Lindemann–Weierstrass theorem

Statement

Let $\alpha_1, \dots, \alpha_n$ be distinct algebraic numbers. Then $e^{\alpha_1}, \dots, e^{\alpha_n}$ are linearly independent over the field of algebraic numbers. That is, if $\beta_1, \dots, \beta_n$ are algebraic numbers with $\beta_1 e^{\alpha_1} + \dots + \beta_n e^{\alpha_n} = 0$, then $\beta_1 = \dots = \beta_n = 0$.

Remarks

A consequence: $e^{\alpha}$ is transcendental for each nonzero algebraic number $\alpha$. In particular, $e$ is transcendental. Another consequence: $\pi$ is transcendental. Otherwise $i\pi$ would be algebraic, and $e^{i\pi} + 1 = 0$ would contradict the theorem. So the circle cannot be squared with ruler and compass.

History

Ferdinand von Lindemann proved in 1882 that $e^{\alpha}$ is transcendental for each nonzero algebraic $\alpha$. Karl Weierstrass proved the general statement in 1885.

Explanations

L1 human

The exponential function sends nonzero algebraic numbers to transcendental numbers. With $\alpha = 1$ this shows that $e$ is transcendental, and with Euler's formula it shows the same for $\pi$.