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# HISTORY OF MATHEMATICS MA 330

UK

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This 72 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 330 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 42 views. For similar materials see /class/228149/ma-330-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15

Refutati 0f Eermgfs Cogj ectugg 7 011 i Eul r39s m 39U O CJ LJC39UOH of Fer39ma 1 a mmwmar 8 mg nlp39eanal Mg bmgWM 6 ambial Wm Wmmmm m m indmm pnm 1 39 amm mm gammy mamame WE mm W Biography Wm mmama m Mm mg m has mmi m WWW TI 3 MW Blogmphy mm ha mle ii Whig mammng 5 NEW 4 A m n 39 M me F Qm fi mm to f 39 e mg Tr m j mamml WE o g m ph y 3 a 9 mail JE 393 i l m i i iminl 511 5 Emmy 11m 9 o ue gnawing mums may mm one Lama mew my ft B though WW W 658 rmnmwag Blogmphy at evall an h r I i fI I Ig 1 Q Mi magi WIme m m ie ESQ 39 mm 6 en et en 9 hes m s fg mh e he r o g m oh y I F quot 49 as H ew mmm 39 a K fm k t Lam m WWW as at ema I s Ma Fm M W m Wamgn Wm am m was r 0 o Notuntil Blaise Pascal wrote him concerning his ideas on probability They had a short correspondence about the theory and that is why they are both credited for it Worrespondences with other a quot mathema i jjammntinuedm i Fermat died domnuary 12 1665 Blogmphy U Ammm PM m mm WW f WWWW man by mm j be mg p g he EL w wa mm m 39 WK rm tth er quot Inmng w t wag m wn w mm m gm Wm mmwm mm mm V quot This work was published in 1679 and became known as Cartiesian geometry Eelfmat39s favorite eld of study was numberWarly Number Theoryw 7 a msmwm mm we Wmmmem mmbmw m lo Farr quot Wklde ugwwfna qw name so mgwm d I I L513 I J IIJ The theorem to jv39r c39r he 39 r I39I f39 g Xi 1 y a h Mg m I W hz m l 1 m I39mrag Fae m m quot ng 39 L W Wth 1 m f H 39 y h F er39rn art Dh wfw j Ultd m hm m t VLT ym 3 V Perm 5 eat my theozem 5 L133 Theorem I Wmcamm Wag anm 191mmerth ammwmo e W PFEGJ mam Wag vmmwg m a i wig mm mm o 39 e o my my i afszk unify L133 Theorem l 5 39 k 3141 N 394 1 39 mm a mmmf g WM 150M m 1 m ang riwe 39 l wr Mm lung W at 4205 quot hagan Perhaps this is what Fermat intended all along 3quot M Born on m 15m Apm 739 Eudra 39 gm i3 LMEWW M m g FE mg g E mp a mm mm dit m m m h mi 2 iiwm B 313 valtzerlewd warms z tgnfmg m Um wm iy E3U at lFAalwayLe had a deep interest in mathematics reading books on the subject from an early age Whilst growing up he had come into contact a V iw vitl1v Johann Bernoulli a faggin friendand Life of ELJJer D Enitaraql mg W27 gram prim 7 am TD Immime Ema M uwwwwu awnrquot w NH mx gnawng n He immuma e aim We of ELJJer i ELM m j 131 w i nzra a avIy a s afievns am m 4 mgismy mm wary p J A mm vie an Ille was su ffering health issues nearly died of fever in 1735 and lost the sight in one eye In 1766 Euler left Berlin to return to St tersburg Lshortly after his return to Russiah lg totally Elma Life of ELJJer a He mm 4amp6 WE mm Wk WWW mm I mama We arr l39gd L y 9 21 Wu amg h ga Q Farmer to ELJJer39 aw Goldbach i S x x t q 1mm 5ch mm mm 1m mamwwmmw o 9W mmmmmrmmm me a mmmmm W m 1mff M E WE do We E m Farmer to ELJJer39 aw Goldbach n mm guitar m mwux am Nmmmmm Wmcan W39 W B mi 551 Elma vaca Maggi 9 W8 Ymmk s WWW WM To Fer39mat to ELJJer aw Goldbach Di m mm mm m 7 mm mm Jagrg Emmy rm lm Wm 22 er e39rm r W if nvesjgated primes that can be vWitte as the sum of two perfect squares Besides 2 all primes are odd numbers When an ggg dumber is gmg d by 4 one eWi r s marhder of either 1 or 3 7 if pgt2 and p is prime p 4k 1 or r p 317 3r Mme w dlle umber k Example of ELJJer s Work on Fermat a r 19 a 1 m i Wags 7f S kw mmw m mmwm w p 391 m EBi I gum to 1m 193 4x48 1 This can be written as I JEEB W ZM39HE 4597 72 2quotquot which is unique 199 4X49 1 not be written as thgsum of Mg l LH EJe