Solution Manual for A History of Mathematics, 3rd Edition

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CHAPTER 11. The answers are given in the answer section of the text. For the Egyptian hieroglyphics,375 is three hundreds, seven tens and five ones, while 4856 is four thousands, eighthundreds, five tens, and six ones. For Babylonian cuneiform, note that 375 = 6×60 + 15while 4856 = 1×3600 + 20×60 + 56.2.134′26841368272′1654418612151050′(multiply by 10)210(double first line)420(double third line)840′(double fourth line)22 2′(halve first line)102′(invert third line)18 2 10933.12142428485614.128256411216

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2Chapter 15. We multiply 10 by 3 30:1330′213154231030′85210The total of the two marked lines is then 7, as desired.6.17248215244312863354Note that the sum of the three last terms in the second column is 99 2 4. We thereforeneed to figure out by what to multiply 7 2 4 8 to give 4 so that we get a total of 100.But since we know from the fourth line that multiplying that value by 8 gives 63, we alsoknow that multiplying it by 63 gives 8. Thus the required number is double 63, which is42 126. Thus the final result of our division is 12 3 42 126.7.17 2 4 8215 2 4′431 2′863′34 3 3 6 1212 398 2 3 3 6 1299 2 4

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Egypt and Mesopotamia38.2÷111112÷2312337 3315 333 337 361 3 6′63 2 3666′121 2 4 6′27612′6 66212 27629.x+17x= 19. Choosex= 7; then 7 +17·7 = 8. Since 19÷8 = 238, the correct answer is238×7 = 1658.10. (x+23x)−13(x+23x) = 10. In this case, the “obvious” choice forxisx= 9. Then 9added to 2/3 of itself is 15, while 1/3 of 15 is 5. When you subtract 5 from 15, you get10. So in this case our “guess” is correct.11. The equation here is (1 +13+14)x= 2. Therefore. we can find the solution by dividing2 by 1 +13+14. We set up that problem:11 2 431 1832 3664 72128 144The sum of the numbers in the right hand column beneath the initial line is 1141144. So weneed to find multipliers giving us3144= 144 72. But 1 3 4 times 144 is 228. It followsthat multiplying 1 3 4 by 228 gives 144 and multiplying by 114 gives 72.Thus, theanswer is 1 6 12 114 228.12. The equation isx+ 2x= 9. This reduces to 3x= 9, so the answer is 3.13. Sincexmust satisfy 100 : 10 =x: 45, we would get thatx=45×10010; the scribe breaksthis up into a sum of two parts,35×10010and10×10010.14. The ratio of the cross section area of a log of 5 handbreadths in diameter to one of 4handbreadths diameter is 52: 42= 25 : 16 = 1916.Thus, 100 logs of 5 handbreadthsdiameter are equivalent to 1916×100 = 15614logs of 4 handbreadths diameter.16. The modern formula for the surface area of a half-cylinder of diameterdand heighthisA=12πdh.Similarly, the modern formula for the surface area of a hemisphere ofdiameterdisA=12πd2. These formulas are identical ifh=d.17. 7/5 = 1; 2413/15 = 0; 5211/24 = 0; 27,3033/50 = 0; 39,36

