Dividing Negative ExponentsDate: 10/14/98 at 21:33:59 From: Zack potts Subject: Division of negative exponents Hi Dr. Math. My question is simple. I can solve the math problem from my math teacher's explanation, but when I asked her why a negative exponent becomes positive when you move it under the division sign, she did not know. Please help me. Thanks, Zack Date: 10/15/98 at 17:25:00 From: Doctor Rick Subject: Re: Division of negative exponents Hi, Zack. You are right, there must be an explanation. Here is the mathematical way of saying "A negative exponent becomes positive when you move it under the division sign": -n 1 x = --- n x This has a reason, but in a way it's a matter of definition. Remember that when you first learned about exponents, they were defined this way, for a whole number n: x^n = x * x * ... * x \____________/ n times You then discovered that: x^n * x^m = x*x*... *x * x*x*...*x = x*x*...*x = x^(n+m) \_______/ \_______/ \_______/ n times m times n+m times Now you know all about negative numbers, and the question comes to your mind (well, let's pretend it did): Does it make any sense to put a negative integer in the exponent? What would it mean? Well, if a negative exponent is going to make any sense, it ought to follow the addition-of-exponents rule that worked for positive numbers. That is: x^n * x^(-m) = x^(n-m) Let's try n = 2 and m = 1. Then we ought to get: x^2 * x^(-1) = x^(2-1) = x^1 = x If we divide both sides by x^2, we get: x^(-1) = x / x^2 = 1/x So in order to make the addition-of-exponents rule hold, we must DEFINE: +--------------+ | x^(-1) = 1/x | +--------------+ Next let's try m = n = 1. We expect to get: x^1 * x^(-1) = x^(1-1) = x^0 But this is: x^1 * x^(-1) = x * 1/x = 1 so we have to DEFINE: +--------------+ | x^0 = 1 | +--------------+ One more time now ... let's let m = n. Then we ought to get: x^n * x^(-n) = x^(n-n) = x^0 = 1 So if we divide both sides by x^n, we have: +---------------+ | x^-n = 1 / x^n | +---------------+ So now we have a DEFINITION for every negative integer exponent. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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