Proof of the impossibility of squaring the circle

Builds on Constructible number Squaring the circle The constructible numbers form a field closed under square roots The degree of a constructible number is a power of two Transcendence of pi Complex conjugate Modulus of a complex number A complex number times its conjugate

Proof

Let $\mathcal{C}$ be the set of the constructible numbers. We argue by contradiction.

By the field theorem, $\mathcal{C}$ is a subfield of $\C$ closed under square roots.
Suppose that $\sqrt{\pi} \in \mathcal{C}$ . Then $\pi = \sqrt{\pi} \cdot \sqrt{\pi} \in \mathcal{C}$ , since $\mathcal{C}$ is closed under products.
Every constructible number is algebraic over $\Q$ . So $\pi$ is algebraic over $\Q$ .
This contradicts the transcendence of $\pi$ . 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 would construct two points $P, Q$ with $\abs{P - Q} = \sqrt{\pi}$ , the side of a square of area $\pi$ .

Let $\overline{\mathcal{C}}$ be the set of the conjugates of the elements of $\mathcal{C}$ . Conjugation preserves the field operations, so $\overline{\mathcal{C}}$ is a subfield of $\C$ .
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.
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 $\bar z = u \in \mathcal{C}$ .
Let $w = P - Q \in \mathcal{C}$ . Then $\bar w \in \mathcal{C}$ , so $w \bar w = \abs{w}^2$ is in $\mathcal{C}$ .
Its square root, the modulus $\abs{w} = \sqrt{\pi}$ , is in $\mathcal{C}$ . This contradicts step 4.
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.

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The degree of a constructible number is a power of two statement
Proof of the impossibility of squaring the circle proof

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2 pages build on this one.

Tag	`26G1QB6G1W`
URI	`m2://26G1QB6G1W`
Kind	proof
Writing	W0 stub
MSC	12F05
Proves	Impossibility of squaring the circle
Uses	Constructible number, Squaring the circle, The constructible numbers form a field closed under square roots, The degree of a constructible number is a power of two, Transcendence of pi, Complex conjugate, Modulus of a complex number, A complex number times its conjugate
Changed	4 November 2026

Proof of the impossibility of squaring the circle

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