[F]
why in an equation we have that (negative)x(negative)=(positive)?
for example: (-2)x(-2)=+4

p.s x=multiplication

The proof that a negative times a negative results in a positive. Good question. Something we take for granted, but rarely think about, why?

To get into the nuts and bolts of this it would help to be familiar with the concept of a ring.

Let me explain what a ring is. It is an albebraic structure that exhibits certain properties.

It has two operations, addition and multiplication(+ and *).

If x, y, and z are in R, then:

1: x+y is closed under addition.
2. x+y=y+x, commutative.
3. (x+y)+z=x+(y+z), associative.
4. x*(y+z)=x*y+x*z, distributive
5. x*(y*z)=(x*y)*z, associative.
6. x+0=x, additive identity.
7. x*y is closed under multiplication.
8. There exists a y such that x+y=0 (additive inverse, y=-x).
9. x*y=y*x ( Commutativity under multiplication).


First, let's prove that 0*a=0 for all a:

0*a=(0+0)*a=0*a+0*a by distibutive property.
By 8, add the additive inverse of 0*a to both sides and we get 0=0*a.

Now, prove that -1*a=-a:

-1*a+a=-1*a+1*a because of identity
-1*a+a=(-1+1)*a because of distributive.
-1*a+a=(-1+1)*a=0*a because of what we showed above.

Hence: -1*a+a=-1*a+1*a=(-1+1)*a=0*a. Therefore, -1*a=-a by the additive identity of a.

Now, if -a*-b=a*b:

-a*b=(-1*a)*(-1*b)=(-1*1)*(a*b).

So, it is enough to show that -1*-1=1. But -1*-1=-(-1). -(-1) is the additive inverse of -1, which is 1 and the proof is complete.


This probably more than you're ready to absorb.
That's where basic group theory comes in. It is helpful in proving these intuitive things we take for granted.