Irrationality of e+pi and e*pi

Date: 09/24/2001 at 00:38:32
From: Andy Weck
Subject: Irrationality of e+pi and e*pi

I have read that it is unknown if either E + Pi or E * Pi is an 
irrational number. However, it is provable that at most one of the 
two numbers is rational. How do you prove this? I thought that the 
proof would involve summing the two numbers in some manner and then 
proving that the result is irrational. (This way you know at least one 
of the addends must be irrational.) But I can't figure out how to 
prove that any combination of the two numbers is irrational.

Date: 09/24/2001 at 10:59:22
From: Doctor Luis
Subject: Re: Irrationality of e+pi and e*pi

Hi Andy,

Thanks for the very interesting question. Like you, I was unable to 
find a combination e+pi and e*pi that proved immediately the 
irrationality of either number. There's hardly anything that can 
easily be proved by using that approach. However, given further 
knowledge about both both pi and e, your claim can be proved rather 
elegantly.

Now, e and pi are rather peculiar numbers. It turns out that, in 
addition to being irrational numbers, they are also transcendental 
numbers. Basically, a number is transcendental if there are no 
polynomials with rational coefficients that have that number as a 
root.

Clearly, p(x) = (x-e)*(x-pi) is a polynomial whose roots are e and pi, 
so its coefficients cannot all be rational, by the definition of 
transcendental numbers. Expanding that expression, we get

     (x-e)*(x-pi) = x^2 - (e+pi)*x + e*pi

This means that 1, -(e+pi), e*pi cannot all be rational. If all the 
coefficients were rational, we would have found a polynomial with 
rational coefficients that had e and pi as roots, and that has been 
proven impossible already. Hermite proved that e is transcendental in 
1873, and Lindemann proved that pi is transcendental in 1882. In fact, 
Lindemann's proof was similar to Hermite's proof and was based on the 
fact that e is also transcendental.

In other words, at most one of e+pi and e*pi is rational. (We know 
that they cannot both be rational, so that's the most we can say).

Great question. I hope this explanation helped!

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