Sutra 1 - By one more than the one before

By one more than the one before

"Ekādhikena Pūrveṇa" is the Sanskrit term for "[by] One more than the previous one". It provides a simple way of calculating values like 1/x9 (e.g: 1/19, 1/29, etc). The sūtra can be used for multiplying as well as dividing algorithms.

Example: let's calculate 1/19. In this case, x = 1 . For the multiplication algorithm (working from right to left), the method is to start by denoting the dividend, 1, as the first (rightmost) digit of the result. Then multiply that digit by 2 (i.e.: x + 1 ), and denote that next digit to its left. If the result of this multiplication was greater than 10, denote (value – 10) and keep the "1" as "carry over" which you'll add to the next digit directly after multiplying.

The preposition "by" means the operations this formula concerns are either multiplication or division. [In case of addition/subtraction preposition "to" or "from" is used.] Thus this formula is used for either multiplication or division. It turns out that it is applicable in both operations.

Note: This sūtra can also be applied to multiplication of numbers with the same first digit and the sum of their last unit digits is 10.

An interesting sub-application of this formula is in computing squares of numbers ending in five. Examples:

35×35 = ((3×3)+3),25 = 12,25 and 125×125 = ((12×12)+12),25 = 156,25

or by the sūtra, multiply "by one more than the previous one."

35×35 = ((3×4),25 = 12,25 and 125×125 = ((12×13),25 = 156,25

The latter portion is multiplied by itself (5 by 5) and the previous portion is square of first digit or first two digit (3×3) or (12×12) and adding the same digit in that figure (3or12) resulting in the answer 1225.

(Proof) This is a simple application of (a + b)² = a² + 2ab + b² when a = 10c and b = 5, i.e.

It can also be applied in multiplications when the last digit is not 5 but the sum of the last digits is the base (10) and the previous parts are the same. Examples:

37 × 33 = (3 × 4),7 × 3 = 12,21

29 × 21 = (2 × 3),9 × 1 = 6,09

This uses (a + b)(a − b) = a² − b² twice combined with the previous result to produce:

(10c + 5 + d)(10c + 5 − d) = (10c + 5)² − d² = 100c(c + 1) + 25 − d² = 100c(c + 1) + (5 + d)(5 − d).

We illustrate this sūtra by its application to conversion of fractions into their equivalent decimal form. Consider fraction 1/19. Using this formula, this can be converted into a decimal form in a single step. This can be done by applying the formula for either a multiplication or division operation, thus yielding two methods.

Method 1: example: using multiplication to calculate 1/19

For 1/19, since 19 is not divisible by 2 or 5, the fractional result is a purely circulating decimal. (If the denominator contains only factors 2 and 5, the result is a purely non-circulating decimal, else it is a mixture of the two: a short non-circulating sequence of digits, followed by an endless repetition.) Each factor of 2 or 5 or 10 in the denominator gives one fixed decimal digit.

So we start with the last digit of the result, being the dividend:

1

Multiply this by "one more", that is, 2 (this is the "key" digit from 'Ekādhikena')

21

Multiplying 2 by 2, followed by multiplying 4 by 2

421 → 8421

Now, multiplying 8 by 2, sixteen

68421

1 ← carry

multiplying 6 by 2 is 12 plus 1 carry gives 13

368421

1 ← carry

Continuing

7368421 → 47368421 → 947368421

1

Now we have 9 digits of the answer. There are a total of 18 digits (= denominator − numerator) in the answer computed by complementing the lower half (with its complement from nine):

052631578

947368421

Thus the result is 1/19 = 0.052631578,947368421 repeating.

1
21
421
8421
68421 (carry 1) – we got 16, so we keep 6 and carry 1
368421 (carry 1) – we get 6*2 + carry 1 = 13, so we keep 3 and carry one

do this to eighteen digits (19–1. If you picked up 1/29,
you'll have to do it till 28 digits). You'll get the following
1/19 = 052631578947368421
     10100111101011000

Run this on your favorite calculator and check the result!

Method 2: example: using division to calculate 1/19

The earlier process can also be done using division instead of multiplication. We start again with 1 (dividend of "1/x9"), dividing by 2 (" x + 1 "). We divide 1 by 2, answer is 0 with remainder 1

result .0

Next 10 divided by 2 is five

.05

Next 5 divided by 2 is 2 with remainder 1

.052

next 12 (remainder,2) divided by 2 is 6

.0526

and so on.

Other fractions can sometimes be converted into the format of "d/x9"; as another example, consider 1/7, this is the same as 7/49 which has 9 as the last digit of the denominator. The previous digit is 4, by one more is 5. So we multiply (or divide) by 5, that is:

…7 → 57 → 857 → 2857 → 42857 → 142857 → .142,857 (stop after 7 − 1 digits)

Carry overs - 3 2 4 1 2