Transcendence of pi

Statement

The number $\pi$ is transcendental over $\mathbb{Q}$.

Remarks

The statement follows from the Lindemann–Weierstrass theorem: $e^\alpha$ is transcendental for every nonzero algebraic number $\alpha$. Indeed, if $\pi$ were algebraic, then $i \pi$ would be algebraic and nonzero, but $e^{i \pi} = -1$ is algebraic. So $\sqrt{\pi}$ is also transcendental, and it is not constructible.

History

Charles Hermite proved the transcendence of $e$ in 1873. Ferdinand von Lindemann proved in 1882 that $e^\alpha$ is transcendental for every nonzero algebraic $\alpha$, and so that $\pi$ is transcendental. Karl Weierstrass proved a more general statement in 1885.

Explanations

L1 human

The number $\pi$ is not a root of any nonzero polynomial with rational coefficients. This is much stronger than the irrationality of $\pi$, and the proof is much harder.