# Schanuel's Conjecture Implies Algebraic Independence of Pi and Euler's Number over the Rationals

## Theorem

Let Schanuel's Conjecture be true.

Then $\pi$ (pi) and $e$ (Euler's number) are algebraically independent over the rational numbers $\Q$.

## Proof

Assume the truth of Schanuel's Conjecture.

Let $z_1 = 1$ and $z_2 = i \pi$.

Note that $z_1$ is wholly real and $z_2$ is wholly imaginary.

Hence, by Wholly Real Number and Wholly Imaginary Number are Linearly Independent over the Rationals, they are linearly independent over $\Q$.

By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_2, e^{z_1}, e^{z_2}}\right)$ has transcendence degree at least $2$ over $\Q$.

That is, the extension field $\Q \left({1, i \pi, e, -1}\right)$ has transcendence degree at least $2$ over $\Q$.

However, $1$ and $-1$ are algebraic.

Therefore $i \pi$ and $e$ must be algebraically independent over $\Q$.

Aiming for a contradiction, suppose $\pi$ and $e$ are not algebraically independent over $\Q$.

Then, then there would be a non-trivial polynomial $g \left({x, y}\right)$ with rational coefficients satisfying:

- $g \left({\pi, e}\right) = 0$

Then, one can construct a non-trivial polynomial $f \left({x, y}\right) = g \left({i x, y}\right) g \left({-i x, y}\right)$ with rational coefficients satisfying:

- $f \left({i \pi, e}\right) = 0$

which is contradictory to the previous statement that no such polynomials exist.

Therefore, if Schanuel's Conjecture is true, then $\pi$ and $e$ are algebraically independent over $\Q$.

$\blacksquare$