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?
 A: 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).
A: 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 $-5\cdot(-\$7)=+\$35$.
A: 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
$$
1 = (-1)(-1).
$$
A: Use the fact that $(-1)(-1)=1$, $(1)(1)=1$, commutativity and assocativity to write $(-a)(-b)=(-1)a(-1)b=(-1)(-1)ab$ $(a,b>0)$.
