Page:The Rhind Mathematical Papyrus, Volume I.pdf/33

Rh To make up $1/undefined$ the author multiplies 17 by 3, getting 51, and, because 3 times 17 is 51, it follows that $1/undefined$ of 17 is $1/undefined$. In the same way, getting 68 as 4 times 17, he knows that $1/undefined$ of 17 is $1/undefined$. Thus his answer is $1/undefined$ $1/undefined$ $1/undefined$.

He has two ways of writing down this third step. There are two multiplications, the multiplier of the second being the reciprocal of the product of the first, and differing from it only in the dot used in writing the reciprocal of a whole number. Thus in getting the $1/undefined$ these multiplications are

3 times 17 is 51 and $1/undefined$ times 17 is $1/undefined$.

But he always omits one of these two numbers, either the multiplier of the second multiplication or the product of the ﬁrst. Thus in this example he might write:

As it happens, the author uses the second form in this case, but in some cases he uses the first. It may be, however, that he always has in mind the first multiplication, even when he puts a dot over the product, and that the fraction written after this multiplication is put in simply to indicate its purpose.

To illustrate B, I will take the case of 13. The papyrus says:

There remains $1/undefined$ $1/undefined$ for which the multipliers are $1/undefined$ and $1/undefined$ and the answer is $1/undefined$ $1/undefined$ $1/undefined$.

An example of AD is the number 25. Here we have simply:

This would seem to indicate that the author first took $1/undefined$, getting $1/undefined$, and