Worked solution: Lindemann's transcendence proof of $\pi$ settles squaring the circle (1882)

Frequently asked questions

If exact squaring is impossible, why do engineers and builders still use approximations like

\pi\approx22/7

Approximate squaring — building a square whose area is close enough to a circle's for practical purposes — has always been easy and remains perfectly fine for engineering. What Lindemann's theorem rules out is an exact construction using straightedge and compass in finitely many steps; no amount of practical approximation changes that mathematical fact.

Source

Ferdinand von Lindemann, 1882

Über die Zahl π

Step 5 of 6: Conclusion:

\pi

is transcendental, so the circle cannot be squared

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In plain words

The chain is now complete. Step 4 showed $\pi$ cannot be algebraic, i.e. it is transcendental: no polynomial with rational coefficients, of any degree whatsoever, has $\pi$ as a root. Since the square of an algebraic number is always algebraic, if $\sqrt{\pi}$ were algebraic then $(\sqrt{\pi})^2=\pi$ would be too — but it isn't. So $\sqrt{\pi}$ is transcendental as well.

And Wantzel's theorem (Step 1) already told us that every straightedge-and-compass constructible number is algebraic. A transcendental number is therefore never on the list of buildable lengths, full stop — settling, after more than two thousand years, that squaring the circle is impossible.

\pi \notin \overline{\mathbb{Q}} \implies \sqrt{\pi} \notin \overline{\mathbb{Q}} \implies \text{squaring the circle is impossible}

Detailed analysis

Step 4 established that $\pi\notin\overline{\mathbb{Q}}$ , where $\overline{\mathbb{Q}}$ denotes the field of algebraic numbers: $\pi$ is transcendental. Since $\overline{\mathbb{Q}}$ is closed under taking square roots (if $\alpha\in\overline{\mathbb{Q}}$ satisfies polynomial $P(x)$ , then $\sqrt{\alpha}$ satisfies $P(x^2)$ , also a nonzero rational polynomial), the contrapositive gives: if $\sqrt{\pi}$ were algebraic, its square $\pi$ would be too. Since $\pi$ is not, $\sqrt{\pi}\notin\overline{\mathbb{Q}}$ as well — $\sqrt{\pi}$ is transcendental.

Step 1 established that every straightedge-and-compass constructible real number is algebraic (Wantzel 1837), i.e. lies in $\overline{\mathbb{Q}}$ . Since $\sqrt{\pi}$ lies outside $\overline{\mathbb{Q}}$ , it cannot be constructed with straightedge and compass. But squaring a unit circle requires constructing exactly a segment of length $\sqrt{\pi}$ (Step 1), so this construction is impossible.

This completes the resolution of squaring the circle, the last of the three classical Greek construction problems (alongside angle trisection and doubling the cube) to be settled, 45 years after Wantzel's 1837 paper and closing a question that had remained open since antiquity.