Finding Square RootsDate: 09/06/98 at 17:48:28 From: jack mardekian Subject: Re: WHY algorithm for finding square root works Dear Dr. Math, I know there is an algorithm for calculating square roots, but I'm interested in finding out why the process works. I am only in 7th grade and am taking Algebra this year. Can you please explain why the algorithm works in simple terms? I'm not quite sure I understand the other explanation given in the FAQ. Thank you. Stacey M. Date: 09/09/98 at 12:00:37 From: Doctor Peterson Subject: Re: WHY algorithm for finding square root works Hi, Stacey. It's hard to really prove the method works without algebra, but maybe I can more or less convince you what's going on by looking closely at an example. You'll need to understand the use of parentheses and the rules for arithmetic, especially the distributive rule, but I'll avoid the letters and try to take you through it slowly. (I'll use a different example to avoid the zero, which could confuse things a little.) Here's the first pass through the algorithm: 1 /------------- \/ 2 64.00 00 00 1 ---- 1 64 Here we had the first digit, giving an answer of 10 (so far); then we subtracted the square of 1, leaving a remainder of 1, and brought down the 64. That's really the same as subtracting the square of 10 from 264, leaving 164. So what we've done so far is: 264 = 10^2 + 164 writing 264 as a square plus a remainder. Now we get the next digit this way: 1 6. /------------- \/ 2 64.00 00 00 1 ---- 20+6=26 | 1 64 1 56 ---- 8 Here we doubled the 10 that was our current approximation, and looked for a digit (the 6) that we could add to it that so that: 164 = (2*10 + 6)*6 + 8 with the smallest possible remainder, 8. Why did we want this? Let's put together what we've found: 264 = 10^2 + 164 = 10^2 + (2*10 + 6)*6 + 8 We can rearrange this to look this way: 264 = 10^2 + 2*10*6 + 6*6 + 8 But the first part of this is exactly what we get when we square 16, which is 10 + 6: 16^2 = (10 + 6)^2 = (10 + 6) * (10 + 6) = 10 * (10 + 6) + 6 * (10 + 6) = 10*10 + 10*6 + 6*10 + 6*6 = 10^2 + 2*10*6 + 6*6 This is really just what we do when we use the usual method to multiply two numbers, and is a trick that is often taught for squaring two-digit numbers: square the tens' digit to get the hundreds of the answer, square the ones' digit to get the ones', and double the product of the tens' and the ones' to get the tens' digit of the answer (then so whatever carries you need). You can even write multiplication out digit by digit like this: 16 16 * 16 * 16 ---- ---- 36 36 = 6*6 96 96 = 6*16 6 60 = 6*10 6 60 = 10*6 16 160 = 10*16 1 100 = 10*10 ---- ---- 256 256 So what we've figured out is that: 264 = 16^2 + 8 which again gives us a square plus a remainder. If you think through that, you can see that when we looked for the value, 6, that would make: 164 = (2*10 + _)*_ + rem we were just looking for the ones' digit that would make 264 = (10 + _)^2 + rem The next pass will look like this: 1 6. 2 /------------- \/ 2 64.00 00 00 1 ---- 20+6=26 | 1 64 1 52 ---- 320+2=322 | 8 00 6 44 ----- 1 56 (Here we have to ignore the decimal point for a while, and imagine we are looking for the square root of 26400.) We've chosen the 2 so that: 800 = (2*160 + 2)*2 + 156 and since 800 comes from the previous remainder, 8, multiplied by 100, we know that: 26400 = 16^2 * 100 + 8 * 100 = 160^2 + 800 Putting this together, we see that: 26400 = 160^2 + (2*160 + 2)*2 + 156 which can be rearranged as before to say: 26400 = 160^2 + 2*160*2 + 2*2 + 156 = (160 + 2)^2 + 156 = 162^2 + 156 Now we've approximated 26400 as a square plus a remainder, which (putting the decimal point back in) means: 264.00 = 16.2^2 + 1.56 Again, we've written 264 as the sum of a square and a still smaller remainder. At each step, if the digits so far form a number N, and the number whose square root we are finding (up to the decimal place we are working on) is M, we are looking for a next digit d so that: M = (10*N)^2 + (20*N + d)*d + rem = (10*N)^2 + 2*(10*N)*d + d^2 + rem = (10*N + d)^2 + rem I think this should help you out. Mostly what I have done is to do this algebra with numbers instead of letters - not a bad way to think through algebra when you aren't sure of it yet. If you have trouble following something I said, just read closely, looking for where each part of an expression came from, and I think you should be able to follow it. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/