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4Chapter 118. 0; 22,30 = 3/80; 08,06 = 27/2000; 04,10 = 5/720; 05,33,20 = 5/5419. Since 3×18 = 54, which is 6 less than 60, it follows that the reciprocal of 18 is 313, or,putting this in sexagesimal notation, 3,20. Since 60 is (178)×32, and78can be expressedas 52,30, the reciprocal of 32 is 1,52,30. Since 60 = 119×54, and19can be expressed as110+190=660+403600= 0; 06,40, the reciprocal of 54 is 1,06,40. Also, because 60 =1516×64,the reciprocal of 64 is1516. Since116= 3,45, we get that1516= 56,15.If the only primedivisors ofnare 2, 3, 5, thennis a regular sexagesimal.20. 25×1,04 = 1,40 + 25,00 = 26,40.18×1,21 = 6,18 + 18,00 = 24,18.50÷18 =50×0; 3,20 = 2; 30 + 0; 16,40 = 2; 46,40.1,21÷32 = 1,21×0; 01,52,30 = 1; 21 +1; 10,12 + 0; 00,40,30 = 2; 31,52,30.21. Since the length of the circumferenceCis given byC= 4a, and becauseC= 6r, it followsthatr=23a. The lengthTof the long transversal is thenT=r√2 = (23a)(1712) =1718a.The lengthtof the short transversal ist= 2(r−t2) = 2a(23−1736) =718a. The areaAofthe barge is twice the difference between the area of a quarter circle and the area of theright triangle formed by the long transversal and two perpendicular radii drawn fromthe two ends of that line. ThusA= 2(C248−r22)= 2(a23−2a29)= 29a2.22. Since the length of the circumferenceCis given byC= 3a, and becauseC= 6r, it followsthatr=a2. The lengthTof the long transversal is thenT=r√3 = (a2)(74) =78a. Thelengthtof the short transversal is twice the distance from the midpoint of the arc tothe center of the long transversal. If we set up our circle so that it is centered on theorigin, the midpoint of the arc has coordinates (r2,√3r2) while the midpoint of the longtransversal has coordinates (r4,√3r4). Thus the length of half of the short transversal isr2and thent=r=a2. The areaAof the bull’s eye is twice the difference between thearea of a third of a circle and the area of the triangle formed by the long transversal andradii drawn from the two ends of that line. ThusA= 2(C236−12r2T)= 2(9a236−12a47a8)= 2a2(14−764)=932a2.23. Ifais the length of one of the quarter-circle arcs defining the concave square, then thediagonal is equal to the diameter of that circle. Since the circumference is equal to 4a,the diameter is one-third of that circumference, or 113a. The transversal is equal to thediagonal of the circumscribing square less the diameter of the circle (which is equal tothe side of the square). Since the diagonal of a square is approximated by 17/12 of theside, the transversal is therefore equal to 5/12 of the diameter, or512 43a=59a.24. 1; 24,51,10 = 1+2460+513600+10216000= 1+0.4+0.0141666666+0.0000462962 = 1.414212963.On the other hand,√2 = 1.414213562. Thus the Babylonian value differs from the truevalue by approximately 0.00004%.25.√3 =√22−1≈2−12·1·12= 2−0; 15 = 1; 45.Since an approximate recipro-cal of 1;45 is 0;34,17.09, we get further that√3 =√(1; 45)2−0; 03,45=1; 45−

