The Rhind Mathematical Papyrus, which is also known as the Ahmes Papyrus, is the major source of our knowledge of the mathematics of ancient Egypt. Although the many ceremonial objects which have been discovered in tombs and elsewhere in Egypt during archaeological explorations provide examples of numerals written in hieroglyphic and hieratic script and while many non-mathematical papyri contain hieratic and demotic writings involving numerals, these usually contain little or no actual mathematics as such.

The Rhind Papyrus dates from approximately 1650 B.C.E. Mr. A. Henry Rhind, a Scottish lawyer, visited Egypt in the mid-nineteenth century on the advice of his physician in hopes that its dry climate would be beneficial to his poor health, and he became very interested in Egyptian antiquities. This interest led Rhind to participate in archaeological excavations in Thebes, an ancient Pharaohnic capital. Rhind purchased the papyrus in Luxor, Egypt, in 1858. In later years, it was willed to the British Museum, where it remains today. A piece missing from the center of the papyrus was located in New York City many years later and was restored to the Rhind Papyrus after 1922.

Recall that Champollion had deciphered Egyptian hieroglyphic script in the period from 1810 to 1830 and that the key to unlocking the secret of this ancient writing, the Rosetta Stone, also contained a large passage written in demotic script, which enabled scholars to interpolate between the two systems and decode hieratic script as well. In 1877, Professor A. Eisenlohr published the first translation of the Rhind Papyrus into a modern language, in his book Ein matematisches handbuch der älten Ägypter (Papyrus Rhind der British Museum), übersetzt und erklärt. T.E. Peet issued an English edition in 1923.

The papyrus is in the form of a scroll about 12-13 inches wide and 18 feet long, written from right to left in hieratic script on both sides of the sheet, in black and red inks. After identifying himself as the writer, the scribe Ahmes (or Ahmose -- since the Egyptians seldom indicated vowels, there is some ambiguity about the pronunciation) begins by saying that he has copied this work from a very old scroll from the period of the Middle Kingdom, a couple of hundred years earlier. Thus the mathematics in this papyrus dates from the same time period as that in another important mathematical scroll, the Golenishev or Moscow Papyrus. Before getting down to business, Ahmes begins with an announcement, that he will provide a "complete and thorough study of all things" and will reveal "the knowledge of all secrets." Then he goes on to the content, consisting of two tables of fractions followed by 84 worked problems.

The first fraction table occupies a large part of the manuscript. For each odd integer n from 5 through 101, it gives a decomposition of twice the unit fraction 1/n into a sum of distinct UNIT FRACTIONS, fractions whose numerator is 1. This table is not just a list of facts; every entry is either derived from scratch or is verified in detail. Notationally, the hieratic symbol for 1/k consisted of the symbol for k with a dot over it. The hieroglyphic equivalent used a small oval over the numeral for the integer. To facilitate working with unit fractions, Otto Neugebauer introduced the notation which is now in general use, n denoting 1/n. (Similarly to Ahmes' double over-dots, Neugebauer used a 3 with two overbars for the important special fraction that we write as 2/3.)

The second fraction table decomposes one tenth of n as a sum of distinct unit fractions, for n = 1, 2, ..., 9. These representations will be referred to several times in the rest of his text.

Evidently, the ancient Egyptians did not consider a fraction whose numerator is greater than 1 to be a number -- only unit fractions, the number 2/3 and a few other fractions of the special form n/(n + 1) were felt to be completed processes. For example, 3/7 is felt to be incomplete, not exactly a number, and so it must be expressed as a sum of genuine fractions when it appears either in the answer to a problem or in a related calculation.

Why is n odd in the doubles table? These people of long ago realized that the case of even n's was easy. As Neugebauer says in his The Exact Sciences in Antiquity, "Here we find that twice 2, 4, 6, 8, etc. are always replaced by 1, 2, 3, 4, respectively." Some of the table decompositions seem reasonable, given the unstated rule that no term in the sum can be repeated. For instance, 2/3 and 1/2 1/6 (addition is understood) were recognized as being interchangeable, and that 2/5 is 1/3 1/15 is understandable to us. But some of the table entries are hard to believe, such as: twice 1/83 is 1/60 1/332 1/415 1/498.

Not only were 2/3 and the 1/n's essentially the only fractions used as numbers, but repetitions were not allowed in the representations of fractions as sums. E.g., 2/7 could not be presented as 1/7 1/7. But, using the twice 1/n table one can express any proper fraction in terms of the basic ones. E.g., 5/7 = 1/7 + 4/7 = 1/7 + 2(2/7). From Table 1, 2&lowast 7 = 4 28; so 2&lowast (2&lowast 7) = 2&lowast 4 28 = 2 14. Hence 5/7 is 2 7 14 , if we list the parts in decreasing size (increasing order of denominators).

Historians of mathematics have devised many explanations of how the twice 1/n and one-tenth n tables were derived. For more information, you may consult the books ESA (by Neugebauer, cited above), Mathematics in Civilization (Resnikoff and Wells), The Rhind Mathematical Papyrus: an ancient Egyptian text (G. Robins and C. Shute; Dover Publications reprint of the 1987 edition published by the Trustees of the British Museum), Mathematics in the Time of the Pharaohs (R.J. Gillings; Dover Publications 1982 reprint of the 1972 edition published by the MIT Press).

Some of Ahmes' problems were straightforward mathematics, such as seeking the solution of a linear equation. E.g., Problem 24 asks to find a quantity if it plus one seventh of it equals nineteen. This was stated partly verbally and partly symbolically, since Ahmes had symbols for "plus" and "minus" and used a word meaning "heap" or "quantity" (pronounced something like 'aha') for the unknown. Some of the linear equations problems are stated as word problems involving the division of commodities such as grain, bread, or beer into equal parts. (Important for those responsible for feeding the construction crews or overseers who built pyramids.) The presentation is usually in three parts: the problem is stated, a brief calculation is done which leads to a solution -- or a solution is pulled out of the air, and a check is done to show that the number found really satisfies the condition. The arithmetical calculations are presented in detail.

