El Papiro Matemático de Rhind ( RMP ; también designado como papiro Museo Británico 10057 y pBM 10058) es uno de los ejemplos más conocidos de las matemáticas del Antiguo Egipto . Lleva el nombre de Alexander Henry Rhind , un anticuario escocés , que compró el papiro en 1858 en Luxor, Egipto ; aparentemente fue encontrado durante excavaciones ilegales en o cerca del Ramesseum . Data de alrededor de 1550 a. C. [1] El Museo Británico, donde ahora se guarda la mayor parte del papiro, lo adquirió en 1865 junto con el Rollo de Cuero Matemático Egipcio , también propiedad de Henry Rhind;[2] hay algunos pequeños fragmentos en poder del Museo de Brooklyn en la ciudad de Nueva York [3] [4] y falta una sección central de 18 cm. Es uno de los dos papiros matemáticos más conocidos junto con el papiro matemático de Moscú . El papiro de Rhind es más grande que el papiro matemático de Moscú, mientras que este último es más antiguo. [3]
Papiro matemático de Rhind | |
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Museo Británico de Londres | |
Fecha | Segundo período intermedio de Egipto |
Lugar de origen | Tebas |
Idioma (s) | Egipcio ( hierático ) |
Tamaño | Primera sección ( BM 10057 ): · Longitud: 295,5 cm (116,3 pulgadas) · Ancho: 32 cm (13 pulgadas ) Segunda sección ( BM 10058 ): · Longitud: 199,5 cm (78,5 pulgadas ) · Ancho: 32 cm (13 pulgadas ) |
El papiro matemático de Rhind se remonta al segundo período intermedio de Egipto . Fue copiado por el escriba Ahmes ( es decir, Ahmose; Ahmes es una transcripción más antigua favorecida por los historiadores de las matemáticas), de un texto ahora perdido del reinado del rey Amenemhat III ( dinastía XII ). Escrito en escritura hierática , este manuscrito egipcio mide 33 cm (13 pulgadas) de alto y consta de varias partes que en total superan los 5 m (16 pies) de largo. El papiro comenzó a transliterarse y traducirse matemáticamente a fines del siglo XIX. El aspecto de la traducción matemática permanece incompleto en varios aspectos. El documento está fechado en el año 33 del rey Hyksos Apophis y también contiene una nota histórica posterior separada en su reverso que probablemente data del período ("Año 11") de su sucesor, Khamudi . [5]
En los primeros párrafos del papiro, Ahmes presenta el papiro como un "cálculo exacto para investigar las cosas y el conocimiento de todas las cosas, misterios ... todos los secretos". Continúa con:
Este libro fue copiado en el año de reinado 33, mes 4 de Akhet , bajo la majestad del Rey del Alto y Bajo Egipto, Awserre, dado vida, de una copia antigua hecha en la época del Rey del Alto y Bajo Egipto Nimaatre. El escriba Ahmose escribe esta copia. [2]
Se han publicado varios libros y artículos sobre el papiro matemático de Rhind, y algunos de ellos se destacan. [3] El Papiro Rhind fue publicado en 1923 por Peet y contiene una discusión del texto que siguió al esquema del Libro I, II y III de Griffith [6] Chace publicó un compendio en 1927-29 que incluía fotografías del texto. [7] Robins y Shute publicaron en 1987 un resumen más reciente del Papiro Rhind.
Libro I - Aritmética y Álgebra
La primera parte del papiro de Rhind consta de tablas de referencia y una colección de 21 problemas aritméticos y 20 algebraicos. Los problemas comienzan con expresiones fraccionarias simples, seguidas de problemas de finalización ( sekem ) y ecuaciones lineales más complicadas ( problemas aha ). [3]
La primera parte del papiro está ocupada por la tabla 2 / n . Las fracciones 2 / n para impar n varía de 3 a 101 son expresadas como sumas de fracciones unitarias . Por ejemplo,. La descomposición de 2 / n en fracciones unitarias nunca es más de 4 términos como, por ejemplo,.
A esta tabla le sigue una tabla diminuta y mucho más pequeña de expresiones fraccionarias para los números del 1 al 9 divididos por 10. Por ejemplo, la división de 7 entre 10 se registra como:
- 7 dividido por 10 produce 2/3 + 1/30
Después de estas dos tablas, el papiro registra 91 problemas en total, que los modernos han designado como problemas (o números) 1-87, incluidos otros cuatro elementos que han sido designados como problemas 7B, 59B, 61B y 82B. Los problemas 1 a 7, 7B y 8 a 40 están relacionados con la aritmética y el álgebra elemental.
Los problemas 1 a 6 calculan las divisiones de un cierto número de barras de pan entre 10 hombres y registran el resultado en fracciones unitarias. Los problemas 7 a 20 muestran cómo multiplicar las expresiones 1 + 1/2 + 1/4 = 7/4 y 1 + 2/3 + 1/3 = 2 por diferentes fracciones. Los problemas 21 a 23 son problemas de terminación, que en notación moderna son simplemente problemas de resta. Los problemas 24 a 34 son problemas "ajá"; estas son ecuaciones lineales . El problema 32, por ejemplo, corresponde (en notación moderna) a resolver x + 1/3 x + 1/4 x = 2 para x. Los problemas 35 a 38 involucran divisiones del heqat, que es una unidad de volumen del antiguo Egipto . A partir de este punto, una variedad de unidades de medida se vuelven mucho más importantes a lo largo del resto del papiro y, de hecho, una consideración importante en el resto del papiro es el análisis dimensional . Los problemas 39 y 40 calculan la división de panes y usan progresiones aritméticas . [2]
Libro II - Geometría
La segunda parte del papiro de Rhind, que son los problemas 41-59, 59B y 60, consiste en problemas de geometría . Peet se refirió a estos problemas como "problemas de medición". [3]
Volúmenes
Los problemas 41 a 46 muestran cómo hallar el volumen de los graneros tanto cilíndricos como rectangulares. En el problema 41, Ahmes calcula el volumen de un granero cilíndrico. Dado el diámetro dy la altura h, el volumen V viene dado por:
En notación matemática moderna (y usando d = 2r) esto da . El término fraccionario 256/81 aproxima el valor de π como 3,1605 ..., un error de menos del uno por ciento.