More of Euler 21nd Fermal J War f F m W m Whirth 3p vie o mm EMIQ il n 39Fr mmrez in Q nggmy itmmrdlca 1 r mn rmm 9f 1 3 Egg am 41 This theorem as we shall see was proved by Euler in 1736 P1ap 2 WW 2gtlt1 3gtltZgtlt1 Proof of L We Fermat Th earem 1 n Ffj m fy mmma n m 4P 1 a a prmm mg mywm a m 51 a Mit Emmi o a 1P 2 up patp 1 pp 1 apiz 2 gtlt1 pp 1X117 2 am a1 I 3gtltZgtlt1 p 1 Proof of LittJe Format Theorem I minahma m awwmammwsa M Qme mwf s Mwow ma a WWWKMWWW Wu WW 70 ame 1p ap 1ap aa1p ap 1ap a a1F a1 J Proof of LittJe Format Theorem m mm quunrm mm 1mm Tm mm 15 maii rghmib j w is WQFSWMQWQYSWWE imam 1 WM 19 v a a as zep easwe a a 1797 x 4h n r u Abs 3 m a as Ur rm A EMT t Euler used Theorem 2 for a 2 21p 21 3quot 3 Hence The result holds for a 3 Repeating this process Euler found that this elds forath thle num So Ts a 7 0M a 39 Egg p Proof of LittJe rermelt Theorem a wax Wmi mommy 45f 1L3 gonnm Cg MAME mmmmm mm mm 3 626 J ni nem 7 l y 3 m a 6 mmmfmalmm lap aaap1 1quot Sinee p iisra prime implies that p mFst divide evenly into either a or aP391 1 or both But by assumption p does not divide evenly w into 3 Thus p diVideSvui tO aP 391 L 397 J in quotW on of Ferrn39eft E gomjecture 1 quotJ 397 717w rgmm a TIAzafgsmmmam 3amp1 WEME39QdQMEnmmfo myawmmmmmg smM wan m r3 ii wmm 161mi Lay imam 4pmmpfggqp mm paw r Kgiihmpsgkmo and p is a prime that is not a factor of a but such that p does divide evenly into a2 1 Then for some whole number k p 4k 1 tgt l3r oof Shown on 88 4 and p isE prime that is not a factor of a but such the p does divide evenly into a4 1 Then for some whole number k p 8k 1 Proof Shown on 88 232 1 4294967297 641 X 6700417 Fermat was wrong By Andrew Rast Hani Madbak Kam Chan prasentation by Alberto Corso MA 330 History of Mathematics a I n I I n a I n wp M Ancient Greeks were enthralled by the symmetries the visual beauty and the legical structure of geometry 0 Particularly intriguing was the manner in which the simple and elementary could serve as foundation for the cemplex and intricate 0 This enchantment with building the complex from the simple was also evident in the Greeks geometric wnstructions o The rules of the game required that all constructions be done only with compass and unmarked straightedge 0 These two fairly unsaphisticated tools allowing the geometer to produce the most perfect uniform onedimensional gure the straight line and the most perfect uniform twodimensional gure the circle must have appealed to the Greek sensibilities for order simplicity and beauty 0 Moreover these constructions were within reach of the technology of the day a Q i I I 9 a Q i u 0 These seemingly unsophisticated tools can produce a rich set of constructions from the bisection of lines and angles tn the drawing of parallels and perpendiculars to the creatian of regular polygons of great beauty A considerably mare challenging problem in ancient Greece was that of the quadrature of a plane figure The quadrature or squaring of a plane gure is the construction using only compass and straightng a square having area equal to that of the original gure If this is the case the gure is said to be quadrable or a Q i I I 9 a Q i 3 Let ABCD an arbitrary rectangle We must construct with compass and straightedge only a square having area equal to that of ABCD B Extend line AD la the right and use the compass to mark eff segment DE with length equal to that of CD Bisect AE at o and with center 0 and radius A O E describe a semicircle as shown At D construct line DH perpendicular to AE where H is the point of intersec on of the perpendicular and the semicircle Construct the desired squpre 2 DEGH I n a u I 3 an Why does the square DFGH have the same area as the rectangle ABCD B Set a b c to be the lengths of segments OH OD and DH