# Algebraic independence of pi and e

 Importance: High ✭✭✭
 Author(s):
 Subject: Number Theory
 Keywords: algebraic independence
 Posted by: porton on: July 8th, 2008

\begin{conjecture} $\pi$ and $e$ are \Def[algebraically independent]{Algebraic_independence} \end{conjecture}

% You may use many features of TeX, such as % arbitrary math (between $...$ and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{http://www.research.att.com/~njas/sequences/}

## Bibliography

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)

* indicates original appearance(s) of problem.

### Schanuel's conjecture

Assuming \Def[Schanuel's conjecture]{Schanuel's conjecture}, one can show that $\pi$ and $e$ are algebraically independent over $\mathbb Q$.

### in which subfield K of which field L?

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

### By definition?

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

### not all transcedentials are algebraically independant

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

### two transcendentals are not necessarily algebraically independen

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

### ?

but that's not a polynomial.