El problema 47 es una tabla con igualdades fraccionarias que representan las diez situaciones en las que la cantidad de volumen físico de "100 heqats cuádruples" se divide por cada uno de los múltiplos de diez, de diez a cien. Los cocientes se expresan en términos de fracciones del ojo de Horus , a veces también utilizando una unidad de volumen mucho más pequeña conocida como "ro cuádruple". El cuádruple heqat y el cuádruple ro son unidades de volumen derivadas de los más simples heqat y ro, de manera que estas cuatro unidades de volumen satisfacen las siguientes relaciones: 1 cuádruple heqat = 4 heqat = 1280 ro = 320 cuádruple ro. Por lo tanto,
- 100/10 heqat cuádruple = 10 heqat cuádruple
- 100/20 heqat cuádruple = 5 heqat cuádruple
- 100/30 cuádruple heqat = (3 + 1/4 + 1/16 + 1/64) cuádruple heqat + (1 + 2/3) cuádruple ro
- 100/40 cuádruple heqat = (2 + 1/2) cuádruple heqat
- 100/50 heqat cuádruple = 2 heqat cuádruple
- 100/60 cuádruple heqat = (1 + 1/2 + 1/8 + 1/32) cuádruple heqat + (3 + 1/3) cuádruple ro
- 100/70 cuádruple heqat = (1 + 1/4 + 1/8 + 1/32 + 1/64) cuádruple heqat + (2 + 1/14 + 1/21 + 1/42) cuádruple ro
- 100/80 cuádruple heqat = (1 + 1/4) cuádruple heqat
- 100/90 cuádruple heqat = (1 + 1/16 + 1/32 + 1/64) cuádruple heqat + (1/2 + 1/18) cuádruple ro
- 100/100 heqat cuádruple = 1 heqat cuádruple [2]
Áreas
Los problemas 48 a 55 muestran cómo calcular una variedad de áreas . El problema 48 es notable porque calcula sucintamente el área de un círculo aproximando π . Específicamente, el problema 48 refuerza explícitamente la convención (usada a lo largo de la sección de geometría) de que "el área de un círculo es igual a la del cuadrado que lo circunscribe en la proporción 64/81". De manera equivalente, el papiro se aproxima a π como 256/81, como ya se señaló anteriormente en la explicación del problema 41.
Otros problemas muestran cómo hallar el área de rectángulos, triángulos y trapezoides.
Pirámides
Los últimos seis problemas están relacionados con las pendientes de las pirámides . Un problema seked es informado por: [8]
- Si una pirámide tiene 250 codos de alto y el lado de su base 360 codos de largo, ¿cuál es su secado ? "
La solución al problema se da como la relación entre la mitad del lado de la base de la pirámide y su altura, o la relación entre el recorrido y el aumento de su cara. En otras palabras, la cantidad encontrada para el seked es la cotangente del ángulo a la base de la pirámide y su cara. [8]
Libro III - Miscelánea
La tercera parte del papiro de Rhind consta del resto de los 91 problemas, que son 61, 61B, 62-82, 82B, 83-84 y "números" 85-87, que son elementos que no son de naturaleza matemática. Esta sección final contiene tablas de datos más complicadas (que con frecuencia involucran fracciones del ojo de Horus), varios problemas pefsu que son problemas algebraicos elementales relacionados con la preparación de alimentos, e incluso un problema divertido (79) que sugiere progresiones geométricas, series geométricas y ciertas problemas posteriores y acertijos de la historia. El problema 79 cita explícitamente, "siete casas, 49 gatos, 343 ratones, 2401 mazorcas de espelta, 16807 hekats". En particular, el problema 79 se refiere a una situación en la que 7 casas contienen cada una siete gatos, que se comen siete ratones, cada uno de los cuales habría comido siete espigas de grano, cada uno de los cuales habría producido siete medidas de grano. La tercera parte del papiro de Rhind es, por tanto, una especie de mezcla, basada en lo que ya se ha presentado. El problema 61 tiene que ver con las multiplicaciones de fracciones. Mientras tanto, el problema 61B da una expresión general para calcular 2/3 de 1 / n, donde n es impar. En notación moderna, la fórmula dada es
La técnica dada en 61B está estrechamente relacionada con la derivación de la tabla 2 / n.
Los problemas 62 a 68 son problemas generales de naturaleza algebraica. Los problemas 69 a 78 son todos problemas pefsu de una forma u otra. Implican cálculos sobre la fuerza del pan y la cerveza, con respecto a ciertas materias primas utilizadas en su producción. [2]
El problema 79 suma cinco términos en una progresión geométrica . Su lenguaje sugiere fuertemente el acertijo y la canción de cuna más modernos " As I was going to St Ives ". [3] Los problemas 80 y 81 calculan las fracciones de ojo de Horus de hinu (o heqats). Los últimos cuatro elementos matemáticos, problemas 82, 82B y 83–84, calculan la cantidad de alimento necesaria para varios animales, como aves y bueyes. [2] Sin embargo, estos problemas, especialmente 84, están plagados de ambigüedad generalizada, confusión y simple inexactitud.
Los últimos tres elementos del papiro de Rhind se designan como "números" del 85 al 87, en oposición a los "problemas", y están esparcidos ampliamente por el reverso o reverso del papiro. Son, respectivamente, una pequeña frase que finaliza el documento (y tiene algunas posibilidades de traducción, que se indican a continuación), un trozo de papel de desecho que no está relacionado con el cuerpo del documento, que se utiliza para mantenerlo unido (pero que contiene palabras y fracciones egipcias que ahora son familiares para el lector del documento), y una pequeña nota histórica que se cree que fue escrita algún tiempo después de la finalización del cuerpo de la escritura del papiro. Se cree que esta nota describe los acontecimientos durante la " dominación de los hicsos ", un período de interrupción externa en la sociedad del antiguo Egipto que está estrechamente relacionado con su segundo período intermedio. Con estas erratas no matemáticas pero histórica y filológicamente intrigantes, la escritura del papiro llega a su fin.
Concordancia de la unidad
Gran parte del material del papiro de Rhind se refiere a las unidades de medida del Antiguo Egipto y especialmente al análisis dimensional utilizado para convertir entre ellas. En la imagen se muestra una concordancia de las unidades de medida utilizadas en el papiro.
Contenido
Esta tabla resume el contenido del Papiro Rhind mediante una paráfrasis moderna concisa. Se basa en la exposición en dos volúmenes del papiro que fue publicada por Arnold Buffum Chace en 1927 y en 1929. [7] En general, el papiro consta de cuatro secciones: una página de título, la tabla 2 / n, una minúscula "tabla 1–9 / 10" y 91 problemas o "números". Estos últimos están numerados del 1 al 87 e incluyen cuatro elementos matemáticos que los modernos han designado como problemas 7B, 59B, 61B y 82B. Mientras tanto, los números 85 a 87 no son elementos matemáticos que forman parte del cuerpo del documento, sino que son respectivamente: una pequeña frase que termina el documento, un trozo de "papel de borrador" que se utiliza para mantener el documento junto (que ya contiene escritura no relacionada), y una nota histórica que se cree que describe un período de tiempo poco después de la finalización del cuerpo del papiro. Estos tres últimos elementos están escritos en áreas dispares del verso (reverso) del papiro , lejos del contenido matemático. Por lo tanto, Chace los diferencia al diseñarlos como números en lugar de problemas , como los otros 88 elementos numerados.
