A Negative Times a Negative
Date: 06/02/98 at 21:24:45 From: El-ad Blech Subject: Integers (multiplying) Hello Dr. Math, I would like to know more about integers. My question is, why do you have to take out the negative sign when you multiply two negative numbers (i.e. -2 x -2 = 4)? Thank You, El-ad Blech
Date: 06/13/98 at 23:31:49 From: Doctor Peterson Subject: Re: Integers (multiplying) Hi, El-ad, This is something a lot of people have trouble understanding. I'll give you an explanation I find helpful. One thing that helps is to realize that negative numbers are just an extension of the number line, and when we try to decide how to do operations on them, our main concern is to keep everything working the way it does when you work with positive numbers. Suppose I make a graph of a line; maybe it represents how far I travel if I keep going forward at a constant speed, so the horizontal axis is time and the vertical axis is distance (or position along a road): D | / 6 + X | / 3 + X |/ 0 +-+-+---- T 0 1 2 This graph shows that I went 3 miles in the first hour, and another 3 in the second hour, so in two hours, I went forward 2 * 3 = 6 miles. I can represent the graph by the equation: D = 3 * T I can continue this graph backward in time, to figure out where I would have been at earlier times. If I do that, it should still follow the same rules as for positive times. As I go from right to left on the graph above, D decreases by 3 each time I decrease T by 1. It should still do that if I continue it for negative values of T, so it will look like this: D | / 6 + X | / 3 + X |/ ------+-+-+-+-+---- T -2-1/| 1 2 X + -3 / | X + -6 / | So for T = -2, D = 3 * (-2) = -6. That part isn't hard to believe. All we're doing is continuing the straight line to the left. In terms of motion, at a time before the present I was at a place behind my starting point. Now let's think about negative speeds. Suppose I'm driving backwards (maybe the wrong way down a one-way street, or with my car in reverse). Then for each hour I drive, I go -3 miles, that is, 3 miles backward: D | 6 + | 3 + | +-+-+---- T |\1 2 -3 + X | \ -6 + X | \ This represents the equation: D = (-3) * T How do I continue this for negative times? Since D increases by 3 each time T decreases by 1, it should look like this, continuing the straight line backwards: D \ | X 6+ \ | X3+ \| ------+-+-+-+-+---- T -2-1 |\1 2 -3+ X | \ -6+ X | \ But that means that at time T = -2, D = (-3) * (-2) = 6 What all this really means is just that a negative sign on either factor reverses the result, and reversing it twice makes it positive again. I hope that helps make it more visible. -Doctor Peterson, The Math Forum Check out our web site! http://mathforum.org/dr.math
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