# Vedic Maths

### VEDIC MATHEMATICS AND ITS APPLICATION [Top]

Vedic mathematics is based on sixteen sūtras and some subsutras as proposed instructions for solving different Mathematical problems. Below is a list of the sūtras and subsutras. Vedic mathematics is apparently a part of Athar-Ved There is however a controversy among vedic scholars as to whether these Sutras are part of the original Vedic text (these sutras are not found in the vedic texts) or some one inserted it as corollary to the vedas at a later stage.

##### SUTRAS AND THEIR MEANING

By one more than the one before.All from 9 and the last from 10.Vertically and Cross-wise Transpose and Apply Samuchaya is the Same, it is Zero One is in Ratio the Other is Zero By Addition and by Subtraction By the Completion or Non-Completion Differential Calculus By the differential Specific and General The Remainders by the Last Digit

The Ultimate and Twice the Penultimate
By One Less than the One Before
The Product of the Sums
All the Multipliers
The subsutras
Proportionately
The Remainder Remains Constant
The First by the First and the Last by the Last
For 7 the Multiplicand is 143
By Osculation
Lessen by the differential
Whatever the differential, lessen by that
amount and set up the Square of the differential
Last Totalling 10
Only the Last Terms
The Sum of the Products
By Alternative Elimination and Retention
By Mere Observation
The Product of the Sums is the Sum of the Products
On the Flag
Upon further simplifying the meaning for practical application, we can write these formulae as follows:
"By one more than the previous one"
"All from 9 and the last from 10"
"Vertically and crosswise (multiplications)"
"Transpose (To move a term from one side of an algebraic equation to the other side, reversing the sign to maintain equality) and apply"
"If the Samuchaya (factor) is the same (on both sides of the equation) then that Samuchaya is equal to zero"
By the Parāvartya rule
"If one is in ratio, the other one is zero."
By the completion or non-completion (of the square, the cube, the fourth power, etc.)

Differential calculus

By and to the extent of shortfall
Specific and general
The remainders by the last digit
"The ultimate (binomial) and twice the penultimate (binomial) equals zero"
"Only the last terms,"
By one less than the one before
The product of the sum
All the multipliers
Subsūtras or corollaries
"Proportionately"
The remainder remains constant
"The first by the first and the last by the last"
For 7 the multiplicand is 143
By osculation( a contact as between two curves or surfaces, at three or more common  points)
Subtract to the extent of the shortfall
"Whatever the extent of the differential, lessen it still further to that very extent; and also set up the square (of the differential)"
"By one more than the previous one"
"Last totalling ten"

### The sum of the products [Top]

"By (alternative) elimination and retention (of the highest and lowest powers)"
By observation only(mental mathematics)
The product of the sum is equal to the sum of the products
On the flag
To check the factorisation of polynomial expressions: "The product of the sum of the coefficients in the factors is equal to the sum of the coefficients in the product." Additionally, this sub-sūtra helps to find unknown factors when some factors are known. This is normally done in algebraic equations for finding unknown quantities.
Some actual mathematical calculations applying the sutras given above:
All from nine and the last from ten
To subtract a number from a large number with 1 as first digit and many zeros after it, we normally borrow 1 from the previous digit and go on subtracting until we come to the first digit on the left hand side. In vedic system we subtract the last right hand digit from 10 and each other digit from 9. For example, to subtract 6789 from say, 10000, following procedure will apply.
10000
-6789
The leftmost three digits of 6789 that is 6,7,8 are each subtracted from 9 giving the figure 321 and the last digit that is 9 is subtracted from 10 ,giving 1 ,so the answer is 3211. Please note that in this system the operation is carried from left to right as against the conventional right to left system used nowadays. However, this method has its own limitations.

### Finding Square of numbers: [Top]

"By whatever amount the figure is short from the base number, reduce it further by the same amount from the figure to be squared and also square the figure of the differential and set it up. As an example, in finding the square of 7 we proceed as follows-
10 is taken as base number for squaring one and two digit numbers. Since 7 is 3 less than 10, decrease it by the differential (7- 3 =4). Thus 4 is the left side of our answer.
On the right hand side putting the square of the shortfall, that is 3, is 9 Putting it after 4, gives us 49, which is the answer.

Similarly, the square of other single digit numbers can be calculated. This method of squaring is simpler than conventional   method for small numbers only. For two digit numbers above 10, we add the differential instead of subtracting.
As an example, let us find out the square of number 15.
Base number is 10, (15-10) is 5 (excess from base number) which is the differential. Add this to 15
Which is 20, this is the left side of our solution. Now square the differential.5 squared is 25, now add in the following manner
20
+25
We get 225, which is the answer.
This method is  generally meant for small numbers, never- the- less, large number calculations can also be carried out. In the vedic era large number calculations may not have been required, although our Rishis evolved the Sanskrit nomenclature of numbers even up to trillions, proving that they had the knowledge of large numbers.
‘By’ one more than the one before

"Ekādhikena Pūrveṇa" means” by one more than the previous one". It gives a simple method of calculating values like 1/x9 (e.g. 1/19, 1/29, etc). This sutra also gives the method of multiplying as well as dividing algorithms.

