The Rhind Mathematical Papyrus ( RMP ; also designated as papyrus British Museum 10057, pBM 10058, and Brooklyn Museum 37.1784Ea-b) is one of the best known examples of ancient Egyptian mathematics .
41-708: Egyptian scribe [REDACTED] A portion of the Rhind Mathematical Papyrus Ahmes ( Ancient Egyptian : jꜥḥ-ms “, a common Egyptian name also transliterated Ahmose ) was an ancient Egyptian scribe who lived towards the end of the Fifteenth Dynasty (and of the Second Intermediate Period ) and the beginning of the Eighteenth Dynasty (and of
82-766: A Scottish antiquarian, purchased two parts of the papyrus in 1858 in Luxor, Egypt ; it was stated to have been found in "one of the small buildings near the Ramesseum ", near Luxor. The British Museum, where the majority of the papyrus is now kept, acquired it in 1865 along with the Egyptian Mathematical Leather Roll , also owned by Henry Rhind. Fragments of the text were independently purchased in Luxor by American Egyptologist Edwin Smith in
123-465: A "quadruple ro". The quadruple heqat and the quadruple ro are units of volume derived from the simpler heqat and ro, such that these four units of volume satisfy the following relationships: 1 quadruple heqat = 4 heqat = 1280 ro = 320 quadruple ro. Thus, Problems 48–55 show how to compute an assortment of areas . Problem 48 is notable in that it succinctly computes the area of a circle by approximating π . Specifically, problem 48 explicitly reinforces
164-507: A certain number of loaves of bread by 10 men and record the outcome in unit fractions. Problems 7–20 show how to multiply the expressions 1 + 1/2 + 1/4 = 7/4, and 1 + 2/3 + 1/3 = 2 by different fractions. Problems 21–23 are problems in completion, which in modern notation are simply subtraction problems. Problems 24–34 are ‘‘aha’’ problems; these are linear equations . Problem 32 for instance corresponds (in modern notation) to solving x + 1/3 x + 1/4 x = 2 for x. Problems 35–38 involve divisions of
205-594: A collection of 21 arithmetic and 20 algebraic problems. The problems start out with simple fractional expressions, followed by completion ( sekem ) problems and more involved linear equations ( aha problems ). The first part of the papyrus is taken up by the 2/ n table . The fractions 2/ n for odd n ranging from 3 to 101 are expressed as sums of unit fractions . For example, 2 15 = 1 10 + 1 30 {\displaystyle {\frac {2}{15}}={\frac {1}{10}}+{\frac {1}{30}}} . The decomposition of 2/ n into unit fractions
246-574: A handful of these stand out. The Rhind Papyrus was published in 1923 by the English Egyptologist T. Eric Peet and contains a discussion of the text that followed Francis Llewellyn Griffith 's Book I, II and III outline. Chace published a compendium in 1927–29 which included photographs of the text. A more recent overview of the Rhind Papyrus was published in 1987 by Robins and Shute. The Rhind Mathematical Papyrus dates to
287-434: A title page, the 2/n table, a tiny "1–9/10 table", and 91 problems, or "numbers". The latter are numbered from 1 through 87 and include four mathematical items which have been designated by moderns as problems 7B, 59B, 61B, and 82B. Numbers 85–87, meanwhile, are not mathematical items forming part of the body of the document, but instead are respectively: a small phrase ending the document, a piece of "scrap-paper" used to hold
328-633: Is a well-known mathematical papyrus, usually referenced together with the Rhind Mathematical Papyrus . The Moscow Mathematical Papyrus is older than the Rhind Mathematical Papyrus, while the latter is the larger of the two. The problems in the Moscow Papyrus follow no particular order, and the solutions of the problems provide much less detail than those in the Rhind Mathematical Papyrus . The papyrus
