Technique of the extraction of root

The technical of drop by drop the is a algorithm making it possible to extract the square Racine of a decimal Number.

Algorithm

To cut out the number in sections of 2 digits starting from the comma
To take the section on the left and to cut off the successive odd numbers to him as long as that is possible
the number of subtractions carried out is the figure on the left of the root
Prendre the last odd number used, to add 1 to him, multiply by 10 and to add 1
To the result of the subtractions carried out at stage 2, to stick the following section
To the number thus obtained, to cut off, as long as that is possible, the odd numbers starting from the odd number obtained at stage 4
the number of subtractions carried out is the following figure of the root
Recommencer starting from stage 4

Remarks :

if the result of the subtractions is 0 and that the remaining sections comprise only of the zeros, one stops calculations and one writes of zero on the right digits already obtained of the root (as much as remaining sections)
if the number of subtractions carried out is 0, one removes 2 with the odd number obtained at stage 4; that gives the last odd number used

Example with number 71214,28

7 is the section on the left

7-1-3 = 3 there is 2 subtractions thus 2 is the figure on the left of the root of 71214,28

3 is the last odd number used. ( 3 +1) ×10+1 = 41

3 is the result of the subtractions. With the following section, one obtains 312

312-41-43-45-47-49-51 = 36 there is 6 subtractions thus 6 is the following figure of the root

51 is the last odd number used. ( 51 + 1) ×10+1 = 521

36 is the result of the subtractions. With the following section, one obtains 3614

3614-521-523-525-527-529-531 = 458 there is 6 subtractions thus 6 is the following figure of the root (266)

531 is the last odd number used. ( 531 +1) ×10+1 = 5321

458 is the result of the subtractions. With the following section, one obtains 45828

45828-5321-5323-5325-5327-5329-5331-5333-5335 = 3204 there is 8 subtractions thus 8 is the following figure of the root (266,8)

5335 is the last odd number used. ( 5335 +1) ×10+1 = 53361. 3204 is the result of the subtractions. With the following section, one obtains 320400

320400-53361-53363-53365-53367-53369-53371 = 204 there is 6 subtractions , 6 is the following figure of the root (266,86)

53371 is the last odd number used. ( 53371 +1) ×10+1 = 533721

204 and the following section gives 20400

one cannot thus carry out subtractions the following figure of the root is 0 (266,860)

533721-2=533719 is the last odd number “used”. The next odd number to use is ( 533719 +1) ×10+1 = 5337201

20400 and the following section gives 2040000

one cannot thus carry out subtractions the following figure of the root is still 0 (266,8600)

5337201-2=5337199 is the last odd number “used”. The next one to use is ( 5337199 +1) ×10+1 = 53372001

2040000 and the following section gives 204000000

204000000-53372001-53372003-53372005 = 43883991 there is 3 subtractions thus 3 is the following figure of the root (266,86003)

and so on…

One can check that 266,86003 ² + 0,0043883991 = 71214,28 because the root having 5 digits after the comma, the difference comprises 10 digits after the comma.

Proof of the algorithm

The algorithm is based on the one hand on the property that the sum of N first odd numbers (of 1 with 2n - 1) is N ², and on the other hand that when one changes section (2 digits), that corresponds to a change of a figure for the root and thus to consider that the root is multiplied by 10; the number is thus (10n) ² which is the sum of the 10n first odd numbers, of 1 with 20n - 1 according to the preceding property.

And like 20n + 1 = ((2n - 1) + 1) × 10 + 1, one has calculations of stage 4 of the algorithm to obtain the odd number to use in the continuation of the subtractions.

Let us notice that when one has a number with comma, one can bring back oneself to an integer by a shift of the comma by section of 2 digits: that corresponds to a shift of the comma of 1 figure for the square root. In practice, the passage of the comma consists in putting a comma in the root (see the example).

For the justification, let us call NR an integer of which one seeks the square root.

The first stage consists in separating NR in sections from 2 digits starting from the figure of the units: NR = A₀A₁… A_k where A_i are the sections of 2 digits except possibly for A₀. NR having (k+1) sections of 2 digits, its square root will be made up of (k+1) figures: c₀c₁… c_k.

The second phase of the algorithm consists in withdrawing first A₀ section the c₀ first odd numbers: A₀ = 1 + 3 +… + (2c₀ - 1) + d₀ with d₀ < 2c₀ + 1 The number of subtractions thus gives the figure corresponding of the square root: it is stage 3 of the algorithm.

Like 1 + 3 +… + (2c₀ - 1) = c₀ ², one thus has A₀ = c₀ ² + d₀. Let us multiply by 100: 100 × A0 = 100 × c₀ ² + 100 × d₀ or A₀00 = (10 × c₀) ² + d₀00. And by adding the following section, one obtains: A₀A₁ = c₀0 ² + d₀A₁ which corresponds to the beginning of stage 5 of the algorithm. Like c₀0 ² = 1 + 3 +… + (20 × c₀ - 1), the next odd number to withdraw is 20 × c₀ + 1 which is calculated starting from 2c₀ - 1: 20 × c₀ + 1 = ((2c₀ - 1) + 1) × 10 - 1 and one finds calculations of stage 4 of the algorithm to determine this number.

Then, one seeks the following odd numbers contained in d₀A₁ from 20 × c₀ + 1 and one obtains c₁ odd numbers: d₀A₁ = (20 × c₀ + 1) +… + (2c₀c₁ - 1) + d₁ with d₁ < 2c₀c₁ + 1. What gives A₀A₁ = c₀0 ² + (20 × c₀ + 1) +… + (2c₀c₁ - 1) + d₁ or A₀A₁ = 1 +… + (20 × c₀ - 1) + (20 × c₀ + 1) +… + (2c₀c₁ - 1) + d₁ one has the sum of the 10 well ´ c0 first numbers odd with the following c1 and thus A₀A₁ = (10 × c₀ + c₁) ² + d₁. The number of subtractions gave the following figure of the square root well.

The particular cases were mentioned in the example: case where one of the c_i is 0 and case where there are nothing any more but sections 00

Alternative

1. This alternative, faster and being able to be carried out of head, goes only if one is sure that the root which one seeks is an integer.

2. To start, divide the number by 100 and locate the result between two squares of consecutive integers. It is thus recommended to know the squares of the integers up to a certain point (20-25).

Examples: 14400/100=144=12 ² \ 24649/100=246,49 15 ² (225) <246,49<16 ² (256)

The exact result (in first example 12) or the lower limit of the framing (in second example 15) gives the first figures of the result. Only the last remainder to be found.

3. Look at the last figure of the initial number; it is easy to check that:

- If it is one 0, the last figure of the result is 0

- If it is one 5, the last figure of the result is 5

- If it is one 1, the last figure of the result is 1 or 9

- If it is one 4, the last figure of the result is 2 or 8

- If it is one 9, the last figure of the result is 3 or 7

- If it is one 6, the last figure of the result is 4 or 6

it cannot be 2,3 or 7 (a number being finished by 2,3 or 7 cannot have a whole root).

If you have the choice between two digits, look at, into 2. , if the number is closer to the lower or higher limit of the framing, and choose consequently smallest or largest of the two digits.

Examples: 14400, the last figure is 0, therefore the result is 120 (12 then 0) \ 24649, the last figure is 9 and 246,49 is closer to 256 than of 225, the result is thus 157 (15 then 7)

4. The results given are, of course exact. However if you test this technique with numbers chosen randomly, you do not astonish to obtain false results and read again the 1.

See too

Method of Héron

Random links: