Introduction to Vedic Mathematics

Dr. Goutam Mukherjee

98300 23972

goutam.kv2@gmail.com

Origin of Counting and Mathematics

• Ancient people used just a line to keep track of number of items
• For example, for each sheep, they will put one | on ground, with a stick.
• Number system |||||||| … gets difficult, so next step was to make groups of say, 5. After writing 4 |-s put a horizontal line on them.
• Ex: |||| is five, so |||| |||| ||| is 13, this way.
• Babylonians used various symbols for One, 10, 100 etc.
• In Roman Number system, the symbols I, V, X, L, C, D and M are used to denote 1, 5, 10, 50, 100, 500 and 1000 respectively. They are additive. However, a lower value will do subtract, if it is placed on the left of a higher value. Eg. DCLXVI = 666, but CDXLIV is 444 etc.
• Roman Numbers are used as Ordinal Number till date.

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Cardinal and Ordinal Numbers

• Cardinal and Ordinal Numbers are two different aspects of usage.
• A Cardinal Number is a number that says how many of something there are, such as one, two, three, four, five. An Ordinal Number is a number that tells the position of something in a list, such as 1st, 2nd, 3rd, 4th, 5th etc. Most ordinal numbers end in "th" except for: �one ⇒ first (1^st), two ⇒ second (2^nd) and three ⇒ third (3^rd).
• Mathematical calculations can be done on Cardinal Numbers.
• Doing the same on Ordinal numbers may not have any meaning.
• There is no point in adding or multiplying two Roll nos., PINs, Aadhar nos. etc.

Positional Number system of India

• It seems the Positional or the Place Value number system was known in India since the Vedic times. But, through Aryabhatta (400 AD), it started spreading widely, became known to Arabs in 6^th Century and was spread to Europe in 7^th Century from them.
• By 8^th Century, the Indian Positional number system became the most dominant Number System throughout the world, but being wrongly named as Arabic number system.
• Nine symbols 1…9 are used to denote numbers. When some place has no Value, a heavy dot or a circle was put, in place under that place holder. That is the origin or so called discovery of Zero, which is actually a Corollary of the Place Value system.

Use of Zero as part of Calculation

• In AD 628, Brahmagupta wrote a book ‘Brahmasphuta Siddhanta’ and developed the rules for Arithmetic involving Zero and Negative Numbers, such as
• any number subtracted from same number gives zero.
• Zero added or subtracted to any number results in that same number.
• Any number multiplied with or divided from zero produces zero.
• Square or Square root of zero is zero.
• However, the issue of Division by Zero was not correctly handled by him.
• In AD 830, Mahavira wrote ‘Ganita Sara Samhita’, there too it was not.
• After AD 1100 Bhaskara wrongly stated that anything divided by Zero gives Infinity, which later got refined & rectified with concept of Limit.

Concept of Base in a Place Value System

• Any Positional or Place Value number system will be associated with a base. In India, then world-wide the number system used has the Base 10. It is said that, ten fingers in hand made 10 as natural base.
• The system with base 10 is known as Decimal system. Various powers of 10 are the Place Value of a position : Unit, ten, hundred, thousand etc. are the names of positions for powers 0, 1, 2, 3 etc. of 10.
• For computers, the most natural number system is Binary, with base 2, digits 0 & 1. Binary, when grouped into 3, gives Octal, base 8, digits 0-7. Again, Binary grouped into 4, gives Hexadecimal, base 16, digits 0-9 and alphabets A, B, C, D, E & F, used as numerical digits carrying values 10 – 15 respectively.

Concept of Place Value and Face Value

• In Place Value system, the Value associated with each Place holder is called the Place value and Digit that occupies that Place holder is the Face value. For example, in a Decimal system, Place values from right to left are Unit (10⁰), Ten (10¹), Hundred (10²) etc. Thus, any number can be expressed as a sum of Place Value multiplied by Face value.
• For example, 756 = 7x100 + 5x10 + 6, where 7, 5 & 6 are face values and 100, 10 & 1 are Place values.
• Negative Place Values denote Fractions expressed in Decimal notations, such as 0.34 will denote 0.34 = 3x10⁻¹ + 4x10⁻².
• In Vedic number system, use of Negative Face value, called Vinculum is also possible. In specific cases, Vinculum makes calculation easier.

Numerical Symbols of World with Codes

• Various States of India used various Numerical Symbols for numbers 0, 1-9. Various other countries too used their own symbols, most of them work in Indian Place Value system. Those are listed in next two slides, with Hex-codes in UNICODE system.
• Among these, English numerical symbols being most common, their ASCII Codes are simply given as Hex 30 to 39 (Decimal 48-57).
• Interestingly, Numerical calculations can be done using those other symbols, too in an MS-Excel sheet, in the same way as can be done using common English numbers.
• Numbers of Chinese and few countries (not listed) do not use Place Value system and they are not accepted as numbers is Excel.
• A symbol can be displayed in MS-Word by writing its Hex-code followed by typing Alt-x immediately next to it.

Numeric Symbols used in India

Numeric Symbols used in other Countries

What is Vedic Mathematics

• Vedic Mathematics is the Intuitional use of 16 Vedic Sutras (Formulae) and 32 sub-sutras (Corollaries) (later only 13 are known or used), from the Vedas, in tackling any mathematical problem.
• Vedic Mathematics is collection of various alternative methods in solving any mathematical problem and develops better control on number system.
• It aims at developing mathematical Intuition in choosing the Best method, in case to case basis, rather than mechanically adhering to only one or few methods.
• Practice of Vedic Mathematics can be a method of Mental Development, in parallel to solving the mathematics at hand.

