Irrational Powers

Date: 8/30/96 at 22:20:49
From: Anonymous
Subject: Irrational numbers to irrational powers

Does an irrational number to the irrational power yield a rational 
number?  If so, what are the numbers?  The two irrational numbers 
don't need to be the same.  

I approached the problem through trial and error, but it is too time 
consuming.    

Date: 9/5/96 at 10:13:58
From: Doctor Jerry
Subject: Re: Irrational numbers to irrational powers

In general, an irrational number to an irrational power yields an 
irrational number.  The mathematics needed to prove this is quite 
complicated.  I'll try to explain a little.

A mathematician named Ferdinand Lindemann proved that if 
x1, x2, ..., xn are distinct algebraic numbers (see below) and 
p1, p2, ...,pn are algebraic numbers, not all zero, then the sum

   p1*e^(x1) + p2*e^(x2) + ... + pn*e^(xn)

cannot be zero.

A number is an algebraic number if it is a root of an equation
    
   a0 + a1*x^1 + a2*x^2 + ... +an*x^n = 0,

where a0, a2, a2, ... , an are rational numbers.  All of the rational 
numbers and many of the irrational numbers are algebraic.  

For example, the number 2^(1/2) is algebraic since it is a root of 
x^2 - 2 = 0.

If we take n = 2, p1 = 1, and x2 = 0, then e^(x1)+ p2 cannot be zero 
if x1 is a nonzero algebraic number and p1 is any algebraic number. 
From this we conclude that e^(x1) cannot be rational, and so must be 
irrational.  The number e is irrational and we may take x1 = 2^(1/2).  
Hence, an irrational number to an irrational power need not be 
rational.  

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 9/5/96 at 10:17:12
From: Doctor Ken
Subject: Re: algebra(rational and irrational numbers)

Hello -

If you're looking for some specific examples of irrational numbers 
x and y such that x^y is rational, you can probably construct some 
yourself.  Here's how you might go about doing it: take your favorite 
irrational number and call it x.  Now ask yourself what number y would 
have to be in order to satisfy x^y = 5 (Think logs!).

Good luck!

-Doctor Ken,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/