Impossibility of squaring the circle

Statement

The number $\sqrt{\pi}$ is not a constructible number. So the circle cannot be squared with straightedge and compass.

Remarks

The proof: every constructible number is algebraic over $\mathbb{Q}$. If $\sqrt{\pi}$ were constructible, then $\pi = (\sqrt{\pi})^2$ would be constructible and so algebraic. This contradicts the transcendence of $\pi$.

History

The impossibility was proved in 1882, as a consequence of Lindemann's proof that $\pi$ is transcendental. It had been known for decades that the transcendence of $\pi$ would imply the impossibility.

Explanations

L1 human

A square with the area of the unit circle has side $\sqrt{\pi}$. Constructible numbers are algebraic, and $\pi$ is not, so this side cannot be constructed.