Irrationality of e+pi and e*piDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/