Open Problem Garden - Algebraic independence of pi and e - Comments

Schanuel's conjecture (re: Algebraic independence of pi and e)

warut — Sun, 29 Apr 2012 13:45:04 +0200

Assuming Schanuel's conjecture, one can show that and are algebraically independent over .

not all transcedentials are algebraically independant (re: Algebraic independence of pi and e)

cubola zaruka — Fri, 05 Aug 2011 08:37:49 +0200

pi and 4-pi are both transcedential and sum to 4, so are not algebraically independant.

two transcendentals are not necessarily algebraically independen (re: Algebraic independence of pi and e)

Anonymous — Thu, 21 Jul 2011 00:19:37 +0200

e and e^2 are both transcendental but (e,e^2) makes the two-variable polynomial f(x,y)=x^2-y equal to zero

By definition? (re: Algebraic independence of pi and e)

Anonymous — Sat, 16 Jul 2011 02:31:03 +0200

I think any two distinct transcendental numbers must be algebraically independent, almost by definition. Since e and pi are transcendental, they must be a. i. No? - David Spector

? (re: Algebraic independence of pi and e)

Jon Noel — Wed, 18 May 2011 01:16:02 +0200

but that's not a polynomial.

in which subfield K of which field L? (re: Algebraic independence of pi and e)

Comet — Wed, 16 Feb 2011 08:29:32 +0100

After all, e to the pi i = -1, so this shows that pi and e are not always algebraically independent.

Algebraic independence of pi and e

porton — Tue, 08 Jul 2008 02:01:30 +0200

Author(s):

Subject: Number Theory

Conjecture and are algebraically independent