Page:The Rhind Mathematical Papyrus, Volume I.pdf/36

20 as his ﬁrst fraction. $1/undefined$ is so small that he has to get $1/undefined$, $1/undefined$, and $1/undefined$ to make 2, and in his answer he has four fractions. If he had halved twice more instead of multiplying by $1/undefined$ he would have had $1/undefined$ for his ﬁrst fraction and one form for his ﬁnal answer would have been $1/undefined$ $1/undefined$ $1/undefined$.

When the author takes $2/3$ and $1/undefined$ in the same example his result in some cases comes most simply by taking $2/3$ ﬁrst and $1/undefined$ at the end. Examples are 47, 53, 79, and 89. Yet in other cases, as of 55, 73, 83, and 95, he apparently multiplied by $1/undefined$ ﬁrst. In the case of 59 he takes $2/3$ twice. The most natural process of obtaining the form given seems to be:

writing for $2/3$, 9 $1/undefined$ $1/undefined$ instead of 9 $2/3$ $1/undefined$, and halving again, so that his fractions will have even denominators, before he takes $1/undefined$ the second time.

The author uses the word gem meaning "ﬁnd" before the first denominator in the case of every number after 41 which is not a multiple of 3, except perhaps 101, and also in the cases of 93 and 99, and he uses it before both denominators for 91. All of these cases, except perhaps 93 and 99, are the more difficult ones, and he does not put in the details but leaves them to the reader.

All of these various cases seem to indicate that there was no definite rule for determining the multipliers to be used, but probably the slow experience of different writers suggested different multipliers for different examples, as they seemed to them the easiest or gave results in the most satisfactory form.

In the table as here reproduced I have put: first, the letter or letters indicating the kind of multipliers employed; second, the number; third, the first fraction of the answer, this being the multiplier that produces a number a little less than 2; then the number a little less than 2 that is produced by this multiplier; the remainder necessary to make 2; and, finally, the answer.