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Descartes' Square Root Method


Date: 10/30/96 at 20:26:35
From: Joseph R. Potvin
Subject: Page 1 of Descartes "Geometry"

Why does Rene Descartes' geometric method for finding square roots 
work? (See page one of his "Geometry".)  


Date: 10/30/96 at 22:25:31
From: Doctor Patrick
Subject: Re: Page 1 of Descartes 

Hi!  Interesting question!  

Descartes is making use of Euclid's proposition VI-13:

To find a mean proportional to two given straight lines. 

(Copied from http://aleph0.clarku.edu/~djoyce/java/elements/    )

Let AB and BC be the two given straight lines. 

It is required to find a mean proportional to AB and BC. Place them in 
a straight line, and describe the semicircle ADC on AC. Draw BD from 
the point B at right angles to the straight line AC, and join AD and
DC.
                                                                       
I.11
Since the angle ADC is an angle in a semicircle, it is right.
                                                                            
III.31
And, since, in the right-angled triangle ADC, BD has been drawn from 
the right angle perpendicular to the base, therefore BD is a mean 
proportional between the segments of the base, AB and BC. 
                                                                            
VI.8, Cor
Therefore, a mean proportional BD has been found to the two given 
straight lines AB and BC. 

Q. E. F.

----End quoted text----

Check out a copy of the Elements (the whole text is online at the site 
mentioned above) to see the references.

In other words, the line dropped from the circumference of the circle 
will be the mean proportional of the lines it cuts off.  How it 
applies to Descartes' method for finding a square root is that the 
line dropped from the circumference, which Descartes says will be the 
root (line IG in my book) is the mean proportional to the line we 
wanted to find the root of (GH) and the unity (FG).  So FG:IG::IG:GH.  
If we rewrote the ratio as a fraction, we would have FG/IG = IG/GH.  
Then FG = (IG*IG)/GH and FG*GH = IG^2.  But since FG is the unity,
FG*GH = GH.  This leaves us with GH = IG^2 or IG is the square root of 
GH, which was what we wanted to prove.

Hope this helps,

-Doctor Patrick,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

    
Associated Topics:
High School Euclidean/Plane Geometry
High School Exponents
High School Geometry

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