Examples of greek arithmetic

Author(s): J. Gilbart Smyly Source: Hermathena, Vol. 19, No. 42 (1920), pp. 105-114 Published by: Trinity College Dublin Stable URL: http://www.jstor.org/stable/23037108 . Accessed: 04/01/2015 15:52 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact support@jstor.org. . Trinity College Dublin is collaborating with JSTOR to digitize, preserve and extend access to Hermathena. http://www.jstor.org This content downloaded from 192.87.31.20 on Sun, 4 Jan 2015 15:52:20 PM All use subject to JSTOR Terms and Conditions

SOME EXAMPLES OF GREEK ARITHMETIC. APYRUS No. 186 of the Papiri della Società Italiana contains two arithmetical problems which have not been adequately dealt with by the editors. They say: “ Poichè non sono conservati tutti i dati, non è possibile controllare la soluzione, secondo cui si avrebbero 8400 spettatori al teatro, e 4860 artabe nel vaso.” Itis, however, easy, in both cases, to reproduce the complete mathematical calculation, although it may be impossible to restore with certainty the exact words. The text of the papyrus, which is assigned to the fourth century A.D., and is written in extremely bad Greek, is as follows :— Recto. deadpov TO pev ávw | m, TO. dé ara Babpov | — Ba-| Opwv px, eüpeiv roc ovs ? X0pYT El TO Oéaôpo lv Ca pw ovvdes To avw BaF] a , là >»4 KaTw, TovreoTıv T | evi Tas Tov Babpuv p| k ? là \ 4 xopyo: To Oeadpov Tw | mu emt TOV Opor| wr TxNPA VTOKELTAL. (Figure. ) Verso. |yvAovv, où Y ÖLauerpos |Babos tnx À, evpeiv ]* xwpyoe: ran THS dia-

[uerpov 15 — ja yilvovraı) £6 rovrwv ot devo- [wevoe -— Jun eri To Babos yx A” [E ov y 69 dca Ti emı Tov a ]repeòs mx’ xwpqoe Enpov Jxwpyor Enpod 6 dw". (Figure.) These problems resemble very closely some of those in the mathematical papyrus of Akhmim (J. Baillet, Mé. de la Mission Arch. Franc. au Catre, ix, 1892), from which we may make some inferences as to the way in which arithmetic was taught in Egyptian schools. It is, I think, quite certain that the Akhmim papyrus is the exercisebook of a schoolboy; it is a collection of disconnected problems, with no method in their arrangement, except that the earlier are somewhat problems, as would be expected easier in than the later a schoolboy’s book. The writer has also made mistakes in his work, which are not mere errors of calculation, but due to the application of wrong methods. A remarkable feature of this papyrus is the use of the word duotws to introduce the solution and often some of the subsidiary steps. The writer was clearly working to a pattern, and we must suppose that, instead of learning general principles, he was taught a separate rule for each class of problems, and givena model example to copy: a method which, common at the present day. of these unfortunately, is only too Behind the solution of each Greek problems there lies a general formula, which we should now represent by algebraical symbols: by the Greeks it was probably expressed in words, accompanied by a particular example. A student, having such an example before his eyes, when working out a similar problem with different numbers, might naturally introduce the several steps of his calculation by the word époiwe. The first problem is a good example of his procedure,

and is also instructive, because the result is wrong; if we can discover the source of the error, we shall learn a good deal about the methods of teaching Greek logistic. AakKos oToyvAovy NV avw Tepiperpos TUXWV K e 4 La en , La) n KdTW Tepiperpos muxwv ¿8 Bados ruyov sl ôpoiws x kat ıB yilvovraı) AB To L av AB y veras) ts Gpoiws > N 4 € ld > N us émi us yl vera) ovs dpoiws ovs éri SL yi(verat) ax£d pepioov As ös elvai us g' in ‘‘There was a circular pit. its lower perimeter was contents). Its upper perimeter was 20 cubits, 12 cubits, its depth 63 cubits (find its Similarly 20 + 12 = 32; 4 of 32 = 16; similarly 16 x16 = 256; similarly 256 x 63 = 1664; divide by 36, so that the result is 464 4.” The difficulty is the final division by 36. Baillet says: ‘ Au lieu d’une division par 12, nous trouvons une division par 36, soit par un nombre 3 fois plus fort. Ou bienilya là une erreur grossière, ou bien il y a un changement d’unité non annoncé, de même que cela semble avoir lieu également dans le problème 2.” The theory of a change of unit is ingenious, but, I think, untenable. It is true that not only in problem 2, where the result is multiplied by 33, but also in problem 5, where the result is divided by 27, there is a change of unit, a most important change, which will be discussed later; but for the present it is sufficient to remark that both 33 and 27 are perfect cubes, and therefore the change is in the linear unit. In this case we should have two units of volume in the ratio 3:1, and the corresponding linear units would be incommensurable. We must have recourse to the alternative that there is a gross error; but I think that the source of this error can be determined. The general problem, of which this is a particular case,

