# Schanuel's Conjecture Implies Transcendence of Pi by Euler's Number

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## Theorem

Let Schanuel's Conjecture be true.

Then $\pi \times e$ is transcendental.

## Proof

Assume the truth of Schanuel's Conjecture.

By Schanuel's Conjecture Implies Algebraic Independence of Pi and Euler's Number over the Rationals, $\pi$ and $e$ are algebraically independent over the rational numbers $\Q$.

That is, no non-trivial polynomials $\map f {x, y}$ with rational coefficients satisfy:

- $\map f {\pi, e} = 0$

Aiming for a contradiction, suppose $\pi \times e$ is algebraic.

Then there would be a non-trivial polynomial $\map g z$ with rational coefficients satisfying:

- $\map g {\pi \times e} = 0$

However, $\map f {x, y} := \map g {x \times y}$ would be a non-trivial polynomial with rational coefficients satisfying:

- $\map f {\pi, e} = 0$

which contradicts the earlier statement that no such polynomials exist.

Therefore, if Schanuel's Conjecture holds, $\pi \times e$ is transcendental.

$\blacksquare$