth changes by$-5\cdot(-\$7)=+\$35$. - add comment Use the fact that$(-1)(-1)=1$,$(1)(1)=1$, commutativity and assocativity to write$(-a)(-b)=(-1)a(-1)b=(-1)(-1)ab(a,b>0)\$.

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