---
tag: 26XC0EM9M6
kind: proof
name: Proof of the transcendence of pi from the Lindemann–Weierstrass theorem
msc: [11J81]
proves: 0G0F7FGCGP
uses: [00SW8B91E7, 0MB4TGMXG6, 0QFTETCHKB, 0JNCBFMPXR, 04R3CY41VJ]
sources:
  - url: https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem
    locator: Transcendence of e and π
---

## Proof

We argue by contradiction.

1. Suppose that $\pi$ is not [[transcendental|00SW8B91E7]] over $\Q$. Then $\pi$ is an [[algebraic number|0MB4TGMXG6]].
2. The number $i$ is algebraic, since it is a root of $x^2 + 1$.
3. The [[algebraic numbers form a field|0QFTETCHKB]], so $\alpha_1 = i\pi$ is algebraic. Also $\alpha_2 = 0$ is algebraic, and $\alpha_1 \neq \alpha_2$ because $\pi \neq 0$.
4. By [[Euler's formula|04R3CY41VJ]], $e^{i\pi} = \cos \pi + i \sin \pi = -1$. Also $e^0 = 1$, from the series of the exponential.
5. With $\beta_1 = \beta_2 = 1$, which are algebraic, we get $\beta_1 e^{\alpha_1} + \beta_2 e^{\alpha_2} = -1 + 1 = 0$.
6. By the [[Lindemann–Weierstrass theorem|0JNCBFMPXR]], this forces $\beta_1 = \beta_2 = 0$. This contradicts $\beta_1 = 1$.
7. So $\pi$ is transcendental over $\Q$.

## Idea

Euler's identity $e^{i\pi} + 1 = 0$ is a linear relation, with algebraic coefficients, between exponentials of $i\pi$ and $0$. The Lindemann–Weierstrass theorem forbids such a relation when the exponents are distinct algebraic numbers. So $i\pi$, and therefore $\pi$, cannot be algebraic.