may be stated in more modern language thus :—F ind the volume of a frustum of a circular cone, given the perimeter (Pj of the upper circle, the perimeter (2) of the lower circle, and the distance (4) between them. It is known that sometimes, as in P. Ox. III. 470, the ancients erroneously substituted for this volume that of a cylinder of the same height standing on a base with a radius equal to half the sum of the given radii. ( The volume of such a cylinder is P+prd P) 2. If we take 7 = 3, the rule as stated to the 2 yr (greek student would be :— Add the perimeters of the upper and lower circles ; divide by 2; square the quotient; multiply by the height, and divide by 12. This is exactly what the writer of the papyrus has except that in done, the last step he has divided by 36 instead of 12. The true value of the volume of a cone was well known to the Greeks. Eudoxus is said to have been the first to discover the proof, but we now know that Archimedes in the "Epodoc (see Hermes, vol. xlii, 1907, p. 245) ascribed much of the theorem merit without to Democritus, who first stated the proof. An accurate formula for the volume of a truncated cone, a kwvoc, kódovpoc, such as is considered in this problem, is given by Heron of Alexandria (Metrica, II. 9); but itis expressed in terms of the diameters instead of the perimeters of the bounding circles, and the value of 7 is taken as 34. In terms of the perimeters the volume is (P* + 2° + Pp) pei: and the rule for calculating it would have been—Add together the perimeters and their product. squares of the Multiply by the height and divide by 36 (= 127). I think it is almost certain that the writer of the papyrus was given both rules, as alternatives, for problems of this kind, and that he in this case adopted the first rule, but in

his last step employed the divisor belonging to the second rule. The second problem of the Akhmin papyrus is very simple, but of great importance, because a change of unit is introduced. TeTpaywvog It is to find measuring dimensions are Io x multiplied the contents of a Onoavpoc 10 x 8 cubits. The three together, giving 800 cubic cubits, and this product is then multiplied by 33. Baillet did not recognize that the symbol here used was that for artabae, but because he supposing thought it stood for “l'unité,” probably wrongly interpreted the symbol for aroura it to stand for artaba. But he guessed that the artaba was the measure intended, for he says, “ Cette mesure serait l’artabe, assimilée au pied cube.” In the second of the problems in the Italian papyrus the same multiplier is employed to reduce cubic cubits to artabae. Wilcken (Ostr. I, p. 752) has pointed out that we can deduce from this the important equation :— 1 cubic ell = 32 artabae. We can go a step further than this, and assert that, at least theoretically, an artaba was produce contained in one cubic foot. the amount of dry Since 32 is the cube of 13, the linear unit must be % of a cubit; and, since a cubit is 13 feet, the linear unit is 1 foot. It is known that artabae of different sizes were used in Ptolemaic times. Wilcken (Osfr. 1, p. 744) enumerates artabae of 40, 30, 29, 26, and 24 choenices. The existence of the first is proved by P. P. II, 25, of the second by Rev. Pap. 30,2. Lhe third (P. Gren. I. 18, 19) was a private measure, which may have been intended to hold 30, but was found, when tested, to contain only 29 choenices: but the reading is doubtful, and Wilcken suggests (Archiv. II, p. 123) that mpde ro Ba(otAucov) x(aAkovv) should be read for

Tpòc ro k0x; the fourth is also doubtful, depending, as it does, upon an uncertain reading of a single ostracon. We are thus left with three artabae of 40, 30, and 24 choenices. It is certain (see P. P. III. App. and P. Lille 1) that zaudia and aora were generally measured by the double royal cubit; but, in one instance (P. P. III, p. 125) it was specially directed that the aoilza were to be measured rw TpiskaldexaTraXaoTut perpwi. Hence we have three different linear measures, proportional to 14, 13, and 12, employed for calculating volumes. If the variation in the size of the artaba depended upon this variation in the linear unit, we should expect to find three artabae proportional to the cubes of these numbers: that is, in the proportion 2744 : 2197: 1728, which is equivalent to 38:1 : 30°5 : 24. This is so close to the ratios of the known artabae 40 : 30: 24, that there can be little doubt that the variation in the size of the artaba was due to the difference in the linear feet by which the vessels were measured. The fifth problem is interesting because it is another example of the application of a wrong method, and because it contains another and different unit of capacity. The problem is:—Given a rectangular channel (&dpové rerpayovog), length 20 cubits, breadth 8 cubits, and depth 3q cubits, to find its contents. He should have multiplied together the three linear dimensions; but, instead of doing SO, Says :—opolws k kal n yiyverar kn, TO L rev Ky ylyverac 10, 16 ¿mi 10 yiyverai pge, pag emt yLO yiyverar Ye pep. KZ, de tva «Ec. Similarly 20 + 8 = 28; + of 28 = 14; 14° = 196; 196 x 3} = 735; divide by 27, so that the result is 271 + 34.” . Thus he uses the formula P + pvy' ) d, as in the first problem following a model intended for a different class of question. But the most important point is that the result in cubic cubits is divided by measured This implies a unit of volume by a cube with a side of 3 cubits: can it be