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Egypt and Mesopotamia5(0; 30)(0; 03,45)(0; 34,17.09) = 1; 45−0; 01,04,17,09 = 1; 43,55,42,51, which we trun-cate to 1;43,55,42 because we know this value is a slight over-approximation.26.v+u= 1; 48 = 145andv−u= 0; 33,20 =59. So 2v= 2; 21,20 andv= 1; 10,40 =10690.Similarly, 2u= 1; 14,40 andu= 0; 37,20 =5690. Multiplying by 90 givesx= 56,d= 106.In the second part,v+u= 2; 05 = 2112andv−u= 0; 28,48 =1225. So 2v= 2; 33,48andv= 1; 16,54 =769600. Similarly, 2u= 1; 36,12 andu= 0; 48,06 =481600. Multiplying by600 givesx= 481,d= 769. Next, ifv=481360andu=319360, thenv+u= 229= 2; 13,20.Finally, ifv=289240andu=161240, thenv+u= 178= 1; 52,30.27. The equations foruandvcan be solved to givev= 1; 22,08,27 =295707216000=9856972000andu= 0; 56,05,57 =201957216000=6731972000.Thus the associated Pythagorean triple is 67319,72000, 98569.28. The two equations arex2+y2= 1525;y=23x+ 5. If we substitute the second equationinto the first and simplify, we get 13x2+ 60x= 13500.The solution is thenx= 30,y= 25.29. If we guess that the length of the rectangle is 60, then the width is 45 and the diagonalis√602+ 452= 75. Since this value is 178times the given value of 40, the correct lengthof the rectangle should be 60÷178= 32. Then the width is 24.30. One way to solve this is to letxandx−600 be the areas of the two fields. Then theequation is23x+12(x−600) = 1100. This reduces to76x= 1400, sox= 1200. The secondfield then has area 600.31. Letxbe the weight of the stone. The equation to solve is thenx−17x−113(x−17x) = 60.We do this using false position twice. First, sety=x−17x. The equation inyis theny−113y= 60.We guessy= 13. Since 13−11313 = 12, instead of 60, we multiply ourguess by 5 to gety= 65. We then solvex−17x= 65. Here we guessx= 7 and calculatethe value of the left side as 6. To get 65, we need to multiply our guess by656= 1016. Soour answer isx= 7×656= 7556gin, or 1mina1556gin.32. We do this in three steps, each using false position. First, setz=x−17x+111(x−17x).The equation forzis thenz−113z= 60. We guess 13 forzand calculate the value ofthe left side to be 12, instead of 60. Thus we must multiply our original guess by 5 andputz= 65. Then sety=x−17x. The equation foryisy+111y= 65. If we now guessy= 11, the result on the left side is 12, instead of 65. So we must multiply our guessby6512to gety=71512= 59712.We now solvex−17x= 59712.If we guessx= 7, theleft side becomes 6 instead of 59712. So to get the correct value, we must multiply 7 by71512/6 =71572. Therefore,x= 7×71572=500572= 693772gin= 1mina93772gin.33. Start with a square of sidexand cut off a strip of widthafrom the right side.Theremaining rectangle then has areax2−ax, orb. This rectangle can then be thought ofas a square of sidex−a/2 that is missing a small square of sidea/2. If one adds backthat small square, then the square of sidex−a/2 has areab+ (a/2)2, so we can findx.34. The equationx−60x= 7 is equivalent tox2−60 = 7xor tox2−7x= 60. The solutionis thenx=√(72)2+ 60 +72=172+72= 12. Thus the two numbers are 12 and 5.35. Given the appropriate coefficients, the equation becomes49a2+a+43a=2318, wherea

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6Chapter 1is the length of the arc. If we scale up by49, we get the equation (49a)2+73(49a) =4681.The algorithm for this type of equation gives49a=√(76)2+4681−76=2518−76=29. Thusa=12.36. The equation is23x2+13x=13. To solve, we scale by23: (23x)2+13(23x) =29. The solutionis23x=√(16)2+29−16=12−16=13. Thusx=12.37. All the triangles in this diagram are similar to one another, and therefore their sidesare all in the ratio 3:4:5.Therefore,AD=45AB= 0; 48×0; 45 = 0; 36 andBD=35AB= 0; 36×0; 45 = 0; 27.Similarly,DE=45AD= 0; 48×0; 36 = 0; 28,48.Also,EF=45DE= 0; 48×0; 28,48 = 0; 23,02,24.ThenDF=35DE= 0; 36×0; 28,48 =0; 17,16,48 andF C=BC−BD−DF= 1; 15−0; 27−0; 17,16,48 = 0; 30,43,12.38. If the circumference is 60, then the radius is 10.Thus, if the distance of the chordfrom the circumference isx, then we have a right triangle of sides 6 and 10−x, withhypotenuse 10. The Pythagorean theorem leads to the equation 62+ (10−x)2= 100,orx2+ 36 = 20x, for which the only valid solution isx= 2.39. The two equations are`+w= 7,`w+12`+13w= 15.To put this into a standardBabylonian form, we can rewrite the second equation in the form (`+13)(w+12) = 1516.We can then rewrite the first equation as (`+13) + (w+12) = 756. Then the Babylonianalgorithm yields`+13=4712+√(4712)2−916=4712+512=133.Therefore,`= 4 and sow= 3.

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CHAPTER 21. 125 =ρκ, 62 =ξβ, 4821 =′δωκα, 23,855 =Mβ′γων2.89=6´γ´ι´η(8/9 = 1/2 + 1/3 + 1/18)3. The answer is in the back of the text (except the last character should be anηinsteadof aβ). The basic idea is that 200/9 = 2229= 22 + 1/6 + 1/18.4. The average ofaandcis 1/4+1/16+1/64. The average ofbanddis 1/2+1/4+1/8+1/16.The product of the two averages is 1/8 + 1/16 + 1/32 + 1/64 + 1/32 + 1/64 + 1/128 +1/256 + 1/128 + 1/256 + 1/512 + 1/1024, or 1/4 + 59/1024. This is slightly less than thegiven answer of 1/4 + 1/16.5. SinceAB=BC; since the two angles atBare equal; and since the angles atAandCareboth right angles, it follows by the angle-side-angle theorem that4EBCis congruentto4SBAand therefore thatSA=EC.6. Because both angles atEare right angles; becauseAEis common to the two triangles;and because the two anglesCAEare equal to one another, it follows by the angle-side-angle theorem that4AETis congruent to4AES. ThereforeSE=ET.7. The distance from the center of the pyramid to the tip of the shadow is 378 + 342 = 720feet. Therefore the height of the pyramid is 6/9 = 2/3 of this value, or 480 feet.8.Tn= 1 + 2 +· · ·+n=n(n+1)2.Therefore the oblong numbern(n+ 1) is double thetriangular numberTn.9.n2=(n−1)n2+n(n+1)2, and the summands are the triangular numbersTn−1andTn.10.8n(n+1)2+ 1 = 4n2+ 4n+ 1 = (2n+ 1)2.11. Supposea2+b2=c2.Supposeais odd. Thena2is odd. Ifbis odd, thenb2is odd andc2is even, socis even. Ifbis even, thenb2is even andc2is odd, socis odd. A similarresult holds ifcis odd.12. Examples using the first formula are (3,4,5), (5,12,13), (7,24,25), (9,40,41), (11,60,61).Examples using the second formula are (8,15,17), (12,35,37), (16,63,65), (20,99,101),(24,143,145).13. Let us assume that the second leg is commensurable to the first and letb,abe numbersrepresenting the two legs (in terms of some unit).We may as well assume thatbandaare relatively prime. Since the hypotenuse is double the first leg, we haveb2+a2=(2a)2= 4a2, orb2= 3a2. Sinceb2is a multiple of 3, it must also be a multiple of 9, sob2= 9c2andb= 3c. Then 9c2= 3a2, ora2= 3c2. This implies thata2is a multiple of9, so thatais a multiple of 3. But then bothaandbare multiples of 3, contradictingthe fact that they are relatively prime.14. Since similar segments are to their corresponding circles in the same ratio, the areas ofsimilar segments are to one another as the squares on the diameters of the circles. Thus,the areas of similar segments are also to one another as the squares on the radii of thecircles. But in similar segments, the triangles formed by the two radii and chords aresimilar triangles. Thus the chord of one segment is to the chord in the similar segmentas the radius of the first circle to the radius of the second. That is, the squares on the7

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8Chapter 2radii are to one another as the squares on the chords.Therefore, the areas of similarsegments are to one another as the squares on their chords.15. By exercise 14, the area of segmentBDis the area of segmentABas the square onBDis the square onAB. But this ratio is equal to 3. Thus, the area of segmentBDis threetimes the area of segmentAB, or is equal to the sum of the areas of segmentsAB,AC,andCD. Therefore, the area of lune is equal to the difference between the area of thelarge segment and the area of segmentBD. But this is equal to the difference betweenthe area of the large segment and the areas of the three small segments, which is in turnequal to the area of the trapezoid. To construct the trapezoid, note that one can certainlyconstruct a line segment equal to√3 times the length of a given line segment. To placethis line segment both parallel to the original one and such that the lines connecting theendpoints of the two segments are each equal to the original line segment, we simplyneed to find the distance between the two segments.And that can be constructed byusing the Pythagorean Theorem applied to the triangle whose hypotenuse is equal to theoriginal segment and one leg of which is equal to half the difference between the new linesegment and the original one. To circumscribe a circle around this trapezoid, note thatone can construct a circle through three points, SayB,A, andC. By the symmetry ofthe trapezoid, this circle will also go through pointD.21. If one equates the times of the two runners, wheredis the distance traveled by Achilles,the equation isd/10 = (d−500)/(1/5). This is equivalent to 49d= 25,000, sod= 510.2yards. Since Achilles is traveling at 10 yards per second, this will take him 51.02 seconds.