369-484: Is different from Wikidata Articles containing Ancient Egyptian-language text Rhind Mathematical Papyrus It is one of two well-known mathematical papyri, along with the Moscow Mathematical Papyrus . The Rhind Papyrus is the larger, but younger, of the two. In the papyrus' opening paragraphs Ahmes presents the papyrus as giving "Accurate reckoning for inquiring into things, and
410-572: Is never more than 4 terms long as in for example: This table is followed by a much smaller, tiny table of fractional expressions for the numbers 1 through 9 divided by 10. For instance the division of 7 by 10 is recorded as: After these two tables, the papyrus records 91 problems altogether, which have been designated by moderns as problems (or numbers) 1–87, including four other items which have been designated as problems 7B, 59B, 61B and 82B. Problems 1–7, 7B and 8–40 are concerned with arithmetic and elementary algebra. Problems 1–6 compute divisions of
451-780: Is reported as follows: The solution to the problem is given as the ratio of half the side of the base of the pyramid to its height, or the run-to-rise ratio of its face. In other words, the quantity found for the seked is the cotangent of the angle to the base of the pyramid and its face. The third part of the Rhind papyrus consists of the remainder of the 91 problems, being 61, 61B, 62–82, 82B, 83–84, and "numbers" 85–87, which are items that are not mathematical in nature. This final section contains more complicated tables of data (which frequently involve Horus eye fractions), several pefsu problems which are elementary algebraic problems concerning food preparation, and even an amusing problem (79) which
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#1732776015182492-452: Is suggestive of geometric progressions, geometric series, and certain later problems and riddles in history. Problem 79 explicitly cites, "seven houses, 49 cats, 343 mice, 2401 ears of spelt, 16807 hekats." In particular problem 79 concerns a situation in which 7 houses each contain seven cats, which all eat seven mice, each of which would have eaten seven ears of grain, each of which would have produced seven measures of grain. The third part of
533-422: Is thought to describe events during the " Hyksos domination", a period of external interruption in ancient Egyptian society which is closely related with its second intermediary period. With these non-mathematical yet historically and philologically intriguing errata, the papyrus's writing comes to an end. Much of the Rhind Papyrus's material is concerned with Ancient Egyptian units of measurement and especially
574-513: Is well known for some of its geometry problems. Problems 10 and 14 compute a surface area and the volume of a frustum respectively. The remaining problems are more common in nature. Problems 2 and 3 are ship's part problems. One of the problems calculates the length of a ship's rudder and the other computes the length of a ship's mast given that it is 1/3 + 1/5 of the length of a cedar log originally 30 cubits long. Aha problems involve finding unknown quantities (referred to as aha , "stack") if
615-682: The Golenishchev Mathematical Papyrus after its first non-Egyptian owner, Egyptologist Vladimir Golenishchev , is an ancient Egyptian mathematical papyrus containing several problems in arithmetic , geometry , and algebra . Golenishchev bought the papyrus in 1892 or 1893 in Thebes . It later entered the collection of the Pushkin State Museum of Fine Arts in Moscow , where it remains today. Based on
656-714: The New Kingdom ). He transcribed the Rhind Mathematical Papyrus , a work of ancient Egyptian mathematics that dates to approximately 1550 BC; he is the earliest contributor to mathematics whose name is known. Ahmes claimed not to be the writer of the work but rather just the scribe. He claimed the material came from an even older document from around 2000 B.C. See also [ edit ] List of ancient Egyptian scribes References [ edit ] ^ "The Rhind Mathematical Papyrus" . britishmuseum.org . Retrieved 2017-09-18 . ^ Pickover, Clifford A. (2009), The Math Book: From Pythagoras to
697-551: The Second Intermediate Period of Egypt . It was copied by the scribe Ahmes (i.e., Ahmose; Ahmes is an older transcription favoured by historians of mathematics) from a now-lost text from the reign of the 12th dynasty king Amenemhat III . It dates to around 1550 BC. The document is dated to Year 33 of the Hyksos king Apophis and also contains a separate later historical note on its verso likely dating from "Year 11" of his successor, Khamudi . Alexander Henry Rhind ,
738-620: The palaeography and orthography of the hieratic text, the text was most likely written down in the 13th Dynasty and based on older material probably dating to the Twelfth Dynasty of Egypt , roughly 1850 BC. Approximately 5.5 m (18 ft) long and varying between 3.8 and 7.6 cm (1.5 and 3 in) wide, its format was divided by the Soviet Orientalist Vasily Vasilievich Struve in 1930 into 25 problems with solutions. It
779-530: The volume of a truncated pyramid : where a and b are the base and top side lengths of the truncated pyramid and h is the height. Researchers have speculated how the Egyptians might have arrived at the formula for the volume of a frustum but the derivation of this formula is not given in the papyrus. Richard J. Gillings gave a cursory summary of the Papyrus' contents. Numbers with overlines denote
820-1106: The 57th Dimension, 250 Milestones in the History of Mathematics , Sterling Publishing Company, Inc., p. 36, ISBN 978-1-4027-5796-9 . ^ Derbyshire, John (2006), Unknown quantity: a real and imaginary history of algebra , National Academies Press, p. 29 , ISBN 978-0-309-09657-7 . ^ O'Connor, John J.; Robertson, Edmund F. , "Ahmes" , MacTutor History of Mathematics Archive , University of St Andrews External links [ edit ] The Ahmes Papyrus Authority control databases [REDACTED] International VIAF FAST National United States Norway Israel Other IdRef Retrieved from " https://en.wikipedia.org/w/index.php?title=Ahmes&oldid=1254562738 " Categories : 16th-century BC people Ancient Egyptian scribes Ancient Egyptian mathematicians Hidden categories: Articles with short description Short description
861-472: The Moscow Mathematical calculates the volume of a frustum . Problem 14 states that a pyramid has been truncated in such a way that the top area is a square of length 2 units, the bottom a square of length 4 units, and the height 6 units, as shown. The volume is found to be 56 cubic units, which is correct. The text of the example runs like this: "If you are told: a truncated pyramid of 6 for
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#1732776015182902-538: The Rhind papyrus is therefore a kind of miscellany, building on what has already been presented. Problem 61 is concerned with multiplications of fractions. Problem 61B, meanwhile, gives a general expression for computing 2/3 of 1/n, where n is odd. In modern notation the formula given is The technique given in 61B is closely related to the derivation of the 2/n table. Problems 62–68 are general problems of an algebraic nature. Problems 69–78 are all pefsu problems in some form or another. They involve computations regarding
