---
tag: 26G1QB6G1W
kind: proof
name: Proof of the impossibility of squaring the circle
msc: [12F05]
proves: 0X7CVDE3BG
uses: [0FMN2MN7TM, 0BSXRER81B, 0PN9WMRWT7, 0SSSM1D475, 0G0F7FGCGP, 0BJV24C4SY, 0KW2X0TFBX, 0ZB7J071WQ]
sources:
  - url: https://en.wikipedia.org/wiki/Squaring_the_circle
    locator: Impossibility
---

## Proof

Let $\mathcal{C}$ be the set of the [[constructible numbers|0FMN2MN7TM]]. We argue by contradiction.

1. By the [[field theorem|0PN9WMRWT7]], $\mathcal{C}$ is a subfield of $\C$ closed under square roots.
2. Suppose that $\sqrt{\pi} \in \mathcal{C}$. Then $\pi = \sqrt{\pi} \cdot \sqrt{\pi} \in \mathcal{C}$, since $\mathcal{C}$ is closed under products.
3. Every constructible number is [[algebraic over $\Q$|0SSSM1D475]]. So $\pi$ is algebraic over $\Q$.
4. This contradicts the [[transcendence of $\pi$|0G0F7FGCGP]]. So $\sqrt{\pi} \notin \mathcal{C}$.

Now we turn to the construction problem. Take the given circle with centre $0$ through $1$, so its area is $\pi$. A [[squaring of the circle|0BSXRER81B]] would construct two points $P, Q$ with $\abs{P - Q} = \sqrt{\pi}$, the side of a square of area $\pi$.

5. Let $\overline{\mathcal{C}}$ be the set of the [[conjugates|0BJV24C4SY]] of the elements of $\mathcal{C}$. Conjugation preserves the field operations, so $\overline{\mathcal{C}}$ is a subfield of $\C$.
6. If $w = \bar z$ with $z \in \mathcal{C}$ and $t^2 = w$, then $\bar t^2 = z$, so $\bar t \in \mathcal{C}$ and $t \in \overline{\mathcal{C}}$. So $\overline{\mathcal{C}}$ is closed under square roots.
7. The field theorem says that $\mathcal{C}$ is the smallest subfield of $\C$ closed under square roots. So $\mathcal{C} \subseteq \overline{\mathcal{C}}$. Each $z \in \mathcal{C}$ is then $\bar u$ for some $u \in \mathcal{C}$, and $\bar z = u \in \mathcal{C}$.
8. Let $w = P - Q \in \mathcal{C}$. Then $\bar w \in \mathcal{C}$, so [[$w \bar w = \abs{w}^2$|0ZB7J071WQ]] is in $\mathcal{C}$.
9. Its square root, the [[modulus|0KW2X0TFBX]] $\abs{w} = \sqrt{\pi}$, is in $\mathcal{C}$. This contradicts step 4.
10. So $\sqrt{\pi}$ is not constructible, and the circle cannot be squared with straightedge and compass.

## Idea

Constructible numbers are algebraic, and they form a field. If $\sqrt{\pi}$ were constructible, its square $\pi$ would be algebraic, which Lindemann's theorem rules out. A squaring of the circle would make the length $\sqrt{\pi}$ constructible.