prove that 2 negative multiples = a positive number?

Question

I'm sorry I mistyped my first question. Just showing it's true isn't a proof =D

Answer 1

Let K be any ordered field, and let a and b be elements of K such that a<0 and b<0. Then ab>0.

Proof:

First note that by the distributive law, (-a)(-b) + (-a)b = (-a)(-b + b) = (-a)*0 = 0. Likewise, ab + (-a)b = (a+(-a))b = 0*b = 0. Since both quantities are equal to 0, (-a)(-b) + (-a)b = ab + (-a)b. Since these are equal, they will remain equal upon subtracting (-a)b from both sides, so (-a)(-b) = ab. Now, since K is an ordered field, the order relation is compatible with addition, so we may add (-a) to both sides of the inequality a<0 to obtain 0<(-a). Similarly, 0<(-b). Since both (-a) and (-b) are greater than zero, their product is also greater than zero (the fact that the product of two positive quantities is positive is an axiom of ordered fields). Therefore (-a)(-b) > 0. However, (-a)(-b) = ab, so ab>0, and the product of two negatives is positive. Q.E.D.

Answer 2

-2 x -2 = +4

Answer 3

-x(-x)=x^2

Think of this: How else would you describe a negative number?