Re: Irrational numbers questions

Dave Seaman wrote:

On Wed, 03 Jan 2007 08:09:15 -0500, G. A. Edgar wrote:

Can it be proven that, with e irrational, also e^N (with N natural) is
irrational? Guess so, but why ..

Let's see ... If a = sqrt(2), then a is irrational, but a^2 is rational.

But e is also transcendental, and from that we can conclude that e^n is
transcendental for each natural number n.

Let's see. e is transcendental because there doesn't exist a polynomial
equation with integer coefficients such that e is a solution. Right?
Now let's set up such an equation for e^n as well, then upon working out
the powers it boils down to the same sort of equation for e. Now suppose
that e^n is not transcendental, then it follows e is not transcendental.
Contradiction. Hence e^n is transcendental. Did I forget something?

Han de Bruijn

.