Example: Let us calculate the value of 1/19. In this case, x = 1. For the multiplication algorithm (working from right to left), the method starts by writing the numerator, 1, as the first (rightmost) digit of the solution. Then multiply that digit by 2 (i.e.: x 1), and write that digit to the left of the already written digit. If the multiplied figure is greater than 10, denote (value – 10) and keep the "1" as "carry over" which is added to the next digit directly after multiplying.

The word "by” in this sutra means the operations that can be carried out by this formula are either multiplication or division. [In case of addition/subtraction the word "to" or "from" is used.] Thus, this formula is used both for multiplication and for division. This sūtra can also be used for multiplication of numbers with the same first digit and the sum of whose last unit digits are 10.
This sutra, multiply "by one more than the previous one." easily finds the square of numbers, which have 5 as unit place value
Example:
For a two digit number-45*45=(4*5),25=20,25
and for a three digit number 205*205=(20*21),25=420,25

Explanation- The last digit is squared (5 *5) and the previous portion is multiplicand of first figure( 4 or 20 in the above  example) one more of the first (that is 4 1 =5)0r (20 1=21) that is  ,4*5 or 20*21and arranged in the manner given in example gives  the solution.

The above method, in modern mathematics, is algebraic expression (a b) 2 = a2 2ab b2

It can also be applied in multiplications when the last digit is not 5 but the sum of the last digits is equal to 10 and the left hand side numbers are same.
Examples:
37 * 33 =3* (3 1)* 7*3= (3 * 4) 7 * 3 = 12, 21
29  *21 =2*(2 1)* 9*1=  (2 * 3)  9 * 1 = 6, 09

Modern mathematics converted this sutra as follows:
(a b)(a − b) = a2 − b2_ twice combined with the previous result to give:
(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 in 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 answer is an infinite decimal series. If the denominator contains only factors ending with 2 and 5, the answer is a finite decimal series, else it is a mixture of the two: a short finite series, followed by an infinite series.

So we start with the last digit of the result, being the dividend:
1
Multiply this by "one more", that is, 2 and place on the left and keep following this step on and on (this is the "key" from 'Ekādhikena')
21
Multiplying 2 by 2, followed by multiplying 4 by 2
421 → 8421
Now, multiplying 8 by 2, we get sixteen (here the number is more than 10, so we have to carry)
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. We have calculated up to 18 digits after decimal by using digits obtained by subtracting numerator from denominator in the answer (computed by complementing the lower half with its complement from nine):
052631578
947368421
Thus the answer is 1/19 = 0.052631578,947368421 repeating (this is an infinite series).Following this method we can go on calculating up to any place we wish. Now using division method, let us calculate the value of 1/19 as given in the sutra.

Method 2-We start again with 1 (dividend of "1/x9"), dividing by 2 (" x 1 "). We divide 1 by 2, answer is 0(actually the answer is not zero, but it is an indication that decimal has to start right now) 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" such as 1/7 can be written as 7/49 and solved following the sutra. The basic ides is to have 9 as the last digit of the denominator.

Multiplication of any number with 11, for example
11*35= 385
(1) The five in the ones place of the answer is taken from the five in 35.
(2) The eight in the answer is the sum of 35 (3 5=8).
(3) The three in the hundreds place of the answer is taken from the three in 35.

However, if in step #2 the sum is greater than 9, the sum's left digit is added to the first digit of the number multiplied by 11. For example:
11*59= 649

(1) The nine in the ones place of the answer is taken from the nine in 59.
(2) The four in the answer is the right digit in the sum of 59 (5 9=14)
(3) The six in the hundreds place of the answer is taken from the sum of the five in 59 and the digit in the tens place from the sum of 59 (5 9=14), 1 is carried over here. (5 1=6)

The steps for multiplying a three-digit number by 11 are as follows:
11*768= 8448
(1) The 8 in the ones place of the answer is taken from the eight in 768.
(2) The 4 in the tens place of the answer is taken from the sum of 8, in the ones place of 768, and 6, in the tens place of 768 (8 6=14). As 14 is greater than 9, the 1 is carried over to step 3.
(3) The 4 in the hundreds place of the answer is taken from the sum of 6, in the tens place of 768, and 7, in the hundreds place of 768, plus the carried 1 from step 2, (6 7 1=14). As 14 is greater than 9, the 1 is carried over to step 4.
(4) The 8 in the thousandths place of the answer is taken from the sum of 7, in the hundreds place of 768, plus the carried 1 from step 3 (7 1=8).

Vertically and crosswise

This formula applies to all cases of multiplication and is very useful in multiplication of one large number by another large number.
For example, to multiply 23 by 12:
2           3
|     *     |
1           2
2*1 (2*2 3*1) 3*2
2            7         6
So 23*12=276.

When any of these calculations exceeds 9 then a carry over is required.
This is the equivalent of (10a b)(10c d)=100ac 10(ad bc) bd in modern mathematics.

### Transpose and apply [Top]

This formula complements "all from nine and the last from ten", which is useful in divisions by large numbers. This formula is useful in cases where the divisor consists of small digits.

