Vedic mathematics (calculation methods)

Under Vedic mathematics is understood calculation rules, which of Bharati Krishna Tirthaji allegedly (1884-1960) from 1911 to 1918 from the Veda were worked out. They were published posthumously in 1965 and are believed to be based on a lost appendix to the Atharvaveda . Traceability to the Veda was contested from the start and Tirthaji was never able to provide any evidence to support his claim. This type of calculation is based on 16 rules. It has similarities with the Trachtenberg high- speed calculation method , as it speeds up some arithmetic calculations.

Critics not only question the term “Vedic”, but also believe that these rules do not deserve the term “ mathematics ”. They point out that there are no sutras of the Vedic period that conform to these rules.

Proponents highlight the speed with which invoices can be executed. They could be used much more efficiently than the arithmetic rules that are generally taught in primary schools. One advantage, for example, is that you only have to master the multiplication table up to 5 in order to be able to multiply all the numbers.

Tirthaji was the abbot (Sankaracharya) of the monastery Govardhana matha in Puri from 1925 until his death .

There is also genuine mathematics handed down from the Vedic literature, see Sulbasutras .

The sutras

1. One more than the one before
2. All of 9 and the last of 10
3. Vertical and crosswise
4. Turn around and apply
5. If the combination is the same, it's zero
6. If one is the ratio, the other is zero
7. With addition and with subtraction
8. At completion or incompletion
9. Different differential and integral computing
10. If incomplete
11. Specific and general
12. The remaining to the last place
13. The last and twice the penultimate
14. One less than the one before
15. The product of the sum
16. All multipliers

The sub-sutras

1. Proportionality
2. The remaining remains constant
3. The first to the first and the last to the last
4. The multiplicant of 7 is 143
5. On contact
6. Subtract the difference
7. Whatever the distance, increase the distance one more time and make the square of the distance
8. Add the last to 10
9. Only the last
10. The sum of the product
11. Alternatively with exclusion and retention
12. With mere observation
13. The product of the sum is the sum of the product
14. With the symbol

application

subtraction

Subtract any number from a power of ten

The second sutra "All of 9 and the last of 10" helps to subtract any number from a natural power of ten. To do this, form the difference to 9 for each digit and the difference to 10 for the last digit.

Example: ${\displaystyle 10.000-4.856\rightarrow 9-4|9-8|9-5|10-6=5.144}$

Subtract any number by completing it

It is easier to subtract numbers from each other if you increase or decrease both by the same amount.

Example: by increasing by 3 you get an even number at the subtrahend and you can easily read the result.

${\displaystyle {\begin{array}{rrrrl}664\\-147\\\hline ?\\\end{array}}}$ ${\displaystyle {\begin{array}{rrrrl}667&(+3)\\-150&(+3)\\\hline 517\\\end{array}}}$

Multiplication of two-digit numbers

Special case: first digits equal, last digits added result in 10

With the Vedic rule “One more than the one before” it is easy to multiply two-digit numbers where the first digits are the same and the last digits add up to 10. The first digit of the number multiplied by its successor gives the first digit of the result. The second digits of the two numbers multiplied together give the last digits of the result.

Example: ${\displaystyle 32\cdot 38=1216}$

front digits: ${\displaystyle 3\cdot (3+1)=12}$

back digits: ${\displaystyle 2\cdot 8=16}$

Squaring numbers ending with 5

Squaring numbers with the final digit 5 ​​is a special case of the previous rule, as it can also be used with three-digit numbers (and more).

Example: ${\displaystyle 125\cdot 125\rightarrow 12\cdot (12+1)|25=15625}$

The real and simple method of squaring with the final digit 5

Example: 35² (35 times 35)

You write the number in two parts ${\displaystyle {\begin{array}{rrrrl}3&&|&&5\\\end{array}}}$

The right part is always 25 (5 times 5) ${\displaystyle {\begin{array}{rrrrl}&&|&&25\\\end{array}}}$

The number on the left is increased by 1 (3 + 1 = 4)

and you get a MULTIPLIER (4)

this is multiplied by the number on the left

4 times 3 = 12 and you get the result ${\displaystyle {\begin{array}{rrrrl}12&&|&&25\\\end{array}}}$

1225

or 75² -> front 7 times (7 + 1) = 56, back 25 -> 5625

if you can still multiply two-digit numbers, 735² is no problem either!

Multiplication of any two-digit numbers

Any two-digit numbers can be multiplied with the Vedic rule “vertically and crosswise”. To do this, the numbers are written one below the other and then the digits are multiplied vertically and multiplied and added crosswise. Carry-overs can occur if intermediate results (which only represent one digit) assume values ​​greater than 9.

