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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
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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