How did Saint Vincent prove that if $\frac{a}{b} = \frac{c}{d}$, then the area of a hyperbola $y = \frac{1}{x}$ from $a$ to $b$ equals the area from $c$ to $d$? What references (pdfs, links, books) are there on this specific subject? Are there any English translations of the relevant work of Saint Vincent?


1 Answer 1


Dhombres's Is One Proof Enough? gives English translation of the Latin text with original figures, and explains why St. Vincent gave two proofs:"the Jesuit mathematician prefers to describe precisely the road, the method of discovery, rather than to show the shortest way to the summit". There is a nice parallel to Archimedes, who gave a heuristic mechanical argument for the quadrature of the parabola in the Method, and then reproved it by the method of exhaustion. Of course St. Vincent could not possibly know of the Method, but his motivation echo's Archimedes's "certain things first became clear to me by a mechanical method, although they had to be demonstrated by geometry afterwards".

Moreover, St Vincent's first proof is also heuristic, and the other parallels Archimedean method of exhaustion quadrature! Burn in Gregory of St Vincent and the Rectangular Hyperbola gives a modernization of it, replacing Apollonian lemmas with coordinate geometry:"Gregory provided two proofs, A and B; proof A used inscribed parallelograms of equal area and of increasing multiplicity, by repeated insertion of geometric means. For the rectangular hyperbola these become inscribed rectangles. Proof A (in its structure) is a recognisable precursor of the later theory of integration, though the existence of the limit is not quite proved. Proof B, which is the basis of the proof described below, bears a remarkable similarity to Archimedes' quadrature of the parabola and is as complete a proof as that of Archimedes, but to my knowledge it has not been reproduced in any English book on the history of mathematics. Gregory's work was based on expert knowledge of the conic sections, and did not use coordinate geometry".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.