Example: ${\displaystyle \textstyle 56\cdot 23=1288}$

${\displaystyle {\begin{array}{rrrl}5&&6&\\\cdot \downarrow &\searrow \!\!\!\!\!\!\nearrow &\cdot \downarrow &\\2&&3&\\\hline 10&\mathbf {2} 7&\mathbf {1} 8&(27=5\cdot 3+2\cdot 6)\\10&\mathbf {2} 8&8&(28=27+1\;(\mathrm {{\ddot {U}}bertrag} ))\\12&8&8&(12=10+2\;(\mathrm {{\ddot {U}}bertrag} ))\\\end{array}}}$

Explanation: The result consists of three parts: . These three numbers are next to each other. This is followed by the dissolving of the transfers from right to left. The 18 has the carry 1, which is added to the 27. The resulting 28 then has the carry 2 (the tens digit), which is added to the adjacent 10. The result is 1288. ${\displaystyle \textstyle 5\cdot 2=10,\;5\cdot 3+2\cdot 6=27,\;6\cdot 3=18}$

Further examples: ${\displaystyle \textstyle 23\cdot 18=414}$

${\displaystyle {\begin{array}{rrrl}2&&3&\\\cdot \downarrow &\searrow \!\!\!\!\!\!\nearrow &\cdot \downarrow &\\1&&8&\\\hline 2&\mathbf {1} 9&\mathbf {2} 4&(19=2\cdot 8+1\cdot 3)\\2&\mathbf {2} 1&4&(21=19+2\;(\mathrm {{\ddot {U}}bertrag} ))\\4&1&4&(4=2+2\;(\mathrm {{\ddot {U}}bertrag} ))\\\end{array}}}$

The example above 735²

${\displaystyle \textstyle 73\cdot 74=}$

${\displaystyle {\begin{array}{rrrl}7&&3&\\\cdot \downarrow &\searrow \!\!\!\!\!\!\nearrow &\cdot \downarrow &\\7&&4&\\\hline 49&\mathbf {4} 9&\mathbf {1} 2&(49=7\cdot 4+7\cdot 3)\\49&\mathbf {5} 0&2&(50=49+1\;(\mathrm {{\ddot {U}}bertrag} ))\\54&0&2&(54=49+5\;(\mathrm {{\ddot {U}}bertrag} ))\\\end{array}}}$

${\displaystyle \textstyle 5402}$

73 times 74 and the 25 -> 540225

735² = 540225

There are also special methods for squaring any number of multiple digits of any size.

Multiplication of numbers that are close to a power of ten

→ See also: Vedic multiplication

Numbers that are just above or below a power of ten can also be multiplied according to the Vedic rule “vertically and crosswise”. The general formula for all cases is x * y = (x + (y-10 ^ n)) * 10 ^ n + (x-10 ^ n) (y-10 ^ n), but the procedure is explained more clearly below .

1st case: Both numbers are just under a power of ten

First you write the two numbers below each other and next to them the difference to the next power of ten (power of ten minus number). The differences are then cross-subtracted from the numbers. The differences are then multiplied together. The result is made up of these two partial results, whereby a carryover must be formed from the first result if there are more than three digits.

Example: ${\displaystyle 998\cdot 889=887222}$

${\displaystyle {\begin{array}{rrr}998&&2\\&\nearrow \!\!\!\!\!\!\searrow &\cdot \downarrow \\889&&111\\\hline (998-111=889-2=)887&&222\\\end{array}}}$

2nd case: Both numbers are just above a power of ten

Similar to the first case, the numbers are written one below the other and next to them the difference to the next power of ten, but again with a positive sign (i.e. number minus the power of ten). The differences are now added crosswise to the numbers and the differences are multiplied together. The result is made up of the two partial results.

Example: ${\displaystyle 105\cdot 102=10710}$

${\displaystyle {\begin{array}{rrr}105&&5\\&\nearrow \!\!\!\!\!\!\searrow &\cdot \downarrow \\102&&2\\\hline (105+2=102+5=)107&&10\\\end{array}}}$

Alternatively, you can proceed exactly as in the first case, but then you have to reckon with negative differences.

${\displaystyle {\begin{array}{rrr}105&&-5\\&\nearrow \!\!\!\!\!\!\searrow &\cdot \downarrow \\102&&-2\\\hline (105-(-2)=102-(-5)=)107&&10\\\end{array}}}$

3rd case: A number above and a number below a power of ten

In this case, negative carryovers must be expected. Otherwise, proceed as in the first case.

