THE SQUARE ROOT ALGORITHM


    Divide the digits into groups of two, starting at the decimal 
point:
           ____________
          V 10'75'80.00

The first group may contain either one or two digits.


    Find the largest integer (call it a) whose square is less 
than or equal to the first group, and write it above that group:

             3
           ____________
          V 10'75'80.00

    Subtract a^2 from that group and bring down the next group, 
as if you were doing long division:

             3
           ____________
          V 10'75'80.00
             9
            _____
             1 75

    Now the fun begins.  Multiply the answer so far by 20, and 
divide it into the number at the bottom.  Call the quotient 
(without the remainder) b, and write it above the second group:

             3  2
           ____________
          V 10'75'80.00
             9
            _____
          60)1 75

(To understand what is going on in this step and the next, stare 
awhile at Figure 5.1 of Katz.  The extra factor of 10 in the 
divisor arises because the digit  a  contributes 10a units on the 
scale of the digit b being considered now.)

    Multiply b by the divisor (60) and add b^2.  Subtract this 
number from the bottom line, and bring down the next group:

             3  2
           ____________
          V 10'75'80.00
             9
            _____
          60)1 75
             1 24
            --------
               51 80

    Multiply the entire answer so far by 20 and divide to get c.  
Multiply c by the divisor and add c^2.

             3  2  8
           ____________
          V 10'75'80.00
             9
            _____
          60)1 75
             1 24
            --------
           640)51 80
               51 84   (= 640 x 8 + 64)

Oops!  We overshot, so we need to reduce c by 1.  (This can't 
happen to a competent student in long division, but it can happen 
here because of the need to accommodate the c^2 term.)

             3  2  7
           ____________
          V 10'75'80.00
             9
            _____
          60)1 75
             1 24
            --------
           640)51 80
               45 29   (= 640 x 7 + 49)
               --------
                6 51 00

    We can keep going to calculate places after the decimal 
point: 

             3  2  7. 9
           ____________
          V 10'75'80.00
             9
            _____
          60)1 75
             1 24
            --------
           640)51 80
               45 29   
               --------
           6540)6 51 00
                5 89 41
                ----------
                  61 59 00,   etc.

However, you must not do this unless you know that the digits 
after the decimal point in the original number really are zero; 
the number of genuinely significant figures in the answer can't 
exceed the number of groups (half the number of significant 
digits) in the original number.  That's because an error in the 
root gets multiplied by the big, accurate part in computing the 
square.

    My calculator says:

           _______
          V 107580 =327.9939

and 

          (327.9)^2 = 107518.41,   (327.99)^2 = 107577.44
     
We can see how the error in the square is diminishing.