Prasad Arithmetic Operations Revisited 1

VEDIC MATHEMATICS : Arithmetic Operations

T. K. Prasadhttp://www.cs.wright.edu/~tkprasad

Prasad Arithmetic Operations Revisited 2

Positional Number SystemTEN-

THOUSANDSTHOUSANDS HUNDREDS TENS UNITS

43210=

4 * 10,000+ 3 * 1,000+ 2 * 100+ 1 * 10

+ 0

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Two Digit Multiplication (above the base) using Vedic Approach

1. Method : Vertically and Crosswise Sutra

2. Correctness and Applicability

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Method: Multiply 13 * 12

• Write the first number to be multiplied and excess over 10 in the first row, and the second number to be multiplied and excess over 10 in the second row.

13 3

12 2

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13 3

12 2

• To determine the 3-digit product:– add crosswise to obtain the left digits

• (13 + 2) = (12 + 3) = 15

– and – multiply the excess vertically to obtain the

right digit. • (3 * 2) = 6

• 13 * 12 = 156

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Another Example

• 12 * 14 =

• 12 2

• 14 4

• 16 8

• 12 * 14 = 168

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Questions

• Why do both crosswise additions yield the same result?

• Why does this method yield the correct answer for this example?

• Does this method always work for any pair of numbers?

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Proof Sketch

• (12 + 4) = (14 + 2) = 16

• Why are they same?• That is, the sum of first number and excess over 10 of the

second number, and ….

• (12 + (14 – 10)) = (12+14 – 10) = (26 – 10) = 16• (14 + (12 – 10)) = (14+12 – 10) = (26 – 10) = 16

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Left digits[CrosswiseAddition]

Left digits[CrosswiseAddition]

Correctness Argument:Two possibilities

• 12 = (10 + 2)

• 14 = (10 + 4)

• 12 * 14

= (10 + 2) * 14

= 10 * 14 + 2 * 14

= 10 * 14 + 2 * (10 + 4)

= 10 * 14 + 2 * 10 + (2 * 4)

= 10 * (14+2) + 8

= 10 * 16 + 8

= 168

• 12 = (10 + 2)

• 14 = (10 + 4)

• 12 * 14

= 12 * (10 + 4)

= 12 * 10 + 12 * 4

= 12 * 10 + (10 + 2) * 4

= 12 * 10 + 10 * 4 + (2 * 4)

= 10 * (12 + 4) + 8

= 10 * 16 + 8

= 168

Right digit[Vertical Product]

Right digit[Vertical Product]

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• 15 * 12

15 5

12 2

17 10

18 0

Another Example

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• 17 * 15

17 7

15 5

22 35

22+3 5

25 5Need proof to feel comfortable!

Yet Another Example

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Method: Multiply 113 * 106

• Write the first number to be multiplied and excess over 100 in the first row, and the second number to be multiplied and excess over 100 in the second row.

113 13

106 6

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113 13

106 6

• To determine the 5-digit product:– add crosswise to obtain the left digits

• (113 + 6) = (106 + 13) = 119

– and – multiply the excess vertically to obtain the

right digits. • (13 * 6) = 78

• 113 * 106 = 11978

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Questions

• Why do both crosswise additions yield the same result?

• Why does this method yield the correct answer for this example?

• Does this method always work for any pair of 3 digit numbers?

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Proof Sketch

• (113 + 6) = (106 + 13) = 119

• Why are they same?

• (113 + (106 – 100)) = (113 + 106 – 100) = 119• (106 + (113 – 100)) = (106 + 113 – 100) = 119

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Right digits[Vertical Product]

Right digits[Vertical Product]

Left digits[CrosswiseAddition]

Left digits[CrosswiseAddition]

Correctness of Product :Two possibilities

• 113 = (100 + 13)

• 106 = (100 + 6)

• 113 * 106

= 113 * (100 + 6)

= 113 * 100 + (100 + 13) * 6

= 113 * 100 + 100 * 6 + (13 * 6)

= 100 * (113 + 6) + 78

= 100 * 119 + 78

= 11978

• 113 = (100 + 13)

• 106 = (100 + 6)

• 113 * 106

= (100 + 13) * 106

= 100 * 106 + 13 * (100 + 6)

= 100 * 106 + 13 * 100 + (13 * 6)

= 100 * (106 + 13) + 78

= 100 * 119 + 78

= 11978

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• 160 * 180 160 60 180 80 240 4800 288 00• Note that, the product of the excess over

100 has more than two digits. However, the weight associated with 240 and 48 are both 100, and hence they can be combined.

Another Example

Breakdown?!

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• 190 * 199

190 90

199 99

289 8910

289+89 10

378 10This approach is validThis approach is valid with suggested

modifications!

Yet Another Example

Breakdown?!

More Shortcuts

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Quick squaring of numbers that end in 5

• 15 * 15= 225

= (1*2) (5*5)

• 75 * 75= 5625

= (7*8) (5*5)

• 95 * 95= 9025

= (9*10) (5*5)

• Proof: Let the two digit number be written as D5.

• D5 * D5

= (D*10 + 5) * (D*10 + 5)

= (D*D*100) + (D*2*50) + 5*5

= (D*(D+1))*100 + 25

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Quick Multiplication : Special Case

• Proof: Let two digit numbers be AB and AC.• AB * AC

= (A*10 + B) * (A*10 + C)

= (A*A*100) + (A*10*(B+C)) + B*C

= (A*A)*100 + (A)*(B+C)*10 + (B*C)• For B+C=10, this reduces to

A*(A+1)*100 + B*C• For A=12, B=8 and C=2, this reduces to

(12)*(13)*100 + 16 = 15616Prasad Arithmetic Operations Revisited 21

Quicking squaring of numbers that begin with 5

• 51 * 51= (5*5+1)*100 + (1*1)

= 2601

• 57 * 57= (5*5+7) *100 + (7*7)

= 3249

• 59 * 59= (5*5+9) *100 + (9*9)

=3481

• Proof: Let the two digit number be written as 5D.

• 5D * 5D

= (50 + D) * (50 + D)

= (25 + D)*100 + (D*D)

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Quick squaring of two digit numbers

• Proof: Let two digit numbers be AB.• AB * AB

= (A*10 + B) * (A*10 + B)

= (A*A)*100 + 2*(A*10)*B + B*B

= (A*A)*100 + 20*(A*B) + (B*B)• For AB=79, this reduces to 4900+20*63+81

= 4981+1260 =6241• For AB=116, this reduces to 12100+20*66+36

= 12136+1320 =13456Prasad Arithmetic Operations Revisited 23

Generalized Multplication Using Working Base

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23 +3

24 +4• To determine the product, choose working base as 20:

– add crosswise to obtain the left digits with weight 20

• (23 + 4) = (24 + 3) = 27

– multiply the excess vertically to obtain the right digits.

• (3 * 4) = 12

• 23 * 24 = 27 * 20 + 12

• = 540 + 12

23 * 24 = 552

723 +23

724 +24• To determine the product, choose working base as 700:

– add crosswise to obtain the left digits with weight 700

• (723 + 24) = (724 + 23) = 747

– multiply the excess vertically to obtain the right digits.

• (23 * 24) = 552

• 723 * 724 = 747 * 700 + 552

• = 522900 + 552

723 * 724 = 523452Prasad Arithmetic Operations Revisited 26

783 -17

775 -25• To determine the product, choose working base as 800:

– add crosswise to obtain the left digits with weight 800

• (783 - 25) = (775 - 17) = 758

– multiply the excess vertically to obtain the right digits.

• (17 * 25) = 425

• 783 * 775 = 758 * 800 + 425

• = 606400 + 425

783 * 775 = 606825Prasad Arithmetic Operations Revisited 27

532 +32

472 -28• To determine the product, choose working base as 1000/2:

– add crosswise to obtain the left digits with wt. 1000/2

• (532 - 28) = (472 + 32) = 504

– multiply the excess vertically to obtain the right digits.

• (+32) * (-28) = 896

• 532 * 472= (504 / 2)*1000 + (104 -1000)

• = 252000 + 104 - 1000

532 * 472= 251104Prasad Arithmetic Operations Revisited 28