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CHAPTER 31. One way to do this is to use I-4. Namely, consider the isosceles triangleABCwith equalsidesABandBCalso as a triangleCBA.Then the trianglesABCandCBAhavetwo sides equal to two sides and the included angles also equal. Thus, by I-4, they arecongruent. Therefore, angleBACis equal to angleBCA, and the theorem is proved.2. Put the point of the compass on the vertexVof the angle and swing equal arcs inter-secting the two legs atAandB. Then place the compass atAandBrespectively andswing equal arcs, intersecting atC. The line segment connectingVtoCthen bisects theangle. To show that this is correct, note that trianglesV ACandV BCare congruentbySSS. Therefore, the two anglesAV CandBV Care equal.3. Let the linesABandCDintersect atE. Then anglesAEBandCEDare both straightangles, angles equal to two right angles. If one subtracts the common angleCEBfromeach of these, the remaining anglesAECandBEDare equal, and these are the verticalangles of the theorem.4. Suppose the three lines have lengtha,b, andc, witha≥b≥c. On the straight lineDHof lengtha+b+c, let the length ofDFbea, the length ofF Gbeb, and the length ofGHbec. Then draw a circle centered onFwith radiusaand another circle centeredatGwith radiusc. LetKbe an intersection point of the two circles. Then connectF KandGK. TriangleF KGwill then be the desired triangle. ForF K=F D, and this haslengtha. AlsoF Ghas lengthb, whileGK=GHand this has lengthc. Note also thatwe must haveb+c > a, for otherwise the two circles would not intersect. Thata+b > canda+c > bis obvious from how we have labeled the three lengths.5. Suppose angleDCEis given, and we want to construct an angle equal to angleDCEatpointAof lineAB. Draw the lineDEso that we now have a triangleDCE. (HereDandEare arbitrary points along the two arms of the given angle.) Then, by the resultof exercise 4, construct a triangleAGF, withAGalong lineAB, whereAG=CE,AF=CD, andF G=DE. By the side-side-side congruence theorem, triangleAGFiscongruent to triangleCED. Therefore, angleF AGis equal to angleDCE, as desired.6. LetABCbe the given triangle. ExtendBCtoDand drawCEparallel toAB. By I–29,anglesBACandACEare equal, as are anglesABCandECD. Therefore angleACDequals the sum of the anglesABCandBAC. If we add angleACBto each of these, weget that the sum of the three interior angles of the triangle is equal to the straight angleBCD. Because this latter angle equals two right angles, the theorem is proved.7. Place the given rectangleBEF Gso thatBEis in a straight line withAB. ExtendF Gto9