When one reads Ahmes' calculations one sees very quickly that MULTIPLICATION of whole numbers was done by a process (DUPLATION) consisting of repeated doublings and one addition. For instance to multiply 43 by 19 (i.e., to find 19 times 43), Ahmes would have done the work in two columns more or less as follows. (To find 43 times 19, the roles would have been interchanged.)

               43      1/     Keep doubling until you reach the last power
       86      2/     of two which is below the multiplier.         
      172      4      Check the rows needed to add up to the multiplier, 
      344      8      in this case 1 + 2 + 16 = 19, and then add the        
      688     16/     corresponding doubles in the other column, in this
      ----    ---     case 43 + 86 + 688 (which was easy to do using 
      817     19      the hieroglyphic notation).  

Multiplying by a fraction was done the same way, which explains the presence of the first table in the Rhind Papyrus. Neugebauer writes "Every multiplication and division which involves fractions leads to the problem of how to double unit fractions."

DIVISION was carried out by a variation of the duplation process. (If the quotient was an integer, the division was very much like duplation; otherwise, adjustments were necessary.) See examples in the text, and note David Burton's remark, "It is extraordinary that to get one third of a number, the Egyptians first found two-thirds of the number and then took one-half of the result."

These multiplication and division processes hinge crucially on the fact that every positive integer is the sum of distinct powers of two. Indeed, although they didn't have the concept, the Egyptians of those days were using base two expansions.

In his article "The Rhind Papyrus" in the four-volume book The World of Mathematics , which he compiled, James R. Newman writes "It is remarkable that the Egyptians, who attained much skill in their arithmetic manipulations, were unable to devise a fresh notation and less cumbersome methods. We are forced to realize how little we understand the circumstances of cultural advance: why societies move -- or perhaps it is jump -- from one orbit to another of intellectual energy, why the science of Egypt "ran its course on narrow lines" and adhered so rigidly to its clumsy rules. Unit fractions continued in use, side by side with improved methods, even among Greek mathematicians. Archimedes, for instance, wrote 1/2 1/4 for 3/4, and
Hero[n] 1/2 1/17 1/34 1/51 for 31/51."

Some of the problems in the Rhind Papyrus translate for us into the form "Find x if x + ax = b." In four of those problems, a is a unit fraction, and in four others it is a sum of two or three unit fractions. The equations where a = 1/n are solved by Ahmes by a method which has become known as Regula Falsi (in Latin), or FALSE POSITION (or False Supposition). In this method, an incorrect but convenient GUESS of the value of the unknown is made, the left side of the equation is evaluated at the guessed value, and then the guess is adjusted by multiplying by a suitable amount. In a sense, Ahmes knew that multiplication distributes over addition. So he understood that solving the equation amounts to dividing b by (1 + a) -- e.g., he knew that "x + (1/n) x = b" is equivalent to [(n + 1)/n] x = b and thus x is b multiplied by n/(n + 1). But he almost never solved a linear equation by a process that we could consider as equivalent to doing that.

Consider the solution of Problem 24, which is to find heap if heap and its one-seventh added together become 19. Ahmes states a linearity or proportionality principle: "As many times as 8 must be multiplied to give 19, just as many times must 7 be multiplied to give the correct answer." That is, guess that heap is 7; then the left side of the equation evaluates to 7 + one-seventh of 7, or 8. We want the left side to be 19, so we change everything proportionally, using the factor which will take us from 8 to 19. That factor is 2 plus 1/4 1/8, which will give us 16 + 2 + 1 = 19. So the solution is (2 1/4 1/8) times 7; but that's too complicated to find by duplation, so Ahmes does the simpler calculation, 7 times (2 1/4 1/8) and finds that heap = 16 1/2 1/8.

The method of false position was used in Europe from the 1200s until algebra notation was firmly established in the late sixteenth century, owing largely to its popularization by Leonardo of Pisa (nowadays known as Fibonacci) as a way to solve problems of commercial arithmetic.

Ahmes's Problem 6 is to find how to divide 9 loaves equally among 10 men. Because of the severe limitation on the ancient Egyptians' use of fractions, this was a difficult problem.

Here is his presentation:
"The doing as it occurs: Make thou the multiplication 2/3 1/5 1/30 times.

        1      2/3    1/5   1/30 
       \2     1 2/3   1/10  1/30       
        4     3 1/2   1/10      
       \8     7 1/5 

Total loaves 9; it, this is."

Ahmes tells the reader the answer without deriving it. (This may have made for complications if one had to divide 9 loaves among 15 men, e.g., since we're given no clue on how to begin.) Then he presents a verification that his answer is correct, in the usual duplation style. In the first step, he doubles
2/3    1/5   1/30 and gets
1 2/3   1/10  1/30 but gives no explanation, not even a reference to the doubles table. (We would get 4/3 2/5 1/15 as the double, then refer to the table to split 2/5 as 1/3 1/15, and continue.)

The next doubling goes from

  1 2/3   1/10  1/30 to 3 1/2 1/10, again without explanation.

(We would get 10/3 1/5 1/15 = 3 1/3 1/5 1/15, but that's not Ahmes's decomposition.) The final doubling gives us 6 + 1 + 1/5 = 7 1/5, which agrees with Ahmes's value.
To verify the correctness of the answer, one adds the multiples corresponding to the checkmarks, namely
1 2/3 1/10 1/30 plus 7 1/5, giving us 8 2/3 1/5 1/10 1/30.
Some simplifying is required to see that this equals 9.