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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.2012-11-07
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    Please specify what type of "numbers" you refer to. The notion of "sign" does not make sense for all types of numbers.2012-11-07
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    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.2012-11-07

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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?2012-11-07
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    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 $-1\cdot -1 = 1$.2012-11-07
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    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?2012-11-07
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    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 $-(-1) = 1$.2012-11-07
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    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 $1$.2012-11-07
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    That seems what I was asking. Thank you a lot!2012-11-07
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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 $-5\cdot(-\$7)=+\$35$.

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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.2012-11-07
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    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 $(-1)(-1)=1$ :)2012-11-07
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    @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.2012-11-07
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    Oh my, didn't notice it. Great one, then. Thank you a lot :)2012-11-07
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Use the fact that $(-1)(-1)=1$, $(1)(1)=1$, commutativity and assocativity to write $(-a)(-b)=(-1)a(-1)b=(-1)(-1)ab$ $(a,b>0)$.