Descartes' Square Root MethodDate: 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/ |
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