Algebraic Proof of Finding out the Decimal Expansion of Vulgar Fractions using Ekadhikena Purvena - Suthra

 

Vedic Mathematics

Suthra 1     :        Ekadhikena Purvena

Application : Recurring Decimal Expansion of Vulgar Fractions with denominator ending with ‘9’. (Algebraic Proof)

Meaning    :        By One More than the Previous One

The Algebraic Proof for the same as follows:

Any Vulgar fraction of the form 1/a9

Can be written as

1/a9           =        1/(a+1)x-1) where x=10

                   =                           1              `

                                 (a+1)x[1-1/(a+1)x]

                   =        1/[(a+1)x] [1+1/(a+1)x]^-1

                   =        1/(a+1)x [1+ 1/(a+1)x + 1/(a+1)²x² +…….]

                   =        1/(a+1)x + 1/(a+1)²x²+1/(a+1)³x³+………………….

                   =        10^-1(1/(a+1) + 10^-2(1/(a+1)² + 10^-3 (1/a+1)³ +………

The series explains the process “Ekadhikena….”

Let us consider 1/19

Using the above algebraic proof, we have

1/19                    =        10^-1(1/(1+() + 10^-2(1/(1+1)² + 10^-3(1/(1+1)³+ ………

                   =        10^-1(0.5) + 10^-2 (0.25) + 10^-3 (0.125) +……….

                   =        0.05 + 0.0025 + 0.000125 + 0.00000625 + ……..

                   =        0.052631…….

Vedic Mathematics : Recurring decimal Expression of Vulgar Fraction with denominator ending with '9'

 

Vedic Mathematics

Suthra 1     :        Ekadhikena Purvena

Application : Recurring Decimal Expansion of Vulgar Fractions with denominator ending with ‘9’.

Meaning    :        By One More than the Previous One

We can apply the same suthra to find out the decimal representation(Recurring Decimal)  of VULGAR fractions with denominator ending with the digit ‘9’.

There is another method to find out the result of the same operation. In the previous post, you have seen by using ‘Division Method’. Here we use a simple technique, ‘Multiplication Method’. Here it goes…..

 

Let us use this suthra to find out the recurring decimal of

In, 1/19 – PREVIOUS of 9 is ‘1’.

And ONE MORE THAN of it is ‘1+1=2’.

Hence ‘2’ is the multiplier of the conversion.

Write the last digit in the numerator as 1

And follow the following steps…..

Step 1 :           1

Step 2 :           21 (multiply 1 by 2, write the result on left)

Step 3 :           421 (multiply 2 by 2, write it on left)

Step 4             :           8421 (multiply 4 by 2, write it on left)

Step 5 :           ₁68421(multiply 8 by 2, write 6 on left, carry over 1)

Step 6 :           ₁368421(multiply 6 by 2,add the carry over,with carry over 1)

Step 7 :           7368421(3 x 2 + 1 = 7)

Step 8 :           ₁47368421(7x2=14, 4 with carry over 1)

Step 9             :           947368421(4x2+1=9)

Step 10:          ₁8947368421(9x2=18, 8 carry over 1)

Step 11:          ₁78947368421(8x2+1=17, 7 with carry over 1)

Step 12:          ₁578947368421(7x2+1=15, 5 carry over 1)

Step 13:          ₁1578947368421(5x2+1=11, 1 carry over 1)

Step 14:          31578947368421(1x2+1=2)

Step 15:          631578947368421(3x2=6)

Step 16:          ₁2631578947368421(6x2=12, 2 carry over 1)

Step 17:          52631578947368421(2x2+1=5)

Step 18:          ₁052631578947368421

Next step onwards, these 18 digits repeat……..like recurring decimal…

Hence 1/19 = 0. 05231578947368421…..

 

Vedic Mathematics - Suthra 01 - Finding the Recurring Decimal Expansions of Vulgar Fraction with Denominator ending with the digit '9'

 

Vedic Mathematics

Suthra 1     :        Ekadhikena Purvena

Application : Recurring Decimal Expansion of Vulgar Fractions with denominator ending with ‘9’.

Meaning    :        By One More than the Previous One

We can apply the same suthra to find out the decimal representation(Recurring Decimal)  of VULGAR fractions with denominator ending with the digit ‘9’.

First Method : Division Method

Let us use this suthra to find out the recurring decimal of

1/19

Here the number of decimal places before repetition is

“Difference of Numerator and Denominator”

i.e., 19 – 1 = 18 places

In this fraction, 1/19, “Purva”(before 9) is 1

Hence “Ekadhikena Purvena” means 1 + 1 =2

While applying “Ekadhikena Purvena”

We have

Step 1 : Divide numerator 1 by 20

            1/20 = 0.1/2 = ₁0 (0 times, 1 remainder)

Step 2 : Divide 10 by 2

             0.0₀5(5 times, o remainder)

Step 3 : Divide 5 by 2

             0.05₁2 (2 times, 1 reminder)

Step 4 : Divide ₁2 I.e., 12 by 2

             0.052₀6(6 times, 0 reminder)

Step 5 : Divide 6 by 2

             0.0526₀3(3 times, 0 reminder)

Step 6 : Divide 3 by 2

             0.05263₁1(1 times, 1 remainder)

Step 7 : Divide 11 by 2

             0.052631₁5(5 times, 1 reminder)

Step 8 : Divide 15 by 2

             0.0526315₁7 (7 times, 1 remainder)

Step 9 : Divide 17 by 2

             0.05263157₁8(8 times, 1 reminder)

Step 10 : Divide 18 by 2

             0.052631578₀9(9 times, 0 remainder)

Step 11 : Divide 9 by 2

            0.0526315789₁4(4 times, 1 remainder)

Step 12 : Divide 14 by 2

            0.05263157894₁7(7 times, 1 remainder)

Step 13 : Divide 7 by 2

            0.052631578947₁3(3 times, 1 reminder)

Step 14 : Divide 13 by 2

            0.0526315789473₁6(6 times, 1 reminder)

Step 15 : Divide 16 by 2(8 times, 0 remainder)

            0.05263157894736₀8

Step 16 : Divide 8 by 2(4 times, 0 remainder)

            0.052631578947368₀4

Step 17 : Divide 4 by 2(2 times, 0 remainder)

            0.0526315789473684₀2

Step 8 : Divide 2 by 2(1 time, 0 remainder)

            0.052631578947368421

Hence 1/19     = 0. 052631578947368421052631578947368421……….