Vedic Mathematics by Bharati Krshnaji Tirthaji

The Vedic Sutras – Meaning in English

1. Ekadhikena Purvena
• One more than the previous
2. Nikhilam Navatah Charamam Dasatah
• All from nine and ultimate from ten
3. Urdhwa Tiryagbhyam
• With upper slant
4. Paravartya Yojayet
• Transpose and Apply

Vedic Sutras – Meaning (contd.)

5. Sunyam Samya Samucchaye
• A common factor is the same which is Zero
6. (Anurupye) Sunyamanyat
• If one is in ratio, the other one is Zero
7. Sankalana-Vyavakalanabhyam
• By addition and by subtraction
8. Purana Apuranabhyam
• By completion or non-completion

Vedic Sutras – Meaning (contd.)

9. Chalana-Kalanabhyam
• Sequential Motion Differentiation
10. Yavadunam
• The Deficiency
11. Vyastisamastih
• Whole as one and one as whole
12. Seshanyan Kena Charamena
• Remainder by the last digit

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Vedic Sutras – Meaning (contd.)

13. Sopantyadwayamantyam
• Ultimate and twice the Penultimate
14. Ekanyunena Purvana
• One less than the Previous
15. Gunita Samucchyah
• The whole product is the same
16. Gunaka Samucchayah
• Collectivity of Multipliers

Important Sub-sutras with Meaning

• Anurupyena – Proportionality
• Adyamadyena Antamantyena – the first by the first last by the last
• Yavadunam Tavadunikrtya Vargancha Yojayet
• Whatever the deficiency subtract that deficit from the number and write alongside the square of the deficit
• Antyayor Dasakepi – Numbers of which the last digits add up to 10
• Antyayoreva – only the last terms

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Sub-sutras – some Meaning (contd.)

• Lopana Sthapanabhyam – by alternate elimination and retention
• Vilokanam - Observation
• Gunita Samucchayah : Samucchaya Gunitah – �Product is the same, same is the product
• Sisyate Sesasamjnah
• Kevalaih Saptakam Gunyat
• Vestanam
• Yavadunam Tavadunam
• Samucchaya Gunitah

Scope of Vedic Mathematics

• Arithmetic Computation
• Multiplication
• Division
• Recurring Decimals
• Divisibility
• Highest Common Factor
• Equations
• Factorization and Differentiation
• Integration

Arithmetical Computation

• The rules of Addition, Subtraction, Multiplication and Division, which are conventionally taught in schools : all have some or other algebraic background.
• Thus, once the inherent algebra is changed, the procedure or method gets changed automatically.
• In Vedic Mathematics, negative Face value is also used, specially for numbers more than 5. For example, 48 can be expressed as: 52̅, with a bar above 2. These are called Vinculum notation, and reduces effort. Detailed in next slide.

Concept of Vinculum:

A number with a Bar above a number is called Vinculum notation in Vedic Mathematics. They denote Negative Face Value of numbers.

(Because of Technical Difficulty in writing a Bar above a number by using Character U+305, in this Presentation we shall write it with a graphic line instead. So, 2̅ = -2.)

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General Number Conversion with Vinculum number and vice versa:

6 = 10 – 4, so in Vinculum notation 6 can be written as 14

97 = 100 – 3, thus it is 103 1431̅4̅3 = 142857

289 = 300 – 11, so it is 311 etc. 313244 = 286756

The sutras ‘Nikhilam Navatascharamam Dasatah’ (All from nine, last from ten) and ‘Ekanyunena purvena’ (One less than the previous)are useful for this conversion.

Vinculum : Application and Purpose

• As was told earlier, Vinculum means negative Face Value. Excepting the sign, it works exactly the same way as positive Face Value.
• Vinculum is applied to reduce effort of calculation.
• Expert and intelligent use of Vinculum will require table of upto 5 x 5 in any work out of Addition, Multiplication, Division etc.

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Addition using Nikhilam Sutra & Vinculum

• Where the digits in the Addends are ≥5, it is easier to convert and add Vinculums and finally convert the resultant Vinculum into actual number:

Numbers to Add Vinculums Explanation

39678 🡪 40322 40000 - 0322

27689 🡪 32311 30000 - 2311

18567 🡪 21433 20000 - 1433

47598 🡪 52402 50000 - 2402

------- ---------------- --------------------

13 3532 🡨 146468 140000 – 6468 = 133532

Addition using Nikhilam Sutra & Vinculum

• Where the digits in the Addends are ≥5, it is easier to convert and add Vinculums and finally convert the resultant Vinculum into actual number:

Numbers to Add Vinculums Explanation

39274 🡪 4 1 3 3 4 40304 – 0301 = 4 0 0 0 3

53749 🡪 5 4 3 5 1 54050 – 1030 = 5 3 0 2 0

28362 🡪 3 2 4 4 2 30402 – 0204 = 3 0 2 0 2

43826 🡪 4 4 2 3 4 44030 – 2040 = 4 2 0 1 0

--------- ---------------- -----------------------------------

165211 🡨 16 5 2 1 1 16 5 5 1 1

Squaring Numbers ending with 5

• (10a + 5)² = 100a² + 2.5.10a + 25 =100 a(a+1) + 25

Therefore Ekadhikena Purvena is applied.

Examples: 15² = 1x2 25 = 225 25² = 2 x 3 25 = 625

35² = 3x4 25 = 1225 45² = 4x5 25 =2025 55² = 5x6 25 = 3025

125² = 12x13 25 = 15625 305² = 30x31 25 = 93025

1225² = 122 x 123 25 = 1500 625 3025² = 302x303 25 = 9150 625

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• Above so called Tricks are used sometimes as shortcuts to various calculations. However, Vedic Mathematics is not meant for Tricks, they have much deeper level of Algebraic interpretations.

Multiplication near a Base (below)-1

• Multiplication of the patterns : 97 x 93, 987 x 988, 9985 x 9970, 9888 x 9997 can be handled in this method.
• A Base is defined as the nearest power of Ten, which must be the same for the multiplicands.
• Let’s take examples : for Base 100

97 -03 : Here, 97 is 3 short of Base, we write 03, maintaining the number of digits.

x 93 -07 : 93 is 7 short of Base.

