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[PDF page 223] CH. VIII] THE DUPLICATION OF THE CUBE. 207 of the one is double that of the other will intersect in another point whose abscissa (or ordinate) will give a solution. If we use analysis this is obvious; for, if the equations of the parabolas arey2 = 2ax and x2 = ay, they intersect in a point whose abscissa is given byx3 = 2a3. It is probable that this method was suggested by the form in which Hippocrates had cast the problem: namely, to find x and y so that a : x = x : y = y : 2a, whence we havex2 = ay and y2 = 2ax. The second solution given by Menaechmus was as follows. Describe a parabola of latus rectuml. Next describe a rectangular hyperbola, the length of whose real axis is4l, and having for its asymptotes the tangentatthevertexoftheparabolaandtheaxisoftheparabola. Then the ordinate and the abscissa of the point of intersection of these curves are the mean proportionals betweenl and 2l. This is at once obvious by analysis. The curves arex2 = ly and xy = 2l2. These cut in a point determined by x3 = 2l3 and y3 = 4l3. Hence l : x = x : y = y : 2l. The solution of Apollonius*, which was given about 220b.c., was as follows. The problem is to find two mean proportionals between two given lines. Construct a rectangleOADB, of which the adjacent sides OA and OB are respectively equal to the two given lines. BisectAB in C. With C ascentredescribeacirclecutting OA producedin aandcut- ting OB produced inb, so thataDb shall be a straight line. If this circle can be so described, it will follow thatOA : Bb = Bb : Aa = Aa : OB, that is, Bb and Aa are the two mean proportionals betweenOA and OB. It is impossible to construct the circle by Euclidean geometry, but Apollonius gave a mechanical way of describing it. The only other construction of antiquity to which I will refer is that given by Diocles and Sporus†. It is as follows. Take two sides of a rectangle OA, OB, equal to the two lines between which the means are sought. Suppose OA to be the greater. With centreO and radius OA describe a circle. LetOB produced cut the circumference inC and let AO produced cut it inD. Find a point E on BC so that if DE cuts AB produced in F and cuts the circumference inG, then F E = EG. If E can be found, thenOE is the first of the means betweenOA and OB. Diocles invented the cissoid in order to determineE, but it can be found equally conveniently by the aid of conics. * Ibid., ed. Torelli, p. 137; ed. Heiberg, vol.iii, pp. 76–79. The solution is given in myHistory of Mathematics, London, 1901, p. 84. † Ibid., ed. Torelli, pp. 138, 139, 141; ed. Heiberg, vol.iii, pp. 78–84, 90–93.

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[PDF page 224] 208 THREE GEOMETRICAL PROBLEMS. [CH. VIII In more modern times several other solutions have been suggested. I may allude in passing to three given by Huygens*, but I will enunciate only those proposed respectively by Vieta, Descartes, Gregory of St Vincent, and Newton. Vieta’s construction is as follows†. Describe a circle, centre O, whose radius is equal to half the length of the larger of the two given lines. In it draw a chord AB equal to the smaller of the two given lines. Produce AB to E so that BE = AB. Through A draw a line AF parallel to OE. Through O draw a line DOCF G, cutting the circumference in D and C, cutting AF in F, and cutting BA produced in G, so that GF = OA. If this line can be drawn then AB : GC = GC : GA = GA : CD. Descartespointedout ‡thatthecurves x2 = ay and x2+y2 = ay+bx cut in a point(x, y) such that a : x = x : y = y : b. Of course this is equivalent to the first solution given by Menaechmus, but Descartes preferred to use a circle rather than a second conic. Gregory’s construction was given in the form of the following the- orem§. The hyperbola drawn through the point of intersection of two sides of a rectangle so as to have the two other sides for its asymptotes meets the circle circumscribing the rectangle in a point whose distances from the asymptotes are the mean proportionals between two adjacent sides of the rectangle. This is the geometrical expression of the propo- sition that the curvesxy = ab and x2 + y2 = ay + bx cut in a point (x, y) such that a : x = x : y = y : b. One of the constructions proposed by Newton is as follows∥. Let OA be the greater of two given lines. BisectOA in B. With centre O and radius OB describe a circle. Take a pointC on the circumference so that BC is equal to the other of the two given lines. FromO draw ODE cutting AC produced in D, and BC produced in E, so that the intercept DE = OB. Then BC : OD = OD : CE = CE : OA. Hence OD and CE are two mean proportionals between any two lines BC and OA. * Opera Varia, Leyden, 1724, pp. 393–396. † Opera Mathematica, ed. Schooten, Leyden, 1646, prop,v, pp. 242–243. ‡ Geometria, bk. iii, ed. Schooten, Amsterdam, 1659, p. 91. § Gregory of St Vincent,Opus Geometricum Quadraturae Circuli, Antwerp, 1647, bk. vi, prop. 138, p. 602. ∥ Arithmetica Universalis, Ralphson’s (second) edition, 1728, p. 242; see also pp. 243, 245.

