MST Topics in History of Mathematics Euclid’s Elements and the Works of Archimedes Paul Yiu Department of Mathematics Florida Atlantic University Summer 2014 June 25 Chapter 0 Greek geometry before Euclid Chronology -586 Babylonian captivity -585 Thales of Miletus; deductive geometry -580 Birth of Pythagoras -540 Pythagorean arithmetic and geometry -430 Elements of Hippocrates of Chios -427 Birth of Plato -420 Incommensurables -399 Death of Socrates -360 Eudoxus on proportion and exhaustion -347 Death of Plato -335 Eudemus: History of Geometry -332 Alexandria founded -323 Death of Alexander -322 Death of Aristotle -300 Euclid’s Elements -225 Apollonius: Conics -212 Death of Archimedes +75 Works of Heron of Alexandria +250 Diophantus: Arithmetica +320 Pappus: Mathematical Collections +485 Death of Proclus 2 Greek geometry before Euclid Thales of Miletus Theorem (Thales). The angle inscribed in semicircle is a right angle. 3 Other theorems attributed to Thales Eudemus’ History (ca -320): 1. A circle is bisected by a diameter. 2. The base angle of an isosceles triangle are equal. (Euclid I.5) 3. The pairs of vertical angles formed by two intersecting lines are equal. (Eu- clid I.15) 4. If two triangles are such that two angles and a side of one are equal respec- tively to two angles and a side of the other, then the triangles are congruent. (Euclid I.26) 4 Greek geometry before Euclid Pythaogas of Samos Proclus, on Euclid I: (p.298) Eudemus the Peripatetic ascribes to the Pythagoreans the discovery of this theorem, that any triangle has internal angles equal to right angles. He says they proved the theorem in question after this fashion. A D E B C Let ABC be a triangle, and through A let DE be drawn parallel to BC. Now since BC, DE are parallel, and the alternative angles are equal, the angle DAB is equal to the angle ABC, and EAC is equal to ACB. Let BAC be added to both. Then the angles DAB, BAC, CAE, that is, the angles DAB, BAE, that is, two right angles, are equal to the three angles of the triangle. Therefore the three angles of the triangle are equal to two right angles. Such is the proof of the Pythagoreans. 5 Iamblichus: On the Pythagorean Life (Thomas, 222–225): It is related of Hippasus that he was a Pythagoreans, and that, owing to his being the first to publish and describe the sphere from the twelve pentagons, he perished at sea for his impiety, but he received credit for the discovery, though really it all belonged to HIM (for in this way they refer to Pythagoras, and they do not call him by his name). 6 Greek geometry before Euclid Euclid, Elements X, Scholium i The Pythagoreans were the first to make inquiry into commensurability, having first discovered it as a result of their observation of numbers; for though the unit is a common measure of all numbers they could not find a common measure of all magnitudes. The reason is that all numbers, of whatsoever kind, however they be divided leave some least part which will not suffer further division; but all magnitudes are divisible ad infinitum and do not leave some part which, being the least possible, will not admit of further division, but that remainder can be divided ad infinitum so as to give an infinite number of parts, of which each can be divided ad infinitum; and in sum, magnitude partakes in division of the principle of the infinite, but in its entirety of the principle of the finite, while number in division partakes of the finite, but in its entirety of the infinite. ... There is a legend that the first of the Pythagoreans who made public the investigation of these matters perished in a shipwreck. 7 Democritus Archimedes, Method: ...incase of those theorems concerning the cone and pyramid of which Eudoxus first discovered the proof, the theorem that the cone is the third part of the cylinder, and the pyramind of the prism, having the same base and equal height, no small share of the credit should be given to Democritus, who was the first to make the assertion with regard to the said figure, though without proof. 