Page:The Rhind Mathematical Papyrus, Volume I.pdf/34

18 then $2/3$, $2/3$ of $1/undefined$ being $1/undefined$. $2/3$ of 2 would be $1 1/3$ and $2/3$ of $1/undefined$ would be $1/undefined$, making $1 2/3$. Having obtained $1 2/3$ he requires $1/undefined$ more to make 2 and the multiplier for this is $1/undefined$. His answer therefore is $1/undefined$ $1/undefined$.

An example of BD is the number 31. Here, supplying the first two lines, which are not given, we have:

There remains $1/undefined$ $1/undefined$, for which the multipliers are obtained by the usual method, and his answer is $1/undefined$ $1/undefined$ $1/undefined$.

For C, I will take first the number 21, which is a multiple of 3. $1/undefined$ of 21 is 14; therefore $1/undefined$ of 21 is $1/undefined$. There remains $1/undefined$; 2 times 21 is 42, so that $1/undefined$ of 21 is $2/3$, and the answer is $1/undefined$ $1 1/2$.

For a multiple of 5 take 65. We may suppose that the author proceded in this way:

There remains $1/undefined$; 3 times 65 is 195, and his answer is $1/undefined$ $1/undefined$. The last line of this multiplication is obtained by taking $1/undefined$ of the numbers in the preceding line, $1/undefined$ of $1/undefined$ being the same as in the case of 25.

I have marked 35, 91, and 101 with E. These cases may be explained in the following way:

We have seen that the relations in this table may be regarded in two ways, either as expressing twice the reciprocal of an odd number, or as showing by what the odd number must be multiplied to make 2. In obtaining them the Egyptian nearly always takes the latter point of view, but in the case of 35 he seems to have considered that he was doubling the reciprocal of 35. To ﬁnd the double of $1/undefined$ he applies this fraction to the number 210, which is 6 times 35. This is indicated by a 6 written in red under 35. The double of 6 is 12, equal to 7 and 5, and these numbers are $1/undefined$ and $1/undefined$ of 210. Therefore his answer is $1/undefined$ $1/undefined$. He writes out his answer in the usual form, but puts under each fraction the amount that it makes when applied to 210. Finally, in the lines below, where he usually puts the multiplications of his solution, he writes the two fractions of his answer with the results of multiplying