943-413: The amount of feed necessary for various animals, such as fowl and oxen. However, these problems, especially 84, are plagued by pervasive ambiguity, confusion, and simple inaccuracy. The final three items on the Rhind papyrus are designated as "numbers" 85–87, as opposed to "problems", and they are scattered widely across the papyrus's back side, or verso. They are, respectively, a small phrase which ends
984-400: The area of a semi-cylinder (Peet). Below we assume that the problem refers to the area of a hemisphere. The text of problem 10 runs like this: "Example of calculating a basket. You are given a basket with a mouth of 4 1/2. What is its surface? Take 1/9 of 9 (since) the basket is half an egg-shell. You get 1. Calculate the remainder which is 8. Calculate 1/9 of 8. You get 2/3 + 1/6 + 1/18. Find
1025-423: The convention (used throughout the geometry section) that "a circle's area stands to that of its circumscribing square in the ratio 64/81." Equivalently, the papyrus approximates π as 256/81, as was already noted above in the explanation of problem 41. Other problems show how to find the area of rectangles, triangles and trapezoids. The final six problems are related to the slopes of pyramids . A seked problem
1066-413: The dimensional analysis used to convert between them. A concordance of units of measurement used in the papyrus is given in the image. This table summarizes the content of the Rhind Papyrus by means of a concise modern paraphrase. It is based upon the two-volume exposition of the papyrus which was published by Arnold Buffum Chace in 1927, and in 1929. In general, the papyrus consists of four sections:
1107-406: The document (and has a few possibilities for translation, given below), a piece of scrap paper unrelated to the body of the document, used to hold it together (yet containing words and Egyptian fractions which are by now familiar to a reader of the document), and a small historical note which is thought to have been written some time after the completion of the body of the papyrus's writing. This note
1148-421: The document together (having already contained unrelated writing), and a historical note which is thought to describe a time period shortly after the completion of the body of the papyrus. These three latter items are written on disparate areas of the papyrus's verso (back side), far away from the mathematical content. Chace therefore differentiates them by styling them as numbers as opposed to problems , like
1189-4510: The following multiplications, write the product as an Egyptian fraction. 9 : ( 1 2 + 1 14 ) S = 1 ; 10 : ( 1 4 + 1 28 ) S = 1 2 ; 11 : 1 7 S = 1 4 {\displaystyle 9:{\bigg (}{\frac {1}{2}}+{\frac {1}{14}}{\bigg )}S=1\;\;\;;\;\;\;10:{\bigg (}{\frac {1}{4}}+{\frac {1}{28}}{\bigg )}S={\frac {1}{2}}\;\;\;;\;\;\;11:{\frac {1}{7}}S={\frac {1}{4}}} 12 : 1 14 S = 1 8 ; 13 : ( 1 16 + 1 112 ) S = 1 8 ; 14 : 1 28 S = 1 16 {\displaystyle 12:{\frac {1}{14}}S={\frac {1}{8}}\;\;\;;\;\;\;13:{\bigg (}{\frac {1}{16}}+{\frac {1}{112}}{\bigg )}S={\frac {1}{8}}\;\;\;;\;\;\;14:{\frac {1}{28}}S={\frac {1}{16}}} 15 : ( 1 32 + 1 224 ) S = 1 16 ; 16 : 1 2 T = 1 ; 17 : 1 3 T = 2 3 {\displaystyle 15:{\bigg (}{\frac {1}{32}}+{\frac {1}{224}}{\bigg )}S={\frac {1}{16}}\;\;\;;\;\;\;16:{\frac {1}{2}}T=1\;\;\;;\;\;\;17:{\frac {1}{3}}T={\frac {2}{3}}} 18 : 1 6 T = 1 3 ; 19 : 1 12 T = 1 6 ; 20 : 1 24 T = 1 12 {\displaystyle 18:{\frac {1}{6}}T={\frac {1}{3}}\;\;\;;\;\;\;19:{\frac {1}{12}}T={\frac {1}{6}}\;\;\;;\;\;\;20:{\frac {1}{24}}T={\frac {1}{12}}} 22 : ( 2 3 + 1 30 ) + x = 1 → x = 1 5 + 1 10 {\displaystyle 