Vedic Mathematical Concepts and Their Application to Unsolved ...

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The following three proofs, based on the application of Vedic Mathematical ..... This factor being free of the length and degree of dimensional regular bodies ac-.
Vedic Mathematical Concepts and Their Application to Unsolved Mathematical Problems: Three Proofs of Fermat's Last Theorem S.K. K a p o o r I n d i a n Institute o f M a h a r i s h i

Vedic

Science and Technology

Preface T h e f o l l o w i n g t h r e e proofs, b a s e d o n the a p p l i c a t i o n o f V e d i c M a t h e m a t i c a l con­ c e p t s , a d d r e s s a f a m o u s u n s o l v e d p r o b l e m o f m a t h e m a t i c s , F e r m a t ' s Last T h e o r e m . T h e f o l l o w i n g p a s s a g e of J a b a l i U p a n i s h a d p r o v i d e d the structural key for d e v e l o p i n g the m u l t i - d i m e n s i o n a l s p a c e s u s e d i n the a r g u m e n t for F e r m a t ' s Last T h e o r e m :

Then Paippaladi asked Lord Jabali, "Tell me, Lord, the secret, supreme reality. What is tattva [existence]? What is jiva [individual life]? What is pashu [the soul]? Who is Ish [the Master]? What are the means to enlightenment?" He said to him, "Very good! Every­ thing that you have asked, I will explain to you, as it is known." Again he said to him, " H o w is it that you know this?" Again he said to him, "from Shadanan." Again he said to him, "How does he then know this?" Again he said to him, "from Ishan." Again he said to him, "How does he know it from him?" And again he said, "from upasana [worship]." In

an

earlier work,

The Rationale of

Оm—Its

Formulation,

Significance and O c c u r -

rence and Applications in Ancient Literature, I c o m m e n t e d on the a b o v e p a s s a g e as folloWS:

The import in the above nine mantras of the Jabali Upanishad is that Paippaladi had asked Jabali Rishi to enlighten him about tattva, jiva, pashu and Ish. Jabali Rishi happily prepared to instruct him about the questions asked. Just as Jabali Rishi was about to be­ gin his discourse, Paippaladi inquired how he Jabali Rishi had achieved enlightenment.

Address correspondence to: S.K. Kapoor, Visiting Professor Indian Institute of Maharishi's Vedic Science and Technology Maharishi Nagar, Ghaziabad, UP 201 307, India Modern Science and Vedic Science, Volume 3, Number 1, 1989 © 1989 Maharishi International University

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On this, the Jabali Rishi disclosed that he was enlightened on the points by Shadanan. On this he further asked from where Shadanan had achieved enlightenment and to this the answer of Jabali Rishi was that Shadanan had received enlightenment from Ishan.

Shadanan to Ishan The insight of descendence of Brahmavidya for the Lower Mathematical domain is that the processing should begin from within the fourth-dimensional domain in terms of its constituent, that is a plane is to be along its diagonal which would be facing North-East (Ishan) and would be leading to the sixth-dimensional domain (Shadanan). Sequential order would emerge because of dedh-devata The processing which was initiated from within the fourth-dimensional domain led to the sixth-dimensional domain. The processing was along the diagonal. The diagonal is greater than either side. The diagonal is also less than the sum of both the sides (of the triangle). This concept is the concept of one and a half units. The processing line of one and a half units is 4x3/2=6. The Upanishadic enlightenment on the point is that the devas are 1, 3/2, 2, 3, and so on (Brihadaranyaka Upanishad). The concept of dedh-devata is the specific processing concept. In terms of this concept the processing along the North-East line (Ishan) which has taken the fourth-dimensional domain to the sixth-dimensional domain sequentially would carry the processing further, naturally to the ninth-dimensional domain as 6x3/2 = 9. T h e s i g n i f i c a n c e o f t h e a b o v e w i t h i n t h e a r i t h m e t i c d o m a i n i s that i n o r d e r t o u n d e r s t a n d t h e s t r u c t u r a l f r a m e s a n d s y s t e m s of n a t u r a l n u m b e r 9 (in U p a n i s h a d i c l a n g u a g e : B r a h m a v i d y a ) w e h a v e t o d e v e l o p t h e u n d e r s t a n d i n g i n t e r m s o f the s t r u c t u r a l f r a m e s a n d s y s t e m s o f n a t u r a l n u m b e r 6 (in U p a n i s h a d i c l a n g u a g e S h a d a nan b e s t o w e d enlightenment upon Jabali Rishi). A n d understanding the structural f r a m e s a n d s y s t e m s of n a t u r a l n u m b e r 6 m u s t be in t e r m s of the s t r u c t u r a l f r a m e s a n d s y s t e m s o f n a t u r a l n u m b e r 4 (in the U p a n i s h a d i c l a n g u a g e I s h a n b e s t o w e d e n l i g h t e n m e n t u p o n S h a d a n a n ) . T h i s r e v e r s e s e q u e n t i a l p r o c e s s c o n t a i n s the s t r u c t u ral k e y . T h e b a s i c V e d i c M a t h e m a t i c a l c o n c e p t s u s e d are that the unity (single-syllable O m ) is p r o c e s s a b l e q u a r t e r by q u a r t e r (Shri Pada processing line) and the fourth q u a r t e r is the i n t e g r a t i o n of t h e first t h r e e q u a r t e r s (Maharishi processing line). As s u c h , I interpret the a b o v e U p a n i s h a d i c p a s s a g e in the following w a y : the structural f r a m e s and s y s t e m s of natural n u m b e r 4 (and h e n c e the f o u r t h - d i m e n s i o n a l d o m a i n ) are to be h a n d l e d as unity, a d m i t t i n g p r o c e s s i n g quarter by quarter, and the fourth

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q u a r t e r (the u n m a n i f e s t e d quarter) is to be p r o c e s s e d as integrating the first three q u a r t e r s ( m a n i f e s t e d q u a r t e r s ) . T h e p r o c e s s i n g w i t h i n the fourth quarter a l o n g the N o r t h - E a s t d i a g o n a l w o u l d lead to the d o m a i n of S h a d a n a n ( a d m i t t i n g structural f r a m e s a n d s y s t e m s of natural n u m b e r 6 and h e n c e a s i x - d i m e n s i o n a l d o m a i n ) . T h e U p a n i s h a d i c c o m m a n d is to r e v e r s e the p r o c e s s , as the p r o c e s s i n g is to be had in the d e s c e n d i n g o r d e r from 9 to 6 and 6 to 4, b u t here in this w o r l d e v e r y t h i n g is h a n g i n g upside

