Prove that minus one times minus one equals one. It’s such a simple statement that many people wouldn’t know where to begin. The hitch is that you’re only given the 9 axioms of algebra. My father worked on this problem for a few hours for his graduate math class this semester at SJSU before he gave up and let me take a shot at it. I feel that I just got lucky and saw the solution after playing around with it for 15 minutes.

A1) a + b = b + a
A2) (a + b) + c = a + (b + c)
A3) 0 + a = a and a + 0 = a
A4) a + (-a) = 0 and (-a) + a = 0
M1) a * b = b * a
M2) (a * b) * c = a * (b * c)
M3) 1 * a = a and a * 1 = a
M4) a * (1/a) = 1 and (1/a) * a = 1
D) a * (b + c) = (a * b) + (a * c)

And on top of those, knowing that 0 * a = 0 would make our lives a little easier, so let us prove that theorem: Start) a = a M3) a = a * 1 A3) a = a * (1+0) D) a = a * 1 + a * 0 M3) a = a + (a * 0) = a + 0 A3) a * 0 = 0 QED: T1) a * 0 = 0

And now, to prove -1 * -1 = 1: Start) -1 * -1 = 1 A3) -1 * -1 + 0 = 1 A4) -1 * -1 + -1 + 1 = 1 M3) -1 * -1 + -1 * 1 + 1 = 1 D) -1 (-1 + 1) + 1 = 1 A4) -1 (0) + 1 = 1 T1) 0 + 1 = 1 A3) 1 = 1 QED! It’s proven!

And there it is. I bet you got pretty worried I wouldn’t make it there towards the end, huh?