‾‾‾‾‾‾‾‾‾

90 21 : Here, 90 is obtained by 97-7 or 93-3, i.e. Cross subtraction. 21 is (-7)x(-3)

Thus, 97 x 93 = 9021.

Multiplication near a Base (below)-2

• Similar example, with higher Bases :

Base 1000 Base 10000 Base 10000

997 -003 9997 -0003 9876 -0124

993 -007 9993 -0007 9987 -0013

‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾

990 021 9990 0021 9863 1612

Thus it is seen that, increasing more 9’s in both multiplicands results in addition of same no. of 9-s in the first part and same no. of 0-s in the second part.

• Algebraic Explanation : Say N stands for the Base, then

(N – a) (N – b) = N (N – a – b) + ab

Here, Cross subtraction is (N – a – b) and ab is the Multiplication in second part.

Multiplication near a Base (above)-1

• Multiplication of the patterns : 107 x 103, 1017 x 1012, 10015 x 10030, 10123 x 10005 can be handled in this method.
• A Base is defined as the nearest power of Ten, which must be the same for the multiplicands.
• Let’s take examples : for Base 100

103 +03 : as 103 is 3 more than the base, 03 maintains no. of digts

107 +07 : 107 is 7 more than base

‾‾‾‾‾‾‾

110 21 : Here 110 = 103 +07 or 107 +03 (cross addition) and 21 = (+03)x(+07)

Thus, 103 x 107 = 11021

Multiplication near a Base (above)-2

• Similar example, with higher Bases :

Base 1000 Base 10000 Base 10000

1007 +007 10007 +0007 10123 +0123

1003 +003 10003 +0003 10005 +0005

---------------- ------------------ -------------------

1010 021 10010 0021 10128 0615

Thus it is seen that, increasing more 0’s in both multiplicands results in addition of same no. of 0-s in the first part and same no. of 0-s in the second part.

• Algebraic Explanation : Say N stands for the Base, then

(N + a) (N + b) = N (N + a + b) + ab

Here, Cross addition is (N + a + b) and ab is the Multiplication in second part.

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Multiplication near a Base (Misc)-1

• Multiplication of the patterns : 97 x 103, 997 x 1012, 9995 x 10030, 99923 x 100005 can be handled in this method.
• A Base is defined as the nearest power of Ten, which must be the same for the multiplicands.
• Let’s take examples : for Base 100

93 -07 : as 93 is 7 less than the base, 07 maintains no. of digts

107 +07 : 107 is 7 more than base

-----------

100 49 : Here 100 = 93 +07 or 107 -07 (cross result) and � -49 (written as 49) = (-07)x(+07). 100 00 – 49 makes it 99 51

Thus, 93 x 107 = 10049 = 9951.

Multiplication near a Base (Misc)-2

• Similar example, with higher Bases :

Base 1000 Base 10000

1013 +013 10003 +0003

993 -007 9993 -0007

-------------- ------------------

1006 091 9996 0021 (becomes 1005 909 and 9995 9979 resp.)

Thus it is seen that, increasing more 0’s & 9’s in multiplicands results in addition of same no. of 0-s or 9-s in the first part and same no. of 0-s in the second part.

• Algebraic Explanation : Say N stands for the Base, then

(N + a) (N – b) = N (N + a – b) – ab

Here, Cross result is (N + a – b) and ab is the Multiplication in second part.

Multiplication near Sub-base

• Consider : 483 ✕ 497

Here we may consider Base 1000, and sub-base 500.

Alternatively, if we consider Base as 100, sub-base will still be 500.

Procedures will be different in these two cases, pl see below.

Base 1000 Base 100 (both sub-base 500)

483 -017 483 -17 According to Base, we have to write the difference.

497 -003 497 -03 Procedure is similar to earlier. But, at the last, we have to

------------ ----------- Divide by 2 (500=1000/2) for first, and Multiply by 5 in the

2) 480 051 5x 480 51 second case (500 = 100 x 5). Ultimately, result becomes

240 051 2400 51 the same, as is supposed to be.

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Multiplication near Sub-base (contd.)

• Consider 235 x 247 with Consider 18 x 45

Base 1000, Base 100,

Sub-base 1000 /4 = 250 Sub-base 100/2 = 50

235 -015 18 -32

247 -003 45 -05

4)232 045 2)13 160 (Carry is written in small)

58 045 6½ 160 = 6 210 (50 + 160 = 210)

8 10 (Carry 2 is added to 6 to get 8)

Note: A half (½) in hundred position means 5 in 10’s position.

Algebra: (mN – a) (mN – b) = mN (mN – a – b) + ab

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General Multiplication by Tiryag method (2x2)

• (10a+b)(10c+d) = 100ac + 10(ad+bc) + bd
• Take example:

43 Left most 4x2 =8. Middle 4x5 + 3x2 =26. Right 3x5 =15

25

------

865 The Unit digits are written in this row

21 The Carry-s are written in this row below left.

------

1075 Final answer is obtained by adding those Carry-s.

General Multiplication by Tiryag method (3x3)

• (100a+10b+c)(100p+10q+r)

= 10000ap + 1000(aq+bp) + 100(ar+bq+cp) + 10(br+cq) + cr

Example: Digits are calculated like this:

314 3x3 =9, 3x5+1x3=18, 3x2+1x5+4x3=23, 1x2+4x5=22, 4x2=8

X 352

--------

98328 Unit’s digits are placed here.

122 Carry-s were written below one digit left

---------

110528 Carry-s are added to get the final Product

General Multiplication by Tiryag method (5x5)

• Discuss this with an Example

23431 Left most digit 2x3 =6, Left but one 2x5+3x3 =19

x 35642 Next 2x6+3x5+4x3 =39, Next 2x4+3x6+4x5+3x3=55

--------- Middle 2x2+3x4+4x6+3x5+1x3 =4+12+24+15+3 =58,

699585602 Next, 3x2+4x4+3x6+1x5 =45, Next, 4x2+3x4+1x6 =26

1355421 Last but one 3x2+1x4=10. Right most 1x2=2

--------------- The “Carry”-s were written below and left. Finally,

835127702 respective “Carry”-s were added to above to get the Product.