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[PDF page 225] CH. VIII] THE TRISECTION OF AN ANGLE. 209 The Trisection of an Angle*. The trisection of an angle is the second of these classical problems, but tradition has not enshrined its origin in romance. The following two constructions are among the oldest and best known of those which have been suggested; they are quoted by Pappus†, but I do not know to whom they were due originally. The first of them is as follows. LetAOB be the given angle. From any point P in OB draw P M perpendicular to OA. Through P draw P Rparallel toOA. On M P take a pointQ so that ifOQ is produced to cut P Rin R then QR = 2 · OP. If this construction can be made, then AOR = 1 3 AOB. The solution depends on determining the position of R. This was effected by a construction which may be expressed analytically thus. Let the given angle be tan−1(b/a). Construct the hyperbola xy = ab, and the circle(x − a)2 + (y − b)2 = 4(a2 + b2). Of the points where they cut, letx be the abscissa which is greatest, then P R = x − a, and tan−1(b/x) = 1 3 tan−1(b/a). The second construction is as follows. LetAOB be the given angle. Take OB = OA, and with centreO and radius OA describe a circle. Produce AO indefinitely and take a pointC on it external to the circle so that if CB cuts the circumference inD then CD shall be equal to OA. Draw OE parallel to CDB. Then, if this construction can be made, AOE = 1 3 AOB. The ancients determined the position of the point C by the aid of the conchoid: it could be also found by the use of the conic sections. I proceed to give a few other solutions; confining myself to those effected by the aid of conics. Among the other constructions given by Pappus‡I may quote the following. Describe a hyperbola whose eccentricity is two. Let its centre be C and its verticesA and A′. Produce CA ′ to S so that A′S = CA ′. On AS describe a segment of a circle to contain the given angle. Let * On the bibliography of the subject see the supplements toL’Intermédiaire des Mathématiciens, Paris, May and June, 1904. † Pappus, Mathematicae Collectiones, bk. iv, props. 32, 33 (ed. Commandino, Bonn, 1670, pp. 97–99). On the application to this problem of the traditional Greek methods of analysis seeGeometrical Analysis, by J. Leslie, Edinburgh, second edition, 1811, pp. 245–247. ‡ Pappus, bk. iv, prop. 34, pp. 99–104.

