Proof of the result expressed in this form.
| 1 | 1⁄41⁄16 | hekat | 12⁄31⁄111⁄221⁄66 | ro |
| 2 | 1⁄21⁄8 | " | 31⁄21⁄111⁄331⁄66 | " |
| 1⁄7 | 1⁄32 | " | 41⁄21⁄22 | " |
The total of the larger portions of these products makes
1⁄21⁄41⁄81⁄161⁄321⁄64 hekat 42⁄3ro,
or 3192⁄3ro.
The smaller fractions
1⁄111⁄111⁄221⁄221⁄331⁄661⁄66
taken as parts of 66, are equal to
6633211.
The total is 22, which is 1⁄3 of 66, and this with 319 % makes 320 ro or 1 hekat, as it should.
There are some peculiarities here in the numerical work. Thus in multiplying 31⁄7 so as to get 1, the author starts by saying that 1⁄22 of 31⁄7 is 1⁄22, and his explanation, written on one side, is literally, "The making it is of 1⁄7 times 22 to find 31⁄7." Also a little further along, in proving his answer, 1⁄61⁄111⁄221⁄66, multiplying it by 31⁄7, he writes as one step, 1⁄7 1⁄22, and explains it by saying in almost the same words, "The making it is of 1⁄22 times 7 to find the fraction above." It is as if he were familiar with the relation of the three numbers 7, 31⁄7 and 22. Possibly he had this relation in some table; or in some previous reckoning he had multiplied 7 "for the finding of 22," which would give him at once 31⁄7. Then he would know that 1⁄7 of 22 is 31⁄7, that 1⁄22 of 31⁄7 is 1⁄7, and that 1⁄22 of 7 is the same as the result of "getting 1 by operating on 31⁄7" that is, it is the number that he gets in this solution, 1⁄61⁄111⁄221⁄66, so that, finally, 1⁄7 of the last expression is equal to 1⁄22.[1]
The fractional expressions in the solution and different proofs are quite complicated and it is possible that he applied them to 66 each time that he had additions to make, but, except in the last case, there is no indication of this.
In the first proof he doubles 1⁄61⁄111⁄221⁄66 and gets 1⁄21⁄111⁄331⁄56. Here twice 1⁄6 is 1⁄3, twice 1⁄11 1⁄61⁄66 (by the table at the beginning) and the 1⁄3 and 1⁄6 make 1⁄2 finally, twice 1⁄221⁄66 make 1⁄111⁄22 In a very similar way in the second and third proofs he doubles the fractional expression 2⁄31⁄111⁄221⁄66
In working out the expression of the result in ro he has 1⁄11 of 320 equal to 291⁄11, and in the proof which follows, 1⁄7 of 101 2⁄31⁄111⁄221⁄66 is given as 141⁄21⁄2; also in the last proof 1⁄7 of 1⁄41⁄16. hekat 12⁄31⁄111⁄221⁄66 ro is given as 1⁄32 hekeat 41⁄21⁄22 ro. These are more difficult than most of the divisions in the papyrus. and the author does not state explicitly how he performed them.
- ↑ See Introduction, page 5, footnote 2.
