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Eudoxes... not only based the method [of exhaustion] on rigorous demonstration... but he actually applied the method to find the volumes (1) of any p… - Thomas Little Heath
" "Eudoxes... not only based the method [of exhaustion] on rigorous demonstration... but he actually applied the method to find the volumes (1) of any pyramid, (2) of the cone, proving (1) that any pyramid is one third part of the prism which has the same base and equal height, and (2) that any cone is one third part of the cylinder which has the same base and equal height. Archimedes, however, tells us the remarkable fact that these two theorems were first discovered by Democritus, though he was not able to prove them (which no doubt means, not that he gave no sort of proof, but that he was not able to establish the propositions by the rigorous methods of Eudoxes. Archimedes adds that we must give no small share of the credit for these theorems to Democritus... another testimony to the marvellous powers, in mathematics as well as in other subjects, of the great man who, in the words of Aristotle, "seems to have thought of everything". ...Democritus wrote on irrationals; he is also said to have discussed the question of two parallel sections of a cone (which were evidently supposed to be indefinitely close together), asking whether we are to regard them as equal or unequal... Democritus was already close on the track of infinitesimals.
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About Thomas Little Heath
Sir Thomas Little Heath (5 October 1861 – 16 March 1940) was a British civil servant, mathematician, classical scholar, historian of ancient Greek mathematics, translator, and mountaineer. Heath translated works of Euclid of Alexandria, Apollonius of Perga, Aristarchus of Samos, and Archimedes of Syracuse into English.
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Thomas Heath (classicist)
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Thomas L. Heath
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Sir Thomas Little Heath
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Additional quotes by Thomas Little Heath
Dr. James Gow did a great service by the publication in 1884 of his Short History of Greek Mathematics, a scholarly and useful work which has held its own and has been quoted with respect and appreciation by authorities on the history of mathematics in all parts of the world. At the date when he wrote, however, Dr. Gow had necessarily to rely upon the works of the pioneers Bretschneider, Hankel, Allman, and (first edition). Since then the subject has been very greatly advanced... scholars and mathematicians... have thrown light on many obscure points. It is therefore high time for the complete story to be rewritten.
Hippocrates also attacked the problem of doubling the cube. ...Hippocrates did not, indeed, solve the problem, but he succeeded in reducing it to another, namely, the problem of finding two mean proportionals in continued proportion between two given straight lines, i.e. finding x, y such that a:x=x:y=y:b, where a, b are the two given straight lines. It is easy to see that, if a:x=x:y=y:b, then b/a = (x/a)<sup>3</sup>, and, as a particular case, if b=2a, x<sup>3</sup>=2a<sup>3</sup>, so that the side of the cube which is double of the cube of side a is found.
The discovery of Hippocrates amounted to the discovery of the fact that from the relation
(1)<math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}</math>it follows that<math>(\frac{a}{x})^3 = [\frac{a}{x} \cdot \frac{x}{y} \cdot \frac{y}{b} =] \frac{a}{b}</math>and if <math>a = 2b</math>, [then <math>(\frac{a}{x})^3 = 2</math>, and]<math>a^3 = 2x^3</math>.The equations (1) are equivalent [by reducing to common denominators or cross multiplication] to the three equations
(2)<math>x^2 = ay, y^2 = bx, xy = ab</math>[or equivalently...<math>y = \frac{x^2}{a}, x = \frac{y^2}{b}, y = \frac{ab}{x}</math> ]and the solutions of Menaechmus described by Eutocius amount to the determination of a point as the intersection of the curves represented in a rectangular system of Cartesian coordinates by any two of the equations (2).
Let AO, BO be straight lines placed so as to form a right angle at O, and of length a, b respectively. Produce BO to x and AO to y.
The first solution now consists in drawing a parabola, with vertex O and axis Ox, such that its parameter is equal to BO or b, and a hyperbola with Ox, Oy as asymptotes such that the rectangle under the distances of any point on the curve from Ox, Oy respectively is equal to the rectangle under AO, BO i.e. to ab. If P be the point of intersection of the parabola and hyperbola, and PN, PM be drawn perpendicular to Ox, Oy, i.e. if PN, PM be denoted by y, x, the coordinates of the point P, we shall have <math>\begin{cases}y^2 = b.ON = b.PM = bx\\ and\\ xy = PM.PN = ab\end{cases}</math>whence<math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math>
In the second solution of Menaechmus we are to draw the parabola described in the first solution and also the parabola whose vertex is O, axis Oy and parameter equal to a. The point P where the two parabolas intersect is given by<math>\begin{cases}y^2 = bx\\x^2 = ay\end{cases}</math>whence, as before,<math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math>
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