identified? should be observed that the 111 problem relates to a GiwpuË, and the answer is the amount of earth or sand excavated. In Ptolemaic times the unit of volume used in measuring such excavations was an aozlzon, and the same volume constructing of material embankments was used in building or called zaubion. In the appendix to Petrie Papyri 111 I proved that an aozlzon is a volume equal to the cube whose side is a royal double cubit. There was at that time no evidence to determine the size of a zaubzon; but subsequently Lille Pap. 1 furnished proof that its volume was equal to that of the aotlion. Naubta and acilia were in Ptolemaic times of the same volume, containing 8 cubic royal cubits. But in Roman times the standard was changed, and the naubion was a cube whose side measured a ¿úvlow (Ox. P. iv. 669). There were, according to the same papyrus, two ¿úla, a BacıAıköv of 3 cubits, and another of 23 cubits: the volume of a naubion measured by the former would be 27 cubic cubits. Hence, having obtained the volume in cubits, the number of nmaubra (or aozlza) is determined by division by 27. We may now return to the Italian problems, and discuss the general formulae of which they are particular examples. The first is to find the number of spectators in a theatre, given the number of seats in the top and bottom rows, and the number of rows. It is assumed that the increase in the number of seats in the rows is uniform from bottom to top. It is easy to construct the general solution: let x be the number of seats in the top row, y the number in the bottom row, z the number of rows. Then if a be the average increase in the number of seats in each row x=y+(r-ı)a, and the total number of seats will be N=" fay (72 - E

eliminating a, we get N=-(x DIR +7). Hence the general rule: add together the number of seats in the top and bottom rows; multiply by the number of rows, and divide the result by 2. If we know any three of the quantities in the equation WIR? N=-(x+9y), we can determine the fourth. N = 8,400, Here x = 80, 2 = 120, and we find that y = 60. Hence we may restore the text with certainty as to the meaning, but hesitation as to the exact words. deadpov. TO pev avw [| Babpov xwpel m, TO dé KaTW Badpov [ È, Ovrwv Ba- Opwv pk eúpetv rool ovs avOpwrovs Y x ld xwpyoer TO Heaöpo [v. 5 oúvOes Te avw Bab pov kai To , [4 x € KATW, Tovreorıv m |Kal È / pp. a ext ras TOV Babpwr p[kx / M sw. rovrwv xupnoeı TO Beadpov tw |]uuov, TovVreorıv ‚mv. 9 10 a € „m Emi TOV Opot| wv mpoßAnuarwv TO a e , \ LA CXYMGA VITOKELTAL, “A theatre. The top row has 80 seats, the lowest has 60; if there are 120 rows, contain. Add the top row and the lowest row, that is 80 and 60; the sum is 16,800. 140; find how many spectators the theatre will multiply by 120, the number of rows; result The theatre will contain half of this, that is 8,400. The figure for similar problems is given below.” I suppose the rw of 1. 5 and the rw of 1. 8 to be mistakes for ro. In 1. 7 èrì rae would be éwira in ordinary Greek. Ihe second is an example of the general problem to

find the number artabae contained in receptacle of given diameter and depth. 113 a cylindrical The volume of a cylinder of diameter 4 and depth % is whe If we take m = 3, We get the rule: Square the diameter, take 2 of the result, then multiply in cubic cubits. by the depth. This is the volume To find the contents in artabae, multiply by 32, the number of artabae in a cubic cubit. We can now reconstruct the text. in the Akhmim papyrus begins The first problem with the words Aakkoe orovyvAovv, and here we have ]yvAovv; we may presume, the Greek being equally bad in both papyri, that the beginning was [Aakkoe o7poy JyvAovv. [Adkkos oTpoy |yvAoîr, où 7) didperpos [rnx(@v) y ro de] Bados ryx(oy) À, edpetv [róvas dptaBas| xwpyoe. Tan THs da- [uérpou eq” éaur Ja yi(verar) $0, TovTwv...... [ ] uy, eri ro Bados anx(av) À [yé(verar) javp . . .]. ov 3%, State Eri Tov [ 6 a |repeòs anx(vs) xwpyoe Enpod [6 de Adkxos| xwpyoi Enpod 6 dwé. ‘ A circular pit, of which the diameteris 8 cubits and the depth 30 cubits: to find how many artabae it will contain. 8 of the diameter by itself; result 64. Multiply the Take ¿of this; result 48. Multiply by the depth 30 cubits; result 1,440. Multiply by 38, because a cubic cubit contains so many artabae. The pit will contain 4,860 artabae.” In 1. 14 the editors read roúrewv oi pevó | nuevo, and say = oi dawöuevo. This cannot be right; such a word has no place in this context. What was written must have been one of the many ways in which “ three-quarters of this is 48” could have been expressed in Greek, such as rovrwv HERMATHENA—VOL, XIX.

To L Kai è’, Or rovrwv Ta Tpla rérapra, OF TOUTWY àpeÂe TO reraprov. In 1. 16 I suppose ea ri to be a mistake for deri. The word Enpod in ll. 17 and 18 is not superfluous, because the sentence means “the amount of dry stuff a cubic cubit, or the receptacle, contains is so many artabae.” Such a vessel might contain wine or water, in which case the writer would have said “a cubic cubit contains so many metretae of fluid.” J. G SMYLY.