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Dividing Negative Exponents


Date: 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/   
    
Associated Topics:
High School Exponents
High School Negative Numbers
Middle School Exponents
Middle School Negative Numbers

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