Números de sección o problema | Declaración del problema o descripción | Solución o descripción | Notas |
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Pagina del titulo | Ahmes se identifica a sí mismo y a sus circunstancias históricas. | "Cálculo exacto. La entrada al conocimiento de todas las cosas existentes y todos los secretos oscuros. Este libro fue copiado en el año 33, en el cuarto mes de la temporada de inundaciones, bajo la majestad del rey del Alto y Bajo Egipto, 'A -user-Re ', dotado de vida, a semejanza de los antiguos escritos hechos en la época del rey del Alto y Bajo Egipto, Ne-ma'et-Re'. Es el escriba Ahmes quien copia este escrito ". | De la portada se desprende claramente que Ahmes identifica tanto su propio período como el período de un texto o textos más antiguos de los que se supone que copió, creando así el Papiro Rhind. El papiro tiene material escrito en ambos lados, es decir, en el anverso y el reverso . Vea la imagen para más detalles. |
Tabla 2 / n | Expresa cada uno de los cocientes desde 2/3 hasta 2/101 (donde el denominador siempre es impar) como fracciones egipcias . | Consulte el artículo de la tabla Rhind Mathematical Papyrus 2 / n para obtener un resumen y las soluciones de esta sección. | A lo largo del papiro, la mayoría de las soluciones se dan como representaciones fraccionarias egipcias particulares de un número real dado. Sin embargo, dado que cada número racional positivo tiene infinitas representaciones como fracción egipcia, estas soluciones no son únicas. También tenga en cuenta que la fracción 2/3 es la única excepción, que se usa además de los números enteros, que Ahmes usa junto con todas las fracciones unitarias racionales (positivas) para expresar las fracciones egipcias. Se puede decir que la tabla 2 / n sigue parcialmente un algoritmo (véase el problema 61B) para expresar 2 / n como una fracción egipcia de 2 términos, cuando n es compuesto. Sin embargo, este algoritmo incipiente se deja de lado en muchas situaciones cuando n es primo. El método de soluciones para la tabla 2 / n, por lo tanto, también sugiere comienzos de la teoría de números , y no meramente aritmética . |
Tabla 1–9 / 10 | Escribe los cocientes desde 1/10 hasta 9/10 como fracciones egipcias. |
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Problems 1–6 | 1, 2, 6, 7, 8 and 9 loaves of bread (respectively, in each problem) are divided among 10 men. In each case, represent each man's share of bread as an Egyptian fraction. |
| The first six problems of the papyrus are simple repetitions of the information already written in the 1–9/10 table, now in the context of story problems. |
7, 7B, 8–20 | Let and . Then for the following multiplications, write the product as an Egyptian fraction. |
| The same two multiplicands (here denoted as S and T) are used incessantly throughout these problems. Also note that Ahmes effectively writes the same problem thrice over (7, 7B, 10), sometimes approaching the same problem with different arithmetic work. |
21–38 | For each of the following linear equations with variable , solve for and express as an Egyptian fraction. |
| Note that problem 31 has an especially onerous solution. Although the statement of problems 21–38 can at times appear complicated (especially in Ahmes' prose), each problem ultimately reduces to a simple linear equation. In some cases, a unit of some kind has been omitted, being superfluous for these problems. These cases are problems 35–38, whose statements and "work" make the first mentions of units of volume known as a heqat and a ro (where 1 heqat = 320 ro), which will feature prominently throughout the rest of the papyrus. For the moment, however, their literal mention and usage in 35–38 is cosmetic. |
39 | 100 bread loaves will be distributed unequally among 10 men. 50 loaves will be divided equally among 4 men so that each of those 4 receives an equal share , while the other 50 loaves will be divided equally among the other 6 men so that each of those 6 receives an equal share . Find the difference of these two shares and express same as an Egyptian fraction. | In problem 39, the papyrus begins to consider situations with more than one variable. | |
40 | 100 loaves of bread are to be divided among five men. The men's five shares of bread are to be in arithmetic progression, so that consecutive shares always differ by a fixed difference, or . Furthermore, the sum of the three largest shares is to be equal to seven times the sum of the two smallest shares. Find and write it as an Egyptian fraction. | Problem 40 concludes the arithmetic/algebraic section of the papyrus, to be followed by the geometry section. After problem 40, there is even a large section of blank space on the papyrus, which visually indicates the end of the section. As for problem 40 itself, Ahmes works out his solution by first considering the analogous case where the number of loaves is 60 as opposed to 100. He then states that in this case the difference is 5 1/2 and that the smallest share is equal to one, lists the others, and then scales his work back up to 100 to produce his result. Although Ahmes does not state the solution itself as it has been given here, the quantity is implicitly clear once he has re-scaled his first step by the multiplication 5/3 x 11/2, to list the five shares (which he does). It bears mentioning that this problem can be thought of as having four conditions: a) five shares sum to 100, b) the shares range from smallest to largest, c) consecutive shares have a constant difference and d) the sum of the three larger shares is equal to seven times the sum of the smaller two shares. Beginning with the first three conditions only, one can use elementary algebra and then consider whether adding the fourth condition yields a consistent result. It happens that once all four conditions are in place, the solution is unique. The problem is therefore a more elaborate case of linear equation solving than what has gone before, verging on linear algebra. | |
41 | Use the volume formula
to calculate the volume of a cylindrical grain silo with a diameter of 9 cubits and a height of 10 cubits. Give the answer in terms of cubic cubits. Furthermore, given the following equalities among other units of volume, 1 cubic cubit = 3/2 khar = 30 heqats = 15/2 quadruple heqats, also express the answer in terms of khar and quadruple heqats. |
| This problem opens up the papyrus's geometry section, and also gives its first factually incorrect result (albeit with a very good approximation of , differing by less than one percent). Other ancient Egyptian volume units such as the quadruple heqat and the khar are later reported in this problem via unit conversion. Problem 41 is therefore also the first problem to treat significantly of dimensional analysis. |
42 | Reuse the volume formula and unit information given in 41 to calculate the volume of a cylindrical grain silo with a diameter of 10 cubits and a height of 10 cubits. Give the answer in terms of cubic cubits, khar, and hundreds of quadruple heqats, where 400 heqats = 100 quadruple heqats = 1 hundred-quadruple heqat, all as Egyptian fractions. |
| Problem 42 is effectively a repetition of 41, performing similar unit conversions at the end. However, although the problem does begin as stated, the arithmetic is considerably more involved, and certain of the given latter fractional terms are not actually present in the original document. However, the context is sufficient to fill in the gaps, and Chace has therefore taken license to add certain fractional terms in his mathematical translation (repeated here) which give rise to an internally consistent solution. |
43 | Use the volume formula
