Page:The Rhind Mathematical Papyrus, Volume I.pdf/35

Rh 35 by them, namely, $1 1/6$ and $2/3$ $1/undefined$. All multiplication work is omitted and what is written could be taken as indicating a proof, especially as the second product is not in a form that could have been obtained by the last step of the process employed in the other cases.

For 91 the same process that he has used for 35 will lead to the form of result that he gives. By this process he would take $1/undefined$ as $1/undefined$ of 910, which is 10. Twice this is 20, equal to 13 and 7, and the fractions that give 13 and 7 are $1/undefined$ and $1/undefined$. Multiplying 91 by these two fractions gives him 1 $1/undefined$ $1/undefined$ and $2/3$ $1/undefined$, together making 2.

In the case of 101 he takes $1/undefined$ itself for his ﬁrst fraction. This makes 1 and he has to find fractions to make up another 1, and so 2 in all. He thinks of the familiar combination $1/undefined$ $1/undefined$ $1/undefined$ and takes for the other fractions of his answer $1/undefined$, $1/undefined$, and $1/undefined$. This example is found entirely in the New York fragments. Eisenlohr and other writers before 1922 supposed that the table went only to 99.

In nearly all cases after 23, except multiples of 3, the author omits the successive multiplications by which he obtains the first fraction of his answer, and though we can tell what these multiplications were from the denominator of this fraction, it is not always easy to say in what order they were taken or just how they were carried through. We have to be guided largely by the form in which he gives the product produced by them. It may be worth while to make one or two suggestions as to the way in which he did this.

With the number 15 he begins “$1/undefined$ 1 $1/undefined$.” He may have obtained this 10 by taking $2/3$ of 15 just as he takes $2/3$ of 21 and of each multiple of 3 after 21, but it is more probable that he notices that he can easily divide by 10, and that this division will give him a number between 1 and 2, and so he does not take $2/3$ of 15 at all. In each case of a multiple of 3 after 15, except 93, he mentions the fact that he takes $2/3$.

With 25 I have supposed that the author took first $1/undefined$ and then $2/3$. He may have taken $2/3$ ﬁrst and then $1/undefined$. $2/3$ of 25 is $16 2/3$. $2/3$ of 10 is $6 2/3$, and therefore $1/undefined$ of $6 2/3$ is $2/3$, and $1/undefined$ of $16 2/3$ is $1 2/3$. Similarly with 53, if he takes $2/3$ and then halves, getting $17 2/3$, he will then say that $1/undefined$ of $16 2/3$ is $1 2/3$ and $1/undefined$ of 1 is $1/undefined$, making 1 $2/3$ $1/undefined$. That seems to be the easiest way to explain the result that he gives.

In the case of 43, if he takes $2/3$ and then proceeds to halve, he will get $28 2/3$, $14 1/3$, and $7 1/6$, and the whole numbers suggest taking $1/undefined$ for a multiplier. There seems no other reason for his using 7 and taking $1/undefined$