vedic math ppt.pptx

8/10/2019 Vedic math ppt.pptx

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FROM -E.C.E(ELECTRONICS

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Inthe Vedic period, records of mathematical activity are mostlyfound in Vedic texts associated with ritual activities. Arithmeticoperations ( ganit ) such as addition, subtraction, multiplication,fractions, squares, cubes, roots are enumerated in Narad Vishnupurana .Examples of geometric knowledge( rekha-ganit) are foundin Sulva Sutras of Baudhyana(800B.C)and

Apasthamba(600B.C) they display the understanding of basicgeometric shape to another of equivalent area, they also containedgeometric solutions of linear equation in single unknown and alsoquadratic equations. Apsthamba's sutra provides a value for thesquare root of 2 accurate to fifth decimal place. and solutions to

general linear equations. Some believe that these results cameabout hit and trail methods or generalizations of observed examples.others believe proofs for these results may have either beendestroyed, or else were transmitted orally through the gurukulsystem, and final results were tabulated in texts.


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1) The use of this sutra in case of

multiplication by 9,999.. is as

follows .

Method :

a) The left hand side digit (digits) is( are) obtained by applying the

ekanyunena purvena i.e. by

deduction 1 from the left side digit

(digits) .

e.g. 7 x 9; 71 = 6 ( L.H.S. digit )

b) The right hand side digit is thecomplement or difference between

the multiplier and the left hand side

digit (digits) . i.e. 7 X 9 R.H.S is 9 -

6 = 3.

c) The two numbers give the

1. EkanyunenaPurvenaThe Sutra Ekanyunena purvena means 'Oneless than the previous or One less than the onebefore.

Algebraic proof :

As any two digit number is of the form ( 10x + y

), we proceed

( 10x + y ) x 99

= ( 10x + y ) x ( 1001 )= 10x . 10210x + 102 .yy

= x . 103 + y . 102( 10x + y )

= x . 103 + ( y1 ) . 102 + [ 102( 10x + y )]

Thus the answer is a four digit number whose

1000th place is x,100th place is

( y - 1 ) and the two digit number which makes

up the 10th and unit place is the

number obtained by subtracting themultiplicand from 100.

Example 1: 8 x 9

Step-1

gives 8 1 = 7 ( L.H.S. Digit )

step-2

gives 9 7 = 2 ( R.H.S. Digit )

Step-3

gives the answer 72


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2.Ekdhikena Prvena

The Sutra Ekdhikena Prvena meansBy one more than the previous one.

1) Squares of numbers ending in 5 :

Now we relate the sutra to the squaring

of numbers ending in 5. Consider

the example 25.

Here the number is 25. We have to find

out the square of the number. For

the number 25, the last digit is 5 andthe 'previous' digit is 2. Hence, one

more than the previous one', that is,

2+1=3. The Sutra, in this context,

gives the procedure to multiply the

previous digit 2 by one more than

itself, that is, by 3'.

It becomes the L.H.S (left hand side) of

the result, that is, 2 X 3 = 6. The

R.H.S (right hand side) of the

result is 5, that is, 25.

Thus 25= [2 X 3 ]25 = 625.

Algebraic proof:

a) Consider (ax + b) a x+ 2abx +b.

This identity for x=10 and b=5 becomes

(10a + 5)= a . 10+ 2. 10a . 5 + 5

= a. 10+ a.10+ 5

= (a+ a ) . 10 +5

= a (a + 1) .10+ 25.

Clearly 10a + 5 represents two-digit numbers. In

such a case the number (10a+5) is of the form

whose L.H.S is a(a+1) and R.H.S is 25, that is,

[a(a+1)]25.

For three digits it is (a10+b10+ 5)

= P (P+1) 10+ 25, where P = 10a+b.

Example : 65= ( . 0 + 6 . 0 + 5) .

It is of the form (ax+bx+c) for a = 1, b = 6, c = 5

and

x =10. It gives the answer P(P+1)/25, where P = 10a

+ b

= 10 X 1+ 6 = 16, the previous. The answer is

= 16 (16+1) / 25

=16x17/25=27225


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3. Antyayor DasakepiThe Sutra signifies numbers of which thelast digits added up give 10. i.e. theSutra works in multiplication of numbers

like: 116 and 114. Further the numbers leftwards to the last digits remain the same.

Example 1 : 47 X 43

See the end digits sum 7 + 3 = 10 ; then by

the sutras Antyayor Dasakepi and

Ekadhikena we have the answer.

Step-1

47 x 43 = ( 4 + ) x4/ 7 x 3

Step-2 = 20 / 2

Step-3 = 202.

Example 5: 395 5+ 5=10, L.H.S.

portion remains the same i.e., 39.

Ekadhikena of 39 gives 40

395 = 395 x 395

= 39 x 40 / 5 x 5

= 1560 / 25

= 156025.


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4.Addition and subtraction

Example 1:

Step 1: 6 + 4 = 10, 1dot ; 0 + 8 = 8;8 + 4 = 12;

1dot and 2 answer under first column total

2 dots.

Step 2: 2+2 ( 2 dots) = 4; 4+9 = 13: 1

dot and 3+0= 3; 3+8 = 11;1dot and 1answer under second column -

total 2 dots.

Step 3: 3+2 ( 2 dots ) = 5; 5+6 = 11:

1dot and +7 = 8; 8+7 = 5;

1dot and 5 under third column as answer -

total 2 dots.

Step 4: 4 + 2 ( 2 dots ) = 6; 6 + 5

=11:1dot and 1+3 = 4; 4+2 = 6. total 1dot in

the fourth 6 column as answer.

Step 5: 1dot in the fourth column

carried over to 5th column (No digits

in it) as 1 Thus answer is from Step5 to

Example 2:

Step 1: 2 < 4. No sudha . 4-2 = 2first digit (form right to left)

Step 2: 9 > 7. Hence a dot onleft of 9 i.e. on 8

Step 3: purak of 9 i.e. 1, addedto upper 7 gives 1+ 7 = 8 second

digit

Step 4:Now means 8 + 1= 9.Step 6:As 9 > 2, once again thesame process: dot on left of i.e.,1

Step 7:purak of 9 i.e. 1, addedto upper 2 gives 1+ 2 = 3, the third

digit.

Step 8:Now 1 means1+1= 2

Step 9:As 2 < 3, we have

3-2 =1, the fourth digit

Thus answer is 1382

Addition Subtraction

3247..

-1892


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Divisions by 5, 9, 11:-

1/5=.2

2/5=.4

3/5=.6 4/5=.8

15/5=3

1/9=.111

2/9=.222

3/9=.333 4/9=.444

15/9=1.66

1/11=.0909

2/11=.1818

3/11=.2727 4/11=.3636

15/11=1.353

5

By 5: Multiply

numerator with 2 and

place the decimal

before the last digit.

By 9: Place point

before last digit. For

every 9 count in

numerator add 0.1

to it and repeat the

number

By 11: Multiply numerator

with 10 and subtract

numerator from it. And

place decimal before two

digits from right.


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vedic math ppt.pptx

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