Example: ${\displaystyle 88\cdot 102=8976}$

${\displaystyle {\begin{array}{rrr}88&&12\\&\nearrow \!\!\!\!\!\!\searrow &\cdot \downarrow \\102&&-2\\\hline (88-(-2)=102-12=)90&&-24\\89&&76\end{array}}}$

Explanation: To make −24 a positive number, add 100 (−24 + 100 = 76). This results in a carry from −1 to 90 (90 - 1 = 89).

Multiplication by 11

To easily multiply a number by 11, write the number twice under each other, shifting it by one digit. Then it is added digit by digit. Carry-overs can occur if intermediate results (which only represent one digit) assume values ​​greater than 9.

Example: ${\displaystyle 423\cdot 11=4653}$

${\displaystyle {\begin{array}{rrrrl}4&2&3&&\\&4&2&3&(+)\\\hline 4&6&5&3&\\\end{array}}}$

Example with carryover: ${\displaystyle 857\cdot 11=9427}$

${\displaystyle {\begin{array}{rrrrl}8&5&7&&\\&8&5&7&(+)\\\hline 8&\mathbf {1} 3&\mathbf {1} 2&7&\\8&\mathbf {1} 4&2&7&(14=13+1(\mathrm {{\ddot {U}}bertrag} ))\\9&4&2&7&(9=8+1(\mathrm {{\ddot {U}}bertrag} ))\\\end{array}}}$

Division by 9 with remainder

The result of dividing by 9 with the remainder can be quickly obtained using the following procedure: The first digit of the result is the first digit of the number that is being divided. The second digit of the result is the sum of the first and second digits of the number. This continues up to the penultimate digit of the number. Carry-overs can occur if intermediate results that only represent one digit assume values ​​greater than 9. The checksum of the number gives the remainder. This can be greater than 9, so that you then have to carry out another division or reduce the remainder by transferring it.

Simple example:

${\displaystyle {\begin{aligned}78:9&=7{\text{, Rest }}7+8\\&=7{\text{, Rest }}15\\&=7+1{\text{, Rest }}15-9\;(\mathrm {{\ddot {U}}bertrag\;des\;Rests} )\\&=8{\text{, Rest }}6\\\end{aligned}}}$

Example:

${\displaystyle {\begin{aligned}123:9&=1|1+2{\text{, Rest }}1+2+3\\&=13{\text{, Rest }}6\\\end{aligned}}}$

Example with carryover:

${\displaystyle {\begin{aligned}791:9&=7|7+9{\text{, Rest }}7+9+1\\&=7|\mathbf {1} 6{\text{, Rest }}17\\&=8|6{\text{, Rest }}17\;(8=7+1(\mathrm {{\ddot {U}}bertrag} ))\\&=86+1{\text{, Rest }}17-9\;(\mathrm {{\ddot {U}}bertrag\;des\;Rests} )\\&=87{\text{, Rest }}8\\\end{aligned}}}$

Fractions

With the sutra "vertical and crosswise" fractions can be added and subtracted. The denominator of the result is the product of the two denominators. The numerator of the result is the numerator of the first fraction times the denominator of the second fraction plus (or minus) the numerator of the second fraction times the denominator of the first fraction. Or in short: numerator 1 times denominator 2 plus (or minus) numerator 2 times denominator 1.

Example of addition: ${\displaystyle {\frac {2}{3}}+{\frac {1}{5}}={\frac {2\cdot 5+1\cdot 3}{3\cdot 5}}={\frac {10+3}{15}}={\frac {13}{15}}}$

Example of subtraction: ${\displaystyle {\frac {6}{7}}-{\frac {2}{3}}={\frac {6\cdot 3-2\cdot 7}{7\cdot 3}}={\frac {18-14}{21}}={\frac {4}{21}}}$

literature

• Shri Bharati Krishna Tirthaji Vedic Mathematics , New Delhi: Motilal Banarassidas, 1965
• Armin Schonard / Cordula Kokot Der Mathknüller, Göppingen, 2011
• SG Dani Myth and reality: on "Vedic Mathematics" , 1993, pdf

Web links

• Vedic Mathematics Academy (English)
• Neither Vedic Nor Mathematics (English)
• Vedic Mathematics - Calculating Tricks of the Ancient Indians

Individual evidence

1. ↑ Calculate like the ancient Indians. In: INDIEN Magazin. 4/08, p. 55.