General Multiplication by Tiryag method (5x5)

2 3 4 3 1 2 3 4 3 1

3 5 6 4 2 4 4 4 4 2

⚊⚊⚊⚊⚊⚊⚊⚊ ―――――――

6 9 9 5 8 5 6 0 2 8 4 4 8 8 6 6 0 2

1 3 5 5 4 2 1 1 1

--------------------- ---------------------

8 3 5 1 2 7 7 0 2 8 4 4 8 8 7 7 0 2

8 3 5 1 2 7 7 0 2

Comparison between Multiplication Methods

• Consider 889 x 898

Urdhva Tirjagabhyam Nikhilam Navatah …

8 8 9 1 1 1 1 889 -111 111 +11

8 9 8 1 1 0 2 898 -102 102 +02

--------------- ---------------- ------------ ------------

64 6 8 5 2 1 2 0 2 3 2 2 787 322 113 22

13 0 4 7 ⇩ 11

2 1 7 9 8 3 2 2 -------------

--------------- 798 322

79 8 3 2 2

Comparing with conventional method

8 8 9 1 1 1 1

8 9 8 1 1 0 2

------------- -----------------

7 1 1 2 2 2 2 2

8 0 0 1 1 1 1 1

71 1 2 1 1 1 1

----------------- ----------------------

7 9 8 3 2 2 1 2 0 2 3 2 2 = 798 322

Conventional … but using Vinculum

Comparison of Multiplication Algebrae

• Given two 3-digit numbers (100a+10b+c) and (100p+10q+r), �the conventional method performs this way:

(100a+10b+c)r is written on first line, then

(100a+10b+c)q is written on 2^nd line, one place shifted left, then

(100a+10b+c)p is written on 3^rd line, one place further shifted left

Then all added.

• However, the Urdhva Tirjagabhyam method does this way:

10000ap+1000(aq+bp)+100(ar+bq+cp)+10(br+cq)+cr

Units are written on top line, carries down-left, finally added.

Tiryag method has been proved to be more efficient through research.

Division : by Nikhilam Method

• Purpose of Division is to find Quotient (q) and Remainder (r) for any given Dividend and Divisor. In Nikhilam Method, deficiency from the nearest power of 10 is worked with.
• Division by 9 : Deficiency from nearest power of 10 is 1

9 ) 3 / 4 9 ) 6 / 1 9 ) 12 / 3 9 ) 25 / 3

1 3 6 1 3 2 7

3 / 7 6 / 7 13 6 27 / 1 / 0

q r q r q r 1 1

28 / 1

q r

Division : by Nikhilam Method (contd.)

• Division by 9 for bigger number :

9 ) 1231 / 2 9 ) 120021 / 2 9 ) 101164 / 9

1 136 / 7 13335 / 6 11239 / 13

1367 / 9 133356 / 8 112403 / 22

1368 / 0 q r 9) 2 / 2

q r 2

112403 2 / 4

112405 / 4

q r

Division : by Nikhilam Method (contd.)

• Division by 8, 7, 6 :

8 ) 2 / 3 8 ) 3 / 1 7 ) 1 / 2 7 ) 2 / 0 6 ) 1 / 1

2 4 2 6 3 3 3 6 4 4

2 / 7 3 / 7 1 / 5 2 / 6 1 / 5 (q / r)

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8 ) 23 / 4 8 ) 34 / 5

2 4 2 6

/ 14 / 20

27 / 18 = 29 / 2 40 / 25 = 43 / 1

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Division : by Nikhilam Method (contd.)

• Division by two or more digit numbers requiring single processing :

89 ) 1 / 11 87 ) 2 / 22 88) 3 / 33

11 11 13 / 26 12 / 36

1 / 22 2 / 48 3 / 69

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888 ) 1 / 234 789 ) 2 / 345

112 112 211 422

1 / 346 2 / 767

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Division : by Nikhilam Method (contd.)

• Division by two or more digit numbers requiring multiple processing:

89 ) 11 / 11 789 ) 21 / 456 789 ) 1 / 731

11 1 / 1 211 4 / 22 211 / 211

/ 22 1055 1 / 942 = 2 / 153

12 / 43 25/1731 🡪 27 / 153

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789 ) 121 / 456 789 ) 3 / 106 789 ) 231 / 456 789 ) 5 / 013

211 21 / 1 211 633 211 42 / 2 211 1055

8 / 44 3 / 739 14 / 77 5 / 1068

2110 3587 789 ) 1 / 068

150 / 3106 🡪 153 / 739 287 / 5013 / 211

🡪 293 / 279 1 / 279

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Division : by Nikhilam Method (contd.)

• Longer complication

654) 341 / 567 654 ) 19 / 799 654 ) 5 / 411 654 ) 2 / 141

346 103 / 8 346 3 / 46 346 / 1730 346 692

48 / 44 / 4152 5/ 2141 2 / 833

/ 17992 22 / 5411 = 3 / 179

492 / 19799 🡪 522 / 179 (Remainder is 179)

Adding quotients of all steps, Quotient = 492 + 22 + 5 + 3 = 522

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• Best Vedic method is “Straight Division” for the above problem.

Division : by Nikhilam… special case

• Division with sub-base technique

Consider 1011 ÷ 23 : 23 x 4 = 92 is near a power of 10.

