Why Does a Negative Times a Negative Make a Positive?Date: 11/25/2003 at 11:36:37 From: Brea Subject: negative integers Why is a negative times a negative a positive? When you answered this question to other people, the way you explained it was over my head. When you have a negative integer, in order to multiply by another negative number why would you have to go to the positive side of the number line? Date: 11/25/2003 at 16:56:03 From: Doctor Ian Subject: Re: negative integers Hi Brea, It's really hard to understand this in terms of a number line. One of the ways math works is that we make up simple ways of thinking about some set of numbers; but those break down when we try to think of other sets of numbers. So then we have to make up new ways of visualizing our expanded sets. The easiest example is fractions. At first, you learn to think of something like 3/4 as meaning "break an item into 4 pieces, and keep 3 of them". That works okay as long as you're dealing with integers. But what would sqrt(2) ------- pi mean in those terms? How would you break something into 'pi' pieces? And how would you keep sqrt(2) of them? Eventually, you have to let go of the 'dividing into pieces' model, and just start thinking of fractions as divisions that you haven't done yet. Similarly, the number line is pretty good for thinking about addition and subtraction, but it's not so good for thinking about multiplication. To be honest, I've never been able to think of a way that is both intuitive and rigorously correct, to explain why the product of two negatives is positive. There are nice ways to visualize the result, but they don't explain why it _has_ to be that way. There is a pretty simple proof, which goes like this: Let a and b be any two real numbers. Consider the number x defined by x = ab + (-a)(b) + (-a)(-b). We can write x = ab + (-a)[ (b) + (-b) ] (factor out -a) = ab + (-a)(0) = ab + 0 = ab. Also, x = [ a + (-a) ]b + (-a)(-b) (factor out b) = 0 * b + (-a)(-b) = 0 + (-a)(-b) = (-a)(-b). So we have x = ab and x = (-a)(-b) Two things that are equal to the same thing are equal to each other, so ab = (-a)(-b) But some people find that too abstract to be convincing. One way we might visualize it is in terms of the coordinate plane. If we have two values, which can be either positive or negative, and represent those values as distances along the horizontal and vertical axes, (+) | | | (-)----- --------(+) | | | (-) then the product of two positives is a positive area, (+) | |.... | . | + . | . (-)----- --------(+) | | | (-) and the product of a negative and a positive is a negative area, (+) | ....| . | . - | . | (-)----- --------(+) | . | - . | . |..... | (-) So what about the remaining possibility? Well, so far, we've seen that changing the sign of _either_ factor changes the sign of the product. That is, in the diagram below, (+) | II | I | (-)----- --------(+) | III | IV | (-) as we move from quadrant I to quadrant II, we change signs; and as we move from quadrant I to quadrant IV, we change signs. So we should also change signs as we move from II to III, or from IV to III, right? But if we do that, we end up with a positive area in quadrant III... which is just another way of saying that the product of two negatives should be a positive. So maybe that's another way to think about it. If we take _anything_, and multiply it by -1, we should end up with a different sign, shouldn't we? So let's start with something like a * b = c where a, b, and c are all positive. If we multiply that by -1, we have to end up with a different sign: -1 * (a * b) = -c and if we multiply by -1 again, we have to end up with a different sign again: -1 * (-1 * (a * b)) = c Does that make sense so far? If we _don't_ do this, then sometimes multiplying by -1 would change the sign, and sometimes it wouldn't. That would be pretty strange. So if c = -1 * (-1 * (a * b)) then, since we can group and order multiplications any way we want, c = -1 * (-1 * (a * b)) = -1 * -1 * a * b = -1 * a * -1 * b = (-1 * a) * (-1 * b) = (-a) * (-b) Do any of these explanations work for you? If not, I can try to find another one that does. But could you let me know either way? Thanks, - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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