Square Root Algorithm
How to Find a Square Root
In order to easily explain the algorithm, I will use the example of 
.
1. Group the digits in pairs working right to left. There may be only one digit at the far left.
2. Find the integer so that it will always have a perfect square without exceeding the first pair. Write this number to the left and on top of the square root sign.
4
3. Now the square the number on top of the square root sign, here it is 
which is 16. Then subtract the result from
the first pair of numbers.
4
-16
2
4. Next, bring down the next pair of numbers along side of the result of the subtraction. Double the number above the square root sign and write the result to the left of the subtraction answer.
4
-16
(8_) 
5. Find a number to fill in the blank so the (eighty-"something") x ("something") is as close to 227 without exceeding it. For this example, the "something" is 2. Write the number 2 in the blank and on top of the square root next to the four.
4
-16
(82)
6. Now calculate 82 x 2 and subtract from 227. Then bring down the next pair of digits, 31.
4 2
-16
(82)
-164
6331
7. Double the amount on top of the square root sign, here it is 42 and repeat steps four and five until the closest approximation is reached.
When the algorithm is carried through for 182731 to one decimal place, the result is the following:
4 2 7 . 4
-16
(82)
-164
(847) 6331
-5929
(8544) 40200
-34176
6024
Derivation of the Algorithm
The underlying principle of the algorithm is 
where x is the approximation and r is the
remainder. When is expanded, the result is:
As more digits are added to x, the closer the approximation comes to the actual square root.
When the number, n, is grouped into pairs, the pairs will not exceed 99. Therefore, the square of the first pair will not exceed one digit, or the number 9. This first approximation is x as shown below.
x
__________
So for a simpler example of 
, the following would be calculated.
1
-1
44
So x = 10 and 
. If 10 is substituted in for x, then:
Now the x needs to be doubled. Remember, since doubling a number is the same as multiplying a number by two, the doubling takes care of the 2(10) in the above equation. When this is calculated, the result is as follows.
x
__________
(2x + _)
For our example, the result is:
1
-1
(20 + _) 44
Since twenty-"something" is the same thing as stating twenty plus "something", we can view this operation either way. For the derivation of the algorithm, it is easier to view this as twenty plus "something".
Next, figure how many times 2x goes into . This can be calculated using polynomial
division to find the answer r. So we add r to 2x and also write an r on the top
of the square root sign. When 2x + r is multiplied by r and subtracted from , the result is the following:
x
__________
(2x + r)
-
___________
0
Our example is calculated like this:
12
-1
(20 + 2) 44
-44
0
The example used, 144, happens to be a perfect square so the last calculation yields a zero. If a number which is not a perfect square is used, then the 2 in the example would have further approximated x and the process would continue until a perfect square was reached, or the closest approximation desired was reached.