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[PDF page 226] 210 THREE GEOMETRICAL PROBLEMS. [CH. VIII the orthogonal bisector ofAS cut this segment inO. With centre O and radius OA or OS describe a circle. Let this circle cut the branch of the hyperbola throughA′ in P. Then SOP = 1 3 SOA. In modern times one of the earliest of the solutions by a direct use of conics was suggested by Descartes, who effected it by the intersection of a circle and a parabola. His construction* is equivalent to finding the points of intersection other than the origin, of the parabolay2 = 1 4 xand the circlex2+y2− 13 4 x+4ay = 0. The ordinates of these points are given by the equation4y3 = 3y − a. The smaller positive root is the sine of one-third of the angle whose sine isa. The demonstration is ingenious. One of the solutions proposed by Newton is practically equivalent to the third one which is quoted above from Pappus. It is as follows†. Let A be the vertex of one branch of a hyperbola whose eccentricity is two, and letS be the focus of the other branch. OnAS describe the segment of a circle containing an angle equal to the supplement of the given angle. Let this circle cut the S branch of the hyperbola inP. Then P AS will be equal to one-third of the given angle. The following elegant solution is due to Clairaut‡. Let AOB be the givenangle. Take OA = OB, andwithcentre O andradius OA describe a circle. Join AB, and trisect it inH, K, so that AH = HK = KB. Bisect the angle AOB by OC cutting AB in L. Then AH = 2 · HL. With focus A, vertex H, and directrixOC, describe a hyperbola. Let the branch of this hyperbola which passes through H cut the cir- cle in P. Draw P M perpendicular to OC and produce it to cut the circle in Q. Then by the focus and directrix property we have AP : P M = AH : HL = 2 : 1 , ∴ AP = 2 · P M = P Q. Hence, by symmetry, AP = P Q = QR. ∴ AOP = P OQ = QOR. I may conclude by giving the solution which Chasles§regards as the most fundamental. It is equivalent to the following proposition. IfOA and OB are the bounding radii of a circular arcAB, then a rectangular hyperbola having OA for a diameter and passing through the point of intersection ofOB with the tangent to the circle atA will pass through * Geometria, bk. iii, ed. Schooten, Amsterdam, 1659, p. 91. † Arithmetica Universalis, problem xlii, Ralphson’s (second) edition, London, 1728, p. 148; see also pp. 243–245. ‡ I believe that this was first given by Clairaut, but I have mislaid my reference. The construction occurs as an example in theGeometry of Conics, by C. Taylor, Cambridge, 1881, No. 308, p. 126. § Traité des sections coniques, Paris, 1865, art. 37, p. 36.

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[PDF page 227] CH. VIII] THE QUADRATURE OF THE CIRCLE. 211 one of the two points of trisection of the arc. The Quadrature of the Circle*. The object of the third of the classical problems was the deter- mination of a side of a square whose area should be equal to that of a given circle. The investigation, previous to the last two hundred years, of this question was fruitful in discoveries of allied theorems, but in more re- cent times it has been abandoned by those who are able to realize what is required. The history of this subject has been treated by compe- tent writers in such detail that I shall content myself with a very brief allusion to it. Archimedes showed† (what possibly was known before) that the problem is equivalent to finding the area of a right-angled triangle whose sides are equal respectively to the perimeter of the circle and the radius of the circle. Half the ratio of these lines is a number, usu- ally denoted by π. That this number is incommensurable had been long suspected, and has been now demonstrated. The earliest analytical proof of it was given by Lambert‡ in 1761; in 1803 Legendre§ extended the proof to show thatπ2 was also incommensurable; and recently Lindemann∥ has * See Montucla’sHistoire des Recherches sur la Quadrature du Cercle, edited by P.L. Lacroix, Paris, 1831; also various articles by A. De Morgan, and especially his Budget of Paradoxes, London, 1872. A popular sketch of the subject has been compiled by H. Schubert,Die Quadratur des Zirkels, Hamburg, 1889; and since the publication of the earlier editions of theseRecreationsProf. F. Rudio of Zurich has given an analysis of the arguments of Archimedes, Huygens, Lambert, and Legendre on the subject, with an introduction on the history of the problem, Leipzig, 1892. † Archimedis Opera, ΚÔκλου µèτρησις , prop. i, ed. Torelli, pp. 203–205; ed. Heiberg, vol. i, pp. 258–261, vol.iii, pp. 269–277. ‡ Mémoires de l’Académie de Berlinfor 1761, Berlin, 1768, pp. 265–322. § Legendre’sGeometry, Brewster’s translation, Edinburgh, 1824, pp. 239–245. ∥ Ueber die Zahl π, Mathematische Annalen, Leipzig, 1882, vol. xx, pp. 213– 225. The proof leads to the conclusion that, ifx is a root of a rational integral algebraical equation, then ex cannot be rational: hence, if πi was the root of such an equation,eπi could not be rational; buteπi is equal to−1, and therefore is rational; hence πi cannot be the root of such an algebraical equation, and therefore neither canπ.

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[PDF page 228] 212 THREE GEOMETRICAL PROBLEMS. [CH. VIII shown that π cannot be the root of a rational algebraical equation. An earlier attempt by James Gregory to give a geometrical demon- stration of this is worthy of notice. Gregory proved* that the ratio of the area of any arbitrary sector to that of the inscribed or circum- scribed polygons is not expressible by a finite number of algebraical terms. Hence he inferred that the quadrature was impossible. This was accepted by Montucla, but it is not conclusive, for it is conceiv- able that some particular sector might be squared, and this particular sector might be the whole circle. In connection with Gregory’s proposition above cited, I may add that Newton † proved that in any closed oval an arbitrary sector bounded by the curve and two radii cannot be expressed in terms of the co-ordinates of the extremities of the arc by a finite number of algebraical terms. The argument is condensed and difficult to follow: the same reasoning would show that a closed oval curve cannot be represented by an algebraical equation in polar co-ordinates. From this proposition no conclusion as to the quadrature of the circle is to be drawn, nor did Newton draw any. In the earlier editions of this work I expressed an opinion that the result presupposed a particular definition of the word oval, but on more careful reflection I think that the conclusion is valid without restriction. With the aid of the quadratrix, or the conchoid, or the cissoid, the quadrature of the circle is easy, but the construction of those curves assumes a knowledge of the value ofπ, and thus the question is begged. I need hardly add that, if π represented merely the ratio of the circumference of a circle to its diameter, the determination of its nu- merical value would have but slight interest. It is however a mere accident that π is defined usually in that way, and it really represents a certain number which would enter into analysis from whatever side the subject was approached. I recollect a distinguished professor explaining how different would be the ordinary life of a race of beings born, as easily they might be, so that the fundamental processes of arithmetic, algebra and geometry were different to those which seem to us so evident, but, he added, it is impossible to conceive of a universe in whiche and π should not exist. * Vera Circuli et Hyperbolae Quadratura, Padua, 1668: this is reprinted in Huy- gens’sOpera Varia, Leyden, 1724, pp. 405–462. † Principia, bk. i, section vi, lemma xxviii.

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[PDF page 229] CH. VIII] APPROXIMATIONS TO THE V ALUE OF π. 213 I have quoted elsewhere an anecdote, which perhaps will bear rep- etition, that illustrates how little the usual definition ofπ suggests its properties. De Morgan was explaining to an actuary what was the chance that a certain proportion of some group of people would at the end of a given time be alive; and quoted the actuarial formula, involving π, which, in answer to a question, he explained stood for the ratio of the circumference of a circle to its diameter. His acquaintance, who had so far listened to the explanation with interest, interrupted him and exclaimed, “My dear friend, that must be a delusion, what can a circle have to do with the number of people alive at the end of a given time?” In reality the fact that the ratio of the length of the circumference of a circle to its diameter is the number denoted byπ does not afford the best analytical definition ofπ, and is only one of its properties. The use of a single symbol to denote this number3.14159 . . .seems to have been introduced about the beginning of the eighteenth century. W. Jones* in 1706 represented it byπ; a few years later†John Bernoulli denoteditby c; Eulerin1734used p, andin1736used c; Chr. Goldback in 1742 usedπ; and after the publication of Euler’sAnalysis the symbol π was generally employed. The numerical value ofπ can be determined by either of two meth- ods with as close an approximation to the truth as is desired. The first of these methods is geometrical. It consists in calculating the perimeters of polygons inscribed in and circumscribed about a cir- cle, and assuming that the circumference of the circle is intermediate between these perimeters‡. The approximation would be closer if the areas and not the perimeters were employed. The second and modern method rests on the determination of converging infinite series forπ. We may say that theπ-calculators who used the first method re- garded π as equivalent to a geometrical ratio, but those who adopted the modern method treated it as the symbol for a certain number which enters into numerous branches of mathematical analysis. It may be interesting if I add here a list of some of the approxi- mations to the value ofπ given by various writers§. This will indicate * Synopsis Palmariorum Matheseos, London, 1706, pp. 243, 263et seq. † See notes by G. Eneström in theBibliotheca Mathematica, Stockholm, 1889, vol. iii, p. 28;Ibid., 1890, vol.iv, p. 22. ‡ The history of this method has been written by K.E.I. Selander,Historik öfver Ludolphska Talet, Upsala, 1868. § For the methods used in classical times and the results obtained, see the notices

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[PDF page 230] 214 THREE GEOMETRICAL PROBLEMS. [CH. VIII incidentally those who have studied the subject to the best advantage. The ancient Egyptians* took 256/81 as the value ofπ, this is equal to 3.1605 . . .; but the rougher approximation of 3 was used by the Babylonians† and by the Jews‡. It is not unlikely that these numbers were obtained empirically. We come next to a long roll of Greek mathematicians who attacked the problem. Whether the researches of the members of the Ionian School, the Pythagoreans, Anaxagoras, Hippias, Antipho, and Bryso led to numerical approximations for the value ofπ is doubtful, and their investigations need not detain us. The quadrature of certain lunes by Hippocrates of Chios is ingenious and correct, but a value ofπ cannot be thence deduced; and it seems likely that the later members of the Athenian School concentrated their efforts on other questions. It is probable that Euclid§, the illustrious founder of the Alexand- rian School, was aware thatπ was greater than3 and less than4, but he did not state the result explicitly. The mathematical treatment of the subject began with Archi- medes, who proved thatπ is less than 3 1 7 and greater than 3 10 71, that is, it lies between 3.1428 . . . and 3.1408 . . .. He established ∥ this by inscribing in a circle and circumscribing about it regular polygons of 96 sides, then determining by geometry the perimeters of these polygons, and finally assuming that the circumference of the circle was intermediate between these perimeters: this leads to the result 6336/2017 1 4 < π < 14688/4673 1 2,1 from which he deduced the lim- of their authors in M. Cantor’sGeschichte der Mathematik, Leipzig, vol.i, 1880. For medieval and modern approximations, see the article by A. De Morgan on the Quadrature of the Circle in vol.xixof thePenny Cyclopaedia, London, 1841; with the additions given by B. de Haan in theVerhandelingen of Amsterdam, 1858, vol.iv, p. 22: the conclusions were tabulated, corrected, and extended by Dr J.W.L. Glaisher in theMessenger of Mathematics, Cambridge, 1873, vol.ii, pp. 119–128; andIbid., 1874, vol.iii, pp. 27–46. * Ein mathematisches Handbuch der alten Aegypter(i.e. the Rhind papyrus), by A. Eisenlohr, Leipzig, 1877, arts. 100–109, 117, 124. † Oppert, Journal Asiatique, August, 1872, and October, 1874. ‡ 1 Kings, ch. 7, ver. 23; 2 Chronicles, ch. 4, ver. 2. § These results can be deduced from Euc. iv, 15, and iv, 8: see also book xii, prop. 16. ∥ Archimedis Opera, ΚÔκλου µèτρησις , prop.iii, ed.Torelli, Oxford, 1792, pp.205– 216; ed. Heiberg, Leipzig, 1880, vol.i, pp. 263-271. 1. Inserted 14688/

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[PDF page 231] CH. VIII] APPROXIMATIONS TO THE V ALUE OF π. 215 its given above. This method is equivalent to using the proposition sin θ < θ < tan θ, where θ = π/96: the values ofsin θ and tan θ were deduced by Archimedes from those ofsin 1 3 π and tan 1 3 π by repeated bisections of the angle. With a polygon ofn sides this process gives a value ofπ correct to at least the integral part of(2 logn−1.19) places of decimals. The result given by Archimedes is correct to2 places of dec- imals. His analysis leads to the conclusion that the perimeters of these polygons for a circle whose diameter is 4970 feet would lie between 15610 feet and 15620 feet—actually it is about15613 feet 9 inches. Apollonius discussed these results, but his criticisms have been lost. Hero of Alexandria gave* the value 3, but he quoted† the result 22/7: possibly the former number was intended only for rough ap- proximations. The only other Greek approximation that I need mention is that given by Ptolemy‡, who asserted thatπ = 3 ◦8′30′′. This is equivalent to taking π = 3 + 8 60 + 30 3600 = 3 17 120 = 3 .141˙6. The Roman surveyors seem to have used3, or sometimes 4, for rough calculations. For closer approximations they often employed3 1 8 instead of 3 1 7, since the fractions then introduced are more convenient in duodecimal arithmetic. On the other hand Gerbert§ recommended the use of 22/7. Before coming to the medieval and modern European mathemati- cians it may be convenient to note the results arrived at in India and the East. Baudhayana∥ took 49/16 as the value ofπ. Arya-Bhata¶, circ. 530, gave62832/20000, which is equal to3.1416. He showed that, ifa is the side of a regular polygon ofn sides inscribed in a circle of unit diameter, and ifb is the side of a regular inscribed polygon of 2n sides, then b2 = 1 2 − 1 2(1 − a2) 1 2. From the side of an inscribed hexagon, he found successively the sides of polygons of12, 24, 48, 96, 192, and 384 sides. The perimeter of the last is given as equalto √ 9.8694, fromwhichhisresultwasobtainedbyapproximation. * Mensurae, ed. Hultsch, Berlin, 1864, p. 188. † Geometria, ed. Hultsch, Berlin, 1864, pp. 115, 136. ‡ Almagest, bk. vi, chap. 7; ed. Halma, vol.i, p. 421. § Œuvres de Gerbert, ed. Olleris, Clermont, 1867, p. 453. ∥ The Sulvasutras by G. Thibaut,Asiatic Society of Bengal, 1875, arts. 26–28. ¶ Leçons de calcul d’Aryabhata, by L. Rodet in theJournal Asiatique, 1879, se- ries 7, vol.xiii, pp. 10, 21.

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[PDF page 232] 216 THREE GEOMETRICAL PROBLEMS. [CH. VIII Brahmagupta*, circ. 650, gave √ 10, which is equal to3.1622 . . .. He is said to have obtained this value by inscribing in a circle of unit diameter regular polygons of 12, 24, 48, and 96 sides, and calculat- ing successively their perimeters, which he found to be √ 9.65, √ 9.81,√ 9.86, √ 9.87 respectively; and to have assumed that as the number of sides is increased indefinitely the perimeter would approximate to √ 10. Bhaskara, circ. 1150, gave two approximations. One †—possibly copied from Arya-Bhata, but said to have been calculated afresh by Archimedes’s method from the perimeters of regular polygons of384 sides—is 3927/1250, which is equal to3.1416: the other‡is 754 240, which is equal to3.141˙6, but it is uncertain whether this was not given only as an approximate value. Among the Arabs the values 22/7, √ 10, and 62832/20000 were given by Alkarisimi§, circ. 830; and no doubt were derived from Indian sources. He described the first as an approximate value, the second as used by geometricians, and the third as used by astronomers. In Chinese works the values3, 22 7 , 157 50 are said to occur: probably the last two results were copied from the Arabs. Returning to European mathematicians, we have the following suc- cessive approximations to the value ofπ: many of those prior to the eighteenth century having been calculated originally with the view of demonstrating the incorrectness of some alleged quadrature. Leonardo of Pisa∥, in the thirteenth century, gave forπ the value 1440/458 1 3 which is equal to3.1418 . . .. In the fifteenth century, Pur- bach¶ gave or quoted the value62832/20000, which is equal to3.141˙6; Cusa believed that the accurate value was3 4( √ 3 + √ 6) which is equal to 3.1423 . . .; and, in 1464, Regiomontanus** is said to have given a value equal to 3.14243. * Algebra...from Brahmegupta and Bhascara, trans. byH.T.Colebrooke, London, 1817, chap. xii, art. 40, p. 308. † Ibid., p. 87. ‡ Ibid., p. 95. § The Algebra of Mohammed ben Musa, ed. by F. Rosen, London, 1831, pp. 71–72. ∥ Boncompagni’sScritti di Leonardo, vol. ii (Practica Geometriae), Rome, 1862, p. 90. ¶ Appendix to theDe Triangulisof Regiomontanus, Basle, 1541, p. 131. ** In his correspondence with Cardinal Cusa,De Quadratura Circuli, Nuremberg, 1533, wherein he proved that Cusa’s result was wrong. I cannot quote the exact reference, but the figures are given by competent writers and I have no doubt are correct.
