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SPECIAL PROCESSES EMPHASIZED
11

which (usually) is not the answer, and get the answer by finding its relation to this number.[1]

False position[2] is used in a number of multiplication problems in which the multiplier and product are given and it is required to find the multiplicand, because by this method it is possible to keep in mind the nature of the quantities involved. Problems 24-27 exhibit the method very clearly although the required multiplicand in these problems is nothing more definite than “a quantity,” and the given product, which must be a quantity of the same kind, is given simply as a number. It is instructive to compare this group of problems with Problems 30-34, which interchange multiplicand and multiplier. The problems of the two groups are problems of the same kind and all of them involve only “a quantity.”

I think, however, that the hekat problems, 35, 37, and 38, show more clearly that this process enables us to keep in mind the nature of the quantities involved, and I will give in detail the reasoning of one of these problems as I understand it.

In Problem 35 a vessel filled 3+1⁄3 times with grain will make 1 hekat (about half a peck). The vessel contains so much grain and the hekat is so much grain. 3+1⁄3 times the former makes the latter. Here we have given the multiplier, 3+1⁄3, and the product, 1 hekat of grain, to find the multiplicand, which will be a certain portion of a hekat of grain. Assume a vessel that itself contains 1 hekat; this filled 3+1⁄3 times will give 3+1⁄3 hekat. Then the one that we have must bear the same relation to the one assumed, that the amount of grain that fills the former 3+1⁄3 times bears to the amount of grain that fills the latter 3+1⁄3 times; that is, that one hekat bears to 3+1⁄3 hekat. Thus we have to find the relation of one hekat to 3+1⁄3 hekat. Making 3+1⁄3 hekat the multiplicand and one hekat the product we get for the multiplier the number 1⁄5 1⁄10. The amount of grain that the given vessel holds is then1⁄5 1⁄10 times the amount of grain that the one assumed holds; that is, it is 1⁄5 1⁄10 times 1 hekat.[3]

  1. ↑ We might notice a point of similarity in this method and the method of applying fractions to a particular number, that in both methods we first assume a particular number or quantity. In Problem 76 the same assumed quantity, namely, 30 loaves, is used for both purposes. First the writer takes 1⁄20 and 1⁄30 as parts of 30 to show that the sum of these fractions is 1⁄12, and then he takes 30 loaves as the basis of a false position process for determining the answer to the problem.
  2. ↑ False position is used in Problems 24-27, 40 and 76. It is used also in Problems 28, 29, and 35-38 as I have explained them.
  3. ↑ It is a little confusing to have 1 hekat come into the solution in two ways.
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