The Egyptians' Method of False PositionDate: 01/16/2003 at 10:46:24 From: Lisa Subject: The Egyptians' method of false position I understand method of false position: if x is a value added to fraction of x/number = some number. Why does the theory work? Is it because it is similar to the way we would solve by making a common denominator, adding the fractions, and cross multiplying, thus solving for x? In a roundabout way the false position method seems to do the same. Any math reason why it works? Date: 01/16/2003 at 12:48:48 From: Doctor Peterson Subject: Re: The Egyptians' method of false position Hi, Lisa. I'm not positive what kind of problem you are referring to; the method of false position can be applied to many problems, and your description is slightly confusing. But I think you mean the kind of problem discussed here: Aha Problems in Egyptian Mathematics http://www.ms.uky.edu/~carl/ma330/html/aha1.html False Position and Backtracking http://www.learner.org/channel/courses/learningmath/algebra/session6/part_b/ In looking for references, I ran across this discussion, which suggests that it is not really false position, but really something close to how we would handle fractions; that makes your analysis correct. Egyptian Algebra and the Method of False Position http://mathforum.org/kb/message.jspa?messageID=1183726 You can read a brief analysis of the method, supporting the interpretation mentioned in the discussion above, at the MacTutor History of Mathematics archive: http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Egyptian_papyri.html It says Problem 24: A quantity added to a quarter of that quantity become 15. What is the quantity? Ahmes uses the "method of false position" which was still a standard method three thousand years later. In modern notation the problem is to solve x + x/4 = 15. Ahmes guesses the answer x = 4. This is to remove the fraction in the x/4 term. Now with x = 4 the expression x + x/4 becomes 5. This is not the correct answer, for the expression is required to equal 15. However, 15 is 3 times 5 so taking 3 times his guess of x = 4, namely x = 12, gives Ahmes the correct result. Another interpretation, favoured by some historians, is that Ahmes thought of the method as dividing x into 4 equal pieces of a size to be determined. Now Ahmes computes x + x/4 getting 5 of these equal pieces. Each piece must now be three so that 5 pieces equals 15. Not very different to our previous way of thinking, but one which is likely to come closer to Ahmes' way of thinking than our former description. Finally Ahmes checks his solution, or proves his answer is correct. He takes x = 4 * 3 = 12. Then x/4 = 3, so x + x/4 = 15 as required. Here is a simple algebraic version of what is being done if the solution is considered to use "false position". Our problem is A quantity and its fifth, added together, give 23. What is the quantity? We take the denominator as our "guess", and do the same operation on it as we are doing to the unknown: 5 + 5/5 = 6 Now we multiply the "guess" by the ratio of the desired result to the result we just got: x = 5(23/6) = 115/6 = 19 1/6 In "false position", the "guess" does not matter, apart from being chosen to make the division easy. If we call it "a", the work looks like this: x = a * 23/(a + a/5) which simplifies to x = 23/(1 + 1/5) = 23/(6/5) = 23*5/6 This works because the original problem is really (1 + 1/5)x = 23 The left side is a linear function f(x) = (1 + 1/5)x so that the value of x is proportional to f(x), and x / f(x) = a / f(a) so that x = a * f(x)/f(a) If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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