When the samuchaya is the same, that samuchaya is zero
This formula is useful in solving several special types of equations mentally. The word samuchaya has various meanings in different applications. For instance, it may mean a term which occurs as a common factor in all the terms concerned. A simple example is equation "12x 3x = 4x 5x". Since "x" occurs as a common factor in all the terms, therefore, x = 0 is a solution. Another meaning may be that samuchaya is a product of independent terms. For instance, in (x 7) (x 9) = (x 3) (x 21), the samuccaya is 7 * 9 = 3 * 21, therefore, x = 0 is a solution. Another meaning is the sum of the denominators of two fractions having the same numerical numerator, for example: 1/ (2x − 1) 1/ (3x − 1) = 0 means we may set the denominators equal to zero, 5x – 2 = 0.

Yet another meaning is "combination" or total. This is commonly used. For instance, if the sum of the numerators and the sum of denominators are the same then that sum is zero. Therefore, 4x 16 = 0 or x = −4.

This meaning ("total") can also be applied in solving quadratic equations. The ‘total” meaning can not only imply sum but also subtraction. For instance when given N1/D1 = N2/D2, if N1 N2 = D1 D2 (as shown earlier) then this sum is zero. Mental cross multiplication reveals that the resulting equation is quadratic (the coefficients of x are different on the two sides). So, if N1 − D1 = N2 − D2 then that samuchaya is also zero. This yields the other root of a quadratic equation.

Yet interpretation of "total" is applied in multi-term RHS (right hand side) and LHS (left hand side). For instance,
Here D1 D2 = D3 D4 = 2x − 16. Thus x = 8.

There are several other cases where samuchaya can be applied with great versatility. For instance, "apparently cubic" or "biquadratic" equations can be easily solved as shown below:
(x − 3)3 (x − 9)3 = 2(x − 6)3.
Note that x − 3 x − 9 = 2 (x − 6). Therefore (x − 6) = 0 or x = 6.
This would not work for the quadratic (x − 3)2 (x − 9)2 = 2(x − 6)2, which has no real or complex solutions.
Consider
Given from above N1 D1 = N2 D2 = 2x 8. Therefore, x = −4.
This formula has been extended further.
If one is in ratio, the other one is zero
This formula is often used to solve simultaneous linear equations which may involve big numbers. But these equations in special cases can be  mentally solved because of a certain ratio between the coefficients. Consider the following example:
6x 7y = 8
19x 14y = 16

Here the ratio of coefficients of y is same as that of the constant terms that is 7/14=8/16, therefore, the "other" variable is zero, as per sutra, i.e., x = 0. Hence, mentally, the solution of the equations is x = 0 and y = 8/7.

Note that it would not work if the coefficient of both the variables and the constant are” in ratio” that is  6/12=7/14=8/16 as given in the two equations below . For then, we will have de facto one equation to find two variables, whereas for finding the value of all variables, number of equations must equal the number of variables. Otherwise, we will have the case of convergent series with an infinite number of solutions.
6x 7y = 8
12x 14y = 16

This formula is easily applicable to more general cases with any number of variables. For instance

ax by cz = a
bx cy az = b
cx ay bz = c
which yields x = 1, y = 0, z = 0.

A corollary says solving "by addition and by subtraction." It is applicable in case of simultaneous linear equations where the x- and y-coefficients are interchanged. For instance:

45x − 23y = 113
23x − 45y = 91
By addition: 68x − 68 y = 204 → 68 (x − y) = 204 → x − y = 3.
By subtraction: 22x 22y = 22 → 22 (x y) = 22 → x y = 1.
Again, by addition, we eliminate the y-terms: 2x = 4, so x = 2.
Or, by subtraction, we eliminate the x-terms: – 2y = 2, and so y = – 1.
The above is the modern mathematical representation of the above sutra.
The most desirable application of Vedic mathematics is in teaching. Vedic mathematical methods are a useful resource for teachers and students, who may find elements of it easier and more accessible to teach and learn than conventional modern mathematics. In particular, these strategies may be an invaluable resource to students who already struggle with mathematics, and could benefit from alternative approaches.
One attempt at incorporating Vedic mathematics into education was made by Mark Gaskell, the head of mathematics at the Maharishi School, Lancashire, England. The school has developed a Vedic mathematics curriculum equivalent to the national one with impressive results. According to Gaskell, the alternative curriculum has resulted in livelier classes, greater student enjoyment and understanding, and improved academic performance. In fact, the first set of students to complete the course were each able to not only pass, but achieve over 80%, on the General Certificate of Secondary Education, a proficiency test taken by all secondary school British students, a year earlier than their seniors in the regular curriculum of year 2000.
If taught properly, there is great potential in how Vedic mathematics can be used to teach, learn and understand mathematics. The most important aspect of including Vedic mathematics in our education system will be to expose our students to the greatness of our culture and also  becoming open to conceptually different mathematical approaches to enhance their lateral intelligence — approaches that could usher in real scientific spiritualism in our country and again producing sage scientists in our country.

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