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10Chapter 3Hso thatAHis parallel toBG. ConnectHBand extend it until it meets the extensionofF EatD. ThroughDdrawDLparallel toF Hand extendGBandHAso they meetDLinMandLrespectively.ThenHDis the diagonal of the rectangleF DLHandso divides it into two equal trianglesHF DandHLD. Because triangleBEDis equalto triangleBMDand also triangleBGHis equal to triangleBAH, it follows that theremainders, namely rectanglesBEF GandABMLare equal.ThusABMLhas beenapplied toABand is equal to the given rectangleBEF G.8. Because trianglesABN,ABC, andANCare similar, we haveBN:AB=AB:BC,soAB2=BN·BC, andNC:AC=AC:BC, soAC2=NC·BC.ThereforeAB2+AC2=BN·BC+NC·BC= (BN+NC)·BC=BC2, and the theorem isproved.9. In this proof, we shall refer to certain propositions in Euclid’s Book I, all of which areproved before Euclid first uses Postulate 5.(That occurs in proposition 29.)First,assume Playfair’s axiom. Suppose linetcrosses linesmandland that the sum of thetwo interior angles (angles 1 and 2 in the diagram) is less than two right angles.Weknow that the sum of angles 1 and 3 is equal to two right angles. Therefore62<63.Now on lineBB′and pointB′construct lineB′C′such that6C′B′B=63 (Proposition23). Therefore, lineB′C′is parallel to linel(Proposition 27). Therefore, by Playfair’saxiom, linemis not parallel to linel. It therefore meetsl. We must show that the twolines meet on the same side asC′. If the meeting pointAis on the opposite side, then62 is an exterior angle to triangleABB′, yet it is smaller than63, one of the interiorangles, contradicting proposition 16. We have therefore derived Euclid’s postulate 5.Second, assume Euclid’s postulate 5. Letlbe a given line andPa point outside the line.Construct the linetperpendicular tolthroughP(Proposition 12). Next, construct thelinemperpendicular to linetatP(Proposition 11). Since the alternate interior anglesformed by linetcrossing linesmandlare both right and therefore are equal, it followsfrom Proposition 27 thatmis parallel tol. Now supposenis any other line throughP.We will show thatnmeetsland is therefore not parallel tol. Let61 be the acute angle

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Euclid11thatnmakes witht.Then the sum of angle 1 and angleP QRis less than two rightangles. By postulate 5, the lines meet.Note that in this proof, we have actually proved the equivalence of Euclid’s Postulate5 to the statement that given a lineland a pointPnot onl, there is at most one linethroughPwhich is parallel tol. The other part of Playfair’s Axiom was proved (in thesecond part above) without use of postulate 5 and was not used at all in the first part.10. One possibility for an algebraic translation: If the line has lengthaand is cut at a pointwith coordinatex, then 4ax+ (a−x)2= (a+x)2. This is a valid identity. Here is ageometric diagram, withAB=ME=aandCB=BD=BK=KR=x:11. IfABCis the given acute-angled triangle andADis perpendicular toBC, then thetheorem states that the square onACis less than the squares onCBandBAby twicethe rectangle contained byCBandBD. If we labelACasb,BAasc, andCBasa,thenBD=ccosB.Thus the theorem can be translated algebraically into the formb2=a2+c2−2accosB, exactly the law of cosines in this case.12. Suppose the diameterCDof a circle with centerEbisects the chordABatF. ThenjoinEAandEB, forming triangleEAB. TrianglesAEFandBEFare congruent byside-side-side (sinceAE=BEare both radii of the circle andFbisectsAB). ThereforeanglesEF AandEF Bare equal. But the sum of those two angles is equal to two rightangles.Hence each is a right angle, as desired.To prove the converse, use the same

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12Chapter 3construction and note that triangleAEBis isosceles, so angleEAFis equal to angleEBF, while both anglesEF AandEF Bare right by hypothesis. It follows that trianglesAEFandBEFare again congruent, this time by angle-angle-side. SoAF=BF, andthe diameter bisects the chord.13. In the circleABC, let the angleBECbe an angle at the center and the angleBACbe anangle at the circumference which cuts off the same arcBC. Connect6EAB. Similarly,6F ECis double6EAC. Therefore the entire6BECis double the entire6BAC. Notethat this argument holds as long as lineEFis within6BEC. If it is not, an analogousargument by subtraction holds.14. Let6BACbe an angle cutting off the diameterBCof the circle.ConnectAto thecenterEof the circle.SinceEB=EA, it follows that6EBA=6EAB.Similarly,6ECA=6EAC. Therefore the sum of6EBAand6ECAis equal to6BAC. But thesum of all three angles equals two right angles. Therefore, twice6BACis equal to tworight angles, and angleBACis itself a right angle.15. Let triangleABCbe given. LetDbe the midpoint ofABandEthe midpoint ofAC.Draw a perpendicular atDtoABand a perpendicular atEtoACand let them meet atpointF(which may be inside or outside the triangle, or on sideBC). Assume first thatFis inside the triangle, and connectF B,F A, andF C. SinceBD=BA, trianglesF DBandF DAare congruent by side-angle-side.ThereforeF B=F A. Similarly, trianglesF EAandF ECare congruent. SoF C=F A. Therefore all three linesF A,F B, andF Care equal, and a circle can be drawn with centerFand radius equal toF A. Thiscircle will circumscribe the given triangle. Finally, note that the identical constructionworks ifFis on lineBCor ifFis outside the triangle.16. LetGbe the center of the given circle andAGDa diameter. With center atDand radiusDG, construct another circle. LetCandEbe the two intersections of the two (equal)circles, and connectDCandDE. ThenDEandDCare two sides of the desired regularhexagon. To find the other four sides, draw the diametersCGFandEGB. ThenCB,