23 x 4 = 92 ) 10 / 11

08 0 / 8

10 / 91 91 – 3 x 23 = 22

x 4 (Remainder will not be affected by this multiplication)

40 / 91 🡪 43 / 22

Division : Paravartya Method

• Paravartya Yojayet means Transpose and Apply
• Ex. Divide 1234 by 112

112 ) 12 / 3 4 1123 ) 13 / 4 5 6

-1 -2 -1 -2 -1-2-3 -1 -2 -3

-1 -2 -2 -4 -6

11 / 0 2 12 / 0 -2 0 🡪 11 / 1103

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Division : Paravartya Method

• Divide 23456 by 123 Divide 4009 by 882

123 ) 2 3 4 / 5 6 882 ) 4 / 0 0 9

-2-3 -4 -6 118 4 +8 -8

2 3 1 +2 -2 4 / 4 8 1

0 0 q = 4, r = 481

2 -1 0 / 8 6

q = 190 , r = 86

Division : Paravartya Method – Special cases

• Divide 2699 by 224 Divide 1699 by 223

2) 224 2 6 / 9 9 2) 223 1 6 / 9 9

112 -2 -4 111½ -1 -1½

-1 -2 -4 -8 -1 -1½ -5 -7½

2) 2 4 / 1 1 2) 1 5 / 2½ 1½

Q= 12, R=11 7½ / 2½ 1½

Q=7, R=111½ + 25 + 1½=138

Note: Remainder(R) remains constant while dividing Quotient by 2.

And : Presence of ½ in Quotient will contribute half of dividend to R.

Division, Rational Fraction and Decimal form

• So far we have seen Division of a Dividend (E) by a Divisor (D) to yield the Quotient (Q) and Remainder (R) : Relation between these are:

E = D x Q + R or E/D = Q + R/D, where R < D

• Depending upon requirements, division may be left at finding Q & R, or we may have to express it in the form as shown above upto R/D,
• or we may express R/D into a Decimal form, for example :

15 ÷ 4 🡪 Q = 3, R=3, or 15/4 = 3¾ = 3.75

• For an expression E/D, where E>D, it can always be expressed as a whole part plus a Fraction of the form N/D, where the Numerator N is less than the Denominator D.

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Fraction : N/D (N<D) to Decimal form

• Any Rational Fraction of the form N/D becomes a Terminating Decimal when the Denominator (D) is purely comprising of factors of 2 & 5 :

Ex. ½= 0.5, ¼= 0.25, ¾=0.75, 1/₈=0.125, 3/₈=0.375, ⁵/₈=0.675,

1/₅=0.2, 3/₅=0.6, 1/₂₅=0.04, 1/₅₀=0.02, 1/₁₀₀=0.01 etc.

• A Rational Fraction, whose Denominator (D) ends in 1,3,7 or 9, becomes a Recurring Decimal :

Ex. 1/3=0.333… 1/7=0.142857…142857… , 1/11=0.0909… etc.

• A Rational Fraction other than the above two categories will have a non-recurring part, followed by a recurring part

Ex. 1/6=0.1666… 8/15=0.5333… 9/14=0.6428571…428571…

Fraction : N/D (N<D) - Transformation

• A fraction of the 3^rd type can be easily transformed by multiplying both numerator(N) and denominator(D) by either a power of 2 or a power of 5, in order to Neutralize D:

Ex. 1/6=5/30= (1/10) 5/3=(1+2/3)/10= 1.666…/10 =0.1666…

Thus, basically the non-terminating part results from a Whole part and the recurring part from a fraction of 2^nd type, viz. D ending in 1,3,7 or 9.

More example: 29/74=145/370=(1/10) (145/37) = (3+34/37)/10 = 0.3+(34/37)/10

Thus, 0.3 + 0.918…918…/10 = 0.3918…918…

29/75=116/300=0.01 x (116/3) = 0.01x(38+2/3)=0.38 + 0.01x 0.666… = 0.38666…

Finding Decimal expression using Nikhilam

• After obtaining Remainder in Nikhilam method, we can put a Decimal point in the Quotient and put a 0 to the Remainder and continue the process, like we do in conventional method. However, the process may become long, in Nikhilam and conventional methods alike.
• Consider 30 ÷ 8: A terminating Decimal

8 ) 3 / 0 …

2 6

3. 6 / 0

12 ( 12-8 = 4)

3. 7 4 / 0

8 (8 – 8 = 0)

3. 7 5 / 0

Once 0 remainder is obtained, the process ends. Ans. 30 ÷ 8 = 3.75

Finding Decimal expression using Nikhilam

• Consider 22 ÷ 7 :

7 ) 2 / 2 3.1428 4 / 0

3 6 12

2 / 8 = 3 / 1 3.14285 5 / 0

3. 1 / 0 3.14285 15

3. 3 3.142857 1 / 0

3.1 3 / 0 3.142857 3

3.1 9 3.142857 1 3 / 0

3.1 3 / 9 repeats

3.14 2 / 0

3.14 / 6

3.142 6 / 0

3.142 18

3.1428 4 / 0

Finding Decimal expression using Nikhilam

• Consider 21313 ÷ 89, where we got Q = 239 & R = 42, we can continue putting 0 after 42. This will be a long recurring decimal:

89 ) 4 / 20 Thus, 42 ÷ 89 = 0.4719101…

11 44

. 4 6 / 40

.4 66

.4 6 106 (106 – 89 = 17)

. 4 7 1 / 70

.4 7 11

.4 7 1 8 / 10

.4 7 1 88

.4 7 1 8 98 ( 98 – 89 = 09)

.4 7 1 9 09 / 00

.4 7 1 9 99 (99 – 89 = 10)

.4 7 1 9 10 1 / 00

Decimal form to Fraction (N/D) form

• For Terminating Decimals, denominator becomes 1 followed by as many zeros as the numbers of digits after Decimal point, then write in minimized form, if required:

0.125 = 125/1000 = 1/8. 5.48 = 548/100 = 137/25.

