the multiplicand is one of the numbers that can be obtained from 1⁄4 1⁄28 by doubling or halving one or more times. In Problem 7, the multiplication is as follows:
| 1 | 1⁄4 1⁄28 |
| 1⁄2 | 1⁄8 1⁄56 |
| 1⁄4 | 1⁄16 1⁄112 |
| Total | 1⁄2. |
1⁄4 1⁄28 is the number which we have found as the value of 2 divided by 7; and in some of the problems 1⁄7 is given as 1⁄2 of this number, 1⁄14 as 1⁄2 of 1⁄7, and so on. In fact, some of these multiplications are carried through in both ways in different problems.[1]
In adding his partial products the Egyptian applies these fractional numbers to the number 28, using the method that I have already explained (page 7), and in Problems 7, 13, 14, and 15 he puts under each fraction the number that it makes when taken as a part of 28. Thus in Problem 7, 1⁄4 and 1⁄28 of 28 are 7 and 1, 1⁄8 and 1⁄56 are 31⁄2 and 1⁄2, and 1⁄16 and 1⁄112 are 1 1⁄2 1⁄4 and 1⁄4. These numbers added together give 14, which is 1⁄2 of 28. Therefore the answer is 1⁄2.
In Problem 15, 1⁄128 and 1⁄896 of 28 are 1⁄8 1⁄16 1⁄32 and 1⁄32, and we have what may seem rather complicated quantities to add, even when we use 28, but they are simpler than the given quantities, being the reciprocals of powers of 2.
In the problems of the other group, 8 and 16-20, the author starts with a single fraction and multiplies by 1 2⁄3 1⁄3, so that the result is simply to double the fraction. The numbers taken are 1⁄4, 1⁄2, 1⁄3, 1⁄6, 1⁄12, and 1⁄24. In this group the fractional numbers are applied to 18, and in the first and last two problems we find placed under the fractions their values as parts of this number. Nearly all of these parts are themselves fractions, but they are the reciprocals of powers of 2.
There are some interesting multiplications in these solutions. In Problem 17 the author has 2⁄3 of 1⁄3 equal to 1⁄6 1⁄18. He may have put this down from memory or have copied it from some table; or he may have applied the rule given in Problem 61 (see below). Then for a half of 1⁄6 1⁄18 he writes 1⁄9 because his table for the division of 2 by odd numbers tells him that 2 times 1⁄9 is 1⁄6 1⁄18.
Problem 61 may also be considered in this connection as it consists of a multiplication table of various fractions. With 61 is associated
- ↑ It is interesting to notice that the multiplicands and partial products in these multiplications can be arranged in a table formed by taking 1⁄2 1⁄14 as the first number and multiplying several times by 1⁄2.