to calculate the volume of a cylindrical grain silo with a diameter of 9 cubits and a height of 6 cubits, directly finding the answer in Egyptian fractional terms of khar, and later in Egyptian fractional terms of quadruple heqats and quadruple ro, where 1 quadruple heqat = 4 heqat = 1280 ro = 320 quadruple ro. |
| Problem 43 represents the first serious mathematical mistake in the papyrus. Ahmes (or the source from which he may have been copying) attempted a shortcut in order to perform both the volume calculation and a unit conversion from cubic cubits to khar all in a single step, to avoid the need to use cubic cubits in an initial result. However, this attempt (which failed due to confusing part of the process used in 41 and 42 with that which was probably intended to be used in 43, giving consistent results by a different method) instead resulted in a new volume formula which is inconsistent with (and worse than) the approximation used in 41 and 42. |
44, 45 | One cubic cubit is equal to 15/2 quadruple heqats. Consider (44) a cubic grain silo with a length of 10 cubits on every edge. Express its volume in terms of quadruple heqats. On the other hand, (45) consider a cubic grain silo which has a volume of 7500 quadruple heqats, and express its edge length in terms of cubits. |
| Problem 45 is an exact reversal of problem 44, and they are therefore presented together here. |
46 | A rectangular prism-grain silo has a volume of 2500 quadruple heqats. Describe its three dimensions in terms of cubits. |
| This problem as stated has infinitely many solutions, but a simple choice of solution closely related to the terms of 44 and 45 is made. |
47 | Divide the physical volume quantity of 100 quadruple heqats by each of the multiples of 10, from 10 through 100. Express the results in Egyptian fractional terms of quadruple heqat and quadruple ro, and present the results in a table. |
| In problem 47, Ahmes is particularly insistent on representing more elaborate strings of fractions as Horus eye fractions, as far as he can. Compare problems 64 and 80 for similar preference of representation. To conserve space, "quadruple" has been shortened to "q." in all cases. |
48 | Compare the area of a circle with diameter 9 to that of its circumscribing square, which also has a side length of 9. What is the ratio of the area of the circle to that of the square? | The statement and solution of problem 48 make explicitly clear this preferred method of approximating the area of a circle, which had been used earlier in problems 41–43. However, it is erroneous. The original statement of problem 48 involves the usage of a unit of area known as the setat, which will shortly be given further context in future problems. For the moment, it is cosmetic. | |
49 | One khet is a unit of length, being equal to 100 cubits. Also, a "cubit strip" is a rectangular strip-measurement of area, being 1 cubit by 100 cubits, or 100 square cubits (or a physical quantity of equal area). Consider a rectangular plot of land measuring 10 khet by 1 khet. Express its area in terms of cubit strips. | - | |
50 | One square khet is a unit of area equal to one setat. Consider a circle with a diameter of 9 khet. Express its area in terms of setat. | Problem 50 is effectively a reinforcement of 48's 64/81 rule for a circle's area, which pervades the papyrus. | |
51 | A triangular tract of land has a base of 4 khet and an altitude of 10 khet. Find its area in terms of setat. | The setup and solution of 51 recall the familiar formula for calculating a triangle's area, and per Chace it is paraphrased as such. However, the papyrus's triangular diagram, previous mistakes, and translation issues present ambiguity over whether the triangle in question is a right triangle, or indeed if Ahmes actually understood the conditions under which the stated answer is correct. Specifically, it is unclear whether the dimension of 10 khet was meant as an altitude (in which case the problem is correctly worked as stated) or whether "10 khet" simply refers to a side of the triangle, in which case the figure would have to be a right triangle in order for the answer to be factually correct and properly worked, as done. These problems and confusions perpetuate themselves throughout 51–53, to the point where Ahmes seems to lose understanding of what he is doing, especially in 53. | |
52 | A trapezoidal tract of land has two bases, being 6 khet and 4 khet. Its altitude is 20 khet. Find its area in terms of setat. | Problem 52's issues are much the same as those of 51. The method of solution is familiar to moderns, and yet circumstances like those in 51 cast doubt over how well Ahmes or his source understood what they were doing. | |
53 | An isosceles triangle (a tract of land, say) has a base equal to 4 1/2 khet, and an altitude equal to 14 khet. Two line segments parallel to the base further partition the triangle into three sectors, being a bottom trapezoid, a middle trapezoid, and a top (similar) smaller triangle. The line segments cut the triangle's altitude at its midpoint (7) and further at a quarter-point (3 1/2) closer to the base, so that each trapezoid has an altitude of 3 1/2 khet, while the smaller similar triangle has an altitude of 7 khet. Find the lengths of the two line segments, where they are the shorter and the longer line segments respectively, and express them in Egyptian fractional terms of khet. Furthermore, find the areas of the three sectors, where they are the large trapezoid, the middle trapezoid, and the small triangle respectively, and express them in Egyptian fractional terms of setat and cubit strips. Use the fact that 1 setat = 100 cubit strips for unit conversions. |
| Problem 53, being more complex, is fraught with many of the same issues as 51 and 52—translation ambiguities and several numerical mistakes. In particular concerning the large bottom trapezoid, Ahmes seems to get stuck on finding the upper base, and proposes in the original work to subtract "one tenth, equal to 1 + 1/4 + 1/8 setat plus 10 cubit strips" from a rectangle being (presumably) 4 1/2 x 3 1/2 (khet). However, even Ahmes' answer here is inconsistent with the problem's other information. Happily the context of 51 and 52, together with the base, mid-line, and smaller triangle area (which are given as 4 + 1/2, 2 + 1/4 and 7 + 1/2 + 1/4 + 1/8, respectively) make it possible to interpret the problem and its solution as has been done here. The given paraphrase therefore represents a consistent best guess as to the problem's intent, which follows Chace. Ahmes also refers to the "cubit strips" again in the course of calculating for this problem, and we therefore repeat their usage here. It bears mentioning that neither Ahmes nor Chace explicitly give the area for the middle trapezoid in their treatments (Chace suggests that this is a triviality from Ahmes' point of view); liberty has therefore been taken to report it in a manner which is consistent with what Chace had thus far advanced. |
54 | There are 10 plots of land. In each plot, a sector is partitioned off such that the sum of the area of these 10 new partitions is 7 setat. Each new partition has equal area. Find the area of any one of these 10 new partitions, and express it in Egyptian fractional terms of setat and cubit strips. |
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55 | There are 5 plots of land. In each plot, a sector is partitioned off such that the sum of the area of these 5 new partitions is 3 setat. Each new partition has equal area. Find the area of any one of these 5 new partitions, and express it in Egyptian fractional terms of setat and cubit strips. |
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56 | 1) The unit of length known as a royal cubit is (and has been, throughout the papyrus) what is meant when we simply refer to a cubit. One royal cubit, or one cubit, is equal to seven palms, and one palm is equal to four fingers. In other words, the following equalities hold: 1 (royal) cubit = 1 cubit = 7 palms = 28 fingers. 2) Consider a right regular square pyramid whose base, the square face is coplanar with a plane (or the ground, say), so that any of the planes containing its triangular faces has the dihedral angle of with respect to the ground-plane (that is, on the interior of the pyramid). In other words, is the angle of the triangular faces of the pyramid with respect to the ground. The seked of such a pyramid, then, having altitude and base edge length , is defined as that physical length such that cot θ {\displaystyle \cot {\theta }} . Put another way, the seked of a pyramid can be interpreted as the ratio of its triangular faces' run per one unit (cubit) rise. Or, for the appropriate right triangle on a pyramid's interior having legs and the perpendicular bisector of a triangular face as the hypotenuse, then the pyramid's seked satisfies . Similar triangles are therefore described, and one can be scaled to the other. 3) A pyramid has an altitude of 250 (royal) cubits, and the side of its base has a length of 360 (royal) cubits. Find its seked in Egyptian fractional terms of (royal) cubits, and also in terms of palms. |
| Problem 56 is the first of the "pyramid problems" or seked problems in the Rhind papyrus, 56–59, 59B and 60, which concern the notion of a pyramid's facial inclination with respect to a flat ground. In this connection, the concept of a seked suggests early beginnings of trigonometry. Unlike modern trigonometry however, note especially that a seked is found with respect to some pyramid, and is itself a physical length measurement, which may be given in terms of any physical length units. For obvious reasons however, we (and the papyrus) confine our attention to situations involving ancient Egygtian units. We have also clarified that royal cubits are used throughout the papyrus, to differentiate them from "short" cubits which were used elsewhere in ancient Egypt. One "short" cubit is equal to six palms. |
57, 58 | The seked of a pyramid is 5 palms and 1 finger, and the side of its base is 140 cubits. Find (57) its altitude in terms of cubits. On the other hand, (58), a pyramid's altitude is 93 + 1/3 cubits, and the side of its base is 140 cubits. Find its seked and express it in terms of palms and fingers. |
| Problem 58 is an exact reversal of problem 57, and they are therefore presented together here. |
59, 59B | A pyramid's (59) altitude is 8 cubits, and its base length is 12 cubits. Express its seked in terms of palms and fingers. On the other hand, (59B), a pyramid's seked is five palms and one finger, and the side of its base is 12 cubits. Express its altitude in terms of cubits. |
| Problems 59 and 59B consider a case similar to 57 and 58, ending with familiar results. As exact reversals of each other, they are presented together here. |
60 | If a "pillar" (that is, a cone) has an altitude of 30 cubits, and the side of its base (or diameter) has a length of 15 cubits, find its seked and express it in terms of cubits. | Ahmes uses slightly different words to present this problem, which lend themselves to translation issues. However, the overall context of the problem, together with its accompanying diagram (which differs from the previous diagrams), leads Chace to conclude that a cone is meant. The notion of seked is easily generalized to the lateral face of a cone; he therefore reports the problem in these terms. Problem 60 concludes the geometry section of the papyrus. Moreover, it is the last problem on the recto (front side) of the document; all later content in this summary is present on the verso (back side) of the papyrus. The transition from 60 to 61 is thus both a thematic and physical shift in the papyrus. | |
61 | Seventeen multiplications are to have their products expressed as Egyptian fractions. The whole is to be given as a table. |
| The syntax of the original document and its repeated multiplications indicate a rudimentary understanding that multiplication is commutative. |
61B | Give a general procedure for converting the product of 2/3 and the reciprocal of any (positive) odd number 2n+1 into an Egyptian fraction of two terms, e.g. with natural p and q. In other words, find p and q in terms of n. |
| Problem 61B, and the method of decomposition that it describes (and suggests) is closely related to the computation of the Rhind Mathematical Papyrus 2/n table. In particular, every case in the 2/n table involving a denominator which is a multiple of 3 can be said to follow the example of 61B. 61B's statement and solution are also suggestive of a generality which most of the rest of the papyrus's more concrete problems do not have. It therefore represents an early suggestion of both algebra and algorithms. |
62 | A bag of three precious metals, gold, silver and lead, has been purchased for 84 sha'ty, which is a monetary unit. All three substances weigh the same, and a deben is a unit of weight. 1 deben of gold costs 12 sha'ty, 1 deben of silver costs 6 sha'ty, and 1 deben of lead costs 3 sha'ty. Find the common weight of any of the three metals in the bag. | Problem 62 becomes a division problem entailing a little dimensional analysis. Its setup involving standard weights renders the problem straightforward. | |
63 | 700 loaves are to be divided unevenly among four men, in four unequal, weighted shares. The shares will be in the respective proportions . Find each share. |
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64 | Recall that the heqat is a unit of volume. Ten heqat of barley are to be distributed among ten men in an arithmetic progression, so that consecutive men's shares have a difference of 1/8 heqats. Find the ten shares and list them in descending order, in Egyptian fractional terms of heqat. |
| Problem 64 is a variant of 40, this time involving an even number of unknowns. For quick modern reference apart from Egyptian fractions, the shares range from 25/16 down through 7/16, where the numerator decreases by consecutive odd numbers. The terms are given as Horus eye fractions; compare problems 47 and 80 for more of this. |
65 | 100 loaves of bread are to be unevenly divided among ten men. Seven of the men receive a single share, while the other three men, being a boatman, a foreman, and a door-keeper, each receive a double share. Express each of these two share amounts as Egyptian fractions. |
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66 | Recall that the heqat is a unit of volume and that one heqat equals 320 ro. 10 heqat of fat are distributed to one person over the course of one year (365 days), in daily allowances of equal amount. Express the allowance as an Egyptian fraction in terms of heqat and ro. | Problem 66 in its original form explicitly states that one year is equal to 365 days, and repeatedly uses the number 365 for its calculations. It is therefore primary historical evidence of the ancient Egyptian understanding of the year. | |
67 | A shepherd had a flock of animals, and had to give a portion of his flock to a lord as tribute. The shepherd was told to give two-thirds OF one-third of his original flock as tribute. The shepherd gave 70 animals. Find the size of the shepherd's original flock. | - | |
68 | Four overseers are in charge of four crews of men, being 12, 8, 6 and 4 men, respectively. Each crewman works at a fungible rate, to produce a single work-product: production (picking, say) of grain. Working on some interval of time, these four gangs collectively produced 100 units, or 100 quadruple heqats of grain, where each crew's work-product will be given to each crew's overseer. Express each crew's output in terms of quadruple heqat. |