• For pure Recurring decimals, put as many 9-s in the denominator, as the number of digits recurring, then minimize :

0.3333… = 3/9 = 1/3. 0.4545..= 45/99 = 5/11

0.142857142857… = 142857/999999 = 1/7

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Decimal form to Fraction (N/D) form

• For mixed kind of decimals, in the denominator put as many 9’s as the number of recurring digits, followed by as many zeros as digits fixed. On the numerator, subtract the fixed part from the total.

0.1666… = (16-1)/90 = 15/90 = 1/6

0.6818181… = (681-6)/990 = 675/990 = 15/22 (45 was common)

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Recurring Decimal : Vedic One Line Method

Recurring Decimal : Vedic One Line Method

• Calculating a Fraction like 1/19, 1/29, 1/49 … are quite long and cumbersome both in the conventional division method and Nikhilam method. Vedic One-line method, arising from the sutra ‘Ekadhikena Purvena’ is easy, having more than one ways and it reveals further properties of the fraction, such as Cyclic property and Nine’s Compliment behavior.
• Vedic One line method is directly applicable for Denominators ending in 9. Let’s start with fraction 1/19:

Vedic One Line Method : case 1/19

• Instead of dividing 1 by 19, we shall start dividing 1 by a digit-divisor dd=(D+1)/10, but in a space way. This comes from the Sutra ‘Ekadhikena Purvena’ meaning one more than the earlier. Here in 19, earlier to 9 is 1 and one more than 1 is 2. So, we start dividing by 2 like this :
• Put the Decimal point. Now, 1 divide by 2, results in quotient q=0, remainder r=1. In the result line we put q, i.e. 0, this becomes .0, put 1 below, a little left. Divide q (0) with this r (1) as ten’s digit for next division. Thus, it becomes 10. Dividing by 2 gives q=5, r=0. Result .05. Divide 5 by 2 gives q=2, r=1 which makes next dividend 12. Result .052. 12 divided by 2 gives q=6, r=0. Result .0526. Detailed:

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Vedic One Line Method (contd.) : case 1/19

• Successive quotients (q) and remainders (r), in the form rq are as follows :
• From beginning of 1/19 : Remainders are written left, in small:�0.10 05 12 06 03 11 15 17 18 09 14 07 13 16 08 04 02 01. At this 18^th digit, we again arrived at 1, from where we started, thus the sequence will repeat.

⦁ ⦁

• Thus, 1/19 = 0.052631578 947368421 05… (length 18)
• It is possible to get the same result from Right to Left, by starting with the same numerator and Multiplying with 2 and adding the carry written left, in small font. E.g. 1x2, 2x4, 2x8, 2x6+1, 2x3+1 … (Pl see above).
• In case of a full cycle like here, digits of one half length are 9’s complement of digits of the other half length. Thus, after evaluating 9 digits, we could have gotten the rest by subtracting them from 9. Pl see above.

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Vedic One Line Method (contd.) : case N/19

• Since with Denominator 19, we get a full cycle, it is very easy to get the result for other numerators. For example, in case of 9/19, the starting digit is 9 and we can follow the same procedures as previous, either division for left to right or multiplication for right to left.
• But, since we know the sequence : 052631578 947368421, alongwith remainders already, just by Vilokanam(observation) we get, 9/19= 0.473684210 526315789… , i.e., sequence is same, but starting point is different. Check 9’s complement property also.
• Exercise: Guess for 4/19, 7/19, 11/19, 15/19 and 18/19 from above.

Vedic One Line Method (contd.) : case 1/29

• Since the Denominator ends in 9, the Digit Divisor will be ‘One more than the previous’ : previous of 9 in 2, one more is 3. Also, dd=(D+1)/10=(29+1)/10=3.
• Divide 1 by 3, q=0, r=1, makes it 10. 10÷3 🡪q=3, r=1, 13÷3🡪14
• This way successive q & r combinations, rq in case of 1/29 becomes:

10 13 14 24 08 22 17 25 18 06 02 20 26 28

19 16 15 05 21 07 12 04 11 23 27 09 03 01 thereafter 10 13 … repeats.

⦁ ⦁

Thus, 1/29 = 0.03448275862068 96551724137931… recurs after 28.

For 29, this recurrence length being 29-1=28, is also a full cycle.

Vedic One Line Method (contd.) : case N/29

• In 1/29 too, we could have started from 1 and kept multiplying by 3, and adding carry, if any for Right to Left method. We see, after half length, i.e. 14 digits the corresponding digits become 9’s complement. For other Numerator N, we can similarly start with N and keep on dividing by 3 (left to right) or multiply by 3 (right to left).
• 1/29 being a full recurring cycle, we can guess for other numerators.

Sequence being 03448275862068 96551724137931, we can guess for say 15/29 as starting from the first 5 in 55, where the remainder was 1. Thus, it was 15, and next will be the first digit.

i.e. 15/29 = 0. 51724137931034 48275862068965…

• Exercise : Guess for 4/29, 7/29, 13/29, 19/29, 23/29, 28/29.

Vedic One Line Method (contd.) : N/a9

• For fractions of the form N/a9, we start dividing N by a+1, as per rule Ekadhikena Purvena, in the way that were illustrated in cases of N/19 and N/29. Here, ‘a’ can be multi-digit as well.
• Let’s have illustration for 1/59: Digit Divisor is (59+1)/10=6.

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• We started dividing 1 by 6 and kept placing the remainder below the quotient.