Pythagoras theorem gives us that a2 b2 02 Observethat D E39CD a b and mab It follows that Area rectangle ABCD ADgtltC D alba b rig b2 02 Area square DFGH a u I I I n a u I a M Given a triangle ABC construct a perpendicular from C meeting AB at point H A H B We know that the area of the triangle ABC is m x CH If we bisect CH at M and construct a rectangle ABDE with m W we obtain a rectangle with the same area as the triangle ABC But we already have seen how to square a rectangle Given a general polygon we can subdivide it into a collection of n A triangles by drawing diagonals eg 723 Now triangles are quadrable We can construct squares with sides 11 a2 a3 and areas A1 A2 A3 Construct a right triangle with legs a1 02 a1 and hypotenuse 11 Next construct a triangle 22 with legs d1 33 and hypotenuse d2 We have d3 d a a a 41 2 A1A2A3 a I Q a l l 3 m 0 Obviously this procedure can be adapted to the situation in which the polygon is divided into any number of triangles By analogous techniques we could likewise square 3 gure whose area is the difference between two squarable39 areas With the previous techniques the Greeks of the 5th century BC could square wildly irregular polygons But this triumph was tempered by the fact that such gures are rectilinear 0 What about the issue of whether gures with curved boundaries curvilinear gures were also quadrable Initially this must have seemed unlikely for there is no obvious means to straighten out curved lines with compass and straightedge The trisection of the angle that is the problem of dividing a given angle into three equal parts 0 The duplication of the cube that is to nd the side of a cube of which the volume is twice that of a given cube the socalled Dalian problem The quadrature of the circle that is ta nd the square of an area equal ta that cf a given circlel 39 41m 0 While a talented geometer Aristotele wrote of him that he seemed in other respects to have been stupid and lacking in sense Legend has it that Hippocrates earned his reputation after being defrauded of his fortune by pirates who apparently took him for an easy mark Needing to make a nancial recovery he traveled to Athens and began teaching He is remembered for Mo important contributions His composition of the rst Elements His quadrature of the lune i 11 1H He is credited Esinee nothing remains today with writing the rst Elements that is the rst expesition developing the theorems of geometry precisely and laglcally from a few given axioms or postulates Whatever merits his book had were to be eclipsed ever a century later by the brilliant Elements of Euclid which essentially rendered Hippocrates writing obselete Still there is reason to believe that Euclid burrowed from his predecessor and thus me much to Hippocrates for his great if lost treatise i 4 1M 0 It must have been quite unexpected when Hippacrates of Chins succeeded ca 440 BC in squaring a curvilinear gure known as a lune 0 We do not have Hippacrates own work but Eudamus account of it fmm arcun d 335 BC Even hare the situation is murky because we do not really have Eudemus account either Rather we have a summary by Simplicius from 530 AD that discussed the writings of Eudemus A lune is a plane figure bounded by two circular arcs For instance Hippocrates did not square all such figures but a particular lune he had carefully constructed the one in the previous slidel His argument rested upon three pneliminary results The Pythagoman Theorem An angle inscribed in a semicircle is right The areas of 2 circles or semicircles are to each other as the squares of their diametem o The rst two of these results were well known long before Hippocrates The last proposition on the other hand is considerably more sophisticated There is widespread doubt that Hippocrates