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There is a method that looks like "long division":

  1. Split the number up into pairs of digits starting at the decimal point; if this results in a lone digit after the decimal point, add a zero, eg if the number was "123.456" it would become "1 23 . 45 60". The first pair at the left hand end may be a single digit. If there is no decimal point, start at the right hand end (where it is hiding). A decimal point and pairs of zeros can be added to the end of any integer if needed, eg "123" could become "1 23 . 00 00"
  2. Start with the first digit or pair of digits.
  3. Find the largest number which when squared is not bigger than this. Write this digit as the divisor and the digit of the quotient above the first digit or pairs of digits.
  4. Square this digit, write it below the digit/pair and subtract (like normal division).
  5. Bring down the next pair of digits
  6. Double the whole quotient so far and multiply it by 10. You now need to find a digit to put in the units of this so that when multiplied by the units digit it is the largest number not larger than the number formed in step 5 (just like normal division, but you are also creating the divisor)
  7. Write this digit as the next digit of the quotient over the pair of digits
  8. Multiply the number in step 6 by its unit digit and write the answer below the number of step 5 and subtract (like normal division).
  9. Repeat steps 5 to 8 until all the digits of the dividend have been brought down and the result of the subtraction in step 8 is zero, or until the required accuracy is reached.
  10. The square root is the quotient

Extra pairs of zeros can be added to the dividend after the decimal point for more accuracy if required, just like normal division.

This is best explained by using an example:

Work out the square root of 7:

In the following example, the numbers in braces (eg {5}) refer to the steps above.

Put a decimal point and (six) zeros after the 7 and then write division by splitting dividend into pairs of digits from the decimal point:

..............2 . ..6...4...5 .....{10}

............--------------

.........2 | 7 . 00 00 00 .....{1, 2, 3} 22 is the highest square not bigger than 7

............. 4 ......................{4} = 2 x 2

.............--

.......46 | 3 ..00 ...............{5}Bring down 00, {6}double 2 = 4 & {7}put 6 in result & on end to make divisor 46*

............. 2...76 ...............{8} = 46 x 6

............. ------

.....524 |..... 2400 ...........{5}Bring down 00, {6}double 26 = 52 & {7}put 4 in result & on end to make divisor 524

.................. 2096 ...........{8}= 524 x 4

.................. -----

..5285 | ....... 30400 .......{5}Bring down 00, {6}double 264 = 528 & {7}put 5 in result & on end to make divisor 5285

.................... 26425 .......{8}= 5285 x 5

.................... -------

...................... 3975 .......{9}

Thus √7 ≈ 2.65

*The step to make the divisor 46 explained slightly fuller, ie where the 6 came from:

2 x 2 = 4, 4 x 10 = 40.

Need to add a digit n to this so that (40 + n) x n is the largest number not bigger than 300 - the result of the subtraction (7-4=3) with the next pair of digits (00) brought down just like in division to give 300. Attempted guesses at the last digit:

(40+5) x 5 = 45 x 5 = 225 - smaller than 300, possible bigger number

(40+7) x 7 = 47 x 7 = 329 - too big

(40+6) x 6 = 46 x 6 = 276 - smaller than 300 - biggest possible.

So 6 is the digit to use to make the divisor (2 x 2) x 10 + 6 = 46 and 6 gets put in the quotient just like in a normal division. Then 46 x 6 = 276 is written below the 300 and subtracted (as in division) to give 300 - 276 = 24.

Then the next pair of digits (00) is brought down to give 2400, the quotient so far is doubled and the process repeats, looking for the units digit again, this time for (2 x 26) x 10 = 520 so that (520 + n) x n is the biggest number not larger than 2400.

When done in binary as opposed to base 10, this process becomes extremely efficient - the finding of the digit to add as the units of the divisor is a choice of 0 or 1! (For a digital computer the doubling and choice is extremely quick and easy.)

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