Algebraic Proof of the Procedure

• Let’s consider the Binomial Expansion of (1-x) ^-1:

(1-x)^-1= 1+x+x²+x³+x⁴+… …

=1+x(1+x(1+x(1+…)…)), where x < 1

• 1/19, 1/29 … 1/a9 can be expressed as 1/(10n-1)

Again, 1/(10n-1) = (1/10n) [ 1-(1/10n)] ^-1

= 1/(10n) [ 1+ (10n)^-1 + (10n)^-2 + (10n)^-3 + … ]

= (10n)^-1 [ 1 + (10n)^-1 [ 1 + (10n)^-1 [ 1 + (10n)^-1 [ 1 + … ]…]]]

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More clearly, in the next slide :

Binomial Expansion of 1/(10n-1)

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∴

Vedic One Line Method (contd.) : Other D

• For Denominators ending in 1,3 or 7, we have to multiply both Numerator and Denominator by 9, 3 and 7 respectively, to get a denominator ending with 9 and follow the same procedure.
• Thus, 1/7 becomes 7/49. We start with 7 and divide by 5:

7÷5 🡪 21 14 42 28 35 07 thereafter repeats. Thus, 1/7=0.142857…

• Similarly, 1/13 becomes 3/39. We start with 3 and divide by 4:

3÷4 🡪 30 27 36 09 12 03 thereafter repeats. Thus, 1/13=0.076923…

• Now, 1/17 becomes 7/119, so we start with 7 and divide by 12:
• 1/17 has 16 Digits.

Vedic One Line Method (contd.) : N/a1

• For Denominators ending in 1, i.e. for fractions like 1/21, 1/31, 1/41 etc. we have to multiply both N & D by 9.
• Thus, 1/21 becomes 9/189 and digit divisor (dd) is 18+1=19. Similarly, 1/31 becomes 9/279, with dd = 28 and 1/41 becomes 9/369 and dd=37. Therefore, Vedic One Line Method needs a change.
• We consider that, 1/21, 1/31, 1/41 etc. are of the form 1/(10n+1). We have Binomial Expansion of (1+x) ^-1:

(1+x)^-1= 1-x+x²-x³+x⁴+… … =1-x(1-x(1-x(1-…)…)), where x < 1

Therefore, 1/(10n+1) = (1/10n) [ 1+(1/10n)] ^-1

= 1/(10n) [ 1 - (10n)^-1 + (10n)^-2 - (10n)^-3 + … ]

= (10n)^-1 [ 1 - (10n)^-1 [ 1 - (10n)^-1 [ 1 - (10n)^-1 [ 1 - … ]…]]]. More clearly,

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Binomial Expansion of 1/(10n+1)

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∴

Vedic One Line Method (contd.) : 1/21

• For Denominators of the form a1, we simply divide by a. Let’s take the case of 1/21, here digit divisor will be (21-1)/10=2.
• We proceed as follows: 1÷2 🡪 q=0, r=1. Considering the Binomial expression, we have to do 10r-q in this case, instead of 10r+q earlier. �10x1-0=10, 10÷2🡪 05, i.e. q=5, r=0. Next, 0x10-5 = -5. -5÷2🡪 q=-3, r=1, because, -3x2+1= -5. Then so far the result line shows :�0.053 . Next 10x1-(-3)=13. 13÷2🡪 q=6, r=1. 0.0536. Next, 10-6=4. 4÷2🡪 q=2, r=0. 0.05362. Next 0x10-2=-2. -2÷2🡪 q=-1, r=0. Result is �0.053621. Next dividend 0x10-(-1)=1. It comes to where from we started. Thus, 1/21 = 0.053621…053621. Clearing the Vinculum:�1/21= 0.047619…047619…

Comparison of Positive and Negative methods

• Excel sheet snapshot of Positive and Negative Vedic one-line methods for 1/21:

dd=Digit Divisor

We started dividing 9 by 19, in

Positive method.

We started dividing 1 by 2, in

Negative method.

Both gives same result for 1/21, after clearing the Negative digits, equivalent to Vinculum in second (5 3 = 47 and 2 1 = 19). It is up to an individual to choose which is suitable for oneself.

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More example of Negative method

• Take 14/41: Start dividing 14 by dd=4: q=3, r=2, we write 23

14÷4🡪 23 (17) 14 (06) 21 (19) 34 (26) 26 (14), back to 14, so repeats.

Thus 14/41 = 0.34146…

• 14/31: 0.451612903225806 4516…

3 1 = 29, 6 2 = 58

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• 35/51: 0.6862745098039215 6862…

7 2 = 68, 1 1 = 09

Vedic One Line Method (contd.) : N/a7

• Denominators ending in 7 can have two different ways:
1. Multiply both N & D by 7, to have Denominator ending in 9, or
2. Multiply both N & D by 3, to have Denominator ending in 1.
• As for example, 8/17 can be expressed as 56/119 or 24/51.
• Positive Method : start dividing 56 by 12

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• Negative Method : start dividing 24 by 5

6 2 = 58, 3 1 = 29

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Exercise on Recurring Decimals

• Find at least first 10 places of the following fractions:
• 13/19, 19/29, 17/39, 34/49, 23/59
• 7/13, 16/23, 17/33, 29/43, 31/53
• 6/7, 11/17, 16/27, 24/37, 33/47
• 10/21, 13/31, 14/41, 23/51, 29/61

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• Find at least last 5 places of recurring decimals in the following:
• 3/19, 5/19, 13/19, 5/29, 15/29, 23/29

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Periodicity of Recurring Decimals

• Periodicity or the Length of the Recurring part in decimals depends upon the Denominator being a factor of “Repeat Nine” numbers of minimum length. i.e. minimum of 9, 99, 999 … etc. For example,
• Denominator D = 3, has periodicity 1, because 3 is a factor of 9. The number 3 is also factor of 99, 999. But 9 is the minimum of all.
• D = 11 has periodicity 2, because 11 is a factor of 99.
• D = 27 or D = 37 has periodicity 3, because they are factors of 999.
• D = 41 has periodicity 5, as it is a factor of 99999.
• D = 7, 13, 21, 39, 63 – all have periodicity 6, all being factors of 999999. This way…

Periodicity of Recurring Decimals

Divisibility Rules : without actual division

1. The numbers 2 and 5 being factors of 10, we have similar rules for 2 and 5:
1. Divisibility by 2ⁿ requires last n digits to be divisible by 2ⁿ.

e.g. 57356 is divisible by 4, because 56 is divisible by 4, etc.