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Euclid13BA,AF, andF Eare the other sides. To demonstrate that we have in fact constructeda regular hexagon, note that all the triangles whose bases are sides of the hexagon andwhose other sides are radii are equilateral; thus all the sides of the hexagon are equaland all the angles of the hexagon are also equal.17. In the circle, inscribe a sideACof an equilateral triangle and a sideABof an equilateralpentagon. Then arcBCis the difference between one-third and one-fifth of the circum-ference of the circle. That is, arcBC=215of the circumference. Thus, if we bisect thatarc atE, then linesBEandECwill each be a side of a regular 15-gon.18. Leta=s1b+r1,b=s2r1+r2,. . .,rk−1=sk+1rk.Thenrkdividesrk−1and thereforealsork−2, . . . , b, a. If there were a greater common divisor ofaandb, it would divider1,r2,. . .,rk. Since it is impossible for a greater number to divide a smaller, we haveshown thatrkis in fact the greatest common divisor ofaandb.19.963 = 1·657 + 306657 = 2·306 + 45306 = 6·45 + 3645 = 1·36 + 936 = 4·9 + 0Therefore, the greatest common divisor of 963 and 657 is 9.4001 = 1·2689 + 13122689 = 2·1312 + 651312 = 20·65 + 1265 = 5·12 + 512 = 2·5 + 25 = 2·2 + 1Therefore, the greatest common divisor of 4001 and 2689 is 1.20.46=7·6+423=7·3+26=1·4+23=1·2+14=2·22=2·1Note that the multiples 7, 1, 2 in the first example equal the multiples 7, 1, 2 in thesecond.

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Solution Manual for A History of Mathematics, 3rd Edition - Page 15 preview image