22:{\bigg (}{\frac {2}{3}}+{\frac {1}{30}}{\bigg )}+x=1\;\;\;\rightarrow \;\;\;x={\frac {1}{5}}+{\frac {1}{10}}} 23 : ( 1 4 + 1 8 + 1 10 + 1 30 + 1 45 ) + x = 2 3 → x = 1 9 + 1 40 {\displaystyle 23:{\bigg (}{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{10}}+{\frac {1}{30}}+{\frac {1}{45}}{\bigg )}+x={\frac {2}{3}}\;\;\;\rightarrow \;\;\;x={\frac {1}{9}}+{\frac {1}{40}}} 24 : x + 1 7 x = 19 → x = 16 + 1 2 + 1 8 {\displaystyle 24:x+{\frac {1}{7}}x=19\;\;\;\rightarrow \;\;\;x=16+{\frac {1}{2}}+{\frac {1}{8}}} 25 : x + 1 2 x = 16 → x = 10 + 2 3 {\displaystyle 25:x+{\frac {1}{2}}x=16\;\;\;\rightarrow \;\;\;x=10+{\frac {2}{3}}} 26 : x + 1 4 x = 15 → x = 12 {\displaystyle 26:x+{\frac {1}{4}}x=15\;\;\;\rightarrow \;\;\;x=12} 27 : x + 1 5 x = 21 → x = 17 + 1 2 {\displaystyle 27:x+{\frac {1}{5}}x=21\;\;\;\rightarrow \;\;\;x=17+{\frac {1}{2}}} 28 : ( x + 2 3 x ) − 1 3 ( x + 2 3 x ) = 10 → x = 9 {\displaystyle 28:{\bigg (}x+{\frac {2}{3}}x{\bigg )}-{\frac {1}{3}}{\bigg (}x+{\frac {2}{3}}x{\bigg )}=10\;\;\;\rightarrow \;\;\;x=9} 29 : 1 3 ( ( x + 2 3 x ) + 1 3 ( x + 2 3 x ) ) = 10 → x = 13 + 1 2 {\displaystyle 29:{\frac {1}{3}}{\Bigg (}{\bigg (}x+{\frac {2}{3}}x{\bigg )}+{\frac {1}{3}}{\bigg (}x+{\frac {2}{3}}x{\bigg )}{\Bigg )}=10\;\;\;\rightarrow \;\;\;x=13+{\frac {1}{2}}} Moscow Mathematical Papyrus The Moscow Mathematical Papyrus , also named
1230-564: The heqat, which is an ancient Egyptian unit of volume. Beginning at this point, assorted units of measurement become much more important throughout the remainder of the papyrus, and indeed a major consideration throughout the rest of the papyrus is dimensional analysis . Problems 39 and 40 compute the division of loaves and use arithmetic progressions . The second part of the Rhind papyrus, being problems 41–59, 59B and 60, consists of geometry problems. Peet referred to these problems as "mensuration problems". Problems 41–46 show how to find
1271-479: The knowledge of all things, mysteries ... all secrets". He continues: This book was copied in regnal year 33, month 4 of Akhet , under the majesty of the King of Upper and Lower Egypt, Awserre, given life, from an ancient copy made in the time of the King of Upper and Lower Egypt Nimaatre. The scribe Ahmose writes this copy. Several books and articles about the Rhind Mathematical Papyrus have been published, and
Ahmes - Misplaced Pages Continue
1312-580: The mid 1860s, were donated by his daughter in 1906 to the New York Historical Society, and are now held by the Brooklyn Museum . An 18 cm (7.1 in) central section is missing. The papyrus began to be transliterated and mathematically translated in the late 19th century. The mathematical-translation aspect remains incomplete in several respects. The first part of the Rhind papyrus consists of reference tables and
1353-1910: The other 88 numbered items. 4 10 = 1 3 + 1 15 ; 5 10 = 1 2 ; 6 10 = 1 2 + 1 10 {\displaystyle {\frac {4}{10}}={\frac {1}{3}}+{\frac {1}{15}}\;\;\;;\;\;\;{\frac {5}{10}}={\frac {1}{2}}\;\;\;;\;\;\;{\frac {6}{10}}={\frac {1}{2}}+{\frac {1}{10}}} 7 10 = 2 3 + 1 30 ; 8 10 = 2 3 + 1 10 + 1 30 ; 9 10 = 2 3 + 1 5 + 1 30 {\displaystyle {\frac {7}{10}}={\frac {2}{3}}+{\frac {1}{30}}\;\;\;;\;\;\;{\frac {8}{10}}={\frac {2}{3}}+{\frac {1}{10}}+{\frac {1}{30}}\;\;\;;\;\;\;{\frac {9}{10}}={\frac {2}{3}}+{\frac {1}{5}}+{\frac {1}{30}}} 6 10 = 1 2 + 1 10 ; 7 10 = 2 3 + 1 30 {\displaystyle {\frac {6}{10}}={\frac {1}{2}}+{\frac {1}{10}}\;\;\;;\;\;\;{\frac {7}{10}}={\frac {2}{3}}+{\frac {1}{30}}} 8 10 = 2 3 + 1 10 + 1 30 ; 9 10 = 2 3 + 1 5 + 1 30 {\displaystyle {\frac {8}{10}}={\frac {2}{3}}+{\frac {1}{10}}+{\frac {1}{30}}\;\;\;;\;\;\;{\frac {9}{10}}={\frac {2}{3}}+{\frac {1}{5}}+{\frac {1}{30}}} S = 1 + 1 / 2 + 1 / 4 = 7 4 {\displaystyle S=1+1/2+1/4={\frac {7}{4}}} and T = 1 + 2 / 3 + 1 / 3 = 2 {\displaystyle T=1+2/3+1/3=2} . Then for
