Swami Bharati Krishna Tirtha's Vedic mathematics

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For the actual mathematics of the Vedic period, see the articles on Sulba Sutras, Ancient Vedic weights and measures and Indian mathematics. Nor should this article be confused with Vedic science.

Vedic mathematics is a system of mental calculation developed by Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaja in the middle 20th century which he claimed he had based on sutras he had found in an appendix of Atharvaveda, an ancient text of the Indian teachings known as the Vedas. He stated that these sutras only appeared in his personal copy of the appendix and not in the generally known appendices; his general editor noted that the style of language of the sutras "point to their discovery by Shri Swamiji himself".

It has some similarities to the Trachtenberg system in that it speeds up some arithmetic calculations. It has applications to more advanced mathematics, such as calculus and linear algebra. The system was first published in the book Vedic Mathematics ISBN 81-208-0164-4 in 1965. The system has since been developed further and there have been several other books released. Tirthaji claims in the book to have authored 16 volumes, one on each Sutra, and that they were subsequently lost in a fire in 1956.[1]

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

Contents

[edit] Sutras

  • By one more than the one before
  • All from 9 and the last from 10
  • Vertically and crosswise
  • Transpose and apply
  • If the Samuccaya is the same it is zero
  • If one is in ratio the other is zero
  • By addition and by subtraction
  • By the completion or non-completion
  • Differential calculus
  • By the deficiency
  • 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 sum
  • All the multipliers

[edit] 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 deficiency
  • Whatever the deficiency lessen by that amount and set up the square of the deficiency
  • Last Totalling 10
  • Only the last terms
  • The sum of the products
  • By alternative elimination and retention
  • By mere observation
  • The product of the sum is the sum of the products
  • On the flag

[edit] All from nine and the last from ten

If you want to subtract 4679 from 10000, you can easily apply the Nikhilam Navatashcaramam Dashatah sutra ("All from 9 and the last from 10"). Each figure in 4679 is subtracted from 9 and the last figure is subtracted from 10, yielding 5321.

[edit] Corollary

"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:
11^2 = (11+1)\cdot 10+1^2 = 121.\,
12^2 = (12+2)\cdot 10+2^2 = 144.\,
14^2 = (14+4)\cdot 10+4^2 = 18\cdot10+16 = 196.\,
25^2 = ((25+5)\cdot 2)\cdot 10+5^2 = 625.\,
35^2 = ((35+5)\cdot 3)\cdot 10+5^2 = 1225.\,
and so on.

This is based on the fact that a2 = (a + b)(a − b) + b2.

[edit] By one more than the one before

"Eka Dikena Parvesam" is the Sanskirt term for "One more than the previous one". It provides a simple way of calculating values like 1/x9 (e.g 1/19, 1/29, etc). Let's take one 1/x9 and calculate. e.g., 1/19. In this case, x=1. The method is to write that x, then multiply it by 2 and keep that digit to its left. If the result of the multiplication is greater than 10, keep (value - 10) and keep the "1" as "carry over" which you'll add to the next digit.

The preposition "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.


Note: This applies to multiplication of numbers with the same first digit and the sum of their last unit digits is 10.


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

35 × 35 = ((3 × 3) + 3),25 = 12,25
125 x 125 = ((12 x 12) + 12),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 (3x3) or (12x12) and adding the same digit in that figure (3or12) resulting in the answer 1225.

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

(10c+5)^2=100c^2+100c+25=100c(c+1)+25.\,

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 (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 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.

[edit] Method 1: Using multiplication

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.) Each factor of 2 or 5 or 10 in the denominalor gives one fixed decimal digit.

So we start with the last digit

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 (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!

[edit] Method 2: Using division

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

[edit] Multiplying by 11

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 the 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 take 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 take from the sum of the five in 59 and the digit in the tens place from the sum of 59 (5+9=14) --> (5+1=6)

[edit] 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.

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 is required.

This is just (10a+b)(10c+d)=100ac+10(ad+bc)+bd.

Video of a simple multiplication

[edit] 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.

[edit] 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 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,

{2x+9 \over 2x+7}={2x+7 \over 2x+9}.

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 x2 are different on the two sides). So, if N1 − D1 = N2 − D2 then that samuccaya is also zero. This yields the other root of a quadratic equation.

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

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

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:

(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 apparently quadratic (x − 3)2 + (x − 9)2 = 2(x − 6)2, which has no real or complex solutions.

Consider

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

Observe: N1 + D1 = N2 + D2 = 2x + 8. Therefore, x = −4.

This formula has been extended further.

[edit] 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 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" variable is zero, i.e., x = 0. Hence, mentally, 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 divide by 2!)

Note that it would not work if both had been "in ratio". For then we have the case of coinciding lines 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.

[edit] Criticism

Critics in India 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, special cases. They further say that such arithmetic as is sped up by application of the sutras can be performed on a computer or calculator anyway, making their knowledge rather irrelevant in the modern world.

They are also 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. [1]

Additionally, the promotion of Vedic Maths as advanced mathematics has drawn critisism from Mathematicians. Although quite useful for practical calculations in daily life, the tricks involved have no connection to higher mathematics or mathematical research and can be explained very simply.

[edit] See Also

  • [2] GCF of two polynomials
  • [3] Separation of a fractional expression into partial fractions
  • [4] Converting fractions to decimal values using auxilliary fractions
  • [5] Calculating the square root by the duplex method

[edit] External links

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