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69 | 1) Consider cooking and food preparation. Suppose that there is a standardized way of cooking, or a production process, which will take volume units, specifically heqats of raw food-material (in particular, some one raw food-material) and produce units of some one finished food product. The pefsu of the (one) finished food product with respect to the (one) raw food-material, then, is defined as the quantity of finished food product units yielded from exactly one heqat of raw food material. In other words, . 2) 3 + 1/2 heqats of meal produce 80 loaves of bread. Find the meal per loaf in heqats and ro, and find the pefsu of these loaves with respect to the meal. Express them as Egyptian fractions. |
| Problem 69 begins the "pefsu" problems, 69–78, in the context of food preparation. Note that the notion of the pefsu assumes some standardized production process without accidents, waste, etc., and only concerns the relationship of one standardized finished food product to one particular raw material. That is, the pefsu is not immediately concerned with matters like production time, or (in any one given case) the relationship of other raw materials or equipment to the production process, etc. Still, the notion of the pefsu is another hint of abstraction in the papyrus, capable of being applied to any binary relationship between a food product (or finished good, for that matter) and a raw material. The concepts that the pefsu entails are thus typical of manufacturing. |
70 | (7 + 1/2 + 1/4 + 1/8) heqats of meal produce 100 loaves of bread. Find the meal per loaf in heqats and ro, and find the pefsu of these loaves with respect to the meal. Express them as Egyptian fractions. |
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71 | 1/2 heqats of besha, a raw material, produces exactly one full des-measure (glass) of beer. Suppose that there is a production process for diluted glasses of beer. 1/4 of the glass just described is poured out, and what has just been poured out is captured and re-used later. This glass, which is now 3/4 full, is then diluted back to capacity with water, producing exactly one full diluted glass of beer. Find the pefsu of these diluted beer glasses with respect to the besha as an Egyptian fraction. | Note that Problem 71 describes intermediate steps in a production process, as well as a second raw material, water. Further note that these are irrelevant to the relationship between the finished unit and the raw material (besha in this case). | |
72 | 100 bread loaves "of pefsu 10" are to be evenly exchanged for loaves "of pefsu 45". Find . | Now that the concept of the pefsu has been established, problems 72–78 explore even exchanges of different heaps of finished foods, having different pefsu. In general however, they assume a common raw material of some kind. Specifically, the common raw material assumed throughout all of 72–78 is called wedyet flour, which is even implicated in the production of beer, so that beer can be exchanged for bread in the latter problems. 74's original statement also mentions "Upper Egyptian barley", but for our purposes this is cosmetic. What problems 72–78 say, then, is really this: equal amounts of raw material are used in two different production processes, to produce two different units of finished food, where each type has a different pefsu. One of the two finished food units is given. Find the other. This can be accomplished by dividing both units (known and unknown) by their respective pefsu, where the units of finished food vanish in dimensional analysis, and only the same raw material is considered. One can then easily solve for x. 72–78 therefore really require that x be given so that equal amounts of raw material are used in two different production processes. | |
73 | 100 bread loaves of pefsu 10 are to be evenly exchanged for loaves of pefsu 15. Find . | - | |
74 | 1000 bread loaves of pefsu 5 are to be divided evenly into two heaps of 500 loaves each. Each heap is to be evenly exchanged for two other heaps, one of loaves of pefsu 10, and the other of loaves of pefsu 20. Find and . |
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75 | 155 bread loaves of pefsu 20 are to be evenly exchanged for loaves of pefsu 30. Find . | - | |
76 | 1000 bread loaves of pefsu 10, one heap, will be evenly exchanged for two other heaps of loaves. The other two heaps each has an equal number of loaves, one being of pefsu 20, the other of pefsu 30. Find . | - | |
77 | 10 des-measure of beer, of pefsu 2, are to be evenly exchanged for bread loaves, of pefsu 5. Find . | - | |
78 | 100 bread loaves of pefsu 10 are to be evenly exchanged for des-measures of beer of pefsu 2. Find . | - | |
79 | An estate's inventory consists of 7 houses, 49 cats, 343 mice, 2401 spelt plants (a type of wheat), and 16807 units of heqat (of whatever substance—a type of grain, suppose). List the items in the estates' inventory as a table, and include their total. |
| Problem 79 has been presented in its most literal interpretation. However, the problem is among the most interesting in the papyrus, as its setup and even method of solution suggests Geometric progression (that is, geometric sequences), elementary understanding of finite series, as well as the St. Ives problem—even Chace cannot help interrupting his own narrative in order to compare problem 79 with the St. Ives nursery rhyme. He also indicates that a suspiciously familiar third instance of these types of problems is to be found in Fibonacci's Liber Abaci. Chace suggests the interpretation that 79 is a kind of savings example, where a certain amount of grain is saved by keeping cats on hand to kill the mice which would otherwise eat the spelt used to make the grain. In the original document, the 2401 term is written as 2301 (an obvious mistake), while the other terms are given correctly; it is therefore corrected here. Moreover, one of Ahmes' methods of solution for the sum suggests an understanding of finite geometric series. Ahmes performs a direct sum, but he also presents a simple multiplication to get the same answer: "2801 x 7 = 19607". Chace explains that since the first term, the number of houses (7) is equal to the common ratio of multiplication (7), then the following holds (and can be generalized to any similar situation):
That is, when the first term of a geometric sequence is equal to the common ratio, partial sums of geometric sequences, or finite geometric series, can be reduced to multiplications involving the finite series having one less term, which does prove convenient in this case. In this instance then, Ahmes simply adds the first four terms of the sequence (7 + 49 + 343 + 2401 = 2800) to produce a partial sum, adds one (2801), and then simply multiplies by 7 to produce the correct answer. |
80 | The hinu is a further unit of volume such that one heqat equals ten hinu. Consider the situations where one has a Horus eye fraction of heqats, and express their conversions to hinu in a table. |
| Compare problems 47 and 64 for other tabular information with repeated Horus eye fractions. |