down

urdhvamulam—Bhagavad-Gita,

15.1). T h i s m e a n s that

w h i l e o u r aim is to p r o c e s s within the u n m a n i f e s t e d quarter in order to go from the f o u r t h - d i m e n s i o n a l d o m a i n of natural n u m b e r 4 to the s i x t h - d i m e n s i o n a l d o m a i n of n a t u r a l n u m b e r 6, we h a v e to p r o c e s s a l o n g the N o r t h - E a s t d i a g o n a l but in the d i r e c tion w h i c h l e a d s from the s i x t h - d i m e n s i o n a l d o m a i n t o w a r d s the f o u r t h - d i m e n s i o n a l d o m a i n . F u r t h e r , the p r o c e s s i n g m u s t b e quarter b y quarter, w h i c h m e a n s that w h e n n a t u r a l n u m b e r 4 is t a k e n as a unity, its quarter w o u l d be o n e , and similarly, w h e n n a t u r a l n u m b e r 6 is t a k e n as a unity, its p r o c e s s i n g unit w o u l d be o n e . H e n c e the p r o c e s s i n g a l o n g the N o r t h - E a s t d i a g o n a l is to be had by d i v i d i n g it into six parts. Out of t h e s e six p a r t s , o n l y o n e part w o u l d be u n m a n i f e s t e d and the r e m a i n i n g five parts w o u l d c o v e r the m a n i f e s t e d part. W i t h this the p r o c e d u r e of structural p r o c e s s i n g s t a n d s as f o l l o w s : Step 1: T h e processing is to be in the fourth quarter of the Om formulation. Step 2: Fourth quarter (fourth c o m p o n e n t ) of the D e v a n a g a r i Om formulation is like a t w o - d i m e n s i o n a l cartesian frame. Step 3: So processing within the fourth quarter (fourth c o m p o n e n t ) a m o u n t s to p r o c e s s i n g on the format of a plane. Step 4: As both the d i m e n s i o n a l lines are s y m m e t r i c , the natural figure of the p l a n e format is a square (to be called the p r o c e s s i n g square). Step 5: For processing of the first three quarters of the manifested d o m a i n s , divide the s q u a r e into 3 x 3 = 9 squares (to be called processing units). Step 6: Therefore, if the length of the square is Z, then the length of the p r o c e s s i n g unit w o u l d b e Z / 3 . Step 7: As the fourth quarter is to be processed as the integration of the first three q u a r t e r s , the length of the processing square should be increased by the length of its p r o c e s s i n g units, that is, the extended p r o c e s s i n g square, as it m a y be called, w o u l d have a length equal to Z + Z/3 = 4 Z / 3 . Step 8: T h e processing square b e i n g of length Z, the manifested part of the NorthEast d i a g o n a l of the extended p r o c e s s i n g square w o u l d be 5 Z / 6 , and the unmanifested part of the North-East diagonal of the said square w o u l d be Z/6. Step 9: T h e a b o v e format d e v e l o p s a square w h o s e length is equal to the diagonal, w h i c h m e a n s we are transcending from the geometrical square to the values-square (as it m a y be called). In this format t r a n s c e n d e n c e is permissible within the format of a g e o metrical square itself in t e r m s of the cross points of the lines parallel to the axes of the t w o - d i m e n s i o n a l cartesian frame of the geometrical square. If we take into account only the c r o s s points of the lines parallel to the a x e s (which lines m a y be called parallel axes) then the cross points on the length or b r e a d t h of the square w o u l d be equal to the cross p o i n t s on any of the t w o diagonals of the square.

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Step 10: T h e effect of extending the processing square into the extended processing s q u a r e upon the values-square of the processing square qua the values-square of the ext e n d e d p r o c e s s i n g square is the structural key for resolving w h y the s u m of t w o regular b o d i e s of N t h - d i m e n s i o n a l space does not constitute another regular body of the Nthd i m e n s i o n a l space. In the present studies I have applied the a b o v e structural key in three different w a y s to p r o v e F e r m a t ' s Last T h e o r e m . T h e s e proofs are submitted not only for the p u r p o s e of s u p p l y i n g a proof of this unsolved t h e o r e m , but also with the aim of appealing to other scholars to approach the main challenges within their disciplines through the V e d i c wisdom. T h e V e d i c perspective on m e t h o d o l o g y integrates objectivity with subjective experie n c e and as such those w h o are trained in objective m e t h o d o l o g y are required only to learn h o w to supplement this approach with their o w n subjective experience of the Vedic Reality. For this I feel highly privileged to be at the feet of His Holiness Shri P a d a Babaji w h o initiated me for the Shri Pada processing line to process this quarter by quarter, and at the feet of H i s H o l i n e s s Maharishi M a h e s h Y o g i w h o initiated me for the M a h a r i shi p r o c e s s i n g line to p r o c e s s the fourth quarter as the integration of the first three quarters. I am also highly obliged to Professor Krishnaji, C h a i r m a n , Indian Institute of M a harishi

V e d i c S c i e n c e and T e c h n o l o g y , for experienced guidance and personal interest

in t h e s e and other studies at the Institute.

Proof by Direct Comparison Overview

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Statement of the Theorem

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Therefore, T is not a rational n u m b e r . N

N

N

Step 5: H e n c e , in the equation U = V + T , V and T are irrational n u m b e r s . T h e r e fore, we h a v e to look for solutions of this equation in the field of real n u m b e r s . Here it N

N

N

N

m a y be relevant to note that U , V , and T are rational n u m b e r s , so U = U , V , = V 1

N

1

N

a n d T = T give rise to the equation U = V + T w h e r e U , V and T are rational n u m 1

1

1

1

1

1

1

b e r s . T h i s has the obvious solution in the field of rational n u m b e r s . H o w e v e r , w h e n the N

N

N

solution is required for U = V + T , w h e r e V and T are irrational n u m b e r s , we have to shift to the field of real n u m b e r s . Step 6: T h e field of real n u m b e r s constitutes the arithmetical c o n t i n u u m w h i c h is e q u i v a l e n t to the linear geometrical c o n t i n u u m of a straight line. N o w the solutions of N

N

N

the e q u a t i o n s U = V + T and M

N .

U

N

= M

fore, the solution of the equation ( M U )

N

N .

V

N

+ M

= (MV)

N

N .

T

N

are directly linked up. T h e r e N

N

+ ( M T ) , w h i c h is nothing but S =

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VEDIC MATHEMATICAL CONCEPTS

p o i n t r e p r e s e n t i n g the n u m b e r z e r o ( 0 ) . As a s e c o n d step, for a g i v e n r a t i o n a l n u m b e r Z, we m a y cut a c l o s e d interval [ O Z ] of length OZ = Z from the line O R . As a t h i r d s t e p , for any rational n u m b e r X < Z, we can cut a c l o s e d interval [ O X ] of l e n g t h OX = X from the interval [ O Z ] of length OZ = Z. 4 . T h e i n t e r v a l [ O Z ] gets d i v i d e d into t w o p a r t s , n a m e l y c l o s e d interval [ O X ] and t h e o n e - s i d e d o p e n interval ( X Z ] . B o t h the parts, n a m e l y [ O X ] and ( X Z ] , d o not h a v e a n y c o m m o n p o i n t . All n u m b e r s r a t i o n a l and irrational less than or e q u a l to X are in [ O X ] , a n d all n u m b e r s rational or irrational g r e a t e r than X are in ( X Z ] . As s u c h , t h e interval ( X Z ] is of an irrational l e n g t h .

Fermat's Last Theorem 5. S t a t e m e n t : "It is i m p o s s i b l e to s e p a r a t e a c u b e into t w o c u b e s , or a b i q u a d r a t e i n t o t w o b i q u a d r a t e s , or in g e n e r a l any p o w e r higher than the s e c o n d into p o w e r s of like d e g r e e . " — F e r m a t n

n

n

6. In m o d e r n mathematical language this is restated as: x + y = z admits a solution in natural n u m b e r s only for n = 2. That m e a n s for given natural n u m b e r s z, n, and x, n

n

with x < z, we cannot find a natural n u m b e r w h o s e nth degree is equal to z - x .

Proof n

n

7. Let V = z - x . n

n

8. z, n, and x are natural n u m b e r s , t h e r e f o r e z and x are also n a t u r a l n u m b e r s . n

n

Let Z = z and X = x . T h e r e f o r e , V = Z - X . 9. As Z and X are n a t u r a l n u m b e r s , so the c l o s e d interval [ O Z ] of l e n g t h Z c u t s a r a t i o n a l l e n g t h on the real line O R . S i m i l a r l y , the c l o s e d interval [ O X ] of l e n g t h X c u t s a r a t i o n a l l e n g t h from the interval [ O Z ] . 10. V = Z - X = [ O Z ] - [ O X ] = ( X Z ] = irrational l e n g t h . 11. H e n c e , V

N " for every n a t u r a l n u m b e r N , a s N

n

i s a natural n u m b e r s o

w o u l d e q u a l a rational l e n g t h and not an irrational l e n g t h . 12. T h e a b o v e p r o v e s the t h e o r e m b u t for the restriction for the n to be

Restrictions for n to be

3.

3

R a t i o n a l e for the r e s t r i c t i o n s : 1 3 . To arrive at the r a t i o n a l e for t h e r e s t r i c t i o n s for n to be

3, we h a v e to go

b a c k to o u r definition of the d i m e n s i o n in t e r m s of its g e o m e t r i c a l r e p r e s e n t a t i o n . 14. T h e definition o f d i m e n s i o n c a n n o t b e e x p r e s s e d e x c e p t w i t h i n the m a n i f e s t ed w o r l d of t h r e e - d i m e n s i o n a l o b j e c t s , t h e r e b y p e r m i t t i n g its a r r a n g e m e n t as a l i n e ar g e o m e t r i c a l c o n t i n u u m of s t r a i g h t l i n e s , e a c h a m a t h e m a t i c a l m o d e l of the arithm e t i c a l c o n t i n u u m o f real n u m b e r s . I n c o n c r e t e t e r m s , g e o m e t r y a c c e p t s d i m e n s i o n as that w h i c h c o m p l e t e l y p e r m i t s r e p r e s e n t a t i o n as a straight line, and the g e o m e t r i cal u n i v e r s e f o l l o w i n g i s the t h r e e - d i m e n s i o n a l s p a c e w i t h the three d i m e n s i o n s n o t h i n g but s t r a i g h t l i n e s . In fact, here lies the r a t i o n a l e and the a n s w e r to the q u e s tion w h y r e s t r i c t i o n s are p l a c e d u p o n n to be

3. H o w e v e r , c o m p r e h e n s i o n of the

m a n i f e s t e d w o r l d c a n n o t be a c c e p t e d as a p r o o f as s u c h of the e x i s t e n c e or o t h e r w i s e o f the h i g h e r - d i m e n s i o n a l s p a c e s . F o r this w e h a v e t o h a v e a p u r e l y m a t h e m a t ical a p p r o a c h . n

1 5 . F o r t h i s , we m a y h a v e insight i n t o the internal s t r u c t u r e of N by a c c e p t i n g it

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to be a r e g u l a r b o d y of l e n g t h N of n t h - d i m e n s i o n a l s p a c e (for t h e p r e s e n t t a k e as a d e f i n i t i o n that d e g r e e n r e p r e s e n t s t h e n u m b e r o f d i m e n s i o n s ) .

The First Dimensional Regular Body 1 6 . T o h a v e p r o p e r i n s i g h t i n t o t h e i n t e r n a l s t r u c t u r e o f h i g h e r - d i m e n s i o n a l regu­ lar b o d i e s , w e w o u l d b e r e q u i r e d t o p i n - p o i n t t h e r a t i o n a l e a s t o w h y t h e v e r y firstl

dimensional regular body ( l ) does not permit division into smaller regular bodies of the first-dimensional space. 1 7 . T h e a n s w e r i s b e c a u s e o f t h e v e r y d e f i n i t i o n o f t h e n a t u r a l n u m b e r 1 . T h e nat­ u r a l n u m b e r 1 i s t h e s m a l l e s t n a t u r a l n u m b e r , s o t h e q u e s t i o n o f t h e e x i s t e n c e o f still smaller natural n u m b e r s d o e s not arise. H e n c e l

1

X

1

1

+ Y , w h e r e X a n d Y a r e natu­

ral n u m b e r s . 18. T h e above situation deserves p r o p e r m a t h e m a t i z a t i o n . Firstly, it takes o n e as u n d e f i n e d . S i m u l t a n e o u s l y i t g i v e s u s t h e f r e e d o m t o a c c e p t any l i n e a r l e n g t h , m a y i t b e e q u a l t o r a t i o n a l o r i r r a t i o n a l u n i t s / n u m b e r s , a s a l i n e a r u n i t and h e n c e " o n e . " 1 9 . U s i n g t h e a b o v e f r e e d o m o f c h o i c e t o a c c e p t a n y l i n e a r l e n g t h a s t h e dimen­ s i o n a l u n i t h e l p s u s t o r e d u c e a n y r e g u l a r b o d y N " t o t h e first r e g u l a r b o d y o f n t h d i m e n s i o n a l s p a c e ( Г ) , w h e r e N = o n e u n i t . T h i s a l s o c a n b e t a k e n a s t h e first regu­ 1

lar b o d y of t h e f i r s t - d i m e n s i o n a l s p a c e ( l ) s i n c e 1

n

1

= 1 = l .

20. T h e r e f o r e , in o r d e r to u n d e r s t a n d t h e internal structural a r r a n g e m e n t s of dimen­ sional r e g u l a r b o d i e s , w e h a v e n o o p t i o n b u t t o c o n s i d e r the structural knot r e s p o n s i b l e for t r a n s f o r m i n g unit linear l e n g t h into a rational fraction p l u s an irrational fraction. 2 1 . D e s i g n a t e t h e p o i n t o f s t u d y a s a structural knot of dimensional regular bodies. W e f o c u s o n t h e s t a t e m e n t t h a t o n e l i n e a r u n i t is e q u a l t o a half-unit r a t i o n a l length plus a half-unit irrational length. 2 2 . B e f o r e w e t a k e u p t h e a b o v e s t a t e m e n t for e l a b o r a t i o n , I w o u l d like t o a d d h e r e t h a t t h e r a t i o n a l e for t h e a b o v e lies i n t h e first p r i n c i p l e (of m a n i f e s t e d w o r l d a d m i t t i n g s t r u c t u r e s ) t h a t t h e s t r u c t u r a l u n i t i s half a s c o m p a r e d t o o n e a s a dimen­ sional unit.

Definition of a Straight Line 2 3 . O n e a t t r i b u t e of t h e s t r a i g h t l i n e is t h a t it h a s a m i d d l e p o i n t . A s e c o n d attrib­ u t e of t h e s t r a i g h t l i n e is t h a t it h a s a m i n i m u m of t h r e e p o i n t s . It m a y n o t be possi­ b l e to g i v e a p r e c i s e d e f i n i t i o n of a s t r a i g h t l i n e in t e r m s of s o m e a t t r i b u t e s . How­ ever, t h e b a s i c attributes of the straight line w o u l d h e l p us settle s o m e practical defini­ t i o n s . T h e p r a c t i c a l d e f i n i t i o n s m a y h a v e w o r t h for a n y d o m a i n o f p r a c t i c a l impor­ tance but the same may not be acceptable to mathematics. A precise mathematical d e f i n i t i o n o f a s t r a i g h t l i n e c a n b e p r o v i d e d b y t h e a r i t h m e t i c a l c o n t i n u u m o f t h e real n u m b e r s . T h e f o r m a t b e n e a t h t h e o r d e r e d d i s p l a y o f t h e set o f r e a l n u m b e r s w o u l d m a t h e m a t i c a l l y qualify t o b e d e s i g n a t e d a s a s t r a i g h t l i n e . A s s u c h t h i s i s a c c e p t e d a s t h e d e f i n i t i o n o f t h e s t r a i g h t l i n e for t h e p u r p o s e o f e x p r e s s i o n s o f d i m e n s i o n a l f r a m e s , a s w e l l a s for t h e p u r p o s e o f c u t t i n g l e n g t h s f r o m a r e g u l a r b o d y .

Structural Unit Versus Dimensional Unit 2 4 . N o w let u s t a k e u p t h e c h a l l e n g e o f t h e s t r u c t u r a l k n o t ( p a r a g r a p h 2 1 ) . H e r e t h e s i t u a t i o n p r o c e e d s i n t w o s t e p s : firstly, i n t e r m s o f t h e a n s w e r t o t h e p o s e r , i f w e

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c a n t r a n s c e n d the t h r e e d i m e n s i o n a l g e o m e t r i c a l u n i v e r s e and h a v e a g e o m e t r i c a l c o n t i n u u m , and s e c o n d l y , in t e r m s of t h e fine c o n n e c t i o n b e t w e e n the structural unit and t h e d i m e n s i o n a l unit m a i n t a i n i n g the ratio of 1/2 : 1.

Square and Cube 2 5 . F o r b o t h t h e s t e p s , initially w e m a y p r o c e e d w i t h the structural c o m m o n n e s s of t h e s q u a r e as a r e g u l a r b o d y of a t w o - d i m e n s i o n a l s p a c e and the c u b e as a r e g u l a r b o d y of a t h r e e - d i m e n s i o n a l s p a c e . 2 6 . T h e s i g n i f i c a n c e of the structural c o m m o n n e s s of a s q u a r e and a c u b e is that t h e r a t i o of area and p e r i m e t e r of the s q u a r e , and v o l u m e and surface area of the c u b e , a d m i t s e q u e n t i a l o r d e r as a function of the d e g r e e of the d i m e n s i o n a l s p a c e . We m a y define the d e g r e e of n - d i m e n s i o n a l s p a c e as n. T h e s e ratios for l e n g t h 'a' of the r e g u l a r b o d i e s is: 2

3

a : 4a a n d a : 6 a 2 7 . T h e a b o v e r a t i o s h a v e the f o r m u l a t i o n

2

a /2na n

: 1 for n = 2 and 3.

n-1

2 8 . T h e above sequential formulation immediately supplies the key to unlock the structural knot of the dimensional regular body. It is that for any dimensional regular n

b o d y the d o m a i n part a (which for the square is the area of the square and for the cube is its v o l u m e ) and the frame part 2na

n-1

(which for the square is its perimeter and for the

c u b e is its surface area) maintain a ratio dependent upon 1/2, the length (a) and the dim e n s i o n a l degree (n). This ratio c o m e s to be a/2n. T h e structural significance is that for any a and any n, there c o m e s into play the factor 1/2. This factor, which may be accepted as a halving-factor, is responsible for providing the required structural knot to bind the structural arrangements to constitute dimensional regular bodies. This is the structural key to unlock the structural knots of dimensional regular bodies of all lengths and all degrees. This factor being free of the length and degree of dimensional regular bodies acquires universal application. H e n c e , even the regular bodies of unit length of all the dimensional spaces of any degree also admit the factor (1/2). By this m e a n s we gain in1

2

3

n

sight into the internal structural arrangements of l , l , 1 . . . 1 .

Framed Domains Sequence 2 9 . B e f o r e w e i n v e s t i g a t e the i n t e r n a l s t r u c t u r e o f r e g u l a r b o d i e s o f unit l e n g t h o f different d i m e n s i o n a l s p a c e s , let us u n d e r s t a n d the s e q u e n c e [a /2na for n = 1,2,3...]. T h i s s e q u e n c e m a y be defined as a f r a m e d d o m a i n s s e q u e n c e and its indiv i d u a l t e r m s a s i n d i v i d u a l framed d o m a i n s o r s i m p l y framed d o m a i n s . T h e r e a s o n for t h e c h o i c e of t e r m i n o l o g y is that t h e r e g u l a r b o d i e s h a v e their d o m a i n part c o n t a i n e d w i t h i n the frame part and t h u s c a n b e d e s i g n a t e d a s framed d o m a i n s . n

n-1

First Framed Domain 3 0 . T h e first m e m b e r of the f r a m e d d o m a i n s s e q u e n c e [a /2na n

that is a /2na n

n-1

n-l

for n = 1,2,3...],

for n = 1, that is a /2xla , that is a/2, is the first f r a m e d d o m a i n . 1

1-1

T h i s i n d i c a t e s that out o f any c l o s e d l i n e a r interval w e can r e m o v e o n e c l o s e d interv a l of half l e n g t h w h i l e the s e c o n d half w o u l d not be a c l o s e d interval. H e n c e t h e s t r u c t u r a l unit is half of the d i m e n s i o n a l unit, w h i c h essentially is the linear unit. O n e m a y r e v e r t b a c k t o this f r a m e d d o m a i n after h a v i n g b e e n t h r o u g h the internal

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s t r u c t u r e o f s e c o n d a n d t h i r d f r a m e d d o m a i n s , w h i c h are t h e w e l l - k n o w n g e o m e t r i ­ c a l f i g u r e s of a s q u a r e a n d a c u b e , to fully r e a l i z e t h i s i m p l i c a t i o n .

Internal Structure of a Square 3 1 . The square is a two-dimensional regular body with the same length on both d i m e n s i o n a l l i n e s . T a k i n g l e n g t h to be a u n i t s , t h e s q u a r e w o u l d h a v e an a r e a a

2

and

a p e r i m e t e r 4a. By s u i t a b l y c h o o s i n g a l i n e a r u n i t , we c a n h a v e t h e l e n g t h of a square equal 1 (linear unit). T h e length 1 (linear unit) would mean a closed interval o f l e n g t h 1 ( l i n e a r u n i t ) . S t r u c t u r a l l y t h i s w o u l d m e a n that w e c a n d i v i d e s a i d c l o s e d i n t e r v a l i n t o t w o i n t e r v a l s , o n e of w h i c h is a c l o s e d i n t e r v a l of l e n g t h 1/2. 3 2 . A l g e b r a i c a l l y we k n o w ( A + B )

2

= A

2

+ AB + BA + В

2

= A

2

2

+ 2 A B + B . Geo­

m e t r i c a l l y , t a k i n g A = [ O X ] , a c l o s e d i n t e r v a l , a n d В = [ O X ) , a o n e - s i d e d o p e n inter­ v a l ( t o b e r e f e r r e d t o a s a n o p e n i n t e r v a l ) , t h e a b o v e a l g e b r a i c e q u a l i t y w o u l d ac­ c o u n t for t h e i n t e r n a l s t r u c t u r e o f t h e s q u a r e o f l e n g t h [ О Х ] .

Structural Break-Up of a Square 3 3 . G e o m e t r i c a l l y t h e c l o s e d i n t e r v a l [ O O ' ] c a n b e d e c o m p o s e d a s t h e c l o s e d in­ t e r v a l [ O O ' ] = [ O X ] + ( X O ' ] = [ O X ] + [ O X ) , w h e r e X is t h e m i d d l e p o i n t of t h e c l o s e d i n t e r v a l [ O O ' ] . T h e s t r u c t u r a l b r e a k - u p of a s q u a r e of l e n g t h 1 ( l i n e a r u n i t ) can be in t e r m s of the structural break-up permissible by the linear unit as a closed i n t e r v a l [ O X ) of 1/2 u n i t l e n g t h a n d o p e n i n t e r v a l [ O X ) of 1/2 u n i t l e n g t h . T h e a b o v e f i g u r e s h o w s t h e s t r u c t u r a l b r e a k - u p of a s q u a r e of l e n g t h [ O O ' ] w i t h X as its m i d d l e point. T h e algebraic equality [OO']

2

= [OX]

2

+ 2 [ O X ] [ O X ) + [OX)

2

when

t r a n s l a t e d i n t o g e o m e t r i c e q u a l i t y , a s h a s b e e n s h o w n i n t h e a b o v e figure, w o u l d di­ v i d e t h e g e o m e t r i c a l s q u a r e i n t o four g e o m e t r i c a l s q u a r e s . O u t o f t h e s e four s q u a r e s , t h e first s q u a r e r e p r e s e n t e d b y [ O X ]

2

is a c o m p l e t e square with an area equal to the

s q u a r e of the length [ O X ] , a p e r i m e t e r equal to 4 times the length of [ O X ] , and the f o u r b o u n d a r y l i n e s a n d four c o r n e r p o i n t s i n t a c t . T h e s e c o n d a n d t h i r d s q u a r e s rep­ r e s e n t e d b y 2 [ O X ] [ O X ) a r e not c o m p l e t e s q u a r e s , s i n c e o n e b o u n d a r y l i n e a n d t w o c o r n e r p o i n t s a r e m i s s i n g . T h e f o u r t h s q u a r e a s w e l l i s not c o m p l e t e , s i n c e its t w o b o u n d a r y lines and three corner points are missing.

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VEDIC MATHEMATICAL CONCEPTS

3 4 . T h e o r i g i n a l s q u a r e m a y b e r e c o n s t i t u t e d i n the f o l l o w i n g four steps a s d e p i c t ed a b o v e . As a first s t e p , out of the four s q u a r e s take the s q u a r e w i t h four b o u n d a r y l i n e s a n d four c o r n e r p o i n t s intact. As a s e c o n d step t a k e o n e m o r e s q u a r e o u t of the r e m a i n i n g t h r e e s q u a r e s w i t h o n l y o n e b o u n d a r y line and t w o c o r n e r p o i n t s m i s s i n g . W h e n t h i s s q u a r e is a d d e d to the s q u a r e of the first s t e p , it c o n s t i t u t e s a r e c t a n g l e of a r e a e q u a l to half the area of the o r i g i n a l s q u a r e . 3 5 . H e r e i t m a y b e r e l e v a n t t o n o t e that g e o m e t r i c a l l y t h e r e r e m a i n s n o v a c u u m a t all w h e n the t w o s q u a r e s o f e q u a l l e n g t h , o n e o f t h e m w i t h its o n e b o u n d a r y line m i s s i n g , are j o i n e d . T h i s is b e c a u s e [ O X ] + [ O X ) = [ O X ] + ( X O ' ] = [ O O ] , t h u s p r o v i d i n g a c o n t i n u u m t h r o u g h o u t t h e b o u n d a r y line a l o n g w h i c h the t w o s q u a r e s are j o i n e d to c o n s t i t u t e a r e c t a n g l e . 3 6 . N o w a s a third s t e p , w e m a y t a k e o n e m o r e s q u a r e w h o s e o n e b o u n d a r y line i s m i s s i n g from the r e m a i n i n g t w o s q u a r e s . W h e n this s q u a r e i s j o i n e d w i t h the r e c t a n g l e c o m p o s e d in the first t w o s t e p s , t h e m i s s i n g b o u n d a r y line, as is s h o w n in the a b o v e figure, b e c o m e s a c o n t i n u u m in t e r m s of the half b o u n d a r y line of t h e inner l e n g t h of the r e c t a n g l e . 3 7 . F o r a fourth step, j o i n the r e m a i n i n g fourth s q u a r e , w h o s e t w o b o u n d a r y lines are m i s s i n g , w i t h the g e o m e t r i c a l figure f o r m e d as a result of the first t h r e e s t e p s . As is e v i d e n t from the a b o v e figure, o n e of t h e m i s s i n g b o u n d a r y lines w o u l d b e c o m e a c o n t i n u u m in t e r m s of t h e u p p e r half of the b o u n d a r y line of the inner l e n g t h of the r e c t a n g l e left u n c o v e r e d until S t e p 3 , w h i l e the s e c o n d m i s s i n g b o u n d a r y line o f this f o u r t h s q u a r e (as d e p i c t e d as S t e p 4 in t h e a b o v e figure) w o u l d b e c o m e a c o n t i n u u m in t e r m s of the i n n e r b o u n d a r y line of t h e third s q u a r e . 3 8 . T h i s internal structural arrangement of the square is significant in several w a y s . T w o of these w h i c h have vital bearing for the present are that the square has nine structural points out of w h i c h eight are symmetrically located around the central ninth point w h e r e all the four squares are joined, and that w h e n out of the square of rational length, a square of rational length is cut out, the r e m a i n i n g portion of the original square consists of three parts, n o n e of w h i c h has all the four boundary lines intact.

Internal Structure of a Cube 3 9 . T h e c u b e is a t h r e e - d i m e n s i o n a l r e g u l a r b o d y w i t h the s a m e l e n g t h on all the t h r e e - d i m e n s i o n a l lines. T a k i n g the length to be a units, the c u b e has v o l u m e a and 3

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MODERN SCIENCE AND VEDIC SCIENCE

2

s u r f a c e a r e a 6a . By s u i t a b l y c h o o s i n g a l i n e a r unit, we c a n h a v e t h e l e n g t h of t h e c u b e = 1 ( l i n e a r u n i t ) . T h e l e n g t h 1 ( l i n e a r u n i t ) w o u l d m e a n a c l o s e d interval of l e n g t h 1 ( l i n e a r u n i t ) . S t r u c t u r a l l y this w o u l d m e a n that w e c a n d i v i d e t h e said c l o s e d i n t e r v a l i n t o t w o i n t e r v a l s , o n e of w h i c h is a c l o s e d interval of l e n g t h 1/2. 4 0 . A l g e b r a i c a l l y we k n o w t h a t ( A + B )

3

3

2

= A + 3A B + ЗАВ

2

3

+ В . Geometrically

t a k i n g A = [ O X ] , a c l o s e d i n t e r v a l , a n d В = [ O X ) , a o n e - s i d e d o p e n i n t e r v a l (to be r e f e r r e d t o a s a n o p e n i n t e r v a l ) , t h e a b o v e a l g e b r a i c e q u a l i t y w o u l d a c c o u n t for t h e internal structure of the cube of length 2 [ O X ] .

Structural Break-Up of a Cube

4 1 . G e o m e t r i c a l l y we h a v e [ O O ' ] = [ O X ] + ( X O ' ] = [ O X ] + [ O X ) , w h e r e X is t h e m i d d l e point of the closed interval [ O O ' ] . T h e structural break-up of a square of length 1 (linear unit) can be had in terms of the structural break-up permissible by t h e l i n e a r u n i t as a c l o s e d i n t e r v a l [ O X ] of 1/2 u n i t l e n g t h a n d an o p e n i n t e r v a l l e n g t h [ O X ) of 1/2 u n i t l e n g t h . T h e a b o v e figure s h o w s t h e s t r u c t u r a l b r e a k - u p of a c u b e o f l e n g t h [ O O ' ] w i t h X a s its m i d d l e p o i n t . T h e a l g e b r a i c e q u a l i t y [ O O ' ] [OX]

3

+ 3[OX]

2

[OX) + 3[OX] [OX)

2

3

=

3

+ [OX) , w h e n translated into a g e o m e t r i c

e q u a l i t y , a s h a s b e e n s h o w n i n t h e a b o v e figure, d i v i d e s t h e g e o m e t r i c a l c u b e i n t o eight geometrical cubes.

3

4 2 . O u t o f t h e a b o v e e i g h t c u b e s , t h e first c u b e , r e p r e s e n t e d b y [ O X ] , i s a com­ p l e t e c u b e h a v i n g a v o l u m e e q u a l t o a c u b e o f t h e l e n g t h [ O X ] , s u r f a c e area e q u a l t o 2

six t i m e s t h e a r e a [ O X ] , a n d all t h e six s u r f a c e s a n d e i g h t c o r n e r p o i n t s i n t a c t .

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VEDIC MATHEMATICAL CONCEPTS

4 6 . T h e division of the c u b e into eight c u b e s of a b o v e description can be used in rev e r s e to reconstitute the original c u b e in t e r m s of the said eight c u b e s . T h i s is similar to the case of the square traced a b o v e in p a r a g r a p h s 34 to 3 7 . T h e said eight c u b e s

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h a v e in all 36 surfaces. W h e n these are c o m b i n e d as a b o v e , twelve surfaces are used to form a c o n t i n u u m w i t h the missing surfaces of said c u b e s . T h e r e m a i n i n g 24 surfaces a c c o u n t for the surface area of the r e - c o m p o s e d original c u b e as is evident from the a b o v e figure.

Summary 4 7 . Before we take up the general Nth term of the framed d o m a i n s sequence, I w o u l d like to s u m m a r i z e the position e m e r g i n g from the above analysis of the first three framed d o m a i n s . T h e s e three cases involve c o m m o n geometrical figures, n a m e l y a c l o s e d interval of a straight line, a square, and a c u b e of rational lengths. 4 8 . W i t h respect to the first framed d o m a i n , it requires a definition of a structural frame for a closed interval. W i t h respect to the square and the c u b e it is o b v i o u s that the area of the square is contained within the perimeter of the square and the v o l u m e of the c u b e is similarly contained within the surface area of the c u b e . As such the p e r i m e t e r and surface area respectively are the framed parts of the square and the c u b e . Structurally the length of the closed interval requires a m i d d l e point and as such we m a y define a point as a zero space figure responsible for p r o v i d i n g a frame for the (linear) length by h a v i n g p l a c e m e n t for the frame point a n y w h e r e in b e t w e e n the interval. ( T o be specific, the p l a c e m e n t for the frame point is not at the b e g i n n i n g or the e n d point of the closed interval.) T h a t w a y only one point (to be called the middle point since it falls m i d w a y b e t w e e n the t w o end points) constitutes the frame of the first framed d o m a i n . 4 9 . F r o m the a b o v e , w e m a y c o n c l u d e (and tabulate a s the figure b e l o w indicates) that w h e n from the first framed d o m a i n , that is, a closed interval of a straight line, a c l o s e d interval of s m a l l e r rational length is cut out, the r e m a i n i n g portion constitutes an irrational length as it is m i s s i n g o n e end point. W h e n from the second framed d o m a i n , that is, a square of rational length, a square of smaller rational length is cut out, the r e m a i n i n g portion constitutes three squares out of w h i c h t w o squares are m i s s i n g o n e b o u n d a r y line, and o n e square is missing t w o b o u n d a r y lines. Similarly, w h e n from the third framed d o m a i n , that is, a c u b e of rational length, a c u b e of smaller rational length is cut out, the r e m a i n i n g portion constitutes seven c u b e s out of w h i c h three c u b e s are m i s s i n g o n e surface, another three c u b e s are m i s s i n g t w o surfaces, and the last, that is, the seventh c u b e , is m i s s i n g three surfaces. 5 0 . T h e following table evidently m a k e s it clear that w h e n the rational length of the regular b o d y of the N t h - d i m e n s i o n (Nth framed d o m a i n ) is divided into rational part (Q) and the r e m a i n i n g irrational part (R), the regular body gets divided into 2 n

d i m e n s i o n a l b o d i e s of the form Q R

N - n

N

N-

, w h e r e n = N, N - l , N - 2 , ... , 2, 1, 0. T h e s e N-

d i m e n s i o n a l b o d i e s , n u m b e r i n g N + l , m a y b e called respectively, first, second,

...

( N + l ) t h - d i m e n s i o n a l b l o c k s of degree N. T h e r e is only one N - d i m e n s i o n a l b o d y of the first-dimensional block. T h e r e are N N - d i m e n s i o n a l bodies of the second-dimensional b l o c k . T h e R - d i m e n s i o n a l b l o c k has ( N . N - l . N - 2 . . . N - R / l . 2 . 3 . . . R ) N - d i m e n s i o n a l b o d i e s . T h e last b l o c k has one N - d i m e n s i o n a l b o d y . 5 1 . As a net result, the following internal arrangements of regular bodies justify our logic and conclusion of p a r a g r a p h s 7 to 12 that w h e n from the rational length, a rational portion is cut out, it leaves behind an irrational length.

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VEDIC MATHEMATICAL CONCEPTS

Degree of Freedom 5 2 . N o w we m a y take up the question w h y restrictions are necessary for n to be 3. In other w o r d s , the question is: w h y for the first and second degree of natural n u m b e r s , that is, for n = 1 and 2, the equation Z = X + Y has solutions. Here it m a y be relevant to note that the division into t w o parts of like degrees is not possible for all Z , n = 1, 2. W h e n Z and n both are equal to one, the division for Z = 1 is not possible, though for all Z 2 and n = 1, the desired division into t w o parts obviously holds by the very definition of the natural n u m b e r s . For e x a m p l e , consider 1, 1 + 1 , 1 + 2 , 1+3 ...,1+s,.... n

n

n

n

n

n

n

1

n

5 3 . For n = 2, every Z is not d e c o m p o s a b l e as X + Y . Illustrations of the point are l , 2 , 3 , 4 . There are infinitely m a n y values of Z for which the desired d e c o m p o s i t i o n is not available. On the contrary, there are infinitely m a n y values of Z for w h i c h the d e sired d e c o m p o s i t i o n holds. O n e class of such values is Z = 5. T h e decomposition for this value is 5 = 3 + 4 . N o w for any value of the form 5Z, a similar relationship also holds, namely (5Z) = (3Z) + (4Z) . 2

2

2

2

2

2

2

2

2

2

5 4 . So for n = 1 and 2, the desired decomposition is not universally permissible t h o u g h the s a m e holds for infinitely m a n y values of Z. T h i s restricted permissibility of the d e c o m p o s i t i o n s for n = 1 holds for all values except Z = 1. A n d for n = 2 for a large n u m b e r (a countable n u m b e r of values of Z) the decomposition holds and for an equally large n u m b e r (a countable n u m b e r of v a l u e s of Z), the s a m e does not hold. 5 5 . T h e reason for the restricted permissibility for n = 1 is that the absolute value of natural n u m b e r 1 and the linear m e a s u r e value of unit length are t w o distinct concepts. T h e natural n u m b e r 1 is the absolute value in the sense that it is not dependent upon any

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MODERN SCIENCE AND VEDIC SCIENCE

d i m e n s i o n a l structures w h i l e the linear m e a s u r e value of unit length is completely dep e n d e n t upon the structural a r r a n g e m e n t of a one-dimensional space, that is a straight line. T h e linear unit w o u l d represent a closed interval of unit length. T h e internal structural restrictions of a closed interval m a k e it impossible for its division into t w o closed intervals. As such the natural n u m b e r 1, as the first degree v a l u e of one, does not admit the desired d e c o m p o s i t i o n . H o w e v e r , the straight line and hence a closed interval, b e i n g a first-dimensional b o d y (figure) has t w o degrees of freedom within a three-dimensional s p a c e . As such, externally the linear unit is free to c o m b i n e itself with a similar unit to constitute a o n e - d i m e n s i o n a l b o d y (figure) of t w o units. T h e role of the origin point of the t h r e e - d i m e n s i o n a l frame is crucial since from it e m a n a t e s dimensional lines and for each of the three-dimensional lines it r e m a i n s a zero value starting point. B e c a u s e of this, it is possible to have linear units simultaneously sprouting from the origin point a l o n g the d i m e n s i o n a l lines. H e n c e , there are t w o distinct planes available for every dim e n s i o n a l line, w h i c h m a y be used by the given linear unit of any of the three d i m e n sional lines. T h i s two-fold freedom for the straight line accounts for the desired d e c o m position for all natural n u m b e r s except Z - l . 5 6 . T u r n i n g to the p l a n e , a s q u a r e , a t w o - d i m e n s i o n a l regular b o d y , has as w e l l , o n e d e g r e e of f r e e d o m in a t h r e e - d i m e n s i o n a l space. It is b e c a u s e of the restricted freedom of the p l a n e that the d e c o m p o s i t i o n is not universally permissible in the case of t w o d i m e n s i o n a l regular b o d i e s . Rather, in their case, within any finite range of natural n u m b e r s , say

the desired d e c o m p o s i t i o n generally w o u l d not be permissible

and in a c o m p a r a t i v e l y very small n u m b e r of values of Z only the desired d e c o m p o s i tion w o u l d hold. T h i s is so also within a t h r e e - d i m e n s i o n a l space, t w o d i m e n s i o n s stand restricted b e c a u s e of the structural format of a plane and it is left with only o n e degree of f r e e d o m . H e r e we m a y c o m p a r e the situation with the fate of a o n e - d i m e n s i o n a l b o d y , that is, a straight line w h o s e structural format has only o n e d i m e n s i o n . It is left w i t h t w o d e g r e e s of freedom in t e r m s of unrestricted t w o d i m e n s i o n s of the threedimensional space. 5 7 . N o w w h e n we c o m e to a three-dimensional body, its structural format restricts all the three dimensional lines and hence we are left with no degree of freedom. It is b e cause of this that it is not possible to duplicate the cube. In the case of a plane, it is possible to duplicate it as a p l a n e h a v i n g one degree of freedom in a three-dimensional space. As such we can m o v e (or pile) a plane (or identical regular b o d i e s of a two-dimensional s p a c e , say square) along the third d i m e n s i o n as is evident from the following figure:

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VEDIC MATHEMATICAL CONCEPTS

5 8 . T h e first b a s i c structural characteristic acquired by the origin point b e c a u s e of the possibility for the free motion of a plane of O X , OY frame along OZ dimensional line is that the origin point gets e m b e d d e d with a plane structure. B e c a u s e of this, it hardly m a t t e r s w h i c h point of the Z axis is chosen as the origin point. T h o u g h the plane structure acquired by the origin point of the dimensional frame does not affect the external freedom for motion of the plane t o w a r d s the OZ axis, the internal structural arrangem e n t of the plane figures is governed by the two-dimensional format. T h e glaring effect of the t w o - d i m e n s i o n a l format for the plane figures is that its every point acquires a t w o - d i m e n s i o n a l structure. B e c a u s e of this even the dimensional lines, as b o u n d a r y lines of plane figures, get inseparably m e r g e d with the plane figure. As such, structurally they too can be dealt with only as plane strips as s h o w n in the figures b e l o w .

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MODERN SCIENCE AND VEDIC SCIENCE

2

5 9 . T h e immediate effect is that w h e n we divide a square N , or as a matter of fact 2

even a rectangle ( N x M ) , of length N, then l units of the X axis, as has b e e n s h o w n 2

a b o v e , w o u l d be N, but the l units of Y axis w o u l d remain only N - l (in the case of a 2

rectangle, only M - l ) . Therefore, the re-assembled position for l = 1 for the X and Y axis would give N x ( N - l ) (and in case of a rectangle N x [ M - l ] ) which is less than N

2

(in the case of a rectangle N x ( M - l ) < N x M ) . T h i s is the reason w h y in a large n u m b e r 2

2

2

of cases it is not possible to d e c o m p o s e Z as X + Y for natural n u m b e r s Z, X, and Y. 6 0 . N o w using one degree of freedom, w h i c h m a y be viewed as the external freedom for the plane figures, the restrictions of the t w o - d i m e n s i o n a l format of the plane figures can be kept in a b e y a n c e . Here it m a y be relevant to note that the proof of the Pythagoras T h e o r e m , w i t h the help of a right-triangle, m a y appear as if we are using only o n e plane. As a matter of fact, we are using m o r e than one plane by considering the sides of a triangle. S u p p o s e the square of the h y p o t e n u s e of a right-angle triangle is split into t w o s q u a r e s w h i c h equal the squares of the perpendicular and base of the said triangle or, as a c o n v e r s e , the reconstruction of a square of the hypotenuse in terms of the squares of the b a s e and perpendicular. T h e n we h a v e to use the internal structural a r r a n g m e n t s permitted by the square as a regular b o d y of the two-dimensional space. Further, we have to use eight s y m m e t r i e s of the square and one degree of freedom (which the square as a t w o - d i m e n s i o n a l b o d y w o u l d have) within a three-dimensional Cartesian frame.

6 1 . A s i s s h o w n a b o v e , o n the t w o - d i m e n s i o n a l format o f O X , O Y dimensional lines we can h a v e plane figure A B C D (which m a y be a square or a rectangle). T h i s figure w o u l d h a v e a d e g r e e of freedom of m o t i o n along the OZ axis. T h i s p r o v i d e s us with the possibility to c h o o s e t w o identical figures, say A B C D and A ' B ' C ' D ' as s h o w n above. T h e diagonal, b e i n g on the o n e - d i m e n s i o n a l format, w o u l d have length equal to any real n u m b e r , w h i c h includes the natural n u m b e r s . T h e t w o - d i m e n s i o n a l figure A C C'A' m a y be a square on a p r o p e r choice of vertices. 6 2 . It is b e c a u s e of the a b o v e freedom of construction for the plane figures in a threed i m e n s i o n a l frame that it b e c o m e s possible to divide the square of the h y p o t e n u s e ( A C

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VEDIC MATHEMATICAL CONCEPTS

C'A') in t e r m s of the squares on the d i m e n s i o n a l lines of the format of the plane figure A B C D and vice-versa. T h o u g h this assignment as such is not taken up here, it may not be out of context to note that the unit square w o u l d be the basic constituent w h i c h w o u l d be used for a division p r o c e s s of one square into t w o squares or the reverse proc e s s of c o m p o s i t i o n of t w o squares into one square. T h e unit square (and as such any s q u a r e ) has eight s y m m e t r i e s , and the three-dimensional regular body as well has eight corner points, and the three-dimensional Cartesian frame cuts the space into eight octants, as s h o w n b e l o w . It is b e c a u s e of this that the division process and the reverse c o m p o s i t i o n p r o c e s s is possible.

6 3 . H e n c e we h a v e justified the restriction of F e r m a t ' s Last T h e o r e m to n ^ 3. 6 4 . N o w we m a y take up the thread of the above logic of internal restrictions of the format and the external freedom, if any is available, for the dimensional regular b o d i e s m o v i n g from t w o - d i m e n s i o n a l b o d i e s to three- and higher-dimensional b o d i e s . C a s e of a Three-Dimensional R e g u l a r Body 3

6 5 . Step 1: Let v o l u m e V = Z c u b i c units. In other w o r d s , the c u b e Z is constituted 3

b y Z unit c u b e s .

6 6 . Step 2: T h e c u b e Z has Z units a l o n g the OX dimensional line as well as Z units a l o n g both of the r e m a i n i n g t w o d i m e n s i o n a l lines.

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MODERN SCIENCE AND VEDIC SCIENCE

6 8 . S t e p 4: Z unit c u b e s are required to constitute a length of Z units along the Y axis and similarly, Z unit c u b e s are required to constitute a length of Z units along the Z axis. 6 9 . Step 5: T h e origin point of the Cartesian frame, b e i n g a dimensionless point, w h e n the X axis and the Y axis e m e r g e from the origin point, the dimensional length along the X axis is not affected by the dimensional length along the Y axis and vice versa.

100

VEDIC MATHEMATICAL CONCEPTS

3

7 2 . S t e p 8: We had started w i t h c u b e Z of length Z and v o l u m e V = Z , but have 3

e n d e d w i t h a c u b e of v o l u m e Z ( Z + 1 ) ( Z + 1 ) w h i c h is greater than Z , hence there is a contradiction.

Conclusion 7 3 . Therefore, the a b o v e contradiction proves that our assumption regarding p e r m i s sibility of the v o l u m e of the c u b e to be separated and handled l x l x l = 1 as a natural 3

unit is w r o n g . H e n c e X c u b i c units cannot be treated as linear units and as such the addition operation of natural n u m b e r s is not a meaningful operation. If cubic units l x l x l are treated as linear units 1, then it w o u l d result in a contradiction. H e n c e , the t h e o r e m for the c u b e s that it is impossible to separate a c u b e into t w o c u b e s . 7 4 . T h e a b o v e logic w o u l d be equally applicable to the bi-quadratic and higher p o w e r s as w e l l , as the natural n u m b e r s addition operation is geometrically linear in nature; the s a m e is m e a n i n g l e s s and it is not applicable to the higher-dimensional units l x l , lxlxl,

lxlxlxl....

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M O D E R N SCIENCE AND VEDIC SCIENCE

75. Z as Z

4

2

( l x l ) and Z

2

( l x l x l x l ) and Z

4

(1) are t w o distinct units. Similarly Z

3

( l x l x l ) and Z

3

(1) a s w e l l

(1) and so on, are pairs of distinct units. T h e addition o p e r a t i o n

w h i c h b i n d s 1 + 1 i s m e a n i n g l e s s for the u n i t s l x l o r l x l x l , o r l x l x l x l and s o o n , since ( l x l ) + ( l x l ) m e a n s a n o p e r a t i o n i n t e r m s o f w h i c h t w o square u n i t s are t o be com­ b i n e d , w h i l e 1 + 1 = 2 is t h e natural n u m b e r s addition o p e r a t i o n . U n l e s s and until we k n o w h o w the s q u a r e unit is t r a n s f o r m a b l e into a linear unit, the linear units addition op­ e r a t i o n w o u l d not b e a p p l i c a b l e . T h e natural n u m b e r s multiplication o p e r a t i o n , w h i c h i s i n t i m a t e l y c o n n e c t e d w i t h the a d d i t i o n o p e r a t i o n , r e m a i n s d e p e n d e n t u p o n the addition o p e r a t i o n only w i t h i n a linear s p a c e . T h e m o m e n t the space s t a n d s c h a n g e d from o n e d i m e n s i o n a l (linear) s p a c e to t w o and h i g h e r - d i m e n s i o n a l s p a c e , the i n d e p e n d e n t char­ acteristics o f t h e m u l t i p l i c a t i o n o p e r a t i o n are displayed. A s such, u n l e s s and until l x l i s suitably

defined,

the

addition

operation

will

remain

only

the

operation

of o n e -

dimensional space.

Geometrical Continuum: Fourth-Dimensional Space 76. N o w we m a y take up the fourth n u m b e r of the framed d o m a i n s sequence n

n-1

[a /2na 4

for n = 1, 2, 3 ... ]. F o r the p r e s e n t it m a y be taken by w a y of definition that

3

[a /8a ] is t h e f o r m u l a t i o n for the regular b o d y of fourth-dimensional space of dimen­ 4

3

s i o n a l length a, c o n t e n t part ( d o m a i n p a r t ) as a , and the frame part as 8a . In t h e con­ 3

text we m a y refer to a c u b e of d i m e n s i o n a l length a, c o n t e n t ( d o m a i n part) as v o l u m e a , and frame part as surface area 6a

2

as t h e t h r e e - d i m e n s i o n a l regular b o d y . F u r t h e r , by

w a y of definition, we m a y take t h e following figure as a g e o m e t r i c a l p r e s e n t a t i o n of the f o u r t h - d i m e n s i o n a l regular b o d y .

Structural Arrangement of Fourth Framed Domain 7 7 . A l g e b r a i c a l l y we k n o w that ( A + B )

4

= A

4

3

2

+ 6 A B + 4A B

2

+ 6AB

3

4

+ B . Geomet­

rically, t a k i n g A = [ O X ] a n d В = [ O X ) as a one-sided o p e n interval (to be referred to as an o p e n interval), in t e r m s of the a b o v e e x p a n s i o n , A + B = [OO'] = [ O X ] + ( X O ' ] = [ O X ] + [ O X ) , w h e n X is a rational n u m b e r and is the m i d d l e point of [ O O ' ] . If [ O X ] = X, t h e n [ O O ' ] = 2 X . 78. The algebraic equality [OO'] 6[OX]

4

= [OX]

4

+ 6[OX]

3

[OX) + 4[OX]

2

[OX)

2

+

3

[ O X ) w o r k s o u t t h e s t r u c t u r a l a r r a n g e m e n t o f t h e fourth f r a m e d d o m a i n . T h i s ,

w h e n t r a n s l a t e d i n t o g e o m e t r y , s h a l l d i v i d e t h e f o u r t h - d i m e n s i o n a l b o d y i n t o seven­ teen fourth-dimensional bodies. 4

7 9 . O u t of t h e s e v e n t e e n f o u r t h - d i m e n s i o n a l b o d i e s , t h e first, n a m e l y [ O X ] , is a c o m p l e t e f o u r t h - d i m e n s i o n a l b o d y in t h e s e n s e that all t h e eight c u b i c c o m p o n e n t s of t h e f r a m e are intact.

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VEDIC MATHEMATICAL CONCEPTS

8 0 . We m a y define a b o d y of such frame with all the eight cubic c o m p o n e n t s intact as a first-dimensional b l o c k of the regular b o d y . 8 1 . As is evident from the above algebraic equality, in all we shall have five types of d i m e n s i o n a l b l o c k s . T h e second, third, fourth, and fifth-dimensional blocks w o u l d be structurally the fourth-dimensional b o d i e s such that their frames are respectively missing 1, 2, 3, and 4 cubic c o m p o n e n t s .

8 2 . Therefore, the internal structural a r r a n g e m e n t of the fourth-dimensional regular b o d y admits a b r e a k - u p as five-dimensional b l o c k s such that these b l o c k s h a v e respectively 1, 6, 4, 6, and 1 fourth-dimensional b o d i e s . Evidently, all the five-dimensional b l o c k s are structurally distinct. N

8 3 . N o w the general b i n o m i a l e x p a n s i o n o f ( A + B ) w o u l d help u s c o n c l u d e that w e N

shall h a v e ( N + l ) - d i m e n s i o n a l b l o c k s . I n this format, 2 N - d i m e n s i o n a l b o d i e s w o u l d N

constitute the regular b o d y ( A + B ) . As for the N - d i m e n s i o n a l space, the format of the d i m e n s i o n a l b l o c k s is to r e m a i n the s a m e , so as to structure a regular body, exactly 2

N

N - d i m e n s i o n a l b o d i e s w o u l d b e required. 8 4 . Therefore, for N N

3 and Z

N

3 (for Z = 1 and 2 the t h e o r e m is o b v i o u s ) , let Z =

N

X + Y w h e r e X, Y, Z, and N are natural n u m b e r s . Further, let Z > X > Y (for X = Y the theorem has b e e n p r o v e d above, and X and Y being general, it remains to p r o v e the N

t h e o r e m w h e n o n e of X or Y is greater than the other). Let Z = X + R. Therefore, X + N

N

N

Y = ( X + R ) . T h e right side o n e x p a n s i o n yields X + N R X N

= X + NRX

N - 1

N - 1

N

+ .... Therefore, X + Y

N

N

+ ... + R .

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N

Therefore, Y = N R X

N - 1

N

+ ... + R . N

N o w the right side is an expression of 2 - 1 N - d i m e n s i o n a l blocks. T h e left side can h a v e divisions into N - d i m e n s i o n a l b l o c k s as Y = A + B . N

N

N

Y = ( A + B ) yields 2 N - d i m e n s i o n a l blocks, while the N - d i m e n s i o n a l b l o c k s o n the N

right side are only 2 - 1 . H e n c e , the left side cannot be equal to the the right side. W i t h this the g e o m e t r i c a l proof of the t h e o r e m stands c o m p l e t e d . 8 5 . N o w if in the light of the a b o v e , we try to find the significance of the internal 1

structural a r r a n g e m e n t of the very first regular b o d y l , w h e r e 1 is equal to an irrational length, then it really w o u l d be e n c h a n t i n g to see h o w Nature w o r k s out 1 = 1/2 (half of a rational unit) + 1/2 (half of an irrational unit). It provides us with a geometrical continuum w h i c h , as a linear c o n t i n u u m , is equivalent to the field of real n u m b e r s w h o s e s u b field is the field of rational n u m b e r s . 8 6 . O b v i o u s l y , the implications are m a n y and far reaching, particularly w h e n they d e m o l i s h m e n t a l b l o c k s of the p h y s i c a l w o r l d and h e l p t r a n s c e n d to four and h i g h e r d i m e n s i o n a l s p a c e s as a g e o m e t r i c a l c o n t i n u u m of spaces w h o s e regular b o d i e s constin

n-1

tute a framed d o m a i n s s e q u e n c e ( a / 2 n a , n = 1, 2, 3 ...).

104