Page:The Rhind Mathematical Papyrus, Volume I.pdf/30

14 to ﬁnd by what the number itself must be multiplied to get 2, and so the problem becomes a multiplication problem of the second kind, a problem in which the given odd number is the multiplicand and 2 is the product. In the ﬁrst eight pages of the papyrus these relations are obtained for all the odd numbers from 3 to 101. Theoretically the result can be put in an inﬁnite number of forms; the forms given are generally the simplest. In the ﬁrst part of the table the relations are worked out in detail in a sufficient number of examples to show the method.

In the papyrus the scribe places his answer to the left of the given number and the “reckoning” below. Hultsch (1895, page 4) and Peet (page 34) both regard the latter as the proof and not the solution. It is true that the interpretation of this as the solution leaves the table without proofs, but the proof would be somewhat like the solution and perhaps the author thought that one process would answer both purposes. The process is introduced by the word seshemet, put in with each number that comes at the top of a page. This word is used elsewhere in the papyrus only to introduce a solution, or the numerical work of a solution after a summary has been given in words. Even if we are to regard the answer as coming before this numerical work, which is placed on lower lines, the arrangement would be like the arrangement in many other places where the numerical work is placed after the solution in words. In these problems the solution is, as I have said, a multiplication of the second kind, while the proof would be a multiplication of the first kind. To carry through the proof the Egyptian would have to multiply the given odd number by the different fractions of his answer and get the product as the sum of the partial products of these multiplications. Take the case of 2 divided by 7. To prove that $1/undefined$ $1/undefined$ times 7 equals 2 he would say: