“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)3, and, as a particular case, if b=2a, x3=2a3, so that the side of the cube which is double of the cube of side a is found.”

— Thomas Little Heath

Achimedes (1920)

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Thomas Little Heath 46

British civil servant and academic 1861–1940

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“Take the case of a famous problem which plays a great part in the history of Greek geometry, the doubling of the cube, or its equivalent, the finding of two mean proportionals in continued proportion between two given straight lines. …if all the recorded solutions are collected together, it is much easier to see the relations, amounting in some cases to substantial identity, between them, and to get a comprehensive view of the history of the problem. I have therefore dealt with this problem in a separate section of the chapter devoted to 'Special Problems,' and I have followed the same course with the other famous problems of squaring the circle and trisecting any angle.”

Thomas Little Heath (1861–1940) British civil servant and academic

A History of Greek Mathematics (1921) Vol. 1. From Thales to Euclid

“While then for a long time everyone was at a loss, Hippocrates of Chios was the first to observe that, if between two straight lines of which the greater is double of the less it were discovered how to find two mean proportionals in continued proportion, the cube would be doubled; and thus he turned the difficulty in the original problem into another difficulty no less than the former. Afterwards, they say, some Delians attempting, in accordance with an oracle, to double one of the altars fell into the same difficulty. And they sent and begged the geometers who were with Plato in the Academy to find for them the required solution. And while they set themselves energetically to work and sought to find two means between two given straight lines, Archytas of Tarentum is said to have discovered them by means of half-cylinders, and Eudoxus by means of the so-called curved lines. It is, however, characteristic of them all that they indeed gave demonstrations, but were unable to make the actual construction or to reach the point of practical application, except to a small extent Menaechmus and that with difficulty.”

Thomas Little Heath (1861–1940) British civil servant and academic

Apollonius of Perga (1896)

“The problem of doubling the cube was henceforth tried exclusively in the form of the problem of the two mean proportionals.”

Thomas Little Heath (1861–1940) British civil servant and academic

p, 125
Achimedes (1920)

“The discovery of Hippocrates amounted to the discovery of the fact that from the relation
(1)\frac{a}{x} = \frac{x}{y} = \frac{y}{b}it follows that(\frac{a}{x})^3 = [\frac{a}{x} \cdot \frac{x}{y} \cdot \frac{y}{b} =] \frac{a}{b}and if a = 2b, [then (\frac{a}{x})^3 = 2, and]a^3 = 2x^3.The equations (1) are equivalent [by reducing to common denominators or cross multiplication] to the three equations
(2)x^2 = ay, y^2 = bx, xy = ab[or equivalently…y = \frac{x^2}{a}, x = \frac{y^2}{b}, y = \frac{ab}{x} ]Doubling the Cube
the 2 solutions of Menaechmusand 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

\begin{cases}y^2 = b. ON = b. PM = bx\\ and\\ xy = PM. PN = ab\end{cases}whence\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.
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.”

Thomas Little Heath (1861–1940) British civil servant and academic

The point P where the two parabolas intersect is given by<center><math>\begin{cases}y^2 = bx\\x^2 = ay\end{cases}</math></center>whence, as before,<center><math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math></center>
Apollonius of Perga (1896)

“By the time of Hippocrates of Chios the scope of Greek geometry was no longer even limited to the Elements; certain special problems were also attacked which were beyond the power of the geometry of the straight line and circle, and which were destined to play a great part in determining the direction taken by Greek geometry in its highest flights. The main problems in question were three: (1) the doubling of the cube, (2) the trisection of any angle, (3) the squaring of the circle; and from the time of Hippocrates onwards the investigation of these problems proceeded pari passu with the completion of the body of the Elements.”

Thomas Little Heath (1861–1940) British civil servant and academic

p, 125
Achimedes (1920)

“Although Cardan reduced his particular equations to forms lacking a term in x^2, it was Vieta who began with the general formx^3 + px^2 + qx + r = 0and made the substitution x = y -\frac{1}{3}p, thus reducing the equation to the formy^3 + 3by = 2c.He then made the substitutionz^3 + yz = b, or y = \frac{b - z^2}{z},which led to the formz^6 + 2cz^2 = b^2,a sextic which he solved as a quadratic.”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, p.465

“The differential equation of the first order
\frac {dy}{dx} = f(x, y)
… prescribes the slope \frac {dy}{dx} at each point of the plane (or at each point of a certain region of the plane we call the field")…. a differential equation of the first order… can be conceived intuitively as a problem about the steady flow of a river: Being given the direction of the flow at each point, find the streamlines…. It leaves open the choice between the two possible directions in the line of a given slope. Thus… we should say specifically "direction of an unoriented straight line" and not merely "direction."”

