quadrature-commensurability-remark

Quadrature and the Commensurability Remark

Summary: §540 — Euler’s closing reflection on chapter 22 and on Book II. The nine problems were “made up in order that the nature of the circle might be penetrated more deeply.” The point is hidden in plain sight: if any of the transcendental quantities computed here had happened to be commensurable with the radius, a quadrature of the circle (or of part of it) would have followed. Since none of them did, Euler stops short of asserting impossibility but acknowledges that no better method is available.

Sources: chapter22, §540 (final paragraph of Book II).

Last updated: 2026-05-12.

---

The remark in full

“I will not pose more questions of this sort, since a method for solving them should be sufficiently clear from the given examples. Besides, these problems were made up in order that the nature of the circle might be penetrated more deeply, since attempts at these quadrature problems have been unsuccessful by all previous methods. If it should happen in the solution of one or another problem, that the arc might be commensurable with the whole circumference, or the sine or tangent of the arc might be constructed from the radius, then indeed quadratures of the circle have been known. For example, if in the solution of problem VI, the sine $DE$ had been equal to $0.6666666=\frac{2}{3}$, rather than $0.6665578$, then a very nice property of the circle could be used, namely, the arc $AE$ could be constructed equal to the straight lines $AD+DE=1+\frac{2}{3}+\sqrt{5/9}$. Even if there is no obvious reason which makes this kind of quadrature of the circle impossible, still, if there should be such a reason, there does not seem to be a better method for solving the problems than that which we have given in this chapter.” — §540

The chapter (and Book II) ends with “END OF THE SECOND BOOK”.

What is being claimed

Three claims, in increasing strength:

1. Method-completeness. The nine examples display the full toolkit (false position, trigonometric reduction, series inversion) for solving the relevant transcendental equations. No new technique is needed.

2. Commensurability would settle quadrature. If even one of the transcendental constants that fell out (the $0.7390847$ of arc-equals-cosine, the $0.6665578$ of arc-equals-versine-plus-sine, the $54^{\circ }18^{\prime }\dots$ angle of equal-area-bisection-problems Problem II, etc.) had been rational, or algebraic from the radius, the arc itself would be expressible as an algebraic combination of straight lines — a quadrature.

3. Impossibility is plausible but unproven. Euler does not declare quadrature impossible. He notes that we have no proof, only no examples — and that if a proof of impossibility were ever obtained, the methods of chapter 22 would still be the best available for approximating the relevant lengths.

The Problem-VI example

The example is chosen with care. In Problem VI (arc-equals-versine-plus-sine), the geometric setup is clean: semicircle, perpendicular ordinate, equation $AE=AD+DE$. The computation produces $DE=0.6665578$. The near-miss with $\frac{2}{3}=0.66\overline{6}$ is striking — about $0.013\%$ off.

If $DE=\frac{2}{3}$ were exact, then $\cos s=-1+\frac{2}{3}\cdot \tan (\text{etc.})$ would yield an algebraic expression for the supplement-arc $BE=41^{\circ }48^{\prime }\dots$ in terms of $1,\frac{2}{3},\sqrt{5/9}$. The chord $AE$ — and hence the related arc — would be the algebraic number $1+\frac{2}{3}+\sqrt{5/9}\approx 2.412$. An algebraic length equal to a circle-arc length is precisely a partial quadrature of the circle: it would let you straighten out the curve into a finite construction with straightedge and compass.

The fact that $DE$ is not $\frac{2}{3}$ — the equation is irreducibly transcendental — is the chapter’s quiet philosophical punchline.

Historical context

In 1748 (publication of the Introductio) the algebraic transcendence of $\pi$ was a century-and-a-quarter from being proved (Lindemann 1882). Euler did not have transcendence theory; the best contemporary belief was that $\pi$ was irrational (Lambert would prove irrationality in 1761), with no proof of transcendence. So Euler’s §540 stops exactly where the available knowledge stopped:

• He knows quadrature is empirically hopeless.
• He knows no proof of impossibility exists.
• He bets, hedgingly, that no quadrature is coming.

The wager paid off — Lindemann’s theorem shows $\pi$ is transcendental, hence not expressible algebraically from the radius, hence the chord-arc identifications Euler dreamt of in §540 are forever ruled out.

Why this is a fitting closer

Book II is, structurally, a long catalogue of curve forms — algebraic curves (chapters 1–20), transcendental curves (chapter 21), and now problems on transcendental curves (chapter 22). The final reflection turns the catalogue back onto the most famous of all curves — the circle — and the oldest of all unsolved problems — its quadrature. Euler does not claim to have solved it; he claims to have illuminated the obstruction.

Related pages

• chapter-22-on-the-solution-to-several-problems-pertaining-to-the-circle — context.
• arc-equals-versine-plus-sine — the example Euler highlights.
• transcendental-curves — the §506 definition that frames the chapter.
• chapter-21-on-transcendental-curves — the catalogue Euler is reflecting on.