81 | Perform "another reckoning of the hinu." That is, express an assortment of Egyptian fractions, many terms of which are also Horus eye fractions, in various terms of heqats, hinu, and ro. | Problem 81's main section is a much larger conversion table of assorted Egyptian fractions, which expands on the idea of problem 80—indeed, it represents one of the largest tabular forms in the entire papyrus. The first part of problem 81 is an exact repetition of the table in problem 80, without the first row which states that 1 heqat = 10 hinu; it is therefore not repeated here. The second part of problem 81, or its "body", is the large table which is given here. The attentive reader will notice two things: several rows repeat identical information, and several forms (but not all) given in both of the "heqat" areas on either side of the table are in fact identical. There are two points worth mentioning, to explain why the table looks the way that it does. For one thing, Ahmes does in fact exactly repeat certain groups of information in different areas of the table, and they are accordingly repeated here. On the other hand, Ahmes also starts out with certain "left-hand" heqat forms, and makes some mistakes in his early calculations. However, in many cases he corrects these mistakes later in his writing of the table, producing a consistent result. Since the present information is simply a re-creation of Chace's translation and interpretation of the papyrus, and since Chace elected to interpret and correct Ahmes' mistakes by substituting the later correct information in certain earlier rows, thereby fixing Ahmes' mistakes and also therefore repeating information in the course of translation, this method of interpretation explains the duplication of information in certain rows. As for the duplication of information in certain columns (1/4 heqat = ... = 1/4 heqat, etc.), this seems simply to have been a convention that Ahmes filled in while considering certain important Horus-eye fractional ratios from both the standpoint of the hinu, and also of the heqat (and their conversions). In short, the various repetitions of information are the result of choices made by Ahmes, his potential source document, and the editorial choices of Chace, in order to present a mathematically consistent translation of the larger table in problem 81. | |
82 | Estimate in wedyet-flour, made into bread, the daily portion of feed for ten fattening geese. To do this, perform the following calculations, expressing the quantities in Egyptian fractional terms of hundreds of heqats, heqats and ro, except where specified otherwise: Begin with the statement that "10 fattening geese eat 2 + 1/2 heqats in one day". In other words, the daily rate of consumption (and initial condition) is equal to 2 + 1/2. Determine the number of heqats which 10 fattening geese eat in 10 days, and in 40 days. Call these quantities and , respectively. Multiply the above latter quantity by 5/3 to express the amount of "spelt", or , required to be ground up. Multiply by 2/3 to express the amount of "wheat", or , required. Divide by 10 to express a "portion of wheat", or , which is to be subtracted from . Find . This is the amount of "grain", (or wedyet flour, it would seem), which is required to make the feed for geese, presumably on the interval of 40 days (which would seem to contradict the original statement of the problem, somewhat). Finally, express again in terms of hundreds of double heqats, double heqats and double ro, where 1 hundred double heqat = 2 hundred heqat = 100 double heqat = 200 heqat = 32000 double ro = 64000 ro. Call this final quantity . |
| Beginning with problem 82, the papyrus becomes increasingly difficult to interpret (owing to mistakes and missing information), to the point of unintelligibility. However, it is yet possible to make some sense of 82. Simply put, there seem to exist established rules, or good estimates, for fractions to be taken of this-or-that food material in a cooking or production process. Ahmes' 82 simply gives expression to some of these quantities, in what is after all declared in the original document to be an "estimate", its somewhat contradictory and confused language notwithstanding. In addition to their strangeness, problems 82, 82B, 83 and 84 are also notable for continuing the "food" train of thought of the recent pefsu problems, this time considering how to feed animals instead of people. Both 82 and 82B make use of the "hundred heqat" unit with regard to t and f; these conventions are cosmetic, and not repeated here. Licence is also taken throughout these last problems (per Chace) to fix numerical mistakes of the original document, to attempt to present a coherent paraphrase. |
82B | Estimate the amount of feed for other geese. That is, consider a situation which is identical to problem 82, with the single exception that the initial condition, or daily rate of consumption, is exactly half as large. That is, let = 1 + 1/4. Find , and especially by using elementary algebra to skip the intermediate steps. |
| Problem 82B is presented in parallel with problem 82, and quickly considers the identical situation where the associated quantities are halved. In both cases, it appears that Ahmes' real goal is to find g_2. Now that he has a "procedure", he feels free to skip 82's onerous steps. One could simply observe that the division by two carries through the entire problem's work, so that g_2 is also exactly half as large as in problem 82. A slightly more thorough approach using elementary algebra would be to backtrack the relationships between the quantities in 82, make the essential observation that g = 14/15 x f, and then perform the unit conversions to transform g into g_2. |
83 | Estimate the feed for various kinds of birds. This is a "problem" with multiple components, which can be interpreted as a series of remarks: Suppose that four geese are cooped up, and their collective daily allowance of feed is equal to one hinu. Express one goose's daily allowance of feed in terms of heqats and ro. Suppose that the daily feed for a goose "that goes into the pond" is equal to 1/16 + 1/32 heqats + 2 ro. Express this same daily allowance in terms of hinu. Suppose that the daily allowance of feed for 10 geese is one heqat. Find the 10-day allowance and the 30-day, or one-month allowance for the same group of animals, in heqats. Finally a table will be presented, giving daily feed portions to fatten one animal of any of the indicated species. |
| Since problem 83's various items are concerned with unit conversions between heqats, ro and hinu, in the spirit of 80 and 81, it is natural to wonder what the table's items become when converted to hinu. The portion shared by the goose, terp-goose and crane is equal to 5/3 hinu, the set-ducks' portion is equal to 1/2 hinu, the ser-gooses' portion is equal to 1/4 hinu (compare the first item in the problem), and the portion shared by the dove and quail is equal to 1/16 + 1/32 hinu. The presence of various Horus eye fractions is familiar from the rest of the papyrus, and the table seems to consider feed estimates for birds, ranging from largest to smallest. The "5/3 hinu" portions at the top of the table, specifically its factor of 5/3, reminds one of the method for finding s in problem 82. Problem 83 makes mention of "Lower-Egyptian grain", or barley, and it also uses the "hundred-heqat" unit in one place; these are cosmetic, and left out of the present statement. |
84 | Estimate the feed for a stable of oxen. |