• Divisibility by 5ⁿ requires last n digits to be divisible by 5ⁿ.

e.g. 543275 is divisible by 25, since 75 is divisible by 25, etc.

3. Divisibility by a Composite number demands, divisibility by its factors. E,g. 42 is divisible by 6, as it is divisible by 2 as well as 3. 135 is divisible by 15, because it is divisible by 3 as well as 5.
4. Divisibility by numbers ending with 1,3,7 and 9 has general method of Osculation, which gets a trivial form in the cases of 3, 9 and 11.

Divisibility Rules (contd.)

• Divisibility by 3 : A number is divisible by 3, if sum of all its digits is divisible by 3. For example, the number 5423451 is divisible by 3, because sum of all its digits, viz. 24 is divisible by 3.
• Divisibility by 9 : A number is divisible by 9, if sum of all its digits is divisible by 9. For example, the number 8423451 is divisible by 9, because sum of all its digits, viz. 27 is divisible by 9.
• Divisibility by 11: A number is divisible by 11, if the difference of sums of alternative digits (say, even and odd positioned digits) is zero or is divisible by 11. For example, the number 934571 is divisible by 11, because (9+4+7) – (3+5+1) = 20 – 9 = 11 is divisible by 11.
• Exercise: Find whether the number 654786 is divisible by 3, 9 or 11.

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Divisibility : Osculation Method

• A two or more digit Integer, N can be expressed in the form:

N = 10R + U, where U is the Unit’s digit and R is the Rest (Comprising of Ten’s digit and beyond).

Osculation is a Vedic Method to create a smaller Integer T = R + mU, where this multiplier m is also called the Osculator. It is being said that, the Integer N is Osculated by m, to get T. The process can be recursive with the same Osculator m, i.e. the Unit’s digit of T can be multiplied by m and added to Rest digits of T.

Osculation is an efficient method for testing Divisibility. It is a matter to determine the required m, given a number N and a divisor D (ending in 1,3,7 or 9). The Osculator m can be found with the following consideration :

Divisibility : Osculation Method : Algebra

• Consider mN – T :

mN – T = m(10R + U) – (R + mU) = 10mR + mU – R – mU

Or mN – T = (10m – 1)R … (1)

Let’s now consider the case of Divisibility by a number, i.e. Divisor D.

If we can find an Osculator m, such that,

(10m – 1) mod D = 0 …(2), that will mean, from Eq. 1:

(mN – T) mod D = 0 …(3)

Thus, once a proper Osculator m is found, our job will be to test T for divisibility by D. Because, T mod D = 0 will imply mN mod D = 0 from Eq. (3), and in turn will lead to N mod D = 0, because, m mod D # 0.

Divisibility : Osculation Method : Example

• [Note: Why? We found an m, so that (10m – 1) mod D = 0, that means, 10m mod D # 0, i.e. m mod D # 0 proved. Now, 10 mod D # 0 implies this Osculation method is only applicable for Divisors, not comprising of factors 2 & 5, i.e. D must end in 1,3,7 or 9.]

Example, Lets test whether 343 is divisible by 7. First we find Osculator for 7. We see, that (10x5 – 1), i.e. 49 is divisible by 7. Thus, Osculator m = 5. Once found m, we execute the following process:� 343 🡪 34 + 3x5 = 49, which is divisible by 7, so 343 is divisible by 7.

Take 64724: 64724 🡪 6472 + 4x5 = 6492 🡪 649 + 2x5 = 659 🡪 �65 + 9x5 = 110 🡪 11 + 0x5 = 11, not divisible by 7. So, by back tracking, 110, 659, 6492 and 64724 : none will be divisible by 7.

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Divisibility : Osculation Method : Example

• Test Divisibility of 11635 by 13. Let’s first find Osculator m for 13:

We find, (10x4 – 1) i.e. 39 is divisible by 13. So, m=4 for 13. We do:

11635 🡪 1163 + 5x4 = 1183 🡪 118 + 3x4 = 130 🡪 13 + 0x4 = 13.

Since it is 13, therefore 130, 1183 and 11635 are divisible by 13.

• Test Divisibility of 41285 by 23. We find, (10x7 – 1) is divisible by 23.

41285 🡪 4128 + 5x7 = 4163 🡪 416 + 3x7 = 437 🡪 43 + 7x7 = 92 🡪

9 + 2x7 = 23. Thus, 92, 437, 4163 and 41285 are divisible by 23.

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Positive, Negative and Minimum Osculator

• If (10m + 1) is divisible by D, all the Algebraic considerations hold good for Osculator –m. i.e. T should be calculated as R – mU. In this case -m can be called a Negative Osculator.
• Moreover, if m is an Osculator to test Divisibility by D, m ± f.D will also an Osculator, where f is any integer. The Osculator with minimum absolute value is called minimum Osculator and most suitably chosen to test divisibility.
• Example : (10x5 – 1) and (10x2 +1) are divisible by 7. Thus, 5 is the minimum Positive Osculator, -2 is minimum Negative Osculator, as well as minimum Osculator. In the positive side 5, 12, 19, … and in the negative side -2, -9, -16, … all can be used for testing divisibility.
• Table on the next page gives list of Osculators.
• Osculator for 3 & 9 is +1 and for 11 is -1. That’s why their rule is simpler.

Table of Positive and Negative Osculators

• We find Positive and Negative Osculator using following Table: