**The fact that a negative times a negative equals a positive can be proven mathematically using algebraic manipulation.** Specifically, it can be proven using the equation x = (a)(b) + (-a)(b) + (-a)(-b), where a and b are any positive numbers and x is unknown, via factoring and reducing the expression.

The proof is as follows. First, by definition -n = (-1)(n) for any number n.

Now, taking the expression x = (a)(b) + (-a)(b) + (-a)(-b) and factoring out (-a) from the last two terms gives x = (a)(b) + (-a)[(b)+(-b)]. Since (b)+(-b) = 0, this yields x = ab.

Similarly, by starting with the same equation, x = (a)(b) + (-a)(b) + (-a)(-b), and factoring out (b) from the first two terms on the right side, gives x = [a+(-a)](b) + (-a)(-b). Since (a)+(-a) = 0, this gives x = (-a)(-b).

Since equality is transitive, meaning that things which are equal to the same are equal to each other, (-a)(-b) = (a)(b). Since (a) and (b) are positive, (a)(b) must be positive. Thus, as (a)(b) = (-a)(-b), (-a)(-b) is also positive. Q.E.D.

This can also be demonstrated by mapping products of a negative number and other chosen numbers on a number line going from positive values to negative values. For example, in mapping products with (-1), one would get that -1(10) = -10, 10 units left of zero, -1(9) = -9, nine units left of zero and one unit right of -10 = -1(10). Continuing in such a fashion, one gets to -1(1) = -1 and -1(0) = 0. Now to find -1(-1), one continues the pattern and moves one unit to the right of -1(0) = 0 to yield -1(-1) = 1.