1394-523: The output of a shoemaker given that he has to cut and decorate sandals. Seven of the twenty-five problems are geometry problems and range from computing areas of triangles, to finding the surface area of a hemisphere (problem 10) and finding the volume of a frustum (a truncated pyramid). The tenth problem of the Moscow Mathematical Papyrus asks for a calculation of the surface area of a hemisphere (Struve, Gillings) or possibly
1435-518: The problems are pefsu problems (see: Egyptian algebra ): 10 of the 25 problems. A pefsu measures the strength of the beer made from a hekat of grain A higher pefsu number means weaker bread or beer. The pefsu number is mentioned in many offering lists. For example, problem 8 translates as: Problems 11 and 23 are Baku problems. These calculate the output of workers. Problem 11 asks if someone brings in 100 logs measuring 5 by 5, then how many logs measuring 4 by 4 does this correspond to? Problem 23 finds
1476-488: The remainder of this 8 after subtracting 2/3 + 1/6 + 1/18. You get 7 + 1/9. Multiply 7 + 1/9 by 4 + 1/2. You get 32. Behold this is its area. You have found it correctly." The solution amounts to computing the area as The formula calculates for the area of a hemisphere, where the scribe of the Moscow Papyrus used 256 81 ≈ 3.16049 {\displaystyle {\frac {256}{81}}\approx 3.16049} to approximate π . The fourteenth problem of
1517-407: The strength of bread and beer, with respect to certain raw materials used in their production. Problem 79 sums five terms in a geometric progression . Its language is strongly suggestive of the more modern riddle and nursery rhyme " As I was going to St Ives ". Problems 80 and 81 compute Horus eye fractions of hinu (or heqats). The last four mathematical items, problems 82, 82B and 83–84, compute
1558-498: The sum of the quantity and part(s) of it are given. The Rhind Mathematical Papyrus also contains four of these type of problems. Problems 1, 19, and 25 of the Moscow Papyrus are Aha problems. For instance, problem 19 asks one to calculate a quantity taken 1 + 1 ⁄ 2 times and added to 4 to make 10. In other words, in modern mathematical notation one is asked to solve 3 2 x + 4 = 10 {\displaystyle {\frac {3}{2}}x+4=10} . Most of
1599-413: The value of π as being 3.1605..., an error of less than one percent. Problem 47 is a table with fractional equalities which represent the ten situations where the physical volume quantity of "100 quadruple heqats" is divided by each of the multiples of ten, from ten through one hundred. The quotients are expressed in terms of Horus eye fractions, sometimes also using a much smaller unit of volume known as
1640-428: The vertical height by 4 on the base by 2 on the top: You are to square the 4; result 16. You are to double 4; result 8. You are to square this 2; result 4. You are to add the 16 and the 8 and the 4; result 28. You are to take 1/3 of 6; result 2. You are to take 28 twice; result 56. See, it is of 56. You will find [it] right" The solution to the problem indicates that the Egyptians knew the correct formula for obtaining
1681-500: The volume of both cylindrical and rectangular granaries. In problem 41 Ahmes computes the volume of a cylindrical granary. Given the diameter d and the height h, the volume V is given by: In modern mathematical notation (and using d = 2r) this gives V = ( 8 / 9 ) 2 d 2 h = ( 256 / 81 ) r 2 h {\displaystyle V=(8/9)^{2}d^{2}h=(256/81)r^{2}h} . The fractional term 256/81 approximates