                        =0. 052631578947368421

Vedic Mathematics - First Suthra (Ekadhikena Purvena) : Algebraic Proof

 

Vedic Mathematics

Suthra 1     :        Ekadhikena Purvena (Algebraic Proof)

Meaning    :        By One More than the Previous One

Let us consider

(ax + b)² = a²x² + 2abx + b²

To find 15²,

Here x= 10 and b = 5

(10a + 5) ²       =          a².10² + 2.10.a.5 + 5²

                        =          a².10² + a.10² + 5²

                        =          (a² + a) .10² + 5²

                        =          a(a+1). 10² + 25

Here, (10a + 5) represents numbers

(10x1+5)          =          15

(10x2+5)          =          25

(10x3+5)          =          35

(10x4+5)          =          45

(10x5+5)          =          55

(10x6+5)          =          65 etc…..

Thus

75²       =         (70+5)²

            Here b = 5 and a =7

            75²       =          7.(7+1)x10² + 5²

                        =          7x8 ×100+ 25

                        =          5625

Similarly try for 85², 95²

Vedic Mathematics : First Sutra : Ekadhikena Purvena : Using Find the Squares of numbers ending with the digit '5'

 

Vedic Mathematics

Suthra 1     :        Ekadhikena Purvena

Meaning    :        By One More than the Previous One

Let us make use of this Sutra to find out the

Squares of numbers ending with 5.

For Example: 15²

We start with 15, we need to find out its square.

(Using the formula – One more than the previous one),

Here we have, previous digit is 1 (purvena)

1 MORE than the PREVIOUS digit is 1 + 1 = 2 (ekadhikena)

Hence

15² = 1x2/25

15² = 125

Similarly

25² = 2x3/25 = 625

35² = 3x4/25 = 1225

45² = 4x5/25 = 2025

55² = 5x6/25 = 3025

65² = 6x7/25 = 4225

Now try to find out the squares of 75², 85², and 95².

In a similar way, we can extend this to three digit numbers like,

105² = 10x11/25 = 11025

115² = 11x12/25 13225

In the next post, you will see how we can verify this by algebraically.

Vedic Mathematics -03 : Sub SUthras

 

Vedic Mathematics

13 Sub Suthras in Vedic Mathematics are

1.         Anurupyena

2.         Sishyate Seshasamjnah

3.         Aaduamaadye naantyamantyena

4.         Kevalaih Saptakam Gunyaat

6.         Yaavadoonam Tavadoonam

7.         Yavadoonam Tavadoonyokrutya vargam cha yojayet

8.         Antyayordashakepi

9.         Antyayoreva

10.       Samucchayugunita

11.       Lopasthapanabhyam

12.       Vilokanam

13.       Gunitasamuchchayah Samuchchyagunita

Each Sub suthras will be explained with examples and applications in the following posts. Keep Reading.......

Vedic Mathematics -02 : 16 Suthras

 

Vedic Mathematics

16 Suthras in Vedic Mathematics are

1.         Ekadhikena Purvena

2.         Nikhilam Navascharatham Dasatah

3.         Urdhva – tiryagbhyam

4.         Paravartya Yojayet

5.         Sunyam Samnyasamuchaye

6.         (Anurupyr) Sunyamanyat

7.         Sankalanam – Vyavakalanabhyam

8.         Puranapuranabhyam

9.         Calana Kalanabhyam

10.       Yavadunam

11.       Vyastisamastih

12.       Sesanyankena Caramena

13.       Sopantyadvayamantyam

14.       Ekanyunena Purvena

15.       Gunitasamucchayah

16.       Gunakasamucchaya

Each Sutras will be explaining with examples and its applications in the forthcoming posts.

Please continue reading....

Vedic Mathematics - Introduction -01

 

Vedic Mathematics

Vedic Mathematics contains 16 Suthras and 13 Sub Suthras (simple mathematical formulae from Vedas). Application of these Suthras improves the computational skills of the learners in a wide area of problems, ensuring both speed and accuracy, firmly based on rational and logical reasoning.

The word ‘Veda’ is derived from the word ‘Vid’ which means – to know without limit.

Dr.NarendraPuri of the Roorke University prepared teaching material based on Vedic Mathematics during 1986-89. A few of his opinions are stated here:

·         Mathematics, derived from Veda, provide one line, mental and superfast methods along with quick cross checking systems.

·         Vedic Mathematics converts a tedious subject into a play full and blissful one which students learn with SMILES.

·         Vedic Mathematics offers a new and entirely different approach to the study of Mathematics based on pattern recognition. It allows for constant expressions of a student’s creativity, and is found to be easier to learn.

·         In this system, for any problem, there is always one general technique applicable to all cases and also a number of special pattern problems. The elements of choice and flexibility at each stage keeps the mind lively and about to develop clarity of thought and intuition, and thereby a holistic development of the human brain automatically takes place.

·         Vedic Mathematics with its special features has the inbuilt potential to solve the psychological problems of Mathematics – Anxiety.

I will be explaining all 16 Suthras and 13 Sub Suthras with its applications in the forthcoming Posts.