Page:The Rhind Mathematical Papyrus, Volume I.pdf/31

Rh Here the sixth line is obtained by halving the preceding line in order to get the $1/undefined$ of 7 called for by the answer that we are trying to prove, but in the solution (and in the papyrus) we have $1/undefined$ as the fraction required to complete to 2 the 1 $1/undefined$ $1/undefined$ in line 3, and we get $1/undefined$ as the multiplier necessary to produce it. That is, in the solution we get the $1/undefined$ from the $1/undefined$, while in the proof we get the $1/undefined$ from the $1/undefined$. The only other way of carrying through the proof would be to interchange multiplier and multiplicand (see page 6). With this interchange the Egyptian would have said:

getting the total, if necessary, by applying his fractions to some number, say 28. This would be like the third step in the solutions of Problems 24-27.

In the latter part of the table the second step is omitted and only the result of the first step is given, but the third step is given for all of the numbers except 3, 35, and 91, and indicates clearly that we have, at least in each case except these, the solution and not the proof.

In the first step of these solutions the author takes small multipliers