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Topic: Can an algorithm be developed to implement new square root method?
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Peter Duveen

Posts: 163
From: New York
Registered: 4/11/12
Can an algorithm be developed to implement new square root method?
Posted: Apr 13, 2012 6:01 AM
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I recently developed this simple method to find a square root, and have not been able to find it anywhere else in the literature. I have had discussions with a couple of computer savvy people, one of whom says an algorithm could not be developed for this method, another who said it could be. I will illustrate this method.

Let's take sqrt 10. Merely multiply and divide by the nearest perfect square not exceeding the number, in this case, 9. Thus, sqrt (10 x 9/9), rewrite as sqrt (9 x 10/9); rewrite as 3 x sqrt 10/9. Now your problem is no longer to find the square root of ten, but to find the square root of 10/9. We simply apply this procedure again and again. Convert 10/9 to a decimal. 3 sqrt 1.11111... Now we must similarly find a perfect square close to but not exceeding 9/10. Ok, 1.05 squared = 1.1025. So we have 3 x sqrt 1.11111...x 1.1025/1.1025 = 3 x 1.05 sqrt 1.11111.../1.1025. = 3 x 1.05 x sqrt 1.00781. Just continue the process. Right off, I can see 1.003 will be a good candidate. When squared, it yields 1.006009. Now I have 3 x 1.05 x 1.003 sqrt 1.00781../1.006009 = 3 x 1.05 x 1.003 sqrt 1.00179.. And our candidate is 1.0008, which squared yields 1.00160064. Now we have 3 x 1.05 x 1.003 x 1.0008 sqrt my candidate is 1.00009, which squared gives 1.0001800081. So we now have 3 x 1.05 x 1.003 x 1.0008 x 1.00009 sqrt 1.0001898.../1.0001800081 = 3.1622621379804 sqrt 1.0001898.../1.0001800081 Now our numbers outside the radical sign increasingly approximate our square root. If you multiply them out and then multiply the result by itself, it will yield 9.99990182930437. You can obtain greater accuracy by continuing the calculations. The merit of this method is its conceptual simplicity. It does not require a knowledge of calculus, and can be carried out with a simple knowledge of long division. I have written about this method at

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