# Vedic mathematics

Note: Vedic mathematics is not to be confused with Vedic physics or any other Vedic Science, which is a kind of Hindu philosophy, and is unrelated to the practice of actual science, (that is, to mathematics or the application of the scientific method.)

Vedic mathematics is a system of mental calculation developed by Shri Bharati Krishna Tirthaji which he claimed he had based on a lost appendix of Atharvaveda, an ancient text of the Indian teachings called Veda. It has some similarities to the Trachtenberg system in that it speeds up some arithmetic calculations. It claims to have applications to more advanced mathematics, such as calculus and linear algebra. The system was first published in the book Vedic Mathematics ISBN 8120801644 in 1965. The system has since been developed further and there have been several other books released.

Critics have questioned whether this subject deserves the name Vedic or indeed mathematics. They point to the lack of evidence of any sutras from the Vedic period consistent with the system, the inconsistency between the topics addressed by the system (such as decimal fractions) and the known mathematics of early India, the substantial extrapolations from a few words of a sutra to complex arithmetic, and the restriction of applications to convenient cases. They have also been worried that it deflects attention from genuine achievements of ancient and modern Indian mathematics and mathematicians, and that its promotion by Hindu nationalists may damage mathematics education in India.

The system is based upon sixteen formulas and their corollaries, some of which are described below.

 Contents

## All from nine and the last from ten

Corollary 1: Whatever the extent of its deficiency, lessen it still further to that very extent; and also set up the square of that deficiency.

For instance, in computing the square of 9 we go through the following steps:

1. The nearest power of 10 to 9 is 10. Therefore, let us take 10 as our base.
2. Since 9 is 1 less than 10, decrease it still further to 8. This is the left side of our answer.
3. On the right hand side put the square of the deficiency, that is 12. Hence the answer is 81.
4. Similarly, 82 = 64, 72 = 49.
5. For numbers above 10, instead of looking at the deficit we look at the surplus. For example:
[itex]11^2 = (11+1)\cdot 10+1^2 = 121.\, [itex]
[itex]12^2 = (12+2)\cdot 10+2^2 = 144.\, [itex]
[itex]14^2 = (14+4)\cdot 10+4^2 = 18\cdot10+16 = 196.\, [itex]
and so on.

This is based on the identities [itex](a+b)(a-b)=a^2-b^2[itex] and [itex](a+b)^2=a^2+2ab+b^2[itex].

## By one more than the one before

The proposition "by" means the operations this formula concerns are either multiplication or division. [ In case of addition/subtraction proposition "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.

An interesting application of this formula is in computing squares of numbers ending in five. Consider:

35 × 35 = (3 × (3 + 1)),25 = 12,25

The latter portion is multiplied by itself (5 by 5) and the previous portion is multiplied by one more than itself (3 by 4) resulting in the answer 1225.

This is a simple application of [itex](a+b)^2=a^2+2ab+b^2[itex] when [itex]a=10c[itex] and [itex]b=5[itex], i.e.

[itex](10c+5)^2=100c^2+100c+25=100c(c+1)+25[itex].

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. Consider:

37 × 33 = (3 × 4),7 × 3 = 12,21
29 × 21 = (2 × 3),9 × 1 = 6,09

This uses [itex](a+b)(a-b)=a^2-b^2[itex] twice combined with the previous result to produce:

[itex](10c+5+d)(10c+5-d)=(10c+5)^2-d^2=100c(c+1)+25-d^2=100c(c+1)+(5+d)(5-d)[itex].

We illustrate this formula 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: using multiplications

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 is a purely non-circulating decimal, else it is a mixture of the two.)

1

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

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:

052631578
947368421

Thus the result is .052631578,947368421

#### Method 2: using divisions

The earlier process can also be done using division instead of multiplication. We divide 1 by 2, answer is 0 with remainder 1

.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.

As another example, consider 1/7, this same as 7/49 which as last digit of the denominator as 9. 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)

3     2      4       1        2

## Vertically and crosswise

This formula applies to all cases of multiplication and is very useful in division of one large number by another large number.

## Transpose and apply

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. This formula can be used to derive the Horner's process of Synthetic Division.

## When the samuccaya is the same, that samuccaya is zero

This formula is useful in solution of several special types of equations that can be solved visually. The word samuccaya 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 samuccaya 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 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,

[itex]{2x+9 \over 2x+7}={2x+7 \over 2x+9}.[itex]

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 N1D1 = 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 x2 are different on the two sides). So, if N1D1 = N2D2 then that samuccaya is also zero. This yield the other root of a quadratic equation.

Yet interpretation of "total" is applied in multi-term RHS and LHS. For instance, consider

[itex]{1 \over x-7}+{1 \over x-9}={1 \over x-6}+{1 \over x-10}.[itex]

Here D1 + D2 = D3 + D4 = 2x − 16. Thus x = 8.

There are several other cases where samuccaya can be applied with great versatility. For instance "apparently cubic" or "biquadratic" equations can be easily solved as shown below:

[itex](x-3)^2+(x-9)^3=2(x-6)^3.[itex]

Note that x − 3 + x − 9 = 2(x − 6). Therefore (x − 6) = 0 or x = 6.

(remark by different author: Note also that:[itex](6-3)^2+(6-9)^3=3^2-3^3=9-27=-18[itex], whereas [itex]2 (6-6)^3=0[itex]

Thus: x=6 is NOT a solution!)

(Another note. This example does work if one considers all exponants to be cubic, in which case [itex](6-3)^3+(6-9)^3[itex] will give [itex]27-27[itex].)

Consider

[itex]{(x+3)^3 \over (x+5)^3}={x+1 \over x+7}.[itex]

Observe: 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 simple equations which may involve big numbers. But these equations in special cases can be visually 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. Therefore, the "other" is zero, i.e., x = 0. Hence the solution of the equations is x = 0 and y = 8/7.

(alternatively:

19x + 14y = 16 is equivalent to:
(19/2)x +7y = 8.

Thus it is obvious that x has to be zero, no ratio needed, just div by2!)

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 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(xy) = 204 => xy = 3.

By subtraction: 22x + 22y = 22 => 22(x + y) = 22 => x + y = 1.

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