8 Greek geometry before Euclid Duplication of the cube Theon of Smyrna (Thomas 257): In his work entitled Platonicus Eratosthenes says that, when the god announced to the Delians by oracle that to get rid of a plague they must construct an altar double of the existing one, their craftsmen fell into great perplexity in trying to find how a solid could be made double of another solid, and they went to ask Plato about it. He told them that the god had given this oracle, not because he wanted an altar of double the size, but he wished, in setting this task before them, to reproach the Greeks for their neglect of mathematics and their contempt for geometry. 9 Eutocius: Commentary on Archimedes’ Sphere and Cylinder, II (Thomas 257–259): It became a subject of inquiry among geometers in what manner one might double the given solid while it remained the same shape, and this problem was called the duplication of the cube; for given a cube, they sought to double it. When all were for a long time at a loss, Hippocrates of Chios first conceived that, if two mean proportionals could be found in continued proportion between two straight lines, of which the greater was double the lesser, the cube would be doubled, so that the puzzle was by him turned into no less a puzzle. After a time, it is related, certain Delians, when attempting to double a certain altar in accordance with an oracle, fell into the same quandary, and sent over to ask the geometers who were with Plato in the Academy to find what they sought. When these men applied themselves diligently and sought to find two mean proportionals between two given straight lines, Archytas of Taras is said to have found them by the half-cylinders, and Eudoxus by the so-called curved lines; but it turned out that all their solutions were theoretical, and they could not give a practical construction and it it to use, except to a certain small extent Menaechmus, and that with difficulty. An easy mechanical solution, was however, found by me, and by means of it I will find, not only two means to the given straight lines, but as many as may be enjoined. 10 Greek geometry before Euclid Nicomedes’ solution Given: Two lines AD and DC at right angles. To construct: Two continued mean proportionals between AD and CD. M A D F B E C K H T Z Construction: Complete the rectangle ADCB and take the midpoints E, F of BC and AB. Join DF to meet CB produced at H. Construct EZ perpendicular to BC so that CZ = AF . Join HZ and construct the parallel through C. On this parallel construct a point T so that ZT meet BC produced at K with TK = CZ. Then AD : AM = AM : CK = CK : CD, and AM, CK are two continued mean proportionals between AD and CD. 11 MAA Focus June/July 2014, 27. x and y are two continuous mean proportionals between 1 and 2: A x 1 1 Y 1 B 1 C y X 1 Z 12 Greek geometry before Euclid Cubic root of 2 by paper folding B. Casselman, If Euclid had been Japanese, Notices of AMS, 54 (2007) 626–628. A paper square ABCD is divided into three strips of equal area by the parallel lines PQand RS. The square is then folded so that C falls on AD and S falls on PQ (as C in the second√ diagram). AC = 3 2 Then CD . A B A B P Q P S C R S R D C D C 13 Trisection of an angle Archimedes [Book of Lemmas, Proposition 8] Given angle AOB with OA = OB (contained in a circle, center O), construct a line through A such that the intercept between the circle and the line BO has the same length as the radius of the circle. ∠AOC = 1 ∠AOB Then 3 . A A C O B 14 Greek geometry before Euclid Trisection of an angle To triangle an angle AOB, pass a line through O such that the intercept between the parallel and the perpendicular at A to the line OB is 2 · OA. Then this line is a trisector of angle AOB. A E M D O B 15 Angle trisection with the use of conics To trisect an angle AOB, construct the hyperbola with focus A, directrix OM, and eccentricity 2, to intersect the arc AB at C. Then ∠AOB =3∠AOC. C K B A M P O 16 Greek geometry before Euclid The quadratrix of Hippias A horizontal line HK (with initial position AB) falls vertically, and a radius OP (with initial position OA) rotates about O, both uniformly and arrive at OC at the same time. The locus of the intersection Q = HK ∩ OP is the quadratrix. A B P Q H K O C 17 Trisection of an angle by the quadratrix B P K Q A O To trisect angle AOB, let OB intersect the quadratrix at P . Trisect the segment OP at K. Construct the parallel through K to OA to intersect the quadratrix at Q. Then OQ is a trisector of angle AOB. 18 Greek geometry before Euclid Angle trisection by paper folding B. Casselman, If Euclid had been Japanese, Notices of AMS, 54 (2007) 626–628.