| 84 is the last problem, or number, comprising the mathematical content of the Rhind papyrus. With regard to 84 itself, Chace echoes Peet: "One can only agree with Peet that 'with this problem the papyrus reaches its limit of unintelligibility and inaccuracy.'"(Chace, V.2, Problem 84). Here, instances of the "hundred heqat" unit have been expressed by "c. heqat" in order to conserve space. The three "cattle" mentioned are described as "common" cattle, to differentiate them from the other animals, and the two headers concerning loaves and "common food" are with respect to heqats. The "fine oxen" at the table's beginning are described as Upper Egyptian oxen, a phrase also removed here for space reasons. Problem 84 seems to suggest a procedure to estimate various food materials and allowances in similar terms as the previous three problems, but the extant information is deeply confused. Still, there are hints of consistency. The problem seems to start out like a conventional story problem, describing a stable with ten animals of four different types. It seems that the four types of animals consume feed, or "loaves" at different rates, and that there are corresponding amounts of "common" food. These two columns of information are correctly summed in the "total" row, however they are followed by two "spelt" items of dubious relationship to the above. These two spelt items are indeed each multiplied by ten to give the two entries in the "10 days" row, once unit conversions are accounted for. The "one month" row items do not seem to be consistent with the previous two, however. Finally, information in "double heqats" (read hundred double heqats, double heqats and double ro for these items) concludes the problem, in a manner reminiscent of 82 and 82B. The two items in the final row are in roughly, but not exactly, the same proportion to one another as the two items in the "one month" row. |
Number 85 | A small group of cursive hieroglyphic signs is written, which Chace suggests may represent the scribe "trying his pen". It appears to be a phrase or sentence of some kind, and two translations are suggested. 1) "Kill vermin, mice, fresh weeds, numerous spiders. Pray the god Re for warmth, wind and high water." 2) Interpret this strange matter, which the scribe wrote ... according to what he knew." | The remaining items 85, 86 and 87, being various errata which are not mathematical in nature, are therefore styled by Chace as "numbers" as opposed to problems. They are also located on areas of the papyrus which are well away from the body of the writing, which had just ended with Problem 84. Number 85, for example, is some distance away from problem 84 on the verso—but not too far away. Its placement on the papyrus therefore suggests a kind of coda, in which case the latter translation, which Chace describes as an example of the "enigmatic writing" interpretation of ancient Egyptian documents, seems most appropriate to its context in the document. | |
Number 86 | Number 86 seems to be from some account, or memorandum, and lists an assortment of goods and quantities, using words familiar from the context of the rest of the papyrus itself. [The original text is a series of lines of writing, which are therefore numbered in the following.] | "1... living forever. List of the food in Hebenti... 2... his brother the steward Ka-mose... 3... of his year, silver, 50 pieces twice in the year... 4... cattle 2, in silver 3 pieces in the year... 5... one twice; that is, 1/6 and 1/6. Now as for one... 6... 12 hinu; that is, silver, 1/4 piece; one... 7... (gold or silver) 5 pieces, their price therefor; fish, 120, twice... 8... year, barley, in quadruple heqat, 1/2 + 1/4 of 100 heqat 15 heqat; spelt, 100 heqat... heqat... 9... barley, in quadruple heqat, 1/2 + 1/4 of 100 heqat 15 heqat; spelt, 1 + 1/2 + 1/4 times 100 heqat 17 heqat... 10... 146 + 1/2; barley, 1 + 1/2 + 1/4 times 100 heqat 10 heqat; spelt, 300 heqat... heqat... 11... 1/2, there was brought wine, 1 ass(load?)... 12... silver 1/2 piece; ... 4; that is, in silver... 13... 1 + 1/4; fat, 36 hinu; that is, in silver... 14... 1 + 1/2 + 1/4 times 100 heqat 21 heqat; spelt, in quadruple heqat, 400 heqat 10 heqat... 15-18 (These lines are repetitions of line 14.)" | Chace indicates that number 86 was pasted onto the far left side of the verso (opposite the later geometry problems on the recto), to strengthen the papyrus. Number 86 can therefore be interpreted as a piece of "scrap paper". |
Number 87 | Number 87 is a brief account of certain events. Chace indicates an (admittedly now dated and possibly changed) scholarly consensus that 87 was added to the papyrus not long after the completion of its mathematical content. He goes on to indicate that the events described in it "took place during the period of the Hyksos domination." | "Year 11, second month of the harvest season. Heliopolis was entered. The first month of the inundation season, 23rd day, the commander (?) of the army (?) attacked (?) Zaru. 25th day, it was heard that Zaru was entered. Year 11, first month of the inundation season, third day. Birth of Set; the majesty of this god caused his voice to be heard. Birth of Isis, the heavens rained." | Number 87 is located toward the middle of the verso, surrounded by a large, blank, unused space. |
Ver también
- List of ancient Egyptian papyri
- Ahmes
- Akhmim wooden tablet
- Ancient Egyptian units of measurement
- As I was going to St. Ives
- Berlin Papyrus 6619
- Arnold Buffum Chace
- Egyptian fraction
- Egyptian Mathematical Leather Roll
- Eye of Horus
- History of mathematics
- Lahun Mathematical Papyri
- Moscow Mathematical Papyrus
- Alexander Henry Rhind
- Rhind Mathematical Papyrus 2/n table
- Seked
Bibliografía
- Chace, Arnold Buffum; et al. (1927). The Rhind Mathematical Papyrus. 1. Oberlin, Ohio: Mathematical Association of America – via Internet Archive.
- Chace, Arnold Buffum; et al. (1929). The Rhind Mathematical Papyrus. 2. Oberlin, Ohio: Mathematical Association of America – via Internet Archive.
- Gillings, Richard J. (1972). Mathematics in the Time of the Pharaohs (Dover reprint ed.). MIT Press. ISBN 0-486-24315-X.
- Robins, Gay; Shute, Charles (1987). The Rhind Mathematical Papyrus: an Ancient Egyptian Text. London: British Museum Publications Limited. ISBN 0-7141-0944-4.
Referencias
- ^ "The Rhind Mathematical Papyrus". britishmuseum.org. Retrieved 2017-09-18.
- ^ a b c d e f Clagett, Marshall (1999). Ancient Egyptian Science, A Source Book. Memoirs of the American Philosophical Society. Volume Three: Ancient Egyptian Mathematics. American Philosophical Society. ISBN 978-0-87169-232-0.
|volume=
has extra text (help) - ^ a b c d e f Spalinger, Anthony (1990). "The Rhind Mathematical Papyrus as a Historical Document". Studien zur Altägyptischen Kultur. Helmut Buske Verlag. 17: 295–337. JSTOR 25150159.
- ^ "Collections: Egyptian, Classical, Ancient Near Eastern Art: Fragments of Rhind Mathematical Papyrus". Brooklyn Museum. Retrieved November 1, 2012.
- ^ cf. Schneider, Thomas (2006). "The Relative Chronology of the Middle Kingdom and the Hyksos Period (Dyns. 12–17)". In Hornung, Erik; Krauss, Rolf; Warburton, David (eds.). Ancient Egyptian Chronology. Handbook of Oriental Studies. Brill. pp. 194–195.
- ^ Peet, Thomas Eric (1923). The Rhind Mathematical Papyrus, British Museum 10057 and 10058. London: The University Press of Liverpool limited and Hodder & Stoughton limited.
- ^ a b Chace, Arnold Buffum (1979) [1927–29]. The Rhind Mathematical Papyrus: Free Translation and Commentary with Selected Photographs, Translations, Transliterations and Literal Translations. Classics in Mathematics Education. 8. 2 vols (Reston: National Council of Teachers of Mathematics Reprinted ed.). Oberlin: Mathematical Association of America. ISBN 0-87353-133-7.
- ^ a b Maor, Eli (1998). Trigonometric Delights. Princeton University Press. p. 20. ISBN 0-691-09541-8.
enlaces externos
- Allen, Don. April 2001. The Ahmes Papyrus and Summary of Egyptian Mathematics.
- Egypt/Texts at Curlie
- British Museum webpage on the Papyrus at the Wayback Machine (archived June 29, 2015).
- O'Connor and Robertson, 2000. Mathematics in Egyptian Papyri.
- Truman State University, Math and Computer Science Division. Mathematics and the Liberal Arts: The Rhind/Ahmes Papyrus at the Wayback Machine (archived January 5, 2013).
- "Rhind Papyrus". MathWorld–A Wolfram Web Resource.
- Williams, Scott W. Mathematicians of the African Diaspora, containing a page on Egyptian Mathematics Papyri.
- BBC audio file A History of the World in 100 Objects. (15 mins)
Preceded by 16: Flood tablet | A History of the World in 100 Objects Object 17 | Succeeded by 18: Minoan Bull-leaper |