Exponents Containing Negative Integers

Date: 12/15/98 at 19:44:53
From: Patrick Harrison
Subject: n^-x,  -n^x,  -n^-x

I simply don't understand how to work exponents that contain negative 
integers. Please explain how I'd work the following types of problems:
      
   N^-X
  -N^ X
  -N^-X

Thank you for your time.

Date: 12/16/98 at 08:55:41
From: Doctor Rick
Subject: Re: n^-x,  -n^x,  -n^-x

Hi, Patrick, welcome to Ask Dr. Math!

First of all, you must be careful about order of operations when using
negative numbers and exponents. "-N^X" means to raise N to the X power
first, then take the negative, which is not what I think you mean. 
I make a habit of using parentheses whenever there MIGHT be confusion 
about what I mean. I would write: N^(-X); (-N)^X; (-N)^(-X).

Any number raised to a negative power is the same as 1 over the number
raised to the positive power:

   -X    1
  N   = ---
        N^X

It is defined this way so that the addition rule for exponents will 
still work with negative exponents. For instance:

  N^X * N^(-X) = N^(X - X) = N^0 = 1

To understand what happens when you raise a negative number to a power, 
you can use the distributive principle of exponents over 
multiplication:

  (AB)^X = (A^X)(B^X)

So,

  (-N)^X = (-1 * N)^X = ((-1)^X)(N^X)

What is (-1)^X? We know that (-1)^2 = 1. Therefore -1 to any EVEN power 
is 1, and -1 to any ODD power is -1. Putting it all together,

  (-N)^X =   N^X    if X is even,
    OR   = -(N^X)   if X is odd.

You can combine the rules for N^(-X) and (-N)^X to get this rule:

  (-N)^(-X) =  1/(N^X)  if X is even,
              -1/(N^X) if X is odd.

I hope this helps.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/