14Chapter 321.33=2·12+911=2·4+312=1·9+34=1·3+19=3·33=3·1It follows that both ratios can be represented by the sequence (2,1,3).22. Since 1−x=x2, we have1 = 1·x+ (1−x) = 1·x+x2x= 1·x2+ (x−x2) = 1·x2+x(1−x) = 1·x2+x3x2= 1·x3+ (x2−x3) = 1·x3+x2(1−x) = 1·x3+x4· · ·Thus 1 :xcan be expressed in the form (1,1,1, . . .).23. Ifdis the diagonal of a square of sides, then the first division givesd= 1s+r.Tounderstand the next steps, it is probably easiest to sets= 1 and deal with the numericalvalues. Therefore,d=√2, andr=√2−1. Our next division givess= 2r+t, wheret= 3−2√2. Geometrically, if we lay offsalong the diagonal, thenris the remainderd−s. Then draw a square of siderwith part of the side of the original square being itsdiagonal. Note that if we now lay offralong the diagonal of that square, the remainderist. In other words,ris the difference between the diagonal of a square of sidesands,whiletis the difference between the diagonal of a square of siderandr. It follows thatif one performs the next division in the process, we will get the same relationship. Thatis,r= 2t+u, where nowuis the difference between the diagonal of a square of sidetandt. Thus, this process will continue indefinitely and the ratiod:sand be expressedas (1,2,2,2, . . .).24. Sincea > b, there is an integral multiplemofa−bwithm(a−b)> c. Letqbe the firstmultiple ofcthat exceedsmb. Thenqc > mb≥(q−1)c, orqc−c≤mb < qc. Sincec < ma−mb, it follows thatqc≤mb+c < ma. But alsoqc > mb. Thus we have amultiple (q) ofcthat is greater than a multiple (m) ofb, while the same multiple (q)ofcis not greater than the same multiple (m) ofa.Thus by definition 7 of Book V,c:b > c:a.25. LetA:B=C:D=E:F. We want to show thatA:B= (A+C+E) : (B+D+F).Take any equimultiplesmAandm(A+C+E) of the first and third and any equimultiplesnBandn(B+D+F) of the second and fourth. Sincem(A+C+E) =mA+mC+mE, andsincen(B+D+F) =nB+nD+nF, and since whenevermA > nB, we havemC > nDandmE > nF, it follows thatmA > nBimplies thatm(A+C+E)> n(B+D+F).Since a similar statement holds for equality and for “less than”, the result follows fromEudoxus’ definition. A modern proof would use the fact thata1bi=b1aifor everyiandthen conclude thata1(b1+b2+· · ·+bn) =b1(a1+a2+· · ·+an).26. Given thata:b=c:d, we want to show thata:c=b:d. So take any equimultiplesma,mbofaandband also equimultiplesnc,ndofcandd.Nowma:mb=a:b=c:d=nc:nd.Thus ifma > nc, thenmb > nd; ifma=nc, thenmb=nd; and ifma < nc, thenmb < nd. Thus, by the definition of equal ratio, we havea:c=b:d.

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Solution Manual for A History of Mathematics, 3rd Edition - Page 16 preview image

Euclid1527. In the diagram, letDG= 8,GE= 4, andDH= 6. ConnectGHand drawEFparalleltoGH. ThenHFis equal to the fourth proportionalx.28. LetAB= 9 andBC= 5. Draw a circle withACas diameter and erect a perpendiculartoACatB, meeting the circle atE. ThenBEis the desired lengthx.29. Suppose the first of the equal and equiangular parallelograms has sides of lengthaandbwhile the second has sides of lengthcandd, each pair surrounding an angle equaltoα.Since the area of a parallelogram is the product of the two sides with the sineof the included angle, we know thatabsinα=cdsinα. It follows thatab=cdor thata:c=d:b, as desired. Conversely, ifa:c=d:band the parallelograms are equiangularwith angleαbetween each pair of given sides, thenab=cdandabsinα=cdsinα, sothe parallelograms are equal.Euclid’s proof is, of course, different from this modernone. Namely, if the two parallelograms areP1=ADBFandP2=BGCE, with equalangles atB, Euclid places them so thatF BandBGare in a straight line as areEBandBD.He then completes the third parallelogramP3=F BEK.SinceP1=P2,we haveP1:P3=P2:P3.ButP1:P3=DB:BE, sinceBFis common, andP2:P3=BG:BF, sinceBEis common. Thus,DB:BE=BG:BF, the desiredconclusion. The converse is proved by reversing the steps.30. Supposea:b=f:gand suppose the numbersc,d,. . .,eare the numbers in continuedproportion betweenaandb. Letr,s,t,. . .,u,vbe the smallest numbers in the sameratio asa,c,d,. . .,e,b.Thenr,vare relatively prime andr:v=a:b=f:g. Itfollows thatf=mr,g=mvfor some integermand that the numbersms,mt,. . .,muare in the same ratio as the original set of numbers.Thus there are at least as manynumbers in continued proportion betweenfandgas there were betweenaandb. Sincethe same argument works starting withfandg, it follows that there are exactly as manynumbers in continued proportion betweenfandgas betweenaandb.Since there isno integer betweennandn+ 1, it follows that there cannot be a mean proportionalbetween any pair of numbers in the ratio (n+ 1) :n.

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