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The Egyptians' Method of False Position

Date: 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 

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 

   False Position and Backtracking

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 

   Egyptian Algebra and the Method of False Position 

You can read a brief analysis of the method, supporting the 
interpretation mentioned in the discussion above, at the MacTutor 
History of Mathematics archive:

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 
Associated Topics:
High School Basic Algebra
High School History/Biography

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