George Pólya (1887–1985) Hungarian mathematician

Mathematical Methods in Science (1977)

“Suppose we a certain Number of things exposed, different each from other, as a, b, c, d, e, &c.; The question is, how many ways the order of these may be varied? as, for instance, how many changes may be Rung upon a certain Number of Bells; or, how many ways (by way of Anagram) a certain Number of (different) Letters may be differently ordered?
Alt.1,21) If the thing exposed be but One, as a, it is certain, that the order can be but one. That is 1.
2) If Two be exposed, as a, b, it is also manifest, that they may be taken in a double order, as ab, ba, and no more. That is 1 x 2 = 2. Alt.3
3) If Three be exposed; as a, b, c: Then, beginning with a, the other two b, c, may (by art. 2,) be disposed according to Two different orders, as bc, cb; whence arise Two Changes (or varieties of order) beginning with a as abc, acb: And, in like manner it may be shewed, that there be as many beginning with b; because the other two, a, c, may be so varied, as bac, bca. And again as many beginning with c as cab, cba. And therefore, in all, Three times Two. That is 1 x 2, x 3 = 6.
Alt.34) If Four be exposed as a, b, c, d; Then, beginning with a, the other Three may (by art. preceeding) be disposed six several ways. And (by the same reason) as many beginning with b, and as many beginning with c, and as many beginning with d. And therefore, in all, Four times six, or 24. That is, the Number answering to the case next foregoing, so many times taken as is the Number of things here exposed. That is 1 x 2 x 3, x 4 = 6 x 4 = 24.
5) And in like manner it may be shewed, that this Number 24 Multiplied by 5, that is 120 = 24 x 5 = 1 x 2 x 3 x 4 x 5, is the number of alternations (or changes of order) of Five things exposed. (Or, the Number of Changes on Five Bells.) For each of these five being put in the first place, the other four will (by art. preceeding) admit of 24 varieties, that is, in all, five times 24. And in like manner, this Number 120 Multiplied by 6, shews the Number of Alternations of 6 things exposed; and so onward, by continual Multiplication by the conse quent Numbers 7, 8, 9, &c.;
6) That is, how many so ever of Numbers, in their natural Consecution, beginning from 1, being continually Multiplied, give us the Number of Alternations (or Change of order) of which so many things are capable as is the last of the Numbers so Multiplied. As for instance, the Number of Changes in Ringing Five Bells, is 1 x 2 x 3 x 4 x 5 = 120. In Six Bells, 1 x 2 x 3 x 4 x 5 x 6 = 120 x 6 = 720. In Seven Bells, 720 x 7 = 5040. In Eight Bells, 5040 x 8 = 40320, And so onward, as far as we please.”

John Wallis (1616–1703) English mathematician

Source: A Discourse of Combinations, Alterations, and Aliquot Parts (1685), Ch.II Of Alternations, or the different Change of Order, in any Number of Things proposed.

“I was looking for a name like the Crickets that meant two things, and from crickets I got to beetles. And I changed [to] B E A because … B E E T L E S didn't mean two things, so I changed … the E to an A. And it meant two things then. … When you said it, people thought of crawly things; and when you read it, it was beat music.”

John Lennon (1940–1980) English singer and songwriter

Pop Chronicles: Show 27 - The British Are Coming! The British Are Coming!: The U.S.A. is invaded by a wave of long-haired English rockers. Part 1 https://digital.library.unt.edu/ark:/67531/metadc19782/m1/#track/4, 24 August 1964 http://www.jpgr.co.uk/pro202.html.

“Let the letters a b c denote the three angular points of a rectilineal triangle. If the point did move continuously over the lines ab, bc, ca, that is, over the perimeter of the figure, it would be necessary for it to move at the point b in the direction ab, and also at the same point b in the direction bc. These motions being diverse, they cannot be simultaneous. There-fore, the moment of presence of the movable point at vertex b, considered as moving in the direction ab, is different from the moment of presence of the movable point at the same vertex b, considered as moving in the same direction bc. But between two moments there is time; therefore, the movable point is present at point b for some time, that is, it rests. Therefore it does not move continuously, which is contrary to the assumption. The same demonstration is valid for motion over any right lines including an assignable angle. Hence a body does not change its direction in continuous motion except by following a line no part of which is straight, that is, a curve, as Leibnitz maintained.”

Immanuel Kant (1724–1804) German philosopher

Kant's Inaugural Dissertation (1770), Section III On The Principles Of The Form Of The Sensible World

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