Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Raising a Negative to an Irrational Power? It Depends

Date: 03/16/2011 at 19:10:28
From: Nick
Subject: Negative Numbers To Irrational Powers

Dear Dr. Math,

Hi, my name is Nick. I am a 10th grader attending Lake Central High
School.

I do things most other normal kids don't. I seem to rather enjoy math a
little too much. I can understand many restrictions and rules in functions
and stuff, but I came across a strange question that none of my math
teachers could answer -- and it still gets me.

The question is as follows:

"When taking a function to an irrational power, the negative set is
excluded even though the possibility of finding a positive number to an
irrational power IS possible. Why is finding a negative to an irrational
power not?"

I have asked this to all of my teachers and my father (they are the best
for my math answers) and even other relatives (though they're not really
any help) -- but no one has had a clear answer.

So, Dr. Math, why can't you find a value for a negative number raised to an
irrational exponent? Or maybe you can; does it deal with logarithmic
functions?

Please try it. For example, type this on your calculator and 
see what you get:

   -3^(sqrt3)
   
I seem to continue to get "Error"; and when writing it out, I am stuck.

This is not only for the square root of three as an exponent, but other
irrationals, such -3^e, and imaginaries, such as -3^i. I know i is the
square root of (-1), but these all seem to not have any real value, so I
am just truly baffled.

Please, don't discard this request.

                                                             -Nick



Date: 03/19/2011 at 01:08:29
From: Doctor Vogler
Subject: Re: Negative Numbers To Irrational Powers

Hi Nick,

Thanks for writing to Dr. Math. I enjoy math more than other adults do,
too, and I am proud of that fact.

There are a number of different ways to approach your question. There is
some discussion of this issue at

Base of an Exponential Function
    http://mathforum.org/library/drmath/view/55604.html 

Graph of y = (-n)^x
    http://mathforum.org/library/drmath/view/66708.html 

Why Is (-n)^fractional Invalid?
    http://mathforum.org/library/drmath/view/62979.html 

y to the x Power
    http://mathforum.org/library/drmath/view/63367.html 

Basically, it all comes down to: What do you mean when you write an
expression like x^y?

The first definition of exponents you have likely seen was the algebraic
definition, where you define x^n for positive integers n as x times itself
n times, then x^(1/n) as a number y such that y^n = x, and x^(p/q) as
(x^p)^(1/q). You might have even seen x^y for irrational numbers y defined
as a limit (if you've encountered limits; they are usually introduced in
calculus or pre-calculus). In this setting, negative numbers to irrational
exponents are not defined because the limit I spoke of generally does not
exist; when it does, it depends on the choice of sequence approaching y.

A different way to define exponents uses calculus and first defines the
natural logarithm ln(x) as an integral, or exp(x) (also written e^x) with
a differential equation, and the other as the inverse function. In
this context,

   x^y = exp(y*ln(x))

But then you have the problem that the integral for ln(x) is only defined
when x is positive, so this definition only works when the base x is
positive.

A third way uses complex analysis and imaginary numbers, where the
function exp(x) can be defined for complex numbers. But then exp(x) has
many inverses, so the natural logarithm ln(x) is a multi-valued function.
Lucky for us, exp(y*ln(x)) is always the same, no matter which value you
pick for ln(x), as long as y is an integer. But if y is rational, then
different choices of ln(x) will give different values for x^y, and the
number of different values you can get is equal to the denominator of y.

For example, the four values for 1^(1/4) are 1, -1, i, and -i. When y is
irrational, then there are infinitely many different values for x^y. If x
is a positive real number, and y is a real irrational number, then exactly
one of those infinitely many different values for x^y is a real number.
But if x is a negative real number, and y is a real irrational number,
then none of those infinitely many different values for x^y is a real
number; all of them are complex numbers.

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Exponents
High School Imaginary/Complex Numbers

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/