
Square Root by Long Division - ColinWright
http://www.solipsys.co.uk/new/SquareRootByLongDivision.html?HN_20141201
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jonsen
Maybe too OT, but here is another proof

    
    
      P = n * (n＋1) * (n＋2) * (n＋3) 
    
         = n*(n＋3) * (n＋1)*(n＋2) 
    
         = (n*n ＋ 3n) * (n*n ＋ 3n ＋ 2) 
    
         = m * (m＋2) 
    
      If P has an integer square root, it must be (m＋1) 
    
      Trying (m＋1) * (m＋1) we get m*m + 2m + 1 = m * (m＋2) + 1 = P + 1

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arh68
I like your proof better! Much simpler to read. The _if_ part confused me a
bit, though

    
    
        k = m + 1
    
        P = m * (m + 2)
    
          = (k-1) * (k+1)
    
          = k^2 - 1

~~~
jonsen
The _if_ part is meant to illustrate the idea, that if P is not a perfect
square, then (m+1)^2 will give the nearest perfect square.

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jandrese
I'm a little confused by the description. He starts out saying:

Let's compute the square root of 677329.

We start by thinking of it as 67 x 10^4 + 7392. That means the square root
will be 800 and a bit (because the square root of 67 is 8 and a bit, and the
square root of 10^4 is 100.)

How did he know to start with 67 x 10^4 + 7392 instead of 6 x 10^5 + 77392?
The Square root of 6 is a bit over 2, but obviously starting with this digit
doesn't get you the correct answer. In the next step he works with 3 digits
instead of two. Knowing how many digits to work with at each step seems a
little magical.

I guess you look for the biggest single digit square factor in the original
number? So the biggest number < 81? Or am I thinking about this wrong?

This is one of those frustrating math examples where the author skips over a
lot of the detail because it is "obvious".

Edit: I've really gotta learn how the markup works on this site.

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jonsen
You should factorize as n*10^m where m is even. Then the square root of the
10^m factor will be 10^(m/2), which then gives the order of the full square
root.

~~~
mturmon
And one way to really internalize this is to start taking square roots (in
your head, of course) of large numbers you see around you. Maybe license plate
numbers if you're waiting in traffic.

If you have extra time, you can do one step of refinement of your original
guess, as done in the OP. ("Take the residual R, and the guessed-at factor p,
and find the largest number N such that 2 __* p __* N <= R")