actually had a valid proof He may well have thought he could prave it but madam scholars generally feel that this theorem presented logical dif culties far beyond what Hippocrates would have been to handle 0 This latter result appeared as Pmposition 2 in Back XI I of Euclid39s Elements 0 Q l I O a Q i A 175 Begin with a semicircle having center 0 and radius AT 2 W A 0 Construct OC perpendicular to AB with point C on the semicircle and draw lines AC and BC Bisect A0 at D and using m as a radius and D as center draw semicircle AEC thus creating lune AECF which is shaded in the diagram Theorem The lune AECF is quadrable 39 1 Hippocrates plan of attack was simple yet brilliant A 0 He rst had to establish that the lune in questicm had precisely the same area as the shaded triangle A00 Mth this behind him he could then apply the known fact that triangles can be squared to conclude that the lune can be squared as well I 1 Note that the angle ACE is right since it is inscribed in a semicircle A O Triangles A00 and 500 are congruent and so AT K We thus apply the Pythagorean theorem to get m2 E2E2 2amp2 Then the semicircle AEC has area of the semicircle ACB Areasemicircle AEC E2 E2 1 Area semicircle ACB mg 2A7 239 I C i I I D I C 39 Thus semicircle AEC has A half area of semicircle AGE 0 New quadrant AFCO has the area of semicirele ACB So Area semicircle AEC Area quadrant AFCO Subtractring the area of their shared region AFCD leaves Area semicircle AEC Area region AFCD Area quadrant AFCO Area region AFCD or lune AECF Area triangle ACO I 0 213 Vl th Hippocrates39 success at squaring the lune Greek mathematicians must have been optimistic about squaring the most perfect curvilinear gure the circle The ancients devoted much time to this problem and some writers attributed an attempt to Hippocrates himself Piecing together the evidence we gather that what follows is the sort of argument some ancient writers had in mind Imagine an arbitrary circle with diameter AB Construct a large circle with center 0 and C diameter CD that is twice AB Within the larger circle inscribe a regular hexagon Using the six segments CEEFFDDGGHCH construct the six semicircles shown in the figure This generates the shaded region composed of the six Iunes I U i I I ll I U 4m i NOW observe that Area hexagon 3 Area circle an AB Area large circle Area 6 lunes The large circle having twice the diameter of AB must have 4 times the area of the smaller circlel Hence Area hexagon 3 Area circle on AB 4 Area circle on AB Area 6 lunes and subtracting 3 Area circle on A3 from both sides of the above equation we get Area circle on AB Area hexagon Area 6 tunes 0 I i l I a a I i 42m Since both the hexagon being a polygon and the Iunes can be squared thus the circle on AB can be squared by the simple process of subtracting areas Unfortunately there is a glaring argument the lunes that Hippocrates squared were not constructed along the side of an hexagon but rather along the side of a square The problem of squaring the circle remained unresolved for about 2000 years At last in 1882 the German mathematician Ferdinand Lindemann proved that the quadrature of the circle is impossible It suf oed to prove that 7r is a transcendental number that is 7139 is not a solution of any polynomial equation with integer ooe cierntsil A very hard proof Overall there are 5 type of lunes that are quadrable 3 types were found by Hippocrates and 2 more kinds were found by Leonhard Euler in 1771 In the 20th century Tschebatorew and Dorodnow proved Wq iad a39b39ew 39 39 o The 5 Iunes of Hippocrates and Euler 0 The three famous problems of antiquity and their solution Constructible numbers Lindernann s proof of the transcendence of 71 I 391 3M

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