
Re: Square root of six
Posted:
Aug 27, 2012 12:09 AM


I looked at your paper and see that for a number like 12.643 you would start with something like 12.643 / 9, make a second guess until you got to a number slightly greater than one and then use essentially a variant of the newton method thereafter. I am still not sure why this is better than just starting with the newton method?
Bob Hansen
On Aug 26, 2012, at 11:53 PM, Robert Hansen <bob@rsccore.com> wrote:
> I agree with your statement, that it is natural for students to ask how you find the square root, but I am not following why you think this method is better than Newton's? If the kids are of arithmetic age then I would use a guess and check method. This will be their first real experience with irrational numbers and interpolation exercises will inch them closer to a notion of real numbers. In your example though you seem to be using algebraic reasoning (which would be in an algebra class) and I think Newton's method is easier to explain (how it works), plus it is an actual algorithm (it can be programmed). There is not a lot of practical benefit from a square root algorithm (as Wayne suggested) but I think there are pedagogical benefits, and the question (how do you find the square root?) deserves an answer. I am not sure how this is the answer though. > > Using your method, how would you find the square root of 12.643? > > Bob Hansen > > > On Aug 26, 2012, at 7:53 PM, Peter Duveen <pduveen@yahoo.com> wrote: > >> It is natural for students to ask how to find the square root of a number, much as they would want to calculate, say, the quotient of two numbers through long division. >> >> We endeavor to work out the square root of six with simple arithmetic. >> >> Find sqrt 6 >> >> sqrt 6 = sqrt (6 x 4/4) >> >> = sqrt ([6/4] x 4) >> >> = sqrt (6/4) x sqrt (4) >> >> = 2 sqrt (6/4) >> >> = 2 sqrt 1.5 >> >> We seek a number whose square is reasonably close to, but does not exceed, 1.5 >> >> 1.2 x 1.2 = 1.44, so >> >> 2 sqrt 1.5 = 2 sqrt (1.5 x 1.44/1.44) >> >> = 2 sqrt (1.5/1.44 x 1.44) >> >> = 2 sqrt 1.5/1.44 x sqrt 1.44 >> >> = 2 x 1.2 x sqrt 1.5/1.44 >> >> Carry out the division: >> >> >> 1.44 1.50000000 >> >> 1.04166666666.... >> = 144 150.000000000000 >> >> It is immediately evident that a number whose square will be reasonably close to 1.041666...., but does not exceed it, is 1.02. That's because the square of 1 plus a small number is approximately 1 plus twice the small number. Conversely, the square root of one plus a small number is 1 plus half the small number. This is a rule of thumb, but the analysis could be made more rigorously. >> >> Thus, 1.02 x 1.02 = 1.0404. >> >> 2 x 1.2 x sqrt 1.5/1.44 >> >> = 2 x 1.2 x sqrt 1.0416666.... >> >> = 2 x 1.2 x 1.02 x sqrt 1.04166666.../1.0404 >> >> = 2 x 1.2 x 1.02 x sqrt 1.001217..etc >> >> By our rule of thumb, we can immediately say that 1.0006 is a number whose square will be close to, but will not exceed, 1.001217..etc >> >> Thus, we have: >> >> 2 x 1.2 x 1.02 x 1.0006 sqrt 1.001217..etc/1.00120036 >> >> Let's see how close the numbers outside the radical sign approach sqrt 6: >> >> 2 x 1.2 x 1.02 x 1.0006 = 2.4494688. >> >> Remembering that the square of our result should be less than 6 by a small amount, we carry out the multiplication, which yields: >> >> 5.99989740217344 >> >> By continuing our calculations, we could have found a number that differs even less from the square root we seek. >> >> This method, which I developed about a year ago, is accessible to those with rudimentary mathematical skills, including long division, multiplication of fractions, the concept of square root, and the rules governing the square roots of products. >> >> The method itself is transparent in terms of the way it works, dependent as it is on the basic rules of mathematics, so it should not be confusing to students. I call this method "Beating the square root out of the radical sign with the number one." >> >> This method is certainly not in wide use, and one would be hard pressed, as I have been, to find it in old text books or anywhere on the internet. I have heard claims that this method is the same as Newton's method, or the Babylonian method, etc. etc., but no clear demonstration of such claims. This method is based on ratios, whereas other methods generally rely on differences. The same approach may be used to also find cube roots and fourth roots, etc.

