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

Frequently asked questions

Isn't

\sqrt{\pi}

just some irrational number, like

\sqrt{2}

? Why treat it as a harder question?

Irrational is not the same as transcendental. $\sqrt{2}$ is irrational but algebraic (root of $x^2-2$ ), and is constructible. $\pi$ turns out to be irrational and transcendental — a much stronger, harder-to-prove property, and it is specifically transcendence, not mere irrationality, that Wantzel's theorem rules out.

Source

Ferdinand von Lindemann, 1882

Über die Zahl π

Step 1 of 6: The task: build a square with the same area as a circle

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In plain words

Give a circle of radius $1$ ; its area is $\pi$ . 'Squaring the circle' means using only straightedge and compass to build a square with that exact same area — which means building a square whose side has length $\sqrt{\pi}$ , since a square of side $s$ has area $s^2$ .

Greek geometers tried this for centuries with clever partial tricks (curves like the quadratrix of Hippias could do it, but those aren't straightedge-and-compass tools). The question sat unresolved for over two thousand years: is $\sqrt{\pi}$ actually one of the numbers a compass can reach?

\text{Area} = \pi r^2 = \pi \implies \text{side of equal square} = \sqrt{\pi}

Detailed analysis

The classical problem, dating to at least the 5th century BCE in Greek mathematics, asks for a straightedge-and-compass construction of a square with the same area as a given circle. Taking the circle to have radius $1$ (area $\pi$ ), the required square has side length $s$ with $s^2=\pi$ , i.e. $s=\sqrt{\pi}$ .

By Wantzel's 1837 theorem (used to settle angle trisection and doubling the cube), every straightedge-and-compass constructible real number lies at the top of a tower of quadratic field extensions over $\mathbb{Q}$ , and in particular is algebraic — a root of some nonzero polynomial with rational coefficients. So squaring the circle is possible only if $\sqrt{\pi}$ is algebraic. Since the square of an algebraic number is algebraic and vice versa (an algebraic $\alpha$ satisfies a polynomial $P$ , and $\alpha^2$ satisfies a related polynomial built from $P$ ), $\sqrt{\pi}$ is algebraic exactly when $\pi$ is.

So the geometric question reduces entirely to a question about one specific real number: is $\pi$ algebraic, or is it transcendental (satisfying no polynomial equation with rational coefficients at all)? Ferdinand von Lindemann answered this in 1882, extending a method Charles Hermite had used nine years earlier to handle $e$ .

Terms in this step

Algebraic number: A number that is a root of some nonzero polynomial with rational (equivalently, integer) coefficients — for instance $\sqrt{2}$ , a root of $x^2-2$ .
Transcendental number: A real or complex number that is not algebraic — it satisfies no polynomial equation with rational coefficients at all, however large the degree.

Knowledge used in this step

Galois theory Ancient Mathematics