Misplaced Pages

#628371

62-502: 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 the knowledge of all things, mysteries ... all secrets". He continues: This book was copied in regnal year 33, month 4 of Akhet , under

124-414: A 1 , … , a n {\displaystyle b,a_{1},\ldots ,a_{n}} are the coefficients , which are often real numbers . The coefficients may be considered as parameters of the equation and may be arbitrary expressions , provided they do not contain any of the variables. To yield a meaningful equation, the coefficients a 1 , … ,

186-407: A n {\displaystyle a_{1},\ldots ,a_{n}} are required to not all be zero. Alternatively, a linear equation can be obtained by equating to zero a linear polynomial over some field , from which the coefficients are taken. The solutions of such an equation are the values that, when substituted for the unknowns, make the equality true. In the case of just one variable, there

248-437: A x + b = 0 , {\displaystyle ax+b=0,} with a ≠ 0 {\displaystyle a\neq 0} . The solution is x = − b a {\displaystyle x=-{\frac {b}{a}}} . A linear equation in two variables x and y can be written as a x + b y + c = 0 , {\displaystyle ax+by+c=0,} where

310-573: A and b are not both 0 . If a and b are real numbers, it has infinitely many solutions. If b ≠ 0 , the equation is a linear equation in the single variable y for every value of x . It has therefore a unique solution for y , which is given by This defines a function . The graph of this function is a line with slope − a b {\displaystyle -{\frac {a}{b}}} and y -intercept − c b . {\displaystyle -{\frac {c}{b}}.} The functions whose graph

372-399: A linear equation in two variables is an equation whose solutions form a line. If b ≠ 0 , the line is the graph of the function of x that has been defined in the preceding section. If b = 0 , the line is a vertical line (that is a line parallel to the y -axis) of equation x = − c a , {\displaystyle x=-{\frac {c}{a}},} which

434-440: A linear equation is an equation that may be put in the form a 1 x 1 + … + a n x n + b = 0 , {\displaystyle a_{1}x_{1}+\ldots +a_{n}x_{n}+b=0,} where x 1 , … , x n {\displaystyle x_{1},\ldots ,x_{n}} are the variables (or unknowns ), and b ,

496-465: 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

558-507: 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

620-593: 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

682-551: 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 the Second Intermediate Period of Egypt . It was copied by the scribe Ahmes (i.e., Ahmose; Ahmes

SECTION 10

#1732773189629

744-422: A line can be expressed simply in terms of a determinant . There are two common ways for that. The equation ( x 2 − x 1 ) ( y − y 1 ) − ( y 2 − y 1 ) ( x − x 1 ) = 0 {\displaystyle (x_{2}-x_{1})(y-y_{1})-(y_{2}-y_{1})(x-x_{1})=0}

806-742: A linear equation form a line in the Euclidean plane, and, conversely, every line can be viewed as the set of all solutions of a linear equation in two variables. This is the origin of the term linear for describing this type of equation. More generally, the solutions of a linear equation in n variables form a hyperplane (a subspace of dimension n − 1 ) in the Euclidean space of dimension n . Linear equations occur frequently in all mathematics and their applications in physics and engineering , partly because non-linear systems are often well approximated by linear equations. This article considers

868-487: A linear equation in n variables are the Cartesian coordinates of the points of an ( n − 1) -dimensional hyperplane in an n -dimensional Euclidean space (or affine space if the coefficients are complex numbers or belong to any field). In the case of three variables, this hyperplane is a plane . If a linear equation is given with a j ≠ 0 , then the equation can be solved for x j , yielding If

930-434: 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

992-464: Is a n -tuple such that substituting each element of the tuple for the corresponding variable transforms the equation into a true equality. For an equation to be meaningful, the coefficient of at least one variable must be non-zero. If every variable has a zero coefficient, then, as mentioned for one variable, the equation is either inconsistent (for b ≠ 0 ) as having no solution, or all n -tuples are solutions. The n -tuples that are solutions of

1054-435: Is a line are generally called linear functions in the context of calculus . However, in linear algebra , a linear function is a function that maps a sum to the sum of the images of the summands. So, for this definition, the above function is linear only when c = 0 , that is when the line passes through the origin. To avoid confusion, the functions whose graph is an arbitrary line are often called affine functions , and

1116-633: 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

1178-494: 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 , a Scottish antiquarian, purchased two parts of

1240-424: Is exactly one line that passes through them. There are several ways to write a linear equation of this line. If x 1 ≠ x 2 , the slope of the line is y 2 − y 1 x 2 − x 1 . {\displaystyle {\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}.} Thus, a point-slope form is By clearing denominators , one gets

1302-410: Is exactly one solution (provided that a 1 ≠ 0 {\displaystyle a_{1}\neq 0} ). Often, the term linear equation refers implicitly to this particular case, in which the variable is sensibly called the unknown . In the case of two variables, each solution may be interpreted as the Cartesian coordinates of a point of the Euclidean plane . The solutions of

