
Re: Square root of six
Posted:
Aug 27, 2012 1:11 PM


Let me see, Bob, if I can answer a few of your questions.
"Using your method, how would you find the square root of 12.643?"
I manually calculated the following product, using the same method as before:
3 x 1.1 x 1.07 x 1.006 x 1.0009 x 1.00008, which yields (with the help of a calculator, as I had a mild bout of number fatigue) 12.642776454660. I approach these calculations always from lesser values, so one can more easily detect mistakes in calculation. If you get something that is more than the number you are looking for, you know for sure you made a mistake in your calculations.
Any number will yield to this method. Say you have sqrt 2,865.3289. Just convert it to sqrt 28.653289 x 10^2 = 10sqrt28.653289 and proceed: 10 x 5 x sqrt28.653289/25...etc etc.
I think Newton's method requires calculus to strictly justify it, at least according to a wikipedia article on it.
My method requires for its justification simple arithmetic. I see it as an arithmetical, not an algebraic, method.
Also, I don't see the numbers as "guesses," but simply numbers whose squares are known, and whose squares are close to, but somewhat less than, the number the square root of which is sought. The image that comes to me is a gradual corralling of the square root, not so much wild guessing. You just keep setting limits, and the square root gets trapped.

