pi+e

[gravenewworld]

Is it algebraic? I remember my professor talking about this probelm, he just swept it under the rug. Are there any proofs that pi+e is algebraic or not?

[jcsd]

according to Wolfram at least one of the numbers pi + e and pi*e must be trancendental and probably both, though trancendence has been proved for neither:

http://mathworld.wolfram.com/TranscendentalNumber.html

[matt grime]

also at least one of pi+e and pi-e must be transcendental (if they were both algebraic, so would be one half their sum and difference, ie pi and e would be algebraic).

[mathwonk]

lets see if wolframs statement is obvious. consider (x-e)(x-pi) = x^2 - (e+pi)x + e*pi.

Now if both e+pi and e*pi were algebraic, then e and pi would be roots of an equation with algebraic coefficients, so wouldn't they both be algebraic?

[shmoe]

Yes, and since pi and e are both transcendental, the result follow. That the roots of a polynomial with algebraic coefficients are algebraic follows from the fact that if F2 is and algebraic extension of F1 and F1 is an algebraic extension of F, then F2 is an algebraic extension of F (F2, F1, F fields). Proof should be in most algebra texts and isn't too difficult.