SECTION 20

#1732773189629

1364-572: 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

1426-417: Is not parallel to an axis and does not pass through the origin cuts the axes into two different points. The intercept values x 0 and y 0 of these two points are nonzero, and an equation of the line is (It is easy to verify that the line defined by this equation has x 0 and y 0 as intercept values). Given two different points ( x 1 , y 1 ) and ( x 2 , y 2 ) , there

1488-531: Is not the graph of a function of x . Similarly, if a ≠ 0 , the line is the graph of a function of y , and, if a = 0 , one has a horizontal line of equation y = − c b . {\displaystyle y=-{\frac {c}{b}}.} There are various ways of defining a line. In the following subsections, a linear equation of the line is given in each case. A non-vertical line can be defined by its slope m , and its y -intercept y 0 (the y coordinate of its intersection with

1550-778: 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

1612-452: 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

1674-470: Is the result of expanding the determinant in the equation The equation ( y 1 − y 2 ) x + ( x 2 − x 1 ) y + ( x 1 y 2 − x 2 y 1 ) = 0 {\displaystyle (y_{1}-y_{2})x+(x_{2}-x_{1})y+(x_{1}y_{2}-x_{2}y_{1})=0} can be obtained by expanding with respect to its first row

1736-422: 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

1798-513: 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

1860-682: 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

1922-620: 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

Rhind Mathematical Papyrus - Misplaced Pages Continue

1984-714: 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

2046-402: The unit fraction having that number as denominator , e.g. 4 ¯ = 1 4 {\displaystyle {\bar {4}}={\frac {1}{4}}} ; unit fractions were common objects of study in ancient Egyptian mathematics. Other mathematical texts from Ancient Egypt include: General papyri: For the 2/n tables see: Linear equations In mathematics ,

2108-530: 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

2170-408: The y -axis). In this case, its linear equation can be written If, moreover, the line is not horizontal, it can be defined by its slope and its x -intercept x 0 . In this case, its equation can be written or, equivalently, These forms rely on the habit of considering a nonvertical line as the graph of a function . For a line given by an equation these forms can be easily deduced from

2232-472: 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

2294-538: 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

2356-412: 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

2418-400: 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

2480-409: The case of a single equation with coefficients from the field of real numbers , for which one studies the real solutions. All of its content applies to complex solutions and, more generally, to linear equations with coefficients and solutions in any field . For the case of several simultaneous linear equations, see system of linear equations . A linear equation in one variable x can be written as

2542-423: 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

Rhind Mathematical Papyrus - Misplaced Pages Continue

2604-409: The determinant in the equation Besides being very simple and mnemonic, this form has the advantage of being a special case of the more general equation of a hyperplane passing through n points in a space of dimension n – 1 . These equations rely on the condition of linear dependence of points in a projective space . A linear equation with more than two variables may always be assumed to have

2666-412: 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:

2728-406: 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

2790-421: 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

2852-405: The equation which is valid also when x 1 = x 2 (for verifying this, it suffices to verify that the two given points satisfy the equation). This form is not symmetric in the two given points, but a symmetric form can be obtained by regrouping the constant terms: (exchanging the two points changes the sign of the left-hand side of the equation). The two-point form of the equation of

2914-4511: 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

2976-523: The form The coefficient b , often denoted a 0 is called the constant term (sometimes the absolute term in old books ). Depending on the context, the term coefficient can be reserved for the a i with i > 0 . When dealing with n = 3 {\displaystyle n=3} variables, it is common to use x , y {\displaystyle x,\;y} and z {\displaystyle z} instead of indexed variables. A solution of such an equation

3038-562: 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

3100-497: The linear functions such that c = 0 are often called linear maps . Each solution ( x , y ) of a linear equation may be viewed as the Cartesian coordinates of a point in the Euclidean plane . With this interpretation, all solutions of the equation form a line , provided that a and b are not both zero. Conversely, every line is the set of all solutions of a linear equation. The phrase "linear equation" takes its origin in this correspondence between lines and equations:

3162-535: 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 a handful of these stand out. The Rhind Papyrus was published in 1923 by the English Egyptologist T. Eric Peet and contains

SECTION 50

#1732773189629

3224-576: 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

3286-1910: 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

3348-523: 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

3410-518: 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

3472-409: The relations A non-vertical line can be defined by its slope m , and the coordinates x 1 , y 1 {\displaystyle x_{1},y_{1}} of any point of the line. In this case, a linear equation of the line is or This equation can also be written for emphasizing that the slope of a line can be computed from the coordinates of any two points. A line that

3534-488: 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

3596-405: 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

3658-498: 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

3720-413: 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

3782-428: 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

SECTION 60

#1732773189629

3844-500: 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

#628371